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 ina 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
x' = (x-vt)/sqrt(12-v^2/c^2)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
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.
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 believeEinstein 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
x'=x-vtground, 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
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
x = x'- m'n'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.
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
On Monday, October 9, 2023 at 11:27:35 AM UTC-7, Robert Winn wrote: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
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
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 thex' = (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.
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 hisIsaac 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
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 forx'=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
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.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
After how many decades you are still confused and in square one.
--
Jan
On Monday, October 9, 2023 at 11:27:35 AM UTC-7, Robert Winn wrote: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
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
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 thex' = (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.
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 hisIsaac 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
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 forx'=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
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.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
After how many decades you are still confused and in square one.Jan
--
Jan
On Monday, October 9, 2023 at 12:25:05 PM UTC-7, JanPB wrote: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
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
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 thex' = (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.
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 hisIsaac 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
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 forx'=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
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.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
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?
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 clockin 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
x' = (x-vt)/sqrt(12-v^2/c^2)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
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.
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 believeEinstein 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
x'=x-vtthe 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
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
x = x'- m'n'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.
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
On 10-Oct-23 5:27 am, Robert Winn wrote: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
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
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 thex' = (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.
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 hisIsaac 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
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 forx'=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
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.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
Don't you think it would have been noticed long ago if you were rightSylvia, what about the Richard Hertz back story to the Einstein catapult to scientific stardom?
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.
Time dilation is the second item in Einstein-EDoMB-1905 Section §4v go by with less face time than his own clock.
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-
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.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.
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-
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
On 10-Oct-23 5:27 am, Robert Winn wrote: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
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
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 thex' = (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.
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 hisIsaac 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
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 forx'=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
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.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
Don't you think it would have been noticed long ago if you were rightWell, 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
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.
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote: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
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
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 thex' = (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.
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 hisIsaac 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
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 forx'=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
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.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
Don't you think it would have been noticed long ago if you were rightSylvia, what about the Richard Hertz back story to the Einstein catapult to scientific stardom?
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.
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote: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
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
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 thex' = (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.
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 hisIsaac 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
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 forx'=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
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.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
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 trillionsDon'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.Sylvia, what about the Richard Hertz back story to the Einstein catapult to scientific stardom?
On Monday, October 9, 2023 at 7:27:30 PM UTC-7, Sylvia Else wrote: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
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
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 thex' = (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.
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 hisIsaac 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
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 forx'=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
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.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
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 aDon't you think it would have been noticed long ago if you were rightWell, 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
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.
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 explainingrelativity.
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 hesaid explained the Michelson-Morley experiment.
x = ctgives 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
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
On 10-Oct-23 5:27 am, Robert Winn wrote: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
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
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 thex' = (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.
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 hisIsaac 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
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 forx'=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
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.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
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.
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.
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.
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.
The Einstein 1905 paper on it is best seen as didactics,
This is why that 'obscure patent clerk' had such an immediate impact:
he explained what 'everybody' knew already,
from an entirely new viewpoint,
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 wasBut, 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
How do you explain the result of the Fizeau experiment, which was
performed half a century before Einstein proposed his theory?
Sylvia Else <syl...@email.invalid> wrote: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
On 10-Oct-23 5:27 am, Robert Winn wrote:[-]
You may want to have a look at Hafele and Keating, for observationsSince 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.
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 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 notI see that you have become more cautious about th experimental basis.
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.
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.
On 10-Oct-23 3:39 pm, Robert Winn wrote: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
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
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 thex' = (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.
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 hisIsaac 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
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 forx'=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
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.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
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 aDon't you think it would have been noticed long ago if you were rightWell, 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
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.
relativity.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
said explained the Michelson-Morley experiment.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
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 wasx = 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
How do you explain the result of the Fizeau experiment, which wasFizeau'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
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 toincrease 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.
On 10-Oct-23 3:39 pm, Robert Winn wrote: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
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
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 thex' = (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.
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 hisIsaac 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
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 forx'=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
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.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
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 aDon't you think it would have been noticed long ago if you were rightWell, 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
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.
relativity.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
said explained the Michelson-Morley experiment.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
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 wasx = 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
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
On Tuesday, October 10, 2023 at 1:23:37 AM UTC-7, J. J. Lodder wrote:transformation equations. There is no length contraction in the Galilean transformation equations. So we have
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 observationsSince 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.
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 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.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.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
None were needed, because whatever could have been refered too
was common knowledge, for the intended readership.
x'=x-vtthis, but that is just the way it is.
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
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.
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.
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 ofthe 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 becausethey are incorrect for this situation (dealing with speeds close to c,
[snip babble]
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.
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:Good to see you, Dono. When did I lose out on the utter cretinism prize?
The eastward travelling plane experiencesYou take the prize. The utter cretinism prize
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.
On 10-Oct-23 9:38 pm, Robert Winn wrote: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.
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
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. 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,
Sylvia.
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:Good to see you, Dono. When did I lose out on the utter cretinism prize?
The eastward travelling plane experiencesYou take the prize. The utter cretinism prize
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.
On Tuesday, October 10, 2023 at 4:50:06 AM UTC-7, Sylvia Else wrote: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.
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
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.
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:You are always in the running. Along with Dick Hertz and Pattycakes Dolan. Lou is just a newcomer.
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:Good to see you, Dono. When did I lose out on the utter cretinism prize?
The eastward travelling plane experiencesYou take the prize. The utter cretinism prize
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.
On 10/10/2023 6:38 AM, Robert Winn wrote:the pipes used by Fizeau.
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
Wrong, there were concerns the pipe diameter affected the light. Later experiments using better equipment CONFIRMED Fizeau's results.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.
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 becausethey 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.
On 10/10/2023 11:49 AM, Robert Winn wrote: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.
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
Well, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.If you never knew anything about it, why did you lie and say it was a version of the MMX?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.
On Tuesday, October 10, 2023 at 8:51:28 AM UTC-7, Robert Winn wrote:One thing I have noticed about science. You never use any equations. Why is that?
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:You are always in the running. Along with Dick Hertz and Pattycakes Dolan. Lou is just a newcomer.
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:Good to see you, Dono. When did I lose out on the utter cretinism prize?
The eastward travelling plane experiencesYou take the prize. The utter cretinism prize
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.
On Tuesday, October 10, 2023 at 8:13:08 AM UTC-7, Volney wrote:the pipes used by Fizeau.
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:No, it was not. While both used light fringes to detect changes of
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.
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
Wrong, there were concerns the pipe diameter affected the light. LaterWell, nobody ever asked me about the Fizeau experiment before. The only thing I knew about it was its name until this morning.
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 becausethey 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.
[snip babble]
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:One thing I have noticed about science. You never use any equations. Why is that?
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:You are always in the running. Along with Dick Hertz and Pattycakes Dolan. Lou is just a newcomer.
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:Good to see you, Dono. When did I lose out on the utter cretinism prize?
The eastward travelling plane experiencesYou take the prize. The utter cretinism prize
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.
On 10/10/2023 7:18 AM, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote:
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 aHow do you explain the result of the Fizeau experiment, which wasAnd so for instance when the water
performed half a century before Einstein proposed his theory?
https://en.wikipedia.org/wiki/Fizeau_experiment
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.
constant c/n.
You are grasping at straws.
On Tuesday, October 10, 2023 at 9:01:09 AM UTC-7, Volney wrote: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.
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
Well, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.If you never knew anything about it, why did you lie and say it was aIf 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.
version of the MMX?
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:Obviously bogus. If that were so, the speed of light in water would get
How do you explain the result of the Fizeau experiment, which wasAnd so for instance when the water
performed half a century before Einstein proposed his theory?
https://en.wikipedia.org/wiki/Fizeau_experiment
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.
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?
On Tuesday, October 10, 2023 at 9:05:35 AM UTC-7, Robert Winn wrote: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
On Tuesday, October 10, 2023 at 8:54:00 AM UTC-7, Dono. wrote:You are not only an imbecile, you are a liar as well.
On Tuesday, October 10, 2023 at 8:51:28 AM UTC-7, Robert Winn wrote:One thing I have noticed about science. You never use any equations. Why is that?
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:You are always in the running. Along with Dick Hertz and Pattycakes Dolan.
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:Good to see you, Dono. When did I lose out on the utter cretinism prize?
The eastward travelling plane experiencesYou take the prize. The utter cretinism prize
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.
Lou is just a newcomer.
On 10/10/2023 12:07 PM, Robert Winn wrote: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.
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
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.As I said, both were interferometers, but that's where the similaritiesWell, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.If you never knew anything about it, why did you lie and say it was aIf you're going to just outright lie, there's not a lot of point to thisI did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name.
interaction.
version of the MMX?
end. And you did lie. Fizeau's experiment was not the MMX.
On Tuesday, October 10, 2023 at 9:44:38 AM UTC-7, Volney wrote: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.
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
Well, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.If you never knew anything about it, why did you lie and say it was aIf 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.
version of the MMX?
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.
: Fizeau's experiment was an early version of the Michelson-Morley experiment using water instead of air as the medium of conducting light.
On 10/10/2023 1:11 PM, Robert Winn wrote: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.
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
I wrote it. So how is using water and using air for the same kind of experiment the same experiment?Well, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.If you never knew anything about it, why did you lie and say it was a >>>> version of the MMX?If you're going to just outright lie, there's not a lot of point to thisI did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name.
interaction.
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?
On Tuesday, October 10, 2023 at 1:23:37?AM UTC-7, J. J. Lodder wrote:[misunderstandings about planes]
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)
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 notI see that you have become more cautious about th experimental basis.
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.
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.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
None were needed, because whatever could have been refered too
was common knowledge, for the intended readership.
does not matter to the Galilean transformation equations.
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
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: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
How do you explain the result of the Fizeau experiment, which wasAnd so for instance when the water
performed half a century before Einstein proposed his theory?
https://en.wikipedia.org/wiki/Fizeau_experiment
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.
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.Because you claimed the light slows by going through 'more water' (your term). Increasing the distance obviously means more water traversed.
Slower with more distance? How so?
On Tuesday, October 10, 2023 at 9:39:13 AM UTC-7, Dono. wrote:That you are liar and an imbecile? Well, it is an established fact.
On Tuesday, October 10, 2023 at 9:05:35 AM UTC-7, Robert Winn wrote:Well, I don't really think so.
On Tuesday, October 10, 2023 at 8:54:00 AM UTC-7, Dono. wrote:You are not only an imbecile, you are a liar as well.
On Tuesday, October 10, 2023 at 8:51:28 AM UTC-7, Robert Winn wrote:One thing I have noticed about science. You never use any equations. Why is that?
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:You are always in the running. Along with Dick Hertz and Pattycakes Dolan.
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:Good to see you, Dono. When did I lose out on the utter cretinism prize?
The eastward travelling plane experiencesYou take the prize. The utter cretinism prize
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.
Lou is just a newcomer.
Den 10.10.2023 14:16, skrev Lou:
No need for relativity to explain Hafael Keating.So clocks at higher altitude in a plane will be subject to
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’.
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 experiencesI see.
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
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?
Den 10.10.2023 14:16, skrev Lou: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
No need for relativity to explain Hafael Keating.So clocks at higher altitude in a plane will be subject to
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’.
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 experiencesI see.
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
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/
Robert Winn <rbw...@gmail.com> wrote: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
On Tuesday, October 10, 2023 at 1:23:37?AM UTC-7, J. J. Lodder wrote:
Sylvia Else <syl...@email.invalid> wrote:
[misunderstandings about planes]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)
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 aI see that you have become more cautious about th experimental basis.
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.
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
I don't want to bother with your fantasies,[1] This is also the reason for the lack of references.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.
None were needed, because whatever could have been refered too
was common knowledge, for the intended readership.
but I hope that it did occur to you
that the Earth, and the atmosphere with it,
is rotating?
Jan
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:That you are liar and an imbecile? Well, it is an established fact.
On Tuesday, October 10, 2023 at 9:05:35 AM UTC-7, Robert Winn wrote:Well, I don't really think so.
On Tuesday, October 10, 2023 at 8:54:00 AM UTC-7, Dono. wrote:You are not only an imbecile, you are a liar as well.
On Tuesday, October 10, 2023 at 8:51:28 AM UTC-7, Robert Winn wrote:One thing I have noticed about science. You never use any equations. Why is that?
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:You are always in the running. Along with Dick Hertz and Pattycakes Dolan.
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:Good to see you, Dono. When did I lose out on the utter cretinism prize?
The eastward travelling plane experiencesYou take the prize. The utter cretinism prize
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.
Lou is just a newcomer.
On Tuesday, October 10, 2023 at 10:08:08 AM UTC-7, Robert Winn wrote:But, Dono, all you ever do is make accusations. You never really say anything.
On Tuesday, October 10, 2023 at 9:39:13 AM UTC-7, Dono. wrote:That you are liar and an imbecile? Well, it is an established fact.
On Tuesday, October 10, 2023 at 9:05:35 AM UTC-7, Robert Winn wrote:Well, I don't really think so.
On Tuesday, October 10, 2023 at 8:54:00 AM UTC-7, Dono. wrote:You are not only an imbecile, you are a liar as well.
On Tuesday, October 10, 2023 at 8:51:28 AM UTC-7, Robert Winn wrote:One thing I have noticed about science. You never use any equations. Why is that?
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:You are always in the running. Along with Dick Hertz and Pattycakes Dolan.
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:Good to see you, Dono. When did I lose out on the utter cretinism prize?
The eastward travelling plane experiencesYou take the prize. The utter cretinism prize
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.
Lou is just a newcomer.
So why did the east going clock in the plane tick slower than
the clock on the ground?
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 thanBecause your Holiest Postulate is such an absurd
the clock on the ground?
that not even you stick to it.
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:But, Dono, all you ever do is make accusations. You never really say anything.
On Tuesday, October 10, 2023 at 9:39:13 AM UTC-7, Dono. wrote:That you are liar and an imbecile? Well, it is an established fact.
On Tuesday, October 10, 2023 at 9:05:35 AM UTC-7, Robert Winn wrote:Well, I don't really think so.
On Tuesday, October 10, 2023 at 8:54:00 AM UTC-7, Dono. wrote:You are not only an imbecile, you are a liar as well.
On Tuesday, October 10, 2023 at 8:51:28 AM UTC-7, Robert Winn wrote:One thing I have noticed about science. You never use any equations. Why is that?
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:You are always in the running. Along with Dick Hertz and Pattycakes Dolan.
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:Good to see you, Dono. When did I lose out on the utter cretinism prize?
The eastward travelling plane experiencesYou take the prize. The utter cretinism prize
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.
Lou is just a newcomer.
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}
On Tuesday, October 10, 2023 at 10:16:10 AM UTC-7, Volney wrote: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.
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
Well, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.If you never knew anything about it, why did you lie and say it was a >>>>>> version of the MMX?If you're going to just outright lie, there's not a lot of point to thisI did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name.
interaction.
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?
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:Because you claimed the light slows by going through 'more water' (your
On 10/10/2023 7:18 AM, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote: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
How do you explain the result of the Fizeau experiment, which wasAnd so for instance when the water
performed half a century before Einstein proposed his theory?
https://en.wikipedia.org/wiki/Fizeau_experiment
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.
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?
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.
No Relativity needed to correctly model Fizeau classically.
On Tuesday, October 10, 2023 at 12:54:44 PM UTC-7, Robert Winn wrote: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.
On Tuesday, October 10, 2023 at 12:20:50 PM UTC-7, Dono. wrote:I say that you are an idiot and a liar. This is a fact.
On Tuesday, October 10, 2023 at 10:08:08 AM UTC-7, Robert Winn wrote:But, Dono, all you ever do is make accusations. You never really say anything.
On Tuesday, October 10, 2023 at 9:39:13 AM UTC-7, Dono. wrote:That you are liar and an imbecile? Well, it is an established fact.
On Tuesday, October 10, 2023 at 9:05:35 AM UTC-7, Robert Winn wrote:Well, I don't really think so.
On Tuesday, October 10, 2023 at 8:54:00 AM UTC-7, Dono. wrote:You are not only an imbecile, you are a liar as well.
On Tuesday, October 10, 2023 at 8:51:28 AM UTC-7, Robert Winn wrote:One thing I have noticed about science. You never use any equations. Why is that?
On Tuesday, October 10, 2023 at 6:54:35 AM UTC-7, Dono. wrote:You are always in the running. Along with Dick Hertz and Pattycakes Dolan.
On Tuesday, October 10, 2023 at 5:16:02 AM UTC-7, Lou wrote:Good to see you, Dono. When did I lose out on the utter cretinism prize?
The eastward travelling plane experiencesYou take the prize. The utter cretinism prize
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.
Lou is just a newcomer.
On 10/10/2023 1:49 PM, Robert Winn wrote: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.
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
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 lightWell, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.If you never knew anything about it, why did you lie and say it was a >>>>>> version of the MMX?If you're going to just outright lie, there's not a lot of point to thisI did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name.
interaction.
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?
On Tuesday, October 10, 2023 at 8:26:09 PM UTC-7, Volney wrote: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.
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
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.Well, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.If you never knew anything about it, why did you lie and say it was a >>>>>>>> version of the MMX?If you're going to just outright lie, there's not a lot of point to thisI did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name.
interaction.
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
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.
On 10/10/2023 1:49 PM, Robert Winn wrote: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.
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
Well, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.If you never knew anything about it, why did you lie and say it was a >>>>>> version of the MMX?If you're going to just outright lie, there's not a lot of point to thisI did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name.
interaction.
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?
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 thanBecause your Holiest Postulate is such an absurd
the clock on the ground?
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.
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 thanBecause your Holiest Postulate is such an absurd
the clock on the ground?
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...
[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.
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.
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote: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
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
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 thex' = (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.
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 hisIsaac 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
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 distancesx'=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
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.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
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 aDon't you think it would have been noticed long ago if you were right >> about this.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
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.
relativity.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
he said explained the Michelson-Morley experiment.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
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. Hisx = 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
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
SylviaNo need for relativity. A classical model does just fine.
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.
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:Because you claimed the light slows by going through 'more water' (your >> term). Increasing the distance obviously means more water traversed.
On 10/10/2023 7:18 AM, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote: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
How do you explain the result of the Fizeau experiment, which was >>>>>> performed half a century before Einstein proposed his theory?And so for instance when the water
https://en.wikipedia.org/wiki/Fizeau_experiment
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.
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?
Grabbing at fantasy straws still. You snipped the formula I cited. There is *no*My God, are you thick!
‘d’ in that formula. Just v of the water.
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?
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote: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
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
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 clockx' = (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
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 hisIsaac 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
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 samex'=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
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.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
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 aDon't you think it would have been noticed long ago if you were right >> about this.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
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.
explaining relativity.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
that he said explained the Michelson-Morley experiment.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
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. Hisx = 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
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
SylviaNo need for relativity. A classical model does just fine.
No, it doesn't. You simply don't know the problems involved.
On 10/11/2023 12:10 AM, Robert Winn wrote: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.
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
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.Well, I did not lie. What I read said that Michelson and Morley got their idea for their interferometer from Fizeau's experiment.If you never knew anything about it, why did you lie and say it was aIf you're going to just outright lie, there's not a lot of point to thisI did not outright lie. The only thing I ever knew about the Fizeau experiment before this morning was its name.
interaction.
version of the MMX?
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
You can't even get that correct! The MMX was an attempt to measure the earth's speed through the ether!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
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'?
On Tuesday, October 10, 2023 at 4:18:05 AM UTC-7, Lou wrote: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
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
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 clockx' = (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
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 hisIsaac 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
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 samex'=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
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.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
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 aDon't you think it would have been noticed long ago if you were right >> about this.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
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.
explaining relativity.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
that he said explained the Michelson-Morley experiment.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
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. Hisx = 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
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.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
SylviaNo need for relativity. A classical model does just fine.
