• #### The Miracle of Special Relativity

From Robert Winn@21:1/5 to All on Mon Oct 9 11:27:33 2023
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a clock in
a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get a
faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe Einstein
was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his clock is
slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on the
ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for x
and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate that
there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From JanPB@21:1/5 to Robert Winn on Mon Oct 9 12:25:03 2023
On Monday, October 9, 2023 at 11:27:35 AM UTC-7, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a clock in
a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get a faster
speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special Relativity
show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on the
ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for x
and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'. So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate that
there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.

After how many decades you are still confused and in square one.

--
Jan

--- SoupGate-Win32 v1.05
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• From Robert Winn@21:1/5 to JanPB on Mon Oct 9 12:39:13 2023
On Monday, October 9, 2023 at 12:25:05 PM UTC-7, JanPB wrote:
On Monday, October 9, 2023 at 11:27:35 AM UTC-7, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a clock
in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get a
faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on
the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for
x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate
that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
After how many decades you are still confused and in square one.

--
Jan

--- SoupGate-Win32 v1.05
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• From Robert Winn@21:1/5 to JanPB on Mon Oct 9 12:47:54 2023
On Monday, October 9, 2023 at 12:25:05 PM UTC-7, JanPB wrote:
On Monday, October 9, 2023 at 11:27:35 AM UTC-7, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a clock
in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get a
faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on
the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for
x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate
that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
After how many decades you are still confused and in square one.

--
Jan
Jan
Thank you for your response. Maybe you would like to take a few moments to show what you think I am confused about. Do you agree with Einstein that the pilot of the airplane and the observer on the ground would get the same speed for the airplane?

--- SoupGate-Win32 v1.05
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• From JanPB@21:1/5 to Robert Winn on Mon Oct 9 15:57:03 2023
On Monday, October 9, 2023 at 12:47:57 PM UTC-7, Robert Winn wrote:
On Monday, October 9, 2023 at 12:25:05 PM UTC-7, JanPB wrote:
On Monday, October 9, 2023 at 11:27:35 AM UTC-7, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a
clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get
a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on
the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for
x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate
that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
After how many decades you are still confused and in square one.

--
Jan
Jan
Thank you for your response. Maybe you would like to take a few moments to show what you think I am confused about. Do you agree with Einstein that the pilot of the airplane and the observer on the ground would get the same speed for the airplane?

No, it would be a waste of time. You never learn.

--
Jan

--- SoupGate-Win32 v1.05
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• From xip14@21:1/5 to All on Mon Oct 9 17:17:02 2023
Time dilation is the second item in Einstein-EDoMB-1905 Section §4

https://www.fourmilab.ch/etexts/einstein/specrel/www/

Lorentz Contraction is the first item.

Time dilation has an equation. Lorentz Contraction does not.

Moving time = stationary time / gamma

gamma is a dimensionless number greater than 1.

So moving time is less than stationary time.

Moving time is a clock which moves, with speed-v, from x = 0 to some stationary milepost-x where an observer is in possession of a likewise stationary clock. This clock is synchronized with origin clock x = 0. The stationary observer sees moving clock-
v go by with less face time than his own clock.

This is time “dilation” in the sense that the moving unit of time, a second, a minute, is expanded in duration, while keeping unit value. Dilation of the unit subtracts from clock face value.

And then it gets really weird. The guy who has the moving clock considers himself stationary at location [ x′ = 0 ] on the axis X′, the entirety of axis X′ moving with speed-v. When this purportedly moving guy considers himself stationary, the so-
called stationary clock appears to him to move with negative speed-v, and that clock suffers time dilation just as his own clock once did.

The clock paradox has been around for 120 years.

--- SoupGate-Win32 v1.05
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• From Sylvia Else@21:1/5 to Robert Winn on Tue Oct 10 13:27:26 2023
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a clock
in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get a
faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on
the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for
x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'. So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate
that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.

Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.

--- SoupGate-Win32 v1.05
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• From patdolan@21:1/5 to Sylvia Else on Mon Oct 9 19:31:20 2023
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a clock
in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get a
faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on
the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for
x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate
that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Sylvia, what about the Richard Hertz back story to the Einstein catapult to scientific stardom?

--- SoupGate-Win32 v1.05
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• From Robert Winn@21:1/5 to All on Mon Oct 9 21:08:24 2023
On Monday, October 9, 2023 at 5:17:04 PM UTC-7, xip14 wrote:
Time dilation is the second item in Einstein-EDoMB-1905 Section §4

https://www.fourmilab.ch/etexts/einstein/specrel/www/

Lorentz Contraction is the first item.

Time dilation has an equation. Lorentz Contraction does not.

Moving time = stationary time / gamma

gamma is a dimensionless number greater than 1.

So moving time is less than stationary time.

Moving time is a clock which moves, with speed-v, from x = 0 to some stationary milepost-x where an observer is in possession of a likewise stationary clock. This clock is synchronized with origin clock x = 0. The stationary observer sees moving clock-
v go by with less face time than his own clock.

This is time “dilation” in the sense that the moving unit of time, a second, a minute, is expanded in duration, while keeping unit value. Dilation of the unit subtracts from clock face value.

And then it gets really weird. The guy who has the moving clock considers himself stationary at location [ x′ = 0 ] on the axis X′, the entirety of axis X′ moving with speed-v. When this purportedly moving guy considers himself stationary, the so-
called stationary clock appears to him to move with negative speed-v, and that clock suffers time dilation just as his own clock once did.

The clock paradox has been around for 120 years.
Yes, I know what scientists think. You seem to understand it better than most scientists, which can be seen from your last statement, "the stationary clock appears to him to move with negative speed -v. This is the part of the Lorentz equation
interpretation I disagree with and have disagreed with since high school. If the moving clock is slower, then the stationary clock does not appear to move with a negative speed -v. It appears to move with a velocity greater than -v. I explained this
with the Galilean transformation equations for the slower clock.
x = x' - m'n' where m' is the velocity of the so-called stationary clock relative to the so-called moving clock. Since the distances do not change from the original Galilean transformation equation
x' = x-vt, we get x-x'=-m'n', x-x'=vt, so -m'n'=vt, m' = -vt/n', meaning that if n' is less time than t, m' is a faster speed than v. This agrees with reality, since if the pilot of an airplane has a slower clock, as Einstein says he will, he will get a
faster speed for the airplane. These equations also work if the moving clock is faster than the stationary clock, as we observe with a GPS satellite. Then the satellite clock results in a slower speed for the satellite.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Sylvia Else on Mon Oct 9 21:39:38 2023
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a clock
in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get a
faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on
the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for
x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate
that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of today do not consider
is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is faster than a clock on
earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves. I am explaining
relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the Lorentz equations that he
said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'. But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of the slower clock
gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book. His problem
was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for it. The
Galilean transformation equations agree with reality if a moving clock is faster or slower.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Sylvia Else@21:1/5 to patdolan on Tue Oct 10 15:21:29 2023
On 10-Oct-23 1:31 pm, patdolan wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a clock
in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get a
faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on
the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for
x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate
that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Sylvia, what about the Richard Hertz back story to the Einstein catapult to scientific stardom?

I try not to contaminate by brain by remembering any of Hertz's nonsense.

Sylvia.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to patdolan on Mon Oct 9 21:44:24 2023
On Monday, October 9, 2023 at 7:31:23 PM UTC-7, patdolan wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a
clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get
a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on
the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for
x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate
that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.

It's not as if Einstein, as a young patent clerk, had some ability to impose his theory on an unwilling world. Experimenters had been taking a very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Sylvia, what about the Richard Hertz back story to the Einstein catapult to scientific stardom?
I have talked to enough scientists to convince me that there is not going to be a scientist in my lifetime who is going to disagree with what other scientists are doing. Scientists since World War II have been given millions, billions, and now trillions
of dollars by governments in return for all agreeing with Einstein. I just go ahead and disagree because I am not a scientist, and no one pays me for using wrong equations.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Sylvia Else@21:1/5 to Robert Winn on Tue Oct 10 19:08:39 2023
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a clock
in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get a
faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on
the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for
x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate
that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of today do not
consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is faster than a
clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves. I am explaining
relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the Lorentz equations that he
said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'. But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of the slower clock
gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book. His problem
was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for it. The
Galilean transformation equations agree with reality if a moving clock is faster or slower.

How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to Sylvia Else on Tue Oct 10 00:46:51 2023
On Tuesday, 10 October 2023 at 04:27:30 UTC+2, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a clock
in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get a
faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on
the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for
x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate
that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a very close look at reality, and had been finding that it was not
behaving in the expected way.

Sorry, lady, it's a nonsensical propaganda bullshit, developed
after.

Einstein provided a solution.

His mumble was not even consistent. Sometimes it
happens that people are following a mumbling mystician,
and this is the case.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to Sylvia Else on Tue Oct 10 10:23:33 2023
Sylvia Else <sylvia@email.invalid> wrote:

On 10-Oct-23 5:27 am, Robert Winn wrote:
[-]
Since distances are the same in all of these Galilean equations, vt = -m'n'. So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as
we observe it to be. They indicate that there is no need for the
miracle Einstein describes. How this relates to electromagnetism, I do
not say, just that these equations describe relativity in the correct manner.

You may want to have a look at Hafele and Keating, for observations
on real clocks on real airplanes. (and the explanations thereof)

Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the
scientific community.

I see that you have become more cautious about th experimental basis.
My take on it: It all started with the experiments
of Weber and Kohlrausch,
who showed that there must be a fundamental velocity
hiding in electricity and magnetism. (equal to the velocity of light)
Maxwell found a wave equation that incorporated this,
with the internal velocity indeed being the velocity of light.

This posed a puzzle: what is this velocity,
and what is it with respect to?
In particular, in which frame should Maxwell's equations be valid?

For 25 years people floundered, without finding the correct solution.
All partial explanations conflicted with other partial explanations.

Einstein's revolutionary flash of insight, sometime in spring 1905,
was that the answer must be:
Maxwell's equations are valid in every inertial frame!
(with the same fundamental velocity for all)

The Einstein 1905 paper on it is best seen as didactics,
explaining how this at first sight perplexing answer is possible.
(with profound implications for the nature of space-time)

This is why that 'obscure patent clerk' had such an immediate impact:
he explained what 'everybody' knew already,
from an entirely new viewpoint,
thereby resolving all problems with electromagnetism at one go. [1]

Jan

[1] This is also the reason for the lack of references.
None were needed, because whatever could have been refered too
was common knowledge, for the intended readership.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to J. J. Lodder on Tue Oct 10 02:41:39 2023
On Tuesday, 10 October 2023 at 10:23:37 UTC+2, J. J. Lodder wrote:

My take on it: It all started with the experiments
of Weber and Kohlrausch,
who showed that there must be a fundamental velocity
hiding in electricity and magnetism. (equal to the velocity of light)
Maxwell found a wave equation that incorporated this,
with the internal velocity indeed being the velocity of light.

This posed a puzzle: what is this velocity,
and what is it with respect to?
In particular, in which frame should Maxwell's equations be valid?

For 25 years people floundered, without finding the correct solution.
All partial explanations conflicted with other partial explanations.

Cut this mystical mumbo jumbo. Speed is a
derivative of 2 coordinates, speed is a result of
a conventiion. That's why it needs postulates.

The Einstein 1905 paper on it is best seen as didactics,

Poor halfbrain was mumbling inconsistently.

This is why that 'obscure patent clerk' had such an immediate impact:
he explained what 'everybody' knew already,
from an entirely new viewpoint,

So did Jesus Christ.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to Sylvia Else on Tue Oct 10 02:29:03 2023
On Tuesday, 10 October 2023 at 10:08:44 UTC+2, Sylvia Else wrote:

But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of the slower clock
gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book. His problem was
that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for it. The Galilean
transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

Do You seriously insist it violates GT?
Maybe it also violates Euclidean geometry

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to J. J. Lodder on Tue Oct 10 03:54:24 2023
On Tuesday, October 10, 2023 at 1:23:37 AM UTC-7, J. J. Lodder wrote:
Sylvia Else <syl...@email.invalid> wrote:

On 10-Oct-23 5:27 am, Robert Winn wrote:
[-]
Since distances are the same in all of these Galilean equations, vt = -m'n'. So the velocity of the ground relative to the airplane would be m' =
-vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate that there is no need for the
miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
You may want to have a look at Hafele and Keating, for observations
on real clocks on real airplanes. (and the explanations thereof)

It's not as if Einstein, as a young patent clerk, had some ability to impose his theory on an unwilling world. Experimenters had been taking a very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the
scientific community.
I see that you have become more cautious about th experimental basis.
My take on it: It all started with the experiments
of Weber and Kohlrausch,
who showed that there must be a fundamental velocity
hiding in electricity and magnetism. (equal to the velocity of light) Maxwell found a wave equation that incorporated this,
with the internal velocity indeed being the velocity of light.

This posed a puzzle: what is this velocity,
and what is it with respect to?
In particular, in which frame should Maxwell's equations be valid?

For 25 years people floundered, without finding the correct solution.
All partial explanations conflicted with other partial explanations.

Einstein's revolutionary flash of insight, sometime in spring 1905,
was that the answer must be:
Maxwell's equations are valid in every inertial frame!
(with the same fundamental velocity for all)

The Einstein 1905 paper on it is best seen as didactics,
explaining how this at first sight perplexing answer is possible.
(with profound implications for the nature of space-time)

This is why that 'obscure patent clerk' had such an immediate impact:
he explained what 'everybody' knew already,
from an entirely new viewpoint,
thereby resolving all problems with electromagnetism at one go. [1]

Jan

[1] This is also the reason for the lack of references.
None were needed, because whatever could have been refered too
was common knowledge, for the intended readership.
Yes, I know all about the Hafele-Keating experiment. If clocks were flown around the earth one way, they were slower than clocks on the ground. If they were flown around the earth the other way, they were faster. That does not matter to the Galilean
transformation equations. There is no length contraction in the Galilean transformation equations. So we have
x'=x-vt
x-x' = vt

x = x' - m'n' n' is the time of a clock that does not agree with t'=t, but has a different rate, slower or faster. m' is the velocity of frame of reference S relative to frame of reference S' as shown by the time of the clock with time n'.
x - x' = -m'n'
-m'n'= vt
m' = -vt/n'
If n' is less than t, m' is a faster velocity. If n' is greater than t, m' is a slower velocity. No length contraction, no miracles, just junior high algebra. So this is the way I will continue to work relativity. I am sorry if scientists get upset
by this, but that is just the way it is.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Sylvia Else on Tue Oct 10 03:38:27 2023
On Tuesday, October 10, 2023 at 1:08:44 AM UTC-7, Sylvia Else wrote:
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a
clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get
a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on
the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for
x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate
that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a >> very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why >> a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of today do not
consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is faster than a
clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves. I am explaining
relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the Lorentz equations that he
said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of the slower clock
gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book. His problem was
that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for it. The Galilean
transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia.
Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light. Fizeau did not get the result he expected. Instead, a smaller speed than the speed of the water was shown to
increase the speed of the light. Later experiments with better apparatus showed that the slight increase in speed was because of the size of the pipes used by Fizeau. So according to scientists, the result was essentially the same as the Michelson-
Morley experiment, which I already explained. Einstein said that the results of the Michelson-Morley experiment were explained by two little equations which he said he extracted from the Lorentz equations, x=ct and x'=ct'. As I explained, these
equations will not work in the Galilean transformation equations because t' is defined to equal t. If there is a slower or faster clock that does not agree with t, the time of the clock that is not moving, then you have to use another set of Galilean
transformation equations with different variables for time and velocity than in the equations that show x'=x-vt because t' is defined to equal t. So, as I explained, I said that velocity as calculated from the time of the moving clock was m' and the
time of the moving clock was n'. So then these Galilean transformation equations calculated from the time of the slower clock would be
x = x' - m'n'
y = y'
z = z'
n = n'
What these equations mean is that the moving frame of reference is being used as a preferred frame, and the time of the clock in S'(x',y',z',n') is being used in both
frames of reference. There is a faster clock in S(x,y,z,t), but the time of that clock is shown by the Galilean transformation equations
x'=x-vt
y'=y
z'=z
t'=t
It takes two sets of Galilean transformation equations to show the times of both clocks.
What scientists do is to say that a second is always the same amount of time, even if the earth rotates a different number of degrees during a second measured by a slower clock. So whatever scientists do with what they call atomic or scientific time is
going to include a length contraction if they use the Lorentz equations. All of these things proposed by scientists are minor miracles which are measured in nanoseconds and extremely small differences in speed concerning anything that can be measured.
I just go ahead and use the Galilean transformation equations, which do not incorporate any of these little miracles. That is not to say that miracles do not occur. I just work the math without them because I believe that a pilot of an airplane with a
slower clock than a clock on the ground will get a faster speed for his airplane than an observer on the ground, or a faster clock in a GPS satellite indicates a slower speed for the satellite than an observer on the ground would get.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Sylvia Else@21:1/5 to Robert Winn on Tue Oct 10 22:50:01 2023
On 10-Oct-23 9:38 pm, Robert Winn wrote:

Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light. Fizeau did not get the result he expected. Instead, a smaller speed than the speed of the water was shown to
increase the speed of the light. Later experiments with better apparatus showed that the slight increase in speed was because of the size of the pipes used by Fizeau.

If you're going to just outright lie, there's not a lot of point to this interaction.

Sylvia.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Sylvia Else on Tue Oct 10 04:18:02 2023
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a
clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get
a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on
the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for
x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate
that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a >> very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why >> a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of today do not
consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is faster than a
clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves. I am explaining
relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the Lorentz equations that he
said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of the slower clock
gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book. His problem was
that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for it. The Galilean
transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia

No need for relativity. A classical model does just fine.
In Fizeau the variable speed is proportional to the the 'extra distance' through
the water that the light has to travel. In the non moving experiment lightspeed
in the water is defined by c/n.
When the water moves relative to the source the lightspeed changes to reflect the increase or decrease in optical density. And so for instance when the water
moves towards the source this can be modelled mathematically as the refractive index of the refractive index for the 'extra distance' travelled.
As the light effectively travels through 'more water' to get the same distance from source to detector.
So for example in a non moving experiment light travels through x distance of water.
When the water moves, light has to travel through an extra distance y of water. So the change in speed from c/n is +- the refractive index of the refractive index
of the extra distance of water travelled as defined by
C+-(V x .67)/n
Where .67 =1-(1-n)

This traditional Fresnel equation can also be expressed as c/n+-v{(1-n)+(1-n)^2}

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Robert Winn on Tue Oct 10 05:16:00 2023
On Tuesday, 10 October 2023 at 11:54:26 UTC+1, Robert Winn wrote:
On Tuesday, October 10, 2023 at 1:23:37 AM UTC-7, J. J. Lodder wrote:
Sylvia Else <syl...@email.invalid> wrote:

On 10-Oct-23 5:27 am, Robert Winn wrote:
[-]
Since distances are the same in all of these Galilean equations, vt = -m'n'. So the velocity of the ground relative to the airplane would be m' =
-vt/n', a faster speed for the airplane if the clock in the airplane is
slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
You may want to have a look at Hafele and Keating, for observations
on real clocks on real airplanes. (and the explanations thereof)

It's not as if Einstein, as a young patent clerk, had some ability to impose his theory on an unwilling world. Experimenters had been taking a very close look at reality, and had been finding that it was not behaving in the expected way. Einstein provided a solution. That is why a young patent clerk was able to get his theory accepted by the scientific community.
I see that you have become more cautious about th experimental basis.
My take on it: It all started with the experiments
of Weber and Kohlrausch,
who showed that there must be a fundamental velocity
hiding in electricity and magnetism. (equal to the velocity of light) Maxwell found a wave equation that incorporated this,
with the internal velocity indeed being the velocity of light.

This posed a puzzle: what is this velocity,
and what is it with respect to?
In particular, in which frame should Maxwell's equations be valid?

For 25 years people floundered, without finding the correct solution.
All partial explanations conflicted with other partial explanations.

Einstein's revolutionary flash of insight, sometime in spring 1905,
was that the answer must be:
Maxwell's equations are valid in every inertial frame!
(with the same fundamental velocity for all)

The Einstein 1905 paper on it is best seen as didactics,
explaining how this at first sight perplexing answer is possible.
(with profound implications for the nature of space-time)

This is why that 'obscure patent clerk' had such an immediate impact:
he explained what 'everybody' knew already,
from an entirely new viewpoint,
thereby resolving all problems with electromagnetism at one go. [1]

Jan

[1] This is also the reason for the lack of references.
None were needed, because whatever could have been refered too
was common knowledge, for the intended readership.
Yes, I know all about the Hafele-Keating experiment. If clocks were flown around the earth one way, they were slower than clocks on the ground. If they were flown around the earth the other way, they were faster. That does not matter to the Galilean
transformation equations. There is no length contraction in the Galilean transformation equations. So we have
x'=x-vt
x-x' = vt

x = x' - m'n' n' is the time of a clock that does not agree with t'=t, but has a different rate, slower or faster. m' is the velocity of frame of reference S relative to frame of reference S' as shown by the time of the clock with time n'.
x - x' = -m'n'
-m'n'= vt
m' = -vt/n'
If n' is less than t, m' is a faster velocity. If n' is greater than t, m' is a slower velocity. No length contraction, no miracles, just junior high algebra. So this is the way I will continue to work relativity. I am sorry if scientists get upset by
this, but that is just the way it is.

No need for relativity to explain Hafael Keating.
Look at classical resonance. It has been known for centuries that
the natural resonant frequency of a system will reduce its frequency
if subject to an external force. So an atom, also confirmed by all observations to date to be a resonant system, will also reduce its frequency when subject to external force. As we see happen where less g force with altitude increases the atoms ‘ticking’.
The same occurs in Hafael Keating. The eastward travelling plane experiences more force than the westward plane relative to the earths Center. Because it travels at a greater speed relative to the earth Center, than the westward plane.
Which means the eastward plane (and its caesium atoms) will tick slower
than the westward plane.
As observed

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Dono.@21:1/5 to Lou on Tue Oct 10 06:54:32 2023
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:
The eastward travelling plane experiences
more force than the westward plane relative to the earths Center. Because it travels at a greater speed relative to the earth Center, than the westward plane.
Which means the eastward plane (and its caesium atoms) will tick slower
than the westward plane.

You take the prize. The utter cretinism prize

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Robert Winn on Tue Oct 10 11:13:01 2023
On 10/10/2023 6:38 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 1:08:44 AM UTC-7, Sylvia Else wrote:

How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia.

Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light.

No, it was not. While both used light fringes to detect changes of
speeds, there the similarities end. Fizeau measured light speed through
moving water while MMX was to detect the speed of earth through a
purported stationary ether.

Fizeau did not get the result he expected. Instead, a smaller speed than the speed of the water was shown to increase the speed of the light. Later experiments with better apparatus showed that the slight increase in speed was because of the size of
the pipes used by Fizeau.

Wrong, there were concerns the pipe diameter affected the light. Later experiments using better equipment CONFIRMED Fizeau's results.

So according to scientists, the result was essentially the same as the Michelson-Morley experiment,

Wrong.

which I already explained. Einstein said that the results of the Michelson-Morley experiment were explained by two little equations which he said he extracted from the Lorentz equations, x=ct and x'=ct'.

No, the Fizeau experiment can be explained by the Lorentzian speed
combination formula, with the two speeds being v (speed of the water)
and c/n (speed of light in [stationary] water). It was one of a couple
of experiments which inspired Einstein to come up with his 1905 paper.

As I explained, these equations will not work in the Galilean transformation equations because
they are incorrect for this situation (dealing with speeds close to c,
in this case c/n). For low speeds the difference between the Lorentzian transformation and the Galilean transformation (simple addition) is too
small to matter.

[snip babble]

I see that in the couple of years that you've been gone, you haven't
learned anything.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Lou on Tue Oct 10 11:47:38 2023
On 10/10/2023 7:18 AM, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:

How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

And so for instance when the water
moves towards the source this can be modelled mathematically as the refractive
index of the refractive index for the 'extra distance' travelled.
As the light effectively travels through 'more water' to get the same distance
from source to detector.

Obviously bogus. If that were so, the speed of light in water would get
slower and slower as it traversed through more and more water, even if stationary. Instead, the speed of light in (stationary) water is a
constant c/n.

You are grasping at straws.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Dono. on Tue Oct 10 08:51:25 2023
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:
The eastward travelling plane experiences
more force than the westward plane relative to the earths Center. Because it
travels at a greater speed relative to the earth Center, than the westward plane.
Which means the eastward plane (and its caesium atoms) will tick slower than the westward plane.
You take the prize. The utter cretinism prize
Good to see you, Dono. When did I lose out on the utter cretinism prize?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Sylvia Else on Tue Oct 10 08:49:20 2023
On Tuesday, October 10, 2023 at 4:50:06 AM UTC-7, Sylvia Else wrote:
On 10-Oct-23 9:38 pm, Robert Winn wrote:

Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light. Fizeau did not get the result he expected. Instead, a smaller speed than the speed of the water was shown to
increase the speed of the light. Later experiments with better apparatus showed that the slight increase in speed was because of the size of the pipes used by Fizeau.
If you're going to just outright lie, there's not a lot of point to this interaction.

Sylvia.
I did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name. Whenever I talk with scientists I try to answer all of their posts. This is how I learn. While scientists never answer anything I say,
they will always find fault with my answers to anything they say. You asked, What about the Fizeau experiment? So I looked up the Fizeau experiment and said, Well, here is what I remember about what I just read concerning the Fizeau experiment. It is
one way to get a bunch of scientists to engage me in conversation. So how about the pilot of the airplane with the slower clock? Do you believe in the miracle Einstein describes or do you believe in the reality we common people live in where if the
pilot of the airplane has a slower clock he will get a faster speed for the airplane than an observer on the ground with a faster clock?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Dono.@21:1/5 to Robert Winn on Tue Oct 10 08:53:58 2023
On Tuesday, October 10, 2023 at 8:51:28 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:
The eastward travelling plane experiences
more force than the westward plane relative to the earths Center. Because it
travels at a greater speed relative to the earth Center, than the westward plane.
Which means the eastward plane (and its caesium atoms) will tick slower than the westward plane.
You take the prize. The utter cretinism prize
Good to see you, Dono. When did I lose out on the utter cretinism prize?

You are always in the running. Along with Dick Hertz and Pattycakes Dolan.
Lou is just a newcomer.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Robert Winn on Tue Oct 10 12:01:03 2023
On 10/10/2023 11:49 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 4:50:06 AM UTC-7, Sylvia Else wrote:
On 10-Oct-23 9:38 pm, Robert Winn wrote:

Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light. Fizeau did not get the result he expected. Instead, a smaller speed than the speed of the water was shown to
increase the speed of the light. Later experiments with better apparatus showed that the slight increase in speed was because of the size of the pipes used by Fizeau.
If you're going to just outright lie, there's not a lot of point to this
interaction.

I did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name.

If you never knew anything about it, why did you lie and say it was a
version of the MMX?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Dono. on Tue Oct 10 09:04:23 2023
On Tuesday, October 10, 2023 at 8:54:00 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 8:51:28 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:
The eastward travelling plane experiences
more force than the westward plane relative to the earths Center. Because it
travels at a greater speed relative to the earth Center, than the westward plane.
Which means the eastward plane (and its caesium atoms) will tick slower
than the westward plane.
You take the prize. The utter cretinism prize
Good to see you, Dono. When did I lose out on the utter cretinism prize?
You are always in the running. Along with Dick Hertz and Pattycakes Dolan. Lou is just a newcomer.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Volney on Tue Oct 10 09:02:23 2023
On Tuesday, October 10, 2023 at 8:13:08 AM UTC-7, Volney wrote:
On 10/10/2023 6:38 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 1:08:44 AM UTC-7, Sylvia Else wrote:

How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia.

Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light.
No, it was not. While both used light fringes to detect changes of
speeds, there the similarities end. Fizeau measured light speed through moving water while MMX was to detect the speed of earth through a
purported stationary ether.
Fizeau did not get the result he expected. Instead, a smaller speed than the speed of the water was shown to increase the speed of the light. Later experiments with better apparatus showed that the slight increase in speed was because of the size of
the pipes used by Fizeau.
Wrong, there were concerns the pipe diameter affected the light. Later experiments using better equipment CONFIRMED Fizeau's results.
So according to scientists, the result was essentially the same as the Michelson-Morley experiment,
Wrong.
which I already explained. Einstein said that the results of the Michelson-Morley experiment were explained by two little equations which he said he extracted from the Lorentz equations, x=ct and x'=ct'.
No, the Fizeau experiment can be explained by the Lorentzian speed combination formula, with the two speeds being v (speed of the water)
and c/n (speed of light in [stationary] water). It was one of a couple
of experiments which inspired Einstein to come up with his 1905 paper.
As I explained, these equations will not work in the Galilean transformation equations because
they are incorrect for this situation (dealing with speeds close to c,
in this case c/n). For low speeds the difference between the Lorentzian transformation and the Galilean transformation (simple addition) is too small to matter.

[snip babble]

I see that in the couple of years that you've been gone, you haven't
learned anything.
Well, nobody ever asked me about the Fizeau experiment before. The only thing I knew about it was its name until this morning. So I looked it up. This is something I have always done with scientists. When they ask me a question, I try to answer it.
Then a bunch of scientists will jump into the conversation with their criticisms of my response. That is how I learn about science. We common people often converse in this manner. Scientists are different. I have asked scientists about Einstein's
description of a miracle for about forty years. None have ever answered me. There have been some who called me names and tried to insult me. So, what do you think about Einstein's description of a miracle? Would a pilot of an airplane with a slower
clock and an observer on the ground with a faster clock get the same speed for the airplane the way Einstein describes with his equations in Special Relativity?
We common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Volney on Tue Oct 10 09:07:31 2023
On Tuesday, October 10, 2023 at 9:01:09 AM UTC-7, Volney wrote:
On 10/10/2023 11:49 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 4:50:06 AM UTC-7, Sylvia Else wrote:
On 10-Oct-23 9:38 pm, Robert Winn wrote:

Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light. Fizeau did not get the result he expected. Instead, a smaller speed than the speed of the water was shown to
increase the speed of the light. Later experiments with better apparatus showed that the slight increase in speed was because of the size of the pipes used by Fizeau.
If you're going to just outright lie, there's not a lot of point to this >> interaction.
I did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name.
If you never knew anything about it, why did you lie and say it was a version of the MMX?
Well, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Dono. on Tue Oct 10 09:05:32 2023
On Tuesday, October 10, 2023 at 8:54:00 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 8:51:28 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:
The eastward travelling plane experiences
more force than the westward plane relative to the earths Center. Because it
travels at a greater speed relative to the earth Center, than the westward plane.
Which means the eastward plane (and its caesium atoms) will tick slower
than the westward plane.
You take the prize. The utter cretinism prize
Good to see you, Dono. When did I lose out on the utter cretinism prize?
You are always in the running. Along with Dick Hertz and Pattycakes Dolan. Lou is just a newcomer.
One thing I have noticed about science. You never use any equations. Why is that?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Robert Winn on Tue Oct 10 12:39:49 2023
On 10/10/2023 12:02 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 8:13:08 AM UTC-7, Volney wrote:
On 10/10/2023 6:38 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 1:08:44 AM UTC-7, Sylvia Else wrote:

How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia.

Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light.
No, it was not. While both used light fringes to detect changes of
speeds, there the similarities end. Fizeau measured light speed through
moving water while MMX was to detect the speed of earth through a
purported stationary ether.
Fizeau did not get the result he expected. Instead, a smaller speed than the speed of the water was shown to increase the speed of the light. Later experiments with better apparatus showed that the slight increase in speed was because of the size of
the pipes used by Fizeau.
Wrong, there were concerns the pipe diameter affected the light. Later
experiments using better equipment CONFIRMED Fizeau's results.
So according to scientists, the result was essentially the same as the Michelson-Morley experiment,
Wrong.
which I already explained. Einstein said that the results of the Michelson-Morley experiment were explained by two little equations which he said he extracted from the Lorentz equations, x=ct and x'=ct'.
No, the Fizeau experiment can be explained by the Lorentzian speed
combination formula, with the two speeds being v (speed of the water)
and c/n (speed of light in [stationary] water). It was one of a couple
of experiments which inspired Einstein to come up with his 1905 paper.
As I explained, these equations will not work in the Galilean transformation equations because
they are incorrect for this situation (dealing with speeds close to c,
in this case c/n). For low speeds the difference between the Lorentzian
transformation and the Galilean transformation (simple addition) is too
small to matter.

[snip babble]

I see that in the couple of years that you've been gone, you haven't
learned anything.
Well, nobody ever asked me about the Fizeau experiment before. The only thing I knew about it was its name until this morning.

So if you knew nothing about it, why did you claim it was an MMX experiment?

[snip babble]

As I stated, you have learned nothing.

Your major mistake is using the Galilean transformation where it doesn't
apply. With the Lorentzian transformation, both distance and length are affected as seen from the other frame, but by the same amount (gamma),
so the speeds are the same (after negation). In other words, x/t =
-x'/t'. Invoking the Galilean transformation, where t'=t, gives the

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Dono.@21:1/5 to Robert Winn on Tue Oct 10 09:39:11 2023
On Tuesday, October 10, 2023 at 9:05:35 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 8:54:00 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 8:51:28 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:
The eastward travelling plane experiences
more force than the westward plane relative to the earths Center. Because it
travels at a greater speed relative to the earth Center, than the westward plane.
Which means the eastward plane (and its caesium atoms) will tick slower
than the westward plane.
You take the prize. The utter cretinism prize
Good to see you, Dono. When did I lose out on the utter cretinism prize?
You are always in the running. Along with Dick Hertz and Pattycakes Dolan. Lou is just a newcomer.
One thing I have noticed about science. You never use any equations. Why is that?

You are not only an imbecile, you are a liar as well.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Volney on Tue Oct 10 09:40:25 2023
On Tuesday, 10 October 2023 at 16:47:41 UTC+1, Volney wrote:
On 10/10/2023 7:18 AM, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:

How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment
And so for instance when the water
moves towards the source this can be modelled mathematically as the refractive
index of the refractive index for the 'extra distance' travelled.
As the light effectively travels through 'more water' to get the same distance
from source to detector.
Obviously bogus. If that were so, the speed of light in water would get slower and slower as it traversed through more and more water, even if stationary. Instead, the speed of light in (stationary) water is a
constant c/n.

You are grasping at straws.

You aren’t just grabbing at straws. You are making up the straws.
Slower with more distance? How so? The distance doesn’t change the
outcome of the formula I cited.There is no ‘d’ in the formula.
The only way to make the light go slower and slower as you imagine
would be to have the water move faster and faster until it reaches c.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Robert Winn on Tue Oct 10 12:44:36 2023
On 10/10/2023 12:07 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:01:09 AM UTC-7, Volney wrote:
On 10/10/2023 11:49 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 4:50:06 AM UTC-7, Sylvia Else wrote:
On 10-Oct-23 9:38 pm, Robert Winn wrote:

Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light. Fizeau did not get the result he expected. Instead, a smaller speed than the speed of the water was shown to
increase the speed of the light. Later experiments with better apparatus showed that the slight increase in speed was because of the size of the pipes used by Fizeau.
If you're going to just outright lie, there's not a lot of point to this >>>> interaction.
I did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name.
If you never knew anything about it, why did you lie and say it was a
version of the MMX?
Well, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.

As I said, both were interferometers, but that's where the similarities
end. And you did lie. Fizeau's experiment was not the MMX.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Lou on Tue Oct 10 12:50:43 2023
On 10/10/2023 12:40 PM, Lou wrote:
On Tuesday, 10 October 2023 at 16:47:41 UTC+1, Volney wrote:
On 10/10/2023 7:18 AM, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:

How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment
And so for instance when the water
moves towards the source this can be modelled mathematically as the refractive
index of the refractive index for the 'extra distance' travelled.
As the light effectively travels through 'more water' to get the same distance
from source to detector.
Obviously bogus. If that were so, the speed of light in water would get
slower and slower as it traversed through more and more water, even if
stationary. Instead, the speed of light in (stationary) water is a
constant c/n.

You are grasping at straws.

You aren’t just grabbing at straws. You are making up the straws.
Slower with more distance? How so?

Because you claimed the light slows by going through 'more water' (your
term). Increasing the distance obviously means more water traversed.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Dono. on Tue Oct 10 10:08:06 2023
On Tuesday, October 10, 2023 at 9:39:13 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 9:05:35 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 8:54:00 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 8:51:28 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:
The eastward travelling plane experiences
more force than the westward plane relative to the earths Center. Because it
travels at a greater speed relative to the earth Center, than the westward plane.
Which means the eastward plane (and its caesium atoms) will tick slower
than the westward plane.
You take the prize. The utter cretinism prize
Good to see you, Dono. When did I lose out on the utter cretinism prize?
You are always in the running. Along with Dick Hertz and Pattycakes Dolan.
Lou is just a newcomer.
One thing I have noticed about science. You never use any equations. Why is that?
You are not only an imbecile, you are a liar as well.
Well, I don't really think so. What do you think about Einstein's miracle? Do you agree with him that the pilot of the airplane and the observer on the ground would get the same speed for the airplane? We common people live in something called
reality where if the pilot has a slower clock, he will get a faster speed for the airplane.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Volney on Tue Oct 10 10:11:41 2023
On Tuesday, October 10, 2023 at 9:44:38 AM UTC-7, Volney wrote:
On 10/10/2023 12:07 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:01:09 AM UTC-7, Volney wrote:
On 10/10/2023 11:49 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 4:50:06 AM UTC-7, Sylvia Else wrote: >>>> On 10-Oct-23 9:38 pm, Robert Winn wrote:

Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light. Fizeau did not get the result he expected. Instead, a smaller speed than the speed of the water was shown to
increase the speed of the light. Later experiments with better apparatus showed that the slight increase in speed was because of the size of the pipes used by Fizeau.
If you're going to just outright lie, there's not a lot of point to this
interaction.
I did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name.
If you never knew anything about it, why did you lie and say it was a
version of the MMX?
Well, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.
As I said, both were interferometers, but that's where the similarities
end. And you did lie. Fizeau's experiment was not the MMX.
Well, it seems you are being untruthful. When did I say that Fizeau's experiment was the Michelson-Morley experiment? The Fizeau experiment was in 1851, and the Michelson -Morley experiment was in 1887.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Robert Winn on Tue Oct 10 13:16:07 2023
On 10/10/2023 1:11 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:44:38 AM UTC-7, Volney wrote:
On 10/10/2023 12:07 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:01:09 AM UTC-7, Volney wrote:
On 10/10/2023 11:49 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 4:50:06 AM UTC-7, Sylvia Else wrote: >>>>>> On 10-Oct-23 9:38 pm, Robert Winn wrote:

Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light. Fizeau did not get the result he expected. Instead, a smaller speed than the speed of the water was shown to
increase the speed of the light. Later experiments with better apparatus showed that the slight increase in speed was because of the size of the pipes used by Fizeau.
If you're going to just outright lie, there's not a lot of point to this >>>>>> interaction.
I did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name.
If you never knew anything about it, why did you lie and say it was a
version of the MMX?
Well, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.

As I said, both were interferometers, but that's where the similarities
end. And you did lie. Fizeau's experiment was not the MMX.

Well, it seems you are being untruthful. When did I say that Fizeau's experiment was the Michelson-Morley experiment? The Fizeau experiment was in 1851, and the Michelson -Morley experiment was in 1887.

You said right here, in this thread:
: Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light.

Do you deny writing that?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Volney on Tue Oct 10 10:49:43 2023
On Tuesday, October 10, 2023 at 10:16:10 AM UTC-7, Volney wrote:
On 10/10/2023 1:11 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:44:38 AM UTC-7, Volney wrote:
On 10/10/2023 12:07 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:01:09 AM UTC-7, Volney wrote:
On 10/10/2023 11:49 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 4:50:06 AM UTC-7, Sylvia Else wrote: >>>>>> On 10-Oct-23 9:38 pm, Robert Winn wrote:

Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light. Fizeau did not get the result he expected. Instead, a smaller speed than the speed of the water was shown
to increase the speed of the light. Later experiments with better apparatus showed that the slight increase in speed was because of the size of the pipes used by Fizeau.
If you're going to just outright lie, there's not a lot of point to this
interaction.
I did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name.
If you never knew anything about it, why did you lie and say it was a >>>> version of the MMX?
Well, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.

As I said, both were interferometers, but that's where the similarities >> end. And you did lie. Fizeau's experiment was not the MMX.

Well, it seems you are being untruthful. When did I say that Fizeau's experiment was the Michelson-Morley experiment? The Fizeau experiment was in 1851, and the Michelson -Morley experiment was in 1887.
You said right here, in this thread:
: Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light.
Do you deny writing that?
I wrote it. So how is using water and using air for the same kind of experiment the same experiment?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From J. J. Lodder@21:1/5 to Robert Winn on Tue Oct 10 20:00:45 2023
Robert Winn <rbwinn3@gmail.com> wrote:

On Tuesday, October 10, 2023 at 1:23:37?AM UTC-7, J. J. Lodder wrote:
Sylvia Else <syl...@email.invalid> wrote:

On 10-Oct-23 5:27 am, Robert Winn wrote:
You may want to have a look at Hafele and Keating, for observations
on real clocks on real airplanes. (and the explanations thereof)

It's not as if Einstein, as a young patent clerk, had some ability to impose his theory on an unwilling world. Experimenters had been taking a very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the scientific community.
I see that you have become more cautious about th experimental basis.
My take on it: It all started with the experiments
of Weber and Kohlrausch,
who showed that there must be a fundamental velocity
hiding in electricity and magnetism. (equal to the velocity of light) Maxwell found a wave equation that incorporated this,
with the internal velocity indeed being the velocity of light.

This posed a puzzle: what is this velocity,
and what is it with respect to?
In particular, in which frame should Maxwell's equations be valid?

For 25 years people floundered, without finding the correct solution.
All partial explanations conflicted with other partial explanations.

Einstein's revolutionary flash of insight, sometime in spring 1905,
was that the answer must be:
Maxwell's equations are valid in every inertial frame!
(with the same fundamental velocity for all)

The Einstein 1905 paper on it is best seen as didactics,
explaining how this at first sight perplexing answer is possible.
(with profound implications for the nature of space-time)

This is why that 'obscure patent clerk' had such an immediate impact:
he explained what 'everybody' knew already,
from an entirely new viewpoint,
thereby resolving all problems with electromagnetism at one go. [1]

Jan

[1] This is also the reason for the lack of references.
None were needed, because whatever could have been refered too
was common knowledge, for the intended readership.
Yes, I know all about the Hafele-Keating experiment. If clocks were flown around the earth one way, they were slower than clocks on the ground. If they were flown around the earth the other way, they were faster. That
does not matter to the Galilean transformation equations.

I don't want to bother with your fantasies,
but I hope that it did occur to you
that the Earth, and the atmosphere with it,
is rotating?

Jan

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Paul B. Andersen@21:1/5 to All on Tue Oct 10 20:10:55 2023
Den 10.10.2023 14:16, skrev Lou:

No need for relativity to explain Hafael Keating.
Look at classical resonance. It has been known for centuries that
the natural resonant frequency of a system will reduce its frequency
if subject to an external force. So an atom, also confirmed by all observations
to date to be a resonant system, will also reduce its frequency when subject to external force. As we see happen where less g force with altitude increases
the atoms ‘ticking’.

So clocks at higher altitude in a plane will be subject to
lower g-force and will tick faster than clocks on the ground, right?

So why did the east going clock in the plane tick slower than
the clock on the ground?

The same occurs in Hafael Keating. The eastward travelling plane experiences more force than the westward plane relative to the earths Center. Because it travels at a greater speed relative to the earth Center, than the westward plane.
Which means the eastward plane (and its caesium atoms) will tick slower
than the westward plane.
As observed

I see.
The g-force on east going clocks is higher than the g-force
on west going clocks, so east going clocks will tick
slower than west going clocks.

Is it a personal experience of yours that you are heavier when
you sit in an east going plane than when you sit in an east going?

--
Paul :-J

https://paulba.no/

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From xip14@21:1/5 to All on Tue Oct 10 11:57:23 2023
The derivation of 1905 uses a version of the temporal Lorentz Transform. Same thing in today’s textbooks:

t′ = gamma × ( t – vx / c² )

Axis X ( unprimed ) is stationary and axis X′ ( frame k of 1905 ) moves with positive speed-v. The stationary observer is at fixed point-x on axis X ( not x = 0 ) whereas the moving clock to be observed ( always at origin [ x′ = 0 ] of axis X′ )
moves to location-x on axis X according to an equation of motion on axis X:

x = vt

Substitute that into the temporal LT:

t′ = gamma × ( t – v² t / c² )

t′ = gamma × t × ( 1 – v²/c² )

t′ = ( gamma × t ) / gamma²

t′ = t / gamma

It wouldn’t work with a plus sign. Why do we have minus signs in the Lorentz Transforms ( temporal and spatial ) when speed-v is positive?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Volney on Tue Oct 10 12:23:42 2023
On Tuesday, 10 October 2023 at 17:50:46 UTC+1, Volney wrote:
On 10/10/2023 12:40 PM, Lou wrote:
On Tuesday, 10 October 2023 at 16:47:41 UTC+1, Volney wrote:
On 10/10/2023 7:18 AM, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:

How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment
And so for instance when the water
moves towards the source this can be modelled mathematically as the refractive
index of the refractive index for the 'extra distance' travelled.
As the light effectively travels through 'more water' to get the same distance
from source to detector.
Obviously bogus. If that were so, the speed of light in water would get >> slower and slower as it traversed through more and more water, even if
stationary. Instead, the speed of light in (stationary) water is a
constant c/n.

You are grasping at straws.

You aren’t just grabbing at straws. You are making up the straws.
Slower with more distance? How so?
Because you claimed the light slows by going through 'more water' (your term). Increasing the distance obviously means more water traversed.

Grabbing at fantasy straws still. You snipped the formula I cited. There is *no*
‘d’ in that formula. Just v of the water. Notice both Fizeau and SR formulae
also have v. Notice also that when the water column moves between the
source and detector although the distance between the source and detector doesn’t change, the distance travelled through the water column does increase.
But the time taken to to travel this extra distance is not defined by the v
of the water relative to the source. As I described and re-quoted below.
In a quote from my last post which you conveniently snipped (notice that contrary to your fantasy, there is no ‘d’ in the formula):

“When the water moves, light has to travel through an extra distance y of water.
So the change in speed from c/n is +- the refractive index of the refractive index
of the extra distance of water travelled as defined by
C+-(V x .67)/n
Where .67 =1-(1-n)
This traditional Fresnel equation can also be expressed as c/n+-v{(1-n)+(1-n)^2} “

No Relativity needed to correctly model Fizeau classically.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Dono.@21:1/5 to Robert Winn on Tue Oct 10 12:20:48 2023
On Tuesday, October 10, 2023 at 10:08:08 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:39:13 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 9:05:35 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 8:54:00 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 8:51:28 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:
The eastward travelling plane experiences
more force than the westward plane relative to the earths Center. Because it
travels at a greater speed relative to the earth Center, than the westward plane.
Which means the eastward plane (and its caesium atoms) will tick slower
than the westward plane.
You take the prize. The utter cretinism prize
Good to see you, Dono. When did I lose out on the utter cretinism prize?
You are always in the running. Along with Dick Hertz and Pattycakes Dolan.
Lou is just a newcomer.
One thing I have noticed about science. You never use any equations. Why is that?
You are not only an imbecile, you are a liar as well.
Well, I don't really think so.
That you are liar and an imbecile? Well, it is an established fact.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Paul B. Andersen on Tue Oct 10 12:50:51 2023
On Tuesday, 10 October 2023 at 19:10:11 UTC+1, Paul B. Andersen wrote:
Den 10.10.2023 14:16, skrev Lou:

No need for relativity to explain Hafael Keating.
Look at classical resonance. It has been known for centuries that
the natural resonant frequency of a system will reduce its frequency
if subject to an external force. So an atom, also confirmed by all observations
to date to be a resonant system, will also reduce its frequency when subject
to external force. As we see happen where less g force with altitude increases
the atoms ‘ticking’.
So clocks at higher altitude in a plane will be subject to
lower g-force and will tick faster than clocks on the ground, right?