No, it doesn't. You simply don't know the problems involved.
--
Jan
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}
On Wednesday, October 11, 2023 at 2:29:54 AM UTC-7, JanPB wrote: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
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
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 clockx' = (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
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 airplaneIsaac 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
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 samex'=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
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.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
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 aDon't you think it would have been noticed long ago if you were rightWell, 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
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.
explaining relativity.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
that he said explained the Michelson-Morley experiment.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
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. Hisx = 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
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
SylviaNo need for relativity. A classical model does just fine.
No, it doesn't. You simply don't know the problems involved.
--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.
Jan
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 ,
Repeating crank claims doesn't make them right, makes you a stubborn 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}
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?
Im assuming that Hafael Keating observed that the eastward clock ticks slower.Close enough.
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 )
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?
On Wednesday, 11 October 2023 at 13:14:44 UTC+1, Robert Winn wrote: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
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
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 clockx' = (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
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 airplaneIsaac 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
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 samex'=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
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.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
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 aDon't you think it would have been noticed long ago if you were rightWell, 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
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.
explaining relativity.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
that he said explained the Michelson-Morley experiment.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
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.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
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
SylviaNo need for relativity. A classical model does just fine.
No, it doesn't. You simply don't know the problems involved.
Anti-relativists are not scientists. They are--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.
Jan
religious wackos.
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.Close enough.
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.
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.
On Wednesday, October 11, 2023 at 6:14:00 AM UTC-7, Lou wrote: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
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
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 clockx' = (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
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 airplaneIsaac 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
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 samex'=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
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.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
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 thanDon't you think it would have been noticed long ago if you were rightWell, 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
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.
explaining relativity.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
equations that he said explained the Michelson-Morley experiment.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
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.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
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
SylviaNo need for relativity. A classical model does just fine.
No, it doesn't. You simply don't know the problems involved.
Anti-relativists are not scientists. They are--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.
Jan
religious wackos.
You sure are
On Wednesday, October 11, 2023 at 6:32:54 AM UTC-7, Dono. wrote: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
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
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 clockx' = (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
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 airplaneIsaac 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
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 samex'=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
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.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.
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 thanDon't you think it would have been noticed long ago if you were rightWell, 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
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.
explaining relativity.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
equations that he said explained the Michelson-Morley experiment.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
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.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
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
SylviaNo need for relativity. A classical model does just fine.
No, it doesn't. You simply don't know the problems involved.
Anti-relativists are not scientists. They are--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.
Jan
religious wackos.
You sure areFizeau 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.
On Wednesday, 11 October 2023 at 14:40:55 UTC+1, Dono. wrote: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
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
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 ax' = (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
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 airplaneIsaac 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
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 thex'=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
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.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.
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 fasterDon't you think it would have been noticed long ago if you were rightWell, 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
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.
explaining relativity.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
equations that he said explained the Michelson-Morley experiment.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
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.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
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
SylviaNo need for relativity. A classical model does just fine.
No, it doesn't. You simply don't know the problems involved.
Anti-relativists are not scientists. They are--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.
Jan
religious wackos.
You pretended c/n+v(1-1/n^2) was discovered by EinsteinYou sure areFizeau 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.
On Wednesday, October 11, 2023 at 6:49:11 AM UTC-7, Lou wrote: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
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
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 ax' = (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
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 theIsaac 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,
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 thex'=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
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.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.
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 isDon't you think it would have been noticed long ago if you were rightWell, 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
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.
am explaining relativity.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
equations that he said explained the Michelson-Morley experiment.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
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.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
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
SylviaNo need for relativity. A classical model does just fine.
No, it doesn't. You simply don't know the problems involved.
Anti-relativists are not scientists. They are--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.
Jan
religious wackos.
You pretended c/n+v(1-1/n^2) was discovered by EinsteinYou sure areFizeau 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.
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.
On Wednesday, October 11, 2023 at 6:53:30 AM UTC-7, Dono. wrote: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
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
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 thanx' = (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
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 theIsaac 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,
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 withx'=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
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.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
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 isDon't you think it would have been noticed long ago if you were rightWell, 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
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.
I am explaining relativity.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.
equations that he said explained the Michelson-Morley experiment.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
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.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
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
SylviaNo need for relativity. A classical model does just fine.
No, it doesn't. You simply don't know the problems involved.
Anti-relativists are not scientists. They are--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.
Jan
religious wackos.
You pretended c/n+v(1-1/n^2) was discovered by EinsteinYou sure areFizeau 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.
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.
On Wednesday, 11 October 2023 at 15:02:49 UTC+1, Dono. wrote: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,
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
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 thanx' = (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
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 theIsaac 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,
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 withx'=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
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.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
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 isDon't you think it would have been noticed long ago if you were rightWell, 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
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.
I am explaining relativity.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.
Lorentz equations that he said explained the Michelson-Morley experiment.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
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 hisx = 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
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
SylviaNo need for relativity. A classical model does just fine.
No, it doesn't. You simply don't know the problems involved.
Anti-relativists are not scientists. They are--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.
Jan
religious wackos.
You pretended c/n+v(1-1/n^2) was discovered by EinsteinYou sure areFizeau 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 that Fizeau’s formula was invented by AlbertNo, 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.
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.
On Wednesday, October 11, 2023 at 7:21:56 AM UTC-7, Lou wrote: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,
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
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 beingx' = (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
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 speedIsaac 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
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 equationsx'=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,
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.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
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 isDon't you think it would have been noticed long ago if you were rightWell, 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
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.
waves. I am explaining relativity.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
Lorentz equations that he said explained the Michelson-Morley experiment.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
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 inx = 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
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
SylviaNo need for relativity. A classical model does just fine.
No, it doesn't. You simply don't know the problems involved.
Anti-relativists are not scientists. They are--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.
Jan
religious wackos.
You pretended c/n+v(1-1/n^2) was discovered by EinsteinYou sure areFizeau 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 are lying, there is no such claim.You pretended that Fizeau’s formula was invented by AlbertNo, 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.
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..Crank,
Prove it.
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!
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:
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).Anti-relativists are not scientists. They are
religious wackos.
You sure are
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 ?
On Wednesday, 11 October 2023 at 15:39:50 UTC+1, Dono. wrote: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,
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
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 beingx' = (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
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 speedIsaac 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
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 equationsx'=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,
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.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
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 isDon't you think it would have been noticed long ago if you were rightWell, 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
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.
waves. I am explaining relativity.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
Lorentz equations that he said explained the Michelson-Morley experiment.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
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 inx = 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
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
SylviaNo need for relativity. A classical model does just fine.
No, it doesn't. You simply don't know the problems involved.
Anti-relativists are not scientists. They are--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.
Jan
religious wackos.
You pretended c/n+v(1-1/n^2) was discovered by EinsteinYou sure areFizeau 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.
Posting (repeatedly) the same imbecilities doesn't make you right. Makes you just a bigger imbecile.You are lying, there is no such claim.You pretended that Fizeau’s formula was invented by AlbertNo, 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.
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..Crank,
Prove it.
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.
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:My God, are you thick!
On 10/10/2023 12:40 PM, Lou wrote:
On Tuesday, 10 October 2023 at 16:47:41 UTC+1, Volney wrote:Because you claimed the light slows by going through 'more water' (your >>>> term). Increasing the distance obviously means more water traversed.
On 10/10/2023 7:18 AM, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote: >>>>>>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.
How do you explain the result of the Fizeau experiment, which was >>>>>>>> performed half a century before Einstein proposed his theory?And so for instance when the water
https://en.wikipedia.org/wiki/Fizeau_experiment
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.
You are grasping at straws.
You aren’t just grabbing at straws. You are making up the straws.
Slower with more distance? How so?
Grabbing at fantasy straws still. You snipped the formula I cited. There is *no*
‘d’ in that formula. Just v of the water.
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.
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.
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:My God, are you thick!
On 10/10/2023 12:40 PM, Lou wrote:Grabbing at fantasy straws still. You snipped the formula I cited. There is *no*
On Tuesday, 10 October 2023 at 16:47:41 UTC+1, Volney wrote:Because you claimed the light slows by going through 'more water' (your >>>> term). Increasing the distance obviously means more water traversed. >>>
On 10/10/2023 7:18 AM, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote: >>>>>>Obviously bogus. If that were so, the speed of light in water would get
How do you explain the result of the Fizeau experiment, which was >>>>>>>> performed half a century before Einstein proposed his theory? >>>>>>>>And so for instance when the water
https://en.wikipedia.org/wiki/Fizeau_experiment
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.
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?
‘d’ in that formula. Just v of the water.
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?So you simply can't support your "traverse more water" claim with a
c/n+-v{(n-1)+(n-1)^2}
Can’t find it? Oh well. Looks like you’ve been spouting BS again.
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 waterWhere is your evidence for this (now changed) claim? Remember, science
creates a more optically dense medium in the experiment.
is built on experimental evidence and scientific observations.
And classically one way to model this is to make this newNow you are making up your own crap formula which conflicts with
value of n is by this formula... c/n+-v{(n-1)+(n-1)^2}
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.
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:
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).Anti-relativists are not scientists. They are
religious wackos.
You sure are
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 EinsteinAdolf?? 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 wasThat's a huge lie. Einstein EXPLICITLY stated that the Fizeau experiment
his own formula ?
was one of four experiments which inspired him to derive SR. He was CREDITING Fizeau's discovery.
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.
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.
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 theframe of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.
The derivation of 1905 uses a version of the temporal Lorentz Transform. Same thing in today’s textbooks:moves to location-x on axis X according to an equation of motion on axis X:
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′ )
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?
So you simply can't support your "traverse more water" claim with a
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.Indeed.
Did I mention air drag?
No.
You did.
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.Congratulations.
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.
You have broken the NG record for most ignorance of elementary physics.
Well done!
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.
Den 10.10.2023 21:52, skrev Robert Winn:frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.
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
Immediately after the ball is dropped the speed of the ball is: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
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/
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:So you simply can't support your "traverse more water" claim with a
On 10/10/2023 3:23 PM, Lou wrote:
On Tuesday, 10 October 2023 at 17:50:46 UTC+1, Volney wrote:My God, are you thick!
On 10/10/2023 12:40 PM, Lou wrote:Grabbing at fantasy straws still. You snipped the formula I cited. There is *no*
On Tuesday, 10 October 2023 at 16:47:41 UTC+1, Volney wrote:Because you claimed the light slows by going through 'more water' (your >>>>>> term). Increasing the distance obviously means more water traversed. >>>>>
On 10/10/2023 7:18 AM, Lou wrote:
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote: >>>>>>>>Obviously bogus. If that were so, the speed of light in water would get
How do you explain the result of the Fizeau experiment, which was >>>>>>>>>> performed half a century before Einstein proposed his theory? >>>>>>>>>>And so for instance when the water
https://en.wikipedia.org/wiki/Fizeau_experiment
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.
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?
‘d’ in that formula. Just v of the water.
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.
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 differentWhere is your evidence for this (now changed) claim? Remember, science
value of n which takes into account the fact that the moving water
creates a more optically dense medium in the experiment.
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.
On Tuesday, October 10, 2023 at 11:23:05 PM UTC-7, Volney wrote: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.
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: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
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:
You can't even get that correct! The MMX was an attempt to measure the
earth's speed through the ether!
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 resultIn 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
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:So you simply can't support your "traverse more water" claim with a
On 10/10/2023 3:23 PM, Lou wrote:
On Tuesday, 10 October 2023 at 17:50:46 UTC+1, Volney wrote:My God, are you thick!
On 10/10/2023 12:40 PM, Lou wrote:Grabbing at fantasy straws still. You snipped the formula I cited. There is *no*
On Tuesday, 10 October 2023 at 16:47:41 UTC+1, Volney wrote: >>>>>>>> On 10/10/2023 7:18 AM, Lou wrote:Because you claimed the light slows by going through 'more water' (your
On Tuesday, 10 October 2023 at 09:08:44 UTC+1, Sylvia Else wrote: >>>>>>>>Obviously bogus. If that were so, the speed of light in water would get
How do you explain the result of the Fizeau experiment, which was >>>>>>>>>> performed half a century before Einstein proposed his theory? >>>>>>>>>>And so for instance when the water
https://en.wikipedia.org/wiki/Fizeau_experiment
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.
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?
term). Increasing the distance obviously means more water traversed. >>>>>
‘d’ in that formula. Just v of the water.
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.
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 waterWhere is your evidence for this (now changed) claim? Remember, science
creates a more optically dense medium in the experiment.
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
your claim.
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
contradicted classical theory.
As confirmed by Fizeau.
Nope. Don't put words in Fizeau's mouth.
by Fizeau : c/n+-v(1-(1/n^2))
Or by my formula: c/n+-v[{(n-1)+(n-1)}^2]
On Wednesday, October 11, 2023 at 11:07:19 AM UTC-7, Paul B. Andersen wrote:frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.
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
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 timedNo, 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
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/
Den 12.10.2023 01:37, skrev Robert Winn:the frame of reference of the airplane than in the frame of reference of the ground because the clock in the airplane is slower.
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
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 byNo, 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
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/
Let's use your original description.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
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/
On Wednesday, 11 October 2023 at 10:29:54 UTC+1, JanPB wrote: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
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
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 thex' = (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.
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 hisIsaac 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
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 distancesx'=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
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.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
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 aDon't you think it would have been noticed long ago if you were right >>>>>> about this.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
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.
relativity.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
he said explained the Michelson-Morley experiment.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
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. Hisx = 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
You don’t understand basic physics. Fact: A more optically dense medium will have a slightly greater refractive index.How do you explain the result of the Fizeau experiment, which wasNo need for relativity. A classical model does just fine.
performed half a century before Einstein proposed his theory?
https://en.wikipedia.org/wiki/Fizeau_experiment
Sylvia
No, it doesn't. You simply don't know the problems involved.
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.
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 ina 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
x' = (x-vt)/sqrt(12-v^2/c^2)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
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.
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 believeEinstein 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
x'=x-vtground, 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
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
x = x'- m'n'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.
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
On Monday, October 9, 2023 at 11:27:35 AM UTC-7, Robert Winn wrote: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
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
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 thex' = (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.
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 hisIsaac 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
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 forx'=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
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.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
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 ajoke/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
On 10/11/2023 6:45 AM, Lou wrote: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
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
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 clockx' = (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
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 hisIsaac 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
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 samex'=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
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.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
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 aDon't you think it would have been noticed long ago if you were right >>>>>> about this.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
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.
explaining relativity.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
that he said explained the Michelson-Morley experiment.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
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. Hisx = 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
You don’t understand basic physics. Fact: A more optically dense medium will have a slightly greater refractive index.How do you explain the result of the Fizeau experiment, which wasNo need for relativity. A classical model does just fine.
performed half a century before Einstein proposed his theory?
https://en.wikipedia.org/wiki/Fizeau_experiment
Sylvia
No, it doesn't. You simply don't know the problems involved.
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.
Again, your data?
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.
Den 12.10.2023 20:04, skrev Robert Winn: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
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.
Note:
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.
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
I don't buy it, Robert.Well, I just look at the equations, Paul. Einstein says that the vertical distance does not change.
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.
I think you are talking nonsense to evade addressing the issue.
So please, answer the questions.
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/
Den 12.10.2023 20:04, skrev Robert Winn: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
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.
Note:
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.
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
I don't buy it, Robert.
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.
I think you are talking nonsense to evade addressing the issue.
So please, answer the questions.
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
relativity says the jet is flying at two different speeds at once,
proving it is self-contradictory nonsense.
Den 12.10.2023 20:04, skrev Robert Winn: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
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.
Note:
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.
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
I don't buy it, Robert.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.
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.
I think you are talking nonsense to evade addressing the issue.
So please, answer the questions.
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/
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.
Den 12.10.2023 20:04, skrev Robert Winn: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
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.
Note:
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.
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
I don't buy it, Robert.Einstein never qualified as a physicist and you're dumber than he was.
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.
I think you are talking nonsense to evade addressing the issue.
So please, answer the questions.
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/
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.
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.
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 waterWhere is your evidence for this (now changed) claim? Remember, science >>>> is built on experimental evidence and scientific observations.
creates a more optically dense medium in the experiment.
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.
Theoretically. But a classical explanation doesn’t rely on magic.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 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.
I didn’t say it was better. I said both worked as well.
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.
And proved that
the optical density of the moving water relative to the source increases
or decreases with+-v. A purely classical phenomena.
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.
Only if you ignore Fizeau.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
contradicted classical theory.
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))
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))
On Friday, 13 October 2023 at 02:52:52 UTC+1, Volney wrote: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
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
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 clockx' = (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
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 hisIsaac 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
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 samex'=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
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.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
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 aDon't you think it would have been noticed long ago if you were right >>>>>>>> about this.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
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.
explaining relativity.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
that he said explained the Michelson-Morley experiment.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
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. Hisx = 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
Look up optical density vs refraction and you get many quotes likeYou don’t understand basic physics. Fact: A more optically dense medium >>> will have a slightly greater refractive index.How do you explain the result of the Fizeau experiment, which wasNo need for relativity. A classical model does just fine.
performed half a century before Einstein proposed his theory?
https://en.wikipedia.org/wiki/Fizeau_experiment
Sylvia
No, it doesn't. You simply don't know the problems involved.
Is today "make up a fact day" or something?
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.I’m not trying to write a textbook on refraction and optical
density?
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.
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.
Again, your data?
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
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.
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
proving it is self-contradictory nonsense.
Den 12.10.2023 20:04, skrev Robert Winn: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
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.
Note:
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.
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
I don't buy it, Robert.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
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.
I think you are talking nonsense to evade addressing the issue.
So please, answer the questions.
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/
In physics, speed is ALWAYS relative to a specified coordinate system.
On 10/13/2023 3:27 AM, Lou wrote: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
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
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 clockx' = (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
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 airplaneIsaac 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
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 samex'=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
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.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
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 aDon't you think it would have been noticed long ago if you were rightWell, 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
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.
explaining relativity.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
that he said explained the Michelson-Morley experiment.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
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. Hisx = 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
Look up optical density vs refraction and you get many quotes likeYou don’t understand basic physics. Fact: A more optically dense mediumHow do you explain the result of the Fizeau experiment, which was >>>>>> performed half a century before Einstein proposed his theory?No need for relativity. A classical model does just fine.
https://en.wikipedia.org/wiki/Fizeau_experiment
Sylvia
No, it doesn't. You simply don't know the problems involved.
will have a slightly greater refractive index.
Is today "make up a fact day" or something?
the following: “ The refractive index of the material is an indicator
of its optical density.”
Yet you need to show that for a speed v, the index of refraction forPerhaps it does, do you have a reference for index of refraction vs.I’m not trying to write a textbook on refraction and optical
density?
density.
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 ofConjecture.
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.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 densityAgain, it's your wacky theory, you provide the evidence.
increasing proportional to observed speed of light in a transparent medium.
Please supply your 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.
Again, your data?
Fizeau experiment.Logical Fallacy alert! Assuming your conclusion. You have to show a
cause and effect relationship.
Proof that moving water relative to a source willYou cannot prove a negative.
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.Sure, Albert was inspired by an old experiment. Nothing wrong with that!
Relativists, in their infinite lack of wisdom,No, cranks make up false beliefs like Einstein stealing formulas because they have nothing else to go on.
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 thatYou are assuming your conclusion again! You only know that Fizeau
the optical density of the moving water does slow the light
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 onSo today is make up a fact day instead.
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 communityNobody "ignored" it. It was an unsolved mystery of physics before Einstein.