No I’m suggesting that this horizontal force is not force due to gravity.
But a seperate force from horizontal acceleration .
Isn’t the formula for this f=ma?

So why did the east going clock in the plane tick slower than
the clock on the ground?

Im assuming that Hafael Keating observed that the eastward clock ticks slower. That’s my reading of the wiki reference.
But to answer ‘why’...I assume the speed relative to the ground is the same for both
planes in the experiment. Let’s call it speed x.
The earth rotates eastward at 1600k/ hr.
So the zero point of reference is the earth observer travelling at 1600 k/hr relative
to the earths Center of mass.
Then relative to this earth center reference, the eastward plane travels at 1600 +x kilometers per hour. And the westward plane travels at 1600-x kilometers
per hour.
The eastward plane therefore experiences greater F than earth observer
And the westward plane lesser F than the the earth observer.
And seeing as a classical model uses resonance as the explanation
for this different ticking rates of the caesium atoms natural resonant frequency.
Then the westward clock has less F from horizontal
speed than the earth observer and the westward plane has a greater F than
the earth observer. Which accounts for the 3 different rates of ticking. (Don’t forget...the earth observers ‘ticking rate’ is in part due to Gravity force
but also in part due to its rotational/ horizontal speed of 1600k/hr )

The same occurs in Hafael Keating. The eastward travelling plane experiences
more force than the westward plane relative to the earths Center. Because it
travels at a greater speed relative to the earth Center, than the westward plane.
Which means the eastward plane (and its caesium atoms) will tick slower than the westward plane.
As observed
I see.
The g-force on east going clocks is higher than the g-force
on west going clocks, so east going clocks will tick
slower than west going clocks.

Sort of. But don’t forget the Gravity force pushing you down is a seperate source of
force from the F force pushing against you as you move horizontally. I’m not sure
what word you relativists prefer but probably inertia or momentum would
be appropriate?

Is it a personal experience of yours that you are heavier when
you sit in an east going plane than when you sit in an east going?

Is it your personal experience that you were lighter when you last
hiked up in the mountains of Norway?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Paul B. Andersen on Tue Oct 10 12:52:26 2023
On Tuesday, October 10, 2023 at 11:10:11 AM UTC-7, Paul B. Andersen wrote:
Den 10.10.2023 14:16, skrev Lou:

No need for relativity to explain Hafael Keating.
Look at classical resonance. It has been known for centuries that
the natural resonant frequency of a system will reduce its frequency
if subject to an external force. So an atom, also confirmed by all observations
to date to be a resonant system, will also reduce its frequency when subject
to external force. As we see happen where less g force with altitude increases
the atoms ‘ticking’.
So clocks at higher altitude in a plane will be subject to
lower g-force and will tick faster than clocks on the ground, right?

So why did the east going clock in the plane tick slower than
the clock on the ground?
The same occurs in Hafael Keating. The eastward travelling plane experiences
more force than the westward plane relative to the earths Center. Because it
travels at a greater speed relative to the earth Center, than the westward plane.
Which means the eastward plane (and its caesium atoms) will tick slower than the westward plane.
As observed
I see.
The g-force on east going clocks is higher than the g-force
on west going clocks, so east going clocks will tick
slower than west going clocks.

Is it a personal experience of yours that you are heavier when
you sit in an east going plane than when you sit in an east going?

--
Paul :-J

https://paulba.no/
Here is something you might want to think about. According to modern interpretation of science, Galileo's principle of equivalence no longer applies. If we drop a ball from the ceiling of an airplane that is flying, the ball is falling faster in the
frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to J. J. Lodder on Tue Oct 10 12:45:52 2023
On Tuesday, October 10, 2023 at 11:00:50 AM UTC-7, J. J. Lodder wrote:
Robert Winn <rbw...@gmail.com> wrote:
On Tuesday, October 10, 2023 at 1:23:37?AM UTC-7, J. J. Lodder wrote:
Sylvia Else <syl...@email.invalid> wrote:

On 10-Oct-23 5:27 am, Robert Winn wrote:
You may want to have a look at Hafele and Keating, for observations
on real clocks on real airplanes. (and the explanations thereof)

It's not as if Einstein, as a young patent clerk, had some ability to impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the scientific community.
I see that you have become more cautious about th experimental basis.
My take on it: It all started with the experiments
of Weber and Kohlrausch,
who showed that there must be a fundamental velocity
hiding in electricity and magnetism. (equal to the velocity of light) Maxwell found a wave equation that incorporated this,
with the internal velocity indeed being the velocity of light.

This posed a puzzle: what is this velocity,
and what is it with respect to?
In particular, in which frame should Maxwell's equations be valid?

For 25 years people floundered, without finding the correct solution. All partial explanations conflicted with other partial explanations.

Einstein's revolutionary flash of insight, sometime in spring 1905,
was that the answer must be:
Maxwell's equations are valid in every inertial frame!
(with the same fundamental velocity for all)

The Einstein 1905 paper on it is best seen as didactics,
explaining how this at first sight perplexing answer is possible.
(with profound implications for the nature of space-time)

This is why that 'obscure patent clerk' had such an immediate impact:
he explained what 'everybody' knew already,
from an entirely new viewpoint,
thereby resolving all problems with electromagnetism at one go. [1]

Jan

[1] This is also the reason for the lack of references.
None were needed, because whatever could have been refered too
was common knowledge, for the intended readership.
Yes, I know all about the Hafele-Keating experiment. If clocks were flown around the earth one way, they were slower than clocks on the ground. If they were flown around the earth the other way, they were faster. That does not matter to the Galilean transformation equations.
I don't want to bother with your fantasies,
but I hope that it did occur to you
that the Earth, and the atmosphere with it,
is rotating?

Jan
You said something about me misunderstanding planes. It is entirely possible. My father was a pilot in World War II. He could probably answer your questions about airplanes better than I could, but he is dead. So exactly what was it that you thought
I misunderstood?
I did understand about the rotation of the earth. In fact, rotation of the earth was what directed my attention to the Galilean transformation equations in the first place. Scientists say that a second is a certain number of transitions of a cesium
atom. But if a cesium atom that is moving relative to earth has more or less transitions than a cesium atom that is not moving relative to earth, such as Hafele and Keating described as finding in their experiment, then you have two different
definitions for a second during the same number of degrees of rotation of the earth.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Dono. on Tue Oct 10 12:53:32 2023
On Tuesday, October 10, 2023 at 12:20:50 PM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 10:08:08 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:39:13 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 9:05:35 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 8:54:00 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 8:51:28 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:
The eastward travelling plane experiences
more force than the westward plane relative to the earths Center. Because it
travels at a greater speed relative to the earth Center, than the westward plane.
Which means the eastward plane (and its caesium atoms) will tick slower
than the westward plane.
You take the prize. The utter cretinism prize
Good to see you, Dono. When did I lose out on the utter cretinism prize?
You are always in the running. Along with Dick Hertz and Pattycakes Dolan.
Lou is just a newcomer.
One thing I have noticed about science. You never use any equations. Why is that?
You are not only an imbecile, you are a liar as well.
Well, I don't really think so.
That you are liar and an imbecile? Well, it is an established fact.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Dono. on Tue Oct 10 12:54:41 2023
On Tuesday, October 10, 2023 at 12:20:50 PM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 10:08:08 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:39:13 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 9:05:35 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 8:54:00 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 8:51:28 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:
The eastward travelling plane experiences
more force than the westward plane relative to the earths Center. Because it
travels at a greater speed relative to the earth Center, than the westward plane.
Which means the eastward plane (and its caesium atoms) will tick slower
than the westward plane.
You take the prize. The utter cretinism prize
Good to see you, Dono. When did I lose out on the utter cretinism prize?
You are always in the running. Along with Dick Hertz and Pattycakes Dolan.
Lou is just a newcomer.
One thing I have noticed about science. You never use any equations. Why is that?
You are not only an imbecile, you are a liar as well.
Well, I don't really think so.
That you are liar and an imbecile? Well, it is an established fact.
But, Dono, all you ever do is make accusations. You never really say anything.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to Paul B. Andersen on Tue Oct 10 14:21:41 2023
On Tuesday, 10 October 2023 at 20:10:11 UTC+2, Paul B. Andersen wrote:

So why did the east going clock in the plane tick slower than
the clock on the ground?

Because your Holiest Postulate is such an absurd
that not even you stick to it.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From mitchrae3323@gmail.com@21:1/5 to Maciej Wozniak on Tue Oct 10 18:47:48 2023
On Tuesday, October 10, 2023 at 2:21:44 PM UTC-7, Maciej Wozniak wrote:
On Tuesday, 10 October 2023 at 20:10:11 UTC+2, Paul B. Andersen wrote:

So why did the east going clock in the plane tick slower than
the clock on the ground?
Because your Holiest Postulate is such an absurd
that not even you stick to it.

How many Cesium atoms are necessary?
What is the atom doing and what could count it?
No. The atomic clock drifts instead...

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Dono.@21:1/5 to Robert Winn on Tue Oct 10 18:53:09 2023
On Tuesday, October 10, 2023 at 12:54:44 PM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 12:20:50 PM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 10:08:08 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:39:13 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 9:05:35 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 8:54:00 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 8:51:28 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:
The eastward travelling plane experiences
more force than the westward plane relative to the earths Center. Because it
travels at a greater speed relative to the earth Center, than the westward plane.
Which means the eastward plane (and its caesium atoms) will tick slower
than the westward plane.
You take the prize. The utter cretinism prize
Good to see you, Dono. When did I lose out on the utter cretinism prize?
You are always in the running. Along with Dick Hertz and Pattycakes Dolan.
Lou is just a newcomer.
One thing I have noticed about science. You never use any equations. Why is that?
You are not only an imbecile, you are a liar as well.
Well, I don't really think so.
That you are liar and an imbecile? Well, it is an established fact.
But, Dono, all you ever do is make accusations. You never really say anything.

I say that you are an idiot and a liar. This is a fact.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Dono.@21:1/5 to Lou on Tue Oct 10 19:55:52 2023
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:

So the change in speed from c/n is +- the refractive index of the refractive index
of the extra distance of water travelled as defined by
C+-(V x .67)/n
Where .67 =1-(1-n)

This traditional Fresnel equation can also be expressed as c/n+-v{(1-n)+(1-n)^2}

Crank ,

Experiment measures c/n+v(1-1/n^2), not the idiocy you posted above.
SR predicts c/n+v(1-1/n^2), in accordance with the experiment.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Robert Winn on Tue Oct 10 23:26:07 2023
On 10/10/2023 1:49 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 10:16:10 AM UTC-7, Volney wrote:
On 10/10/2023 1:11 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:44:38 AM UTC-7, Volney wrote:
On 10/10/2023 12:07 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:01:09 AM UTC-7, Volney wrote:
On 10/10/2023 11:49 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 4:50:06 AM UTC-7, Sylvia Else wrote: >>>>>>>> On 10-Oct-23 9:38 pm, Robert Winn wrote:

Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light. Fizeau did not get the result he expected. Instead, a smaller speed than the speed of the water was shown
to increase the speed of the light. Later experiments with better apparatus showed that the slight increase in speed was because of the size of the pipes used by Fizeau.
If you're going to just outright lie, there's not a lot of point to this
interaction.
I did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name.
If you never knew anything about it, why did you lie and say it was a >>>>>> version of the MMX?
Well, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.

As I said, both were interferometers, but that's where the similarities >>>> end. And you did lie. Fizeau's experiment was not the MMX.

Well, it seems you are being untruthful. When did I say that Fizeau's experiment was the Michelson-Morley experiment? The Fizeau experiment was in 1851, and the Michelson -Morley experiment was in 1887.

You said right here, in this thread:
: Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light.
Do you deny writing that?

I wrote it. So how is using water and using air for the same kind of experiment the same experiment?

What language was "Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light." written in? Klingon? What does it mean in English?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Lou on Tue Oct 10 23:23:24 2023
On 10/10/2023 3:23 PM, Lou wrote:
On Tuesday, 10 October 2023 at 17:50:46 UTC+1, Volney wrote:
On 10/10/2023 12:40 PM, Lou wrote:
On Tuesday, 10 October 2023 at 16:47:41 UTC+1, Volney wrote:
On 10/10/2023 7:18 AM, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:

How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment
And so for instance when the water
moves towards the source this can be modelled mathematically as the refractive
index of the refractive index for the 'extra distance' travelled.
As the light effectively travels through 'more water' to get the same distance
from source to detector.
Obviously bogus. If that were so, the speed of light in water would get >>>> slower and slower as it traversed through more and more water, even if >>>> stationary. Instead, the speed of light in (stationary) water is a
constant c/n.

You are grasping at straws.

You aren’t just grabbing at straws. You are making up the straws.
Slower with more distance? How so?
Because you claimed the light slows by going through 'more water' (your
term). Increasing the distance obviously means more water traversed.

Grabbing at fantasy straws still. You snipped the formula I cited. There is *no*
‘d’ in that formula. Just v of the water.

My God, are you thick!

The "d" would be the length of a tube in a trivial experiment which
would measure the speed of light in a tube of length "d". If more water
slowed the light, then, trivially, we'd see light going through a tube
with a small length "d" to be faster than light through a tube with a
medium length d (because the light has to traverse more water), but it
would be faster than light with a large value "d", because the latter
has it traverse even more water.

Or are you claiming that longer tubes don't have more water to traverse
while faster ones do? Or something equally idiotic?

[snip remaining crap]

No Relativity needed to correctly model Fizeau classically.

No, classic Fizeau results would be a fringe shift consistent with the
light beams moving at c/n+v and c/n-v.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Dono. on Tue Oct 10 21:00:07 2023
On Tuesday, October 10, 2023 at 6:53:12 PM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 12:54:44 PM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 12:20:50 PM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 10:08:08 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:39:13 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 9:05:35 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 8:54:00 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 8:51:28 AM UTC-7, Robert Winn wrote:
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:
The eastward travelling plane experiences
more force than the westward plane relative to the earths Center. Because it
travels at a greater speed relative to the earth Center, than the westward plane.
Which means the eastward plane (and its caesium atoms) will tick slower
than the westward plane.
You take the prize. The utter cretinism prize
Good to see you, Dono. When did I lose out on the utter cretinism prize?
You are always in the running. Along with Dick Hertz and Pattycakes Dolan.
Lou is just a newcomer.
One thing I have noticed about science. You never use any equations. Why is that?
You are not only an imbecile, you are a liar as well.
Well, I don't really think so.
That you are liar and an imbecile? Well, it is an established fact.
But, Dono, all you ever do is make accusations. You never really say anything.
I say that you are an idiot and a liar. This is a fact.
Yes, you certainly did say that I am an idiot and a liar. It is a fact that you said that. The problem you have is that it does not really mean anything when you say it.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Volney on Tue Oct 10 21:10:12 2023
On Tuesday, October 10, 2023 at 8:26:09 PM UTC-7, Volney wrote:
On 10/10/2023 1:49 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 10:16:10 AM UTC-7, Volney wrote:
On 10/10/2023 1:11 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:44:38 AM UTC-7, Volney wrote:
On 10/10/2023 12:07 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:01:09 AM UTC-7, Volney wrote: >>>>>> On 10/10/2023 11:49 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 4:50:06 AM UTC-7, Sylvia Else wrote:
On 10-Oct-23 9:38 pm, Robert Winn wrote:

Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light. Fizeau did not get the result he expected. Instead, a smaller speed than the speed of the water was
shown to increase the speed of the light. Later experiments with better apparatus showed that the slight increase in speed was because of the size of the pipes used by Fizeau.
If you're going to just outright lie, there's not a lot of point to this
interaction.
I did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name.
If you never knew anything about it, why did you lie and say it was a >>>>>> version of the MMX?
Well, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.

As I said, both were interferometers, but that's where the similarities >>>> end. And you did lie. Fizeau's experiment was not the MMX.

Well, it seems you are being untruthful. When did I say that Fizeau's experiment was the Michelson-Morley experiment? The Fizeau experiment was in 1851, and the Michelson -Morley experiment was in 1887.

You said right here, in this thread:
: Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light.
Do you deny writing that?

I wrote it. So how is using water and using air for the same kind of experiment the same experiment?
What language was "Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light." written in? Klingon? What does it mean in English?
From what I read about Fizeau's experiment, he was measuring the speed of light through moving water with the expectation that the speed of the water would be added to the speed of the light. Michelson and Morley were measuring the speed of light
through air with the expectation that the speed of the air relative to their interferometer would be added to the speed of light in air that was not moving. In English, water is the name given to H2O. Air is a combination of gases including oxygen,
carbon dioxide, and nitrogen. That is what those words mean in English. If you have difficulty understand more words in English, just ask at any time.

--- SoupGate-Win32 v1.05
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• From Volney@21:1/5 to Robert Winn on Wed Oct 11 02:23:01 2023
On 10/11/2023 12:10 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 8:26:09 PM UTC-7, Volney wrote:
On 10/10/2023 1:49 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 10:16:10 AM UTC-7, Volney wrote:
On 10/10/2023 1:11 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:44:38 AM UTC-7, Volney wrote:
On 10/10/2023 12:07 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:01:09 AM UTC-7, Volney wrote: >>>>>>>> On 10/10/2023 11:49 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 4:50:06 AM UTC-7, Sylvia Else wrote:
On 10-Oct-23 9:38 pm, Robert Winn wrote:

Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light. Fizeau did not get the result he expected. Instead, a smaller speed than the speed of the water was
shown to increase the speed of the light. Later experiments with better apparatus showed that the slight increase in speed was because of the size of the pipes used by Fizeau.
If you're going to just outright lie, there's not a lot of point to this
interaction.
I did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name.
If you never knew anything about it, why did you lie and say it was a >>>>>>>> version of the MMX?
Well, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.

As I said, both were interferometers, but that's where the similarities >>>>>> end. And you did lie. Fizeau's experiment was not the MMX.

Well, it seems you are being untruthful. When did I say that Fizeau's experiment was the Michelson-Morley experiment? The Fizeau experiment was in 1851, and the Michelson -Morley experiment was in 1887.

You said right here, in this thread:
: Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light.
Do you deny writing that?

I wrote it. So how is using water and using air for the same kind of experiment the same experiment?

What language was "Fizeau's experiment was an early version of the
Michelson-Morley experiment using water instead of air as the medium of
conducting light." written in? Klingon? What does it mean in English?

From what I read about Fizeau's experiment, he was measuring the speed of light through moving water with the expectation that the speed of the water would be added to the speed of the light. Michelson and Morley were measuring the speed of light
through air with the expectation that the speed of the air relative to their interferometer would be added to the speed of light in air that was not moving.

You can't even get that correct! The MMX was an attempt to measure the
earth's speed through the ether!

In English, water is the name given to H2O. Air is a combination of gases including oxygen, carbon dioxide, and nitrogen. That is what those words mean in English. If you have difficulty understand more words in English, just ask at any time.

So why did you claim that 'Fizeau's experiment was an early version of
the Michelson-Morley experiment'?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to Volney on Tue Oct 10 23:39:14 2023
On Wednesday, 11 October 2023 at 05:26:09 UTC+2, Volney wrote:
On 10/10/2023 1:49 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 10:16:10 AM UTC-7, Volney wrote:
On 10/10/2023 1:11 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:44:38 AM UTC-7, Volney wrote:
On 10/10/2023 12:07 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:01:09 AM UTC-7, Volney wrote: >>>>>> On 10/10/2023 11:49 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 4:50:06 AM UTC-7, Sylvia Else wrote:
On 10-Oct-23 9:38 pm, Robert Winn wrote:

Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light. Fizeau did not get the result he expected. Instead, a smaller speed than the speed of the water was
shown to increase the speed of the light. Later experiments with better apparatus showed that the slight increase in speed was because of the size of the pipes used by Fizeau.
If you're going to just outright lie, there's not a lot of point to this
interaction.
I did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name.
If you never knew anything about it, why did you lie and say it was a >>>>>> version of the MMX?
Well, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.

As I said, both were interferometers, but that's where the similarities >>>> end. And you did lie. Fizeau's experiment was not the MMX.

Well, it seems you are being untruthful. When did I say that Fizeau's experiment was the Michelson-Morley experiment? The Fizeau experiment was in 1851, and the Michelson -Morley experiment was in 1887.

You said right here, in this thread:
: Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light.
Do you deny writing that?

I wrote it. So how is using water and using air for the same kind of experiment the same experiment?
What language was "Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light." written in?

And what language was yours "setting to 9 192 631 774 is
setting to 9 192 631 770"?

--- SoupGate-Win32 v1.05
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• From Maciej Wozniak@21:1/5 to J. J. Lodder on Wed Oct 11 01:52:19 2023
On Wednesday, 11 October 2023 at 10:19:39 UTC+2, J. J. Lodder wrote:
mitchr...@gmail.com <mitchr...@gmail.com> wrote:
On Tuesday, October 10, 2023 at 2:21:44?PM UTC-7, Maciej Wozniak wrote:
On Tuesday, 10 October 2023 at 20:10:11 UTC+2, Paul B. Andersen wrote:

So why did the east going clock in the plane tick slower than
the clock on the ground?
Because your Holiest Postulate is such an absurd
that not even you stick to it.

How many Cesium atoms are necessary?
One, in principle. (but not in practice for cesium)
For trapped ions one will do.
What is the atom doing and what could count it?
Its thing. And nothing as yet.
The trapped ion itself will become the next clock.
No. The atomic clock drifts instead...
All clocks always drift.

A pity your idiot guru didn't know that.

--- SoupGate-Win32 v1.05
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• From J. J. Lodder@21:1/5 to mitchr...@gmail.com on Wed Oct 11 10:19:33 2023
mitchr...@gmail.com <mitchrae3323@gmail.com> wrote:

On Tuesday, October 10, 2023 at 2:21:44?PM UTC-7, Maciej Wozniak wrote:
On Tuesday, 10 October 2023 at 20:10:11 UTC+2, Paul B. Andersen wrote:

So why did the east going clock in the plane tick slower than
the clock on the ground?
Because your Holiest Postulate is such an absurd
that not even you stick to it.

How many Cesium atoms are necessary?

One, in principle. (but not in practice for cesium)
For trapped ions one will do.

What is the atom doing and what could count it?

Its thing. And nothing as yet.
The trapped ion itself will become the next clock.

No. The atomic clock drifts instead...

All clocks always drift.
The question is by how much,

Jan

--- SoupGate-Win32 v1.05
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• From JanPB@21:1/5 to J. J. Lodder on Wed Oct 11 02:28:26 2023
On Tuesday, October 10, 2023 at 1:23:37 AM UTC-7, J. J. Lodder wrote:

[1] This is also the reason for the lack of references.
None were needed, because whatever could have been refered too
was common knowledge, for the intended readership.

It's also worth pointing out that papers with no references in them
were quite common in those days. If one leafs through the Annalen
der Physik from that time, one sees many of such papers there.
So Einstein's was not at all unusual in that sense.

--
Jan

--- SoupGate-Win32 v1.05
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• From J. J. Lodder@21:1/5 to JanPB on Wed Oct 11 12:02:47 2023
JanPB <filmart@gmail.com> wrote:

On Tuesday, October 10, 2023 at 1:23:37?AM UTC-7, J. J. Lodder wrote:

[1] This is also the reason for the lack of references.
None were needed, because whatever could have been refered too
was common knowledge, for the intended readership.

It's also worth pointing out that papers with no references in them
were quite common in those days. If one leafs through the Annalen
der Physik from that time, one sees many of such papers there.
So Einstein's was not at all unusual in that sense.

Indeed. References only when needed.
Like I said in several previous postings,
physcists were completely mad in those days.
They had the crazy idea that scientific publications
served to communicate with collegues.
And worse, they actually read each others papers!
(like they would read personal lettters)

What's even worse, the breakthrough idea
that the worth of their collegues could be determined
by counting the number of citations hadn't occurred to them.
They judged each other by reading those publications,
and understanding them, and by seeing who had something to say.

And of course they actually met each other, when possible.
Europe wasn't big, you could easily travel from one capital to another
by train in less than a day, or two days at worst.

I have seen estimates to the effect
that there were about a thousand professional physicists
in all of Europe, by the year 1900.
There can have been no more than a few dozen specialised theoretical
physicists at most.
But of course experimentalists were competent
in the theories that were of relevance to them too,

Jan

--- SoupGate-Win32 v1.05
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• From JanPB@21:1/5 to Lou on Wed Oct 11 02:29:52 2023
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a
clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get
a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock
on the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances
for x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate
that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right >> about this.

It's not as if Einstein, as a young patent clerk, had some ability to >> impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why >> a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of today do not
consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is faster than a
clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves. I am explaining
relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the Lorentz equations that
he said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of the slower
clock gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book. His
problem was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for it.
The Galilean transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia
No need for relativity. A classical model does just fine.

No, it doesn't. You simply don't know the problems involved.

--
Jan

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• From Lou@21:1/5 to Dono. on Wed Oct 11 03:20:21 2023
On Wednesday, 11 October 2023 at 03:55:55 UTC+1, Dono. wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:

So the change in speed from c/n is +- the refractive index of the refractive index
of the extra distance of water travelled as defined by
C+-(V x .67)/n
Where .67 =1-(1-n)

This traditional Fresnel equation can also be expressed as c/n+-v{(1-n)+(1-n)^2}
Crank ,

Experiment measures c/n+v(1-1/n^2), not the idiocy you posted above.
SR predicts c/n+v(1-1/n^2), in accordance with the experiment.

Fizeau discovered c/n+v(1-1/n^2) in 1851
Albert just copied it and pretended it was SR.

In my last post I incorrectly typed 1-n. It should have been n-1.
Giving the corrected version as:
c/n+-v{(n-1)+(n-1)^2}

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• From Lou@21:1/5 to Volney on Wed Oct 11 03:37:16 2023
On Wednesday, 11 October 2023 at 04:23:31 UTC+1, Volney wrote:
On 10/10/2023 3:23 PM, Lou wrote:
On Tuesday, 10 October 2023 at 17:50:46 UTC+1, Volney wrote:
On 10/10/2023 12:40 PM, Lou wrote:
On Tuesday, 10 October 2023 at 16:47:41 UTC+1, Volney wrote:
On 10/10/2023 7:18 AM, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:

How do you explain the result of the Fizeau experiment, which was >>>>>> performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment
And so for instance when the water
moves towards the source this can be modelled mathematically as the refractive
index of the refractive index for the 'extra distance' travelled. >>>>> As the light effectively travels through 'more water' to get the same distance
from source to detector.
Obviously bogus. If that were so, the speed of light in water would get >>>> slower and slower as it traversed through more and more water, even if >>>> stationary. Instead, the speed of light in (stationary) water is a
constant c/n.

You are grasping at straws.

You aren’t just grabbing at straws. You are making up the straws.
Slower with more distance? How so?
Because you claimed the light slows by going through 'more water' (your >> term). Increasing the distance obviously means more water traversed.

Grabbing at fantasy straws still. You snipped the formula I cited. There is *no*
‘d’ in that formula. Just v of the water.
My God, are you thick!

The "d" would be the length of a tube in a trivial experiment which
would measure the speed of light in a tube of length "d". If more water slowed the light, then, trivially, we'd see light going through a tube
with a small length "d" to be faster than light through a tube with a
medium length d (because the light has to traverse more water), but it
would be faster than light with a large value "d", because the latter
has it traverse even more water.

Low IQ nonsense as usual from a relativist. Where’s the d in this formula? c/n+-v{(n-1)+(n-1)^2}
Can’t find it? Oh well. Looks like you’ve been spouting BS again.

Or are you claiming that longer tubes don't have more water to traverse while faster ones do? Or something equally idiotic?

No...you are making these ridiculous claims. Not me.
You still haven’t grasped basic physics. It doesn’t matter how long
the tube of water is. As long as the water is always at the same v
in the tube, the light will travel through it at a certain speed
(ie at a certain refractive index.)
And that refractive index isn’t n in water but a slightly different
value of n which takes into account the fact that the moving water
creates a more optically dense medium in the experiment.
And classically one way to model this is to make this new
value of n is by this formula... c/n+-v{(n-1)+(n-1)^2}
Or if you prefer the Fizeau original formula...that works as well too.
All I’ve done is explain how the different speeds in moving water
can be explained classically without having to pull out a nonsense BS Relativistic explanation.

--- SoupGate-Win32 v1.05
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• From Lou@21:1/5 to JanPB on Wed Oct 11 03:45:58 2023
On Wednesday, 11 October 2023 at 10:29:54 UTC+1, JanPB wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a
clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get
a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the
airplane. Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock
on the ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the
clock on the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same
distances for x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They
indicate that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right >> about this.

It's not as if Einstein, as a young patent clerk, had some ability to >> impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of today do not
consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is faster than a
clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves. I am
explaining relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the Lorentz equations
that he said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of the slower
clock gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book. His
problem was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for it.
The Galilean transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia
No need for relativity. A classical model does just fine.

No, it doesn't. You simply don't know the problems involved.

You don’t understand basic physics. Fact: A more optically dense medium
will have a slightly greater refractive index. And if the water moves
relative to the source..a classical model predicts this extra +-v
will increase (or decrease)the optical density of the water as it moves relative to the source.
And Fizeaus original formula ( that SR stole)models this change in density.
As does my own version: c/n+-v{(n-1)+(n-1)^2}
Proving the only people who don’t have any idea are the relativist
fantasists who still think the sun rotates around the earth.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Volney on Wed Oct 11 05:12:26 2023
On Tuesday, October 10, 2023 at 11:23:05 PM UTC-7, Volney wrote:
On 10/11/2023 12:10 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 8:26:09 PM UTC-7, Volney wrote:
On 10/10/2023 1:49 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 10:16:10 AM UTC-7, Volney wrote:
On 10/10/2023 1:11 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:44:38 AM UTC-7, Volney wrote: >>>>>> On 10/10/2023 12:07 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:01:09 AM UTC-7, Volney wrote: >>>>>>>> On 10/10/2023 11:49 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 4:50:06 AM UTC-7, Sylvia Else wrote:
On 10-Oct-23 9:38 pm, Robert Winn wrote:

Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light. Fizeau did not get the result he expected. Instead, a smaller speed than the speed of the water was
shown to increase the speed of the light. Later experiments with better apparatus showed that the slight increase in speed was because of the size of the pipes used by Fizeau.
If you're going to just outright lie, there's not a lot of point to this
interaction.
I did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name.
If you never knew anything about it, why did you lie and say it was a
version of the MMX?
Well, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.

As I said, both were interferometers, but that's where the similarities
end. And you did lie. Fizeau's experiment was not the MMX.

Well, it seems you are being untruthful. When did I say that Fizeau's experiment was the Michelson-Morley experiment? The Fizeau experiment was in 1851, and the Michelson -Morley experiment was in 1887.

You said right here, in this thread:
: Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light.
Do you deny writing that?

I wrote it. So how is using water and using air for the same kind of experiment the same experiment?

What language was "Fizeau's experiment was an early version of the
Michelson-Morley experiment using water instead of air as the medium of >> conducting light." written in? Klingon? What does it mean in English?

From what I read about Fizeau's experiment, he was measuring the speed of light through moving water with the expectation that the speed of the water would be added to the speed of the light. Michelson and Morley were measuring the speed of light
through air with the expectation that the speed of the air relative to their interferometer would be added to the speed of light in air that was not moving.
You can't even get that correct! The MMX was an attempt to measure the earth's speed through the ether!
In English, water is the name given to H2O. Air is a combination of gases including oxygen, carbon dioxide, and nitrogen. That is what those words mean in English. If you have difficulty understand more words in English, just ask at any time.
So why did you claim that 'Fizeau's experiment was an early version of
the Michelson-Morley experiment'?
Well, as a common person, not a scientist, I just take note of the similarities. The experimenters in both cases believed that there was a medium through which light was conducted. In the case of the Fizeau experiment, the medium was water, in
Michelson-Morley it was air. In both experiments it was expected that the speed of the medium relative to the measuring device would affect the result obtained for c, the speed of light. In both experiments the result obtained was not the result
expected. Then there was the fact that Michelson and Morley said that the idea for their interferometer came from Fizeau's experiment. But, as I said, I had no idea of what Fizeau's experiment was until I read something about it yesterday morning.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to JanPB on Wed Oct 11 05:14:42 2023
On Wednesday, October 11, 2023 at 2:29:54 AM UTC-7, JanPB wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a
clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get
a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the
airplane. Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock
on the ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the
clock on the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same
distances for x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They
indicate that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right >> about this.

It's not as if Einstein, as a young patent clerk, had some ability to >> impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of today do not
consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is faster than a
clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves. I am
explaining relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the Lorentz equations
that he said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of the slower
clock gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book. His
problem was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for it.
The Galilean transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia
No need for relativity. A classical model does just fine.

No, it doesn't. You simply don't know the problems involved.

--
Jan
I know of one problem involved. So when do you think a scientist is going to answer the question I asked about Einstein's description of a miracle? My prediction is never.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Dono.@21:1/5 to Lou on Wed Oct 11 06:11:57 2023
On Wednesday, October 11, 2023 at 3:20:23 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 03:55:55 UTC+1, Dono. wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:

So the change in speed from c/n is +- the refractive index of the refractive index
of the extra distance of water travelled as defined by
C+-(V x .67)/n
Where .67 =1-(1-n)

This traditional Fresnel equation can also be expressed as c/n+-v{(1-n)+(1-n)^2}
Crank ,

Experiment measures c/n+v(1-1/n^2), not the idiocy you posted above.

In my last post I incorrectly typed 1-n. It should have been n-1.
Giving the corrected version as:
c/n+-v{(n-1)+(n-1)^2}

Repeating crank claims doesn't make them right, makes you a stubborn crank.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Robert Winn on Wed Oct 11 06:13:58 2023
On Wednesday, 11 October 2023 at 13:14:44 UTC+1, Robert Winn wrote:
On Wednesday, October 11, 2023 at 2:29:54 AM UTC-7, JanPB wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined
a clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would
get a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the
airplane. Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock
on the ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I
believe Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane
if his clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the
clock on the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same
distances for x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They
indicate that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of today do not
consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is faster than a
clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves. I am
explaining relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the Lorentz equations
that he said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of the slower
clock gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book. His
problem was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for it.
The Galilean transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia
No need for relativity. A classical model does just fine.

No, it doesn't. You simply don't know the problems involved.

--
Jan
I know of one problem involved. So when do you think a scientist is going to answer the question I asked about Einstein's description of a miracle? My prediction is never.

Einstein and his relativist followers are not scientists. They are
religious wackos.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Dono. on Wed Oct 11 06:18:28 2023
On Wednesday, 11 October 2023 at 14:12:00 UTC+1, Dono. wrote:
On Wednesday, October 11, 2023 at 3:20:23 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 03:55:55 UTC+1, Dono. wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:

So the change in speed from c/n is +- the refractive index of the refractive index
of the extra distance of water travelled as defined by
C+-(V x .67)/n
Where .67 =1-(1-n)

This traditional Fresnel equation can also be expressed as c/n+-v{(1-n)+(1-n)^2}
Crank ,

Experiment measures c/n+v(1-1/n^2), not the idiocy you posted above.
In my last post I incorrectly typed 1-n. It should have been n-1.
Giving the corrected version as:
c/n+-v{(n-1)+(n-1)^2}
Repeating crank claims doesn't make them right, makes you a stubborn crank.

Said the kettle to the pot.
Answer the question crank. Did Fizeau come up with c/n+v(1-1/n^2).
or did Albert?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Paul B. Andersen@21:1/5 to All on Wed Oct 11 15:22:08 2023
Den 10.10.2023 21:50, skrev Lou:
On Tuesday, 10 October 2023 at 19:10:11 UTC+1, Paul B. Andersen wrote:
Den 10.10.2023 14:16, skrev Lou:

No need for relativity to explain Hafael Keating.
Look at classical resonance. It has been known for centuries that
the natural resonant frequency of a system will reduce its frequency
if subject to an external force. So an atom, also confirmed by all observations
to date to be a resonant system, will also reduce its frequency when subject
to external force. As we see happen where less g force with altitude increases
the atoms ‘ticking’.

So clocks at higher altitude in a plane will be subject to
lower g-force and will tick faster than clocks on the ground, right?

No I’m suggesting that this horizontal force is not force due to gravity. But a seperate force from horizontal acceleration .
Isn’t the formula for this f=ma?

I see. The horizontal force driving the plane at constant
speed through the air will give the plane a constant
horizontal acceleration. Sounds reasonable, doesn't it?

Im assuming that Hafael Keating observed that the eastward clock ticks slower.
That’s my reading of the wiki reference.
But to answer ‘why’...I assume the speed relative to the ground is the same for both
planes in the experiment. Let’s call it speed x.
The earth rotates eastward at 1600k/ hr.
Close enough.

So the zero point of reference is the earth observer travelling at 1600 k/hr relative
to the earths Center of mass.
Then relative to this earth center reference, the eastward plane travels at 1600 +x kilometers per hour. And the westward plane travels at 1600-x kilometers
per hour.

A reasonable speed for x is 800 km/h.

The east going plane travels at 2400 km/h in the ECI-frame.
The ground clock travels at 1600 km/h in the ECI-frame.
The west going plane travels at 800 km/h in the ECI-frame.

The eastward plane therefore experiences greater F than earth observer
And the westward plane lesser F than the the earth observer.
And seeing as a classical model uses resonance as the explanation
for this different ticking rates of the caesium atoms natural resonant frequency.
Then the westward clock has less F from horizontal
speed than the earth observer and the westward plane has a greater F than
the earth observer. Which accounts for the 3 different rates of ticking. (Don’t forget...the earth observers ‘ticking rate’ is in part due to Gravity force
but also in part due to its rotational/ horizontal speed of 1600k/hr )

I see.
Since the east going clock runs slower than the ground clock,
and the ground clock runs slower than the west going clock,
it is obvious that it is the speed in the ECI-frame that is
the major cause for the different clock rates.

So we can conclude:
Since the speed in the ECI-frame is higher for the east going
clock than for the ground clock, the horizontal force to drive
the east going plane through the air at 800 km/h is higher than
the force to drive the ground clock through the air at 0 km/h.

Since the speed in the ECI-frame is higher for the ground clock
than for the west going clock, the horizontal force to drive
the ground clock through the air at 0 km/h is higher than the force
to drive the west going plane through the air at 800 km/h.

The east going clock is more compressed than the ground clock,
and will run slower than the ground clock.
The ground clock is more compressed than the west going clock,
and will run slower than the west going clock.

Right?

But don’t forget the Gravity force pushing you down is a seperate source of force from the F force pushing against you as you move horizontally. I’m not sure
what word you relativists prefer but probably inertia or momentum would
be appropriate?

Air drag.

---

BTW, I think you should tell the airlines that when going from New York
to Paris, they would use less fuel if they travel westwards.

https://paulba.no/pdf/H&K_like.pdf

--
Paul

https://paulba.no/

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Dono.@21:1/5 to Lou on Wed Oct 11 06:32:52 2023
On Wednesday, October 11, 2023 at 6:14:00 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 13:14:44 UTC+1, Robert Winn wrote:
On Wednesday, October 11, 2023 at 2:29:54 AM UTC-7, JanPB wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I
imagined a clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the
pilot would get a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations
of Special Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the
airplane. Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock
on the ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I
believe Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane
if his clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the
clock on the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same
distances for x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They
indicate that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not >> behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the >> scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of today do not
consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is faster than a
clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves. I am
explaining relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the Lorentz equations
that he said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of the
slower clock gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book.
His problem was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for
it. The Galilean transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia
No need for relativity. A classical model does just fine.

No, it doesn't. You simply don't know the problems involved.

--
Jan
I know of one problem involved. So when do you think a scientist is going to answer the question I asked about Einstein's description of a miracle? My prediction is never.
Anti-relativists are not scientists. They are
religious wackos.

You sure are

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Paul B. Andersen on Wed Oct 11 06:43:53 2023
On Wednesday, 11 October 2023 at 14:21:24 UTC+1, Paul B. Andersen wrote:
Den 10.10.2023 21:50, skrev Lou:
On Tuesday, 10 October 2023 at 19:10:11 UTC+1, Paul B. Andersen wrote:
Den 10.10.2023 14:16, skrev Lou:

No need for relativity to explain Hafael Keating.
Look at classical resonance. It has been known for centuries that
the natural resonant frequency of a system will reduce its frequency
if subject to an external force. So an atom, also confirmed by all observations
to date to be a resonant system, will also reduce its frequency when subject
to external force. As we see happen where less g force with altitude increases
the atoms ‘ticking’.

So clocks at higher altitude in a plane will be subject to
lower g-force and will tick faster than clocks on the ground, right?

No I’m suggesting that this horizontal force is not force due to gravity.
But a seperate force from horizontal acceleration .
Isn’t the formula for this f=ma?

I see. The horizontal force driving the plane at constant
speed through the air will give the plane a constant
horizontal acceleration. Sounds reasonable, doesn't it?

Im assuming that Hafael Keating observed that the eastward clock ticks slower.
That’s my reading of the wiki reference.
But to answer ‘why’...I assume the speed relative to the ground is the same for both
planes in the experiment. Let’s call it speed x.
The earth rotates eastward at 1600k/ hr.
Close enough.

So the zero point of reference is the earth observer travelling at 1600 k/hr relative
to the earths Center of mass.
Then relative to this earth center reference, the eastward plane travels at
1600 +x kilometers per hour. And the westward plane travels at 1600-x kilometers
per hour.

A reasonable speed for x is 800 km/h.

The east going plane travels at 2400 km/h in the ECI-frame.
The ground clock travels at 1600 km/h in the ECI-frame.
The west going plane travels at 800 km/h in the ECI-frame.

The eastward plane therefore experiences greater F than earth observer
And the westward plane lesser F than the the earth observer.
And seeing as a classical model uses resonance as the explanation
for this different ticking rates of the caesium atoms natural resonant frequency.
Then the westward clock has less F from horizontal
speed than the earth observer and the westward plane has a greater F than the earth observer. Which accounts for the 3 different rates of ticking. (Don’t forget...the earth observers ‘ticking rate’ is in part due to Gravity force
but also in part due to its rotational/ horizontal speed of 1600k/hr )

I see.
Since the east going clock runs slower than the ground clock,
and the ground clock runs slower than the west going clock,
it is obvious that it is the speed in the ECI-frame that is
the major cause for the different clock rates.

So we can conclude:
Since the speed in the ECI-frame is higher for the east going
clock than for the ground clock, the horizontal force to drive
the east going plane through the air at 800 km/h is higher than
the force to drive the ground clock through the air at 0 km/h.

Since the speed in the ECI-frame is higher for the ground clock
than for the west going clock, the horizontal force to drive
the ground clock through the air at 0 km/h is higher than the force
to drive the west going plane through the air at 800 km/h.

The east going clock is more compressed than the ground clock,
and will run slower than the ground clock.
The ground clock is more compressed than the west going clock,
and will run slower than the west going clock.

Right?

But don’t forget the Gravity force pushing you down is a seperate source of
force from the F force pushing against you as you move horizontally. I’m not sure
what word you relativists prefer but probably inertia or momentum would
be appropriate?

Air drag.

Air Drag!!😂🤣You relativists. Such purveyors of BS.
Did I mention air drag?
No.
You did.
You forgot. The earth rotates.
I know in relativity land that you guys think the earth doesn’t rotate. But sorry...it does.
In which case the eastward plane travels faster relative to the earth Center frame.
And the westward plane slower
If you knew any maths...then even if the planes speed was 800k/hr it would still
be less than the earth observers speed 1600 k/hr speed.
Which means that the eastward plane travels at a higher speed than the earth observer, and the westward plane at a lower speed than the observer.
And seeing as f=ma then the force on the westward plane is less than the
earth observer and the force on the eastward plane is higher.
Which in turn means that due to f=ma the eastward travelling clock will
run slower and the westward clock will run faster than the earth observers clock
Due to mechanical resonance.
As observed in Hafael,Keating.
Do the maths Paul.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Dono.@21:1/5 to Dono. on Wed Oct 11 06:40:53 2023
On Wednesday, October 11, 2023 at 6:32:54 AM UTC-7, Dono. wrote:
On Wednesday, October 11, 2023 at 6:14:00 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 13:14:44 UTC+1, Robert Winn wrote:
On Wednesday, October 11, 2023 at 2:29:54 AM UTC-7, JanPB wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I
imagined a clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the
pilot would get a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations
of Special Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the
airplane. Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock
on the ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I
believe Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane
if his clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of
the clock on the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same
distances for x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They
indicate that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the >> scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of today do
not consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is faster than
a clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves. I am
explaining relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the Lorentz
equations that he said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of the
slower clock gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book.
His problem was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for
it. The Galilean transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia
No need for relativity. A classical model does just fine.

No, it doesn't. You simply don't know the problems involved.

--
Jan
I know of one problem involved. So when do you think a scientist is going to answer the question I asked about Einstein's description of a miracle? My prediction is never.
Anti-relativists are not scientists. They are
religious wackos.

You sure are

Fizeau expected the result to be c/n /pm v. Instead he got a very different result: c/n /pm v(1-1/n^2).
The result contradicts Newtonian kinematics and confirms SR. Crank Lou continues to froth at the mouth.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Dono. on Wed Oct 11 06:49:09 2023
On Wednesday, 11 October 2023 at 14:40:55 UTC+1, Dono. wrote:
On Wednesday, October 11, 2023 at 6:32:54 AM UTC-7, Dono. wrote:
On Wednesday, October 11, 2023 at 6:14:00 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 13:14:44 UTC+1, Robert Winn wrote:
On Wednesday, October 11, 2023 at 2:29:54 AM UTC-7, JanPB wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I
imagined a clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the
pilot would get a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations
of Special Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the
airplane. Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock
on the ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I
believe Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane
if his clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of
the clock on the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same
distances for x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be.
They indicate that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of today do
not consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is faster than
a clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves. I am
explaining relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the Lorentz
equations that he said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of the
slower clock gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book.
His problem was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for
it. The Galilean transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia
No need for relativity. A classical model does just fine.