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 accuratelyCannot prove a negative. And it is you who has no data like indexes of refraction changing with v.
model the observations made in Fizeau.
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.
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 indexBut you claimed before that the amount of water traversed affected the
of the medium changes...the speed changes. But distance travelled
will not affect the refractive index or speed of light in any medium.
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 showingCorrect. Fizeau stumbled across the relativistic speed combination
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.
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.Assuming your conclusion. Logical fallacy.
And believe me, if you actuallyYour evidence of this?
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 speedEvidence?
isnt c/n +-v.
I supplied a formula to model this change in speed.A "model" based on zero evidence? Why would anyone care about THAT?
Or youWhich is not the same as yours. Which one do you believe is correct?
can refer to Fizeau’s formula.
(hint: it can't be both).
Either one accurately predicts how muchNo, both cannot be the correct formula. And you cannot use Fizeau's
light speed slows relative to the source if the water is at v.
formula as proof of varying indexes of refraction without the logical fallacy of assuming your conclusion.
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 waterBut you have no evidence of that.
if the water moves in the tube.
And the reason is that the opticalAssertions are not evidence.
density of the water increases relative to the source if the water is at v.
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 toAgain, your wacky claim, you provide your evidence.
the source in Fizeau when the water is at v?
No.
Thought not.
Nope. Fizeau didn't get his expected classical c/n ± v.Theoretically. But a classical explanation doesn’t rely on magic.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 works just as well.
Seeing as both Fizeau’s and my formulaNeither got the classical c/n ± v.
can accurately predict this classical effect.
Also...you forgot...the formula SR uses...is stolen from classical theoryHow?
As per usual for relativists.
Fizeau's equation was DERIVED from his results. Your formula is, well,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.
you know...
How could it "work as well"? Why even bother with it?I didn’t say it was better. I said both worked as well.
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.
And proved thatNo, you need to show measurements of n vs. speed of water, and reference
the optical density of the moving water relative to the source increases or decreases with+-v. A purely classical phenomena.
the experimental data and its author in any such beliefs. Instead, you
just made it up.
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 movesOf course. Fizeau knew the speed of light in water was c/n. Which is why
at v in the tube.
You can’t.
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.Assuming your conclusion.
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 irreleventHow could it if you don't even know what the change is at a given
because Fizeau’s formula also correctly predicts the optical density changes when water is at v.
velocity v?
Assuming the conclusion is no basis for any theory.Only if you ignore Fizeau.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
contradicted classical theory.
No. Classical theory predicts light speed changes in water if the water moves in the tube at v. And succesfully predicted for classical theoryHow could it if you don't even know how n changes with v (if it does)?
by Fizeau : c/n+-v(1-(1/n^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 measuresThat's not the classical formula. c/n ± v is.
the change in lightspeed ,...he correctly models it with the classical formula c/n+-v(1-(1/n^2))
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.
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?
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.)
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 ina 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
x' = (x-vt)/sqrt(12-v^2/c^2)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
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.
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 believeEinstein 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
x'=x-vtground, 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
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
x = x'- m'n'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.
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
On Friday, 13 October 2023 at 13:43:32 UTC+2, Paul B. Andersen wrote: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
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.
Note:
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.
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
I don't buy it, Robert.
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.
I think you are talking nonsense to evade addressing the issue.
So please, answer the questions.
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:Your tales of observers were nothing but funny even
If the observer in the airplane had a clock which ran
too slow, the observer in the plane and the observer
300 years ago. Face it, trash, you have no clue
about observation process.
On Friday, 13 October 2023 at 13:43:32 UTC+2, Paul B. Andersen wrote: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
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.
Note:
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.
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
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 don't buy it, Robert.
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.
I think you are talking nonsense to evade addressing the issue.
So please, answer the questions.
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:Your tales of observers were nothing but funny even
If the observer in the airplane had a clock which ran
too slow, the observer in the plane and the observer
300 years ago. Face it, trash, you have no clue
about observation process.
On Monday, October 9, 2023 at 11:27:35 AM UTC-7, Robert Winn wrote: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
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
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 thex' = (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.
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 hisIsaac 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
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 forx'=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
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.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
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
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
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.
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,
Sure, assuming he started from rest RELATIVE TO SOME INERTIAL FRAME.
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 (she) does not age instantaneously, [...]
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.
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!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
--
Paul
https://paulba.no/
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.)
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,
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.
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 meMutual 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
x'=x-vtclock.
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
x=ct, x' = cn'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
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
Notice his formula doesn’t do so well for other planets.
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
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.
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 meMutual 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
moving clock.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
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 thex=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
I would ignore Einsteins “predictions” for the mercury anomalous preccession.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
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 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 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}
As follows:
Planet. Classical
Merc. 43.24
V. 8.33
E. 4.49
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.
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.
On 10/16/23 12:27 PM, Tom Roberts wrote:
(I, Mike Fontenot, 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.
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.
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, [...]
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.
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.
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.
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.
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.
Den 16.10.2023 04:54, 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
On Sunday, October 15, 2023 at 1:53:29 PM UTC-7, Paul B. Andersen wrote:I didn't really write this for you, Robert.
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 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'sEinstein 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
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>Send it to Congress. They will send you some money. Just tell them you believe in Einstein's theory.
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
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 didn't really write this for you, Robert.
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 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!
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.
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.
Den 17.10.2023 05:51, skrev Robert Winn: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
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
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/
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 slowerIt 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.
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:
https://www.kevinaylward.co.uk/gr/xht/twinsparadox/twinsparadox.xht
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 slowerIt 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.
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/Einstein definitely said the moving clock was slower.Enough now, Robert.
--
Paul
https://paulba.no/
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 slowerIt 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
Den 16.10.2023 14:54, skrev Lou:
I would ignore Einsteins “predictions” for the mercury anomalous preccession.See:
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.
https://paulba.no/pdf/GRPerihelionAdvance.pdf
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 2.5https://arxiv.org/pdf/0802.0176.pdf
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)
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 perehilionMercury:
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}
Perihelion distance r = 4.60011E10 m
Solar radius R = 696340E3 m
If the equation is r/(r+3R)^2 we getSorry lost in translation from paper to google post
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?
Please explain!
As follows:
Planet. Classical
Merc. 43.24
V. 8.33
E. 4.49
Abjit Biswas is a well known crank, LooLoo. The third column is a fake.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
On Wednesday, October 18, 2023 at 7:00:14 AM UTC-7, Lou wrote:
Abjit Biswas is a well known crank, LooLoo. The third column is a fake.https://arxiv.org/pdf/0802.0176.pdfPlanet. 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
On 10/17/23 12:31 PM, Kevin Aylward wrote:
A correct resolution of the twins paradox, without acceleration or
frame switching is here:
https://www.kevinaylward.co.uk/gr/xht/twinsparadox/twinsparadox.xht
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.
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:
Maybe. But then again I didnt cite that paper in my post.Abjit Biswas is a well known crank, LooLoo. The third column is a fake.https://arxiv.org/pdf/0802.0176.pdfPlanet. 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
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.
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?
Please explain!
Sorry lost in translation from paper to google post
It should read 1/(r+3R)^2
You will get 4.324 x 10-16 for mercury
And if you calculate all 4 planets you will see the progression clearly.
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:
https://www.kevinaylward.co.uk/gr/xht/twinsparadox/twinsparadox.xht
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.
On 10/17/23 12:31 PM, Kevin Aylward wrote:
A correct resolution of the twins paradox, without acceleration or frame
switching is here:
https://www.kevinaylward.co.uk/gr/xht/twinsparadox/twinsparadox.xht
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.
Den 17.10.2023 21:14, skrev Mike Fontenot:
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.
Den 18.10.2023 16:00, skrev Lou:There exists no purer ignorance than relativity.
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}
If r and R are in kmMercury:Sorry lost in translation from paper to google post
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?
Please explain!
It should read 1/(r+3R)^2
You will get 4.324 x 10-16 for mercury
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?
https://paulba.no/pdf/GRPerihelionAdvance.pdf
--
Paul
https://paulba.no/
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:
LooLooMaybe. But then again I didnt cite that paper in my post.Abjit Biswas is a well known crank, LooLoo. The third column is a fake.https://arxiv.org/pdf/0802.0176.pdfPlanet. 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
Look at your post, you are getting nuttier and nuttier.
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}
If r and R are in kmMercury:Sorry lost in translation from paper to google post
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?
Please explain!
It should read 1/(r+3R)^2
You will get 4.324 x 10-16 for mercury
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?
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.
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.Close enough.
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.
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.
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'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.
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.Close enough.
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.
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.
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 andThe 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?
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,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.
On Thursday, October 12, 2023 at 8:35:11 PM UTC-7, Laurence Clark Crossen wrote: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
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
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 thex' = (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.
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 hisIsaac 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
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 forx'=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
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.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
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 hasYes, 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
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 equationscannot 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
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.
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
It should read 1/(r+3R)^2
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.
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.
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.
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.
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 meMutual 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
moving clock.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
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 thex=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
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 didI 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 perehilionI 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
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
On Thursday, October 19, 2023 at 11:35:59 AM UTC-7, Laurence Clark Crossen wrote:you haven't exhibited any knowledge at all paul the heckler.
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...
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.Close enough.
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.
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.
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 andThe 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?
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,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.
I have only assumed the speed of the planes. Hafael Keating dont supplyGranted, 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
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.)
Regarding your point about the speeds relative to earth etc.
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.
On Thursday, October 19, 2023 at 11:35:59 AM UTC-7, Laurence Clark Crossen wrote:Heckling is a characteristic of the envious who are preoccupied with tearing down others instead of developing themselves.
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...
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.Close enough.
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.
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.
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 andThe 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?
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,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.
I have only assumed the speed of the planes. Hafael Keating dont supplyYou may like Xiaochun Mei's articles especially on Mercury's anomalous precession.
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.)
Regarding your point about the speeds relative to earth etc.
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.
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.Close enough.
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.
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.
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 andThe 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?
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,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.
I have only assumed the speed of the planes. Hafael Keating dont supply"Kelly also managed to get a memo that Hafele wrote a year after the experiment which recorded Hafele’s internal reports. Hafele frankly admitted: “Most people (himself included) do not think that the time becoming fast of these atomic clocks means
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.)
Regarding your point about the speeds relative to earth etc.
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.
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.Close enough.
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.
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.
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 theThe 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?
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.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.
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 andI 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.)
Regarding your point about the speeds relative to earth etc.
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 1600Granted, 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
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.
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
It should read 1/(r+3R)^2
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.
Did you miss this?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
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?
No answer?
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. NoticeGobbledegook!
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.
Did you try to say something about the logic that shows
that your 'formula' is nonsense?
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:
δφ Perihelion advance per orbit [rad/orbit]
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.
The advance per orbit will:
Increase with the mass of the star.
Decrease with the semi major axis.
Increase with the eccentricity.
According to your equation 1/(r+3R)² the advance is independent of
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 understandingI see.
of the mechanism of how orbital paths are tugged at by the Sun
as they make their nearest approach to the sun.
You claim Newton was wrong since it according to Newton
is no perihelion advance of a single planet orbiting a star.
We can agree about that.
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?So can you please show the equations for your theory
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?
The observer on the ground feels push from gravity. But they are also being subjected to
a constant horizontal force and are being pushed around in a circle around the Center
of the earth at 1600kph.
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.
On Thursday, October 19, 2023 at 3:24:09 PM UTC-7, Laurence Clark Crossen wrote:
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.My dead dog knows more physics than you do, Larry...
On October 19, Lou wrote:
The observer on the ground feels push from gravity. But they are also being subjected toAn object in orbit around the earth feels a tangential force?
a constant horizontal force and are being pushed around in a circle around the Center
of the earth at 1600kph.
--
Rich
The observer on the ground feels push from gravity. But they are also
being subjected to a constant horizontal force and are being pushed
around in a circle around the Center of the earth at 1600kph.
An object in orbit around the earth feels a tangential force?
Depends on what you mean by tangential force.
Does the object head off in a tangent ?
No it doesn’t. It always seems to change direction and rotate
in a circle around the earths axis.
Presumably this is because the earths gravity pulls by the
same amount of force that is pushing the object tangentially
out into the universe. Some sort of equilibrium has been achieved
dictated by force.
If you want a more detailed description of what happens then
probably using a LeSage push gravity concept would explain it all.
The observer on the ground feels push from gravity.
But they are also being subjected to a constant horizontal force and
are being pushed around in a circle around the Center of the earth at 1600kph.
On Thursday, October 19, 2023 at 3:24:09 PM UTC-7, Laurence Clark Crossen wrote:Maybe your dog could make comments with substance.
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.My dead dog knows more physics than you do, Larry...
On Thursday, 19 October 2023 at 21:33:27 UTC+1, Paul B. Andersen wrote:
Den 18.10.2023 16:00, skrev Lou:
Sorry lost in translation from paper to google post
It should read 1/(r+3R)^2
You will get 4.324 x 10-16 for mercury
And if you calculate all 4 planets you will see the progression clearly.
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?
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:
δφ Perihelion advance per orbit [rad/orbit]
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 [m/s]
The equation shows the advance for one orbit
irrespective of which planet it is.
According to your equation 1/(r+3R)² the advance is independent of
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.
There is no area (ie square km) in 1/(r+3R)².
You made that up. You are so obsessed with your maths
that it has made you incapable of understanding fundamental
physics. Forget area.
All that formula does is give a mathematical relationship between
perehilion distance and how the suns mass is spread across its volume
Notice the observed perehilion increases for each planet the farther
out it is. So r is actually representing not just distance but
mass and speed. And R isn’t just distance...it also represents
mass and volume and diameter of the sun.
Anyways you can object all you want about how much you hate
classical theory being able to not only explain anomalous preccesion..
but to predict it better than Relativity.
The one and only fact that matters is....the formula works.
Better than GR.
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.
Anomalous preccession is caused by suns mass spread across its
volume. And Newton’s formula forgets to account for this effect at perehilion.
Den 20.10.2023 02:01, skrev Lou:
The observer on the ground feels push from gravity.Right.
The ground is exerting an upwards force F = gm on the observer,
where m is the mass of the observer and g is the upwards gravitational acceleration.
But they are also being subjected to a constant horizontal force and
are being pushed around in a circle around the Center of the earth at 1600kph.
You are right, of course.
Everybody knows that if you have wheels on your chair,
the brakes must be on if you want it to stay in front
of your desk.
On October 20, Lou wrote:
The observer on the ground feels push from gravity. But they are also >>> being subjected to a constant horizontal force and are being pushed
around in a circle around the Center of the earth at 1600kph.
An object in orbit around the earth feels a tangential force?
Depends on what you mean by tangential force.It's called centripetal force.
Does the object head off in a tangent ?
No it doesn’t. It always seems to change direction and rotate
in a circle around the earths axis.
Presumably this is because the earths gravity pulls by the
same amount of force that is pushing the object tangentially
out into the universe. Some sort of equilibrium has been achieved
dictated by force.
If you want a more detailed description of what happens then
probably using a LeSage push gravity concept would explain it all.
Den 20.10.2023 02:37, skrev Lou:
On Thursday, 19 October 2023 at 21:33:27 UTC+1, Paul B. Andersen wrote:
Den 18.10.2023 16:00, skrev Lou:
Sorry lost in translation from paper to google post
It should read 1/(r+3R)^2
You will get 4.324 x 10-16 for mercury
And if you calculate all 4 planets you will see the progression clearly.
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?
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:
δφ Perihelion advance per orbit [rad/orbit]
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 [m/s]
Note that the dimensions are the same on both sides
of the equality sign. Radians is a dimensionless number.
dimensionless number = [m³/kg⋅s²]⋅[kg]/([m]⋅[m/s]⋅[m/s])
= m³⋅kg⋅s²/(m³⋅kg⋅s²) = dimensionless number
The equation shows the advance for one orbit
irrespective of which planet it is.
According to your equation 1/(r+3R)² the advance is independent of
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.
There is no area (ie square km) in 1/(r+3R)².I stand corrected! :-J
You made that up. You are so obsessed with your maths
that it has made you incapable of understanding fundamental
physics. Forget area.
(r+3R)² isn't an area!
And since r and R are in km, the non area isn't in square km.
One can but admire your capability to understand basic physics.
All that formula does is give a mathematical relationship between perehilion distance and how the suns mass is spread across its volume Notice the observed perehilion increases for each planet the fartherI see. So:
out it is. So r is actually representing not just distance but
mass and speed. And R isn’t just distance...it also represents
mass and volume and diameter of the sun.
r = 4.6001136690E7 km represents the mass 3.301132E23 kg
and the speed 47.87 km/s
r = 10.747582129E7 km represents the mass 4.867453E24 kg
and the speed 35.0 km/s
r = 14.709844432E7 km represents the mass 6.045809E24 kg
and the speed 29.8 km/s
r = 20.664951765E7 km represents the mass 6.417094E24 kg
and the speed 24.1 km/s
But how can these represented masses and speeds affect
the result when they don't appear in your formula?
But since you say so, they must affect the result.
So the dimension of r is (distance x mass x speed)
In SI units: the dimension of r is [kg⋅km²/s]
and the dimensions in your formula are:
arcsecs = s²/(kg²⋅km⁴)
Which seems very reasonable, doesn't it?
Anyways you can object all you want about how much you hateSure! Well done!
classical theory being able to not only explain anomalous preccesion..
but to predict it better than Relativity.
The one and only fact that matters is....the formula works.
Better than GR.
You didn't answer this question!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.
The GR equation is very simple, it gives the perihelion advance
for one single orbit.
But your equation give the perihelion advance after a number of orbits,
so what is it in the r that say how many orbits the advance is for?
Does r also represent a number so that:
r = 4.6001136690E7 km represents 415.2 orbits
r = 10.747582129E7 km represents 162.5 orbits
r = 14.709844432E7 km represents 100 orbits
r = 20.664951765E7 km represents 53.2 orbits
As I am incapable of understanding fundamental
physics, I don't understand how you from r can
see how many orbits the planet has to do before
the advance is as your formula shows.
So can you, who ARE capable of understanding basic physics
please explain it?
Anomalous preccession is caused by suns mass spread across itsI see. You claim that Newtons F = GMm/r² is wrong
volume. And Newton’s formula forgets to account for this effect at perehilion.
because it doesn't account for the volume of the Sun.
That means that you can't use Newton to calculate
the orbits of planets.
What 'classical' theory are you then referring to?
All you need to calculate orbits according to Newton is
F = ma and F = GMm/r².
But your theory must be based on F = ma and F = f(G,M,m,r,Rsun)
Can you please show what the function f(G,M,m,r,Rsun) is,
and show how it leads your formula?
Should be a piece of cake for someone with your ability
to understand basic physics!
Den 20.10.2023 02:01, skrev Lou:
The observer on the ground feels push from gravity.Right.
The ground is exerting an upwards force F = gm on the observer,
where m is the mass of the observer and g is the upwards gravitational acceleration.
And since you are capable of understanding fundamental
physics, you know that the dimensions must be the same on
both sides of the equal sign, don't you?
Here we have force = acceleration X mass,
In the SI-system N = (m/s²)X(kg)
It would be rather silly if we wrote something like
arcsec/century = 1/km², don't you agree?
But they are also being subjected to a constant horizontal force andYou are right, of course.
are being pushed around in a circle around the Center of the earth at 1600kph.
Everybody knows that if you have wheels on your chair,
the brakes must be on if you want it to stay in front
of your desk.
[...]
adding more weight or mass to a resonant system causes its natural
resonant frequency to decrease.
On Thursday, October 19, 2023 at 6:59:42 PM UTC-7, Paul Alsing wrote:
On Thursday, October 19, 2023 at 3:24:09 PM UTC-7, Laurence Clark Crossen wrote:
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.