No, it doesn't. You simply don't know the problems involved.

--
Jan
I know of one problem involved. So when do you think a scientist is going to answer the question I asked about Einstein's description of a miracle? My prediction is never.
Anti-relativists are not scientists. They are
religious wackos.

You sure are
Fizeau expected the result to be c/n /pm v. Instead he got a very different result: c/n /pm v(1-1/n^2).
The result contradicts Newtonian kinematics and confirms SR. Crank Lou continues to froth at the mouth.

You pretended c/n+v(1-1/n^2) was discovered by Adolf Einstein
Answer the question . Did Fizeau discover c/n+v(1-1/n^2) in 1851.
Or did Einstein, the low IQ plagiarist, steal it from Fizeau and pretend it was
his own formula ?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Dono.@21:1/5 to Lou on Wed Oct 11 06:53:28 2023
On Wednesday, October 11, 2023 at 6:49:11 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 14:40:55 UTC+1, Dono. wrote:
On Wednesday, October 11, 2023 at 6:32:54 AM UTC-7, Dono. wrote:
On Wednesday, October 11, 2023 at 6:14:00 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 13:14:44 UTC+1, Robert Winn wrote:
On Wednesday, October 11, 2023 at 2:29:54 AM UTC-7, JanPB wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I
imagined a clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the
pilot would get a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations
of Special Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for
the airplane. Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a
clock on the ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which
I believe Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane
if his clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time
of the clock on the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the
same distances for x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be.
They indicate that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of today
do not consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is faster
than a clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves. I am
explaining relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the Lorentz
equations that he said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of the
slower clock gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book.
His problem was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for
it. The Galilean transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia
No need for relativity. A classical model does just fine.

No, it doesn't. You simply don't know the problems involved.

--
Jan
I know of one problem involved. So when do you think a scientist is going to answer the question I asked about Einstein's description of a miracle? My prediction is never.
Anti-relativists are not scientists. They are
religious wackos.

You sure are
Fizeau expected the result to be c/n /pm v. Instead he got a very different result: c/n /pm v(1-1/n^2).
The result contradicts Newtonian kinematics and confirms SR. Crank Lou continues to froth at the mouth.
You pretended c/n+v(1-1/n^2) was discovered by Einstein

No, I didn't. I simple pointed out the idiocy in the formula that you posted. Repeatedly.
As predicted, you are now just frothing at the mouth.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Dono.@21:1/5 to Dono. on Wed Oct 11 07:02:46 2023
On Wednesday, October 11, 2023 at 6:53:30 AM UTC-7, Dono. wrote:
On Wednesday, October 11, 2023 at 6:49:11 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 14:40:55 UTC+1, Dono. wrote:
On Wednesday, October 11, 2023 at 6:32:54 AM UTC-7, Dono. wrote:
On Wednesday, October 11, 2023 at 6:14:00 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 13:14:44 UTC+1, Robert Winn wrote:
On Wednesday, October 11, 2023 at 2:29:54 AM UTC-7, JanPB wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I
imagined a clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the
pilot would get a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations
of Special Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for
the airplane. Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a
clock on the ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another,
which I believe Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the
airplane if his clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time
of the clock on the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the
same distances for x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be.
They indicate that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of
today do not consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is
faster than a clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves. I
am explaining relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the Lorentz
equations that he said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of
the slower clock gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book.
His problem was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for
it. The Galilean transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia
No need for relativity. A classical model does just fine.

No, it doesn't. You simply don't know the problems involved.

--
Jan
I know of one problem involved. So when do you think a scientist is going to answer the question I asked about Einstein's description of a miracle? My prediction is never.
Anti-relativists are not scientists. They are
religious wackos.

You sure are
Fizeau expected the result to be c/n /pm v. Instead he got a very different result: c/n /pm v(1-1/n^2).
The result contradicts Newtonian kinematics and confirms SR. Crank Lou continues to froth at the mouth.
You pretended c/n+v(1-1/n^2) was discovered by Einstein

No, I didn't. I simply pointed out the idiocy in the formula that you posted. Repeatedly.
As predicted, you are now just frothing at the mouth.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Dono. on Wed Oct 11 07:21:53 2023
On Wednesday, 11 October 2023 at 15:02:49 UTC+1, Dono. wrote:
On Wednesday, October 11, 2023 at 6:53:30 AM UTC-7, Dono. wrote:
On Wednesday, October 11, 2023 at 6:49:11 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 14:40:55 UTC+1, Dono. wrote:
On Wednesday, October 11, 2023 at 6:32:54 AM UTC-7, Dono. wrote:
On Wednesday, October 11, 2023 at 6:14:00 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 13:14:44 UTC+1, Robert Winn wrote:
On Wednesday, October 11, 2023 at 2:29:54 AM UTC-7, JanPB wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock
I imagined a clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the
pilot would get a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations
of Special Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed
for the airplane. Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than
a clock on the ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another,
which I believe Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the
airplane if his clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the
time of the clock on the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with
the same distances for x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to
be. They indicate that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of
today do not consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is
faster than a clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves.
I am explaining relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the Lorentz
equations that he said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of
the slower clock gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book.
His problem was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for
it. The Galilean transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia
No need for relativity. A classical model does just fine.

No, it doesn't. You simply don't know the problems involved.

--
Jan
I know of one problem involved. So when do you think a scientist is going to answer the question I asked about Einstein's description of a miracle? My prediction is never.
Anti-relativists are not scientists. They are
religious wackos.

You sure are
Fizeau expected the result to be c/n /pm v. Instead he got a very different result: c/n /pm v(1-1/n^2).
The result contradicts Newtonian kinematics and confirms SR. Crank Lou continues to froth at the mouth.
You pretended c/n+v(1-1/n^2) was discovered by Einstein

No, I didn't. I simply pointed out the idiocy in the formula that you posted. Repeatedly.
As predicted, you are now just frothing at the mouth.

You pretended that Fizeau’s formula was invented by Albert
It wasnt. Fizeau thought of it in 1851
Here’s your quote: “ Experiment measures c/n+v(1-1/n^2)
SR predicts c/n+v(1-1/n^2), in accordance with the experiment. “

And if you think my formula c/n+-v{(n-1)+(n-1)^2} does not also correctly predict Fizeau..
Prove it.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Dono.@21:1/5 to Lou on Wed Oct 11 07:39:47 2023
On Wednesday, October 11, 2023 at 7:21:56 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 15:02:49 UTC+1, Dono. wrote:
On Wednesday, October 11, 2023 at 6:53:30 AM UTC-7, Dono. wrote:
On Wednesday, October 11, 2023 at 6:49:11 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 14:40:55 UTC+1, Dono. wrote:
On Wednesday, October 11, 2023 at 6:32:54 AM UTC-7, Dono. wrote:
On Wednesday, October 11, 2023 at 6:14:00 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 13:14:44 UTC+1, Robert Winn wrote:
On Wednesday, October 11, 2023 at 2:29:54 AM UTC-7, JanPB wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving
clock I imagined a clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane,
the pilot would get a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the
equations of Special Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed
for the airplane. Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than
a clock on the ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another,
which I believe Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the
airplane if his clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the
time of the clock on the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with
the same distances for x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to
be. They indicate that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of
today do not consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is
faster than a clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves.
I am explaining relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the
Lorentz equations that he said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time
of the slower clock gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his
book. His problem was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need
for it. The Galilean transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia
No need for relativity. A classical model does just fine.

No, it doesn't. You simply don't know the problems involved.

--
Jan
I know of one problem involved. So when do you think a scientist is going to answer the question I asked about Einstein's description of a miracle? My prediction is never.
Anti-relativists are not scientists. They are
religious wackos.

You sure are
Fizeau expected the result to be c/n /pm v. Instead he got a very different result: c/n /pm v(1-1/n^2).
The result contradicts Newtonian kinematics and confirms SR. Crank Lou continues to froth at the mouth.
You pretended c/n+v(1-1/n^2) was discovered by Einstein

No, I didn't. I simply pointed out the idiocy in the formula that you posted. Repeatedly.
As predicted, you are now just frothing at the mouth.
You pretended that Fizeau’s formula was invented by Albert

You are lying, there is no such claim.

Here’s your quote: “ Experiment measures c/n+v(1-1/n^2)
SR predicts c/n+v(1-1/n^2), in accordance with the experiment. “

"SR predicts " is not "Fizeau’s formula was invented by Albert". Are you as incompetent in English language as you are in physics? Don't answer, it was a rhetorical question.

And if you think my formula c/n+-v{(n-1)+(n-1)^2} does not also correctly predict Fizeau..
Prove it.

Crank,

You are as incompetent in terms of basic algebra as you are in terms of physics. In your demented brain:

n-1+(n-1)^2=1-1/n^2

Way to go, LouLou!

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Dono. on Wed Oct 11 08:18:51 2023
On Wednesday, 11 October 2023 at 15:39:50 UTC+1, Dono. wrote:
On Wednesday, October 11, 2023 at 7:21:56 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 15:02:49 UTC+1, Dono. wrote:
On Wednesday, October 11, 2023 at 6:53:30 AM UTC-7, Dono. wrote:
On Wednesday, October 11, 2023 at 6:49:11 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 14:40:55 UTC+1, Dono. wrote:
On Wednesday, October 11, 2023 at 6:32:54 AM UTC-7, Dono. wrote:
On Wednesday, October 11, 2023 at 6:14:00 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 13:14:44 UTC+1, Robert Winn wrote:
On Wednesday, October 11, 2023 at 2:29:54 AM UTC-7, JanPB wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving
clock I imagined a clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane,
the pilot would get a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the
equations of Special Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster
speed for the airplane. Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being
slower than a clock on the ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one
another, which I believe Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed
for the airplane if his clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t,
the time of the clock on the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations
with the same distances for x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we
have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it
to be. They indicate that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists
of today do not consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is
faster than a clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic
waves. I am explaining relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the
Lorentz equations that he said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for
time of the slower clock gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in
his book. His problem was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no
need for it. The Galilean transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia
No need for relativity. A classical model does just fine.

No, it doesn't. You simply don't know the problems involved.

--
Jan
I know of one problem involved. So when do you think a scientist is going to answer the question I asked about Einstein's description of a miracle? My prediction is never.
Anti-relativists are not scientists. They are
religious wackos.

You sure are
Fizeau expected the result to be c/n /pm v. Instead he got a very different result: c/n /pm v(1-1/n^2).
The result contradicts Newtonian kinematics and confirms SR. Crank Lou continues to froth at the mouth.
You pretended c/n+v(1-1/n^2) was discovered by Einstein

No, I didn't. I simply pointed out the idiocy in the formula that you posted. Repeatedly.
As predicted, you are now just frothing at the mouth.
You pretended that Fizeau’s formula was invented by Albert
You are lying, there is no such claim.
Here’s your quote: “ Experiment measures c/n+v(1-1/n^2)
SR predicts c/n+v(1-1/n^2), in accordance with the experiment. “

"SR predicts " is not "Fizeau’s formula was invented by Albert". Are you as incompetent in English language as you are in physics? Don't answer, it was a rhetorical question.
And if you think my formula c/n+-v{(n-1)+(n-1)^2} does not also correctly predict Fizeau..
Prove it.
Crank,

You are as incompetent in terms of basic algebra as you are in terms of physics. In your demented brain:

n-1+(n-1)^2=1-1/n^2

Way to go, LouLou!

Nice try big boy. You won round 1 on brackets. But you know sweet f
all on physics.
But now prove c/n+- v[{(1.33-1)+(1.33-1)}^2] doesnt correctly
model Fizeau results.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Lou on Wed Oct 11 12:03:25 2023
On 10/11/2023 9:49 AM, Lou wrote:

On Wednesday, 11 October 2023 at 14:40:55 UTC+1, Dono. wrote:
On Wednesday, October 11, 2023 at 6:32:54 AM UTC-7, Dono. wrote:
On Wednesday, October 11, 2023 at 6:14:00 AM UTC-7, Lou wrote:

Anti-relativists are not scientists. They are
religious wackos.

You sure are
Fizeau expected the result to be c/n /pm v. Instead he got a very different result: c/n /pm v(1-1/n^2).
The result contradicts Newtonian kinematics and confirms SR. Crank Lou continues to froth at the mouth.

You pretended c/n+v(1-1/n^2) was discovered by Adolf Einstein

Adolf?? Is there a Godwin here?

Answer the question . Did Fizeau discover c/n+v(1-1/n^2) in 1851.

Yes, he did.

Or did Einstein, the low IQ plagiarist, steal it from Fizeau and pretend it was
his own formula ?

That's a huge lie. Einstein EXPLICITLY stated that the Fizeau experiment
was one of four experiments which inspired him to derive SR. He was
CREDITING Fizeau's discovery.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Dono.@21:1/5 to Lou on Wed Oct 11 08:31:44 2023
On Wednesday, October 11, 2023 at 8:18:53 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 15:39:50 UTC+1, Dono. wrote:
On Wednesday, October 11, 2023 at 7:21:56 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 15:02:49 UTC+1, Dono. wrote:
On Wednesday, October 11, 2023 at 6:53:30 AM UTC-7, Dono. wrote:
On Wednesday, October 11, 2023 at 6:49:11 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 14:40:55 UTC+1, Dono. wrote:
On Wednesday, October 11, 2023 at 6:32:54 AM UTC-7, Dono. wrote:
On Wednesday, October 11, 2023 at 6:14:00 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 13:14:44 UTC+1, Robert Winn wrote:
On Wednesday, October 11, 2023 at 2:29:54 AM UTC-7, JanPB wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote:
On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving
clock I imagined a clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane,
the pilot would get a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the
equations of Special Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster
speed for the airplane. Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being
slower than a clock on the ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one
another, which I believe Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed
for the airplane if his clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t,
the time of the clock on the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations
with the same distances for x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we
have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe
it to be. They indicate that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists
of today do not consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is
faster than a clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic
waves. I am explaining relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the
Lorentz equations that he said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for
time of the slower clock gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in
his book. His problem was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no
need for it. The Galilean transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia
No need for relativity. A classical model does just fine.

No, it doesn't. You simply don't know the problems involved.

--
Jan
I know of one problem involved. So when do you think a scientist is going to answer the question I asked about Einstein's description of a miracle? My prediction is never.
Anti-relativists are not scientists. They are
religious wackos.

You sure are
Fizeau expected the result to be c/n /pm v. Instead he got a very different result: c/n /pm v(1-1/n^2).
The result contradicts Newtonian kinematics and confirms SR. Crank Lou continues to froth at the mouth.
You pretended c/n+v(1-1/n^2) was discovered by Einstein

No, I didn't. I simply pointed out the idiocy in the formula that you posted. Repeatedly.
As predicted, you are now just frothing at the mouth.
You pretended that Fizeau’s formula was invented by Albert
You are lying, there is no such claim.
Here’s your quote: “ Experiment measures c/n+v(1-1/n^2)
SR predicts c/n+v(1-1/n^2), in accordance with the experiment. “

"SR predicts " is not "Fizeau’s formula was invented by Albert". Are you as incompetent in English language as you are in physics? Don't answer, it was a rhetorical question.
And if you think my formula c/n+-v{(n-1)+(n-1)^2} does not also correctly predict Fizeau..
Prove it.
Crank,

You are as incompetent in terms of basic algebra as you are in terms of physics. In your demented brain:

n-1+(n-1)^2=1-1/n^2

Way to go, LouLou!

But now prove c/n+- v[{(1.33-1)+(1.33-1)}^2] doesnt correctly
model Fizeau results.
Posting (repeatedly) the same imbecilities doesn't make you right. Makes you just a bigger imbecile.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Lou on Wed Oct 11 12:09:10 2023
On 10/11/2023 6:37 AM, Lou wrote:
On Wednesday, 11 October 2023 at 04:23:31 UTC+1, Volney wrote:
On 10/10/2023 3:23 PM, Lou wrote:
On Tuesday, 10 October 2023 at 17:50:46 UTC+1, Volney wrote:
On 10/10/2023 12:40 PM, Lou wrote:
On Tuesday, 10 October 2023 at 16:47:41 UTC+1, Volney wrote:
On 10/10/2023 7:18 AM, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote: >>>>>>
How do you explain the result of the Fizeau experiment, which was >>>>>>>> performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment
And so for instance when the water
moves towards the source this can be modelled mathematically as the refractive
index of the refractive index for the 'extra distance' travelled. >>>>>>> As the light effectively travels through 'more water' to get the same distance
from source to detector.
Obviously bogus. If that were so, the speed of light in water would get >>>>>> slower and slower as it traversed through more and more water, even if >>>>>> stationary. Instead, the speed of light in (stationary) water is a >>>>>> constant c/n.

You are grasping at straws.

You aren’t just grabbing at straws. You are making up the straws.
Slower with more distance? How so?
Because you claimed the light slows by going through 'more water' (your >>>> term). Increasing the distance obviously means more water traversed.

Grabbing at fantasy straws still. You snipped the formula I cited. There is *no*
‘d’ in that formula. Just v of the water.
My God, are you thick!

The "d" would be the length of a tube in a trivial experiment which
would measure the speed of light in a tube of length "d". If more water
slowed the light, then, trivially, we'd see light going through a tube
with a small length "d" to be faster than light through a tube with a
medium length d (because the light has to traverse more water), but it
would be faster than light with a large value "d", because the latter
has it traverse even more water.

Low IQ nonsense as usual from a relativist. Where’s the d in this formula? c/n+-v{(n-1)+(n-1)^2}
Can’t find it? Oh well. Looks like you’ve been spouting BS again.

So you simply can't support your "traverse more water" claim with a
simple thought experiment that would support/refute it. Why not admit
that your belief simply won't work?

And that refractive index isn’t n in water but a slightly different
value of n which takes into account the fact that the moving water
creates a more optically dense medium in the experiment.

Where is your evidence for this (now changed) claim? Remember, science
is built on experimental evidence and scientific observations.

And classically one way to model this is to make this new
value of n is by this formula... c/n+-v{(n-1)+(n-1)^2}

Now you are making up your own crap formula which conflicts with
Fizeau's results (as well as SR)?

Or if you prefer the Fizeau original formula...that works as well too.

Except that it's different.

All I’ve done is explain how the different speeds in moving water
can be explained classically without having to pull out a nonsense BS Relativistic explanation.

Except you have no evidence to support your (changing) claims.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Volney on Wed Oct 11 09:27:25 2023
On Wednesday, 11 October 2023 at 17:09:14 UTC+1, Volney wrote:
On 10/11/2023 6:37 AM, Lou wrote:
On Wednesday, 11 October 2023 at 04:23:31 UTC+1, Volney wrote:
On 10/10/2023 3:23 PM, Lou wrote:
On Tuesday, 10 October 2023 at 17:50:46 UTC+1, Volney wrote:
On 10/10/2023 12:40 PM, Lou wrote:
On Tuesday, 10 October 2023 at 16:47:41 UTC+1, Volney wrote:
On 10/10/2023 7:18 AM, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote: >>>>>>
How do you explain the result of the Fizeau experiment, which was >>>>>>>> performed half a century before Einstein proposed his theory? >>>>>>>>
https://en.wikipedia.org/wiki/Fizeau_experiment
And so for instance when the water
moves towards the source this can be modelled mathematically as the refractive
index of the refractive index for the 'extra distance' travelled. >>>>>>> As the light effectively travels through 'more water' to get the same distance
from source to detector.
Obviously bogus. If that were so, the speed of light in water would get
slower and slower as it traversed through more and more water, even if
stationary. Instead, the speed of light in (stationary) water is a >>>>>> constant c/n.

You are grasping at straws.

You aren’t just grabbing at straws. You are making up the straws. >>>>> Slower with more distance? How so?
Because you claimed the light slows by going through 'more water' (your >>>> term). Increasing the distance obviously means more water traversed. >>>
Grabbing at fantasy straws still. You snipped the formula I cited. There is *no*
‘d’ in that formula. Just v of the water.
My God, are you thick!

The "d" would be the length of a tube in a trivial experiment which
would measure the speed of light in a tube of length "d". If more water >> slowed the light, then, trivially, we'd see light going through a tube
with a small length "d" to be faster than light through a tube with a
medium length d (because the light has to traverse more water), but it
would be faster than light with a large value "d", because the latter
has it traverse even more water.

Low IQ nonsense as usual from a relativist. Where’s the d in this formula?
c/n+-v{(n-1)+(n-1)^2}
Can’t find it? Oh well. Looks like you’ve been spouting BS again.
So you simply can't support your "traverse more water" claim with a
simple thought experiment that would support/refute it. Why not admit
that your belief simply won't work?

Fizeau is not a thought experiment. Or any version of it.
But it (light beam from source)does traverse more water!
Or less depending on the direction of v.
Why do you think they move the water through the tube?

And that refractive index isn’t n in water but a slightly different value of n which takes into account the fact that the moving water
creates a more optically dense medium in the experiment.
Where is your evidence for this (now changed) claim? Remember, science
is built on experimental evidence and scientific observations.

I have no evidence that water moves through the tube in Fizeau to
change the observed speed of the light in the experiment? Troll.
Another fact free claim from a relativist.
Notice that any description of Fizeau (including wiki) specifically
states that to get the observed effects...one must have water
move through the column/tube.

And classically one way to model this is to make this new
value of n is by this formula... c/n+-v{(n-1)+(n-1)^2}
Now you are making up your own crap formula which conflicts with
Fizeau's results (as well as SR)?

SR stole Fizeau formula. Albert was a plagiarist.
And the formula I suggested...gave a prediction consistent
with the observations. So does Fizeau’s formula.
But Whats important is that the optical
density of the water changes. If it moves relative to the source.
And Fizeau proves this
.
And you have no evidence to contradict this fact.

Or if you prefer the Fizeau original formula...that works as well too.
Except that it's different.

Of course it’s a different formula. There are different ways to model
optical density in moving water. The maths can change...but
the theory doesn’t.
What’s important is that Fizeau observations are predicted by classical theory
as differences in optical densities of moving mediums compared to when
they don’t move relative to the source.
As confirmed by Fizeau.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Volney on Wed Oct 11 09:51:57 2023
On Wednesday, 11 October 2023 at 17:03:28 UTC+1, Volney wrote:
On 10/11/2023 9:49 AM, Lou wrote:

On Wednesday, 11 October 2023 at 14:40:55 UTC+1, Dono. wrote:
On Wednesday, October 11, 2023 at 6:32:54 AM UTC-7, Dono. wrote:
On Wednesday, October 11, 2023 at 6:14:00 AM UTC-7, Lou wrote:

Anti-relativists are not scientists. They are
religious wackos.

You sure are
Fizeau expected the result to be c/n /pm v. Instead he got a very different result: c/n /pm v(1-1/n^2).
The result contradicts Newtonian kinematics and confirms SR. Crank Lou continues to froth at the mouth.

You pretended c/n+v(1-1/n^2) was discovered by Adolf Einstein
Adolf?? Is there a Godwin here?

My apologies. I’ll try harder to ignore the filth you guys regularly dish out.

Answer the question . Did Fizeau discover c/n+v(1-1/n^2) in 1851.
Yes, he did.
Or did Einstein, the low IQ plagiarist, steal it from Fizeau and pretend it was
his own formula ?
That's a huge lie. Einstein EXPLICITLY stated that the Fizeau experiment
was one of four experiments which inspired him to derive SR. He was CREDITING Fizeau's discovery.

Max von Laue. SR is a broad church.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Paul B. Andersen@21:1/5 to All on Wed Oct 11 19:54:04 2023
Den 11.10.2023 15:43, skrev Lou:
On Wednesday, 11 October 2023 at 14:21:24 UTC+1, Paul B. Andersen wrote:
Den 10.10.2023 21:50, skrev Lou:

I assume the speed relative to the ground is the same for both
planes in the experiment. Let’s call it speed x.
The earth rotates eastward at 1600k/ hr.

Close enough.

So the zero point of reference is the earth observer travelling at 1600 k/hr relative
to the earths Center of mass.
Then relative to this earth center reference, the eastward plane travels at >>> 1600 +x kilometers per hour. And the westward plane travels at 1600-x kilometers
per hour.

A reasonable speed for x is 800 km/h.

The east going plane travels at 2400 km/h in the ECI-frame.
The ground clock travels at 1600 km/h in the ECI-frame.
The west going plane travels at 800 km/h in the ECI-frame.

But don’t forget the Gravity force pushing you down is a seperate source of
force from the F force pushing against you as you move horizontally. I’m not sure
what word you relativists prefer but probably inertia or momentum would
be appropriate?

Air drag.

Air Drag!!😂🤣You relativists. Such purveyors of BS.
Did I mention air drag?
No.
You did.

Indeed.
The force that is pushing against the two aeroplanes
as they move horizontally through the air at 800 km/h
is called "air drag".

You forgot. The earth rotates.
I know in relativity land that you guys think the earth doesn’t rotate. But sorry...it does.
In which case the eastward plane travels faster relative to the earth Center frame.
And the westward plane slower
If you knew any maths...then even if the planes speed was 800k/hr it would still
be less than the earth observers speed 1600 k/hr speed.
Which means that the eastward plane travels at a higher speed than the earth observer, and the westward plane at a lower speed than the observer.
And seeing as f=ma then the force on the westward plane is less than the earth observer and the force on the eastward plane is higher.
Which in turn means that due to f=ma the eastward travelling clock will
run slower and the westward clock will run faster than the earth observers clock
Due to mechanical resonance.
As observed in Hafael,Keating.

Congratulations.
You have broken the NG record for most ignorance of elementary physics.
Well done!

Do the maths Paul.

https://paulba.no/pdf/H&K_like.pdf

--
Paul

https://paulba.no/

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Paul B. Andersen@21:1/5 to All on Wed Oct 11 20:08:03 2023
Den 10.10.2023 21:52, skrev Robert Winn:
Here is something you might want to think about. According to modern interpretation of science, Galileo's principle of equivalence no longer applies. If we drop a ball from the ceiling of an airplane that is flying, the ball is falling faster in the
frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.

Immediately after the ball is dropped the speed of the ball is:
1. In the rest frame of the airplane the speed of the ball is zero.
2: In the the ground frame the speed of the ball is equal to
the speed of the airplane.

Which is fastest?

--
Paul

https://paulba.no/

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Paul B. Andersen@21:1/5 to All on Wed Oct 11 20:44:25 2023
Den 10.10.2023 20:57, skrev xip14:
The derivation of 1905 uses a version of the temporal Lorentz Transform. Same thing in today’s textbooks:

t′ = gamma × ( t – vx / c² )

Axis X ( unprimed ) is stationary and axis X′ ( frame k of 1905 ) moves with positive speed-v. The stationary observer is at fixed point-x on axis X ( not x = 0 ) whereas the moving clock to be observed ( always at origin [ x′ = 0 ] of axis X′ )
moves to location-x on axis X according to an equation of motion on axis X:

x = vt

Substitute that into the temporal LT:

t′ = gamma × ( t – v² t / c² )

t′ = gamma × t × ( 1 – v²/c² )

t′ = ( gamma × t ) / gamma²

t′ = t / gamma

It wouldn’t work with a plus sign. Why do we have minus signs in the Lorentz Transforms ( temporal and spatial ) when speed-v is positive?

γ = 1/√(1−v²/c²)
__________________________________

--|----------------> x' -> v
--|----------------> x

t' = γ(t - vx/c²)
x' = γ(x - vt)

t = γ(t' + vx'/c²)
x = γ(x' + vt')
_________________________________

--|----------------> x'
--|----------------> x -> v

t' = γ(t + vx/c²)
x' = γ(x + vt)

t = γ(t' - vx'/c²)
x = γ(x' - vt')
__________________________________

x'<--------|---------- -> v
-----------|---------> x

t' = γ(t - vx/c²)
x' = γ(-x + vt)

t = γ(t' - vx'/c²)
x = γ(-x' + vt')
_________________________________

x'<--------|----------
-----------|---------> x -> v

t' = γ(t + vx/c²)
x' = γ(-x - vt)

t = γ(t' + vx'/c²)
x = γ(-x' - vt')
_________________________________

--
Paul

https://paulba.no/

--- SoupGate-Win32 v1.05
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• From Maciej Wozniak@21:1/5 to All on Wed Oct 11 13:17:45 2023
So you simply can't support your "traverse more water" claim with a

And what you can support yoiur "setting to 9 192 631 774 is
setting to 9 192 631 770", stupid Mike?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Paul B. Andersen on Wed Oct 11 13:28:26 2023
On Wednesday, 11 October 2023 at 18:53:19 UTC+1, Paul B. Andersen wrote:
Den 11.10.2023 15:43, skrev Lou:
On Wednesday, 11 October 2023 at 14:21:24 UTC+1, Paul B. Andersen wrote:
Den 10.10.2023 21:50, skrev Lou:

I assume the speed relative to the ground is the same for both
planes in the experiment. Let’s call it speed x.
The earth rotates eastward at 1600k/ hr.

Close enough.

So the zero point of reference is the earth observer travelling at 1600 k/hr relative
to the earths Center of mass.
Then relative to this earth center reference, the eastward plane travels at
1600 +x kilometers per hour. And the westward plane travels at 1600-x kilometers
per hour.

A reasonable speed for x is 800 km/h.

The east going plane travels at 2400 km/h in the ECI-frame.
The ground clock travels at 1600 km/h in the ECI-frame.
The west going plane travels at 800 km/h in the ECI-frame.
But don’t forget the Gravity force pushing you down is a seperate source of
force from the F force pushing against you as you move horizontally. I’m not sure
what word you relativists prefer but probably inertia or momentum would >>> be appropriate?

Air drag.

Air Drag!!😂🤣You relativists. Such purveyors of BS.
Did I mention air drag?
No.
You did.
Indeed.
The force that is pushing against the two aeroplanes
as they move horizontally through the air at 800 km/h
is called "air drag".

Air drag has nothing to do with the fact that the earth rotates eastward
Nor does air drag have anything to do with the fact that this eastward rotation of the earth means that the earth observer rotates around the Center of The earth at 1600k/hr, the westward plane travels at 1600-800k/hr, and the eastward plane travels at 1600+800k/hr relative to the earth Center.
That’s why they launch earth orbital bound rockets eastward ...the earths eastward rotation adds to the escape velocity. And you pretend you know
basic physics!!

https://spaceplace.nasa.gov/launch-windows/en/

Fact is that what Einstein pretended was time dilation effects from SR and GR is
actually just external Force from acceleration acting on resonant atomic systems
forcing them to slow down or speed up their natural resonant frequencies. Resonance: a classical phenomena. Understood and documented for centuries.
No magic relativity needed.

You forgot. The earth rotates.
I know in relativity land that you guys think the earth doesn’t rotate. But
sorry...it does.
In which case the eastward plane travels faster relative to the earth Center frame.
And the westward plane slower
If you knew any maths...then even if the planes speed was 800k/hr it would still
be less than the earth observers speed 1600 k/hr speed.
Which means that the eastward plane travels at a higher speed than the earth
observer, and the westward plane at a lower speed than the observer.
And seeing as f=ma then the force on the westward plane is less than the earth observer and the force on the eastward plane is higher.
Which in turn means that due to f=ma the eastward travelling clock will run slower and the westward clock will run faster than the earth observers clock
Due to mechanical resonance.
As observed in Hafael,Keating.
Congratulations.
You have broken the NG record for most ignorance of elementary physics.
Well done!

Yes Paul. And yet you don’t think the earth rotates on its axis!!

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Dono.@21:1/5 to Lou on Wed Oct 11 13:39:09 2023
On Wednesday, October 11, 2023 at 1:28:28 PM UTC-7, Lou wrote:

Fact is that what Einstein pretended was time dilation effects from SR and GR is
actually just external Force from acceleration acting on resonant atomic systems
forcing them to slow down or speed up their natural resonant frequencies.

Dumbotron

It is a know fact that acceleration does not affect time passage. Look up "the Clock Hypothesis" for the experiments confirming this FACT. You have been pantsed. Again.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Paul B. Andersen on Wed Oct 11 16:37:34 2023
On Wednesday, October 11, 2023 at 11:07:19 AM UTC-7, Paul B. Andersen wrote:
Den 10.10.2023 21:52, skrev Robert Winn:
Here is something you might want to think about. According to modern interpretation of science, Galileo's principle of equivalence no longer applies. If we drop a ball from the ceiling of an airplane that is flying, the ball is falling faster in the
frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.

Immediately after the ball is dropped the speed of the ball is:
1. In the rest frame of the airplane the speed of the ball is zero.
2: In the the ground frame the speed of the ball is equal to
the speed of the airplane.

Which is fastest?

--
Paul

https://paulba.no/
No, Paul, No. When we say something is falling, it is going toward the earth from a height. Whether is falls in a straight line or a parabolic one because of forward momentum, its distance from the earth decreases at the same rate. But so that you do
not get confused again, let's say that we drop it to the ground from the airplane. If the clock in the airplane is slower than a clock on the ground, the object will appear to be falling faster when timed by the clock in the airplane than when timed by
the clock on the ground if the clock in the airplane is slower as Einstein describes.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Lou on Thu Oct 12 00:47:56 2023
On 10/11/2023 12:27 PM, Lou wrote:
On Wednesday, 11 October 2023 at 17:09:14 UTC+1, Volney wrote:
On 10/11/2023 6:37 AM, Lou wrote:
On Wednesday, 11 October 2023 at 04:23:31 UTC+1, Volney wrote:
On 10/10/2023 3:23 PM, Lou wrote:
On Tuesday, 10 October 2023 at 17:50:46 UTC+1, Volney wrote:
On 10/10/2023 12:40 PM, Lou wrote:
On Tuesday, 10 October 2023 at 16:47:41 UTC+1, Volney wrote:
On 10/10/2023 7:18 AM, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote: >>>>>>>>
How do you explain the result of the Fizeau experiment, which was >>>>>>>>>> performed half a century before Einstein proposed his theory? >>>>>>>>>>
https://en.wikipedia.org/wiki/Fizeau_experiment
And so for instance when the water
moves towards the source this can be modelled mathematically as the refractive
index of the refractive index for the 'extra distance' travelled. >>>>>>>>> As the light effectively travels through 'more water' to get the same distance
from source to detector.
Obviously bogus. If that were so, the speed of light in water would get
slower and slower as it traversed through more and more water, even if >>>>>>>> stationary. Instead, the speed of light in (stationary) water is a >>>>>>>> constant c/n.

You are grasping at straws.

You aren’t just grabbing at straws. You are making up the straws. >>>>>>> Slower with more distance? How so?
Because you claimed the light slows by going through 'more water' (your >>>>>> term). Increasing the distance obviously means more water traversed. >>>>>
Grabbing at fantasy straws still. You snipped the formula I cited. There is *no*
‘d’ in that formula. Just v of the water.
My God, are you thick!

The "d" would be the length of a tube in a trivial experiment which
would measure the speed of light in a tube of length "d". If more water >>>> slowed the light, then, trivially, we'd see light going through a tube >>>> with a small length "d" to be faster than light through a tube with a
medium length d (because the light has to traverse more water), but it >>>> would be faster than light with a large value "d", because the latter
has it traverse even more water.

Low IQ nonsense as usual from a relativist. Where’s the d in this formula?
c/n+-v{(n-1)+(n-1)^2}
Can’t find it? Oh well. Looks like you’ve been spouting BS again.
So you simply can't support your "traverse more water" claim with a
simple thought experiment that would support/refute it. Why not admit
that your belief simply won't work?

Fizeau is not a thought experiment. Or any version of it.

Idiot, the thought experiment is water filled tubes with different
lengths "d", which should (if your claim is correct) slow down light
more for longer "d" because the light traverses more water!

But it (light beam from source)does traverse more water!

Just like in a longer tube! How come we don't measure slower light from
longer water filled tubes?

Or less depending on the direction of v.
Why do you think they move the water through the tube?

Duh-h-h-h... The speed of light in the water is c/n, relative to the
water. Relative to the lab, it should be (according to cranks) c/n ±v
but instead Fizeau measured it as c/n ±v(1-1/n^2). Now you, desperately flailing around, are coming up with bogus excuses like there is more
water to traverse when the water is moving but not when a water filled
tube is longer!

And that refractive index isn’t n in water but a slightly different
value of n which takes into account the fact that the moving water
creates a more optically dense medium in the experiment.
Where is your evidence for this (now changed) claim? Remember, science
is built on experimental evidence and scientific observations.

I have no evidence that water moves through the tube in Fizeau to
change the observed speed of the light in the experiment? Troll.

You have no evidence the index of refraction in water is changed by
motion. It is a desperate flailing to try to come up with some other explanation than one of the first glimmers of SR.

Another fact free claim from a relativist.

The fact free claim is that the index of refraction changes due to
motion. Remember, evidence is king in physics. Got any?

Notice that any description of Fizeau (including wiki) specifically
states that to get the observed effects...one must have water
move through the column/tube.

Exactly. It was an early demonstration of the SR speed combination formula.

And classically one way to model this is to make this new
value of n is by this formula... c/n+-v{(n-1)+(n-1)^2}

Now you are making up your own crap formula which conflicts with
Fizeau's results (as well as SR)?

SR stole Fizeau formula. Albert was a plagiarist.

Nope. Fizeau's experiment INSPIRED Albert. It was an unsolved mystery
until SR could explain it.

And the formula I suggested...gave a prediction consistent
with the observations. So does Fizeau’s formula.

So which body orifice did you pull your formula from? Why do you feel it
is better than Fizeau's formula? Remember, you must provide evidence of

But Whats important is that the optical
density of the water changes. If it moves relative to the source.
And Fizeau proves this

Fizeau never proved anything about optical density. Neither did you. You
made the whole thing up, and it is laughable when considered varying
length tubes of water also provide different amounts of water to traverse.
.
And you have no evidence to contradict this fact.

Your claim is NOT a fact. In fact it is up to you to show water speed
produces increased density which changes n, or for your alternate
formula. Remember, science is based on scientific observations and
experimental data.

Or if you prefer the Fizeau original formula...that works as well too.
Except that it's different.

And not worthless like your bogus formula.

Of course it’s a different formula. There are different ways to model optical density in moving water. The maths can change...but
the theory doesn’t.

What theory? You have provided no evidence of anything!

What’s important is that Fizeau observations are predicted by classical theory
as differences in optical densities of moving mediums compared to when
they don’t move relative to the source.

Nope. Classical theory predicts c/n ± v. Fizeau's observations

As confirmed by Fizeau.

Nope. Don't put words in Fizeau's mouth.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Robert Winn on Thu Oct 12 00:54:08 2023
On 10/11/2023 8:12 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 11:23:05 PM UTC-7, Volney wrote:
On 10/11/2023 12:10 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 8:26:09 PM UTC-7, Volney wrote:
On 10/10/2023 1:49 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 10:16:10 AM UTC-7, Volney wrote:
On 10/10/2023 1:11 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:44:38 AM UTC-7, Volney wrote: >>>>>>>> On 10/10/2023 12:07 PM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 9:01:09 AM UTC-7, Volney wrote: >>>>>>>>>> On 10/10/2023 11:49 AM, Robert Winn wrote:
On Tuesday, October 10, 2023 at 4:50:06 AM UTC-7, Sylvia Else wrote:
On 10-Oct-23 9:38 pm, Robert Winn wrote:

From what I read about Fizeau's experiment, he was measuring the speed of light through moving water with the expectation that the speed of the water would be added to the speed of the light. Michelson and Morley were measuring the speed of light
through air with the expectation that the speed of the air relative to their interferometer would be added to the speed of light in air that was not moving.
You can't even get that correct! The MMX was an attempt to measure the
earth's speed through the ether!

No comment?

In English, water is the name given to H2O. Air is a combination of gases including oxygen, carbon dioxide, and nitrogen. That is what those words mean in English. If you have difficulty understand more words in English, just ask at any time.

So why did you claim that 'Fizeau's experiment was an early version of
the Michelson-Morley experiment'?

Well, as a common person, not a scientist, I just take note of the similarities. The experimenters in both cases believed that there was a medium through which light was conducted. In the case of the Fizeau experiment, the medium was water, in
Michelson-Morley it was air. In both experiments it was expected that the speed of the medium relative to the measuring device would affect the result obtained for c, the speed of light. In both experiments the result obtained was not the result
expected. Then there was the fact that Michelson and Morley said that the idea for their interferometer came from Fizeau's experiment. But, as I said, I had no idea of what Fizeau's experiment was until I read something about it yesterday morning.

You didn't answer the question why you thought they were the same
experiment. But your replies are incoherent and you don't even know what
the MMX even was, so I don't expect an answer beyond these mumbles of yours.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Volney on Thu Oct 12 04:32:32 2023
On Thursday, 12 October 2023 at 05:48:17 UTC+1, Volney wrote:
On 10/11/2023 12:27 PM, Lou wrote:
On Wednesday, 11 October 2023 at 17:09:14 UTC+1, Volney wrote:
On 10/11/2023 6:37 AM, Lou wrote:
On Wednesday, 11 October 2023 at 04:23:31 UTC+1, Volney wrote:
On 10/10/2023 3:23 PM, Lou wrote:
On Tuesday, 10 October 2023 at 17:50:46 UTC+1, Volney wrote:
On 10/10/2023 12:40 PM, Lou wrote:
On Tuesday, 10 October 2023 at 16:47:41 UTC+1, Volney wrote: >>>>>>>> On 10/10/2023 7:18 AM, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote: >>>>>>>>
How do you explain the result of the Fizeau experiment, which was >>>>>>>>>> performed half a century before Einstein proposed his theory? >>>>>>>>>>
https://en.wikipedia.org/wiki/Fizeau_experiment
And so for instance when the water
moves towards the source this can be modelled mathematically as the refractive
index of the refractive index for the 'extra distance' travelled. >>>>>>>>> As the light effectively travels through 'more water' to get the same distance
from source to detector.
Obviously bogus. If that were so, the speed of light in water would get
slower and slower as it traversed through more and more water, even if
stationary. Instead, the speed of light in (stationary) water is a >>>>>>>> constant c/n.

You are grasping at straws.

You aren’t just grabbing at straws. You are making up the straws. >>>>>>> Slower with more distance? How so?
Because you claimed the light slows by going through 'more water' (your
term). Increasing the distance obviously means more water traversed. >>>>>
Grabbing at fantasy straws still. You snipped the formula I cited. There is *no*
‘d’ in that formula. Just v of the water.
My God, are you thick!

The "d" would be the length of a tube in a trivial experiment which >>>> would measure the speed of light in a tube of length "d". If more water >>>> slowed the light, then, trivially, we'd see light going through a tube >>>> with a small length "d" to be faster than light through a tube with a >>>> medium length d (because the light has to traverse more water), but it >>>> would be faster than light with a large value "d", because the latter >>>> has it traverse even more water.

Low IQ nonsense as usual from a relativist. Where’s the d in this formula?
c/n+-v{(n-1)+(n-1)^2}
Can’t find it? Oh well. Looks like you’ve been spouting BS again.
So you simply can't support your "traverse more water" claim with a
simple thought experiment that would support/refute it. Why not admit
that your belief simply won't work?

Fizeau is not a thought experiment. Or any version of it.

Idiot, the thought experiment is water filled tubes with different
lengths "d", which should (if your claim is correct) slow down light
more for longer "d" because the light traverses more water!

You don’t understand refractive index. If the refractive index
of the medium changes...the speed changes. But distance travelled
will not affect the refractive index or speed of light in any medium.
So..How could distance travelled affect the predicted speed in the water
in my formula if there is no “d” in my formula?
In other words doesn’t matter how long the tube is. Because my formula,
(and Fizeau’s) will always predict the same speed change in
moving water. As long as that water travelled at the same v relative
to the source. Regardless of distance travelled.
That’s how refractive index works.

But it (light beam from source)does traverse more water!

Just like in a longer tube! How come we don't measure slower light from longer water filled tubes?

Why should we? Fizeau doesn’t predict it. Nor do I.
You have misunderstood something.

Or less depending on the direction of v.
Why do you think they move the water through the tube?

Duh-h-h-h... The speed of light in the water is c/n, relative to the
water. Relative to the lab, it should be (according to cranks) c/n ±v

Maybe relativistic cranks...but I just finished telling you and showing
you a formula saying it wasn’t at c/n+-v. Can’t you read?
And not only that we’ve known it isn’t c/n+-v since 1851.

but instead Fizeau measured it as c/n ±v(1-1/n^2). Now you, desperately flailing around, are coming up with bogus excuses like there is more
water to traverse when the water is moving but not when a water filled
tube is longer!

Liar. I never said that. You either misunderstood me or made this fantasy
claim up.

And that refractive index isn’t n in water but a slightly different >>> value of n which takes into account the fact that the moving water
creates a more optically dense medium in the experiment.
Where is your evidence for this (now changed) claim? Remember, science
is built on experimental evidence and scientific observations.

I have no evidence that water moves through the tube in Fizeau to
change the observed speed of the light in the experiment? Troll.

You have no evidence the index of refraction in water is changed by
motion. It is a desperate flailing to try to come up with some other explanation than one of the first glimmers of SR.

I have no evidence!! 😂💩 Nonsense.
I just cited Fizeau experiment as my evidence. And believe me, if you actually tried reading the various reference on the experiment you would see it
confirms what I say.
Which is: That the faster the water moves in the tube. The slower the light speed is observed relative to the source. And that this change in speed
isnt c/n +-v. I supplied a formula to model this change in speed. Or you
can refer to Fizeau’s formula. Either one accurately predicts how much
light speed slows relative to the source if the water is at v.

Another fact free claim from a relativist.

The fact free claim is that the index of refraction changes due to
motion. Remember, evidence is king in physics. Got any?

Yes. Fizeau. Notice it observed that the speed of light slows down in water
if the water moves in the tube. And the reason is that the optical
density of the water increases relative to the source if the water is at v.
As observed.
Any evidence that there is no change in light-speed relative to
the source in Fizeau when the water is at v?
No.
Thought not.

Notice that any description of Fizeau (including wiki) specifically
states that to get the observed effects...one must have water
move through the column/tube.

Exactly. It was an early demonstration of the SR speed combination formula.

Theoretically. But a classical explanation doesn’t rely on magic.
And works just as well. Seeing as both Fizeau’s and my formula
can accurately predict this classical effect.
Also...you forgot...the formula SR uses...is stolen from classical theory
As per usual for relativists.

And classically one way to model this is to make this new
value of n is by this formula... c/n+-v{(n-1)+(n-1)^2}

Now you are making up your own crap formula which conflicts with
Fizeau's results (as well as SR)?

SR stole Fizeau formula. Albert was a plagiarist.

Nope. Fizeau's experiment INSPIRED Albert. It was an unsolved mystery
until SR could explain it.

And the formula I suggested...gave a prediction consistent
with the observations. So does Fizeau’s formula.

So which body orifice did you pull your formula from? Why do you feel it
is better than Fizeau's formula? Remember, you must provide evidence of

I didn’t say it was better. I said both worked as well. And proved that
the optical density of the moving water relative to the source increases
or decreases with+-v. A purely classical phenomena.
No Relativity needed.

But Whats important is that the optical
density of the water changes. If it moves relative to the source.
And Fizeau proves this

Fizeau never proved anything about optical density. Neither did you. You made the whole thing up, and it is laughable when considered varying
length tubes of water also provide different amounts of water to traverse.

If you think I’m wrong..prove that Fizeau does not observe any change in lightspeed in water relative to the source. Even when the water moves
at v in the tube.
You can’t.

And you have no evidence to contradict this fact.

Your claim is NOT a fact. In fact it is up to you to show water speed produces increased density which changes n, or for your alternate
formula. Remember, science is based on scientific observations and experimental data.

I did prove it. My evidence is ...the Fizeau experiment.

Or if you prefer the Fizeau original formula...that works as well too. >> Except that it's different.

And not worthless like your bogus formula.

My formula predicts as well as Fizeau’s. But that’s irrelevent
because Fizeau’s formula also correctly predicts the optical density
changes when water is at v.

Of course it’s a different formula. There are different ways to model optical density in moving water. The maths can change...but
the theory doesn’t.

What theory? You have provided no evidence of anything!

Only if you ignore Fizeau.

What’s important is that Fizeau observations are predicted by classical theory
as differences in optical densities of moving mediums compared to when they don’t move relative to the source.

Nope. Classical theory predicts c/n ± v. Fizeau's observations

No. Classical theory predicts light speed changes in water if the water
moves in the tube at v. And succesfully predicted for classical theory
by Fizeau : c/n+-v(1-(1/n^2))
Or by my formula: c/n+-v[{(n-1)+(n-1)}^2]

As confirmed by Fizeau.

Nope. Don't put words in Fizeau's mouth.

I didn’t. I only quoted his experiment where he not only measures
the change in lightspeed ,...he correctly models it with the classical
formula c/n+-v(1-(1/n^2))

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Dono.@21:1/5 to Lou on Thu Oct 12 07:19:59 2023
On Thursday, October 12, 2023 at 4:32:34 AM UTC-7, Lou wrote:
And succesfully predicted for classical theory
by Fizeau : c/n+-v(1-(1/n^2))

Crank,

Classical theory (Newton) predicts c/n +v . NOT c/n+-v(1-1/n^2)

Or by my formula: c/n+-v[{(n-1)+(n-1)}^2]

Repeating the posting of "your" crank formula doesn't make it true. It just make you a stubborn crank.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Paul B. Andersen@21:1/5 to All on Thu Oct 12 16:23:46 2023
Den 12.10.2023 01:37, skrev Robert Winn:
On Wednesday, October 11, 2023 at 11:07:19 AM UTC-7, Paul B. Andersen wrote:
Den 10.10.2023 21:52, skrev Robert Winn:
Here is something you might want to think about. According to modern interpretation of science, Galileo's principle of equivalence no longer applies. If we drop a ball from the ceiling of an airplane that is flying, the ball is falling faster in the
frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.

Immediately after the ball is dropped the speed of the ball is:
1. In the rest frame of the airplane the speed of the ball is zero.
2: In the the ground frame the speed of the ball is equal to
the speed of the airplane.

Which is fastest?

--
Paul

https://paulba.no/
No, Paul, No. When we say something is falling, it is going toward the earth from a height. Whether is falls in a straight line or a parabolic one because of forward momentum, its distance from the earth decreases at the same rate. But so that you
do not get confused again, let's say that we drop it to the ground from the airplane. If the clock in the airplane is slower than a clock on the ground, the object will appear to be falling faster when timed by the clock in the airplane than when timed
by the clock on the ground if the clock in the airplane is slower as Einstein describes.

The ball is falling from the ceiling of an airplane, and will
fall a distance h and hit the floor in the airplane.

_____________________________________________________________________

In the airplane's rest frame we have:
---------------------------------------
The ball will fall along a straight vertical line, and
will hit the floor with the vertical velocity u_v = √(2gh).

The speed of the ball when it hit the floor is u = √(2gh)
The ball will hit the floor at the time t' = √(2h/g)
after it was dropped.

This time is measured with coordinate clocks in the airplane's
rest frame. _______________________________________________________________________

In the ground frame, we have:
-------------------------------
The ball will have a constant horizontal speed v,
and the vertical speed when it hits the floor is u_v = √(2gh)

The speed of the ball when it hit the floor is u' = √(v²+2gh)
The ball will hit the floor at the time t' = √(2h/g)
after it was dropped.

This time is measured with coordinate clocks in the ground frame. __________________________________________________________________________

If use the values
g = 9.8 m/s², v = 230 m/s (828.00 km/h), h = 2.5 m, we get :

The speed of the ball when it hits the floor is:
In the rest frame of the airplane: u = 7 m/s, t = 0.7142 s
In the ground frame: u= 230.11 m/s = 828.38 km/h, t = 0.7142 s

This answers are the same for SR and NM.
(We ignore the curvature of spacetime because
the height difference is only 2.5 m)

So the speed is obviously much higher in the ground frame
than in the airplane's rest frame. But the vertical speed
component is the same in both frames.

----------------

However, according to SR we will have mutual time dilation.
Let's see what it is:
We have two events, E0 = ball dropped, E1 = ball hits floor

The coordinates of event E0 are:
Airplane frame: t'₀ = 0, x'₀ = 0
Ground frame: t₀ = 0, x₀ = 0

The coordinates of event E1 are:
Airplane frame: t'₁ = √(2h/g) = 0, x'₁ = 0
Ground frame: t₁ = √(2h/g) = 0.7142 s ,
x₁ = v⋅√(2h/g) = 164.29 m

#1:
We will now find the rate of a clock in the airplane
observed in the ground frame:

We will use the Lorentz transform.
γ = 1/√(1−v²/c²) = (1+2.9E-13)

A clock at x'₀ will be adjacent to x₀
at the time t₀ and it will show t'₀. (trivial, all zero)

The clock at x'₀ will be adjacent to x₁
at the time t₁ and it will show:
t'= γ(t₁-v⋅x₁/c²) = √(2h/g)/γ

So the rate of the clock at x'₀ is:
f = t'/t₁ = 1/γ = 1 - 2.9E-13

So the a clock in the airplane will appear
to run slow when observed in the ground frame. ____________________________________________________

#2:
We will now find the rate of a clock on the ground
observed in the rest frame of the airplane:

Since a clock at x₀ in the ground frame must be
adjacent to two different clocks in the airplane,
we must imagine a clock at x'₂ = - v⋅√(2h/g)

A clock at x₀ will be adjacent to x'₀
at the time t'₀ and it will show t₀. (trivial, all zero)

The clock at x₀ will be adjacent to x'₂
at the time t'₁ and it will show:
t = γ(t'₁+v⋅x'₂/c²) = √(2h/g)/γ

So the rate of the clock at x'₀ is:
f = t/t'₁ = 1/γ = 1 - 2.9E-13

So a clock on the ground will appear
to run slow when observed in the airplane frame.

Mutual time dilation!

https://paulba.no/pdf/Mutual_time_dilation.pdf

--
Paul

https://paulba.no/

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Paul B. Andersen on Thu Oct 12 11:04:28 2023
On Thursday, October 12, 2023 at 7:23:00 AM UTC-7, Paul B. Andersen wrote:
Den 12.10.2023 01:37, skrev Robert Winn:
On Wednesday, October 11, 2023 at 11:07:19 AM UTC-7, Paul B. Andersen wrote:
Den 10.10.2023 21:52, skrev Robert Winn:
Here is something you might want to think about. According to modern interpretation of science, Galileo's principle of equivalence no longer applies. If we drop a ball from the ceiling of an airplane that is flying, the ball is falling faster in
the frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.

Immediately after the ball is dropped the speed of the ball is:
1. In the rest frame of the airplane the speed of the ball is zero.
2: In the the ground frame the speed of the ball is equal to
the speed of the airplane.

Which is fastest?

--
Paul

https://paulba.no/
No, Paul, No. When we say something is falling, it is going toward the earth from a height. Whether is falls in a straight line or a parabolic one because of forward momentum, its distance from the earth decreases at the same rate. But so that you do
not get confused again, let's say that we drop it to the ground from the airplane. If the clock in the airplane is slower than a clock on the ground, the object will appear to be falling faster when timed by the clock in the airplane than when timed by
the clock on the ground if the clock in the airplane is slower as Einstein describes.

The ball is falling from the ceiling of an airplane, and will
fall a distance h and hit the floor in the airplane.

_____________________________________________________________________

In the airplane's rest frame we have: ---------------------------------------
The ball will fall along a straight vertical line, and
will hit the floor with the vertical velocity u_v = √(2gh).

The speed of the ball when it hit the floor is u = √(2gh)
The ball will hit the floor at the time t' = √(2h/g)
after it was dropped.

This time is measured with coordinate clocks in the airplane's
rest frame. _______________________________________________________________________

In the ground frame, we have:
-------------------------------
The ball will have a constant horizontal speed v,
and the vertical speed when it hits the floor is u_v = √(2gh)

The speed of the ball when it hit the floor is u' = √(v²+2gh)
The ball will hit the floor at the time t' = √(2h/g)
after it was dropped.

This time is measured with coordinate clocks in the ground frame. __________________________________________________________________________

If use the values
g = 9.8 m/s², v = 230 m/s (828.00 km/h), h = 2.5 m, we get :

The speed of the ball when it hits the floor is:
In the rest frame of the airplane: u = 7 m/s, t = 0.7142 s
In the ground frame: u= 230.11 m/s = 828.38 km/h, t = 0.7142 s

This answers are the same for SR and NM.
(We ignore the curvature of spacetime because
the height difference is only 2.5 m)

So the speed is obviously much higher in the ground frame
than in the airplane's rest frame. But the vertical speed
component is the same in both frames.

----------------

However, according to SR we will have mutual time dilation.
Let's see what it is:
We have two events, E0 = ball dropped, E1 = ball hits floor

The coordinates of event E0 are:
Airplane frame: t'₀ = 0, x'₀ = 0
Ground frame: t₀ = 0, x₀ = 0

The coordinates of event E1 are:
Airplane frame: t'₁ = √(2h/g) = 0, x'₁ = 0
Ground frame: t₁ = √(2h/g) = 0.7142 s ,
x₁ = v⋅√(2h/g) = 164.29 m

#1:
We will now find the rate of a clock in the airplane
observed in the ground frame:

We will use the Lorentz transform.
γ = 1/√(1−v²/c²) = (1+2.9E-13)

A clock at x'₀ will be adjacent to x₀
at the time t₀ and it will show t'₀. (trivial, all zero)

The clock at x'₀ will be adjacent to x₁
at the time t₁ and it will show:
t'= γ(t₁-v⋅x₁/c²) = √(2h/g)/γ

So the rate of the clock at x'₀ is:
f = t'/t₁ = 1/γ = 1 - 2.9E-13

So the a clock in the airplane will appear
to run slow when observed in the ground frame. ____________________________________________________

#2:
We will now find the rate of a clock on the ground
observed in the rest frame of the airplane:

Since a clock at x₀ in the ground frame must be
adjacent to two different clocks in the airplane,
we must imagine a clock at x'₂ = - v⋅√(2h/g)

A clock at x₀ will be adjacent to x'₀
at the time t'₀ and it will show t₀. (trivial, all zero)

The clock at x₀ will be adjacent to x'₂
at the time t'₁ and it will show:
t = γ(t'₁+v⋅x'₂/c²) = √(2h/g)/γ

So the rate of the clock at x'₀ is:
f = t/t'₁ = 1/γ = 1 - 2.9E-13

So a clock on the ground will appear
to run slow when observed in the airplane frame.

Mutual time dilation!

https://paulba.no/pdf/Mutual_time_dilation.pdf

--
Paul

https://paulba.no/
Well, no, Paul. Einstein says that we have a slower clock in the flying airplane. Hafele and Keating say it might be faster, but it could be slower. So we will go with Einstein's original idea. The clock is slower. If t is the time of a clock on the
ground, the during a time of t= 1 sec., the earth rotates a certain number of degrees on its axis. Mars rotates a certain number of degrees on its axis, Jupiter rotates a certain number of degrees on its axis, etc., etc. The clock in the airplane shows
less time than a second when the clock on the ground shows a second and the earth, Mars, Jupiter, etc., are all rotating these varying number of degrees. So what you are telling me now is that if a scientist looks at the clock in the airplane, the clock
on the ground shows less time than the clock on the airplane. What do the earth, Mars, Jupiter, etc. do? Do they all rotate backward to agree with the clock on the ground every time a scientist looks at the clock in the airplane? Now scientists have
two miracles. I am not opposed to scientists having miracles, but you have to understand that we common people are required to live in something called reality where the planets would just go on rotating the same way they were before the scientist
looked at the clock in the airplane. Or maybe scientists are trying to say that the planets would go on rotating the way they did before, but the clock on the ground would rotate backward to agree with the equations scientists use. It would still just
be another miracle. I have to agree with Galileo and Isaac Newton that if the clock in the airplane was slower, as seen from the airplane, the clock on the ground would be faster. I think scientists are either using the wrong equations for their
calculations of electromagnetism and electromagnetic waves, or else they are misinterpreting the equations they have.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Lou on Thu Oct 12 13:53:57 2023
On 10/11/2023 6:45 AM, Lou wrote:
On Wednesday, 11 October 2023 at 10:29:54 UTC+1, JanPB wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote: >>>>>> On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a
clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get
a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock
on the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances
for x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate
that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right >>>>>> about this.

It's not as if Einstein, as a young patent clerk, had some ability to >>>>>> impose his theory on an unwilling world. Experimenters had been taking a >>>>>> very close look at reality, and had been finding that it was not
behaving in the expected way. Einstein provided a solution. That is why >>>>>> a young patent clerk was able to get his theory accepted by the
scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of today do not
consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is faster than a
clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves. I am explaining
relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the Lorentz equations that
he said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of the slower
clock gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book. His
problem was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for it.
The Galilean transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia
No need for relativity. A classical model does just fine.

No, it doesn't. You simply don't know the problems involved.

You don’t understand basic physics. Fact: A more optically dense medium will have a slightly greater refractive index.

Is today "make up a fact day" or something?

Perhaps it does, do you have a reference for index of refraction vs.
density?

And if the water moves
relative to the source..a classical model predicts this extra +-v
will increase (or decrease)the optical density of the water as it moves relative to the source.

And Fizeaus original formula ( that SR stole)

No, Fizeau's experiment helped inspire Einstein.

models this change in density.

How could it, without any data?

And even if it did, classical physics would ADD the v component to the
alleged density contribution.

As does my own version: c/n+-v{(n-1)+(n-1)^2}

It is make up a fact day!

Proving

No proofs in physics, only disproofs.

the only people who don’t have any idea are the relativist
fantasists who still think the sun rotates around the earth.

Huh? Besides, it is the anti-relativity cranks who are living in the
past (well over 100 years now), not scientists.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Laurence Clark Crossen@21:1/5 to Robert Winn on Thu Oct 12 20:35:09 2023
On Monday, October 9, 2023 at 11:27:35 AM UTC-7, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a clock in
a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get a faster
speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special Relativity
show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on the
ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for x
and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'. So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate that
there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Yes, relativity does not describe physical reality. Einstein was very confused. I don't know why you can't "learn" (be indoctrinated = believe in fairy tales). It's been noticed all along by many excellent scientists who have shown that relativity is a
joke/swindle (e.g. Essen). It is amusing that they can accept "really weird" "science." Einstein had much funding behind him. I wonder what speed the satellite is really moving. People who pretend to do miracles are wizards, not scientists. Science has
no need for "miracles" (relativity).

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Laurence Clark Crossen on Thu Oct 12 22:22:00 2023
On Thursday, October 12, 2023 at 8:35:11 PM UTC-7, Laurence Clark Crossen wrote:
On Monday, October 9, 2023 at 11:27:35 AM UTC-7, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a clock
in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get a
faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on
the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for
x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate
that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Yes, relativity does not describe physical reality. Einstein was very confused. I don't know why you can't "learn" (be indoctrinated = believe in fairy tales). It's been noticed all along by many excellent scientists who have shown that relativity is a
joke/swindle (e.g. Essen). It is amusing that they can accept "really weird" "science." Einstein had much funding behind him. I wonder what speed the satellite is really moving. People who pretend to do miracles are wizards, not scientists. Science has
no need for "miracles" (relativity).
Well, I have always wondered about this. I read Einstein's book, and he gave a fairly good explanation of the Galilean transformation equations, which were used to describe relativity before 1887. Then he said, the Galilean transformation equations
cannot describe the results of the Michelson-Morley experiment. It appeared to me that they could. What you have to do is believe what the equations say and follow the axioms of algebra. If there is an equation that says t'=t, then t' cannot be used
to represent the time of a clock that is slower than a clock that shows t. You have to use an entirely different set of Galilean transformation equations with different variables for time and velocity than v, t, and t' in the equations, x' = x - vt. and
t'=t, the same way you would use two different sets of Galilean transformation equations for time based on the rotation of earth and time based on the rotation of Mars. You may get a close approximation doing what Lorentz and Einstein did, but, why do
it that way?
But, then too, why call what these guys are doing relativity? If their equations have something to do with electromagnetic fields or electromagnetic waves, call it that, but Galileo's equations are the ones that work for relativity.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Volney on Fri Oct 13 00:27:43 2023
On Friday, 13 October 2023 at 02:52:52 UTC+1, Volney wrote:
On 10/11/2023 6:45 AM, Lou wrote:
On Wednesday, 11 October 2023 at 10:29:54 UTC+1, JanPB wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote: >>>>>> On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a
clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get
a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the
airplane. Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock
on the ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the
clock on the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same
distances for x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They
indicate that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right >>>>>> about this.

It's not as if Einstein, as a young patent clerk, had some ability to >>>>>> impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not >>>>>> behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the >>>>>> scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of today do not
consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is faster than a
clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves. I am
explaining relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the Lorentz equations
that he said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of the slower
clock gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book. His
problem was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for it.
The Galilean transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia
No need for relativity. A classical model does just fine.

No, it doesn't. You simply don't know the problems involved.

You don’t understand basic physics. Fact: A more optically dense medium will have a slightly greater refractive index.

Is today "make up a fact day" or something?

Look up optical density vs refraction and you get many quotes like
the following: “ The refractive index of the material is an indicator
of its optical density.”

Perhaps it does, do you have a reference for index of refraction vs. density?

I’m not trying to write a textbook on refraction and optical
density. All I’m doing is pointing out that in the Fizeau experiment
when the water flows at v in the tube, the lightspeed of the beam
travelling the distance between source and interference detector is
slower than when the water doesn’t flow.
Obviously c/n of water no longer applies. This slowing down of
lightspeed in moving water is best described classically as the optical
density and thus refractive index of light in the moving water increases.
As modelled by Fizeau’s or my formula.
If you have any proof that there is no relationship between optical density increasing proportional to observed speed of light in a transparent medium. Please supply your evidence.
But you haven’t. Obviously you haven’t a clue about refraction.

And if the water moves
relative to the source..a classical model predicts this extra +-v
will increase (or decrease)the optical density of the water as it moves relative to the source.

Fizeau experiment. Proof that moving water relative to a source will
increase the optical density of the medium and slow the speed.
Any proof that Fizeau doesn’t observe a change in speed in
the light in moving water?
No

And Fizeaus original formula ( that SR stole)

No, Fizeau's experiment helped inspire Einstein.

Fizeau’s formula was made in 1851, years before the serial plagiarist
Albert was born. Relativists, in their infinite lack of wisdom,
think that Albert thought up Fizeau’s formula.

models this change in density.

How could it, without any data?

If you want to ignore Fizeau’s experimental data showing that
the optical density of the moving water does slow the light ...then go ahead Nothing new for relativists to ignore data. SR and GR were built on
the assumption that all observed data must be ignored.

And even if it did, classical physics would ADD the v component to the alleged density contribution.

Before1851. But only a desperate relativist would pretend the physics community
ignored Fizeau’s results after that date.

As does my own version: c/n+-v{(n-1)+(n-1)^2}

It is make up a fact day!

Says the fact free relativist who can’t prove the formula doesn’t accurately
model the observations made in Fizeau.
But if you don’t like my formula ,...use Fizeau’s formula.
It also correctly predicts that light speed slows relative to an increase in optical
density of the moving water.

Proving

No proofs in physics, only disproofs.

OK ..disprove my claim that increasing optical density in moving
water will decrease the observed lightspeed through the water.

the only people who don’t have any idea are the relativist
fantasists who still think the sun rotates around the earth.

Huh? Besides, it is the anti-relativity cranks who are living in the
past (well over 100 years now), not scientists.

‘Relativist’: Another word for a delusional fantasist who cannot
accept any empirical data.

--- SoupGate-Win32 v1.05
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• From Paul B. Andersen@21:1/5 to All on Fri Oct 13 13:44:19 2023
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--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Paul B. Andersen on Fri Oct 13 09:25:27 2023
On Friday, October 13, 2023 at 4:43:32 AM UTC-7, Paul B. Andersen wrote:
Den 12.10.2023 20:04, skrev Robert Winn:
On Thursday, October 12, 2023 at 7:23:00 AM UTC-7, Paul B. Andersen wrote:

Den 10.10.2023 21:52, skrev Robert Winn:
If we drop a ball from the ceiling of an airplane that is flying, the ball is falling faster in the frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.

The ball is falling from the ceiling of an airplane, and will
fall a distance h and hit the floor in the airplane.

_____________________________________________________________________

In the airplane's rest frame we have:
---------------------------------------
The ball will fall along a straight vertical line, and
will hit the floor with the vertical velocity u_v = √(2gh).

The speed of the ball when it hit the floor is u = √(2gh)
The ball will hit the floor at the time t' = √(2h/g)
after it was dropped.

This time is measured with coordinate clocks in the airplane's
rest frame.
_______________________________________________________________________ >>
In the ground frame, we have:
-------------------------------
The ball will have a constant horizontal speed v,
and the vertical speed when it hits the floor is u_v = √(2gh)

The speed of the ball when it hit the floor is u' = √(v²+2gh)
The ball will hit the floor at the time t' = √(2h/g)
after it was dropped.

This time is measured with coordinate clocks in the ground frame.
__________________________________________________________________________

If use the values
g = 9.8 m/s², v = 230 m/s (828.00 km/h), h = 2.5 m, we get :

The speed of the ball when it hits the floor is:
In the rest frame of the airplane: u = 7 m/s, t = 0.7142 s
In the ground frame: u= 230.11 m/s = 828.38 km/h, t = 0.7142 s

This answers are the same for SR and NM.
(We ignore the curvature of spacetime because
the height difference is only 2.5 m)

So the speed is obviously much higher in the ground frame
than in the airplane's rest frame. But the vertical speed
component is the same in both frames.
Note:
The vertical speed component is the same in both frames.

Original question:
In which frame is the speed of the ball fastest?

Second question:
In which frame is the vertical component of the ball's velocity fastest?
Well, no, Paul. Einstein says that we have a slower clock in the flying airplane. Hafele and Keating say it might be faster, but it could be slower. So we will go with Einstein's original idea. The clock is slower. If t is the time of a clock on the
ground, the during a time of t= 1 sec., the earth rotates a certain number of degrees on its axis. Mars rotates a certain number of degrees on its axis, Jupiter rotates a certain number of degrees on its axis, etc., etc. The clock in the airplane shows
less time than a second when the clock on the ground shows a second and the earth, Mars, Jupiter, etc., are all rotating these varying number of degrees. So what you are telling me now is that if a scientist looks at the clock in the airplane, the clock
on the ground shows less time than the clock on the airplane. What do the earth, Mars, Jupiter, etc. do? Do they all rotate backward to agree with the clock on the ground every time a scientist looks at the clock in the airplane? Now scientists have two
miracles. I am not opposed to scientists having miracles, but you have to understand that we common people are required to live in something called reality where the planets would just go on rotating the same way they were before the scientist looked at
the clock in the airplane. Or maybe scientists are trying to say that the planets would go on rotating the way they did before, but the clock on the ground would rotate backward to agree with the equations scientists use. It would still just be another
miracle. I have to agree with Galileo and Isaac Newton that if the clock in the airplane was slower, as seen from the airplane, the clock on the ground would be faster. I think scientists are either using the wrong equations for their calculations of
electromagnetism and electromagnetic waves, or else they are misinterpreting the equations they have.
You are ignorant of physics, but you are nowhere near as stupid
as the above babble indicates. If you were, you wouldn't be able
to handle a job like - say welding.

Hint:
The ball "is in" both frames of reference of course,
it is not one ball in each frame. One reality!
And since the frames are moving relative to each other
along the horizontal axis, the horizontal component of
the ball's velocity must be different, while the vertical
component must be equal in the two frames.

If you measure the vertical velocity with two clocks with
different rates, you will get two different results, but
the vertical speed would still be the same in both frames.

With other words:
If the observer in the airplane had a clock which ran
too slow, the observer in the plane and the observer
on the ground would disagree about what the speed was,
but they would agree that the speed was the same in
the two frames.

--
Paul

https://paulba.no/
Well, I just look at the equations, Paul. Einstein says that the vertical distance does not change.
y'=y
Einstein says that the clock in the airplane is slower than the clock on the ground.
t'=(t-vx/c^2)/sqrt(1-v^2/c^2)
That was what he said this equation means. Now, as you mention, I am just an uneducated, ignorant common person, but I did serve four years on an aircraft carrier in the Vietnam War working on radar equipment, so I do have some experience with
electromagnetic waves. The equipment I worked on was from World War II, so it used vacuum tubes instead of the electronics used today. But the electromagnetic theory was the same back then as it is now. I became a welder after I got out of the Navy
because I liked welding better than working on electronics, and there really was no demand for electronics technicians who had never gone to electronics school and had just worked on vacuum tube radar repeaters that real electronics technicians did not
want to work on. But I did pick up a little knowledge about electromagnetic waves because that is what radar signals are. I also know that if the engineers who designed the radar equipment used the Lorentz equations in calculating their designs, it
does not mean the Lorentz equations were exactly right because we electronics technicians had to calibrate the equipment all the time by adjusting variable resistors and variable capacitors when the times and distances did not match up on the radar
screen, much as scientists do when they adjust a clock in a GPS satellite to match the time of a cesium clock on the surface of the earth. Adjusting a clock to match another clock does not really prove anything except that the two clocks did not agree
with each other before one was adjusted.
What I believe is that you are using t'=t with the vertical height in your equations and t'=(t-vx/c^2)/sqrt(1-v^2/c^2) with the horizontal distance or you have a length contraction in the vertical direction, which would disagree with your equation y'=y.
This is just a suspicion at this point, but I will go through the equations and give my opinion when I have made a closer examination.
I am having some difficulty in visualizing your last statement about disagreeing concerning what the speed was, but agreeing that the speed was the same in two frames. That seems to me to be just an illogical statement trying to justify Einstein's
miracle. But I will try to go through the equations in more detail and get back to you. I am a little busy right now, so it may take some time. Anyway, thank you for your response and the information about how scientists are getting their result.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to Paul B. Andersen on Fri Oct 13 12:35:20 2023
On Friday, 13 October 2023 at 13:43:32 UTC+2, Paul B. Andersen wrote:
Den 12.10.2023 20:04, skrev Robert Winn:
On Thursday, October 12, 2023 at 7:23:00 AM UTC-7, Paul B. Andersen wrote:

Den 10.10.2023 21:52, skrev Robert Winn:
If we drop a ball from the ceiling of an airplane that is flying, the ball is falling faster in the frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.

The ball is falling from the ceiling of an airplane, and will
fall a distance h and hit the floor in the airplane.

_____________________________________________________________________

In the airplane's rest frame we have:
---------------------------------------
The ball will fall along a straight vertical line, and
will hit the floor with the vertical velocity u_v = √(2gh).

The speed of the ball when it hit the floor is u = √(2gh)
The ball will hit the floor at the time t' = √(2h/g)
after it was dropped.

This time is measured with coordinate clocks in the airplane's
rest frame.
_______________________________________________________________________ >>
In the ground frame, we have:
-------------------------------
The ball will have a constant horizontal speed v,
and the vertical speed when it hits the floor is u_v = √(2gh)

The speed of the ball when it hit the floor is u' = √(v²+2gh)
The ball will hit the floor at the time t' = √(2h/g)
after it was dropped.

This time is measured with coordinate clocks in the ground frame.
__________________________________________________________________________

If use the values
g = 9.8 m/s², v = 230 m/s (828.00 km/h), h = 2.5 m, we get :

The speed of the ball when it hits the floor is:
In the rest frame of the airplane: u = 7 m/s, t = 0.7142 s
In the ground frame: u= 230.11 m/s = 828.38 km/h, t = 0.7142 s

This answers are the same for SR and NM.
(We ignore the curvature of spacetime because
the height difference is only 2.5 m)

So the speed is obviously much higher in the ground frame
than in the airplane's rest frame. But the vertical speed
component is the same in both frames.
Note:
The vertical speed component is the same in both frames.

Original question:
In which frame is the speed of the ball fastest?

Second question:
In which frame is the vertical component of the ball's velocity fastest?
Well, no, Paul. Einstein says that we have a slower clock in the flying airplane. Hafele and Keating say it might be faster, but it could be slower. So we will go with Einstein's original idea. The clock is slower. If t is the time of a clock on the
ground, the during a time of t= 1 sec., the earth rotates a certain number of degrees on its axis. Mars rotates a certain number of degrees on its axis, Jupiter rotates a certain number of degrees on its axis, etc., etc. The clock in the airplane shows
less time than a second when the clock on the ground shows a second and the earth, Mars, Jupiter, etc., are all rotating these varying number of degrees. So what you are telling me now is that if a scientist looks at the clock in the airplane, the clock
on the ground shows less time than the clock on the airplane. What do the earth, Mars, Jupiter, etc. do? Do they all rotate backward to agree with the clock on the ground every time a scientist looks at the clock in the airplane? Now scientists have two
miracles. I am not opposed to scientists having miracles, but you have to understand that we common people are required to live in something called reality where the planets would just go on rotating the same way they were before the scientist looked at
the clock in the airplane. Or maybe scientists are trying to say that the planets would go on rotating the way they did before, but the clock on the ground would rotate backward to agree with the equations scientists use. It would still just be another
miracle. I have to agree with Galileo and Isaac Newton that if the clock in the airplane was slower, as seen from the airplane, the clock on the ground would be faster. I think scientists are either using the wrong equations for their calculations of
electromagnetism and electromagnetic waves, or else they are misinterpreting the equations they have.
You are ignorant of physics, but you are nowhere near as stupid
as the above babble indicates. If you were, you wouldn't be able
to handle a job like - say welding.

Hint:
The ball "is in" both frames of reference of course,
it is not one ball in each frame. One reality!
And since the frames are moving relative to each other
along the horizontal axis, the horizontal component of
the ball's velocity must be different, while the vertical
component must be equal in the two frames.

If you measure the vertical velocity with two clocks with
different rates, you will get two different results, but
the vertical speed would still be the same in both frames.

With other words:
If the observer in the airplane had a clock which ran
too slow, the observer in the plane and the observer

Your tales of observers were nothing but funny even
300 years ago. Face it, trash, you have no clue

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Tom Roberts@21:1/5 to Laurence Clark Crossen on Fri Oct 13 16:14:30 2023
On 10/13/23 3:55 PM, Laurence Clark Crossen wrote:
relativity says the jet is flying at two different speeds at once,

When you don't know what the words you use actually mean, you end up
spewing nonsense like this.

In physics, speed is ALWAYS relative to a specified coordinate system.
There are an infinite number of coordinate systems that are valid where
the jet is located, so it actually has an infinite number of speeds
relative to different coordinate systems. Not only that, but YOU,
YOURSELF also have an infinite number of speeds relative to different coordinate systems.

You REALLY need to learn very basic physics before attempting to write
about it. All you do is display how stupid and ignorant you are.

Tom Roberts

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Laurence Clark Crossen@21:1/5 to Paul B. Andersen on Fri Oct 13 13:55:02 2023
On Friday, October 13, 2023 at 4:43:32 AM UTC-7, Paul B. Andersen wrote:
Den 12.10.2023 20:04, skrev Robert Winn:
On Thursday, October 12, 2023 at 7:23:00 AM UTC-7, Paul B. Andersen wrote:

Den 10.10.2023 21:52, skrev Robert Winn:
If we drop a ball from the ceiling of an airplane that is flying, the ball is falling faster in the frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.

The ball is falling from the ceiling of an airplane, and will
fall a distance h and hit the floor in the airplane.

_____________________________________________________________________

In the airplane's rest frame we have:
---------------------------------------
The ball will fall along a straight vertical line, and
will hit the floor with the vertical velocity u_v = √(2gh).

The speed of the ball when it hit the floor is u = √(2gh)
The ball will hit the floor at the time t' = √(2h/g)
after it was dropped.

This time is measured with coordinate clocks in the airplane's
rest frame.
_______________________________________________________________________ >>
In the ground frame, we have:
-------------------------------
The ball will have a constant horizontal speed v,
and the vertical speed when it hits the floor is u_v = √(2gh)

The speed of the ball when it hit the floor is u' = √(v²+2gh)
The ball will hit the floor at the time t' = √(2h/g)
after it was dropped.

This time is measured with coordinate clocks in the ground frame.
__________________________________________________________________________

If use the values
g = 9.8 m/s², v = 230 m/s (828.00 km/h), h = 2.5 m, we get :

The speed of the ball when it hits the floor is:
In the rest frame of the airplane: u = 7 m/s, t = 0.7142 s
In the ground frame: u= 230.11 m/s = 828.38 km/h, t = 0.7142 s

This answers are the same for SR and NM.
(We ignore the curvature of spacetime because
the height difference is only 2.5 m)

So the speed is obviously much higher in the ground frame
than in the airplane's rest frame. But the vertical speed
component is the same in both frames.
Note:
The vertical speed component is the same in both frames.

Original question:
In which frame is the speed of the ball fastest?

Second question:
In which frame is the vertical component of the ball's velocity fastest?
Well, no, Paul. Einstein says that we have a slower clock in the flying airplane. Hafele and Keating say it might be faster, but it could be slower. So we will go with Einstein's original idea. The clock is slower. If t is the time of a clock on the
ground, the during a time of t= 1 sec., the earth rotates a certain number of degrees on its axis. Mars rotates a certain number of degrees on its axis, Jupiter rotates a certain number of degrees on its axis, etc., etc. The clock in the airplane shows
less time than a second when the clock on the ground shows a second and the earth, Mars, Jupiter, etc., are all rotating these varying number of degrees. So what you are telling me now is that if a scientist looks at the clock in the airplane, the clock
on the ground shows less time than the clock on the airplane. What do the earth, Mars, Jupiter, etc. do? Do they all rotate backward to agree with the clock on the ground every time a scientist looks at the clock in the airplane? Now scientists have two
miracles. I am not opposed to scientists having miracles, but you have to understand that we common people are required to live in something called reality where the planets would just go on rotating the same way they were before the scientist looked at
the clock in the airplane. Or maybe scientists are trying to say that the planets would go on rotating the way they did before, but the clock on the ground would rotate backward to agree with the equations scientists use. It would still just be another
miracle. I have to agree with Galileo and Isaac Newton that if the clock in the airplane was slower, as seen from the airplane, the clock on the ground would be faster. I think scientists are either using the wrong equations for their calculations of
electromagnetism and electromagnetic waves, or else they are misinterpreting the equations they have.
You are ignorant of physics, but you are nowhere near as stupid
as the above babble indicates. If you were, you wouldn't be able
to handle a job like - say welding.

Hint:
The ball "is in" both frames of reference of course,
it is not one ball in each frame. One reality!
And since the frames are moving relative to each other
along the horizontal axis, the horizontal component of
the ball's velocity must be different, while the vertical
component must be equal in the two frames.

If you measure the vertical velocity with two clocks with
different rates, you will get two different results, but
the vertical speed would still be the same in both frames.

With other words:
If the observer in the airplane had a clock which ran
too slow, the observer in the plane and the observer
on the ground would disagree about what the speed was,
but they would agree that the speed was the same in
the two frames.

--
Paul

https://paulba.no/
Speaking of evading the issue, relativity says the jet is flying at two different speeds at once, proving it is self-contradictory nonsense. That is the issue. That disproves relativity.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From RichD@21:1/5 to Lou on Fri Oct 13 14:34:51 2023
On October 10, Lou wrote:
The same occurs in Hafael Keating. The eastward travelling plane experiences more force than the westward plane relative to the earths Center. Because it travels at a greater speed relative to the earth Center, than the westward plane.

You mean, it travels at a higher wind speed, hence greater air resistance?

--
Rich

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Laurence Clark Crossen@21:1/5 to Paul B. Andersen on Fri Oct 13 16:51:56 2023
On Friday, October 13, 2023 at 4:43:32 AM UTC-7, Paul B. Andersen wrote:
Den 12.10.2023 20:04, skrev Robert Winn:
On Thursday, October 12, 2023 at 7:23:00 AM UTC-7, Paul B. Andersen wrote:

Den 10.10.2023 21:52, skrev Robert Winn:
If we drop a ball from the ceiling of an airplane that is flying, the ball is falling faster in the frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.

The ball is falling from the ceiling of an airplane, and will
fall a distance h and hit the floor in the airplane.

_____________________________________________________________________

In the airplane's rest frame we have:
---------------------------------------
The ball will fall along a straight vertical line, and
will hit the floor with the vertical velocity u_v = √(2gh).

The speed of the ball when it hit the floor is u = √(2gh)
The ball will hit the floor at the time t' = √(2h/g)
after it was dropped.

This time is measured with coordinate clocks in the airplane's
rest frame.
_______________________________________________________________________ >>
In the ground frame, we have:
-------------------------------
The ball will have a constant horizontal speed v,
and the vertical speed when it hits the floor is u_v = √(2gh)

The speed of the ball when it hit the floor is u' = √(v²+2gh)
The ball will hit the floor at the time t' = √(2h/g)
after it was dropped.

This time is measured with coordinate clocks in the ground frame.
__________________________________________________________________________

If use the values
g = 9.8 m/s², v = 230 m/s (828.00 km/h), h = 2.5 m, we get :

The speed of the ball when it hits the floor is:
In the rest frame of the airplane: u = 7 m/s, t = 0.7142 s
In the ground frame: u= 230.11 m/s = 828.38 km/h, t = 0.7142 s

This answers are the same for SR and NM.
(We ignore the curvature of spacetime because
the height difference is only 2.5 m)

So the speed is obviously much higher in the ground frame
than in the airplane's rest frame. But the vertical speed
component is the same in both frames.
Note:
The vertical speed component is the same in both frames.

Original question:
In which frame is the speed of the ball fastest?

Second question:
In which frame is the vertical component of the ball's velocity fastest?
Well, no, Paul. Einstein says that we have a slower clock in the flying airplane. Hafele and Keating say it might be faster, but it could be slower. So we will go with Einstein's original idea. The clock is slower. If t is the time of a clock on the
ground, the during a time of t= 1 sec., the earth rotates a certain number of degrees on its axis. Mars rotates a certain number of degrees on its axis, Jupiter rotates a certain number of degrees on its axis, etc., etc. The clock in the airplane shows
less time than a second when the clock on the ground shows a second and the earth, Mars, Jupiter, etc., are all rotating these varying number of degrees. So what you are telling me now is that if a scientist looks at the clock in the airplane, the clock
on the ground shows less time than the clock on the airplane. What do the earth, Mars, Jupiter, etc. do? Do they all rotate backward to agree with the clock on the ground every time a scientist looks at the clock in the airplane? Now scientists have two
miracles. I am not opposed to scientists having miracles, but you have to understand that we common people are required to live in something called reality where the planets would just go on rotating the same way they were before the scientist looked at
the clock in the airplane. Or maybe scientists are trying to say that the planets would go on rotating the way they did before, but the clock on the ground would rotate backward to agree with the equations scientists use. It would still just be another
miracle. I have to agree with Galileo and Isaac Newton that if the clock in the airplane was slower, as seen from the airplane, the clock on the ground would be faster. I think scientists are either using the wrong equations for their calculations of
electromagnetism and electromagnetic waves, or else they are misinterpreting the equations they have.
You are ignorant of physics, but you are nowhere near as stupid
as the above babble indicates. If you were, you wouldn't be able
to handle a job like - say welding.

Hint:
The ball "is in" both frames of reference of course,
it is not one ball in each frame. One reality!
And since the frames are moving relative to each other
along the horizontal axis, the horizontal component of
the ball's velocity must be different, while the vertical
component must be equal in the two frames.

If you measure the vertical velocity with two clocks with
different rates, you will get two different results, but
the vertical speed would still be the same in both frames.

With other words:
If the observer in the airplane had a clock which ran
too slow, the observer in the plane and the observer
on the ground would disagree about what the speed was,
but they would agree that the speed was the same in
the two frames.

--
Paul

https://paulba.no/
Einstein never qualified as a physicist and you're dumber than he was.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Lou on Sat Oct 14 01:50:28 2023
On 10/12/2023 7:32 AM, Lou wrote:
On Thursday, 12 October 2023 at 05:48:17 UTC+1, Volney wrote:
On 10/11/2023 12:27 PM, Lou wrote:
On Wednesday, 11 October 2023 at 17:09:14 UTC+1, Volney wrote:

So you simply can't support your "traverse more water" claim with a
simple thought experiment that would support/refute it. Why not admit
that your belief simply won't work?

Fizeau is not a thought experiment. Or any version of it.

Idiot, the thought experiment is water filled tubes with different
lengths "d", which should (if your claim is correct) slow down light
more for longer "d" because the light traverses more water!

You don’t understand refractive index. If the refractive index
of the medium changes...the speed changes. But distance travelled
will not affect the refractive index or speed of light in any medium.

But you claimed before that the amount of water traversed affected the
index of refraction. (of course you changed that once I pointed that out
to you)

Or less depending on the direction of v.
Why do you think they move the water through the tube?

Duh-h-h-h... The speed of light in the water is c/n, relative to the
water. Relative to the lab, it should be (according to cranks) c/n ±v

Maybe relativistic cranks...

Nope. Relativity predicts a formula close to what Fizeau actually found.
It was Fizeau who predicted (but did not measure) the classical c/n ± v
speed.

but I just finished telling you and showing
you a formula saying it wasn’t at c/n+-v. Can’t you read?
And not only that we’ve known it isn’t c/n+-v since 1851.

Correct. Fizeau stumbled across the relativistic speed combination
formula which was close to Fizeau's result of c/n ± v(1-1/n²).

but instead Fizeau measured it as c/n ±v(1-1/n^2). Now you, desperately
flailing around, are coming up with bogus excuses like there is more
water to traverse when the water is moving but not when a water filled
tube is longer!

Liar. I never said that.

You did, but changed it later once I pointed out the effects of tubes
with different length "d".

And that refractive index isn’t n in water but a slightly different >>>>> value of n which takes into account the fact that the moving water
creates a more optically dense medium in the experiment.
Where is your evidence for this (now changed) claim? Remember, science >>>> is built on experimental evidence and scientific observations.

I have no evidence that water moves through the tube in Fizeau to
change the observed speed of the light in the experiment? Troll.

You have no evidence the index of refraction in water is changed by
motion. It is a desperate flailing to try to come up with some other
explanation than one of the first glimmers of SR.

I have no evidence!! 😂💩 Nonsense.

Yes, your claims are all nonsense.

I just cited Fizeau experiment as my evidence.

And believe me, if you actually
tried reading the various reference on the experiment you would see it confirms what I say.
Which is: That the faster the water moves in the tube. The slower the light speed is observed relative to the source.

And that this change in speed
isnt c/n +-v.

Evidence?

I supplied a formula to model this change in speed.

A "model" based on zero evidence? Why would anyone care about THAT?

Or you
can refer to Fizeau’s formula.

Which is not the same as yours. Which one do you believe is correct?
(hint: it can't be both).

Either one accurately predicts how much
light speed slows relative to the source if the water is at v.

No, both cannot be the correct formula. And you cannot use Fizeau's
formula as proof of varying indexes of refraction without the logical

Another fact free claim from a relativist.

The fact free claim is that the index of refraction changes due to
motion. Remember, evidence is king in physics. Got any?

Yes. Fizeau.

Bzzzzt. Assuming the conclusion.

Notice it observed that the speed of light slows down in water
if the water moves in the tube.

But you have no evidence of that.

And the reason is that the optical
density of the water increases relative to the source if the water is at v.

Assertions are not evidence.

As observed.

You have a table of indexes of water vs. speed? Why not show it!

Any evidence that there is no change in light-speed relative to
the source in Fizeau when the water is at v?
No.
Thought not.

Notice that any description of Fizeau (including wiki) specifically
states that to get the observed effects...one must have water
move through the column/tube.

Exactly. It was an early demonstration of the SR speed combination formula. >>>
Theoretically. But a classical explanation doesn’t rely on magic.
And works just as well.

Nope. Fizeau didn't get his expected classical c/n ± v.

Seeing as both Fizeau’s and my formula
can accurately predict this classical effect.

Neither got the classical c/n ± v.

Also...you forgot...the formula SR uses...is stolen from classical theory

How?

As per usual for relativists.

And classically one way to model this is to make this new
value of n is by this formula... c/n+-v{(n-1)+(n-1)^2}

Now you are making up your own crap formula which conflicts with
Fizeau's results (as well as SR)?

SR stole Fizeau formula. Albert was a plagiarist.

Nope. Fizeau's experiment INSPIRED Albert. It was an unsolved mystery
until SR could explain it.

And the formula I suggested...gave a prediction consistent
with the observations. So does Fizeau’s formula.

Fizeau's equation was DERIVED from his results. Your formula is, well,
you know...

So which body orifice did you pull your formula from? Why do you feel it
is better than Fizeau's formula? Remember, you must provide evidence of

I didn’t say it was better. I said both worked as well.

How could it "work as well"? Why even bother with it?

And proved that
the optical density of the moving water relative to the source increases
or decreases with+-v. A purely classical phenomena.

No, you need to show measurements of n vs. speed of water, and reference
the experimental data and its author in any such beliefs. Instead, you

If you think I’m wrong..prove that Fizeau does not observe any change in lightspeed in water relative to the source. Even when the water moves
at v in the tube.
You can’t.

Of course. Fizeau knew the speed of light in water was c/n. Which is why
he expected the classical c/n ± v.

And you have no evidence to contradict this fact.

Your claim is NOT a fact. In fact it is up to you to show water speed
produces increased density which changes n, or for your alternate
formula. Remember, science is based on scientific observations and
experimental data.

I did prove it. My evidence is ...the Fizeau experiment.

Or if you prefer the Fizeau original formula...that works as well too. >>>> Except that it's different.

And not worthless like your bogus formula.

My formula predicts as well as Fizeau’s.

How can different formulas work equally as well?

But that’s irrelevent
because Fizeau’s formula also correctly predicts the optical density changes when water is at v.

How could it if you don't even know what the change is at a given
velocity v?

Of course it’s a different formula. There are different ways to model
optical density in moving water. The maths can change...but
the theory doesn’t.

What theory? You have provided no evidence of anything!

Only if you ignore Fizeau.

Assuming the conclusion is no basis for any theory.

What’s important is that Fizeau observations are predicted by classical theory
as differences in optical densities of moving mediums compared to when
they don’t move relative to the source.