My dead dog knows more physics than you do, Larry...
Maybe your dog could make comments with substance.
On 10/19/23 7:01 PM, Lou wrote:
[...]
Your GUESS that an applied force affects the tick rate of an atomic
clock is just plain wrong, refuted by several experiments referenced
below, and also by simple observation of a pendulum clock.
Your GUESS does not work for a pendulum clock -- its tick rate scales as sqrt(g) which is proportional to 1/r, not 1/r^2.
The coup de gras for your GUESS: a pendulum clock ticks SLOWER at higher altitude, while an atomic clock ticks FASTER.
(The former is due to changing the timekeeping mechanism
of the pendulum clock, while the latter is due to
"gravitational time dilation".)
You REALLY need to learn very basic physics
On Friday, 20 October 2023 at 19:36:10 UTC+1, Paul B. Andersen wrote:
Den 20.10.2023 02:01, skrev Lou:
The observer on the ground feels push from gravity.
But they are also being subjected to a constant horizontal force and
are being pushed around in a circle around the Center of the earth at 1600kph.
You are right, of course.
Everybody knows that if you have wheels on your chair,
the brakes must be on if you want it to stay in front
of your desk.
I think focussing on maths is not a good idea for a theorist.
You can’t see the Norwegian wood for the trees.
You forgot...it isn’t just *you* travelling at 1600 kph in a circle
around the Center of the earth.
It’s the whole room!! Fact is it’s the whole planet!!
Wake up and smell the coffee Paul.
They do have coffee up there in the Norwegian Arctic do they?
Or maybe you have to make do with Lingonberry and birchbark tea.
On Friday, 20 October 2023 at 19:27:39 UTC+1, Paul B. Andersen wrote:
Den 20.10.2023 02:37, skrev Lou:
On Thursday, 19 October 2023 at 21:33:27 UTC+1, Paul B. Andersen wrote: >>>>>> Den 18.10.2023 16:00, skrev Lou:
Sorry lost in translation from paper to google post
It should read 1/(r+3R)^2
You will get 4.324 x 10-16 for mercury
And if you calculate all 4 planets you will see the progression clearly.
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.
All that formula does is give a mathematical relationship between
perehilion distance and how the suns mass is spread across its volume
Notice the observed perehilion increases for each planet the farther
out it is. So r is actually representing not just distance but
mass and speed. And R isn’t just distance...it also represents
mass and volume and diameter of the sun.
You don’t understand how 1/(r+3R)² does a better job of modelling perehilion advance than relativities version?
Or you don’t want to understand?
Anomalous preccession is caused by suns mass spread across its
volume. And Newton’s formula forgets to account for this effect at perehilion.
Well, when you ditch magic and go for basic physics...then maybe you
will understand how 1/(r+3R)² accurately models the anomolous
preccesion of planets.
Den 21.10.2023 00:20, skrev Lou:
On Friday, 20 October 2023 at 19:36:10 UTC+1, Paul B. Andersen wrote:
Den 20.10.2023 02:01, skrev Lou:
The observer on the ground feels push from gravity.
But they are also being subjected to a constant horizontal force and
are being pushed around in a circle around the Center of the earth at 1600kph.
You are right, of course.
Everybody knows that if you have wheels on your chair,
the brakes must be on if you want it to stay in front
of your desk.
I think focussing on maths is not a good idea for a theorist.Please tell me one thing, Lou.
You can’t see the Norwegian wood for the trees.
You forgot...it isn’t just *you* travelling at 1600 kph in a circle around the Center of the earth.
It’s the whole room!! Fact is it’s the whole planet!!
Wake up and smell the coffee Paul.
They do have coffee up there in the Norwegian Arctic do they?
Or maybe you have to make do with Lingonberry and birchbark tea.
You sit on a chair with wheels.
Do you _really_ believe that you are subjected to a constant
horizontal force pushing you in the eastward direction?
If yes, why don't you accelerate eastwards according
to Newton's law a = F/m?
Den 21.10.2023 00:06, skrev Lou:
On Friday, 20 October 2023 at 19:27:39 UTC+1, Paul B. Andersen wrote:
Den 20.10.2023 02:37, skrev Lou:
On Thursday, 19 October 2023 at 21:33:27 UTC+1, Paul B. Andersen wrote: >>>>>> Den 18.10.2023 16:00, skrev Lou:
Sorry lost in translation from paper to google post
It should read 1/(r+3R)^2
You will get 4.324 x 10-16 for mercury
And if you calculate all 4 planets you will see the progression clearly.
Utter nonsense, but funny nonsense. :-DAll that formula does is give a mathematical relationship between
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.
perehilion distance and how the suns mass is spread across its volume >>> Notice the observed perehilion increases for each planet the farther
out it is. So r is actually representing not just distance but
mass and speed. And R isn’t just distance...it also represents
mass and volume and diameter of the sun.
You don’t understand how 1/(r+3R)² does a better job of modelling perehilion advance than relativities version?I do indeed understand.
Or you don’t want to understand?
It is very obvious how you arrived at the equation.
You noticed that the perihelion advance in "/century of
the three inner planets were approximately proportional
to 1/r^2 where r is the perihelion distance of the planets.
But to make the function fit the three points better, you
wrote: 1/(r + fudge-factor)^2 and by trying you found
that the fudge factor should be 3 solar diameters.
What you forgot is that your equation should be:
(1E17 km²/arcsec)/(r + 3R)² = number of arcsec per century
This isn't physics, it is simple curve fitting.
And your equation doesn't do very well:
https://paulba.no/paper/Clemence.pdf
Observed anomalous precession:
Mercury: (42.56 ± 0.94) "/century
https://arxiv.org/pdf/0802.0176.pdf
Observed anomalous precession:
Venus: 8.6247 ± 0.0005) "/century
Earth: 3.8387 ± 0.0004) "/century
https://paulba.no/pdf/GRPerihelionAdvance.pdf
Simulated values shown below:
All numbers are "/century
Planet Observed GR Lou Simulert
Mercury 42.56 42.9827 43.240 42.979
Venus 8.6247 8.6249 8.330 8.624
Earth 3.8387 3.8389 4.493 3.838
Mars ? 1.3510 2.295 1.351
To make a better curve fit, you could have set:
Mercury: r₁ = 4.6001136690E7 km
Venus: r₂ = 10.747582129E7 km
Earth: r₃ = 14.709844432E7 km
Find the constants a, b and c by solving the three equations:
a⋅r₁² + b⋅r₁ + c = 42.56"
a⋅r₂² + b⋅r₂ + c = 8.6247"
a⋅r₃² + b⋅r₃ + c = 3.8387"
The dimension of a: "/km², of b: "/km, of c: " .
Then the equation: arcsec = a⋅r² + b⋅r + c
would give an exact fit of the observed values for the three planets.
But like your equation, it would not work at all for other than
the three planets. Because it isn't physics at all.
A physic equation, like δφ = 6πGM/a(1−e²)c²
can predict the perihelion advance for ANY planet orbiting ANY star.
Meaningless babble.Anomalous preccession is caused by suns mass spread across its
volume. And Newton’s formula forgets to account for this effect at perehilion.
Your equations is not physics.
But you are right when saying that Newton was wrong.
F = GMm/r² can't produce a perihelion advance.
GR can. And you can see an approximated equation for
the gravitational acceleration of GR here: https://paulba.no/PerihelionAdvance.html
Note that in addition to the centripetal acceleration
there is an acceleration along the velocity vector
which will be responsible for the rotation of the elliptic
orbit.
Well, when you ditch magic and go for basic physics...then maybe youCurve fitting isn't basic physics, it is basic math.
will understand how 1/(r+3R)² accurately models the anomolous
preccesion of planets.
But you did obviously not know what you were doing.
On 10/19/23 7:01 PM, Lou wrote:
[...]
Your GUESS that an applied force affects the tick rate of an atomic
clock is just plain wrong, refuted by several experiments referenced
below, and also by simple observation of a pendulum clock.
Your GUESS does not work for a pendulum clock -- its tick rate scales as sqrt(g) which is proportional to 1/r, not 1/r^2.
The coup de gras for your GUESS: a pendulum clock ticks SLOWER at higher altitude, while an atomic clock ticks FASTER.
(The former is due to changing the timekeeping mechanism
of the pendulum clock, while the latter is due to
"gravitational time dilation".)
You REALLY need to learn very basic physics before attempting to write
about it. Note also the Newtonian physics has no "time dilation" of any kind.
Here are two formal experiments that show that your basic claim that a
force applied to a clock makes it tick slower is refuted experimentally:
[1 g = 9.8 m/s^2, a convenient unit of acceleration.]
- Sherwin, "Some Recent Experimental Tests of the 'Clock
Paradox'", Phys. Rev. 129 no. 1 (1960), pg 17.
He discusses some Moessbauer experiments that support
the thesis that the tick rate of a clock is independent
of its acceleration (~10^16 g), and depends only on
its speed.
- Bailey et al., "Measurements of relativistic time dilation
for positive and negative muons in a circular orbit,"
Nature 268 (July 28, 1977) pg 301.
- Bailey et al., Nuclear Physics B 150 pg 1–79 (1979).
They observed that muons in their storage ring had a
lifetime consistent with inertially moving muons at
the same speed; their muons were subject to the
enormous acceleration of 10^18 g.
Elsewhere you said:
adding more weight or mass to a resonant system causes its natural resonant frequency to decrease.
This is true for mass, but is true for weight ONLY when gravity provides
the restoring force (e.g. in a pendulum clock). There is no way to
change the mass of the timekeeping resonant system of an atomic clock.
As mentioned above, a pendulum clock and an atomic clock respond QUITE DIFFERENTLY to a change in altitude.
Then the best evidence is...GPS. The natural resonant frequency of
Caesium atoms at 20000 km is increased when G force decreases.
At a rate which is r. Not r^2.
As predicted by classical theory.
On Saturday, 21 October 2023 at 10:17:05 UTC+1, Paul B. Andersen wrote:
Den 20.10.2023 02:01, skrev Lou:
The observer on the ground feels push from gravity.
But they are also being subjected to a constant horizontal force and >>>>> are being pushed around in a circle around the Center of the earth at 1600kph.
Please tell me one thing, Lou.
You sit on a chair with wheels.
Do you _really_ believe that you are subjected to a constant
horizontal force pushing you in the eastward direction?
If yes, why don't you accelerate eastwards according
to Newton's law a = F/m?
The whole room is accelerating. The house is accelerating And
so is the earth underneath its foundations.
Den 22.10.2023 14:54, skrev Lou:
On Saturday, 21 October 2023 at 10:17:05 UTC+1, Paul B. Andersen wrote:Sure.
The whole room is accelerating. The house is accelerating AndPlease tell me one thing, Lou.Den 20.10.2023 02:01, skrev Lou:
The observer on the ground feels push from gravity.
But they are also being subjected to a constant horizontal force and >>>>> are being pushed around in a circle around the Center of the earth at 1600kph.
You sit on a chair with wheels.
Do you _really_ believe that you are subjected to a constant
horizontal force pushing you in the eastward direction?
If yes, why don't you accelerate eastwards according
to Newton's law a = F/m?
so is the earth underneath its foundations.
According to Newton there is a centripetal acceleration
which make you, your house and the earth underneath its
foundations go in a circle. This acceleration is VERTICAL,
towards the centre of the Earth.
But you, Lou, wrote:
"The observer on the ground is also subjected to
a constant HORIZONTAL force."
So please answer my question:
You sit on a chair with wheels.
Do you _really_ believe that you are subjected to a constant
HORIZONTAL force pushing you in the eastward direction?
If yes, why don't you accelerate HORIZONTALLY eastwards
according to Newton's law a = F/m?
Please don't evade the question by restating that there is
a VERTICAL acceleration.
Then the best evidence is...GPS.
The natural resonant frequency of
Caesium atoms at 20000 km is increased when G force decreases.
At a rate which is r. Not r^2.
As predicted by classical theory.
On 10/22/2023 9:40 AM, Lou wrote:
Then the best evidence is...GPS.GPS is the best evidence of general relativity, since GR is specifically invoked in its design.
The natural resonant frequency ofNope. Gravitational force varies as 1/r^2, and is about 1/10th the
Caesium atoms at 20000 km is increased when G force decreases.
At a rate which is r. Not r^2.
ground value. Other GNS's have different orbits and thus different
offsets. They also vary according to 1/r, not 1/r^2.
As predicted by classical theory.I assume you mean Newtonian theory (relativity is a classical theory
these days), but Newtonian theory states gravitational forces vary as 1/r^2.
You tried grasping at some sort of "Newtonian 1/r variation" straw in
your desperate attempt to invalidate GR. Unfortunately for you, no such straw exists...you're going down for the third time... blub blub blub...
On Sunday, 22 October 2023 at 18:38:47 UTC+1, Paul B. Andersen wrote:
But you, Lou, wrote:
"The observer on the ground is also subjected to
a constant HORIZONTAL force."
So please answer my question:
You sit on a chair with wheels.
Do you _really_ believe that you are subjected to a constant
HORIZONTAL force pushing you in the eastward direction?
If yes, why don't you accelerate HORIZONTALLY eastwards
according to Newton's law a = F/m?
Please don't evade the question by restating that there is
a VERTICAL acceleration.
Yes I can see your point and you can get into semantics if you’d prefer. But I’m just trying to look at what actually is observed.
Why does NASA send its rockets out eastward to get an additional
1600kph speed boost if there is no eastward “force” to give this additional 1600kph to its rockets?
The horizontal force is there by virtue of being attached to the
horizontal 1600 kph spin of earths surface
Or if the earth somehow stopped spinning eastward, why does the generally accepted wisdom state that you, your chair, the room and house
would keep on spinning eastward at 1600kph?
What pushes the NASA rocket or you and the house eastward if there
is no eastward force pushing you?
On Saturday, 21 October 2023 at 21:22:50 UTC+1, Paul B. Andersen wrote:
Den 21.10.2023 00:06, skrev Lou:
You don’t understand how 1/(r+3R)² does a better job of modelling
perehilion advance than relativities version?
Or you don’t want to understand?
I do indeed understand.
It is very obvious how you arrived at the equation.
You noticed that the perihelion advance in "/century of
the three inner planets were approximately proportional
to 1/r^2 where r is the perihelion distance of the planets.
But to make the function fit the three points better, you
wrote: 1/(r + fudge-factor)^2 and by trying you found
that the fudge factor should be 3 solar diameters.
And your equation doesn't do very well:
https://paulba.no/paper/Clemence.pdf
Observed anomalous precession:
Mercury: (42.56 ± 0.94) "/century
https://arxiv.org/pdf/0802.0176.pdf
Observed anomalous precession:
Venus: 8.6247 ± 0.0005) "/century
Earth: 3.8387 ± 0.0004) "/century
Wrong
Earth is 5
Venus is 8.
Your “observed” data were made up by relativists to cover up the fact that
GR failed miserably.
Poor Paul. Doesn’t do physics.
Cant do physics seeimg as he’s spent all his life fiddling data with
fancy formulae.
Your website is full of nonsense.
Take one claim that you make that a classical model cannot
predict a path difference for Sagnac supposedly because
one has to take aether into account.!! Nonsense
You forgot... classical physics dumped the aether in 1898.
Well Before Einstein published his ridiculous ether for every observer nonsense.
If you do your calculations properly based on physics( not magic)
you will find A classical model DOES predict a path difference.
On 10/22/2023 9:40 AM, Lou wrote:
Then the best evidence is...GPS.GPS is the best evidence of general relativity, since GR is specifically invoked in its design.
The natural resonant frequency ofNope. Gravitational force varies as 1/r^2,
Caesium atoms at 20000 km is increased when G force decreases.
At a rate which is r. Not r^2.
On Sunday, 22 October 2023 at 20:54:32 UTC+2, Volney wrote:
On 10/22/2023 9:40 AM, Lou wrote:
Stupid Mike, your idiot guru has shown that there is no gravitationalThen the best evidence is...GPS.GPS is the best evidence of general relativity, since GR is specifically invoked in its design.
The natural resonant frequency ofNope. Gravitational force varies as 1/r^2,
Caesium atoms at 20000 km is increased when G force decreases.
At a rate which is r. Not r^2.
force. No surprise, of course, you don't know, samely as that
idiot Andersen.
On Sunday, 22 October 2023 at 19:54:32 UTC+1, Volney wrote:
On 10/22/2023 9:40 AM, Lou wrote:
Then the best evidence is...GPS.GPS is the best evidence of general relativity, since GR is specifically
invoked in its design.
The natural resonant frequency ofNope. Gravitational force varies as 1/r^2, and is about 1/10th the
Caesium atoms at 20000 km is increased when G force decreases.
At a rate which is r. Not r^2.
ground value. Other GNS's have different orbits and thus different
offsets. They also vary according to 1/r, not 1/r^2.
As predicted by classical theory.I assume you mean Newtonian theory (relativity is a classical theory
these days), but Newtonian theory states gravitational forces vary as 1/r^2. >>
You tried grasping at some sort of "Newtonian 1/r variation" straw in
your desperate attempt to invalidate GR. Unfortunately for you, no such
straw exists...you're going down for the third time... blub blub blub...
You are the one grasping at straws. I showed you how shadow classical
gravity was proportional to r not r^2.
In your desperation I notice you snipped the bit you
couldn’t answer. Here it is again:
On 10/22/2023 4:03 PM, Lou wrote:
On Sunday, 22 October 2023 at 19:54:32 UTC+1, Volney wrote:
On 10/22/2023 9:40 AM, Lou wrote:
Then the best evidence is...GPS.GPS is the best evidence of general relativity, since GR is specifically >> invoked in its design.
The natural resonant frequency ofNope. Gravitational force varies as 1/r^2, and is about 1/10th the
Caesium atoms at 20000 km is increased when G force decreases.
At a rate which is r. Not r^2.
ground value. Other GNS's have different orbits and thus different
offsets. They also vary according to 1/r, not 1/r^2.
As predicted by classical theory.I assume you mean Newtonian theory (relativity is a classical theory
these days), but Newtonian theory states gravitational forces vary as 1/r^2.
You tried grasping at some sort of "Newtonian 1/r variation" straw in
your desperate attempt to invalidate GR. Unfortunately for you, no such >> straw exists...you're going down for the third time... blub blub blub...
You are the one grasping at straws. I showed you how shadow classical gravity was proportional to r not r^2.I ignored the shadow because it is irrelevant. We're discussing the
In your desperation I notice you snipped the bit you
couldn’t answer. Here it is again:
effects of gravity going as inverse squares, not anything about shadows. Besides, for large distances (where the approximation sin(x) ~= 1 for
small intercepted angle x is good) it goes as an inverse square
function. For close distances, sin(x) ~= 1 doesn't hold, it will be a
more complex function involving the angle subtended by the shading
object, and it's not worth bothering to search/solve as it is irrelevant.
[snip irrelevancies]
Den 22.10.2023 20:20, skrev Lou:
On Sunday, 22 October 2023 at 18:38:47 UTC+1, Paul B. Andersen wrote:Please answer the question!
But you, Lou, wrote:
"The observer on the ground is also subjected to
a constant HORIZONTAL force."
So please answer my question:
You sit on a chair with wheels.
Do you _really_ believe that you are subjected to a constant
HORIZONTAL force pushing you in the eastward direction?
If yes, why don't you accelerate HORIZONTALLY eastwards
according to Newton's law a = F/m?
Please don't evade the question by restating that there is
a VERTICAL acceleration.
It is still:
You sit on a chair with wheels.
Do you _really_ believe that you are subjected to a constant
HORIZONTAL force pushing you in the eastward direction?
If yes, why don't you accelerate HORIZONTALLY eastwards
according to Newton's law a = F/m?
Please don't evade the question yet again by stating irrelevancies
like the below.