Nope. Classical theory predicts c/n ± v. Fizeau's observations

No. Classical theory predicts light speed changes in water if the water
moves in the tube at v. And succesfully predicted for classical theory
by Fizeau : c/n+-v(1-(1/n^2))

How could it if you don't even know how n changes with v (if it does)?

As confirmed by Fizeau.

Nope. Don't put words in Fizeau's mouth.

I didn’t. I only quoted his experiment where he not only measures
the change in lightspeed ,...he correctly models it with the classical formula c/n+-v(1-(1/n^2))

That's not the classical formula. c/n ± v is.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Volney@21:1/5 to Lou on Sat Oct 14 01:25:20 2023
On 10/13/2023 3:27 AM, Lou wrote:
On Friday, 13 October 2023 at 02:52:52 UTC+1, Volney wrote:
On 10/11/2023 6:45 AM, Lou wrote:
On Wednesday, 11 October 2023 at 10:29:54 UTC+1, JanPB wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:
On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote: >>>>>>>> On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a
clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get
a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the
airplane. Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock
on the ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the
clock on the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same
distances for x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They
indicate that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right >>>>>>>> about this.

It's not as if Einstein, as a young patent clerk, had some ability to >>>>>>>> impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not >>>>>>>> behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the >>>>>>>> scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of today do not
consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is faster than a
clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves. I am
explaining relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the Lorentz equations
that he said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of the slower
clock gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book. His
problem was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for it.
The Galilean transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia
No need for relativity. A classical model does just fine.

No, it doesn't. You simply don't know the problems involved.

You don’t understand basic physics. Fact: A more optically dense medium >>> will have a slightly greater refractive index.

Is today "make up a fact day" or something?

Look up optical density vs refraction and you get many quotes like
the following: “ The refractive index of the material is an indicator
of its optical density.”

Perhaps it does, do you have a reference for index of refraction vs.
density?

I’m not trying to write a textbook on refraction and optical
density.

Yet you need to show that for a speed v, the index of refraction for
light moving in the same direction becomes n', while light in the
reverse direction has an index of refraction n''. Which you can look up
in your magic tables of indexes of refraction vs. speeds.

Since you haven't done that, all you have is a conjecture that the speed
of water changes its index of refraction by motion, in exactly the right
way to match Fizeau's results.

All I’m doing is pointing out that in the Fizeau experiment
when the water flows at v in the tube, the lightspeed of the beam
travelling the distance between source and interference detector is
slower than when the water doesn’t flow.

Obviously

"Obviously" is a word which doesn't belong in physics discussions like
this. You have to show what you think is "obvious".

c/n of water no longer applies.

Where is your table how the index of refraction of water varies with its
speed? No handwaving it away as being "obvious". Show your evidence.

This slowing down of
lightspeed in moving water is best described classically as the optical density and thus refractive index of light in the moving water increases.

Conjecture.

As modelled by Fizeau’s or my formula.

No, Fizeau's formula is what was measured. Classical theory predicts
speed of c/n + v. Your formula is still warm from being found in some best-unnamed bodily orifice.

If you have any proof that there is no relationship between optical density increasing proportional to observed speed of light in a transparent medium. Please supply your evidence.

Again, it's your wacky theory, you provide the evidence.

And if the water moves
relative to the source..a classical model predicts this extra +-v
will increase (or decrease)the optical density of the water as it moves
relative to the source.

Fizeau experiment.

cause and effect relationship.

Proof that moving water relative to a source will
increase the optical density of the medium and slow the speed.
Any proof that Fizeau doesn’t observe a change in speed in
the light in moving water?
No

You cannot prove a negative.

And Fizeaus original formula ( that SR stole)

No, Fizeau's experiment helped inspire Einstein.

Fizeau’s formula was made in 1851, years before the serial plagiarist Albert was born.

Sure, Albert was inspired by an old experiment. Nothing wrong with that!

Relativists, in their infinite lack of wisdom,
think that Albert thought up Fizeau’s formula.

No, cranks make up false beliefs like Einstein stealing formulas because
they have nothing else to go on.

models this change in density.

How could it, without any data?

If you want to ignore Fizeau’s experimental data showing that
the optical density of the moving water does slow the light

You are assuming your conclusion again! You only know that Fizeau
measured a difference in the speed of light which differed from the
expected classical formula c/n+v.

If you find alternate indexes of refraction and actually apply them to
Fizeau's experiment and do the math, and it happens to come out right,
you'll offer an alternate explanation (but not a proof, of course).

Nothing new for relativists to ignore data. SR and GR were built on
the assumption that all observed data must be ignored.

So today is make up a fact day instead.

And even if it did, classical physics would ADD the v component to the
alleged density contribution.

Before1851. But only a desperate relativist would pretend the physics community
ignored Fizeau’s results after that date.

Nobody "ignored" it. It was an unsolved mystery of physics before Einstein.

As does my own version: c/n+-v{(n-1)+(n-1)^2}

It is make up a fact day!

Says the fact free relativist who can’t prove the formula doesn’t accurately
model the observations made in Fizeau.

Cannot prove a negative. And it is you who has no data like indexes of refraction changing with v.

But if you don’t like my formula ,...use Fizeau’s formula.

Which we know doesn't match the classical c/n+v predicted speeds.

It also correctly predicts that light speed slows relative to an increase in optical
density of the moving water.

You got that backwards. You need to show how light speed slows due to an increase in velocity to show you have an alternate explanation for
Fizeau. So far, nothing from you.

Proving

No proofs in physics, only disproofs.

OK ..disprove my claim that increasing optical density in moving
water will decrease the observed lightspeed through the water.

We have no such data so far. You have nothing to support your claim.

the only people who don’t have any idea are the relativist
fantasists who still think the sun rotates around the earth.

Huh? Besides, it is the anti-relativity cranks who are living in the
past (well over 100 years now), not scientists.

‘Relativist’: Another word for a delusional fantasist who cannot
accept any empirical data.

'Relativist' is a crank word for a scientist who understands science.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to Tom Roberts on Sat Oct 14 02:08:46 2023
On Friday, 13 October 2023 at 23:14:43 UTC+2, Tom Roberts wrote:
On 10/13/23 3:55 PM, Laurence Clark Crossen wrote:
relativity says the jet is flying at two different speeds at once,
When you don't know what the words you use actually mean

Stop lying, trash, you admit sometimes that they don't
mean what your moronic church is trying to force them to.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Paul B. Andersen on Sat Oct 14 12:56:37 2023
On Friday, October 13, 2023 at 4:43:32 AM UTC-7, Paul B. Andersen wrote:
Den 12.10.2023 20:04, skrev Robert Winn:
On Thursday, October 12, 2023 at 7:23:00 AM UTC-7, Paul B. Andersen wrote:

Den 10.10.2023 21:52, skrev Robert Winn:
If we drop a ball from the ceiling of an airplane that is flying, the ball is falling faster in the frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.

The ball is falling from the ceiling of an airplane, and will
fall a distance h and hit the floor in the airplane.

_____________________________________________________________________

In the airplane's rest frame we have:
---------------------------------------
The ball will fall along a straight vertical line, and
will hit the floor with the vertical velocity u_v = √(2gh).

The speed of the ball when it hit the floor is u = √(2gh)
The ball will hit the floor at the time t' = √(2h/g)
after it was dropped.

This time is measured with coordinate clocks in the airplane's
rest frame.
_______________________________________________________________________ >>
In the ground frame, we have:
-------------------------------
The ball will have a constant horizontal speed v,
and the vertical speed when it hits the floor is u_v = √(2gh)

The speed of the ball when it hit the floor is u' = √(v²+2gh)
The ball will hit the floor at the time t' = √(2h/g)
after it was dropped.

This time is measured with coordinate clocks in the ground frame.
__________________________________________________________________________

If use the values
g = 9.8 m/s², v = 230 m/s (828.00 km/h), h = 2.5 m, we get :

The speed of the ball when it hits the floor is:
In the rest frame of the airplane: u = 7 m/s, t = 0.7142 s
In the ground frame: u= 230.11 m/s = 828.38 km/h, t = 0.7142 s

This answers are the same for SR and NM.
(We ignore the curvature of spacetime because
the height difference is only 2.5 m)

So the speed is obviously much higher in the ground frame
than in the airplane's rest frame. But the vertical speed
component is the same in both frames.
Note:
The vertical speed component is the same in both frames.

Original question:
In which frame is the speed of the ball fastest?

Second question:
In which frame is the vertical component of the ball's velocity fastest?
Well, no, Paul. Einstein says that we have a slower clock in the flying airplane. Hafele and Keating say it might be faster, but it could be slower. So we will go with Einstein's original idea. The clock is slower. If t is the time of a clock on the
ground, the during a time of t= 1 sec., the earth rotates a certain number of degrees on its axis. Mars rotates a certain number of degrees on its axis, Jupiter rotates a certain number of degrees on its axis, etc., etc. The clock in the airplane shows
less time than a second when the clock on the ground shows a second and the earth, Mars, Jupiter, etc., are all rotating these varying number of degrees. So what you are telling me now is that if a scientist looks at the clock in the airplane, the clock
on the ground shows less time than the clock on the airplane. What do the earth, Mars, Jupiter, etc. do? Do they all rotate backward to agree with the clock on the ground every time a scientist looks at the clock in the airplane? Now scientists have two
miracles. I am not opposed to scientists having miracles, but you have to understand that we common people are required to live in something called reality where the planets would just go on rotating the same way they were before the scientist looked at
the clock in the airplane. Or maybe scientists are trying to say that the planets would go on rotating the way they did before, but the clock on the ground would rotate backward to agree with the equations scientists use. It would still just be another
miracle. I have to agree with Galileo and Isaac Newton that if the clock in the airplane was slower, as seen from the airplane, the clock on the ground would be faster. I think scientists are either using the wrong equations for their calculations of
electromagnetism and electromagnetic waves, or else they are misinterpreting the equations they have.
You are ignorant of physics, but you are nowhere near as stupid
as the above babble indicates. If you were, you wouldn't be able
to handle a job like - say welding.

Hint:
The ball "is in" both frames of reference of course,
it is not one ball in each frame. One reality!
And since the frames are moving relative to each other
along the horizontal axis, the horizontal component of
the ball's velocity must be different, while the vertical
component must be equal in the two frames.

If you measure the vertical velocity with two clocks with
different rates, you will get two different results, but
the vertical speed would still be the same in both frames.

With other words:
If the observer in the airplane had a clock which ran
too slow, the observer in the plane and the observer
on the ground would disagree about what the speed was,
but they would agree that the speed was the same in
the two frames.

--
Paul

https://paulba.no/
I can't quite follow your reasoning. As far as I can tell, you are saying that you use the time of the clock on the ground, t, in both frames of reference for the vertical component, so the time of the vertical component remains the same. What happened
to t'. the time of the clock in the airplane?
It looks to me as though you are doing what Einstein seemed to do quite a bit, reverting back to the Galilean transformation equations whenever you have a question your interpretation of the Lorentz equations can't answer.
The way I would work the problem with the Galilean transformation equations is t is the time of the clock on the ground, n' is the time of the clock in the airplane.
y'=y
y/t is the average speed of the ball as computed by the clock on earth
y'/n' is the average speed of the ball as computed by the clock in the airplane If n' is less time than t, then the ball is falling faster as seen from the airplane.
I don't need to do your trick of switching to the time of the other frame of reference. The position of the ball shows that the clock on the airplane is slower. An observer on the airplane using the clock on the airplane gets a faster speed for the
airplane and a faster speed for the ball, which is what reality also shows us.

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• From Mike Fontenot@21:1/5 to Tom Roberts on Sat Oct 14 13:45:12 2023
On 10/13/23 3:14 PM, Tom Roberts wrote:

In physics, speed is ALWAYS relative to a specified coordinate system.

According to a person who is accelerating (according to his attached accelerometer), what is his speed as a function of his age? I think
there is a unique answer to that question, given that he STARTED
accelerating from rest at some specified instant in his life.

For example, if I started from rest (according to me) when I was ten
years old, and accelerated at 1 ly/y thereafter, I then will have a well-defined speed at each instant in my life thereafter.

--- SoupGate-Win32 v1.05
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• From Lou@21:1/5 to Volney on Sat Oct 14 14:02:07 2023
On Saturday, 14 October 2023 at 06:25:26 UTC+1, Volney wrote:
On 10/13/2023 3:27 AM, Lou wrote:
On Friday, 13 October 2023 at 02:52:52 UTC+1, Volney wrote:
On 10/11/2023 6:45 AM, Lou wrote:
On Wednesday, 11 October 2023 at 10:29:54 UTC+1, JanPB wrote:
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote: >>>>>> On 10-Oct-23 3:39 pm, Robert Winn wrote:
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote: >>>>>>>> On 10-Oct-23 5:27 am, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined
a clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would
get a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the
airplane. Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock
on the ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I
believe Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane
if his clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the
clock on the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same
distances for x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They
indicate that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Don't you think it would have been noticed long ago if you were right

It's not as if Einstein, as a young patent clerk, had some ability to
impose his theory on an unwilling world. Experimenters had been taking a
very close look at reality, and had been finding that it was not >>>>>>>> behaving in the expected way. Einstein provided a solution. That is why
a young patent clerk was able to get his theory accepted by the >>>>>>>> scientific community.

Sylvia.
Well, I know all about that Sylvia. Scientists before 1887 used the Galilean transformation equations. Isaac Newton used absolute time, which shows that all clocks working correctly would agree with each other. What scientists of today do not
consider is that both Galileo and Newton were good enough at following the axioms of algebra that if they had been told, Experiment has shown that a moving clock is slower than a clock that is not moving, or A clock in a GPS satellite is faster than a
clock on earth because of the effects of gravitation, they could have worked the problem. All they had to do was keep the velocities straight, something scientists of today do not do because they use speed instead of velocity.
I am not an experimenter or a scientist. I am a welder with a high school education, but I can follow the axioms of algebra well enough to work the problem of relativity. As I said before, I am not explaining electromagnetic waves. I am
explaining relativity.
If you want to discuss the Michelson-Morley experiment we can do that. I can explain that experiment using the Galilean transformation equations I showed here. Einstein used two little equations he said he extracted from the Lorentz equations
that he said explained the Michelson-Morley experiment.
x = ct
x' = ct'
These two equations will not work with the Galilean transformation equations because t'=t. So we say that the time of the slower clock is n'.
x'=x-vt
cn' = ct - vt
n' = t - vt/c
n' = t - vct/c^2
which is obviously where Lorentz got the numerator for his equation for t'.
But, as I said, I have not believed scientists since I figured the problem in high school and saw that a slower clock would result in a faster velocity as computed from the time of the slower clock. Anyway, this equation for time of the slower
clock gives the same speed for something moving to several decimal places until you get to very fast velocities. But this interpretation of the Galilean transformation equations seems to me to be what Einstein was trying to explain in his book. His
problem was that he was using the Lorentz equations, which show the same speed from either frame of reference, which obviously does not agree with reality. But if scientists want to have a miracle, I really have no objection. I just see no need for it.
The Galilean transformation equations agree with reality if a moving clock is faster or slower.
How do you explain the result of the Fizeau experiment, which was >>>>>> performed half a century before Einstein proposed his theory?

https://en.wikipedia.org/wiki/Fizeau_experiment

Sylvia
No need for relativity. A classical model does just fine.

No, it doesn't. You simply don't know the problems involved.

You don’t understand basic physics. Fact: A more optically dense medium
will have a slightly greater refractive index.

Is today "make up a fact day" or something?

Look up optical density vs refraction and you get many quotes like
the following: “ The refractive index of the material is an indicator
of its optical density.”

Perhaps it does, do you have a reference for index of refraction vs.
density?

I’m not trying to write a textbook on refraction and optical
density.
Yet you need to show that for a speed v, the index of refraction for
light moving in the same direction becomes n', while light in the
reverse direction has an index of refraction n''. Which you can look up
in your magic tables of indexes of refraction vs. speeds.

Since you haven't done that, all you have is a conjecture that the speed
of water changes its index of refraction by motion, in exactly the right
way to match Fizeau's results.

It’s not conjecture. This is observed. Notice in Fizeau when the
water doesn’t move its c/n.
When the water does move at v...the light takes longer to travel the
same distance.
Unfortunately you are unable to read reference on the experiment.

All I’m doing is pointing out that in the Fizeau experiment
when the water flows at v in the tube, the lightspeed of the beam travelling the distance between source and interference detector is
slower than when the water doesn’t flow.

Obviously
"Obviously" is a word which doesn't belong in physics discussions like
this. You have to show what you think is "obvious".

Obviously you are so upset that I’ve pointed out that you know SFA about Fizeau, that instead you harp on about irrelevent points like ‘obviously’. It’s obvious isn’t it?

c/n of water no longer applies.
Where is your table how the index of refraction of water varies with its speed? No handwaving it away as being "obvious". Show your evidence.

If you don’t believe the light takes longer to travel the same distance in Fizeau experiment when the water is moving...then I suggest you
try studying something other than physics. Knitting maybe ?

This slowing down of
lightspeed in moving water is best described classically as the optical density and thus refractive index of light in the moving water increases.
Conjecture.
As modelled by Fizeau’s or my formula.
No, Fizeau's formula is what was measured. Classical theory predicts
speed of c/n + v. Your formula is still warm from being found in some best-unnamed bodily orifice.

Classical theory predicted c/n +-v before 1851.
After 1851 classical theory predicted light travelled at different
speeds but not by +-v. Try reading Fizeau paper instead of hanging
around public toilets.

If you have any proof that there is no relationship between optical density
increasing proportional to observed speed of light in a transparent medium.
Again, it's your wacky theory, you provide the evidence.
And if the water moves
relative to the source..a classical model predicts this extra +-v
will increase (or decrease)the optical density of the water as it moves >>> relative to the source.

Fizeau experiment.
cause and effect relationship.

If one wants people to pretend that angels pull stars and the sun across the sky
as relativity does then yes..one can ignore the observations,
But I prefer classical empirical data over relativistic delusions.

Proof that moving water relative to a source will
increase the optical density of the medium and slow the speed.
Any proof that Fizeau doesn’t observe a change in speed in
the light in moving water?
No
You cannot prove a negative.

And Fizeaus original formula ( that SR stole)

No, Fizeau's experiment helped inspire Einstein.

Fizeau’s formula was made in 1851, years before the serial plagiarist Albert was born.
Sure, Albert was inspired by an old experiment. Nothing wrong with that!
Relativists, in their infinite lack of wisdom,
think that Albert thought up Fizeau’s formula.
No, cranks make up false beliefs like Einstein stealing formulas because they have nothing else to go on.
models this change in density.

How could it, without any data?

If you want to ignore Fizeau’s experimental data showing that
the optical density of the moving water does slow the light
You are assuming your conclusion again! You only know that Fizeau
measured a difference in the speed of light which differed from the
expected classical formula c/n+v.

Fizeau did observe a difference in speeds ..yes.
Notice that Einstein also only *assumed* it was because magic aliens distorted space and time. Theories are always assumptions. Whether it’s religious Wacko theories like SR or more reputable theories like classical.

If you find alternate indexes of refraction and actually apply them to Fizeau's experiment and do the math, and it happens to come out right, you'll offer an alternate explanation (but not a proof, of course).
Nothing new for relativists to ignore data. SR and GR were built on
the assumption that all observed data must be ignored.
So today is make up a fact day instead.

Today! You relativists do it everyday.

And even if it did, classical physics would ADD the v component to the
alleged density contribution.

Before1851. But only a desperate relativist would pretend the physics community
ignored Fizeau’s results after that date.
Nobody "ignored" it. It was an unsolved mystery of physics before Einstein.

😂🤣😂 Sure Volney. And now Albert has told us magic goblins moving
in different time dimensions distort the fabric of the universe proving 1+2=5 ..show all the world that the Fizeau experiment is no longer a mystery. 🤣🤣😅
Make my day pal.

As does my own version: c/n+-v{(n-1)+(n-1)^2}

It is make up a fact day!

Says the fact free relativist who can’t prove the formula doesn’t accurately
model the observations made in Fizeau.
Cannot prove a negative. And it is you who has no data like indexes of refraction changing with v.
But if you don’t like my formula ,...use Fizeau’s formula.
Which we know doesn't match the classical c/n+v predicted speeds.

You don’t understand physics. Classical theory has used Fizeau formula
to accurately predict speeds in the experiment since 1851. But relativists, being
dishonest plagiarists, think that the Fizeau formula was invented by Albert. Even though in fact it was formulated decades before the serial con artist Albert was even born.

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• From Lou@21:1/5 to Volney on Sat Oct 14 14:44:14 2023
On Saturday, 14 October 2023 at 06:50:31 UTC+1, Volney wrote:
Two posts from Boloney for the price of one!

On 10/12/2023 7:32 AM, Lou wrote:
On Thursday, 12 October 2023 at 05:48:17 UTC+1, Volney wrote:
On 10/11/2023 12:27 PM, Lou wrote:
On Wednesday, 11 October 2023 at 17:09:14 UTC+1, Volney wrote:

So you simply can't support your "traverse more water" claim with a >>>> simple thought experiment that would support/refute it. Why not admit >>>> that your belief simply won't work?

Fizeau is not a thought experiment. Or any version of it.

Idiot, the thought experiment is water filled tubes with different
lengths "d", which should (if your claim is correct) slow down light
more for longer "d" because the light traverses more water!

You don’t understand refractive index. If the refractive index
of the medium changes...the speed changes. But distance travelled
will not affect the refractive index or speed of light in any medium.
But you claimed before that the amount of water traversed affected the
index of refraction. (of course you changed that once I pointed that out
to you)

Obviously, whats obvious is you are unable to understand the obvious facts
in the experiment. Here’s Fizeau for dummies like yourself:
Light traverses the experiment from source to detector (distance d) in x amount
of time when the water is not moving. When the water column moves the light takes
a slightly longer amount of time to travel the same distance d.
If the water moves at v during the time the light traverses distance d in the experiment...
then the light pulse has to travel through a longer column of water than
it does when the water isn’t moving.
If you can prove any of the above is false...please send me a reply from

Or less depending on the direction of v.
Why do you think they move the water through the tube?

Duh-h-h-h... The speed of light in the water is c/n, relative to the
water. Relative to the lab, it should be (according to cranks) c/n ±v

Maybe relativistic cranks...
Nope. Relativity predicts a formula close to what Fizeau actually found.
It was Fizeau who predicted (but did not measure) the classical c/n ± v speed.

Wow...you are definitely delusional. Let me guess...Fizeau’s 1851 paper originally didn’t have the correct formula w+=c/n+v(1-(1/n^2)) in it.
But in 1905 Einstein devised the formula w+=c/n+v(1-(1/n^2)) and
then built a time machine...travelled back in time to 1851 to the publishers office and inserted the formula in the typesetting of Fizeau’s paper just before it was sent off to the printers.
Is that roughly what you relativists think happened?

but I just finished telling you and showing
you a formula saying it wasn’t at c/n+-v. Can’t you read?
And not only that we’ve known it isn’t c/n+-v since 1851.
Correct. Fizeau stumbled across the relativistic speed combination
formula which was close to Fizeau's result of c/n ± v(1-1/n²).

😂🤣😅

but instead Fizeau measured it as c/n ±v(1-1/n^2). Now you, desperately >> flailing around, are coming up with bogus excuses like there is more
water to traverse when the water is moving but not when a water filled
tube is longer!

Liar. I never said that.
You did, but changed it later once I pointed out the effects of tubes
with different length "d".

I notice so far you can’t. Because if you tried...
You’ll find out that in fact you were just too stupid to understand
what I said. Your mistake is *Obvious* isn’t it?

And that refractive index isn’t n in water but a slightly different >>>>> value of n which takes into account the fact that the moving water >>>>> creates a more optically dense medium in the experiment.
Where is your evidence for this (now changed) claim? Remember, science >>>> is built on experimental evidence and scientific observations.

I have no evidence that water moves through the tube in Fizeau to
change the observed speed of the light in the experiment? Troll.

You have no evidence the index of refraction in water is changed by
motion. It is a desperate flailing to try to come up with some other
explanation than one of the first glimmers of SR.

I have no evidence!! 😂💩 Nonsense.
Yes, your claims are all nonsense.
I just cited Fizeau experiment as my evidence.
And believe me, if you actually
tried reading the various reference on the experiment you would see it confirms what I say.
Which is: That the faster the water moves in the tube. The slower the light
speed is observed relative to the source.
And that this change in speed
isnt c/n +-v.
Evidence?

So now you say I need evidence to prove that light didn’t travel at
c/n+-v in Fizeau? Things aren’t going well for you today old boy aren’t they?

I supplied a formula to model this change in speed.
A "model" based on zero evidence? Why would anyone care about THAT?
Or you
can refer to Fizeau’s formula.
Which is not the same as yours. Which one do you believe is correct?
(hint: it can't be both).
Either one accurately predicts how much
light speed slows relative to the source if the water is at v.
No, both cannot be the correct formula. And you cannot use Fizeau's
formula as proof of varying indexes of refraction without the logical fallacy of assuming your conclusion.

Sorry I forgot. Relativists strictly forbid any use of any empirical observations or any classical formula devised to correctly
model the data. Unless....relativists can steal the classical formula
and falsely pretend Einstein invented it.

Another fact free claim from a relativist.

The fact free claim is that the index of refraction changes due to
motion. Remember, evidence is king in physics. Got any?

Yes. Fizeau.
Bzzzzt. Assuming the conclusion.
Notice it observed that the speed of light slows down in water
if the water moves in the tube.
But you have no evidence of that.

Let me get this straight then. You and other delusional relativists
sincerely believe that in the Fizeau experiment the light takes the
same amount of time to travel the distance between detector plane
and source.
Regardless of whether or not the water is moving ?
Obviously, You are delusional.

And the reason is that the optical
density of the water increases relative to the source if the water is at v.
Assertions are not evidence.

I thought relativists only accepted assertions and never allow any
evidence to get in the way of their deluded fantastical assumptions?

As observed.

You have a table of indexes of water vs. speed? Why not show it!
Any evidence that there is no change in light-speed relative to
the source in Fizeau when the water is at v?
No.
Thought not.

Notice that any description of Fizeau (including wiki) specifically
states that to get the observed effects...one must have water
move through the column/tube.

Exactly. It was an early demonstration of the SR speed combination formula.

Theoretically. But a classical explanation doesn’t rely on magic.
And works just as well.
Nope. Fizeau didn't get his expected classical c/n ± v.
Seeing as both Fizeau’s and my formula
can accurately predict this classical effect.
Neither got the classical c/n ± v.
Also...you forgot...the formula SR uses...is stolen from classical theory
How?

Oh yes, we mustn’t forget the children’s fairy tale about how Einstein travelled back in time and added in this formula:
w+=c/n+v(1-(1/n^2)) into Fizeau’s 1851paper.

As per usual for relativists.

And classically one way to model this is to make this new
value of n is by this formula... c/n+-v{(n-1)+(n-1)^2}

Now you are making up your own crap formula which conflicts with
Fizeau's results (as well as SR)?

SR stole Fizeau formula. Albert was a plagiarist.

Nope. Fizeau's experiment INSPIRED Albert. It was an unsolved mystery
until SR could explain it.

And the formula I suggested...gave a prediction consistent
with the observations. So does Fizeau’s formula.
Fizeau's equation was DERIVED from his results. Your formula is, well,
you know...

So which body orifice did you pull your formula from? Why do you feel it >> is better than Fizeau's formula? Remember, you must provide evidence of >> your claim.

I didn’t say it was better. I said both worked as well.
How could it "work as well"? Why even bother with it?
And proved that
the optical density of the moving water relative to the source increases or decreases with+-v. A purely classical phenomena.
No, you need to show measurements of n vs. speed of water, and reference
the experimental data and its author in any such beliefs. Instead, you
If you think I’m wrong..prove that Fizeau does not observe any change in lightspeed in water relative to the source. Even when the water moves
at v in the tube.
You can’t.
Of course. Fizeau knew the speed of light in water was c/n. Which is why
he expected the classical c/n ± v.

And you have no evidence to contradict this fact.

Your claim is NOT a fact. In fact it is up to you to show water speed
produces increased density which changes n, or for your alternate
formula. Remember, science is based on scientific observations and
experimental data.

I did prove it. My evidence is ...the Fizeau experiment.

Or if you prefer the Fizeau original formula...that works as well too. >>>> Except that it's different.

And not worthless like your bogus formula.

My formula predicts as well as Fizeau’s.
How can different formulas work equally as well?
But that’s irrelevent
because Fizeau’s formula also correctly predicts the optical density changes when water is at v.
How could it if you don't even know what the change is at a given
velocity v?

But I do...Fizeau told us in 1851 with his formula w+=c/n+v(1-(1/n^2))
Or are you still pretending Albert discovered the formula in 1905
and travelled back in time to 1851 and inserted the formula into
Fizeau’s paper?

Of course it’s a different formula. There are different ways to model >>> optical density in moving water. The maths can change...but
the theory doesn’t.

What theory? You have provided no evidence of anything!

Only if you ignore Fizeau.
Assuming the conclusion is no basis for any theory.

What’s important is that Fizeau observations are predicted by classical theory
as differences in optical densities of moving mediums compared to when >>> they don’t move relative to the source.

Nope. Classical theory predicts c/n ± v. Fizeau's observations

No. Classical theory predicts light speed changes in water if the water moves in the tube at v. And succesfully predicted for classical theory
by Fizeau : c/n+-v(1-(1/n^2))
How could it if you don't even know how n changes with v (if it does)?
As confirmed by Fizeau.

Nope. Don't put words in Fizeau's mouth.

I didn’t. I only quoted his experiment where he not only measures
the change in lightspeed ,...he correctly models it with the classical formula c/n+-v(1-(1/n^2))
That's not the classical formula. c/n ± v is.

Don’t you ever even get a bit tired of pretending that Fizeau didnt write the formula
w+=c/n+v(1-(1/n^2)) in his 1851 paper?

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• From Tom Roberts@21:1/5 to Mike Fontenot on Sat Oct 14 17:24:47 2023
On 10/14/23 2:45 PM, Mike Fontenot wrote:
On 10/13/23 3:14 PM, Tom Roberts wrote:
In physics, speed is ALWAYS relative to a specified coordinate
system.

According to a person who is accelerating (according to his attached accelerometer), what is his speed as a function of his age?

It depends upon which coordinate system you refer the speed to. And it
depends on his acceleration history.

I think there is a unique answer to that question, given that he
STARTED accelerating from rest at some specified instant in his
life.

Nope, your "thinking" is wrong. At any instant in his life, his speed
depends upon which coordinate system you refer the speed to. (You may be implicitly thinking "relative to the inertial coordinates from which he
started at rest", but YOU MUST SAY THAT.)

You repeatedly omit important conditions and qualifications in your
realize you are confused. For instance, just saying "starting from rest"
is not sufficient, as it does not specify which coordinates defined
"rest". You REALLY need to learn how to be more precise in your

Tom Roberts

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• From Mike Fontenot@21:1/5 to Tom Roberts on Sat Oct 14 17:19:30 2023
On 10/14/23 4:24 PM, Tom Roberts wrote:
On 10/14/23 2:45 PM, Mike Fontenot wrote:

(I, Mike Fontenot, wrote):

According to a person who is accelerating (according to his attached
accelerometer), what is his speed as a function of his age?

> Tom Roberts wrote:

It depends upon which coordinate system you refer the speed to. And it depends on his acceleration history.

I (Mike Fontenot wrote:

I think there is a unique answer to that question, given that he
STARTED accelerating from rest at some specified instant in his life.

Tom Roberts wrote:

Nope, your "thinking" is wrong. At any instant in his life, his speed
depends upon which coordinate system you refer the speed to. (You may be implicitly thinking "relative to the inertial coordinates from which he started at rest", but YOU MUST SAY THAT.)

I disagree. I think the accelerating person will INNATELY feel that his
speed was initially zero, and then became non-zero and continually
increasing when he started his constant acceleration.

That is analogous to the feeling that an inertial person (she) has: that
she IS stationary (i.e., that she regards her speed to be zero), and
that other inertial people (not stationary wrt her) are moving.

Similarly, in the twin "paradox", the traveling twin (he) truly BELIEVES
that his home twin (she) instantaneously gets much older when he instantaneously reverses course. And if he ever instantaneously does a
Dirac delta-function acceleration in the direction AWAY from her (when
they are separated by a large distance), he will truly BELIEVE that she instantaneously gets YOUNGER during that instantaneous turnaround. And
he is RIGHT, even though she doesn't agree (and neither do other
inertial people moving with respect to her). In special relativity,
different people are RIGHT, even though they disagree with each other!

You (Tom Roberts) have developed such a MUSHY view of special relativity
that there's no MEANINGFULNESS left in it. If you were actually in a high-speed rocket well away from the earth, you would probably decide to
read, even though they would disagree with your own rate of ageing!

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• From mitchrae3323@gmail.com@21:1/5 to Robert Winn on Sat Oct 14 20:12:47 2023
On Monday, October 9, 2023 at 11:27:35 AM UTC-7, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a clock in
a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get a faster
speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special Relativity
show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on the
ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for x
and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'. So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate that
there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.

How do you know a miracal is real. How do you measure the difference between absolute and relative orders?

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Maciej Wozniak on Sun Oct 15 06:23:19 2023
On Friday, October 13, 2023 at 12:35:22 PM UTC-7, Maciej Wozniak wrote:
On Friday, 13 October 2023 at 13:43:32 UTC+2, Paul B. Andersen wrote:
Den 12.10.2023 20:04, skrev Robert Winn:
On Thursday, October 12, 2023 at 7:23:00 AM UTC-7, Paul B. Andersen wrote:

Den 10.10.2023 21:52, skrev Robert Winn:
If we drop a ball from the ceiling of an airplane that is flying, the ball is falling faster in the frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.

The ball is falling from the ceiling of an airplane, and will
fall a distance h and hit the floor in the airplane.

_____________________________________________________________________ >>
In the airplane's rest frame we have:
---------------------------------------
The ball will fall along a straight vertical line, and
will hit the floor with the vertical velocity u_v = √(2gh).

The speed of the ball when it hit the floor is u = √(2gh)
The ball will hit the floor at the time t' = √(2h/g)
after it was dropped.

This time is measured with coordinate clocks in the airplane's
rest frame.
_______________________________________________________________________ >>
In the ground frame, we have:
-------------------------------
The ball will have a constant horizontal speed v,
and the vertical speed when it hits the floor is u_v = √(2gh)

The speed of the ball when it hit the floor is u' = √(v²+2gh)
The ball will hit the floor at the time t' = √(2h/g)
after it was dropped.

This time is measured with coordinate clocks in the ground frame.
__________________________________________________________________________

If use the values
g = 9.8 m/s², v = 230 m/s (828.00 km/h), h = 2.5 m, we get :

The speed of the ball when it hits the floor is:
In the rest frame of the airplane: u = 7 m/s, t = 0.7142 s
In the ground frame: u= 230.11 m/s = 828.38 km/h, t = 0.7142 s

This answers are the same for SR and NM.
(We ignore the curvature of spacetime because
the height difference is only 2.5 m)

So the speed is obviously much higher in the ground frame
than in the airplane's rest frame. But the vertical speed
component is the same in both frames.
Note:
The vertical speed component is the same in both frames.

Original question:
In which frame is the speed of the ball fastest?

Second question:
In which frame is the vertical component of the ball's velocity fastest?
Well, no, Paul. Einstein says that we have a slower clock in the flying airplane. Hafele and Keating say it might be faster, but it could be slower. So we will go with Einstein's original idea. The clock is slower. If t is the time of a clock on
the ground, the during a time of t= 1 sec., the earth rotates a certain number of degrees on its axis. Mars rotates a certain number of degrees on its axis, Jupiter rotates a certain number of degrees on its axis, etc., etc. The clock in the airplane
shows less time than a second when the clock on the ground shows a second and the earth, Mars, Jupiter, etc., are all rotating these varying number of degrees. So what you are telling me now is that if a scientist looks at the clock in the airplane, the
clock on the ground shows less time than the clock on the airplane. What do the earth, Mars, Jupiter, etc. do? Do they all rotate backward to agree with the clock on the ground every time a scientist looks at the clock in the airplane? Now scientists
have two miracles. I am not opposed to scientists having miracles, but you have to understand that we common people are required to live in something called reality where the planets would just go on rotating the same way they were before the scientist
looked at the clock in the airplane. Or maybe scientists are trying to say that the planets would go on rotating the way they did before, but the clock on the ground would rotate backward to agree with the equations scientists use. It would still just be
another miracle. I have to agree with Galileo and Isaac Newton that if the clock in the airplane was slower, as seen from the airplane, the clock on the ground would be faster. I think scientists are either using the wrong equations for their
calculations of electromagnetism and electromagnetic waves, or else they are misinterpreting the equations they have.
You are ignorant of physics, but you are nowhere near as stupid
as the above babble indicates. If you were, you wouldn't be able
to handle a job like - say welding.

Hint:
The ball "is in" both frames of reference of course,
it is not one ball in each frame. One reality!
And since the frames are moving relative to each other
along the horizontal axis, the horizontal component of
the ball's velocity must be different, while the vertical
component must be equal in the two frames.

If you measure the vertical velocity with two clocks with
different rates, you will get two different results, but
the vertical speed would still be the same in both frames.

With other words:
If the observer in the airplane had a clock which ran
too slow, the observer in the plane and the observer
Your tales of observers were nothing but funny even
300 years ago. Face it, trash, you have no clue

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Maciej Wozniak on Sun Oct 15 06:54:34 2023
On Friday, October 13, 2023 at 12:35:22 PM UTC-7, Maciej Wozniak wrote:
On Friday, 13 October 2023 at 13:43:32 UTC+2, Paul B. Andersen wrote:
Den 12.10.2023 20:04, skrev Robert Winn:
On Thursday, October 12, 2023 at 7:23:00 AM UTC-7, Paul B. Andersen wrote:

Den 10.10.2023 21:52, skrev Robert Winn:
If we drop a ball from the ceiling of an airplane that is flying, the ball is falling faster in the frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.

The ball is falling from the ceiling of an airplane, and will
fall a distance h and hit the floor in the airplane.

_____________________________________________________________________ >>
In the airplane's rest frame we have:
---------------------------------------
The ball will fall along a straight vertical line, and
will hit the floor with the vertical velocity u_v = √(2gh).

The speed of the ball when it hit the floor is u = √(2gh)
The ball will hit the floor at the time t' = √(2h/g)
after it was dropped.

This time is measured with coordinate clocks in the airplane's
rest frame.
_______________________________________________________________________ >>
In the ground frame, we have:
-------------------------------
The ball will have a constant horizontal speed v,
and the vertical speed when it hits the floor is u_v = √(2gh)

The speed of the ball when it hit the floor is u' = √(v²+2gh)
The ball will hit the floor at the time t' = √(2h/g)
after it was dropped.

This time is measured with coordinate clocks in the ground frame.
__________________________________________________________________________

If use the values
g = 9.8 m/s², v = 230 m/s (828.00 km/h), h = 2.5 m, we get :

The speed of the ball when it hits the floor is:
In the rest frame of the airplane: u = 7 m/s, t = 0.7142 s
In the ground frame: u= 230.11 m/s = 828.38 km/h, t = 0.7142 s

This answers are the same for SR and NM.
(We ignore the curvature of spacetime because
the height difference is only 2.5 m)

So the speed is obviously much higher in the ground frame
than in the airplane's rest frame. But the vertical speed
component is the same in both frames.
Note:
The vertical speed component is the same in both frames.

Original question:
In which frame is the speed of the ball fastest?

Second question:
In which frame is the vertical component of the ball's velocity fastest?
Well, no, Paul. Einstein says that we have a slower clock in the flying airplane. Hafele and Keating say it might be faster, but it could be slower. So we will go with Einstein's original idea. The clock is slower. If t is the time of a clock on
the ground, the during a time of t= 1 sec., the earth rotates a certain number of degrees on its axis. Mars rotates a certain number of degrees on its axis, Jupiter rotates a certain number of degrees on its axis, etc., etc. The clock in the airplane
shows less time than a second when the clock on the ground shows a second and the earth, Mars, Jupiter, etc., are all rotating these varying number of degrees. So what you are telling me now is that if a scientist looks at the clock in the airplane, the
clock on the ground shows less time than the clock on the airplane. What do the earth, Mars, Jupiter, etc. do? Do they all rotate backward to agree with the clock on the ground every time a scientist looks at the clock in the airplane? Now scientists
have two miracles. I am not opposed to scientists having miracles, but you have to understand that we common people are required to live in something called reality where the planets would just go on rotating the same way they were before the scientist
looked at the clock in the airplane. Or maybe scientists are trying to say that the planets would go on rotating the way they did before, but the clock on the ground would rotate backward to agree with the equations scientists use. It would still just be
another miracle. I have to agree with Galileo and Isaac Newton that if the clock in the airplane was slower, as seen from the airplane, the clock on the ground would be faster. I think scientists are either using the wrong equations for their
calculations of electromagnetism and electromagnetic waves, or else they are misinterpreting the equations they have.
You are ignorant of physics, but you are nowhere near as stupid
as the above babble indicates. If you were, you wouldn't be able
to handle a job like - say welding.

Hint:
The ball "is in" both frames of reference of course,
it is not one ball in each frame. One reality!
And since the frames are moving relative to each other
along the horizontal axis, the horizontal component of
the ball's velocity must be different, while the vertical
component must be equal in the two frames.

If you measure the vertical velocity with two clocks with
different rates, you will get two different results, but
the vertical speed would still be the same in both frames.

With other words:
If the observer in the airplane had a clock which ran
too slow, the observer in the plane and the observer
Your tales of observers were nothing but funny even
300 years ago. Face it, trash, you have no clue
I think what Paul is saying is that t'=t is used for the vertical component and t' = (t-vx/c^2)/sqrt(1-v^2/c^2) is being used for the horizontal component. What does this clock look like?
I still like the idea of basing the times of clocks on rotations of planets. That way we can visualize the planets rotating backward every time a scientist looks down from an airplane to see the time of a clock on earth. Back in the time of Galileo and
Newton, time was all based on the rotation of the earth. One day was one rotation of the earth, an hour was 1/24th of a day, a minute was 1/60th of an hour, and a second was 1/60th of a minute. Then scientists got the idea of saying that a second was a
certain number of transitions of a cesium isotope atom. I think that if we could better visualize how transitions of cesium isotope atoms relate to rotations of planets, we could make great advances in science.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to mitchr...@gmail.com on Sun Oct 15 07:35:31 2023
On Saturday, October 14, 2023 at 8:12:49 PM UTC-7, mitchr...@gmail.com wrote:
On Monday, October 9, 2023 at 11:27:35 AM UTC-7, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a clock
in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get a
faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on
the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for
x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate
that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
How do you know a miracal is real. How do you measure the difference between absolute and relative orders?
Well, Isaac Newton used absolute time, the idea that all clocks that are working correctly in the universe would agree with one another. Newton was pretty good with mathematics. I think if he had ever been told that clocks were speeding up or slowing
down because of differences in gravitation or other reasons scientists give today, he would have just done what I do, use a different set of Galilean transformation equations with different variable for time and velocity to show what the faster or slower
clock was doing. So here are the equations Einstein was using for absolute time.
x'=x-vt
y'=y
z'=z
t'=t
Where scientists went wrong when they had a clock with a different rate was that they did not seem to believe the last equation, t'=t. What that equation means is that the time of a clock that shows t is being used in both frames of reference. This set
of Galilean equations says nothing whatsoever about a clock that shows a different time than t, other than where the clock would be. In order to show the time of the clock, you have to use another set of Galilean transformation equations with different
variables for time and velocity, as shown by the time of the faster or slower clock. Since Einstein's slower clock is moving, we use inverse Galilean transformation equations. Now the inverse equations with t'=t would be
x = x' -v't'
y = y'
z = z'
t = t'
-v = v'
So we just say that n' is the time of Einstein's slower clock and that m' is the velocity of S(x,y,z,n) relative to S'(x',y',z',n') according to the time of the slower clock.
x = x' - m'n'
y = y'
z = z'
n = n'
-vt/n' = m'
This way the pilot of an airplane with a slower clock will get a faster speed for the airplane than an observer on the ground who is using a faster clock to time the flight of the airplane, instead of both observers getting the same speed for the
airplane the way the Lorentz equations show. It took me a long time to figure out these equations, like about forty years, but a junior high algebra student should be able to figure them out in a few minutes.
So there is no need for Einstein's miracle in reality. That is how we know it is a real miracle, like the planets turning backward when scientists look at a clock on the ground.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to Tom Roberts on Sun Oct 15 09:59:42 2023
On Sunday, 15 October 2023 at 18:36:15 UTC+2, Tom Roberts wrote:

Only if he is an idiot. Any twin who understands relativity KNOWS the
home twin does not age instantaneously, and the change is only in their
age when considered simultaneously in his own new inertial frame. But

But anyone can check GPS, your absurd tales have nothing in
common with the real clocks, real seconds, real time.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Tom Roberts@21:1/5 to Mike Fontenot on Sun Oct 15 11:36:03 2023
On 10/14/23 6:19 PM, Mike Fontenot wrote:
On 10/14/23 4:24 PM, Tom Roberts wrote:
Nope, your "thinking" is wrong. At any instant in his life, his
speed depends upon which coordinate system you refer the speed to.
(You may be implicitly thinking "relative to the inertial
coordinates from which he started at rest", but YOU MUST SAY
THAT.)

I disagree. I think the accelerating person will INNATELY feel that
his speed was initially zero, and then became non-zero and
continually increasing when he started his constant acceleration.

Sure, assuming he started from rest RELATIVE TO SOME INERTIAL FRAME. But
you did not say that. You keep having internal thoughts that you do not
are not clairvoyant or mind readers.

The fact that you omit important information and don't realize it shows

That is analogous to the feeling that an inertial person (she) has:
that she IS stationary (i.e., that she regards her speed to be
zero), and that other inertial people (not stationary wrt her) are
moving.

Again, if you say so, then it's OK. But you do not say so and expect

Similarly, in the twin "paradox", the traveling twin (he) truly
BELIEVES that his home twin (she) instantaneously gets much older
when he instantaneously reverses course.

Only if he is an idiot. Any twin who understands relativity KNOWS the
home twin does not age instantaneously, and the change is only in their
age when considered simultaneously in his own new inertial frame. But
his own inertial frame does not affect the home twin in any way.

This shows how wishy-washy your own thoughts are. And, of course, we

And if he ever instantaneously does a Dirac delta-function
acceleration in the direction AWAY from her (when they are separated
by a large distance), he will truly BELIEVE that she instantaneously
gets YOUNGER during that instantaneous turnaround.

Again, only if he is an idiot. Same as above.

And he is RIGHT,

No, he is not. YOU are confusing "age of the home twin" with "how the
traveling twin PERCEIVES the home twin's age simultaneously in his
various inertial frames".

Tom Roberts

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Mike Fontenot@21:1/5 to Tom Roberts on Sun Oct 15 11:13:21 2023
On 10/15/23 10:36 AM, Tom Roberts wrote:

Sure, assuming he started from rest RELATIVE TO SOME INERTIAL FRAME.