Yes I can see your point and you can get into semantics if you’d prefer. But I’m just trying to look at what actually is observed.--
Why does NASA send its rockets out eastward to get an additional
1600kph speed boost if there is no eastward “force” to give this additional 1600kph to its rockets?
The horizontal force is there by virtue of being attached to the horizontal 1600 kph spin of earths surface
Or if the earth somehow stopped spinning eastward, why does the generally accepted wisdom state that you, your chair, the room and house
would keep on spinning eastward at 1600kph?
What pushes the NASA rocket or you and the house eastward if there
is no eastward force pushing you?
Paul
https://paulba.no/
Den 22.10.2023 15:10, skrev Lou:
On Saturday, 21 October 2023 at 21:22:50 UTC+1, Paul B. Andersen wrote:
Den 21.10.2023 00:06, skrev Lou:
You don’t understand how 1/(r+3R)² does a better job of modelling
perehilion advance than relativities version?
Or you don’t want to understand?
But your equation 1/(r+3R)² is nonsense.I do indeed understand.
It is very obvious how you arrived at the equation.
You noticed that the perihelion advance in "/century of
the three inner planets were approximately proportional
to 1/r^2 where r is the perihelion distance of the planets.
But to make the function fit the three points better, you
wrote: 1/(r + fudge-factor)^2 and by trying you found
that the fudge factor should be 3 solar diameters.
You can't seriously claim that:
1/(r+3R)² = 4.3240E-16 per square km
means that the perihelion advance for mercury is 43.240 "/century
You could have made your equation meaningful like this:
(1E17 km²/arcsec)/(r + 3R)² = number of arcsec per century
And your equation doesn't do very well:
https://paulba.no/paper/Clemence.pdf
Observed anomalous precession:
Mercury: (42.56 ± 0.94) "/century
https://arxiv.org/pdf/0802.0176.pdf
Observed anomalous precession:
Venus: 8.6247 ± 0.0005) "/century
Earth: 3.8387 ± 0.0004) "/century
Wrong
Earth is 5
Venus is 8.
Your “observed” data were made up by relativists to cover up the fact that
GR failed miserably.
You think the correct observed values are
43.1, 8.0 and 5.0 for the planets Mercury,
Venus and Earth respectively.
These are the values you have tried to make your equation
produce. But your equation is a pretty bad job as a curve fitting
equation. Producing 43.240" in stead of 43.1, and 8.3302
in stead of 8.0, and 4.4929 in stead of 5.0 isn't very impressive.
That is because you didn't know what you were doing when you
experimented with your fudge-factor.
You could have done a much better job to make your
equation produce the values you wanted.
This is how:
Mercury: r₁ = 4.6001136690E7 km
Venus: r₂ = 10.747582129E7 km
Earth: r₃ = 14.709844432E7 km
Find the constants a, b and c by solving the three equations:
a⋅r₁² + b⋅r₁ + c = 43.1"
a⋅r₂² + b⋅r₂ + c = 8.0"
a⋅r₃² + b⋅r₃ + c = 5.0"
The dimension of a: "/km², of b: "/km, of c: " .
Then the equation: arcsec = a⋅r² + b⋅r + c
would give an exact fit of the three values you wanted.
But like your equation, it would not work at all for other than
the three planets. Because it isn't physics at all.
This has obviously nothing to do with physics; the values are what you wanted them to be, which may not have anything to do with reality.
A physic equation, like δφ = 6πGM/a(1−e²)c²
can predict the perihelion advance for ANY planet orbiting ANY star.
And you can't make the prediction fit your wishes.
Poor Paul. Doesn’t do physics.Your lethal arguments are noted.
Cant do physics seeimg as he’s spent all his life fiddling data with fancy formulae.
Your website is full of nonsense.
Take one claim that you make that a classical model cannotWhat are you trying to say?
predict a path difference for Sagnac supposedly because
one has to take aether into account.!! Nonsense
You forgot... classical physics dumped the aether in 1898.
Well Before Einstein published his ridiculous ether for every observer nonsense.
If you do your calculations properly based on physics( not magic)
you will find A classical model DOES predict a path difference.
It is a fact that the Sagnac experiment can be explained
with Galilean relativity and a stationary ether.
Sagnac thought he had proved the existence of the ether.
It is a fact that the Sagnac experiment can be explained by SR.
The Sagnac experiment doesn't falsify the existence of an ether,
and it doesn't falsify SR.
Didn't you know that?
I did. There is a horizontal force acting on your chair, the room your house etc.
My evidence is that NASA uses this very same force to add an extra 1600kph launch speed to their rockets.
Now answer this question...if there is no horizontal eastward force acting on
your chair house, NASA rockets etc..then where does the force come from to supply this extra 1600kph velocity to NASA rockets?
It is still:Yes Paul. Obviously you think NASA is imagining that their rockets are getting an extra 1600 kph horizontal boost when launched eastward.
You sit on a chair with wheels.
Do you _really_ believe that you are subjected to a constant
HORIZONTAL force pushing you in the eastward direction?
If yes, why don't you accelerate HORIZONTALLY eastwards
according to Newton's law a = F/m?
Please don't evade the question yet again by stating irrelevancies
like the below.
On Sunday, 22 October 2023 at 21:30:59 UTC+1, Paul B. Andersen wrote:
Den 22.10.2023 15:10, skrev Lou:
On Saturday, 21 October 2023 at 21:22:50 UTC+1, Paul B. Andersen wrote:
Den 21.10.2023 00:06, skrev Lou:
You don’t understand how 1/(r+3R)² does a better job of modelling >>>>> perehilion advance than relativities version?
Or you don’t want to understand?
I do indeed understand.
It is very obvious how you arrived at the equation.
You noticed that the perihelion advance in "/century of
the three inner planets were approximately proportional
to 1/r^2 where r is the perihelion distance of the planets.
But to make the function fit the three points better, you
wrote: 1/(r + fudge-factor)^2 and by trying you found
that the fudge factor should be 3 solar diameters.
But your equation 1/(r+3R)² is nonsense.
You can't seriously claim that:
1/(r+3R)² = 4.3240E-16 per square km
means that the perihelion advance for mercury is 43.240 "/century
! !
My formula more accurately predicts the preccession
of mercury than GR can.
You could have made your equation meaningful like this:
(1E17 km²/arcsec)/(r + 3R)² = number of arcsec per century
You can if you want. I’ll stick to my original one.
It can be translated into arc seconds, arc minutes, per century , per year , per orbit
etc.
Dont forget...Even Alberts formula has to get converted
mathematically seperately to whatever format is needed. His version
doesn’t automatically supply every variation of yrs,days ,orbits ,seconds centuries,minutes etc
On Sunday, 22 October 2023 at 21:39:36 UTC+1, Paul B. Andersen wrote:
Den 22.10.2023 20:20, skrev Lou:
On Sunday, 22 October 2023 at 18:38:47 UTC+1, Paul B. Andersen wrote:Please answer the question!
But you, Lou, wrote:
"The observer on the ground is also subjected to
a constant HORIZONTAL force."
So please answer my question:
You sit on a chair with wheels.
Do you _really_ believe that you are subjected to a constant
HORIZONTAL force pushing you in the eastward direction?
If yes, why don't you accelerate HORIZONTALLY eastwards
according to Newton's law a = F/m?
Please don't evade the question by restating that there is
a VERTICAL acceleration.
I did. There is a horizontal force acting on your chair, the room your house etc.
My evidence is that NASA uses this very same force to add an extra 1600kph launch speed to their rockets.
Now answer this question...if there is no horizontal eastward force acting on your chair house, NASA rockets etc..then where does the force come from to supply this extra 1600kph velocity to NASA rockets?
Yes Paul. Obviously you think NASA is imagining that their rockets are getting an extra 1600 kph horizontal boost when launched eastward.
How could they be so stupid!! There is no horizontal force acting on
their rockets to give an extra 1600kph velocity on lift off!!!
Could you email them and tell them to smarten up?
No wonder none of their rockets make it to orbit.
On Monday, October 23, 2023 at 4:29:35 AM UTC-5, Lou wrote:
I did. There is a horizontal force acting on your chair, the room your house etc.You are getting your terminology completely wrong. That is not
My evidence is that NASA uses this very same force to add an extra 1600kph launch speed to their rockets.
Now answer this question...if there is no horizontal eastward force acting on
your chair house, NASA rockets etc..then where does the force come from to supply this extra 1600kph velocity to NASA rockets?
It is still:Yes Paul. Obviously you think NASA is imagining that their rockets are getting an extra 1600 kph horizontal boost when launched eastward.
You sit on a chair with wheels.
Do you _really_ believe that you are subjected to a constant
HORIZONTAL force pushing you in the eastward direction?
If yes, why don't you accelerate HORIZONTALLY eastwards
according to Newton's law a = F/m?
Please don't evade the question yet again by stating irrelevancies
like the below.
a FORCE that is acting on you sitting in your chair.
If you don't get your basic physics terminology correct, how can you
expect to communicate with others?
Den 23.10.2023 11:29, skrev Lou:
On Sunday, 22 October 2023 at 21:39:36 UTC+1, Paul B. Andersen wrote:
Den 22.10.2023 20:20, skrev Lou:
On Sunday, 22 October 2023 at 18:38:47 UTC+1, Paul B. Andersen wrote: >>>>Please answer the question!
But you, Lou, wrote:
"The observer on the ground is also subjected to
a constant HORIZONTAL force."
So please answer my question:
You sit on a chair with wheels.
Do you _really_ believe that you are subjected to a constant
HORIZONTAL force pushing you in the eastward direction?
If yes, why don't you accelerate HORIZONTALLY eastwards
according to Newton's law a = F/m?
Please don't evade the question by restating that there is
a VERTICAL acceleration.
I did. There is a horizontal force acting on your chair, the room your house etc.So your answer to the first question is yes,
But you have not answered the second question:
If a force F is pushing you eastwards, why do you not
ACCELERATE eastwards according to Newton's law a = F/m?
My evidence is that NASA uses this very same force to add an extra 1600kph launch speed to their rockets.Your ignorance is really amazing. :-D
Now answer this question...if there is no horizontal eastward force acting on
your chair house, NASA rockets etc..then where does the force come from to supply this extra 1600kph velocity to NASA rockets?
Yes, your velocity in the non rotating ECI frame is
CONSTANT 1600 km/h in the horizontal eastwards direction.
It is CONSTANT because no horizontal force is acting on you.
Newtons 1. law.
If a horizontal, eastwards force was acting on you, your speed
would increase, you would move faster that your room and hit the wall.
Yes Paul. Obviously you think NASA is imagining that their rockets are getting an extra 1600 kph horizontal boost when launched eastward.In the ECI frame, the rocket is moving eastwards at 1600 km/h
before it is launched, when no force is acting on it.
So when the rocket fires, it is obviously best to accelerate eastwards, where you start with the speed 1600 km/h eastwards.
If you still haven't got it, you never will.
How could they be so stupid!! There is no horizontal force acting onI can only repeat: Your ignorance is amazing.
their rockets to give an extra 1600kph velocity on lift off!!!
Could you email them and tell them to smarten up?
No wonder none of their rockets make it to orbit.
Den 23.10.2023 11:21, skrev Lou:
On Sunday, 22 October 2023 at 21:30:59 UTC+1, Paul B. Andersen wrote:
Den 22.10.2023 15:10, skrev Lou:
On Saturday, 21 October 2023 at 21:22:50 UTC+1, Paul B. Andersen wrote: >>
Den 21.10.2023 00:06, skrev Lou:
You don’t understand how 1/(r+3R)² does a better job of modelling >>>>> perehilion advance than relativities version?
Or you don’t want to understand?
I do indeed understand.
It is very obvious how you arrived at the equation.
You noticed that the perihelion advance in "/century of
the three inner planets were approximately proportional
to 1/r^2 where r is the perihelion distance of the planets.
But to make the function fit the three points better, you
wrote: 1/(r + fudge-factor)^2 and by trying you found
that the fudge factor should be 3 solar diameters.
But your equation 1/(r+3R)² is nonsense.
You can't seriously claim that:
1/(r+3R)² = 4.3240E-16 per square km
means that the perihelion advance for mercury is 43.240 "/century
! !Your formula is a curve fitting function which give the values you have designed it to show. But you have only managed to give a number
My formula more accurately predicts the preccession
of mercury than GR can.
which is 1E-17 smaller that you wanted it be, and with
the dimension 1/km^2 in stead of arcsec.
But it would still be a curve fitting equationYou could have made your equation meaningful like this:
(1E17 km²/arcsec)/(r + 3R)² = number of arcsec per century
which only works (but not very well) for the three
planets it was designed to fit.
You can if you want. I’ll stick to my original one.You mean that 1/(r+3R)² = 4.3240E-16 per square km
It can be translated into arc seconds, arc minutes, per century , per year , per orbit
etc.
can be interpreted as arc seconds, arc minutes, per century ,
per year , per orbit etc.
Dont forget...Even Alberts formula has to get convertedThe GR equation δφ = 6πGM/a(1−e²)c²
mathematically seperately to whatever format is needed. His version doesn’t automatically supply every variation of yrs,days ,orbits ,seconds centuries,minutes etc
give radians per orbit which can't be interpreted as anything else.
On Monday, 23 October 2023 at 21:11:53 UTC+1, Paul B. Andersen wrote:
Den 23.10.2023 11:21, skrev Lou:
On Sunday, 22 October 2023 at 21:30:59 UTC+1, Paul B. Andersen wrote:
Den 22.10.2023 15:10, skrev Lou:
On Saturday, 21 October 2023 at 21:22:50 UTC+1, Paul B. Andersen wrote: >>>>
Den 21.10.2023 00:06, skrev Lou:
You don’t understand how 1/(r+3R)² does a better job of modelling >>>>>>> perehilion advance than relativities version?
Or you don’t want to understand?
I do indeed understand.
It is very obvious how you arrived at the equation.
You noticed that the perihelion advance in "/century of
the three inner planets were approximately proportional
to 1/r^2 where r is the perihelion distance of the planets.
But to make the function fit the three points better, you
wrote: 1/(r + fudge-factor)^2 and by trying you found
that the fudge factor should be 3 solar diameters.
But your equation 1/(r+3R)² is nonsense.
You can't seriously claim that:
1/(r+3R)² = 4.3240E-16 per square km
means that the perihelion advance for mercury is 43.240 "/century
! !
My formula more accurately predicts the preccession
of mercury than GR can.
Your formula is a curve fitting function which give the values you have
designed it to show. But you have only managed to give a number
which is 1E-17 smaller that you wanted it be, and with
the dimension 1/km^2 in stead of arcsec.
You could have made your equation meaningful like this:
(1E17 km²/arcsec)/(r + 3R)² = number of arcsec per century
But it would still be a curve fitting equation
which only works (but not very well) for the three
planets it was designed to fit.
You can if you want. I’ll stick to my original one.
It can be translated into arc seconds, arc minutes, per century , per year , per orbit
etc.
You mean that 1/(r+3R)² = 4.3240E-16 per square km
can be interpreted as arc seconds, arc minutes, per century ,
per year , per orbit etc.
Its not square km. That’s your fantasy.
It is a ratio. Do all the known planets. You get 4 numbers that
show exactly the proportionate difference of preccession there
is between all 4.
Anyways. You can’t change the fact ..it accurately predicts
the anomalous preccession . Better than GR. because it uses
the only 2 important parameters involved with the anomaly
Perehilion and Solar radius.
Einsteins didn’t work because the anomaly has nothing to do with
Orbital eccentricity or the speed of light. Etc. His is a fake.
"The observer on the ground is also subjected toa constant HORIZONTAL force."
Do you _really_ believe that you are subjected to a constant
HORIZONTAL force pushing you in the eastward direction?
If yes, why don't you accelerate HORIZONTALLY eastwards
according to Newton's law a = F/m?
I did. There is a horizontal force acting on your chair, the room your house etc.
So your answer to the first question is yes,
But you have not answered the second question:
If a force F is pushing you eastwards, why do you not
ACCELERATE eastwards according to Newton's law a = F/m?
My evidence is that NASA uses this very same force to add an extra 1600kph >> launch speed to their rockets.supply this extra 1600kph velocity to NASA rockets?
Now answer this question...if there is no horizontal eastward force acting on
your chair house, NASA rockets etc..then where does the force come from to
Yes, your velocity in the non rotating ECI frame is
CONSTANT 1600 km/h in the horizontal eastwards direction.
It is CONSTANT because no horizontal force is acting on you.
Newtons 1. law.
If a horizontal, eastwards force was acting on you, your speed
would increase, you would move faster that your room and hit the wall.
Den 24.10.2023 11:19, skrev Lou:
On Monday, 23 October 2023 at 21:11:53 UTC+1, Paul B. Andersen wrote:
Den 23.10.2023 11:21, skrev Lou:
On Sunday, 22 October 2023 at 21:30:59 UTC+1, Paul B. Andersen wrote: >>>>> Den 22.10.2023 15:10, skrev Lou:
On Saturday, 21 October 2023 at 21:22:50 UTC+1, Paul B. Andersen
wrote:
Den 21.10.2023 00:06, skrev Lou:
You don’t understand how 1/(r+3R)² does a better job of modelling >>>>>>>> perehilion advance than relativities version?
Or you don’t want to understand?
I do indeed understand.
It is very obvious how you arrived at the equation.
You noticed that the perihelion advance in "/century of
the three inner planets were approximately proportional
to 1/r^2 where r is the perihelion distance of the planets.
But to make the function fit the three points better, you
wrote: 1/(r + fudge-factor)^2 and by trying you found
that the fudge factor should be 3 solar diameters.
But your equation 1/(r+3R)² is nonsense.
You can't seriously claim that:
1/(r+3R)² = 4.3240E-16 per square km
means that the perihelion advance for mercury is 43.240 "/century
! !
My formula more accurately predicts the preccession
of mercury than GR can.
Your formula is a curve fitting function which give the values you have
designed it to show. But you have only managed to give a number
which is 1E-17 smaller that you wanted it be, and with
the dimension 1/km^2 in stead of arcsec.
You could have made your equation meaningful like this:
(1E17 km²/arcsec)/(r + 3R)² = number of arcsec per century
But it would still be a curve fitting equation
which only works (but not very well) for the three
planets it was designed to fit.
You can if you want. I’ll stick to my original one.
It can be translated into arc seconds, arc minutes, per century ,
per year , per orbit
etc.
You mean that 1/(r+3R)² = 4.3240E-16 per square km
can be interpreted as arc seconds, arc minutes, per century ,
per year , per orbit etc.
Its not square km. That’s your fantasy.
It is a ratio. Do all the known planets. You get 4 numbers that
show exactly the proportionate difference of preccession there
is between all 4.
Anyways. You can’t change the fact ..it accurately predicts
the anomalous preccession . Better than GR. because it uses
the only 2 important parameters involved with the anomaly
Perehilion and Solar radius.
Einsteins didn’t work because the anomaly has nothing to do with
Orbital eccentricity or the speed of light. Etc. His is a fake.
What amazes me, is how it is possible to be
so ignorant of your own ignorance.
Have you never wondered why the rest of the world
doesn't agree with you?
What amazes me, is how it is possible to be
so ignorant of your own ignorance.
Have you never wondered why the rest of the world
doesn't agree with you?
On Monday, 23 October 2023 at 22:04:42 UTC+1, Paul B. Andersen wrote:
Yes, your velocity in the non rotating ECI frame is
CONSTANT 1600 km/h in the horizontal eastwards direction.
It is CONSTANT because no horizontal force is acting on you.
Newtons 1. law.
Your ignorance is doubly amazing.