Not "SOME" inertial frame. THE inertial frame in which he was
stationary immediately before he started his constant acceleration. He
has no other choice!

Then, I (Mike Fontenot) said:

Similarly, in the twin "paradox", the traveling twin (he) truly
BELIEVES that his home twin (she) instantaneously gets much older when
he instantaneously reverses course.

And Tom Roberts responded:

Only if he is an idiot. Any twin who understands relativity KNOWS the
home twin (she) does not age instantaneously, [...]

Not true. According to HIM, she DOES age instantaneously during his instantaneous turnaround. There is no other possibility FOR HIM.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Paul B. Andersen@21:1/5 to All on Sun Oct 15 22:54:21 2023
µmmmmDen 13.10.2023 18:25, skrev Robert Winn:
On Friday, October 13, 2023 at 4:43:32 AM UTC-7, Paul B. Andersen wrote:
Den 12.10.2023 20:04, skrev Robert Winn:
On Thursday, October 12, 2023 at 7:23:00 AM UTC-7, Paul B. Andersen wrote:

Den 10.10.2023 21:52, skrev Robert Winn:
If we drop a ball from the ceiling of an airplane that is flying, the ball is falling faster in the frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.

I realize now that most of what I have written in this thread
is very imprecise, and something is plain wrong.
(I tend to be a bit too fast sometimes.)

The big error I made was too claim that the vertical speed of
the ball would be the same when measured in the two frames.
That is wrong. The vertical speed in the ground frame is
slower by 1/γ compared to the airplane frame.

Which was what Robert claimed, but for the wrong reason.
The clock in the airplane and the clock on the ground run
at the same rate.

So let me start from the beginning, and do it properly this time.
There may be typos, but not too many, I hope.

First, remember what a frame of reference is.
The concept you must know is coordinate time.
You can image a three dimensional array of clocks.
These clocks are synchronous, which mean that at
any time all the clocks are showing the same.

All frames of reference are equal, and since we
use second as the time unit, the clocks (and coordinate time)
are SI-clocks. And since the clocks in all frames of reference
are equal SI-clocks, THEY ALL RUN AT THE SAME RATE.

Let's call the rest frame of the airplane K'(t',x',y'),
and let's call the ground frame K(t,x,y).

We have clocks showing coordinate time at the origin of the frames.

The events of interest is E₀, the ball is dropped from the ceiling,
and E₁, the ball hits the floor in the airplane.

(use fixed width font!)

At E₀ we have:
y,y'
|
o-h
|
|
K':-------------------|--------------------> x' -> v
0
K :-------------------|---------------------> x
0
Fig. 1.
---------
The x-axis and the x'-axis are coincident.
At t = t' = 0 The origins and the y-axis and the y'-axis are
also coincident. We choose z = z' = 0 for all t, and forget the z axis.

The coordinates of the ball at event E₀ are:
In K: t = t₀ = 0 s, x = x₀ = 0 m, y = h
In K': t' = t₀' = 0 s, x' = x₀' = 0 m, y' = h
______

At E₁ we have:

When the ball hits the floor, t' = t₁' = √(2h/g)
and x' = x₁' = 0 m

y y'
| |
| |
| |
| |
K':----------|----------o--------> x' -> v
x₂' 0
K :----------|----------|---------> x
0 x₁
Fig. 2.
---------

So the coordinates of the ball at the event E₁ are:
In K':
t' = t₁' = √(2h/g)
x' = x₁' = 0 m
In K:
t₁ = γ(t₁'+ v⋅x₁'/c²) = γ⋅√(2h/g)
x₁ = γ(x₁'+ v⋅t₁') = γ⋅v⋅√(2h/g)

We can find the velocity of the ball:
=====================================
In the airplane frame:
----------------------
The horizontal velocity component is Vh' = x₁'/t₁' = 0
The vertical velocity component: Vv'(t') = gt'
The vertical velocity component when the ball hits the floor:
Vv'(t₁') = gt₁'= √(2gh)

In the ground frame:
---------------------
The horizontal velocity component is
Vh = x₁/t₁ = γ⋅v⋅√(2h/g)/γ⋅√(2h/g) = v

The acceleration g of the ball transformed to to K is a = g/γ²
The vertical velocity component: Vv(t) = a⋅t = (g/γ²)⋅t
The vertical velocity component when the ball hit the floor:
Vv(t₁) = (g/γ²)⋅t₁ = (g/γ²)⋅γ⋅√(2h/g) = √(2gh)/γ

The apparent rate of clocks in K' as observed in K. ===================================================
When clock at the origin of K' shows t₀' = 0, it will
be adjacent to a coordinate clock in K showing t₀ = 0.
When this clock shows t₁' = √(2h/g) it will be adjacent
to a coordinate clock in K showing γ⋅√(2h/g).

So as observed from K, the rate of the clock in K' appears to be:
f' = (t₁'-t₀')/(t₁-t₀) = √(2h/g)/γ⋅√(2h/g) = 1/γ

Note that (t₁'-t₀') is a proper time because both t₀' and t₁'
are read off the same clock, while t₀ and t₁ are read off two
different coordinate clocks

The moving clock appears to run slow.

The apparent rate of clocks in K as observed in K'. ===================================================
When clock at the origin of K shows t₀ = 0, it will
be adjacent to a coordinate clock in K showing t₀' = 0.

See fig2:
We can define an event E₂:
When this clock at x₂ = 0 shows t₂ = √(2h/g) it will be adjacent
to a coordinate clock in K'.
The coordinates in K' are:
t₂' = γ(t₂ - v⋅x₂/c²) = γ⋅√(2h/g)
x₂' = γ(x₂ - v⋅t₂) = -γ⋅v⋅√(2h/g)

So as observed from K, the rate of the clock in K' appears to be:
f = (t₂-t₀)/(t₂'-t₀') = √(2h/g)/γ⋅√(2h/g) = 1/γ

Note that (t₂-t₀) is a proper time because both t₀ and t₂
are read off the same clock, while t₀' and t₂' are read off two
different coordinate clocks

The moving clock appears to run slow.

Mutual time dilation!

--
Paul

https://paulba.no/

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Paul B. Andersen on Sun Oct 15 19:54:55 2023
On Sunday, October 15, 2023 at 1:53:29 PM UTC-7, Paul B. Andersen wrote:
µmmmmDen 13.10.2023 18:25, skrev Robert Winn:
On Friday, October 13, 2023 at 4:43:32 AM UTC-7, Paul B. Andersen wrote:
Den 12.10.2023 20:04, skrev Robert Winn:
On Thursday, October 12, 2023 at 7:23:00 AM UTC-7, Paul B. Andersen wrote:

Den 10.10.2023 21:52, skrev Robert Winn:
If we drop a ball from the ceiling of an airplane that is flying, the ball is falling faster in the frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.

I realize now that most of what I have written in this thread
is very imprecise, and something is plain wrong.
(I tend to be a bit too fast sometimes.)

The big error I made was too claim that the vertical speed of
the ball would be the same when measured in the two frames.
That is wrong. The vertical speed in the ground frame is
slower by 1/γ compared to the airplane frame.

Which was what Robert claimed, but for the wrong reason.
The clock in the airplane and the clock on the ground run
at the same rate.

So let me start from the beginning, and do it properly this time.
There may be typos, but not too many, I hope.

First, remember what a frame of reference is.
The concept you must know is coordinate time.
You can image a three dimensional array of clocks.
These clocks are synchronous, which mean that at
any time all the clocks are showing the same.

All frames of reference are equal, and since we
use second as the time unit, the clocks (and coordinate time)
are SI-clocks. And since the clocks in all frames of reference
are equal SI-clocks, THEY ALL RUN AT THE SAME RATE.

Let's call the rest frame of the airplane K'(t',x',y'),
and let's call the ground frame K(t,x,y).

We have clocks showing coordinate time at the origin of the frames.

The events of interest is E₀, the ball is dropped from the ceiling,
and E₁, the ball hits the floor in the airplane.

(use fixed width font!)

At E₀ we have:
y,y'
|
o-h
|
|
K':-------------------|--------------------> x' -> v
0
K :-------------------|---------------------> x
0
Fig. 1.
---------
The x-axis and the x'-axis are coincident.
At t = t' = 0 The origins and the y-axis and the y'-axis are
also coincident. We choose z = z' = 0 for all t, and forget the z axis.

The coordinates of the ball at event E₀ are:
In K: t = t₀ = 0 s, x = x₀ = 0 m, y = h
In K': t' = t₀' = 0 s, x' = x₀' = 0 m, y' = h
______

At E₁ we have:

When the ball hits the floor, t' = t₁' = √(2h/g)
and x' = x₁' = 0 m

y y'
| |
| |
| |
| |
K':----------|----------o--------> x' -> v
x₂' 0
K :----------|----------|---------> x
0 x₁
Fig. 2.
---------

So the coordinates of the ball at the event E₁ are:
In K':
t' = t₁' = √(2h/g)
x' = x₁' = 0 m
In K:
t₁ = γ(t₁'+ v⋅x₁'/c²) = γ⋅√(2h/g)
x₁ = γ(x₁'+ v⋅t₁') = γ⋅v⋅√(2h/g)

We can find the velocity of the ball:
=====================================
In the airplane frame:
----------------------
The horizontal velocity component is Vh' = x₁'/t₁' = 0
The vertical velocity component: Vv'(t') = gt'
The vertical velocity component when the ball hits the floor:
Vv'(t₁') = gt₁'= √(2gh)

In the ground frame:
---------------------
The horizontal velocity component is
Vh = x₁/t₁ = γ⋅v⋅√(2h/g)/γ⋅√(2h/g) = v

The acceleration g of the ball transformed to to K is a = g/γ²
The vertical velocity component: Vv(t) = a⋅t = (g/γ²)⋅t
The vertical velocity component when the ball hit the floor:
Vv(t₁) = (g/γ²)⋅t₁ = (g/γ²)⋅γ⋅√(2h/g) = √(2gh)/γ

The apparent rate of clocks in K' as observed in K. ===================================================
When clock at the origin of K' shows t₀' = 0, it will
be adjacent to a coordinate clock in K showing t₀ = 0.
When this clock shows t₁' = √(2h/g) it will be adjacent
to a coordinate clock in K showing γ⋅√(2h/g).

So as observed from K, the rate of the clock in K' appears to be:
f' = (t₁'-t₀')/(t₁-t₀) = √(2h/g)/γ⋅√(2h/g) = 1/γ

Note that (t₁'-t₀') is a proper time because both t₀' and t₁'
are read off the same clock, while t₀ and t₁ are read off two
different coordinate clocks

The moving clock appears to run slow.

The apparent rate of clocks in K as observed in K'. ===================================================
When clock at the origin of K shows t₀ = 0, it will
be adjacent to a coordinate clock in K showing t₀' = 0.

See fig2:
We can define an event E₂:
When this clock at x₂ = 0 shows t₂ = √(2h/g) it will be adjacent
to a coordinate clock in K'.
The coordinates in K' are:
t₂' = γ(t₂ - v⋅x₂/c²) = γ⋅√(2h/g)
x₂' = γ(x₂ - v⋅t₂) = -γ⋅v⋅√(2h/g)

So as observed from K, the rate of the clock in K' appears to be:
f = (t₂-t₀)/(t₂'-t₀') = √(2h/g)/γ⋅√(2h/g) = 1/γ

Note that (t₂-t₀) is a proper time because both t₀ and t₂
are read off the same clock, while t₀' and t₂' are read off two different coordinate clocks

The moving clock appears to run slow.

Mutual time dilation!

--
Paul

https://paulba.no/
Well, that is just another miracle that scientists have imagined. I don't really see the need for it. One of the proofs that scientists used to say that Einstein's Theory was true was the perihelion of Mercury. Isaac Newton's absolute time
interpretation did not quite explain where Mercury was. But, as I said, if Newton had been told that More gravitation would result in a slower clock, I think he could have done the mathematics. In looking at the Lorentz equations, it seemed obvious to
me how part of them was derived. If we take two sets of Galilean transformation equations,
x'=x-vt
y'=y
z'=z
t'=t

x = x' - m'n'
y = y'
z - z'
n = n'
Since there is no length contraction in the Galilean transformation equations, we can say
x - x' = vt
x -x' = -m'n'
m'n' = -vt
Einstein said that the Lorentz equations satisfy the results of the Michelson-Morley experiment because x = ct, x' = ct', where t' was the time of the moving clock. With two sets of Galilean transformation equations we have n' as the time of the moving
clock.
x=ct, x' = cn'
So by either clock, light is traveling at c.
x'=x-vt
cn'=ct-vt
n' = t - vt/c = t - vct/c^2 = t-vx/c^2, Which you might recognize as the numerator of Lorentz's equation for t'. At the speed of Mercury in its orbit, the denominator of Lorentz's equation is irrelevant. If something is traveling at the speed of
Mercury, the difference between n' in two sets of Galilean transformation equations and t' in Lorentz's equations is the same to several decimal places. Scientists might have used General Relativity rather than Special in making the calculations for the
orbit of Mercury, but the planet Mercury shows something about relativity. Using the time of the third planet from the sun as a preferred time for the solar system does not really make sense. Mercury moves at 30 miles per second, earth moves at 20
miles per second. The further you get from the sun, the slower the planets are in their orbits. Then you have the orbits of the moons and satellites orbiting earth, etc. It seems to me that Newton was correct with his idea of absolute time to the
extent that there would be a rate of time to which the rates of time of all planets could be converted, but it would be a different time from the time of the third planet from the sun. It seems to me that there would be a time common to the entire solar
system to which the times of orbiting planets could be converted to agree with Newton's equations. Scientists, of course, are not going to be interested in anything but the miracles that Einstein and his disciples describe, but I think that times of
clocks could be used to gain a better understanding of gravitation. Scientists are paid trillions of dollars from governments to say that Einstein's equations are correct. I do not really see any reason to think anything is going to change any time
soon.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to Paul B. Andersen on Sun Oct 15 22:44:48 2023
On Sunday, 15 October 2023 at 22:53:29 UTC+2, Paul B. Andersen wrote:
µmmmmDen 13.10.2023 18:25, skrev Robert Winn:
On Friday, October 13, 2023 at 4:43:32 AM UTC-7, Paul B. Andersen wrote:
Den 12.10.2023 20:04, skrev Robert Winn:
On Thursday, October 12, 2023 at 7:23:00 AM UTC-7, Paul B. Andersen wrote:

Den 10.10.2023 21:52, skrev Robert Winn:
If we drop a ball from the ceiling of an airplane that is flying, the ball is falling faster in the frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.

I realize now that most of what I have written in this thread
is very imprecise, and something is plain wrong.
(I tend to be a bit too fast sometimes.)

Taking it sinply - you're just a fanatic idiot brainwashed
by inconsistent religion.

All frames of reference are equal, and since we
use second as the time unit, the clocks (and coordinate time)
are SI-clocks. And since the clocks in all frames of reference
are equal SI-clocks,

Keep dreaming. Anyone can check GPS, your ideological
absurd is not going to be treaten seriously when it comes
to serious measurements.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Robert Winn on Mon Oct 16 05:54:35 2023
On Monday, 16 October 2023 at 03:54:57 UTC+1, Robert Winn wrote:
On Sunday, October 15, 2023 at 1:53:29 PM UTC-7, Paul B. Andersen wrote:
µmmmmDen 13.10.2023 18:25, skrev Robert Winn:
On Friday, October 13, 2023 at 4:43:32 AM UTC-7, Paul B. Andersen wrote:
Den 12.10.2023 20:04, skrev Robert Winn:
On Thursday, October 12, 2023 at 7:23:00 AM UTC-7, Paul B. Andersen wrote:

Den 10.10.2023 21:52, skrev Robert Winn:
If we drop a ball from the ceiling of an airplane that is flying, the ball is falling faster in the frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.

I realize now that most of what I have written in this thread
is very imprecise, and something is plain wrong.
(I tend to be a bit too fast sometimes.)

The big error I made was too claim that the vertical speed of
the ball would be the same when measured in the two frames.
That is wrong. The vertical speed in the ground frame is
slower by 1/γ compared to the airplane frame.

Which was what Robert claimed, but for the wrong reason.
The clock in the airplane and the clock on the ground run
at the same rate.

So let me start from the beginning, and do it properly this time.
There may be typos, but not too many, I hope.

First, remember what a frame of reference is.
The concept you must know is coordinate time.
You can image a three dimensional array of clocks.
These clocks are synchronous, which mean that at
any time all the clocks are showing the same.

All frames of reference are equal, and since we
use second as the time unit, the clocks (and coordinate time)
are SI-clocks. And since the clocks in all frames of reference
are equal SI-clocks, THEY ALL RUN AT THE SAME RATE.

Let's call the rest frame of the airplane K'(t',x',y'),
and let's call the ground frame K(t,x,y).

We have clocks showing coordinate time at the origin of the frames.

The events of interest is E₀, the ball is dropped from the ceiling,
and E₁, the ball hits the floor in the airplane.

(use fixed width font!)

At E₀ we have:
y,y'
|
o-h
|
|
K':-------------------|--------------------> x' -> v
0
K :-------------------|---------------------> x
0
Fig. 1.
---------
The x-axis and the x'-axis are coincident.
At t = t' = 0 The origins and the y-axis and the y'-axis are
also coincident. We choose z = z' = 0 for all t, and forget the z axis.

The coordinates of the ball at event E₀ are:
In K: t = t₀ = 0 s, x = x₀ = 0 m, y = h
In K': t' = t₀' = 0 s, x' = x₀' = 0 m, y' = h
______

At E₁ we have:

When the ball hits the floor, t' = t₁' = √(2h/g)
and x' = x₁' = 0 m

y y'
| |
| |
| |
| |
K':----------|----------o--------> x' -> v
x₂' 0
K :----------|----------|---------> x
0 x₁
Fig. 2.
---------

So the coordinates of the ball at the event E₁ are:
In K':
t' = t₁' = √(2h/g)
x' = x₁' = 0 m
In K:
t₁ = γ(t₁'+ v⋅x₁'/c²) = γ⋅√(2h/g)
x₁ = γ(x₁'+ v⋅t₁') = γ⋅v⋅√(2h/g)

We can find the velocity of the ball: =====================================
In the airplane frame:
----------------------
The horizontal velocity component is Vh' = x₁'/t₁' = 0
The vertical velocity component: Vv'(t') = gt'
The vertical velocity component when the ball hits the floor:
Vv'(t₁') = gt₁'= √(2gh)

In the ground frame:
---------------------
The horizontal velocity component is
Vh = x₁/t₁ = γ⋅v⋅√(2h/g)/γ⋅√(2h/g) = v

The acceleration g of the ball transformed to to K is a = g/γ²
The vertical velocity component: Vv(t) = a⋅t = (g/γ²)⋅t
The vertical velocity component when the ball hit the floor:
Vv(t₁) = (g/γ²)⋅t₁ = (g/γ²)⋅γ⋅√(2h/g) = √(2gh)/γ

The apparent rate of clocks in K' as observed in K. ===================================================
When clock at the origin of K' shows t₀' = 0, it will
be adjacent to a coordinate clock in K showing t₀ = 0.
When this clock shows t₁' = √(2h/g) it will be adjacent
to a coordinate clock in K showing γ⋅√(2h/g).

So as observed from K, the rate of the clock in K' appears to be:
f' = (t₁'-t₀')/(t₁-t₀) = √(2h/g)/γ⋅√(2h/g) = 1/γ

Note that (t₁'-t₀') is a proper time because both t₀' and t₁'
are read off the same clock, while t₀ and t₁ are read off two different coordinate clocks

The moving clock appears to run slow.

The apparent rate of clocks in K as observed in K'. ===================================================
When clock at the origin of K shows t₀ = 0, it will
be adjacent to a coordinate clock in K showing t₀' = 0.

See fig2:
We can define an event E₂:
When this clock at x₂ = 0 shows t₂ = √(2h/g) it will be adjacent
to a coordinate clock in K'.
The coordinates in K' are:
t₂' = γ(t₂ - v⋅x₂/c²) = γ⋅√(2h/g)
x₂' = γ(x₂ - v⋅t₂) = -γ⋅v⋅√(2h/g)

So as observed from K, the rate of the clock in K' appears to be:
f = (t₂-t₀)/(t₂'-t₀') = √(2h/g)/γ⋅√(2h/g) = 1/γ

Note that (t₂-t₀) is a proper time because both t₀ and t₂
are read off the same clock, while t₀' and t₂' are read off two different coordinate clocks

The moving clock appears to run slow.

Mutual time dilation!

--
Paul

https://paulba.no/
Well, that is just another miracle that scientists have imagined. I don't really see the need for it. One of the proofs that scientists used to say that Einstein's Theory was true was the perihelion of Mercury. Isaac Newton's absolute time
interpretation did not quite explain where Mercury was. But, as I said, if Newton had been told that More gravitation would result in a slower clock, I think he could have done the mathematics. In looking at the Lorentz equations, it seemed obvious to me
how part of them was derived. If we take two sets of Galilean transformation equations,
x'=x-vt
y'=y
z'=z
t'=t
x = x' - m'n'
y = y'
z - z'
n = n'
Since there is no length contraction in the Galilean transformation equations, we can say
x - x' = vt
x -x' = -m'n'
m'n' = -vt
Einstein said that the Lorentz equations satisfy the results of the Michelson-Morley experiment because x = ct, x' = ct', where t' was the time of the moving clock. With two sets of Galilean transformation equations we have n' as the time of the moving
clock.
x=ct, x' = cn'
So by either clock, light is traveling at c.
x'=x-vt
cn'=ct-vt
n' = t - vt/c = t - vct/c^2 = t-vx/c^2, Which you might recognize as the numerator of Lorentz's equation for t'. At the speed of Mercury in its orbit, the denominator of Lorentz's equation is irrelevant. If something is traveling at the speed of
Mercury, the difference between n' in two sets of Galilean transformation equations and t' in Lorentz's equations is the same to several decimal places. Scientists might have used General Relativity rather than Special in making the calculations for the
orbit of Mercury, but the planet Mercury shows something about relativity. Using the time of the third planet from the sun as a preferred time for the solar system does not really make sense. Mercury moves at 30 miles per second, earth moves at 20 miles
per second. The further you get from the sun, the slower the planets are in their orbits. Then you have the orbits of the moons and satellites orbiting earth, etc. It seems to me that Newton was correct with his idea of absolute time to the extent that
there would be a rate of time to which the rates of time of all planets could be converted, but it would be a different time from the time of the third planet from the sun. It seems to me that there would be a time common to the entire solar system to
which the times of orbiting planets could be converted to agree with Newton's equations. Scientists, of course, are not going to be interested in anything but the miracles that Einstein and his disciples describe, but I think that times of clocks could
be used to gain a better understanding of gravitation. Scientists are paid trillions of dollars from governments to say that Einstein's equations are correct. I do not really see any reason to think anything is going to change any time soon.

I would ignore Einsteins “predictions” for the mercury anomalous preccession.
He only knew the amount for mercury and fiddled his formula to match that observed.
Notice his formula doesn’t do so well for other planets. A fact relativists like to ignore.
To start with Mars preccesion rate is very unstable and is hard to calculate. Which is why they pretend Mars preccession rate is 1.3 arc seconds per C
When it probably is more like 2.5
The other planets preccesion rates fare less well under GR
Venus observed is 8 (GR predicted 8.6)
Earth observed is 5 (GR predicts 3.8)

However if one uses a more correct classical formula 1/(r+3R)^2 based on perehilion
not semi major axis as Albert incorrectly did. Then classical theory predicts more accurately
than GR. {where r is perehilion distance and R is radius of sun}
As follows:
Planet. Obs. GR Classical
Merc. 43.1. 43.5 43.24
V. 8. 8.6. 8.33
E. 5. 3.87. 4.49

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Mike Fontenot@21:1/5 to All on Mon Oct 16 07:53:53 2023
One thing you've never come to terms with, Tom, is that for an inertial observer, they have no choice but to believe that their conclusions
about simultaneity-at-a-distance are fully real and meaningful. Because
those conclusions are based ONLY on the fact that the speed of light is
186,000 miles per second in their reference frame. If their conclusions
about simultaneity-at-a-distance aren't correct, then their assumption
that light travels at 186,000 miles per second in their frame can't be
correct, in which case, special relativity can't be correct.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Dono.@21:1/5 to Lou on Mon Oct 16 08:23:23 2023
On Monday, October 16, 2023 at 5:54:37 AM UTC-7, Lou wrote:

Notice his formula doesn’t do so well for other planets.

Actually, it does. You are lying again, crank.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to Tom Roberts on Mon Oct 16 11:41:00 2023
On Monday, 16 October 2023 at 20:27:58 UTC+2, Tom Roberts wrote:

Not true. According to HIM, she DOES age instantaneously during his instantaneous turnaround. There is no other possibility FOR HIM.
Nonsense. If the traveling twin understands relativity, he understands
that his current rest frame has NOTHING WHATSOEVER to do with the home
twin (who is at rest in a different inertial frame). So the most logical
AND USEFUL way

Stop fucking, trash, your way is exactly as itiotic as it looks
like. And anyone can check GPS, noone is going to use it.
Capital letters won't help.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Lou on Mon Oct 16 11:28:50 2023
On Monday, October 16, 2023 at 5:54:37 AM UTC-7, Lou wrote:
On Monday, 16 October 2023 at 03:54:57 UTC+1, Robert Winn wrote:
On Sunday, October 15, 2023 at 1:53:29 PM UTC-7, Paul B. Andersen wrote:
µmmmmDen 13.10.2023 18:25, skrev Robert Winn:
On Friday, October 13, 2023 at 4:43:32 AM UTC-7, Paul B. Andersen wrote:
Den 12.10.2023 20:04, skrev Robert Winn:
On Thursday, October 12, 2023 at 7:23:00 AM UTC-7, Paul B. Andersen wrote:

Den 10.10.2023 21:52, skrev Robert Winn:
If we drop a ball from the ceiling of an airplane that is flying, the ball is falling faster in the frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.

I realize now that most of what I have written in this thread
is very imprecise, and something is plain wrong.
(I tend to be a bit too fast sometimes.)

The big error I made was too claim that the vertical speed of
the ball would be the same when measured in the two frames.
That is wrong. The vertical speed in the ground frame is
slower by 1/γ compared to the airplane frame.

Which was what Robert claimed, but for the wrong reason.
The clock in the airplane and the clock on the ground run
at the same rate.

So let me start from the beginning, and do it properly this time.
There may be typos, but not too many, I hope.

First, remember what a frame of reference is.
The concept you must know is coordinate time.
You can image a three dimensional array of clocks.
These clocks are synchronous, which mean that at
any time all the clocks are showing the same.

All frames of reference are equal, and since we
use second as the time unit, the clocks (and coordinate time)
are SI-clocks. And since the clocks in all frames of reference
are equal SI-clocks, THEY ALL RUN AT THE SAME RATE.

Let's call the rest frame of the airplane K'(t',x',y'),
and let's call the ground frame K(t,x,y).

We have clocks showing coordinate time at the origin of the frames.

The events of interest is E₀, the ball is dropped from the ceiling, and E₁, the ball hits the floor in the airplane.

(use fixed width font!)

At E₀ we have:
y,y'
|
o-h
|
|
K':-------------------|--------------------> x' -> v
0
K :-------------------|---------------------> x
0
Fig. 1.
---------
The x-axis and the x'-axis are coincident.
At t = t' = 0 The origins and the y-axis and the y'-axis are
also coincident. We choose z = z' = 0 for all t, and forget the z axis.

The coordinates of the ball at event E₀ are:
In K: t = t₀ = 0 s, x = x₀ = 0 m, y = h
In K': t' = t₀' = 0 s, x' = x₀' = 0 m, y' = h
______

At E₁ we have:

When the ball hits the floor, t' = t₁' = √(2h/g)
and x' = x₁' = 0 m

y y'
| |
| |
| |
| |
K':----------|----------o--------> x' -> v
x₂' 0
K :----------|----------|---------> x
0 x₁
Fig. 2.
---------

So the coordinates of the ball at the event E₁ are:
In K':
t' = t₁' = √(2h/g)
x' = x₁' = 0 m
In K:
t₁ = γ(t₁'+ v⋅x₁'/c²) = γ⋅√(2h/g)
x₁ = γ(x₁'+ v⋅t₁') = γ⋅v⋅√(2h/g)

We can find the velocity of the ball: =====================================
In the airplane frame:
----------------------
The horizontal velocity component is Vh' = x₁'/t₁' = 0
The vertical velocity component: Vv'(t') = gt'
The vertical velocity component when the ball hits the floor:
Vv'(t₁') = gt₁'= √(2gh)

In the ground frame:
---------------------
The horizontal velocity component is
Vh = x₁/t₁ = γ⋅v⋅√(2h/g)/γ⋅√(2h/g) = v

The acceleration g of the ball transformed to to K is a = g/γ²
The vertical velocity component: Vv(t) = a⋅t = (g/γ²)⋅t
The vertical velocity component when the ball hit the floor:
Vv(t₁) = (g/γ²)⋅t₁ = (g/γ²)⋅γ⋅√(2h/g) = √(2gh)/γ

The apparent rate of clocks in K' as observed in K. ===================================================
When clock at the origin of K' shows t₀' = 0, it will
be adjacent to a coordinate clock in K showing t₀ = 0.
When this clock shows t₁' = √(2h/g) it will be adjacent
to a coordinate clock in K showing γ⋅√(2h/g).

So as observed from K, the rate of the clock in K' appears to be:
f' = (t₁'-t₀')/(t₁-t₀) = √(2h/g)/γ⋅√(2h/g) = 1/γ

Note that (t₁'-t₀') is a proper time because both t₀' and t₁' are read off the same clock, while t₀ and t₁ are read off two different coordinate clocks

The moving clock appears to run slow.

The apparent rate of clocks in K as observed in K'. ===================================================
When clock at the origin of K shows t₀ = 0, it will
be adjacent to a coordinate clock in K showing t₀' = 0.

See fig2:
We can define an event E₂:
When this clock at x₂ = 0 shows t₂ = √(2h/g) it will be adjacent to a coordinate clock in K'.
The coordinates in K' are:
t₂' = γ(t₂ - v⋅x₂/c²) = γ⋅√(2h/g)
x₂' = γ(x₂ - v⋅t₂) = -γ⋅v⋅√(2h/g)

So as observed from K, the rate of the clock in K' appears to be:
f = (t₂-t₀)/(t₂'-t₀') = √(2h/g)/γ⋅√(2h/g) = 1/γ

Note that (t₂-t₀) is a proper time because both t₀ and t₂
are read off the same clock, while t₀' and t₂' are read off two different coordinate clocks

The moving clock appears to run slow.

Mutual time dilation!

--
Paul

https://paulba.no/
Well, that is just another miracle that scientists have imagined. I don't really see the need for it. One of the proofs that scientists used to say that Einstein's Theory was true was the perihelion of Mercury. Isaac Newton's absolute time
interpretation did not quite explain where Mercury was. But, as I said, if Newton had been told that More gravitation would result in a slower clock, I think he could have done the mathematics. In looking at the Lorentz equations, it seemed obvious to me
how part of them was derived. If we take two sets of Galilean transformation equations,
x'=x-vt
y'=y
z'=z
t'=t
x = x' - m'n'
y = y'
z - z'
n = n'
Since there is no length contraction in the Galilean transformation equations, we can say
x - x' = vt
x -x' = -m'n'
m'n' = -vt
Einstein said that the Lorentz equations satisfy the results of the Michelson-Morley experiment because x = ct, x' = ct', where t' was the time of the moving clock. With two sets of Galilean transformation equations we have n' as the time of the
moving clock.
x=ct, x' = cn'
So by either clock, light is traveling at c.
x'=x-vt
cn'=ct-vt
n' = t - vt/c = t - vct/c^2 = t-vx/c^2, Which you might recognize as the numerator of Lorentz's equation for t'. At the speed of Mercury in its orbit, the denominator of Lorentz's equation is irrelevant. If something is traveling at the speed of
Mercury, the difference between n' in two sets of Galilean transformation equations and t' in Lorentz's equations is the same to several decimal places. Scientists might have used General Relativity rather than Special in making the calculations for the
orbit of Mercury, but the planet Mercury shows something about relativity. Using the time of the third planet from the sun as a preferred time for the solar system does not really make sense. Mercury moves at 30 miles per second, earth moves at 20 miles
per second. The further you get from the sun, the slower the planets are in their orbits. Then you have the orbits of the moons and satellites orbiting earth, etc. It seems to me that Newton was correct with his idea of absolute time to the extent that
there would be a rate of time to which the rates of time of all planets could be converted, but it would be a different time from the time of the third planet from the sun. It seems to me that there would be a time common to the entire solar system to
which the times of orbiting planets could be converted to agree with Newton's equations. Scientists, of course, are not going to be interested in anything but the miracles that Einstein and his disciples describe, but I think that times of clocks could
be used to gain a better understanding of gravitation. Scientists are paid trillions of dollars from governments to say that Einstein's equations are correct. I do not really see any reason to think anything is going to change any time soon.
I would ignore Einsteins “predictions” for the mercury anomalous preccession.
He only knew the amount for mercury and fiddled his formula to match that observed.
Notice his formula doesn’t do so well for other planets. A fact relativists
like to ignore.
To start with Mars preccesion rate is very unstable and is hard to calculate.
Which is why they pretend Mars preccession rate is 1.3 arc seconds per C When it probably is more like 2.5
The other planets preccesion rates fare less well under GR
Venus observed is 8 (GR predicted 8.6)
Earth observed is 5 (GR predicts 3.8)

However if one uses a more correct classical formula 1/(r+3R)^2 based on perehilion
not semi major axis as Albert incorrectly did. Then classical theory predicts more accurately
than GR. {where r is perehilion distance and R is radius of sun}
As follows:
Planet. Obs. GR Classical
Merc. 43.1. 43.5 43.24
V. 8. 8.6. 8.33
E. 5. 3.87. 4.49
I have contemplated this for some time. One thing I wondered about was if the rates of clocks varied according to their distance from the sun, then why would the time of the third planet from the sun be the significant time? It seems as though a time
associated with the sun itself or an absence of gravitation from the sun would be the time that would be significant and would be a time to which these other rates of time would be converted before using Newton's equations. At any rate I never did
believe Eddington's proof that Einstein's Relativity was correct.

--- SoupGate-Win32 v1.05
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• From Paul B. Andersen@21:1/5 to All on Mon Oct 16 21:08:49 2023
Den 16.10.2023 14:54, skrev Lou:

I would ignore Einsteins “predictions” for the mercury anomalous preccession.
He only knew the amount for mercury and fiddled his formula to match that observed.
Notice his formula doesn’t do so well for other planets. A fact relativists like to ignore.

See:
Chapter 3. Equation (8), Table 3.

Mercury 42.98"/century

To start with Mars preccesion rate is very unstable and is hard to calculate.

Equation (8) works equally well for all planets.
Mars: 1.35"/century

Which is why they pretend Mars preccession rate is 1.3 arc seconds per C
When it probably is more like 2.5
The other planets preccesion rates fare less well under GR
Venus observed is 8 (GR predicted 8.6)
Earth observed is 5 (GR predicts 3.8)

https://arxiv.org/pdf/0802.0176.pdf
See Table 1:
Venus observed is 8.6247"/century (GR predicted 8.6247"/century)
Earth observed is 3.8387"/century (GR predicts 3.8387"/century)

However if one uses a more correct classical formula r 1/(r+3R)^2 based on perehilion
not semi major axis as Albert incorrectly did. Then classical theory predicts more accurately
than GR. {where r is perehilion distance and R is radius of sun}

Mercury:
Perihelion distance r = 4.60011E10 m
Solar radius R = 696340E3 m

If the equation is r/(r+3R)^2 we get
the number 1.9890E-11 1/m

What does this number mean?
How do you get the numbers below?
Is the equation wrong?
In that case, what should it be?

As follows:
Planet. Classical
Merc. 43.24
V. 8.33
E. 4.49

--
Paul

https://paulba.no/

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• From Tom Roberts@21:1/5 to Mike Fontenot on Mon Oct 16 13:27:44 2023
On 10/15/23 12:13 PM, Mike Fontenot wrote:
On 10/15/23 10:36 AM, Tom Roberts wrote:
Sure, assuming he started from rest RELATIVE TO SOME INERTIAL
FRAME.

Not "SOME" inertial frame. THE inertial frame in which he was
stationary immediately before he started his constant acceleration.
He has no other choice!

That is "some inertial frame".

But, of course, he might not have started from an inertial frame at all,
which is why YOU MUST SPECIFY THE COMPLETE PHYSICAL SITUATION.

Then, I (Mike Fontenot) said:
Similarly, in the twin "paradox", the traveling twin (he) truly
BELIEVES that his home twin (she) instantaneously gets much
older when he instantaneously reverses course.

And Tom Roberts responded:
Only if he is an idiot. Any twin who understands relativity KNOWS
the home twin (she) does not age instantaneously, [...]

Not true. According to HIM, she DOES age instantaneously during his instantaneous turnaround. There is no other possibility FOR HIM.

Nonsense. If the traveling twin understands relativity, he understands
that his current rest frame has NOTHING WHATSOEVER to do with the home
twin (who is at rest in a different inertial frame). So the most logical
AND USEFUL way for the traveling twin to determine the home twin's age
"right now" is to compute the current time in the home twin's rest frame
and use that to determine the home twin's age. After all, that is the
ONLY frame that has relevance for the home twin, including their age.
This has several properties that your approach does not:
1) the home twin's age is monotonically increasing at the
standard rate (1 sec per sec).
2) the home twin's age never jumps, either forward or backward.
3) if the traveling twin returns home, this calculation yields
the correct value as they approach home.

Note that any notion of "aging" that does not satisfy all three of these
does not deserve to be called "aging", because those are part and parcel
of what we mean by a person's age. This of course includes your approach.

One thing you've never come to terms with, Tom, is that for an
inertial observer, they have no choice but to believe that their
conclusions about simultaneity-at-a-distance are fully real and
meaningful.

Nonsense. Any observer who understands relativity KNOWS that their
current inertial frame has NOTHING WHATSOEVER to do with a person at
rest in some other frame.

I don't dispute that the traveling twin can compute things as you
suggest. But I do dispute that the result can be called "the home twin's current age", or "the home twin's current age according to the traveling
twin". Because that computation does not deserve to be called "aging" at
all -- it is something else, and whatever it is, it is essentially useless.

I quit. I have no interest in continuing to argue about words.

Tom Roberts

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• From Mike Fontenot@21:1/5 to Mike Fontenot on Mon Oct 16 14:05:12 2023
On 10/16/23 1:21 PM, Mike Fontenot wrote:
On 10/16/23 12:27 PM, Tom Roberts wrote:

Nonsense. Any observer who understands relativity KNOWS that their
current inertial frame has NOTHING WHATSOEVER to do with a person at
rest in some other frame.

(I, Mike Fontenot, wrote:)
It has EVERYTHING to do with their CORRECT conclusion about the home
twin's current age.  Different inertial observers (moving wrt her) will disagree about the home twin's current age.  And any inertial observer
(he) moving wrt her who adopts HER view will be rejecting the fact that
light travels at 186,000 miles per second in HIS frame.  And that means
he is rejecting special relativity itself.

Your "take" on special relativity is nothing but useless mush.  You have learned NOTHING from your studies.

(And then I [Mike Fontenot] added:)

Tom, I think you should stick to something simpler than special
relativity, like general relativity or cosmology.

--- SoupGate-Win32 v1.05
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• From Mike Fontenot@21:1/5 to Tom Roberts on Mon Oct 16 13:21:26 2023
On 10/16/23 12:27 PM, Tom Roberts wrote:
On 10/15/23 12:13 PM, Mike Fontenot wrote:
On 10/15/23 10:36 AM, Tom Roberts wrote:
Sure, assuming he started from rest RELATIVE TO SOME INERTIAL FRAME.

Not "SOME" inertial frame.  THE inertial frame in which he was
stationary immediately before he started his constant acceleration. He
has no other choice!

That is "some inertial frame".

My point there was that it is the only reasonable inertial frame for him
to use.

But, of course, he might not have started from an inertial frame at all, [...]

How could he NOT start from some inertial frame? Has he ALWAYS been accelerating (and his mother, and her mother ...) ? That doesn't sound
like a scenario of any importance to me.

Then, I (Mike Fontenot) said:
Similarly, in the twin "paradox", the traveling twin (he) truly
BELIEVES that his home twin (she) instantaneously gets much
older when he instantaneously reverses course.

And Tom Roberts responded:
Only if he is an idiot. Any twin who understands relativity KNOWS the
home twin (she) does not age instantaneously, [...]

Not true.  According to HIM, she DOES age instantaneously during his
instantaneous turnaround.  There is no other possibility FOR HIM.

Nonsense. If the traveling twin understands relativity, he understands
that his current rest frame has NOTHING WHATSOEVER to do with the home
twin (who is at rest in a different inertial frame). So the most logical
AND USEFUL way for the traveling twin to determine the home twin's age
"right now" is to compute the current time in the home twin's rest frame
and use that to determine the home twin's age.

NOT true at all! During EACH of the traveler's two inertial legs of his
trip, he is stationary with a perpetually-inertial person (a "PIP").
Each of those PIP's disagree with the home twin's view of her age then,
and they CANNOT decide to accept HER view, without contradicting what
they conclude purely from the fact that light travels at 186,000 miles
per second in their inertial frame. And if they reject the fact that
that light in their frame travels at 186,000 miles per second, they are rejecting the theory of special relativity itself. The traveler spends
MANY years of his life being stationary (at different times) with
respect to EACH of those PIP's ... it makes no sense at all for him to
disagree with them during those long periods of time.

One thing you've never come to terms with, Tom, is that for an
inertial observer, they have no choice but to believe that their
conclusions about simultaneity-at-a-distance are fully real and
meaningful.

Nonsense. Any observer who understands relativity KNOWS that their
current inertial frame has NOTHING WHATSOEVER to do with a person at
rest in some other frame.

It has EVERYTHING to do with their CORRECT conclusion about the home
twin's current age. Different inertial observers (moving wrt her) will disagree about the home twin's current age. And any inertial observer
(he) moving wrt her who adopts HER view will be rejecting the fact that
light travels at 186,000 miles per second in HIS frame. And that means
he is rejecting special relativity itself.

Your "take" on special relativity is nothing but useless mush. You have learned NOTHING from your studies.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Paul B. Andersen@21:1/5 to the errors I on Mon Oct 16 22:25:16 2023
Den 16.10.2023 04:54, skrev Robert Winn:
On Sunday, October 15, 2023 at 1:53:29 PM UTC-7, Paul B. Andersen wrote:

I realize now that most of what I have written in this thread
is very imprecise, and something is plain wrong.
(I tend to be a bit too fast sometimes.)

The big error I made was too claim that the vertical speed of
the ball would be the same when measured in the two frames.
That is wrong. The vertical speed in the ground frame is
slower by 1/γ compared to the airplane frame.

Which was what Robert claimed, but for the wrong reason.
The clock in the airplane and the clock on the ground run
at the same rate.

So let me start from the beginning, and do it properly this time.
There may be typos, but not too many, I hope.

First, remember what a frame of reference is.
The concept you must know is coordinate time.
You can image a three dimensional array of clocks.
These clocks are synchronous, which mean that at
any time all the clocks are showing the same.

All frames of reference are equal, and since we
use second as the time unit, the clocks (and coordinate time)
are SI-clocks. And since the clocks in all frames of reference
are equal SI-clocks, THEY ALL RUN AT THE SAME RATE.

Let's call the rest frame of the airplane K'(t',x',y'),
and let's call the ground frame K(t,x,y).

We have clocks showing coordinate time at the origin of the frames.

The events of interest is E₀, the ball is dropped from the ceiling,
and E₁, the ball hits the floor in the airplane.

(use fixed width font!)

At E₀ we have:
y,y'
|
o-h
|
|
K':-------------------|--------------------> x' -> v
0
K :-------------------|---------------------> x
0
Fig. 1.
---------
The x-axis and the x'-axis are coincident.
At t = t' = 0 The origins and the y-axis and the y'-axis are
also coincident. We choose z = z' = 0 for all t, and forget the z axis.

The coordinates of the ball at event E₀ are:
In K: t = t₀ = 0 s, x = x₀ = 0 m, y = h
In K': t' = t₀' = 0 s, x' = x₀' = 0 m, y' = h
______

At E₁ we have:

When the ball hits the floor, t' = t₁' = √(2h/g)
and x' = x₁' = 0 m

y y'
| |
| |
| |
| |
K':----------|----------o--------> x' -> v
x₂' 0
K :----------|----------|---------> x
0 x₁
Fig. 2.
---------

So the coordinates of the ball at the event E₁ are:
In K':
t' = t₁' = √(2h/g)
x' = x₁' = 0 m
In K:
t₁ = γ(t₁'+ v⋅x₁'/c²) = γ⋅√(2h/g)
x₁ = γ(x₁'+ v⋅t₁') = γ⋅v⋅√(2h/g)

We can find the velocity of the ball:
=====================================
In the airplane frame:
----------------------
The horizontal velocity component is Vh' = x₁'/t₁' = 0
The vertical velocity component: Vv'(t') = gt'
The vertical velocity component when the ball hits the floor:
Vv'(t₁') = gt₁'= √(2gh)

In the ground frame:
---------------------
The horizontal velocity component is
Vh = x₁/t₁ = γ⋅v⋅√(2h/g)/γ⋅√(2h/g) = v

The acceleration g of the ball transformed to to K is a = g/γ²
The vertical velocity component: Vv(t) = a⋅t = (g/γ²)⋅t
The vertical velocity component when the ball hit the floor:
Vv(t₁) = (g/γ²)⋅t₁ = (g/γ²)⋅γ⋅√(2h/g) = √(2gh)/γ

The apparent rate of clocks in K' as observed in K.
===================================================
When clock at the origin of K' shows t₀' = 0, it will
be adjacent to a coordinate clock in K showing t₀ = 0.
When this clock shows t₁' = √(2h/g) it will be adjacent
to a coordinate clock in K showing γ⋅√(2h/g).

So as observed from K, the rate of the clock in K' appears to be:
f' = (t₁'-t₀')/(t₁-t₀) = √(2h/g)/γ⋅√(2h/g) = 1/γ

Note that (t₁'-t₀') is a proper time because both t₀' and t₁'
are read off the same clock, while t₀ and t₁ are read off two
different coordinate clocks

The moving clock appears to run slow.

The apparent rate of clocks in K as observed in K'.
===================================================
When clock at the origin of K shows t₀ = 0, it will
be adjacent to a coordinate clock in K showing t₀' = 0.

See fig2:
We can define an event E₂:
When this clock at x₂ = 0 shows t₂ = √(2h/g) it will be adjacent
to a coordinate clock in K'.
The coordinates in K' are:
t₂' = γ(t₂ - v⋅x₂/c²) = γ⋅√(2h/g)
x₂' = γ(x₂ - v⋅t₂) = -γ⋅v⋅√(2h/g)

So as observed from K, the rate of the clock in K' appears to be:
f = (t₂-t₀)/(t₂'-t₀') = √(2h/g)/γ⋅√(2h/g) = 1/γ

Note that (t₂-t₀) is a proper time because both t₀ and t₂
are read off the same clock, while t₀' and t₂' are read off two
different coordinate clocks

The moving clock appears to run slow.