In fact: You and I are not going eastward. We are both accelerating in a circle at 1600kph. Around the earths core. Thanks to gravity pushing
us down. You will never win the argument. Because the fact is that the earth surface
is rotating at 1600kph and supplying the horizontal force to us.
If a horizontal, eastwards force was acting on you, your speed
would increase, you would move faster that your room and hit the wall.
You not only forgot about Gravity
You also forgot, I wont hit the wall because it’s also going at 1600kph.
I would need extra force on top of my 1600 to hit the wall.
Den 24.10.2023 11:33, skrev Lou:
On Monday, 23 October 2023 at 22:04:42 UTC+1, Paul B. Andersen wrote:
Yes, your velocity in the non rotating ECI frame is
CONSTANT 1600 km/h in the horizontal eastwards direction.
It is CONSTANT because no horizontal force is acting on you.
Newtons 1. law.
Your ignorance is doubly amazing.
In fact: You and I are not going eastward. We are both accelerating in a circle at 1600kph. Around the earths core. Thanks to gravity pushing
us down. You will never win the argument. Because the fact is that the earth surface
is rotating at 1600kph and supplying the horizontal force to us.
If a horizontal, eastwards force was acting on you, your speed
would increase, you would move faster that your room and hit the wall.
You not only forgot about GravityI have snipped most of our conversation,
You also forgot, I wont hit the wall because it’s also going at 1600kph. I would need extra force on top of my 1600 to hit the wall.
but the above sums it up pretty well.
-----------------
You are claiming to give a "classical" (Newtonian)
explanation of the H&K experiment, but you seem
to be ignorant of elementary "classical" physics.
Below is the elementary Newtonian physics relevant
to this thread.
I will advice you not to dispute what the rest of
the world has known for centuries.
According to Newton's gravitation there is a force acting on you.
F = GMm/r²,
where G is the gravitational constant, M is the mass of the Earth,
m is your mass, and r is the radius of the Earth.
The direction of this force is downwards towards the centre of
the Earth.
In the non rotating Earth centred frame of reference, your velocity
is tangential to the surface (horizontal) and towards the east.
Your speed is constant v = 465.1⋅cos(latitude) m/s because
no force with direction along the velocity is acting on you.
So there is only one force acting on you. It is perpendicular
to your velocity in the ECI frame (vertical) and is giving you
a centripetal acceleration GMm/r² which is making you go
in a circle.
On Tuesday, 24 October 2023 at 20:14:53 UTC+1, Paul B. Andersen wrote:
You are claiming to give a "classical" (Newtonian)
explanation of the H&K experiment, but you seem
to be ignorant of elementary "classical" physics.
Below is the elementary Newtonian physics relevant
to this thread.
I will advice you not to dispute what the rest of
the world has known for centuries.
Pure and total nonsense.
According to Newton's gravitation there is a force acting on you.
F = GMm/r²,
where G is the gravitational constant, M is the mass of the Earth,
m is your mass, and r is the radius of the Earth.
The direction of this force is downwards towards the centre of
the Earth.
In the non rotating Earth centred frame of reference, your velocity
is tangential to the surface (horizontal) and towards the east.
Your speed is constant v = 465.1⋅cos(latitude) m/s because
no force with direction along the velocity is acting on you.
You are experiencing force from gravity downwards yes.
But you would experience that same amount of force even if the earth wasnt spinning!
So therefore seeing as the earth also spins eastward...this eastward
rotation also pushes you eastward with a seperate force. The only reason
why you don’t go flying off into space...is because the seperate downwards g force prevents you from doing so
And proof that your claim is nonsense is that NASA uses this seperate eastward
force to enable greater escape velocity against the downward force of earths gravity.
So there is only one force acting on you. It is perpendicular
to your velocity in the ECI frame (vertical) and is giving you
a centripetal acceleration GMm/r² which is making you go
in a circle.
Only one gravitational force down yes.
But the spinning earth also pushes you eastward.
And that is a seperate force.
Otherwise storing energy in spinning flywheel wouldn’t
work.
On October 24, Paul B. Andersen wrote:
What amazes me, is how it is possible to be
so ignorant of your own ignorance.
Actually, it's logically consistent.
If one holds an opinion on a topic, and becomes aware of his ignorance on that topic, he would retract that opinion until he's better informed. Therefore,
if he's ignorant, it's logical that he remains ignorant of his ignorance. (what if he's ignorant, yet holds a correct opinion? )
If it's something arcane, like Italian wine, then it's common to be ignorant, and self-aware.
Have you never wondered why the rest of the world
doesn't agree with you?
They said the same about Christopher Columbus -
Have you never wondered why the rest of the world
doesn't agree with you?
Two words.
Dunning. Kruger.
Yes Paul. It is you who forgot Newton’s first law:
“Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to *change its state by the action of an external force*. This tendency to resist changes in a state of motion is inertia.”
Notice that if there were no gravity from earth then you and the table would travel in a straight line at a uniform speed. And no longer be subject to the Horizontal eastward 1600kph force from the planet below you.
But there is another external force...it’s called gravity. And it’s constantly changing your direction of motion and preventing you from
leaving the surface of the planet and be freed of that constant 1600 kph force imparted to you from the planet beneath you.
So your mistake is to think you and your table are travelling at a uniform speed in a straight line.
You aren’t. You are rotating at 1600 kph in a circle.
And as I’ve already said, but you conveniently snipped, NASA uses
this tangental force from the earth on its rocket at liftoff to
get an additional 1600 kph velocity.
An additional 1600 velocity that you, Volney and a Rich D pretend
doesn’t exist!!😂🐿🌰🥜
On October 23, Paul B. Andersen wrote:
"The observer on the ground is also subjected toa constant HORIZONTAL force."
Do you _really_ believe that you are subjected to a constant
HORIZONTAL force pushing you in the eastward direction?
If yes, why don't you accelerate HORIZONTALLY eastwards
according to Newton's law a = F/m?
I did. There is a horizontal force acting on your chair, the room your house etc.
So your answer to the first question is yes,
But you have not answered the second question:
If a force F is pushing you eastwards, why do you not
ACCELERATE eastwards according to Newton's law a = F/m?
My evidence is that NASA uses this very same force to add an extra 1600kphsupply this extra 1600kph velocity to NASA rockets?
launch speed to their rockets.
Now answer this question...if there is no horizontal eastward force acting on
your chair house, NASA rockets etc..then where does the force come from to
Yes, your velocity in the non rotating ECI frame is
CONSTANT 1600 km/h in the horizontal eastwards direction.
It is CONSTANT because no horizontal force is acting on you.
Newtons 1. law.
If a horizontal, eastwards force was acting on you, your speedLou is confused by the friction between a heavy table and the floor. His mental model is that the table is naturally stationary, but subjected to a horizontal force by the rotating earth.
would increase, you would move faster that your room and hit the wall.
The interesting question here is how many physics students are confused
in the same way, after finishing the course. Or perhaps understood correctly,
but later forgot and reverted to previous ideas.
This experiment was done some years go, in economics. Freshmen, without previous study of the subject, were tested. They failed badly. Then, following
an introductory course, given the same test. They had learned much. Eight months later, tested again... and reverted to their previous misconceptions! Only a fraction of the coursework was retained in long term memory.
On 10/24/2023 6:04 AM, Paul B. Andersen wrote:
What amazes me, is how it is possible to be so ignorant of your ownTwo words. Dunning. Kruger.
ignorance. Have you never wondered why the rest of the world doesn't
agree with you?
Den 24.10.2023 21:56, skrev Lou:
On Tuesday, 24 October 2023 at 20:14:53 UTC+1, Paul B. Andersen wrote:I expected you to dispute it, though.
You are claiming to give a "classical" (Newtonian)
explanation of the H&K experiment, but you seem
to be ignorant of elementary "classical" physics.
Below is the elementary Newtonian physics relevant
to this thread.
I will advice you not to dispute what the rest of
the world has known for centuries.
Your problem isn't only your total ignorance of physics.
More severe is your ignorance of the fact that you are
ignorant of physics.
Pure and total nonsense.
According to Newton's gravitation there is a force acting on you.
F = GMm/r²,
where G is the gravitational constant, M is the mass of the Earth,
m is your mass, and r is the radius of the Earth.
The direction of this force is downwards towards the centre of
the Earth.
In the non rotating Earth centred frame of reference, your velocity
is tangential to the surface (horizontal) and towards the east.
Your speed is constant v = 465.1⋅cos(latitude) m/s because
no force with direction along the velocity is acting on you.
You are experiencing force from gravity downwards yes.
But you would experience that same amount of force even if the earth wasnt spinning!
So therefore seeing as the earth also spins eastward...this eastward rotation also pushes you eastward with a seperate force. The only reason why you don’t go flying off into space...is because the seperate downwards
g force prevents you from doing so
And proof that your claim is nonsense is that NASA uses this seperate eastward
force to enable greater escape velocity against the downward force of earths gravity.
So there is only one force acting on you. It is perpendicular
to your velocity in the ECI frame (vertical) and is giving you
a centripetal acceleration GMm/r² which is making you go
in a circle.
Only one gravitational force down yes.Quite.
But the spinning earth also pushes you eastward.
And that is a seperate force.
Otherwise storing energy in spinning flywheel wouldn’t
work.
Newton's first law of motion is wrong.
I think we leave it at that! :-D
Den 25.10.2023 10:47, skrev Lou:
Yes Paul. It is you who forgot Newton’s first law:
“Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to *change its state by the action of an external force*. This tendency to resist changes in a state of motion is inertia.”
Notice that if there were no gravity from earth then you and the table wouldA fine lecture about Newton's first law. Well done!
travel in a straight line at a uniform speed. And no longer be subject to the
Horizontal eastward 1600kph force from the planet below you.
But there is another external force...it’s called gravity. And it’s constantly changing your direction of motion and preventing you from leaving the surface of the planet and be freed of that constant 1600 kph force imparted to you from the planet beneath you.
So your mistake is to think you and your table are travelling at a uniform speed in a straight line.
You aren’t. You are rotating at 1600 kph in a circle.
And as I’ve already said, but you conveniently snipped, NASA uses
this tangental force from the earth on its rocket at liftoff to
get an additional 1600 kph velocity.
An additional 1600 velocity that you, Volney and a Rich D pretend doesn’t exist!!😂🐿🌰🥜
I knew that if there were no gravity, I would keep moving
in a straight line at 1600 km/h relative to the Earth,
and not be subject to a horizontal eastward force.
What I didn't know, but now have learned, is that because it
is a gravitational force perpendicular to my velocity relative
to the Earth, I am subject to a horizontal eastward 1600kph force from
the planet below me.
Thanks for teaching me.
BTW, how many Newtons is a 1600 kph force?
What I didn't know, but now have learned, is that because it
is a gravitational force perpendicular to my velocity relative
to the Earth, I am subject to a horizontal eastward 1600kph force from
the planet below me.
On Wednesday, 25 October 2023 at 10:43:26 UTC+1, Paul B. Andersen wrote:
What I didn't know, but now have learned, is that because it
is a gravitational force perpendicular to my velocity relative
to the Earth, I am subject to a horizontal eastward 1600kph force from
the planet below me.
Just noticed the sneaky dishonest phrasing of this paragraph.
You pretended that I had said the downward force of gravity was
the source of the 1600kph horizontal force that NASA uses to give its rockets an extra 1600kph boost on takeoff.
I said nothing of the sort.
travel in a straight line at a uniform speed. And no longer besubject to the
Horizontal eastward 1600kph force from the planet below you.
But there is another external force...it’s called gravity. And it’s constantly changing your direction of motion and preventing you from
leaving the surface of the planet and be freed of that constant 1600 kph force imparted to you from the planet beneath you.
Lies as usual from a relativist.
On Wednesday, 25 October 2023 at 10:43:26 UTC+1, Paul B. Andersen wrote:
A fine lecture about Newton's first law. Well done!
I knew that if there were no gravity, I would keep moving
in a straight line at 1600 km/h relative to the Earth,
and not be subject to a horizontal eastward force.
What I didn't know, but now have learned, is that because it
is a gravitational force perpendicular to my velocity relative
to the Earth, I am subject to a horizontal eastward 1600kph force from
the planet below me.
Finally. You realised that the force from the rotation of the earth that NASA uses
Isn’t imaginary!!
Phew. Looks like NASA rockets can make escape velocity. Fortunately
Paul finally, hopefully, realised that the earth is rotating.
Thanks for teaching me.
No problem. Baby steps for baby minds.
Den 25.10.2023 13:28, skrev Lou:
On Wednesday, 25 October 2023 at 10:43:26 UTC+1, Paul B. Andersen wrote:
A fine lecture about Newton's first law. Well done!
I knew that if there were no gravity, I would keep moving
in a straight line at 1600 km/h relative to the Earth,
and not be subject to a horizontal eastward force.
What I didn't know, but now have learned, is that because it
is a gravitational force perpendicular to my velocity relative
to the Earth, I am subject to a horizontal eastward 1600kph force from
the planet below me.
Finally. You realised that the force from the rotation of the earth that NASA uses
Isn’t imaginary!!
Phew. Looks like NASA rockets can make escape velocity. Fortunately
Paul finally, hopefully, realised that the earth is rotating.
Thanks for teaching me.
No problem. Baby steps for baby minds.
I have now learned that I am subject to a horizontal
eastward 1600kph force from the planet below me.
But I have no clear idea of how strong this force is.
Is it 1600kp = 15690N? Is the h in kph a typo?
On Wednesday, 25 October 2023 at 18:29:30 UTC+1, Paul B. Andersen wrote:
Den 25.10.2023 13:28, skrev Lou:
On Wednesday, 25 October 2023 at 10:43:26 UTC+1, Paul B. Andersen wrote: >>>> A fine lecture about Newton's first law. Well done!I have now learned that I am subject to a horizontal
I knew that if there were no gravity, I would keep moving
in a straight line at 1600 km/h relative to the Earth,
and not be subject to a horizontal eastward force.
What I didn't know, but now have learned, is that because it
is a gravitational force perpendicular to my velocity relative
to the Earth, I am subject to a horizontal eastward 1600kph force from >>>> the planet below me.
Finally. You realised that the force from the rotation of the earth that NASA uses
Isn’t imaginary!!
Phew. Looks like NASA rockets can make escape velocity. Fortunately
Paul finally, hopefully, realised that the earth is rotating.
Thanks for teaching me.
No problem. Baby steps for baby minds.
eastward 1600kph force from the planet below me.
Really! Now if you had studied physics instead of accounting
you would have realised that not only is earth spinning ( yes the earth actually rotates believe it or not) at about 1600kph ...NASA also uses the the horizontal force of this angular momentum to assist its rockets in liftoff to achieve escape velocity!!
Of course you relativists don’t think the earth rotates. So I suppose NASA is lying about launching rockets to the east.
But I have no clear idea of how strong this force is.
Let me guess. You are subject to a horizontal force but don’t
know how strong it is? I suppose in relativity lala land if you don’t
know how strong a force is then..it doesn’t exist!
Is it 1600kp = 15690N? Is the h in kph a typo?
Kilometers per hour. Kph. What did you think h represented?
Den 25.10.2023 14:00, skrev Lou:
On Wednesday, 25 October 2023 at 10:43:26 UTC+1, Paul B. Andersen wrote:
What I didn't know, but now have learned, is that because it
is a gravitational force perpendicular to my velocity relative
to the Earth, I am subject to a horizontal eastward 1600kph force from
the planet below me.
Just noticed the sneaky dishonest phrasing of this paragraph.Of course, you didn't, sorry.
You pretended that I had said the downward force of gravity was
the source of the 1600kph horizontal force that NASA uses to give its rockets an extra 1600kph boost on takeoff.
I said nothing of the sort.
|Den 25.10.2023 10:47, skrev Lou: |> Notice that if there were no
gravity from earth then you and the table would
travel in a straight line at a uniform speed. And no longer besubject to the
Horizontal eastward 1600kph force from the planet below you.You clearly stated that gravity had nothing to do
But there is another external force...it’s called gravity. And it’s constantly changing your direction of motion and preventing you from leaving the surface of the planet and be freed of that constant 1600 kph force imparted to you from the planet beneath you.
with the horizontal 1600kph force, the horizontal force
would be there without gravity.
Le 25/10/2023 à 21:02, Lou a écrit :
On Wednesday, 25 October 2023 at 18:29:30 UTC+1, Paul B. Andersen wrote:
Den 25.10.2023 13:28, skrev Lou:
On Wednesday, 25 October 2023 at 10:43:26 UTC+1, Paul B. Andersen wrote: >>>> A fine lecture about Newton's first law. Well done!I have now learned that I am subject to a horizontal
I knew that if there were no gravity, I would keep moving
in a straight line at 1600 km/h relative to the Earth,
and not be subject to a horizontal eastward force.
What I didn't know, but now have learned, is that because it
is a gravitational force perpendicular to my velocity relative
to the Earth, I am subject to a horizontal eastward 1600kph force from >>>> the planet below me.
Finally. You realised that the force from the rotation of the earth that NASA uses
Isn’t imaginary!!
Phew. Looks like NASA rockets can make escape velocity. Fortunately
Paul finally, hopefully, realised that the earth is rotating.
Thanks for teaching me.
No problem. Baby steps for baby minds.
eastward 1600kph force from the planet below me.
Really! Now if you had studied physics instead of accounting
you would have realised that not only is earth spinning ( yes the earth actually rotates believe it or not) at about 1600kph ...NASA also uses the the horizontal force of this angular momentum to assist its rockets in liftoff to achieve escape velocity!!
Of course you relativists don’t think the earth rotates. So I suppose NASA
is lying about launching rockets to the east.
But I have no clear idea of how strong this force is.
Let me guess. You are subject to a horizontal force but don’t
know how strong it is? I suppose in relativity lala land if you don’t know how strong a force is then..it doesn’t exist!
Is it 1600kp = 15690N? Is the h in kph a typo?
Kilometers per hour. Kph. What did you think h represented?*facepalm*
Thanks for the laugh, Paul ! Lou is definitely a gem when
it comes to cranks here !
What I didn't know, but now have learned, is that because it
is a gravitational force perpendicular to my velocity relative
to the Earth, I am subject to a horizontal eastward 1600kph force from
the planet below me.
You pretended that I had said the downward force of gravity was
the source of the 1600kph horizontal force that NASA uses to give its rockets an extra 1600kph boost on takeoff.
I said nothing of the sort.
You forgot, or don’t realise, that the earth rotates.
And like all rotating masses stores energy which can be imparted
as a force tangental to the rotation.
Obviously, like Paul and other relativists, Python never accepted the Copernican Revolution and still thinks the sun rotates around the earth.
On Wednesday, 25 October 2023 at 18:29:30 UTC+1, Paul B. Andersen wrote:
I have now learned that I am subject to a horizontal
eastward 1600kph force from the planet below me.
Really! Now if you had studied physics instead of accounting
you would have realised that not only is earth spinning ( yes the earth actually rotates believe it or not) at about 1600kph ...NASA also uses the the horizontal force of this angular momentum to assist its rockets in liftoff to achieve escape velocity!!
Of course you relativists don’t think the earth rotates. So I suppose NASA is lying about launching rockets to the east.
But I have no clear idea of how strong this force is.
Let me guess. You are subject to a horizontal force but don’t
know how strong it is? I suppose in relativity lala land if you don’t
know how strong a force is then..it doesn’t exist!
Is it 1600kp = 15690N? Is the h in kph a typo?
Kilometers per hour. Kph. What did you think h represented?
On October 25, Lou wrote:
What I didn't know, but now have learned, is that because it
is a gravitational force perpendicular to my velocity relative
to the Earth, I am subject to a horizontal eastward 1600kph force from
the planet below me.
You pretended that I had said the downward force of gravity wasLebron James stands on the bed of a flatbed truck, traveling 100 mph.
the source of the 1600kph horizontal force that NASA uses to give its rockets an extra 1600kph boost on takeoff.
I said nothing of the sort.
You forgot, or don’t realise, that the earth rotates.
And like all rotating masses stores energy which can be imparted
as a force tangental to the rotation.
Aliens have vacuumed away the earth's atmosphere, so he's comfortable.
He paints an X on the spot beneath his feet. He then jumps straight up
10 meters (LBJ can do that). Taking account of the earth's rotation:
I) i) He lands on the X.
ii) He lands in front of the X, due to the horizontal force of his initial velocity and acceleration.
iii) He lands behind the X, due to the the truck's forward movement beneath him.
II) i) He spends equal time rising and falling.
ii) He spends more time rising, as his initial velocity holds him up,
until he starts to fall.
iii) He spends more time falling, as the earth falls away from him,
due to its rotation.
Den 25.10.2023 21:02, skrev Lou:
On Wednesday, 25 October 2023 at 18:29:30 UTC+1, Paul B. Andersen wrote:
I have now learned that I am subject to a horizontal
eastward 1600kph force from the planet below me.
Really! Now if you had studied physics instead of accounting
you would have realised that not only is earth spinning ( yes the earth actually rotates believe it or not) at about 1600kph ...NASA also uses the the horizontal force of this angular momentum to assist its rockets in liftoff to achieve escape velocity!!
Of course you relativists don’t think the earth rotates. So I suppose NASA
is lying about launching rockets to the east.
But I have no clear idea of how strong this force is.
Let me guess. You are subject to a horizontal force but don’t
know how strong it is? I suppose in relativity lala land if you don’t know how strong a force is then..it doesn’t exist!
Is it 1600kp = 15690N? Is the h in kph a typo?
Kilometers per hour. Kph. What did you think h represented?Honestly, Lou:
Do you REALLY not understand why we all find this hilarious?
On Wednesday, 25 October 2023 at 19:04:00 UTC+1, Paul B. Andersen wrote:
Den 25.10.2023 14:00, skrev Lou:
On Wednesday, 25 October 2023 at 10:43:26 UTC+1, Paul B. Andersen wrote: >>>> What I didn't know, but now have learned, is that because it
is a gravitational force perpendicular to my velocity relative
to the Earth, I am subject to a horizontal eastward 1600kph force from >>>> the planet below me.
Just noticed the sneaky dishonest phrasing of this paragraph.
You pretended that I had said the downward force of gravity was
the source of the 1600kph horizontal force.
Den 25.10.2023 10:47, skrev Lou:
Notice that if there were no gravity from earth then you and the table would
travel in a straight line at a uniform speed. And no longer be subject to the
Horizontal eastward 1600kph force from the planet below you.
Den 25.10.2023 10:47, skrev Lou: >>>>> But there is another external force...it’s called gravity. And it’s
constantly changing your direction of motion and preventing you from >>>>> leaving the surface of the planet and be freed of that constant 1600 kph >>>>> force imparted to you from the planet beneath you.
Exactly. Gravity’s downward force doesn’t have anything to do
with the horizontal 1600 kph force.
A horizontal force due to earths
1600kph eastward rotation, which NASA uses to assist its rockets
to counter gravity’s downward force in achieving escape velocity.
As stated repeatedly above in my quotes.
Are you feeling Ok?
Have you got echophelia?
On Tuesday, 24 October 2023 at 20:14:53 UTC+1, Paul B. Andersen wrote:Your response to this was "Pure and total nonsense."
According to Newton's gravitation there is a force acting on you.
F = GMm/r²,
where G is the gravitational constant, M is the mass of the Earth,
m is your mass, and r is the radius of the Earth.
The direction of this force is downwards towards the centre of
the Earth.
In the non rotating Earth centred frame of reference, your velocity
is tangential to the surface (horizontal) and towards the east.
Your speed is constant v = 465.1⋅cos(latitude) m/s because
no force with direction along the velocity is acting on you.
So there is only one force acting on you. It is perpendicular
to your velocity in the ECI frame (vertical) and is giving you
a centripetal acceleration GMm/r² which is making you go
in a circle.
Now answer this question please. Assuming that after the aliens had vacuumed all the atmosphere away, they then hovered their ship 2 meters above the equator but not
rotating with the equator so they were in the ECI frame and watching the earth
spin by them at 1600kph.
If someone standing on the earths surface at the equator threw a rock straight up 2 m into
the air gently just when the spaceship passed overhead...
i) would the rock hit the spaceship at 1600 kph and destroy the spaceship?
Or
iI) would the rock gently bounce off the spaceship and fall straight back down onto the ground?
According to Paul, You and Monty Python the answer would be ii)
I don’t understand why you think a point ( or you on the earths surface) on the circumference of a rotating disc, which is rotating at a steady uniform speed ( as you do on the earths surface)...is considered by you
Rich D and Monty Python to *not be* accelerating. When all my
reference tells me if an object is travelling at a constant speed in a circle it *is* considered to be accelerating. https://en.wikipedia.org/wiki/Angular_acceleration
On Tuesday, 24 October 2023 at 20:14:53 UTC+1, Paul B. Andersen wrote:
According to Newton's gravitation there is a force acting on you.
F = GMm/r²,
where G is the gravitational constant, M is the mass of the Earth,
m is your mass, and r is the radius of the Earth.
The direction of this force is downwards towards the centre of
the Earth.
In the non rotating Earth centred frame of reference, your velocity
is tangential to the surface (horizontal) and towards the east.
Your speed is constant v = 465.1⋅cos(latitude) m/s because
no force with direction along the velocity is acting on you.
So there is only one force acting on you. It is perpendicular
to your velocity in the ECI frame (vertical) and is giving you
a centripetal acceleration GMm/r² which is making you go
in a circle.
On Friday, 27 October 2023 at 13:19:27 UTC+2, Paul B. Andersen wrote:
What other to say, poor halfbrain? You don't even knowOn Tuesday, 24 October 2023 at 20:14:53 UTC+1, Paul B. Andersen wrote:Your response to this was "Pure and total nonsense."
According to Newton's gravitation there is a force acting on you.
F = GMm/r²,
where G is the gravitational constant, M is the mass of the Earth,
m is your mass, and r is the radius of the Earth.
The direction of this force is downwards towards the centre of
the Earth.
In the non rotating Earth centred frame of reference, your velocity |>> is tangential to the surface (horizontal) and towards the east.
Your speed is constant v = 465.1⋅cos(latitude) m/s because
no force with direction along the velocity is acting on you.
So there is only one force acting on you. It is perpendicular
to your velocity in the ECI frame (vertical) and is giving you
a centripetal acceleration GMm/r² which is making you go
in a circle.
that in most cases "towards east" and "tangential to
the surface" are excluding each other..
And when on a pole? You're an idiot, but, well,
it was obvious before.
So there is only one force acting on you. It is perpendicular
to your velocity in the ECI frame (vertical) and is giving you
a centripetal acceleration GMm/r² which is making you go
in a circle.
Den 27.10.2023 13:20, skrev Paul B. Andersen:
Typo. The centripetal acceleration is GM/r² = g. (Newtonian gravitation)
So there is only one force acting on you. It is perpendicular
to your velocity in the ECI frame (vertical) and is giving you
a centripetal acceleration GMm/r² which is making you go
in a circle.
Den 26.10.2023 11:41, skrev Lou:
I don’t understand why you think a point ( or you on the earths surface) on the circumference of a rotating disc, which is rotating at a steady uniform speed ( as you do on the earths surface)...is considered by you Rich D and Monty Python to *not be* accelerating. When all myWhat you think I think has nothing to do with what I think,
reference tells me if an object is travelling at a constant speed in a circle
it *is* considered to be accelerating. https://en.wikipedia.org/wiki/Angular_acceleration
because either you don't understand anything of what I write,
or you do not read it.
On Tuesday, 24 October 2023 at 20:14:53 UTC+1, Paul B. Andersen wrote:Your response to this was "Pure and total nonsense."
According to Newton's gravitation there is a force acting on you.
F = GMm/r²,
where G is the gravitational constant, M is the mass of the Earth,
m is your mass, and r is the radius of the Earth.
The direction of this force is downwards towards the centre of
the Earth.
In the non rotating Earth centred frame of reference, your velocity
is tangential to the surface (horizontal) and towards the east.
Your speed is constant v = 465.1⋅cos(latitude) m/s because
no force with direction along the velocity is acting on you.
So there is only one force acting on you. It is perpendicular
to your velocity in the ECI frame (vertical) and is giving you
a centripetal acceleration GMm/r² which is making you go
in a circle.
Den 26.10.2023 11:33, skrev Lou:
Now answer this question please. Assuming that after the aliens had vacuumed
all the atmosphere away, they then hovered their ship 2 meters above the equator but not
rotating with the equator so they were in the ECI frame and watching the earth
spin by them at 1600kph.
If someone standing on the earths surface at the equator threw a rock straight up 2 m into
the air gently just when the spaceship passed overhead...
i) would the rock hit the spaceship at 1600 kph and destroy the spaceship? Or
iI) would the rock gently bounce off the spaceship and fall straight back down onto the ground?
According to Paul, You and Monty Python the answer would be ii)It would be interesting to see your reasoning
that led you to believe that I would answer ii)
Is your reasoning that if no force is acting on the rock,
it can't destroy the spaceship? :-D
Den 25.10.2023 21:02, skrev Lou:
On Wednesday, 25 October 2023 at 18:29:30 UTC+1, Paul B. Andersen wrote:
I have now learned that I am subject to a horizontal
eastward 1600kph force from the planet below me.
Really! Now if you had studied physics instead of accounting
you would have realised that not only is earth spinning ( yes the earth
actually rotates believe it or not) at about 1600kph ...NASA also uses
the
the horizontal force of this angular momentum to assist its rockets in
liftoff to achieve escape velocity!!
Of course you relativists don’t think the earth rotates. So I suppose
NASA
is lying about launching rockets to the east.
But I have no clear idea of how strong this force is.
Let me guess. You are subject to a horizontal force but don’t
know how strong it is? I suppose in relativity lala land if you don’t
know how strong a force is then..it doesn’t exist!
Is it 1600kp = 15690N? Is the h in kph a typo?
Kilometers per hour. Kph. What did you think h represented?
Honestly, Lou:
Do you REALLY not understand why we all find this hilarious?
When all my
reference tells me if an object is travelling at a constant speed in a circle it *is* considered to be accelerating. https://en.wikipedia.org/wiki/Angular_acceleration
On 10/26/2023 4:21 AM, Paul B. Andersen wrote:
Den 25.10.2023 21:02, skrev Lou:
On Wednesday, 25 October 2023 at 18:29:30 UTC+1, Paul B. Andersen wrote: >>> I have now learned that I am subject to a horizontal
eastward 1600kph force from the planet below me.
Really! Now if you had studied physics instead of accounting
you would have realised that not only is earth spinning ( yes the earth >> actually rotates believe it or not) at about 1600kph ...NASA also uses
the
the horizontal force of this angular momentum to assist its rockets in
liftoff to achieve escape velocity!!
Of course you relativists don’t think the earth rotates. So I suppose >> NASA
is lying about launching rockets to the east.
But I have no clear idea of how strong this force is.
Let me guess. You are subject to a horizontal force but don’t
know how strong it is? I suppose in relativity lala land if you don’t >> know how strong a force is then..it doesn’t exist!
Is it 1600kp = 15690N? Is the h in kph a typo?
Kilometers per hour. Kph. What did you think h represented?
Honestly, Lou:
Do you REALLY not understand why we all find this hilarious?
It's funny how Lou thinks everything is a force, whether gravitational potential or the eastward velocity of the rotating earth. He never
learned that if the units are wrong, the answer is automatically wrong! 😂
On Friday, 27 October 2023 at 12:19:27 UTC+1, Paul B. Andersen wrote:
On Tuesday, 24 October 2023 at 20:14:53 UTC+1, Paul B. Andersen wrote:
According to Newton's gravitation there is a force acting on you.
F = GMm/r²,
where G is the gravitational constant, M is the mass of the Earth,
m is your mass, and r is the radius of the Earth.
The direction of this force is downwards towards the centre of
the Earth.
In the non rotating Earth centred frame of reference, your velocity
is tangential to the surface (horizontal) and towards the east.
Your speed is constant v = 465.1⋅cos(latitude) m/s because
no force with direction along the velocity is acting on you.
So there is only one force acting on you. It is perpendicular
to your velocity in the ECI frame (vertical) and is giving you
a centripetal acceleration GM/r² which is making you go
in a circle.
Your response to this was "Pure and total nonsense."
Exactly. You pretended in the above statement that a spinning mass, be
it the earth gets that spin from its own downward push of gravity,
Pure total nonsense.
On 10/26/2023 5:41 AM, Lou wrote:
When all myHahahaha!! Another failure by Lou! Angular acceleration has nothing to
reference tells me if an object is travelling at a constant speed in a circle
it *is* considered to be accelerating. https://en.wikipedia.org/wiki/Angular_acceleration
do with the rotating earth (other than the tiny variations in the length
of the day)! Angular acceleration is the rate that a spinning object
spins faster or slower. The earth does neither!
FAIL.
Den 27.10.2023 14:50, skrev Lou:
On Friday, 27 October 2023 at 12:19:27 UTC+1, Paul B. Andersen wrote:
On Tuesday, 24 October 2023 at 20:14:53 UTC+1, Paul B. Andersen wrote: >> |>>
According to Newton's gravitation there is a force acting on you.
F = GMm/r²,
where G is the gravitational constant, M is the mass of the Earth,
m is your mass, and r is the radius of the Earth.
The direction of this force is downwards towards the centre of
the Earth.
In the non rotating Earth centred frame of reference, your velocity >> |>> is tangential to the surface (horizontal) and towards the east.
Your speed is constant v = 465.1⋅cos(latitude) m/s because
no force with direction along the velocity is acting on you.
So there is only one force acting on you. It is perpendicular
to your velocity in the ECI frame (vertical) and is giving you
a centripetal acceleration GM/r² which is making you go
in a circle.
Your response to this was "Pure and total nonsense."
Exactly. You pretended in the above statement that a spinning mass, beI see.
it the earth gets that spin from its own downward push of gravity,
The fact that the downward push from gravity make you stay on
the ground and make you move in a circle around Earth's center, means
that it is the downward push from gravity that makes the Earth spin.
A logical conclusion, isn't it?
On Friday, 27 October 2023 at 12:47:52 UTC+1, Paul B. Andersen wrote:
Den 26.10.2023 11:33, skrev Lou:
It would be interesting to see your reasoning
Now answer this question please. Assuming that after the aliens had vacuumed
all the atmosphere away, they then hovered their ship 2 meters above the equator but not
rotating with the equator so they were in the ECI frame and watching the earth
spin by them at 1600kph.
If someone standing on the earths surface at the equator threw a rock straight up 2 m into
the air gently just when the spaceship passed overhead...
i) would the rock hit the spaceship at 1600 kph and destroy the spaceship? >>> Or
iI) would the rock gently bounce off the spaceship and fall straight back down onto the ground?
According to Paul, You and Monty Python the answer would be ii)
that led you to believe that I would answer ii)
Is your reasoning that if no force is acting on the rock,
it can't destroy the spaceship? :-D
My reasoning!!?
More like ...Your reasoning!
After all you, Rich D and Monty Python all think
that there is no force acting on you standing on surface of earth.
If this were true then your stone you throw up gently in the air would
ALSO have no force acting on it.
So that when the spaceship in the ECI frame watches you throw the stone
as you rotate below the spaceship...that stone, according to you, has no tangental ( horizontal in your frame) force. And should therefore cause no damage to the alien spaceship.
An odd conclusion you make because if you guys pretend you and
your stone have no force acting on them....Then how on earth does it
get its 1600kph rotational speed in the alien ships ECI frame?
On Friday, 27 October 2023 at 16:46:33 UTC+1, Volney wrote:speed is constant, its velocity is not constant: velocity, a vector quantity, depends on both the body's speed and its direction of travel. This changing velocity indicates the presence of an acceleration.”
On 10/26/2023 5:41 AM, Lou wrote:
When all myHahahaha!! Another failure by Lou! Angular acceleration has nothing to
reference tells me if an object is travelling at a constant speed in a circle
it *is* considered to be accelerating. https://en.wikipedia.org/wiki/Angular_acceleration
do with the rotating earth (other than the tiny variations in the length of the day)! Angular acceleration is the rate that a spinning object
spins faster or slower. The earth does neither!
FAIL.Sorry pal...You failed to check the reference
Wiki: Uniform circular motion:
“In physics, uniform circular motion describes the motion of a body traversing a circular path at a constant speed. Since the body describes circular motion, its distance from the axis of rotation remains constant at all times. Though the body's
On Friday, 27 October 2023 at 18:51:44 UTC+1, Paul B. Andersen wrote:
Den 27.10.2023 14:50, skrev Lou:
On Friday, 27 October 2023 at 12:19:27 UTC+1, Paul B. Andersen wrote:
On Tuesday, 24 October 2023 at 20:14:53 UTC+1, Paul B. Andersen wrote: >>>> |>>
According to Newton's gravitation there is a force acting on you.
F = GMm/r²,
where G is the gravitational constant, M is the mass of the Earth, >>>> |>> m is your mass, and r is the radius of the Earth.
The direction of this force is downwards towards the centre of
the Earth.
In the non rotating Earth centred frame of reference, your velocity >>>> |>> is tangential to the surface (horizontal) and towards the east.
Your speed is constant v = 465.1⋅cos(latitude) m/s because
no force with direction along the velocity is acting on you.
Nobody in their right mind can interpret the above to mean
So there is only one force acting on you. It is perpendicular
to your velocity in the ECI frame (vertical) and is giving you
a centripetal acceleration GM/r² which is making you go
in a circle.
Exactly. You pretended in the above statement that a spinning mass, be
it the earth gets that spin from its own downward push of gravity,
I see.
The fact that the downward push from gravity make you stay on
the ground and make you move in a circle around Earth's center, means
that it is the downward push from gravity that makes the Earth spin.
A logical conclusion, isn't it?
If you think the downward push of gravity is actually a horizontal push
as you pretend above...then physics is not your strong point.
Den 27.10.2023 22:04, skrev Lou:
On Friday, 27 October 2023 at 18:51:44 UTC+1, Paul B. Andersen wrote:A downwards force F = GMm/r² is acting on you,
Den 27.10.2023 14:50, skrev Lou:
On Friday, 27 October 2023 at 12:19:27 UTC+1, Paul B. Andersen wrote:
On Tuesday, 24 October 2023 at 20:14:53 UTC+1, Paul B. Andersen wrote:
According to Newton's gravitation there is a force acting on you. >>>> |>> F = GMm/r²,
where G is the gravitational constant, M is the mass of the Earth, >>>> |>> m is your mass, and r is the radius of the Earth.
The direction of this force is downwards towards the centre of
the Earth.
Den 27.10.2023 14:59, skrev Lou:
On Friday, 27 October 2023 at 12:47:52 UTC+1, Paul B. Andersen wrote:
Den 26.10.2023 11:33, skrev Lou:
It would be interesting to see your reasoning
Now answer this question please. Assuming that after the aliens had vacuumed
all the atmosphere away, they then hovered their ship 2 meters above the equator but not
rotating with the equator so they were in the ECI frame and watching the earth
spin by them at 1600kph.
If someone standing on the earths surface at the equator threw a rock straight up 2 m into
the air gently just when the spaceship passed overhead...
i) would the rock hit the spaceship at 1600 kph and destroy the spaceship?
Or
iI) would the rock gently bounce off the spaceship and fall straight back down onto the ground?