I didn't really write this for you, Robert.
I know you are not able to read it.

But since these posts are archived, I wanted to correct
the errors I wrote earlier in this thread.

Mutual time dilation!

Well, that is just another miracle that scientists have imagined. I don't really see the need for it. One of the proofs that scientists used to say that Einstein's Theory was true was the perihelion of Mercury.

The issue isn't if SR is 'true'. The issue is what you
claim "modern science" (SR/GR) predicts.

What SR predict isn't a matter of opinion, it is a matter of fact!
And you are simply wrong!

What I responded to is this statement of yours:
Robert Winn wrote:
| According to modern interpretation of science,
| Galileo's principle of equivalence no longer applies.
| If we drop a ball from the ceiling of an airplane that
| is flying, the ball is falling faster in the frame of
| reference of the airplane than in the frame of reference
| of the ground because the clock in the airplane is slower.

So you believe that "according to modern interpretation of science"
(SR), your speed relative to an aircraft can make the clock in
the aircraft run slow.
That is of course an idiotic idea. An arbitrary observer's
speed relative to the airplane can't affect the clock in the aircraft
or the speed of the falling ball measured with that clock.

But the speed of the observer can affect the observer's observations
of the clock and the speed of the ball. In our case, the ground
observer's speed relative to the aircraft will make the clock in
the aircraft appear to run slow, and the vertical speed of the ball
will appear to be slow.

And the aircraft's speed relative to the clock on the ground
will make the clock on the ground appear to run slow when observed
in the plain.

This should be obvious to any thinking person:
An observer's state of motion cant affect the observed object,
but the observers state of motion can affect the observer's
observations of the observed object.

If SR said otherwise, you would never have heard of it,
because it would be inconsistent and dead.

<snip nonsense>

Scientists are paid trillions of dollars from governments to say that Einstein's equations are correct. I do not really see any reason to think anything is going to change any time soon.

To whom do I send the bill? :-D

--
Paul

https://paulba.no/

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Paul B. Andersen on Mon Oct 16 20:51:28 2023
On Monday, October 16, 2023 at 1:24:22 PM UTC-7, Paul B. Andersen wrote:
Den 16.10.2023 04:54, skrev Robert Winn:
On Sunday, October 15, 2023 at 1:53:29 PM UTC-7, Paul B. Andersen wrote:

I realize now that most of what I have written in this thread
is very imprecise, and something is plain wrong.
(I tend to be a bit too fast sometimes.)

The big error I made was too claim that the vertical speed of
the ball would be the same when measured in the two frames.
That is wrong. The vertical speed in the ground frame is
slower by 1/γ compared to the airplane frame.

Which was what Robert claimed, but for the wrong reason.
The clock in the airplane and the clock on the ground run
at the same rate.

So let me start from the beginning, and do it properly this time.
There may be typos, but not too many, I hope.

First, remember what a frame of reference is.
The concept you must know is coordinate time.
You can image a three dimensional array of clocks.
These clocks are synchronous, which mean that at
any time all the clocks are showing the same.

All frames of reference are equal, and since we
use second as the time unit, the clocks (and coordinate time)
are SI-clocks. And since the clocks in all frames of reference
are equal SI-clocks, THEY ALL RUN AT THE SAME RATE.

Let's call the rest frame of the airplane K'(t',x',y'),
and let's call the ground frame K(t,x,y).

We have clocks showing coordinate time at the origin of the frames.

The events of interest is E₀, the ball is dropped from the ceiling,
and E₁, the ball hits the floor in the airplane.

(use fixed width font!)

At E₀ we have:
y,y'
|
o-h
|
|
K':-------------------|--------------------> x' -> v
0
K :-------------------|---------------------> x
0
Fig. 1.
---------
The x-axis and the x'-axis are coincident.
At t = t' = 0 The origins and the y-axis and the y'-axis are
also coincident. We choose z = z' = 0 for all t, and forget the z axis. >>
The coordinates of the ball at event E₀ are:
In K: t = t₀ = 0 s, x = x₀ = 0 m, y = h
In K': t' = t₀' = 0 s, x' = x₀' = 0 m, y' = h
______

At E₁ we have:

When the ball hits the floor, t' = t₁' = √(2h/g)
and x' = x₁' = 0 m

y y'
| |
| |
| |
| |
K':----------|----------o--------> x' -> v
x₂' 0
K :----------|----------|---------> x
0 x₁
Fig. 2.
---------

So the coordinates of the ball at the event E₁ are:
In K':
t' = t₁' = √(2h/g)
x' = x₁' = 0 m
In K:
t₁ = γ(t₁'+ v⋅x₁'/c²) = γ⋅√(2h/g)
x₁ = γ(x₁'+ v⋅t₁') = γ⋅v⋅√(2h/g)

We can find the velocity of the ball:
=====================================
In the airplane frame:
----------------------
The horizontal velocity component is Vh' = x₁'/t₁' = 0
The vertical velocity component: Vv'(t') = gt'
The vertical velocity component when the ball hits the floor:
Vv'(t₁') = gt₁'= √(2gh)

In the ground frame:
---------------------
The horizontal velocity component is
Vh = x₁/t₁ = γ⋅v⋅√(2h/g)/γ⋅√(2h/g) = v

The acceleration g of the ball transformed to to K is a = g/γ²
The vertical velocity component: Vv(t) = a⋅t = (g/γ²)⋅t
The vertical velocity component when the ball hit the floor:
Vv(t₁) = (g/γ²)⋅t₁ = (g/γ²)⋅γ⋅√(2h/g) = √(2gh)/γ

The apparent rate of clocks in K' as observed in K.
===================================================
When clock at the origin of K' shows t₀' = 0, it will
be adjacent to a coordinate clock in K showing t₀ = 0.
When this clock shows t₁' = √(2h/g) it will be adjacent
to a coordinate clock in K showing γ⋅√(2h/g).

So as observed from K, the rate of the clock in K' appears to be:
f' = (t₁'-t₀')/(t₁-t₀) = √(2h/g)/γ⋅√(2h/g) = 1/γ

Note that (t₁'-t₀') is a proper time because both t₀' and t₁'
are read off the same clock, while t₀ and t₁ are read off two
different coordinate clocks

The moving clock appears to run slow.

The apparent rate of clocks in K as observed in K'.
===================================================
When clock at the origin of K shows t₀ = 0, it will
be adjacent to a coordinate clock in K showing t₀' = 0.

See fig2:
We can define an event E₂:
When this clock at x₂ = 0 shows t₂ = √(2h/g) it will be adjacent
to a coordinate clock in K'.
The coordinates in K' are:
t₂' = γ(t₂ - v⋅x₂/c²) = γ⋅√(2h/g)
x₂' = γ(x₂ - v⋅t₂) = -γ⋅v⋅√(2h/g)

So as observed from K, the rate of the clock in K' appears to be:
f = (t₂-t₀)/(t₂'-t₀') = √(2h/g)/γ⋅√(2h/g) = 1/γ

Note that (t₂-t₀) is a proper time because both t₀ and t₂
are read off the same clock, while t₀' and t₂' are read off two
different coordinate clocks

The moving clock appears to run slow.

I didn't really write this for you, Robert.
I know you are not able to read it.

But since these posts are archived, I wanted to correct
the errors I wrote earlier in this thread.

Mutual time dilation!

Well, that is just another miracle that scientists have imagined. I don't really see the need for it. One of the proofs that scientists used to say that Einstein's Theory was true was the perihelion of Mercury.
The issue isn't if SR is 'true'. The issue is what you
claim "modern science" (SR/GR) predicts.

What SR predict isn't a matter of opinion, it is a matter of fact!
And you are simply wrong!

What I responded to is this statement of yours:
Robert Winn wrote:
| According to modern interpretation of science,
| Galileo's principle of equivalence no longer applies.
| If we drop a ball from the ceiling of an airplane that
| is flying, the ball is falling faster in the frame of
| reference of the airplane than in the frame of reference
| of the ground because the clock in the airplane is slower.
So you believe that "according to modern interpretation of science"
(SR), your speed relative to an aircraft can make the clock in
the aircraft run slow.
The clock in the airplane does have an actual rate. Einstein said the clock was slower than a clock on the ground. Scientists put clocks in airplanes making transcontinental flights and said that when the clocks arrived at their destinations, they
showed less time than an identical clock on the ground. Then Hafele and Keating put clocks in jet airplanes that were going from one continent to another and said if the airplane went around the world one way, the clock on the airplane would show less
time than a clock on the ground, if it went around the world the other way, it showed more time. None of that makes any difference to the Galilean transformation equations. Whatever the faster or slower clock reads, you just make another set of
Galilean transformation equations with different variables for time and velocity than was used for the clock on the ground, and the distances remain the same. Einstein was the one who set up the problem. He said the clock on the airplane was slower, so
that is the way I worked the problem.
That is of course an idiotic idea. An arbitrary observer's
speed relative to the airplane can't affect the clock in the aircraft
or the speed of the falling ball measured with that clock.

But the speed of the observer can affect the observer's observations
of the clock and the speed of the ball. In our case, the ground
observer's speed relative to the aircraft will make the clock in
the aircraft appear to run slow, and the vertical speed of the ball
will appear to be slow.
Einstein did not say the clock appeared slower. He said it was slower. Then he gave an equation that showed what the time of the slower clock would be when the time of the clock on the ground was t. Then the scientists who put clocks in airplanes
making transcontinental flights said the clocks showed less time after their flights than a clock on the ground. Then Hafele and Keating said that if they flew clocks in jet airplanes around the world one way, they showed less time than a clock on the
ground. If they flew a clock around the world the other way, it showed more time than a clock on the ground. So let's just go with Einstein's original idea, that a moving clock would be slower.
OK, then according to Einstein, we have a clock on an airplane that shows (t-vx/c^2)/sqrt(1-v^2/c^2) when a clock on the ground shows t. The airplane travels 100 miles. The pilot of the airplane looks at his clock and sees that he traveled 100 miles in
a time of (t-vx/c^2)/sqrt(1-v^2/c^2). The observer on the ground sees that the airplane traveled a distance of 100 miles in a time of t. 100/[(t-vx/c^2)/sqrt(1-v^2/c^2)] is a faster speed than 100/t because t is more time than (t-vx/c^2)/sqrt(1-v^2/
c^2). But the Lorentz equations show that the observer on the ground and the pilot of the airplane get the same speed for the airplane, which, as I pointed out before, is a mathematical description of a miracle that scientists have imagined unto
themselves. We common people live in something called reality where if the pilot of the airplane has a slower clock, he gets a faster speed for the airplane.
The same principle is true with regard to the ball. If the ball falls ten feet according to the time of the clock on the ground, its average speed is 10/t. If it falls ten feet according to the time of the clock on the airplane, its average speed is 10/
[(t-vx/c^2)/sqrt(1-v^2/c^2)], a faster speed because t is more time than (t-vx/c^2)/sqrt(1-v^2/c^2).

And the aircraft's speed relative to the clock on the ground
will make the clock on the ground appear to run slow when observed
in the plain.

plane.
No, Paul, what I was pointing out is that there is something called reality. We common people live in reality. If the clock in the airplane is slower, it is slower in both frames of reference, and the pilot of the airplane gets a faster speed for the
airplane in both frames of reference.
This should be obvious to any thinking person:
An observer's state of motion cant affect the observed object,
but the observers state of motion can affect the observer's
observations of the observed object.

If SR said otherwise, you would never have heard of it,
because it would be inconsistent and dead.

So you are saying that anyone who believes that the pilot gets a faster speed for the airplane is dead? I think you would have to kill a lot of people before that would be true.

<snip nonsense>
Scientists are paid trillions of dollars from governments to say that Einstein's equations are correct. I do not really see any reason to think anything is going to change any time soon.
To whom do I send the bill? :-D
Send it to Congress. They will send you some money. Just tell them you believe in Einstein's theory.
Tell them that you believe that the battle to save Einstein's theory is as important as the war in Ukraine or the war between Israel and Hamas.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to Paul B. Andersen on Mon Oct 16 21:22:44 2023
On Monday, 16 October 2023 at 22:24:22 UTC+2, Paul B. Andersen wrote:
Den 16.10.2023 04:54, skrev Robert Winn:
On Sunday, October 15, 2023 at 1:53:29 PM UTC-7, Paul B. Andersen wrote:

I realize now that most of what I have written in this thread
is very imprecise, and something is plain wrong.
(I tend to be a bit too fast sometimes.)

The big error I made was too claim that the vertical speed of
the ball would be the same when measured in the two frames.
That is wrong. The vertical speed in the ground frame is
slower by 1/γ compared to the airplane frame.

Which was what Robert claimed, but for the wrong reason.
The clock in the airplane and the clock on the ground run
at the same rate.

So let me start from the beginning, and do it properly this time.
There may be typos, but not too many, I hope.

First, remember what a frame of reference is.
The concept you must know is coordinate time.
You can image a three dimensional array of clocks.
These clocks are synchronous, which mean that at
any time all the clocks are showing the same.

All frames of reference are equal, and since we
use second as the time unit, the clocks (and coordinate time)
are SI-clocks. And since the clocks in all frames of reference
are equal SI-clocks, THEY ALL RUN AT THE SAME RATE.

Let's call the rest frame of the airplane K'(t',x',y'),
and let's call the ground frame K(t,x,y).

We have clocks showing coordinate time at the origin of the frames.

The events of interest is E₀, the ball is dropped from the ceiling,
and E₁, the ball hits the floor in the airplane.

(use fixed width font!)

At E₀ we have:
y,y'
|
o-h
|
|
K':-------------------|--------------------> x' -> v
0
K :-------------------|---------------------> x
0
Fig. 1.
---------
The x-axis and the x'-axis are coincident.
At t = t' = 0 The origins and the y-axis and the y'-axis are
also coincident. We choose z = z' = 0 for all t, and forget the z axis. >>
The coordinates of the ball at event E₀ are:
In K: t = t₀ = 0 s, x = x₀ = 0 m, y = h
In K': t' = t₀' = 0 s, x' = x₀' = 0 m, y' = h
______

At E₁ we have:

When the ball hits the floor, t' = t₁' = √(2h/g)
and x' = x₁' = 0 m

y y'
| |
| |
| |
| |
K':----------|----------o--------> x' -> v
x₂' 0
K :----------|----------|---------> x
0 x₁
Fig. 2.
---------

So the coordinates of the ball at the event E₁ are:
In K':
t' = t₁' = √(2h/g)
x' = x₁' = 0 m
In K:
t₁ = γ(t₁'+ v⋅x₁'/c²) = γ⋅√(2h/g)
x₁ = γ(x₁'+ v⋅t₁') = γ⋅v⋅√(2h/g)

We can find the velocity of the ball:
=====================================
In the airplane frame:
----------------------
The horizontal velocity component is Vh' = x₁'/t₁' = 0
The vertical velocity component: Vv'(t') = gt'
The vertical velocity component when the ball hits the floor:
Vv'(t₁') = gt₁'= √(2gh)

In the ground frame:
---------------------
The horizontal velocity component is
Vh = x₁/t₁ = γ⋅v⋅√(2h/g)/γ⋅√(2h/g) = v

The acceleration g of the ball transformed to to K is a = g/γ²
The vertical velocity component: Vv(t) = a⋅t = (g/γ²)⋅t
The vertical velocity component when the ball hit the floor:
Vv(t₁) = (g/γ²)⋅t₁ = (g/γ²)⋅γ⋅√(2h/g) = √(2gh)/γ

The apparent rate of clocks in K' as observed in K.
===================================================
When clock at the origin of K' shows t₀' = 0, it will
be adjacent to a coordinate clock in K showing t₀ = 0.
When this clock shows t₁' = √(2h/g) it will be adjacent
to a coordinate clock in K showing γ⋅√(2h/g).

So as observed from K, the rate of the clock in K' appears to be:
f' = (t₁'-t₀')/(t₁-t₀) = √(2h/g)/γ⋅√(2h/g) = 1/γ

Note that (t₁'-t₀') is a proper time because both t₀' and t₁'
are read off the same clock, while t₀ and t₁ are read off two
different coordinate clocks

The moving clock appears to run slow.

The apparent rate of clocks in K as observed in K'.
===================================================
When clock at the origin of K shows t₀ = 0, it will
be adjacent to a coordinate clock in K showing t₀' = 0.

See fig2:
We can define an event E₂:
When this clock at x₂ = 0 shows t₂ = √(2h/g) it will be adjacent
to a coordinate clock in K'.
The coordinates in K' are:
t₂' = γ(t₂ - v⋅x₂/c²) = γ⋅√(2h/g)
x₂' = γ(x₂ - v⋅t₂) = -γ⋅v⋅√(2h/g)

So as observed from K, the rate of the clock in K' appears to be:
f = (t₂-t₀)/(t₂'-t₀') = √(2h/g)/γ⋅√(2h/g) = 1/γ

Note that (t₂-t₀) is a proper time because both t₀ and t₂
are read off the same clock, while t₀' and t₂' are read off two
different coordinate clocks

The moving clock appears to run slow.

I didn't really write this for you, Robert.
I know you are not able to read it.

But since these posts are archived, I wanted to correct
the errors I wrote earlier in this thread.

Mutual time dilation!

Well, that is just another miracle that scientists have imagined. I don't really see the need for it. One of the proofs that scientists used to say that Einstein's Theory was true was the perihelion of Mercury.
The issue isn't if SR is 'true'. The issue is what you
claim "modern science" (SR/GR) predicts.

What SR predict isn't a matter of opinion, it is a matter of fact!

Only such an idiot can believe such a nonsensical lie.

But the speed of the observer can affect the observer's observations
of the clock and the speed of the ball. In our case, the ground
observer's speed relative to the aircraft will make the clock in
the aircraft appear to run slow, and the vertical speed of the ball
will appear to be slow.

And the aircraft's speed relative to the clock on the ground
will make the clock on the ground appear to run slow when observed
in the plain.

We have GPS now, so anyone can check your tales have
nothing in common with real observers, real clocks, real time.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Maciej Wozniak@21:1/5 to Paul B. Andersen on Tue Oct 17 06:06:28 2023
On Tuesday, 17 October 2023 at 14:58:32 UTC+2, Paul B. Andersen wrote:

In the H&K experiment the clock in the west going aeroplane
gained 275 ns on the clock on the ground, while the clock in
the east going aeroplane lost 40 ns on the ground clock.

The toys of your bunch of idiots may act as stupid
as you want them to, the serious clocks don't. Anyone
can check GPS.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Paul B. Andersen@21:1/5 to All on Tue Oct 17 14:59:27 2023
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--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Paul B. Andersen on Tue Oct 17 09:29:03 2023
On Tuesday, October 17, 2023 at 5:58:32 AM UTC-7, Paul B. Andersen wrote:
Den 17.10.2023 05:51, skrev Robert Winn:
The clock in the airplane does have an actual rate. Einstein said the clock was slower than a clock on the ground. Scientists put clocks in airplanes making transcontinental flights and said that when the clocks arrived at their destinations, they
showed less time than an identical clock on the ground. Then Hafele and Keating put clocks in jet airplanes that were going from one continent to another and said if the airplane went around the world one way, the clock on the airplane would show less
time than a clock on the ground, if it went around the world the other way, it showed more time. None of that makes any difference to the Galilean transformation equations. Whatever the faster or slower clock reads, you just make another set of Galilean
transformation equations with different variables for time and velocity than was used for the clock on the ground, and the distances remain the same. Einstein was the one who set up the problem. He said the clock on the airplane was slower, so that is
the way I worked the problem.
https://paulba.no/paper/Hafele_Keating.pdf

In the H&K experiment the clock in the west going aeroplane
gained 275 ns on the clock on the ground, while the clock in
the east going aeroplane lost 40 ns on the ground clock.

Did Einstein say that a clock in an east going airplane
was slower than a clock on the ground, but a clock in
a west going airplane was faster than a clock on the ground?

No, he said neither.

https://paulba.no/pdf/H&K_like.pdf
All three clocks are equal, and run at their normal rate.

Look:
One clock is travelling westwards, another clock is travelling
eastwards, both clocks are travelling at the same altitude,
and at the same constant speed relative to the ground.
The third clock is stationary at the ground.

The clocks are Cs clocks, and the reference 'oscillator' is
the photon associated with the hyperfine transition of the Cs atom.
The frequency of the photon is 9192631770 Hz.
In an atomic clock the Cs atoms are free-falling.
So how do you think that the Cs atom can 'feel' if it is moving
westwards or eastwards or is stationary, and can 'feel' at
which altitude it is, so that it can change it's rate accordingly?

It can't. So the frequency of the 'oscillator' is always 9192631770 Hz.

What you and other cranks can't understand is:
When two equal clocks which always run at their proper rate
travel between the same two events, along different paths
through space-time, their proper time (what the clocks show)
may be different.

And if you think this is impossible, the Hafele & Keating
is the experiment that shows it is possible.
Einstein did not say the clock appeared slower. He said it was slower.
It is quite common that physicists say that that a clock at higher
altitude is running faster, or that a moving clock is running slower.
But this is only sloppy language, they know that the clocks are running
at the same rate, but are omitting the "apparent", or "as observed
from the ground frame". They assume that the readers will know this,
which they do if they know the physics of the last century.

But cranks stuck in the 19. century won't.

https://paulba.no/pdf/Mutual_time_dilation.pdf

Einstein knew about mutual time dilation.

Do your really think that he meant that two clocks
both can run slower than the other?

--
Paul

https://paulba.no/

Einstein definitely said the moving clock was slower. Read his book. Then scientists claimed to have experimented with cesium clocks on airplanes showing that one clock showed less time than the other. Now as an ignorant common person, I have no way
of knowing exactly what scientists lie about with regard to their experiments. I just work the math that would describe what they are saying, and it all relates back to what I originally thought back in high school. A person with a slower clock gets
faster speeds for anything moving if he uses the time of that clock. As I said, Isaac Newton was fairly good at mathematics, and he would not work the problem the way Einstein did. He would have worked it the way I did if someone had told him, We have
proven by experiment that a moving clock is slower. Newton used the Galilean transformation equations, which show that if a clock in a frame of reference is slower, it does not appear faster from the other frame of reference or any of these other
miracles that scientists tell us about.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Paul B. Andersen@21:1/5 to All on Tue Oct 17 19:23:39 2023
Den 17.10.2023 18:29, skrev Robert Winn:
On Tuesday, October 17, 2023 at 5:58:32 AM UTC-7, Paul B. Andersen wrote:
Den 17.10.2023 05:51, skrev Robert Winn:

Einstein did not say the clock appeared slower. He said it was slower
It is quite common that physicists say that that a clock at higher
altitude is running faster, or that a moving clock is running slower.
But this is only sloppy language, they know that the clocks are running
at the same rate, but are omitting the "apparent", or "as observed
from the ground frame". They assume that the readers will know this,
which they do if they know the physics of the last century.

But cranks stuck in the 19. century won't.

Einstein definitely said the moving clock was slower.

Enough now, Robert.

--
Paul

https://paulba.no/

--- SoupGate-Win32 v1.05
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• From Kevin Aylward@21:1/5 to All on Tue Oct 17 19:31:29 2023
"Mike Fontenot" wrote in message news:9201e490-ae72-1e28-829b-46a885aadb9e@comcast.net...

Only if he is an idiot. Any twin who understands relativity KNOWS the
home twin (she) does not age instantaneously, [...]

Not true. According to HIM, she DOES age instantaneously during his >instantaneous turnaround. There is no other possibility FOR HIM.

A correct resolution of the twins paradox, without acceleration or frame switching is here:

-- Kevin Aylward
http://www.kevinaylward.co.uk/gr/index.html http://www.kevinaylward.co.uk/qm/index.html

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• From Mike Fontenot@21:1/5 to Kevin Aylward on Tue Oct 17 13:14:42 2023
On 10/17/23 12:31 PM, Kevin Aylward wrote:

A correct resolution of the twins paradox, without acceleration or frame switching is here:

That link you provided sounds like complete gibberish to me.

The explanation of the twin "paradox" is simple: The two twins MUST
agree about their respective ages at their reunion, and each twin,
whenever they are NOT accelerating, correctly concludes (via the time
dilation equation (TDE)) that the other twin is ageing more slowly, by
the gamma factor

gamma = 1 / [sqrt { 1 - ( v * v ) } ] .

The only way that can be true is if the traveling twin (he) concludes
that the home twin (she) instantaneously ages by a given, definite large
amount during his instantaneous velocity change.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Robert Winn@21:1/5 to Paul B. Andersen on Tue Oct 17 15:34:49 2023
On Tuesday, October 17, 2023 at 10:22:43 AM UTC-7, Paul B. Andersen wrote:
Den 17.10.2023 18:29, skrev Robert Winn:
On Tuesday, October 17, 2023 at 5:58:32 AM UTC-7, Paul B. Andersen wrote:
Den 17.10.2023 05:51, skrev Robert Winn:

Einstein did not say the clock appeared slower. He said it was slower
It is quite common that physicists say that that a clock at higher
altitude is running faster, or that a moving clock is running slower.
But this is only sloppy language, they know that the clocks are running >> at the same rate, but are omitting the "apparent", or "as observed
from the ground frame". They assume that the readers will know this,
which they do if they know the physics of the last century.

But cranks stuck in the 19. century won't.

Einstein definitely said the moving clock was slower.
Enough now, Robert.

--
Paul

https://paulba.no/
Let us now consider a seconds-clock which is permanently situated at the origin (x'=0) of K'. t' =0 and t'=`1 are two successive ticks of this clock. The first and fourth equations of the Lorentz transformation give for these two ticks t=0 and t' = 1/
sqrt(1-v^2/c^2)As judged from K, the clock is moving with the velocity v; as judged from this reference-body, the time which elapses between two strokes of the clock is not one second, but 1/sqrt(1-v^2/c^2) seconds, a somewhat larger time. As a
consequence of its motion the clock goes more slowly than when at rest. (Relativity, the Special and General Theories, A. Einstein)

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• From Maciej Wozniak@21:1/5 to All on Tue Oct 17 22:33:55 2023
On Tuesday, October 17, 2023 at 5:58:32 AM UTC-7, Paul B. Andersen wrote:
Den 17.10.2023 05:51, skrev Robert Winn:

Einstein did not say the clock appeared slower. He said it was slower
It is quite common that physicists say that that a clock at higher
altitude is running faster, or that a moving clock is running slower.
But this is only sloppy language, they know that the clocks are running >> at the same rate

It's just that even they are not stupid enough to stick to
their absurd beliefs.

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• From Lou@21:1/5 to Paul B. Andersen on Wed Oct 18 07:00:07 2023
On Monday, 16 October 2023 at 20:07:55 UTC+1, Paul B. Andersen wrote:
Den 16.10.2023 14:54, skrev Lou:

I would ignore Einsteins “predictions” for the mercury anomalous preccession.
He only knew the amount for mercury and fiddled his formula to match that observed.
Notice his formula doesn’t do so well for other planets. A fact relativists
like to ignore.
See:
Chapter 3. Equation (8), Table 3.

Mercury 42.98"/century
To start with Mars preccesion rate is very unstable and is hard to calculate.
Equation (8) works equally well for all planets.
Mars: 1.35"/century
Which is why they pretend Mars preccession rate is 1.3 arc seconds per C When it probably is more like 2.5
The other planets preccesion rates fare less well under GR
Venus observed is 8 (GR predicted 8.6)
Earth observed is 5 (GR predicts 3.8)
https://arxiv.org/pdf/0802.0176.pdf
See Table 1:
Venus observed is 8.6247"/century (GR predicted 8.6247"/century)
Earth observed is 3.8387"/century (GR predicts 3.8387"/century)

However if one uses a more correct classical formula r 1/(r+3R)^2 based on perehilion
not semi major axis as Albert incorrectly did. Then classical theory predicts more accurately
than GR. {where r is perehilion distance and R is radius of sun}
Mercury:
Perihelion distance r = 4.60011E10 m
Solar radius R = 696340E3 m

Planet. Obs.— GR ——Classical
Merc—-43.1——43.5 —-43.24
Venus—8———-8.6——-8.33
Earth— 5———- 3.87—-4.49

If the equation is r/(r+3R)^2 we get
the number 1.9890E-11 1/m

What does this number mean?
How do you get the numbers below?
Is the equation wrong?
In that case, what should it be?

As follows:
Planet. Classical
Merc. 43.24
V. 8.33
E. 4.49
Sorry lost in translation from paper to google post
You will get 4.324 x 10-16 for mercury
And if you calculate all 4 planets you will see the progression clearly.

And where I got *observed*...
Table 1 ‘observed’
43.1000 ± 0.5000 mercury
8.0000 ± 5.0000 Venus
5.0000 ± 1.0000 earth
From:
G Nyambuya On the perehilion progression of planetary orbits Oxford Academic (Mon. Not. R. Astron. Soc. 403, 1381–1391 (2010) doi:10.1111/j.1365-2966.2009.16196.x

--- SoupGate-Win32 v1.05
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• From Dono.@21:1/5 to Lou on Wed Oct 18 08:25:56 2023
On Wednesday, October 18, 2023 at 7:00:14 AM UTC-7, Lou wrote:

https://arxiv.org/pdf/0802.0176.pdf

Planet. Obs.— GR ——Classical
Merc—-43.1——43.5 —-43.24
Venus—8———-8.6——-8.33
Earth— 5———- 3.87—-4.49
If the equation is r/(r+3R)^2 we get
the number 1.9890E-11 1/m
Abjit Biswas is a well known crank, LooLoo. The third column is a fake.

--- SoupGate-Win32 v1.05
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• From Lou@21:1/5 to Dono. on Wed Oct 18 10:16:22 2023
On Wednesday, 18 October 2023 at 16:25:59 UTC+1, Dono. wrote:
On Wednesday, October 18, 2023 at 7:00:14 AM UTC-7, Lou wrote:

https://arxiv.org/pdf/0802.0176.pdf
Planet. Obs.— GR ——Classical
Merc—-43.1——43.5 —-43.24
Venus—8———-8.6——-8.33
Earth— 5———- 3.87—-4.49
If the equation is r/(r+3R)^2 we get
the number 1.9890E-11 1/m
Abjit Biswas is a well known crank, LooLoo. The third column is a fake.

Maybe. But then again I didnt cite that paper in my post.
You did. And pretended it was me.
Second column is the fake.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Paul B. Andersen@21:1/5 to All on Wed Oct 18 20:31:47 2023
Den 17.10.2023 21:14, skrev Mike Fontenot:
On 10/17/23 12:31 PM, Kevin Aylward wrote:

A correct resolution of the twins paradox, without acceleration or
frame switching is here:

That link you provided sounds like complete gibberish to me.

The explanation of the twin "paradox" is simple:  The two twins MUST
agree about their respective ages at their reunion, and each twin,
whenever they are NOT accelerating, correctly concludes (via the time dilation equation (TDE)) that the other twin is ageing more slowly, by
the gamma factor

gamma = 1 / [sqrt { 1  -  ( v * v ) } ] .

The only way that can be true is if the traveling twin (he) concludes
that the home twin (she) instantaneously ages by a given, definite large amount during his instantaneous velocity change.

Right. That's because the acceleration at turnaround is infinite.
With a finite acceleration it becomes clear what's happening.

My simulation of the "twin paradox" make it possible to
see what happens from both twins point of view.

https://paulba.no/twins.html

Below is a screenshot of a run of the simulation.
The travelling twin (B) is accelerating (~2g) in the beginning,
at the turnaround and at the end so that she is stationary
relative to the home twin (A) when she is back.

The plots show the travelling twins view.
The plot "Clock A as observed by B " (lower left)
shows the ageing of A as observed by B.
Note the fast (steep) ageing when the rocket is burning at turnaround.
With instant change of velocity, the plot would be vertical (instant)
at turnaround.

https://paulba.no/temp/Twins_simulation_run.pdf

https://paulba.no/pdf/TwinsByMetric.pdf

--
Paul

https://paulba.no/

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• From Dono.@21:1/5 to Lou on Wed Oct 18 12:29:58 2023
On Wednesday, October 18, 2023 at 10:16:24 AM UTC-7, Lou wrote:
On Wednesday, 18 October 2023 at 16:25:59 UTC+1, Dono. wrote:
On Wednesday, October 18, 2023 at 7:00:14 AM UTC-7, Lou wrote:

https://arxiv.org/pdf/0802.0176.pdf
Planet. Obs.— GR ——Classical
Merc—-43.1——43.5 —-43.24
Venus—8———-8.6——-8.33
Earth— 5———- 3.87—-4.49
If the equation is r/(r+3R)^2 we get
the number 1.9890E-11 1/m
Abjit Biswas is a well known crank, LooLoo. The third column is a fake.
Maybe. But then again I didnt cite that paper in my post.

LooLoo

Look at your post, you are getting nuttier and nuttier.

--- SoupGate-Win32 v1.05
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• From Mike Fontenot@21:1/5 to Paul Andersen on Wed Oct 18 13:30:51 2023
I (Mike Fontenot) wrote:

The explanation of the twin "paradox" is simple:  The two twins MUST
agree about their respective ages at their reunion, and each twin,
whenever they are NOT accelerating, correctly concludes (via the time
dilation equation (TDE)) that the other twin is ageing more slowly, by
the gamma factor

gamma = 1 / [sqrt { 1  -  ( v * v ) } ] .

The only way that can be true is if the traveling twin (he) concludes
that the home twin (she) instantaneously ages by a given, definite
large amount during his instantaneous velocity change.

Then, Paul Andersen wrote:

Right. That's because the acceleration at turnaround is infinite.
With a finite acceleration it becomes clear what's happening.

The link below is to a paper I wrote long ago (now contained in a viXra
paper), that gives both the finite acceleration solution and the
instantaneous turnaround solution:

https://vixra.org/pdf/2106.0122v1.pdf

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Paul B. Andersen@21:1/5 to All on Wed Oct 18 22:38:53 2023
Den 18.10.2023 16:00, skrev Lou:
On Monday, 16 October 2023 at 20:07:55 UTC+1, Paul B. Andersen wrote:
Den 16.10.2023 14:54, skrev Lou:

I would ignore Einsteins “predictions” for the mercury anomalous preccession.
He only knew the amount for mercury and fiddled his formula to match that >>> observed.
Notice his formula doesn’t do so well for other planets. A fact relativists
like to ignore.
Which is why they pretend Mars preccession rate is 1.3 arc seconds per C >>> When it probably is more like 2.5
The other planets preccesion rates fare less well under GR
Venus observed is 8 (GR predicted 8.6)
Earth observed is 5 (GR predicts 3.8)
However if one uses a more correct classical formula r 1/(r+3R)^2 based on perehilion
not semi major axis as Albert incorrectly did. Then classical theory predicts more accurately
than GR. {where r is perehilion distance and R is radius of sun}

Mercury:
Perihelion distance r = 4.60011E10 m
Solar radius R = 696340E3 m

If the equation is r/(r+3R)^2 we get
the number 1.9890E-11 1/m

What does this number mean?
How do you get the numbers below?
Is the equation wrong?
In that case, what should it be?

Sorry lost in translation from paper to google post
You will get 4.324 x 10-16 for mercury

If r and R are in km

And if you calculate all 4 planets you will see the progression clearly.

Quite.
Mercury = 4.3240E-16 1/km²
Venus = 8.3302E-17 1/km²
Earth = 4.4929E-17 1/km²
Mars = 2.2951E-17 1/km²

So if 4.3240E-16 means that Mercury's perihelion advances
43.3240" during 415.2 orbits.
Venus' perihelion will advance 8.3302" during 162.5 orbits
Earth's perihelion will advance 4.4929" during 100 orbits
Mars' perihelion will advance 2.2951" during 53.2 orbits

Don't you see how ridiculous this is?
An equation for the perihelion advance of a planet,
which contains orbital data for the planet, must
obviously give the advance per orbit, not per century.
And why is the radius of the Sun in the equation?
It doesn't affect the planet's orbit at all.

"More correct classical formula " indeed! :-D

How did you fail to see that the formula is nonsense?
Pure ignorance?

--
Paul

https://paulba.no/

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• From Maciej Wozniak@21:1/5 to Paul B. Andersen on Wed Oct 18 14:00:35 2023
On Wednesday, 18 October 2023 at 20:30:50 UTC+2, Paul B. Andersen wrote:
Den 17.10.2023 21:14, skrev Mike Fontenot:
On 10/17/23 12:31 PM, Kevin Aylward wrote:

A correct resolution of the twins paradox, without acceleration or
frame switching is here:

That link you provided sounds like complete gibberish to me.

The explanation of the twin "paradox" is simple: The two twins MUST
agree about their respective ages at their reunion, and each twin,
whenever they are NOT accelerating, correctly concludes (via the time dilation equation (TDE)) that the other twin is ageing more slowly, by
the gamma factor

gamma = 1 / [sqrt { 1 - ( v * v ) } ] .

The only way that can be true is if the traveling twin (he) concludes
that the home twin (she) instantaneously ages by a given, definite large amount during his instantaneous velocity change.

Right. That's because the acceleration at turnaround is infinite.
With a finite acceleration it becomes clear what's happening.

My simulation of the "twin paradox" make it possible to
see what happens from both twins point of view.

https://paulba.no/twins.html

Below is a screenshot of a run of the simulation.
The travelling twin (B) is accelerating (~2g) in the beginning,
at the turnaround and at the end so that she is stationary
relative to the home twin (A) when she is back.

Fortunately we have GPS now, so we can be
absolutely sure that all these nonsensical tales
have nothing in common with real clocks or real time.

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• From Richard Hachel@21:1/5 to All on Wed Oct 18 22:07:50 2023
Le 17/10/2023 à 21:14, Mike Fontenot a écrit :
On 10/17/23 12:31 PM, Kevin Aylward wrote:

A correct resolution of the twins paradox, without acceleration or frame
switching is here:

That link you provided sounds like complete gibberish to me.

The explanation of the twin "paradox" is simple: The two twins MUST
agree about their respective ages at their reunion, and each twin,
whenever they are NOT accelerating, correctly concludes (via the time dilation equation (TDE)) that the other twin is ageing more slowly, by
the gamma factor

gamma = 1 / [sqrt { 1 - ( v * v ) } ] .

The only way that can be true is if the traveling twin (he) concludes
that the home twin (she) instantaneously ages by a given, definite large amount during his instantaneous velocity change.

Mais vous êtes des francs malades, les mecs!

Moi, je vous le dis, vous êtes des malades!!!

R.H.

--- SoupGate-Win32 v1.05
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• From Richard Hachel@21:1/5 to All on Wed Oct 18 22:20:11 2023
Le 18/10/2023 à 20:30, "Paul B. Andersen" a écrit :
Den 17.10.2023 21:14, skrev Mike Fontenot:

Right. That's because the acceleration at turnaround is infinite.

? ? ?

WHAT you say? ? ?

With a finite acceleration it becomes clear what's happening.

My simulation of the "twin paradox" make it possible to
see what happens from both twins point of view.

Paul, you are not serious.

What I explain here is still less extravagant.

http://news2.nemoweb.net/?DataID=AP9K8PWdQ8VGGn12U_lCq6JgpEU@jntp

http://news2.nemoweb.net/?DataID=ANyXQtwi2IZvxhCVx4aCbLfLYmg@jntp

R.H.

--
Ce message a été posté avec Nemo : <http://news2.nemoweb.net/?DataID=E9IRTmbc11ZrW50PqACuU8OeM-g@jntp>

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• From Laurence Clark Crossen@21:1/5 to Paul B. Andersen on Wed Oct 18 16:10:30 2023
On Wednesday, October 18, 2023 at 1:37:56 PM UTC-7, Paul B. Andersen wrote:
Den 18.10.2023 16:00, skrev Lou:
On Monday, 16 October 2023 at 20:07:55 UTC+1, Paul B. Andersen wrote:
Den 16.10.2023 14:54, skrev Lou:

I would ignore Einsteins “predictions” for the mercury anomalous preccession.
He only knew the amount for mercury and fiddled his formula to match that
observed.
Notice his formula doesn’t do so well for other planets. A fact relativists
like to ignore.
Which is why they pretend Mars preccession rate is 1.3 arc seconds per C >>> When it probably is more like 2.5
The other planets preccesion rates fare less well under GR
Venus observed is 8 (GR predicted 8.6)
Earth observed is 5 (GR predicts 3.8)
However if one uses a more correct classical formula r 1/(r+3R)^2 based on perehilion
not semi major axis as Albert incorrectly did. Then classical theory predicts more accurately
than GR. {where r is perehilion distance and R is radius of sun}

Mercury:
Perihelion distance r = 4.60011E10 m
Solar radius R = 696340E3 m

If the equation is r/(r+3R)^2 we get
the number 1.9890E-11 1/m

What does this number mean?
How do you get the numbers below?
Is the equation wrong?
In that case, what should it be?
Sorry lost in translation from paper to google post
You will get 4.324 x 10-16 for mercury
If r and R are in km
And if you calculate all 4 planets you will see the progression clearly.
Quite.
Mercury = 4.3240E-16 1/km²
Venus = 8.3302E-17 1/km²
Earth = 4.4929E-17 1/km²
Mars = 2.2951E-17 1/km²

So if 4.3240E-16 means that Mercury's perihelion advances
43.3240" during 415.2 orbits.
Venus' perihelion will advance 8.3302" during 162.5 orbits
Earth's perihelion will advance 4.4929" during 100 orbits
Mars' perihelion will advance 2.2951" during 53.2 orbits

Don't you see how ridiculous this is?
An equation for the perihelion advance of a planet,
which contains orbital data for the planet, must
obviously give the advance per orbit, not per century.
And why is the radius of the Sun in the equation?
It doesn't affect the planet's orbit at all.

"More correct classical formula " indeed! :-D

How did you fail to see that the formula is nonsense?
Pure ignorance?

--
Paul

https://paulba.no/
There exists no purer ignorance than relativity.

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• From Lou@21:1/5 to Dono. on Thu Oct 19 01:54:00 2023
On Wednesday, 18 October 2023 at 20:30:01 UTC+1, Dono. wrote:
On Wednesday, October 18, 2023 at 10:16:24 AM UTC-7, Lou wrote:
On Wednesday, 18 October 2023 at 16:25:59 UTC+1, Dono. wrote:
On Wednesday, October 18, 2023 at 7:00:14 AM UTC-7, Lou wrote:

https://arxiv.org/pdf/0802.0176.pdf
Planet. Obs.— GR ——Classical
Merc—-43.1——43.5 —-43.24
Venus—8———-8.6——-8.33
Earth— 5———- 3.87—-4.49
If the equation is r/(r+3R)^2 we get
the number 1.9890E-11 1/m
Abjit Biswas is a well known crank, LooLoo. The third column is a fake.
Maybe. But then again I didnt cite that paper in my post.
LooLoo

Look at your post, you are getting nuttier and nuttier.

No. Try looking again at the record.
Notice that citation was made by *Paul* in his post. Which
was then inadvertantly included in my post when I replied to him
You had better tell Paul that he is citing crank papers to
substantiate relativity.
Nothing to do with me seeing as I disagree with his posts. And
the cranks he cites.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Paul B. Andersen on Thu Oct 19 01:50:31 2023
On Wednesday, 18 October 2023 at 21:37:56 UTC+1, Paul B. Andersen wrote:
Den 18.10.2023 16:00, skrev Lou:
On Monday, 16 October 2023 at 20:07:55 UTC+1, Paul B. Andersen wrote:
Den 16.10.2023 14:54, skrev Lou:

I would ignore Einsteins “predictions” for the mercury anomalous preccession.
He only knew the amount for mercury and fiddled his formula to match that
observed.
Notice his formula doesn’t do so well for other planets. A fact relativists
like to ignore.
Which is why they pretend Mars preccession rate is 1.3 arc seconds per C >>> When it probably is more like 2.5
The other planets preccesion rates fare less well under GR
Venus observed is 8 (GR predicted 8.6)
Earth observed is 5 (GR predicts 3.8)
However if one uses a more correct classical formula r 1/(r+3R)^2 based on perehilion
not semi major axis as Albert incorrectly did. Then classical theory predicts more accurately
than GR. {where r is perehilion distance and R is radius of sun}

Mercury:
Perihelion distance r = 4.60011E10 m
Solar radius R = 696340E3 m

If the equation is r/(r+3R)^2 we get
the number 1.9890E-11 1/m

What does this number mean?
How do you get the numbers below?
Is the equation wrong?
In that case, what should it be?
Sorry lost in translation from paper to google post
You will get 4.324 x 10-16 for mercury
If r and R are in km
And if you calculate all 4 planets you will see the progression clearly.
Quite.
Mercury = 4.3240E-16 1/km²
Venus = 8.3302E-17 1/km²
Earth = 4.4929E-17 1/km²
Mars = 2.2951E-17 1/km²

So if 4.3240E-16 means that Mercury's perihelion advances
43.3240" during 415.2 orbits.
Venus' perihelion will advance 8.3302" during 162.5 orbits
Earth's perihelion will advance 4.4929" during 100 orbits
Mars' perihelion will advance 2.2951" during 53.2 orbits

Don't you see how ridiculous this is?
An equation for the perihelion advance of a planet,
which contains orbital data for the planet, must
obviously give the advance per orbit, not per century.

Your logic is biased flawed and incorrect at best. Notice
Einstein didn’t even do the calculations for the preccession.
Various others did using 1/r^2. But they added to the calculation
by translating the oroginal values into various different formats
Of arcseconds, radians, per orbit per century etc.
If neccesary my original calculations for century could be
presented in any format correctly. Howso?
Because my formula isn’t based on fakery and wild goblins
in the 17th dimension as relativity is...but on a simple understanding
of the mechanism of how orbital paths are tugged at by the Sun
as they make their nearest approach to the sun.
The same happens in galaxy rotation curves.
They don’t fit Newtonian assumptions...not because of dark
matter....but because the same Newtonian formulae for orbital speed
are incorrectly placing all the mass at the theoretical Center.
And why is the radius of the Sun in the equation?
It doesn't affect the planet's orbit at all.

Why is radius of sun in the formula?
Because the source of the anomalous precession comes physically
from the planet as it makes its closest approach to the Sun.
That’s when the suns mass subtends the largest angle in the sky
and where Newton’s assumption that all the mass of the sun is
assumed to be at its Center for orbital speed and path calculations
is shown to be an incorrect assumption.
This additional 1/r^2 ( which Einstein also did but incorrectly at
the semi major axis) corrects the failure of Newton’s formulae
for orbital mechanics.

"More correct classical formula " indeed! :-D

How did you fail to see that the formula is nonsense?

A formula that accurately predicts the observations is nonsense!!
That’s nonsense Paul. You just don’t like the fact that a classical
model can predict preccesion rates more accurately than relativity.

Pure ignorance?

Sour grapes Paul. It correctly predicts the anomalous preccession.
Better than Relativity. I’ve supplied the formula, it supplies the predictions
and those are confirmed by the data.
If you don’t like it...prove its predictions are incorrect.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Paul Alsing@21:1/5 to Laurence Clark Crossen on Thu Oct 19 11:50:09 2023
On Thursday, October 19, 2023 at 11:35:59 AM UTC-7, Laurence Clark Crossen wrote:

Lou, I wonder if you have published any papers or books because I would like to read them as I agree with much of what you say and you know more about physics than I do.