According to Paul, You and Monty Python the answer would be ii)
that led you to believe that I would answer ii)
Is your reasoning that if no force is acting on the rock,
it can't destroy the spaceship? :-D
My reasoning!!?
More like ...Your reasoning!
After all you, Rich D and Monty Python all think
that there is no force acting on you standing on surface of earth.
If this were true then your stone you throw up gently in the air would
ALSO have no force acting on it.
So that when the spaceship in the ECI frame watches you throw the stone
as you rotate below the spaceship...that stone, according to you, has no tangental ( horizontal in your frame) force. And should therefore cause no damage to the alien spaceship.
An odd conclusion you make because if you guys pretend you and
your stone have no force acting on them....Then how on earth does it
get its 1600kph rotational speed in the alien ships ECI frame?
Thanks for demonstrating your reasoning.
I have got it now.
The rock in my hand will move with the speed 1600 km/s in
the ECI frame. The spaceship is stationary in the ECI frame,
2 m above the ground. I hold the rock 2 m above the ground,
and let it go just before it hits the spaceship. If it is no
horizontal force pushing the rock, it will stop and its horizontal
speed will be zero, and it will fall down behind me, and will never
hit the spaceship.
Right?
On Friday, October 27, 2023 at 1:02:04 PM UTC-7, Lou wrote:speed is constant, its velocity is not constant: velocity, a vector quantity, depends on both the body's speed and its direction of travel. This changing velocity indicates the presence of an acceleration.”
On Friday, 27 October 2023 at 16:46:33 UTC+1, Volney wrote:
On 10/26/2023 5:41 AM, Lou wrote:
When all myHahahaha!! Another failure by Lou! Angular acceleration has nothing to do with the rotating earth (other than the tiny variations in the length of the day)! Angular acceleration is the rate that a spinning object spins faster or slower. The earth does neither!
reference tells me if an object is travelling at a constant speed in a circle
it *is* considered to be accelerating. https://en.wikipedia.org/wiki/Angular_acceleration
FAIL.Sorry pal...You failed to check the reference
Wiki: Uniform circular motion:
“In physics, uniform circular motion describes the motion of a body traversing a circular path at a constant speed. Since the body describes circular motion, its distance from the axis of rotation remains constant at all times. Though the body's
You first are talking about angular acceleration and now you are talking about uniform circular motion! You do understand that these are two different things, right?20velocity%2C%20%CF%89%2C%20is,angular%20velocity%20does%20not%20change.
https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_Introductory_Physics_-_Building_Models_to_Describe_Our_World_(Martin_Neary_Rinaldo_and_Woodman)/06%3A_Applying_Newtons_Laws/6.03%3A_Uniform_circular_motion#:~:text=The%20angular%
The velocity vector, v , is always tangent to the circle.
The acceleration vector, a, is always perpendicular to the velocity vector, because the magnitude of the velocity vector does not change.
The acceleration vector, a, always points towards the center of the circle. The acceleration vector has magnitude a=v2/R .
The angular velocity, ω, is related to the magnitude of the velocity vector by v=ωR and is constant.
and, finally...
The angular acceleration, α, is zero for uniform circular motion, since the angular velocity does not change.
It is quite clear that you do not know what you do not know, so not much has changed...
On Friday, October 27, 2023 at 1:02:04 PM UTC-7, Lou wrote:
On Friday, 27 October 2023 at 16:46:33 UTC+1, Volney wrote:
On 10/26/2023 5:41 AM, Lou wrote:
When all my
reference tells me if an object is travelling at a constant speed in a circle
it *is* considered to be accelerating. https://en.wikipedia.org/wiki/Angular_acceleration
Den 27.10.2023 22:04, skrev Lou:
On Friday, 27 October 2023 at 18:51:44 UTC+1, Paul B. Andersen wrote:A downwards force F = GMm/r² is acting on you,
Den 27.10.2023 14:50, skrev Lou:
On Friday, 27 October 2023 at 12:19:27 UTC+1, Paul B. Andersen wrote:
On Tuesday, 24 October 2023 at 20:14:53 UTC+1, Paul B. Andersen wrote:
According to Newton's gravitation there is a force acting on you. >>>> |>> F = GMm/r²,
where G is the gravitational constant, M is the mass of the Earth, >>>> |>> m is your mass, and r is the radius of the Earth.
The direction of this force is downwards towards the centre of
the Earth.
keeping you on the ground.
You are moving at at a speed v = 465.1⋅cos(latitude) m/s
In the non rotating Earth centred frame of reference, your velocity >>>> |>> is tangential to the surface (horizontal) and towards the east. >>>> |>> Your speed is constant v = 465.1⋅cos(latitude) m/s because
no force with direction along the velocity is acting on you.
in the ECI-frame because the Earth is spinning.
If you were standing on horizontal ice, and no horizontal force
could be mediated from the ice to your feet, you would still move horizontally along with the ground with the speed.
So NO HORIZONTAL FORCE is acting on you.
On Saturday, October 28, 2023 at 7:14:53 AM UTC-7, Lou wrote:
On Saturday, 28 October 2023 at 01:29:40 UTC+1, Paul Alsing wrote:
On Friday, October 27, 2023 at 1:02:04 PM UTC-7, Lou wrote:
On Friday, 27 October 2023 at 16:46:33 UTC+1, Volney wrote:
On 10/26/2023 5:41 AM, Lou wrote:
I have looked back at my original post (above)to see what your problem is. You can’t read.When all my
reference tells me if an object is travelling at a constant speed in a circle
it *is* considered to be accelerating. https://en.wikipedia.org/wiki/Angular_acceleration
Notice I very clearly said “ if an object is travelling at a constant speed in a circle
it *is* considered to be accelerating. “
The link then should then have been : https://en.wikipedia.org/wiki/Circular_motion
Unfortunately you are either too stupid or too dishonest
to have realised that when I said:
“ if an object is travelling at a constant speed in a circle
it *is* considered to be accelerating. “
I meant it.At least you have admitted that you posted an incorrect link... which is all I was claiming.
On Saturday, 28 October 2023 at 01:29:40 UTC+1, Paul Alsing wrote:
On Friday, October 27, 2023 at 1:02:04 PM UTC-7, Lou wrote:
On Friday, 27 October 2023 at 16:46:33 UTC+1, Volney wrote:
On 10/26/2023 5:41 AM, Lou wrote:
I have looked back at my original post (above)to see what your problem is. You can’t read.When all my
reference tells me if an object is travelling at a constant speed in a circle
it *is* considered to be accelerating. https://en.wikipedia.org/wiki/Angular_acceleration
Notice I very clearly said “ if an object is travelling at a constant speed in a circle
it *is* considered to be accelerating. “
The link then should then have been : https://en.wikipedia.org/wiki/Circular_motion
Unfortunately you are either too stupid or too dishonest
to have realised that when I said:
“ if an object is travelling at a constant speed in a circle
it *is* considered to be accelerating. “
I meant it.
A correct resolution of the twins paradox, without acceleration or frame
switching is here:
https://www.kevinaylward.co.uk/gr/xht/twinsparadox/twinsparadox.xht
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.
On 10/17/23 12:31 PM, Kevin Aylward wrote:
A correct resolution of the twins paradox, without acceleration or frame
switching is here:
https://www.kevinaylward.co.uk/gr/xht/twinsparadox/twinsparadox.xht
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.
When all my reference tells me if an object is travelling at a constant speed in a circle
it *is* considered to be accelerating.
On October 28, Lou wrote:
So the spacecraft, before launch, is accelerating, as it revolvesWhen all my reference tells me if an object is travelling at a constant speed in a circle
it *is* considered to be accelerating.
around the earth's center. And f =ma.
Therefore, when it launches, it uses this acceleration and force
to get a boost eastward? That's your position?
When all my reference tells me if an object is travelling at a constant speed in a circle
it *is* considered to be accelerating.
So the spacecraft, before launch, is accelerating, as it revolves
around the earth's center. And f =ma.
Therefore, when it launches, it uses this acceleration and force
to get a boost eastward? That's your position?
I refer to what’s observed. Spin an object around in a circle then release it.
It flies off at a uniform speed in a straight line tangental to the direction of spin.
On October 30, Lou wrote:
When all my reference tells me if an object is travelling at a constant speed in a circle
it *is* considered to be accelerating.
So the spacecraft, before launch, is accelerating, as it revolves
around the earth's center. And f =ma.
Therefore, when it launches, it uses this acceleration and force
to get a boost eastward? That's your position?
I refer to what’s observed. Spin an object around in a circle then release it.Yes, Newton's first law.
It flies off at a uniform speed in a straight line tangental to the direction of spin.
And in your model, the tangential speed at which it flies off is due to
a horizontal force? Like NASA's satellite launches?
So the spacecraft, before launch, is accelerating, as it revolves
around the earth's center. And f =ma.
Therefore, when it launches, it uses this acceleration and force
to get a boost eastward? That's your position?
Spin an object around in a circle then release it.
It flies off at a uniform speed in a straight line tangental to the direction of spin.
Yes, Newton's first law.
And in your model, the tangential speed at which it flies off is due to
a horizontal force? Like NASA's satellite launches?
And in your model the stone doesnt fly off at a tangent when released
and NASA rockets get no additional velocity boost into orbit from
launching eastward?
On October 30, Lou wrote
So the spacecraft, before launch, is accelerating, as it revolves
around the earth's center. And f =ma.
Therefore, when it launches, it uses this acceleration and force
to get a boost eastward? That's your position?
Spin an object around in a circle then release it.
It flies off at a uniform speed in a straight line tangental to the direction of spin.
Yes, Newton's first law.
And in your model, the tangential speed at which it flies off is due to >> a horizontal force? Like NASA's satellite launches?
And in your model the stone doesnt fly off at a tangent when releasedForce, acceleration, velocity, gravity, orbit, rotation... these are separate
and NASA rockets get no additional velocity boost into orbit from launching eastward?
concepts. You muck them into a tangle of spaghetti.
The acceleration due to earth's rotation doesn't affect the launch.
On October 30, Lou wrote
So the spacecraft, before launch, is accelerating, as it revolves
around the earth's center. And f =ma.
Therefore, when it launches, it uses this acceleration and force
to get a boost eastward? That's your position?
Spin an object around in a circle then release it.
It flies off at a uniform speed in a straight line tangental to the direction of spin.
Yes, Newton's first law.
And in your model, the tangential speed at which it flies off is due to >> a horizontal force? Like NASA's satellite launches?
And in your model the stone doesnt fly off at a tangent when releasedForce, acceleration, velocity, gravity, orbit, rotation... these are separate
and NASA rockets get no additional velocity boost into orbit from launching eastward?
concepts. You muck them into a tangle of spaghetti.
The acceleration due to earth's rotation doesn't affect the launch.
If the closest to an inertial frame in this case is the ECI frame, then according to Newton it needs *force* to change speed or direction.
Notice that you on the earth surface are continually changing direction
as you rotate in the ECI frame. If you say that no force acts on you
or the rocket before launch then how do you change direction in
the ECI frame without the help of any external force?
According to Newton's gravitation there is a force acting on you.
F = GMm/r²,
where G is the gravitational constant, M is the mass of the Earth,
m is your mass, and r is the radius of the Earth.
The direction of this force is downwards towards the centre of
the Earth.
In the non rotating Earth centred frame of reference, your velocity
is tangential to the surface (horizontal) and towards the east.
Your speed is constant v = 465.1⋅cos(latitude) m/s because
no force with direction along the velocity is acting on you.
So there is only one force acting on you. It is perpendicular
to your velocity in the ECI frame (vertical) and is giving you
a centripetal acceleration GM/r² which is making you go
in a circle.
Den 01.11.2023 11:16, skrev Lou:
If the closest to an inertial frame in this case is the ECI frame, then according to Newton it needs *force* to change speed or direction.Well said, Lou!
Notice that you on the earth surface are continually changing direction
as you rotate in the ECI frame. If you say that no force acts on you
or the rocket before launch then how do you change direction in
the ECI frame without the help of any external force?
When you know this, why did you then call
the following "Pure and total nonsense"?
On October 24, Paul B. Andersen wrote:
According to Newton's gravitation there is a force acting on you.
F = GMm/r²,
where G is the gravitational constant, M is the mass of the Earth,
m is your mass, and r is the radius of the Earth.
The direction of this force is downwards towards the centre of
the Earth.
In the non rotating Earth centred frame of reference, your velocity
is tangential to the surface (horizontal) and towards the east.
Your speed is constant v = 465.1⋅cos(latitude) m/s because
no force with direction along the velocity is acting on you.
So there is only one force acting on you. It is perpendicular
to your velocity in the ECI frame (vertical) and is giving you
a centripetal acceleration GM/r² which is making you go
in a circle.
Den 01.11.2023 11:16, skrev Lou:
If the closest to an inertial frame in this case is the ECI frame, then according to Newton it needs *force* to change speed or direction.Well said, Lou!
Notice that you on the earth surface are continually changing direction
as you rotate in the ECI frame. If you say that no force acts on you
or the rocket before launch then how do you change direction in
the ECI frame without the help of any external force?
When you know this, why did you then call
the following "Pure and total nonsense"?
On October 24, Paul B. Andersen wrote:
According to Newton's gravitation there is a force acting on you.
F = GMm/r²,
where G is the gravitational constant, M is the mass of the Earth,
m is your mass, and r is the radius of the Earth.
The direction of this force is downwards towards the centre of
the Earth.
In the non rotating Earth centred frame of reference, your velocity
is tangential to the surface (horizontal) and towards the east.
Your speed is constant v = 465.1⋅cos(latitude) m/s because
no force with direction along the velocity is acting on you.
So there is only one force acting on you. It is perpendicular
to your velocity in the ECI frame (vertical) and is giving you
a centripetal acceleration GM/r² which is making you go
in a circle.
On Wednesday, 1 November 2023 at 13:11:19 UTC, Paul B. Andersen wrote:
Den 01.11.2023 11:16, skrev Lou:
If the closest to an inertial frame in this case is the ECI frame, then according to Newton it needs *force* to change speed or direction. Notice that you on the earth surface are continually changing direction as you rotate in the ECI frame. If you say that no force acts on youWell said, Lou!
or the rocket before launch then how do you change direction in
the ECI frame without the help of any external force?
When you know this, why did you then call
the following "Pure and total nonsense"?
On October 24, Paul B. Andersen wrote:
According to Newton's gravitation there is a force acting on you.
F = GMm/r²,
where G is the gravitational constant, M is the mass of the Earth,
m is your mass, and r is the radius of the Earth.
The direction of this force is downwards towards the centre of
the Earth.
In the non rotating Earth centred frame of reference, your velocity
is tangential to the surface (horizontal) and towards the east.
Your speed is constant v = 465.1⋅cos(latitude) m/s because
no force with direction along the velocity is acting on you.
I said your statement was pure and total nonsense because it is.So there is only one force acting on you. It is perpendicular
to your velocity in the ECI frame (vertical) and is giving you
a centripetal acceleration GM/r² which is making you go
in a circle.
Your above quoted piece of nonsense you wrote has nothing to
do with what I have been posting and in particular what I just posted to Rick.
You pretend g force is r^2.
I explicitly have said it’s r. (As relativity does)
You said there is no horizontal force.
I said there is. ( because Newton says there is for any rotating object)
You said there is only one downward force acting on you and the rocket.
I have consistently said there are 2 forces. One horizontal due
to rotation acceleration and one vertical due to gravity.
Only an insane person like yourself could pretend your nonsense
bears ANY similarity to what I have been saying.
Your nonsense isn’t even consistent with Newton.
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 ina 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
x' = (x-vt)/sqrt(12-v^2/c^2)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
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.
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 believeEinstein 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
x'=x-vtground, 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
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
x = x'- m'n'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.
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
El miércoles, 1 de noviembre de 2023 a las 10:42:37 UTC-3, Lou escribió:
On Wednesday, 1 November 2023 at 13:11:19 UTC, Paul B. Andersen wrote:
Den 01.11.2023 11:16, skrev Lou:
If the closest to an inertial frame in this case is the ECI frame, thenWell said, Lou!
according to Newton it needs *force* to change speed or direction. Notice that you on the earth surface are continually changing direction
as you rotate in the ECI frame. If you say that no force acts on you or the rocket before launch then how do you change direction in
the ECI frame without the help of any external force?
When you know this, why did you then call
the following "Pure and total nonsense"?
On October 24, Paul B. Andersen wrote:
According to Newton's gravitation there is a force acting on you.
F = GMm/r²,
where G is the gravitational constant, M is the mass of the Earth,
m is your mass, and r is the radius of the Earth.
The direction of this force is downwards towards the centre of
the Earth.
In the non rotating Earth centred frame of reference, your velocity
is tangential to the surface (horizontal) and towards the east.
Your speed is constant v = 465.1⋅cos(latitude) m/s because
no force with direction along the velocity is acting on you.
I said your statement was pure and total nonsense because it is.So there is only one force acting on you. It is perpendicular
to your velocity in the ECI frame (vertical) and is giving you
a centripetal acceleration GM/r² which is making you go
in a circle.
Your above quoted piece of nonsense you wrote has nothing to
do with what I have been posting and in particular what I just posted to Rick.
You pretend g force is r^2.
I explicitly have said it’s r. (As relativity does)
You said there is no horizontal force.
I said there is. ( because Newton says there is for any rotating object) You said there is only one downward force acting on you and the rocket.
I have consistently said there are 2 forces. One horizontal due
to rotation acceleration and one vertical due to gravity.
Only an insane person like yourself could pretend your nonsense
bears ANY similarity to what I have been saying.
Your nonsense isn’t even consistent with Newton.
"there are 2 forces. One horizontal due to rotation acceleration and one vertical due to gravity".
Amazing and very funny nonsense!!!!!
On Wednesday, 1 November 2023 at 13:11:19 UTC, Paul B. Andersen wrote:
Den 01.11.2023 11:16, skrev Lou:
If the closest to an inertial frame in this case is the ECI frame, then
according to Newton it needs *force* to change speed or direction.
Notice that you on the earth surface are continually changing direction
as you rotate in the ECI frame. If you say that no force acts on you
or the rocket before launch then how do you change direction in
the ECI frame without the help of any external force?
Well said, Lou!
When you know this, why did you then call
the following "Pure and total nonsense"?
On October 24, Paul B. Andersen wrote:
According to Newton's gravitation there is a force acting on you.
F = GMm/r²,
where G is the gravitational constant, M is the mass of the Earth,
m is your mass, and r is the radius of the Earth.
The direction of this force is downwards towards the centre of
the Earth.
In the non rotating Earth centred frame of reference, your velocity
is tangential to the surface (horizontal) and towards the east.
Your speed is constant v = 465.1⋅cos(latitude) m/s because
no force with direction along the velocity is acting on you.
So there is only one force acting on you. It is perpendicular
to your velocity in the ECI frame (vertical) and is giving you
a centripetal acceleration GM/r² which is making you go
in a circle.
I said your statement was pure and total nonsense because it is.
Den 01.11.2023 14:42, skrev Lou:
On Wednesday, 1 November 2023 at 13:11:19 UTC, Paul B. Andersen wrote:You need the VERTICAL *force* F = GMm/r² to change the direction
Den 01.11.2023 11:16, skrev Lou:
If the closest to an inertial frame in this case is the ECI frame, then >>> according to Newton it needs *force* to change speed or direction. Correct.
of the HORIZONTAL velocity.
Correct again.Notice that you on the earth surface are continually changing direction >>> as you rotate in the ECI frame. If you say that no force acts on you
or the rocket before launch then how do you change direction in
the ECI frame without the help of any external force?
According to relativity, the light beam in the MMX has two different speeds at once, that relative to absolute space and that relative to the Earth. Therefore, it is self-contradictory nonsense.
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