Cranks supporting cranks.

"Everybody" knows more physics than you do, Larry, and "almost everybody" knows more than Lou...

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Laurence Clark Crossen@21:1/5 to Lou on Thu Oct 19 11:35:57 2023
On Wednesday, October 11, 2023 at 6:43:55 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 14:21:24 UTC+1, Paul B. Andersen wrote:
Den 10.10.2023 21:50, skrev Lou:
On Tuesday, 10 October 2023 at 19:10:11 UTC+1, Paul B. Andersen wrote:
Den 10.10.2023 14:16, skrev Lou:

No need for relativity to explain Hafael Keating.
Look at classical resonance. It has been known for centuries that
the natural resonant frequency of a system will reduce its frequency >>> if subject to an external force. So an atom, also confirmed by all observations
to date to be a resonant system, will also reduce its frequency when subject
to external force. As we see happen where less g force with altitude increases
the atoms ‘ticking’.

So clocks at higher altitude in a plane will be subject to
lower g-force and will tick faster than clocks on the ground, right?

No I’m suggesting that this horizontal force is not force due to gravity.
But a seperate force from horizontal acceleration .
Isn’t the formula for this f=ma?

I see. The horizontal force driving the plane at constant
speed through the air will give the plane a constant
horizontal acceleration. Sounds reasonable, doesn't it?

Im assuming that Hafael Keating observed that the eastward clock ticks slower.
That’s my reading of the wiki reference.
But to answer ‘why’...I assume the speed relative to the ground is the same for both
planes in the experiment. Let’s call it speed x.
The earth rotates eastward at 1600k/ hr.
Close enough.

So the zero point of reference is the earth observer travelling at 1600 k/hr relative
to the earths Center of mass.
Then relative to this earth center reference, the eastward plane travels at
1600 +x kilometers per hour. And the westward plane travels at 1600-x kilometers
per hour.

A reasonable speed for x is 800 km/h.

The east going plane travels at 2400 km/h in the ECI-frame.
The ground clock travels at 1600 km/h in the ECI-frame.
The west going plane travels at 800 km/h in the ECI-frame.

The eastward plane therefore experiences greater F than earth observer And the westward plane lesser F than the the earth observer.
And seeing as a classical model uses resonance as the explanation
for this different ticking rates of the caesium atoms natural resonant frequency.
Then the westward clock has less F from horizontal
speed than the earth observer and the westward plane has a greater F than
the earth observer. Which accounts for the 3 different rates of ticking. (Don’t forget...the earth observers ‘ticking rate’ is in part due to Gravity force
but also in part due to its rotational/ horizontal speed of 1600k/hr )

I see.
Since the east going clock runs slower than the ground clock,
and the ground clock runs slower than the west going clock,
it is obvious that it is the speed in the ECI-frame that is
the major cause for the different clock rates.

So we can conclude:
Since the speed in the ECI-frame is higher for the east going
clock than for the ground clock, the horizontal force to drive
the east going plane through the air at 800 km/h is higher than
the force to drive the ground clock through the air at 0 km/h.

Since the speed in the ECI-frame is higher for the ground clock
than for the west going clock, the horizontal force to drive
the ground clock through the air at 0 km/h is higher than the force
to drive the west going plane through the air at 800 km/h.

The east going clock is more compressed than the ground clock,
and will run slower than the ground clock.
The ground clock is more compressed than the west going clock,
and will run slower than the west going clock.

Right?

But don’t forget the Gravity force pushing you down is a seperate source of
force from the F force pushing against you as you move horizontally. I’m not sure
what word you relativists prefer but probably inertia or momentum would be appropriate?

Air drag.
Air Drag!!😂🤣You relativists. Such purveyors of BS.
Did I mention air drag?
No.
You did.
You forgot. The earth rotates.
I know in relativity land that you guys think the earth doesn’t rotate. But
sorry...it does.
In which case the eastward plane travels faster relative to the earth Center frame.
And the westward plane slower
If you knew any maths...then even if the planes speed was 800k/hr it would still
be less than the earth observers speed 1600 k/hr speed.
Which means that the eastward plane travels at a higher speed than the earth observer, and the westward plane at a lower speed than the observer.
And seeing as f=ma then the force on the westward plane is less than the earth observer and the force on the eastward plane is higher.
Which in turn means that due to f=ma the eastward travelling clock will
run slower and the westward clock will run faster than the earth observers clock
Due to mechanical resonance.
As observed in Hafael,Keating.
Do the maths Paul.
Lou, I wonder if you have published any papers or books because I would like to read them as I agree with much of what you say and you know more about physics than I do. I agree that H/K is not correctly explained by relativity and cannot be. However, I'
m not sure you are correct about the speeds of the planes. I believe you are wrong because of Galileo's concept of shared speed. Each plane shares the velocity of the source or starting point so they have the same velocity relative to the surface and to
the atmosphere. There is no difference in force.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Lou@21:1/5 to Laurence Clark Crossen on Thu Oct 19 12:52:08 2023
On Thursday, 19 October 2023 at 19:35:59 UTC+1, Laurence Clark Crossen wrote:
On Wednesday, October 11, 2023 at 6:43:55 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 14:21:24 UTC+1, Paul B. Andersen wrote:
Den 10.10.2023 21:50, skrev Lou:
On Tuesday, 10 October 2023 at 19:10:11 UTC+1, Paul B. Andersen wrote:
Den 10.10.2023 14:16, skrev Lou:

No need for relativity to explain Hafael Keating.
Look at classical resonance. It has been known for centuries that >>> the natural resonant frequency of a system will reduce its frequency >>> if subject to an external force. So an atom, also confirmed by all observations
to date to be a resonant system, will also reduce its frequency when subject
to external force. As we see happen where less g force with altitude increases
the atoms ‘ticking’.

So clocks at higher altitude in a plane will be subject to
lower g-force and will tick faster than clocks on the ground, right?

No I’m suggesting that this horizontal force is not force due to gravity.
But a seperate force from horizontal acceleration .
Isn’t the formula for this f=ma?

I see. The horizontal force driving the plane at constant
speed through the air will give the plane a constant
horizontal acceleration. Sounds reasonable, doesn't it?

Im assuming that Hafael Keating observed that the eastward clock ticks slower.
That’s my reading of the wiki reference.
But to answer ‘why’...I assume the speed relative to the ground is the same for both
planes in the experiment. Let’s call it speed x.
The earth rotates eastward at 1600k/ hr.
Close enough.

So the zero point of reference is the earth observer travelling at 1600 k/hr relative
to the earths Center of mass.
Then relative to this earth center reference, the eastward plane travels at
1600 +x kilometers per hour. And the westward plane travels at 1600-x kilometers
per hour.

A reasonable speed for x is 800 km/h.

The east going plane travels at 2400 km/h in the ECI-frame.
The ground clock travels at 1600 km/h in the ECI-frame.
The west going plane travels at 800 km/h in the ECI-frame.

The eastward plane therefore experiences greater F than earth observer And the westward plane lesser F than the the earth observer.
And seeing as a classical model uses resonance as the explanation
for this different ticking rates of the caesium atoms natural resonant frequency.
Then the westward clock has less F from horizontal
speed than the earth observer and the westward plane has a greater F than
the earth observer. Which accounts for the 3 different rates of ticking.
(Don’t forget...the earth observers ‘ticking rate’ is in part due to Gravity force
but also in part due to its rotational/ horizontal speed of 1600k/hr )

I see.
Since the east going clock runs slower than the ground clock,
and the ground clock runs slower than the west going clock,
it is obvious that it is the speed in the ECI-frame that is
the major cause for the different clock rates.

So we can conclude:
Since the speed in the ECI-frame is higher for the east going
clock than for the ground clock, the horizontal force to drive
the east going plane through the air at 800 km/h is higher than
the force to drive the ground clock through the air at 0 km/h.

Since the speed in the ECI-frame is higher for the ground clock
than for the west going clock, the horizontal force to drive
the ground clock through the air at 0 km/h is higher than the force
to drive the west going plane through the air at 800 km/h.

The east going clock is more compressed than the ground clock,
and will run slower than the ground clock.
The ground clock is more compressed than the west going clock,
and will run slower than the west going clock.

Right?

But don’t forget the Gravity force pushing you down is a seperate source of
force from the F force pushing against you as you move horizontally. I’m not sure
what word you relativists prefer but probably inertia or momentum would
be appropriate?

Air drag.
Air Drag!!😂🤣You relativists. Such purveyors of BS.
Did I mention air drag?
No.
You did.
You forgot. The earth rotates.
I know in relativity land that you guys think the earth doesn’t rotate. But
sorry...it does.
In which case the eastward plane travels faster relative to the earth Center frame.
And the westward plane slower
If you knew any maths...then even if the planes speed was 800k/hr it would still
be less than the earth observers speed 1600 k/hr speed.
Which means that the eastward plane travels at a higher speed than the earth
observer, and the westward plane at a lower speed than the observer.
And seeing as f=ma then the force on the westward plane is less than the earth observer and the force on the eastward plane is higher.
Which in turn means that due to f=ma the eastward travelling clock will run slower and the westward clock will run faster than the earth observers clock
Due to mechanical resonance.
As observed in Hafael,Keating.
Do the maths Paul.
Lou, I wonder if you have published any papers or books because I would like to read them as I agree with much of what you say and you know more about physics than I do. I agree that H/K is not correctly explained by relativity and cannot be. However,
I'm not sure you are correct about the speeds of the planes. I believe you are wrong because of Galileo's concept of shared speed. Each plane shares the velocity of the source or starting point so they have the same velocity relative to the surface and
to the atmosphere. There is no difference in force.

I have only assumed the speed of the planes. Hafael Keating dont supply
any info on plane speeds. So I looked up passenger jet speeds on Google
and it’s actually around 550 mph.(880). But this is google groups not Journal of physics. And H-K was written in 1971 so I opted for an even 800.
If I was to publish, which I don’t, seeing as the pro relativist editors
only publish Harry Potter fantasy science, I would probably have to find the exact
1971 airspeed. Problem is landing, take off, routes etc are also
all variable and complex. Notice even H-K admit relativity predictions have large
errors of 10%! (And the final total lost observed for the eastward plane is 1/3 more than
predicted by relativity!! That’s an error of 33% for Einsteins theory.
Not surprising relativists don’t mention this failure.)

My take on it is as follows: The earth rotates eastward at 1600kph.
Relative to the earth Center. (Ie if you were in solar frame watching
the earth it would be rotating at 1600kph eastward.) That HAS to be
taken into account. Especially considering it’s considered “acceleration” in current physics.
So a plane that flies at 800 kph eastward has to be travelling at 800+1600=2400 relative to the earth Center. Thats why NASA
launches rockets always to east. They use rocket thrust+ earths
1600kph rotational velocity to achieve higher orbital escape
velocity. Which means the westward plane is technically only
travelling at1600-800=800 relative to the earth center.
So relative to the earths mass that there-must be three speeds.
And using this assumption does give a good approximation to
the observed various tick rates. So my hunch is...if the theory
is based as best as possible on classical physics and gives an
accurate prediction...it must be a correct assumption I make
about the speeds and its relationship to force etc.
So...Assuming plane speed 800, and assuming that earth rotates
around in its own Center at 1600kph:

Observer rotational speed travelling east must be at 1600
East plane then has to be at 2400
West plane at 800
I can’t think of why it could be anything else. But would
be interested to see why you think it couldn’t.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Laurence Clark Crossen@21:1/5 to Robert Winn on Thu Oct 19 13:26:15 2023
On Thursday, October 12, 2023 at 10:22:02 PM UTC-7, Robert Winn wrote:
On Thursday, October 12, 2023 at 8:35:11 PM UTC-7, Laurence Clark Crossen wrote:
On Monday, October 9, 2023 at 11:27:35 AM UTC-7, Robert Winn wrote:
Back when I was in high school, our physics teacher was explaining Einstein's Special Theory of Relativity to us. He said that Einstein had proven that a moving clock is slower than a clock that is not moving. For the moving clock I imagined a
clock in a flying airplane, and for the clock that was not moving, a clock on the ground. It was obvious to me that if the pilot of the airplane had a slower clock than an observer on the ground had to time the flight of the airplane, the pilot would get
a faster speed for the airplane than an observer on the ground would get using the faster clock on the ground to time the flight of the airplane. Then I read Einstein's book on the subject and was surprised to discover that the equations of Special
Relativity show that the pilot of the airplane and the observer on the ground would get the same speed for the airplane.
x' = (x-vt)/sqrt(12-v^2/c^2)
y' = y
z' = z
t' = )t-vx/c^2)/sqrt(1-v^2/c^2)
inverse equations
x = (x' + vt')/sqrt(1-v^2/c^2)
y = y'
z = z'
t = (t' + vx')/(1-v^2/c^2)
v is the speed of the airplane and is the same speed as seen from either frame of reference. But we common people live in something called reality where if the pilot of the airplane has a slower clock, he will get a faster speed for the airplane.
Special Relativity is nothing more than a mathematical description of a miracle that scientists have imagined to themselves. What would be the correct equations for what Einstein described, a clock in a flying airplane being slower than a clock on the
ground?
Isaac Newton used Galileo's equations for relativity, as did all scientists until 1887. Newton's interpretation of relativity used absolute time, the idea that all clocks that were working correctly would agree with one another, which I believe
Einstein was trying to imitate in saying that the speed of the airplane would be seen the same from either frame of reference, but obviously, reality shows something different. The pilot of the airplane would get a faster speed for the airplane if his
clock is slower. The Galilean transformation equations are
x'=x-vt
y'=y
z'=z
t'=t
inverse equations
x = x' - v't'
y = y'
z = z'
t = t'
v' = -v
Notice the two little horizontal marks between t and t'. They are called an equal sign and mean that t and t' are the same number of seconds. If there is a clock in the moving frame of reference that shows less time than t, the time of the clock on
the ground, the time of that clock is not shown in these Galilean transformation equations. In order to show the time of the slower clock in the airplane, you would have to use another set of Galilean transformation equations with the same distances for
x and x', but with different variables for time and velocity. So suppose we say that the velocity of the airplane according to the time of the slower clock on the airplane is m' and the time of the slower clock is n'. Then we have
x = x'- m'n'
y = y'
z = z'
n = n'
inverse equations
x' = x - mn
y'=y
z'=z
n'=n
m' = -m
Since distances are the same in all of these Galilean equations, vt = -m'n'.
So the velocity of the ground relative to the airplane would be m' = -vt/n', a faster speed for the airplane if the clock in the airplane is slower than a clock on the ground. These equations describe reality as we observe it to be. They indicate
that there is no need for the miracle Einstein describes. How this relates to electromagnetism, I do not say, just that these equations describe relativity in the correct manner.
Yes, relativity does not describe physical reality. Einstein was very confused. I don't know why you can't "learn" (be indoctrinated = believe in fairy tales). It's been noticed all along by many excellent scientists who have shown that relativity is
a joke/swindle (e.g. Essen). It is amusing that they can accept "really weird" "science." Einstein had much funding behind him. I wonder what speed the satellite is really moving. People who pretend to do miracles are wizards, not scientists. Science has
no need for "miracles" (relativity).
Well, I have always wondered about this. I read Einstein's book, and he gave a fairly good explanation of the Galilean transformation equations, which were used to describe relativity before 1887. Then he said, the Galilean transformation equations
cannot describe the results of the Michelson-Morley experiment. It appeared to me that they could. What you have to do is believe what the equations say and follow the axioms of algebra. If there is an equation that says t'=t, then t' cannot be used to
represent the time of a clock that is slower than a clock that shows t. You have to use an entirely different set of Galilean transformation equations with different variables for time and velocity than v, t, and t' in the equations, x' = x - vt. and t'=
t, the same way you would use two different sets of Galilean transformation equations for time based on the rotation of earth and time based on the rotation of Mars. You may get a close approximation doing what Lorentz and Einstein did, but, why do it
that way?
But, then too, why call what these guys are doing relativity? If their equations have something to do with electromagnetic fields or electromagnetic waves, call it that, but Galileo's equations are the ones that work for relativity.
The Lorentz transformations basically negate relative motion, and that is unnecessary because the speed of light is determined by the medium and by the relative motion of the sink or observer.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Paul B. Andersen@21:1/5 to All on Thu Oct 19 22:34:26 2023
Den 19.10.2023 10:50, skrev Lou:
On Wednesday, 18 October 2023 at 21:37:56 UTC+1, Paul B. Andersen wrote:
Den 18.10.2023 16:00, skrev Lou:

Sorry lost in translation from paper to google post
You will get 4.324 x 10-16 for mercury

If r and R are in km

And if you calculate all 4 planets you will see the progression clearly.

Quite.
Mercury = 4.3240E-16 1/km²
Venus = 8.3302E-17 1/km²
Earth = 4.4929E-17 1/km²
Mars = 2.2951E-17 1/km²

So if 4.3240E-16 means that Mercury's perihelion advances
43.3240" during 415.2 orbits.
Venus' perihelion will advance 8.3302" during 162.5 orbits
Earth's perihelion will advance 4.4929" during 100 orbits
Mars' perihelion will advance 2.2951" during 53.2 orbits

Did you miss this?

You claim that your equation 1/(r+3R)² means:

Mercury:
The equation give 4.3240E-16 per square km, and you claim that
this means that the perihelion advance is 43.240" after 415.2 orbits

Venus:
The equation give 8.3302E-17 per square km, and you claim that
this means that the perihelion advance is 8.3302" after 162.5 orbits

Earth:
The equation give 4.4929E-17 per square km, and you claim that
this means that the perihelion advance is 4.4929" after 100 orbits.

Mars:
The equation give 2.2951E-17 per square km, and you claim that
this means that the perihelion advance is 2.2951" after 53.2 orbits.

So please explain how you can know how many orbits the planet
must go before the calculated advance is reached?

Don't you see how ridiculous this is?
An equation for the perihelion advance of a planet,
which contains orbital data for the planet, must
obviously give the advance per orbit, not per century.

Don't you think it is ridiculous that your equation give
the advance after a number of orbits which isn't in your formula?

Your logic is biased flawed and incorrect at best. Notice
Einstein didn’t even do the calculations for the preccession.
Various others did using 1/r^2. But they added to the calculation
by translating the oroginal values into various different formats
Of arcseconds, radians, per orbit per century etc.

Gobbledegook!
Did you try to say something about the logic that shows

This is GR's prediction for the perihelion advance of ANY planet
orbiting any star.
(Ignoring the mass of planet relative to the mass of star)

δφ = 6πGM/a(1−e²)c²
where:
G gravitational constant [m³/kg⋅s²]
M mass of star [kg]
a semi major axis of orbit [m]
e eccentricity of orbit
c speed of light

The equation shows the advance for one orbit
irrespective of which planet it is.

Increase with the mass of the star.
Decrease with the semi major axis.
Increase with the eccentricity.

the mass of the star and the eccentricity of the orbit.
And what's more, it shows the advance as a number per square km,
and the advance is after a number of orbits which are not
included in the equation. You have to guess.

Why do you guess the number of orbits per century?
Why not per decennium or millennium?
There is nothing in the equation to determine which.

And from where did you get the factor 1E17 km²⋅century per arcsec
you have to multiply the number per square km with to get "/century ?

Because my formula is based on a simple understanding
of the mechanism of how orbital paths are tugged at by the Sun
as they make their nearest approach to the sun.

I see.
You claim Newton was wrong since it according to Newton
is no perihelion advance of a single planet orbiting a star.

So which theory of physics, based on a simple understanding
of the mechanism of how orbital paths are tugged at by the Sun
as they make their nearest approach to the sun, are you
referring to?

Can you show the equations which will show how a planet will
orbit the star?

And why is the radius of the Sun in the equation?
It doesn't affect the planet's orbit at all.

Why is radius of sun in the formula?
Because the source of the anomalous precession comes physically
from the planet as it makes its closest approach to the Sun.
That’s when the suns mass subtends the largest angle in the sky
and where Newton’s assumption that all the mass of the sun is
assumed to be at its Center for orbital speed and path calculations
is shown to be an incorrect assumption.
This additional 1/r^2 ( which Einstein also did but incorrectly at
the semi major axis) corrects the failure of Newton’s formulae
for orbital mechanics.

where the radius of the star affects the orbits of planets,
but the mass of the star doesn't?

--
Paul

https://paulba.no/

--- SoupGate-Win32 v1.05
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• From Laurence Clark Crossen@21:1/5 to Robert Winn on Thu Oct 19 13:58:38 2023
On Monday, October 16, 2023 at 11:28:53 AM UTC-7, Robert Winn wrote:
On Monday, October 16, 2023 at 5:54:37 AM UTC-7, Lou wrote:
On Monday, 16 October 2023 at 03:54:57 UTC+1, Robert Winn wrote:
On Sunday, October 15, 2023 at 1:53:29 PM UTC-7, Paul B. Andersen wrote:
µmmmmDen 13.10.2023 18:25, skrev Robert Winn:
On Friday, October 13, 2023 at 4:43:32 AM UTC-7, Paul B. Andersen wrote:
Den 12.10.2023 20:04, skrev Robert Winn:
On Thursday, October 12, 2023 at 7:23:00 AM UTC-7, Paul B. Andersen wrote:

Den 10.10.2023 21:52, skrev Robert Winn:
If we drop a ball from the ceiling of an airplane that is flying, the ball is falling faster in the frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.

I realize now that most of what I have written in this thread
is very imprecise, and something is plain wrong.
(I tend to be a bit too fast sometimes.)

The big error I made was too claim that the vertical speed of
the ball would be the same when measured in the two frames.
That is wrong. The vertical speed in the ground frame is
slower by 1/γ compared to the airplane frame.

Which was what Robert claimed, but for the wrong reason.
The clock in the airplane and the clock on the ground run
at the same rate.

So let me start from the beginning, and do it properly this time. There may be typos, but not too many, I hope.

First, remember what a frame of reference is.
The concept you must know is coordinate time.
You can image a three dimensional array of clocks.
These clocks are synchronous, which mean that at
any time all the clocks are showing the same.

All frames of reference are equal, and since we
use second as the time unit, the clocks (and coordinate time)
are SI-clocks. And since the clocks in all frames of reference
are equal SI-clocks, THEY ALL RUN AT THE SAME RATE.

Let's call the rest frame of the airplane K'(t',x',y'),
and let's call the ground frame K(t,x,y).

We have clocks showing coordinate time at the origin of the frames.

The events of interest is E₀, the ball is dropped from the ceiling, and E₁, the ball hits the floor in the airplane.

(use fixed width font!)

At E₀ we have:
y,y'
|
o-h
|
|
K':-------------------|--------------------> x' -> v
0
K :-------------------|---------------------> x
0
Fig. 1.
---------
The x-axis and the x'-axis are coincident.
At t = t' = 0 The origins and the y-axis and the y'-axis are
also coincident. We choose z = z' = 0 for all t, and forget the z axis.

The coordinates of the ball at event E₀ are:
In K: t = t₀ = 0 s, x = x₀ = 0 m, y = h
In K': t' = t₀' = 0 s, x' = x₀' = 0 m, y' = h
______

At E₁ we have:

When the ball hits the floor, t' = t₁' = √(2h/g)
and x' = x₁' = 0 m

y y'
| |
| |
| |
| |
K':----------|----------o--------> x' -> v
x₂' 0
K :----------|----------|---------> x
0 x₁
Fig. 2.
---------

So the coordinates of the ball at the event E₁ are:
In K':
t' = t₁' = √(2h/g)
x' = x₁' = 0 m
In K:
t₁ = γ(t₁'+ v⋅x₁'/c²) = γ⋅√(2h/g)
x₁ = γ(x₁'+ v⋅t₁') = γ⋅v⋅√(2h/g)

We can find the velocity of the ball: =====================================
In the airplane frame:
----------------------
The horizontal velocity component is Vh' = x₁'/t₁' = 0
The vertical velocity component: Vv'(t') = gt'
The vertical velocity component when the ball hits the floor: Vv'(t₁') = gt₁'= √(2gh)

In the ground frame:
---------------------
The horizontal velocity component is
Vh = x₁/t₁ = γ⋅v⋅√(2h/g)/γ⋅√(2h/g) = v

The acceleration g of the ball transformed to to K is a = g/γ²
The vertical velocity component: Vv(t) = a⋅t = (g/γ²)⋅t
The vertical velocity component when the ball hit the floor:
Vv(t₁) = (g/γ²)⋅t₁ = (g/γ²)⋅γ⋅√(2h/g) = √(2gh)/γ

The apparent rate of clocks in K' as observed in K. ===================================================
When clock at the origin of K' shows t₀' = 0, it will
be adjacent to a coordinate clock in K showing t₀ = 0.
When this clock shows t₁' = √(2h/g) it will be adjacent
to a coordinate clock in K showing γ⋅√(2h/g).

So as observed from K, the rate of the clock in K' appears to be:
f' = (t₁'-t₀')/(t₁-t₀) = √(2h/g)/γ⋅√(2h/g) = 1/γ

Note that (t₁'-t₀') is a proper time because both t₀' and t₁' are read off the same clock, while t₀ and t₁ are read off two different coordinate clocks

The moving clock appears to run slow.

The apparent rate of clocks in K as observed in K'. ===================================================
When clock at the origin of K shows t₀ = 0, it will
be adjacent to a coordinate clock in K showing t₀' = 0.

See fig2:
We can define an event E₂:
When this clock at x₂ = 0 shows t₂ = √(2h/g) it will be adjacent to a coordinate clock in K'.
The coordinates in K' are:
t₂' = γ(t₂ - v⋅x₂/c²) = γ⋅√(2h/g)
x₂' = γ(x₂ - v⋅t₂) = -γ⋅v⋅√(2h/g)

So as observed from K, the rate of the clock in K' appears to be:
f = (t₂-t₀)/(t₂'-t₀') = √(2h/g)/γ⋅√(2h/g) = 1/γ

Note that (t₂-t₀) is a proper time because both t₀ and t₂
are read off the same clock, while t₀' and t₂' are read off two different coordinate clocks

The moving clock appears to run slow.

Mutual time dilation!

--
Paul

https://paulba.no/
Well, that is just another miracle that scientists have imagined. I don't really see the need for it. One of the proofs that scientists used to say that Einstein's Theory was true was the perihelion of Mercury. Isaac Newton's absolute time
interpretation did not quite explain where Mercury was. But, as I said, if Newton had been told that More gravitation would result in a slower clock, I think he could have done the mathematics. In looking at the Lorentz equations, it seemed obvious to me
how part of them was derived. If we take two sets of Galilean transformation equations,
x'=x-vt
y'=y
z'=z
t'=t
x = x' - m'n'
y = y'
z - z'
n = n'
Since there is no length contraction in the Galilean transformation equations, we can say
x - x' = vt
x -x' = -m'n'
m'n' = -vt
Einstein said that the Lorentz equations satisfy the results of the Michelson-Morley experiment because x = ct, x' = ct', where t' was the time of the moving clock. With two sets of Galilean transformation equations we have n' as the time of the
moving clock.
x=ct, x' = cn'
So by either clock, light is traveling at c.
x'=x-vt
cn'=ct-vt
n' = t - vt/c = t - vct/c^2 = t-vx/c^2, Which you might recognize as the numerator of Lorentz's equation for t'. At the speed of Mercury in its orbit, the denominator of Lorentz's equation is irrelevant. If something is traveling at the speed of
Mercury, the difference between n' in two sets of Galilean transformation equations and t' in Lorentz's equations is the same to several decimal places. Scientists might have used General Relativity rather than Special in making the calculations for the
orbit of Mercury, but the planet Mercury shows something about relativity. Using the time of the third planet from the sun as a preferred time for the solar system does not really make sense. Mercury moves at 30 miles per second, earth moves at 20 miles
per second. The further you get from the sun, the slower the planets are in their orbits. Then you have the orbits of the moons and satellites orbiting earth, etc. It seems to me that Newton was correct with his idea of absolute time to the extent that
there would be a rate of time to which the rates of time of all planets could be converted, but it would be a different time from the time of the third planet from the sun. It seems to me that there would be a time common to the entire solar system to
which the times of orbiting planets could be converted to agree with Newton's equations. Scientists, of course, are not going to be interested in anything but the miracles that Einstein and his disciples describe, but I think that times of clocks could
be used to gain a better understanding of gravitation. Scientists are paid trillions of dollars from governments to say that Einstein's equations are correct. I do not really see any reason to think anything is going to change any time soon.
I would ignore Einsteins “predictions” for the mercury anomalous preccession.
He only knew the amount for mercury and fiddled his formula to match that observed.
Notice his formula doesn’t do so well for other planets. A fact relativists
like to ignore.
To start with Mars preccesion rate is very unstable and is hard to calculate.
Which is why they pretend Mars preccession rate is 1.3 arc seconds per C When it probably is more like 2.5
The other planets preccesion rates fare less well under GR
Venus observed is 8 (GR predicted 8.6)
Earth observed is 5 (GR predicts 3.8)

However if one uses a more correct classical formula 1/(r+3R)^2 based on perehilion
not semi major axis as Albert incorrectly did. Then classical theory predicts more accurately
than GR. {where r is perehilion distance and R is radius of sun}
As follows:
Planet. Obs. GR Classical
Merc. 43.1. 43.5 43.24
V. 8. 8.6. 8.33
E. 5. 3.87. 4.49
I have contemplated this for some time. One thing I wondered about was if the rates of clocks varied according to their distance from the sun, then why would the time of the third planet from the sun be the significant time? It seems as though a time
associated with the sun itself or an absence of gravitation from the sun would be the time that would be significant and would be a time to which these other rates of time would be converted before using Newton's equations. At any rate I never did
believe Eddington's proof that Einstein's Relativity was correct.
I believe physics should employ this definition of time derived from the speed formula: Time= Distance/ Speed, or more generally, Time= Change/Rate. For time itself to change all rates of change would have to change in concert and relative motion per se
is not going to do that. Nothing is. Time is universal.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Laurence Clark Crossen@21:1/5 to Paul Alsing on Thu Oct 19 15:24:07 2023
On Thursday, October 19, 2023 at 11:50:12 AM UTC-7, Paul Alsing wrote:
On Thursday, October 19, 2023 at 11:35:59 AM UTC-7, Laurence Clark Crossen wrote:

Lou, I wonder if you have published any papers or books because I would like to read them as I agree with much of what you say and you know more about physics than I do.
Cranks supporting cranks.

"Everybody" knows more physics than you do, Larry, and "almost everybody" knows more than Lou...
you haven't exhibited any knowledge at all paul the heckler.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Laurence Clark Crossen@21:1/5 to Lou on Thu Oct 19 15:35:32 2023
On Thursday, October 19, 2023 at 12:52:10 PM UTC-7, Lou wrote:
On Thursday, 19 October 2023 at 19:35:59 UTC+1, Laurence Clark Crossen wrote:
On Wednesday, October 11, 2023 at 6:43:55 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 14:21:24 UTC+1, Paul B. Andersen wrote:
Den 10.10.2023 21:50, skrev Lou:
On Tuesday, 10 October 2023 at 19:10:11 UTC+1, Paul B. Andersen wrote:
Den 10.10.2023 14:16, skrev Lou:

No need for relativity to explain Hafael Keating.
Look at classical resonance. It has been known for centuries that >>> the natural resonant frequency of a system will reduce its frequency
if subject to an external force. So an atom, also confirmed by all observations
to date to be a resonant system, will also reduce its frequency when subject
to external force. As we see happen where less g force with altitude increases
the atoms ‘ticking’.

So clocks at higher altitude in a plane will be subject to
lower g-force and will tick faster than clocks on the ground, right?

No I’m suggesting that this horizontal force is not force due to gravity.
But a seperate force from horizontal acceleration .
Isn’t the formula for this f=ma?

I see. The horizontal force driving the plane at constant
speed through the air will give the plane a constant
horizontal acceleration. Sounds reasonable, doesn't it?

Im assuming that Hafael Keating observed that the eastward clock ticks slower.
That’s my reading of the wiki reference.
But to answer ‘why’...I assume the speed relative to the ground is the same for both
planes in the experiment. Let’s call it speed x.
The earth rotates eastward at 1600k/ hr.
Close enough.

So the zero point of reference is the earth observer travelling at 1600 k/hr relative
to the earths Center of mass.
Then relative to this earth center reference, the eastward plane travels at
1600 +x kilometers per hour. And the westward plane travels at 1600-x kilometers
per hour.

A reasonable speed for x is 800 km/h.

The east going plane travels at 2400 km/h in the ECI-frame.
The ground clock travels at 1600 km/h in the ECI-frame.
The west going plane travels at 800 km/h in the ECI-frame.

The eastward plane therefore experiences greater F than earth observer
And the westward plane lesser F than the the earth observer.
And seeing as a classical model uses resonance as the explanation for this different ticking rates of the caesium atoms natural resonant frequency.
Then the westward clock has less F from horizontal
speed than the earth observer and the westward plane has a greater F than
the earth observer. Which accounts for the 3 different rates of ticking.
(Don’t forget...the earth observers ‘ticking rate’ is in part due to Gravity force
but also in part due to its rotational/ horizontal speed of 1600k/hr )

I see.
Since the east going clock runs slower than the ground clock,
and the ground clock runs slower than the west going clock,
it is obvious that it is the speed in the ECI-frame that is
the major cause for the different clock rates.

So we can conclude:
Since the speed in the ECI-frame is higher for the east going
clock than for the ground clock, the horizontal force to drive
the east going plane through the air at 800 km/h is higher than
the force to drive the ground clock through the air at 0 km/h.

Since the speed in the ECI-frame is higher for the ground clock
than for the west going clock, the horizontal force to drive
the ground clock through the air at 0 km/h is higher than the force
to drive the west going plane through the air at 800 km/h.

The east going clock is more compressed than the ground clock,
and will run slower than the ground clock.
The ground clock is more compressed than the west going clock,
and will run slower than the west going clock.

Right?

But don’t forget the Gravity force pushing you down is a seperate source of
force from the F force pushing against you as you move horizontally. I’m not sure
what word you relativists prefer but probably inertia or momentum would
be appropriate?

Air drag.
Air Drag!!😂🤣You relativists. Such purveyors of BS.
Did I mention air drag?
No.
You did.
You forgot. The earth rotates.
I know in relativity land that you guys think the earth doesn’t rotate. But
sorry...it does.
In which case the eastward plane travels faster relative to the earth Center frame.
And the westward plane slower
If you knew any maths...then even if the planes speed was 800k/hr it would still
be less than the earth observers speed 1600 k/hr speed.
Which means that the eastward plane travels at a higher speed than the earth
observer, and the westward plane at a lower speed than the observer.
And seeing as f=ma then the force on the westward plane is less than the earth observer and the force on the eastward plane is higher.
Which in turn means that due to f=ma the eastward travelling clock will run slower and the westward clock will run faster than the earth observers clock
Due to mechanical resonance.
As observed in Hafael,Keating.
Do the maths Paul.
Lou, I wonder if you have published any papers or books because I would like to read them as I agree with much of what you say and you know more about physics than I do. I agree that H/K is not correctly explained by relativity and cannot be. However,
I'm not sure you are correct about the speeds of the planes. I believe you are wrong because of Galileo's concept of shared speed. Each plane shares the velocity of the source or starting point so they have the same velocity relative to the surface and
to the atmosphere. There is no difference in force.
I have only assumed the speed of the planes. Hafael Keating dont supply
any info on plane speeds. So I looked up passenger jet speeds on Google
and it’s actually around 550 mph.(880). But this is google groups not Journal
of physics. And H-K was written in 1971 so I opted for an even 800.
If I was to publish, which I don’t, seeing as the pro relativist editors only publish Harry Potter fantasy science, I would probably have to find the exact
1971 airspeed. Problem is landing, take off, routes etc are also
all variable and complex. Notice even H-K admit relativity predictions have large
errors of 10%! (And the final total lost observed for the eastward plane is 1/3 more than
predicted by relativity!! That’s an error of 33% for Einsteins theory.
Not surprising relativists don’t mention this failure.)

My take on it is as follows: The earth rotates eastward at 1600kph.
Relative to the earth Center. (Ie if you were in solar frame watching
the earth it would be rotating at 1600kph eastward.) That HAS to be
taken into account. Especially considering it’s considered “acceleration”
in current physics.
So a plane that flies at 800 kph eastward has to be travelling at 800+1600=2400 relative to the earth Center. Thats why NASA
launches rockets always to east. They use rocket thrust+ earths
1600kph rotational velocity to achieve higher orbital escape
velocity. Which means the westward plane is technically only
travelling at1600-800=800 relative to the earth center.
So relative to the earths mass that there-must be three speeds.
And using this assumption does give a good approximation to
the observed various tick rates. So my hunch is...if the theory
is based as best as possible on classical physics and gives an
accurate prediction...it must be a correct assumption I make
about the speeds and its relationship to force etc.
So...Assuming plane speed 800, and assuming that earth rotates
around in its own Center at 1600kph:

Observer rotational speed travelling east must be at 1600
East plane then has to be at 2400
West plane at 800
I can’t think of why it could be anything else. But would
be interested to see why you think it couldn’t.
Granted, it is acceleration relative to the center of the Earth aiding rockets; this might explain the puzzling fact of supposed time dilation in one direction and time contraction in the other contradicting relativity. It cannot be relative motion per
se, causing the instrumental error of the atomic clocks. As you say, it could be relative to gravity and the Earth's center. The whole experiment seems so slipshod to me that it can't mean anything, and relativity doesn't predict both time dilation and
contraction as I have H-K acknowledging that this was puzzling. Yes, since the rockets share the velocity of the spinning Earth they reach orbit better launched towards the East.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Laurence Clark Crossen@21:1/5 to Paul Alsing on Thu Oct 19 15:39:00 2023
On Thursday, October 19, 2023 at 11:50:12 AM UTC-7, Paul Alsing wrote:
On Thursday, October 19, 2023 at 11:35:59 AM UTC-7, Laurence Clark Crossen wrote:

Lou, I wonder if you have published any papers or books because I would like to read them as I agree with much of what you say and you know more about physics than I do.
Cranks supporting cranks.

"Everybody" knows more physics than you do, Larry, and "almost everybody" knows more than Lou...
Heckling is a characteristic of the envious who are preoccupied with tearing down others instead of developing themselves.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Laurence Clark Crossen@21:1/5 to Lou on Thu Oct 19 16:30:41 2023
On Thursday, October 19, 2023 at 12:52:10 PM UTC-7, Lou wrote:
On Thursday, 19 October 2023 at 19:35:59 UTC+1, Laurence Clark Crossen wrote:
On Wednesday, October 11, 2023 at 6:43:55 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 14:21:24 UTC+1, Paul B. Andersen wrote:
Den 10.10.2023 21:50, skrev Lou:
On Tuesday, 10 October 2023 at 19:10:11 UTC+1, Paul B. Andersen wrote:
Den 10.10.2023 14:16, skrev Lou:

No need for relativity to explain Hafael Keating.
Look at classical resonance. It has been known for centuries that >>> the natural resonant frequency of a system will reduce its frequency
if subject to an external force. So an atom, also confirmed by all observations
to date to be a resonant system, will also reduce its frequency when subject
to external force. As we see happen where less g force with altitude increases
the atoms ‘ticking’.

So clocks at higher altitude in a plane will be subject to
lower g-force and will tick faster than clocks on the ground, right?

No I’m suggesting that this horizontal force is not force due to gravity.
But a seperate force from horizontal acceleration .
Isn’t the formula for this f=ma?

I see. The horizontal force driving the plane at constant
speed through the air will give the plane a constant
horizontal acceleration. Sounds reasonable, doesn't it?

Im assuming that Hafael Keating observed that the eastward clock ticks slower.
That’s my reading of the wiki reference.
But to answer ‘why’...I assume the speed relative to the ground is the same for both
planes in the experiment. Let’s call it speed x.
The earth rotates eastward at 1600k/ hr.
Close enough.

So the zero point of reference is the earth observer travelling at 1600 k/hr relative
to the earths Center of mass.
Then relative to this earth center reference, the eastward plane travels at
1600 +x kilometers per hour. And the westward plane travels at 1600-x kilometers
per hour.

A reasonable speed for x is 800 km/h.

The east going plane travels at 2400 km/h in the ECI-frame.
The ground clock travels at 1600 km/h in the ECI-frame.
The west going plane travels at 800 km/h in the ECI-frame.

The eastward plane therefore experiences greater F than earth observer
And the westward plane lesser F than the the earth observer.
And seeing as a classical model uses resonance as the explanation for this different ticking rates of the caesium atoms natural resonant frequency.
Then the westward clock has less F from horizontal
speed than the earth observer and the westward plane has a greater F than
the earth observer. Which accounts for the 3 different rates of ticking.
(Don’t forget...the earth observers ‘ticking rate’ is in part due to Gravity force
but also in part due to its rotational/ horizontal speed of 1600k/hr )

I see.
Since the east going clock runs slower than the ground clock,
and the ground clock runs slower than the west going clock,
it is obvious that it is the speed in the ECI-frame that is
the major cause for the different clock rates.

So we can conclude:
Since the speed in the ECI-frame is higher for the east going
clock than for the ground clock, the horizontal force to drive
the east going plane through the air at 800 km/h is higher than
the force to drive the ground clock through the air at 0 km/h.

Since the speed in the ECI-frame is higher for the ground clock
than for the west going clock, the horizontal force to drive
the ground clock through the air at 0 km/h is higher than the force
to drive the west going plane through the air at 800 km/h.

The east going clock is more compressed than the ground clock,
and will run slower than the ground clock.
The ground clock is more compressed than the west going clock,
and will run slower than the west going clock.

Right?

But don’t forget the Gravity force pushing you down is a seperate source of
force from the F force pushing against you as you move horizontally. I’m not sure
what word you relativists prefer but probably inertia or momentum would
be appropriate?

Air drag.
Air Drag!!😂🤣You relativists. Such purveyors of BS.
Did I mention air drag?
No.
You did.
You forgot. The earth rotates.
I know in relativity land that you guys think the earth doesn’t rotate. But
sorry...it does.
In which case the eastward plane travels faster relative to the earth Center frame.
And the westward plane slower
If you knew any maths...then even if the planes speed was 800k/hr it would still
be less than the earth observers speed 1600 k/hr speed.
Which means that the eastward plane travels at a higher speed than the earth
observer, and the westward plane at a lower speed than the observer.
And seeing as f=ma then the force on the westward plane is less than the earth observer and the force on the eastward plane is higher.
Which in turn means that due to f=ma the eastward travelling clock will run slower and the westward clock will run faster than the earth observers clock
Due to mechanical resonance.
As observed in Hafael,Keating.
Do the maths Paul.
Lou, I wonder if you have published any papers or books because I would like to read them as I agree with much of what you say and you know more about physics than I do. I agree that H/K is not correctly explained by relativity and cannot be. However,
I'm not sure you are correct about the speeds of the planes. I believe you are wrong because of Galileo's concept of shared speed. Each plane shares the velocity of the source or starting point so they have the same velocity relative to the surface and
to the atmosphere. There is no difference in force.
I have only assumed the speed of the planes. Hafael Keating dont supply
any info on plane speeds. So I looked up passenger jet speeds on Google
and it’s actually around 550 mph.(880). But this is google groups not Journal
of physics. And H-K was written in 1971 so I opted for an even 800.
If I was to publish, which I don’t, seeing as the pro relativist editors only publish Harry Potter fantasy science, I would probably have to find the exact
1971 airspeed. Problem is landing, take off, routes etc are also
all variable and complex. Notice even H-K admit relativity predictions have large
errors of 10%! (And the final total lost observed for the eastward plane is 1/3 more than
predicted by relativity!! That’s an error of 33% for Einsteins theory.
Not surprising relativists don’t mention this failure.)

My take on it is as follows: The earth rotates eastward at 1600kph.
Relative to the earth Center. (Ie if you were in solar frame watching
the earth it would be rotating at 1600kph eastward.) That HAS to be
taken into account. Especially considering it’s considered “acceleration”
in current physics.
So a plane that flies at 800 kph eastward has to be travelling at 800+1600=2400 relative to the earth Center. Thats why NASA
launches rockets always to east. They use rocket thrust+ earths
1600kph rotational velocity to achieve higher orbital escape
velocity. Which means the westward plane is technically only
travelling at1600-800=800 relative to the earth center.
So relative to the earths mass that there-must be three speeds.
And using this assumption does give a good approximation to
the observed various tick rates. So my hunch is...if the theory
is based as best as possible on classical physics and gives an
accurate prediction...it must be a correct assumption I make
about the speeds and its relationship to force etc.
So...Assuming plane speed 800, and assuming that earth rotates
around in its own Center at 1600kph:

Observer rotational speed travelling east must be at 1600
East plane then has to be at 2400
West plane at 800
I can’t think of why it could be anything else. But would
be interested to see why you think it couldn’t.
You may like Xiaochun Mei's articles especially on Mercury's anomalous precession.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)
• From Laurence Clark Crossen@21:1/5 to Lou on Thu Oct 19 16:53:35 2023
On Thursday, October 19, 2023 at 12:52:10 PM UTC-7, Lou wrote:
On Thursday, 19 October 2023 at 19:35:59 UTC+1, Laurence Clark Crossen wrote:
On Wednesday, October 11, 2023 at 6:43:55 AM UTC-7, Lou wrote:
On Wednesday, 11 October 2023 at 14:21:24 UTC+1, Paul B. Andersen wrote:
Den 10.10.2023 21:50, skrev Lou:
On Tuesday, 10 October 2023 at 19:10:11 UTC+1, Paul B. Andersen wrote:
Den 10.10.2023 14:16, skrev Lou:

No need for relativity to explain Hafael Keating.
Look at classical resonance. It has been known for centuries that >>> the natural resonant frequency of a system will reduce its frequency
if subject to an external force. So an atom, also confirmed by all observations
to date to be a resonant system, will also reduce its frequency when subject
to external force. As we see happen where less g force with altitude increases
the atoms ‘ticking’.

So clocks at higher altitude in a plane will be subject to
lower g-force and will tick faster than clocks on the ground, right?

No I’m suggesting that this horizontal force is not force due to gravity.
But a seperate force from horizontal acceleration .
Isn’t the formula for this f=ma?

I see. The horizontal force driving the plane at constant
speed through the air will give the plane a constant
horizontal acceleration. Sounds reasonable, doesn't it?

Im assuming that Hafael Keating observed that the eastward clock ticks slower.
That’s my reading of the wiki reference.
But to answer ‘why’...I assume the speed relative to the ground is the same for both
planes in the experiment. Let’s call it speed x.
The earth rotates eastward at 1600k/ hr.
Close enough.

So the zero point of reference is the earth observer travelling at 1600 k/hr relative
to the earths Center of mass.
Then relative to this earth center reference, the eastward plane travels at
1600 +x kilometers per hour. And the westward plane travels at 1600-x kilometers
per hour.

A reasonable speed for x is 800 km/h.

The east going plane travels at 2400 km/h in the ECI-frame.
The ground clock travels at 1600 km/h in the ECI-frame.
The west going plane travels at 800 km/h in the ECI-frame.

The eastward plane therefore experiences greater F than earth observer
And the westward plane lesser F than the the earth observer.
And seeing as a classical model uses resonance as the e