On October 25, Lou wrote:
Obviously, like Paul and other relativists, Python never accepted the Copernican Revolution and still thinks the sun rotates around the earth.It does. So do all the heavens, on a daily and annual basis, revolve around the earth. This has been observed for thousands of years. That's science: observations, a/k/a reality.
Then, being human, we invent "explanations" for those observations. Which
is the other part of science. And being human, with limited neural capacity, we choose the simplest among competing explanations. (look up Ockham)
Why are we built this way? It's complicated, ask Darwin -
Astronomers found Copernicus simpler than Ptolemy. The rest is history -
So yeah, the sun rotates around the earth. Just watch the sky, to verify this truth -
On Wednesday, 25 October 2023 at 23:08:08 UTC+1, RichD wrote:
On October 25, Lou wrote:
Obviously, like Paul and other relativists, Python never accepted theIt does. So do all the heavens, on a daily and annual basis, revolve around >> the earth. This has been observed for thousands of years. That's science:
Copernican Revolution and still thinks the sun rotates around the earth.
observations, a/k/a reality.
Then, being human, we invent "explanations" for those observations. Which
is the other part of science. And being human, with limited neural capacity, >> we choose the simplest among competing explanations. (look up Ockham)
Why are we built this way? It's complicated, ask Darwin -
Astronomers found Copernicus simpler than Ptolemy. The rest is history -
So yeah, the sun rotates around the earth. Just watch the sky, to verify this truth -
It’s odd then , don’t you think, that despite admitting the earth does rotate
at 1600kph ( at the equator) you and the others don’t think that a person standing
on the earths surface and rotating with the earth at 1600 kph is not considered to
be accelerating, when all the reference says that an object travelling at a constant
speed in a circle is considered to be accelerating.
And I also think it’s odd that if you do accept that the earth is rotating eastward around
it’s axis at 1600 kph then why do you think this eastward rotation cannot be
used by NASA to give its rockets an extra boost in speed to reach escape velocity.
When in fact even NASA admits it does use this extra 1600 kph velocity to boost
it’s rockets into orbit.
And finally I think it’s odd that despite well accepted wisdom about how a spinning
disc can store energy and release it tangentallly as it rotates you still do not think
any tangental force in the direction of its rotation on the atoms at its edge will
be present. Despite the fact that NASA itself uses this tangental eastward force
to assist its rocket launches.
https://en.wikipedia.org/wiki/Flywheel_energy_storage
Le 26/10/2023 à 12:05, Lou a écrit :
On Wednesday, 25 October 2023 at 23:08:08 UTC+1, RichD wrote:
On October 25, Lou wrote:
Obviously, like Paul and other relativists, Python never accepted the >>> Copernican Revolution and still thinks the sun rotates around the earth. >> It does. So do all the heavens, on a daily and annual basis, revolve aroundthe earth. This has been observed for thousands of years. That's science: >> observations, a/k/a reality.
Then, being human, we invent "explanations" for those observations. Which >> is the other part of science. And being human, with limited neural capacity,
we choose the simplest among competing explanations. (look up Ockham)
Why are we built this way? It's complicated, ask Darwin -
Astronomers found Copernicus simpler than Ptolemy. The rest is history - >>
So yeah, the sun rotates around the earth. Just watch the sky, to verify this truth -
It’s odd then , don’t you think, that despite admitting the earth does rotateSure Lou, rotating motion implied an acceleration. But not a tangential
at 1600kph ( at the equator) you and the others don’t think that a person standing
on the earths surface and rotating with the earth at 1600 kph is not considered to
be accelerating, when all the reference says that an object travelling at a constant
speed in a circle is considered to be accelerating.
one: a centripetal one. Do the math. This is basic Newtonian physics.
And I also think it’s odd that if you do accept that the earth is rotating eastward aroundWhat is used is an initial velocity wrt to ECI frame (the one that make sense when trying to obtain orbital speed), it not a force.
it’s axis at 1600 kph then why do you think this eastward rotation cannot be
used by NASA to give its rockets an extra boost in speed to reach escape velocity.
When in fact even NASA admits it does use this extra 1600 kph velocity to boost
it’s rockets into orbit.
And finally I think it’s odd that despite well accepted wisdom about how a spinning
disc can store energy and release it tangentallly as it rotates you still do not think
any tangental force in the direction of its rotation on the atoms at its edge will
be present. Despite the fact that NASA itself uses this tangental eastward force
to assist its rocket launches.
https://en.wikipedia.org/wiki/Flywheel_energy_storageTangential velocity does not mean tangential *force*.
Confusing both, as you do, is a kind of weird demented (not even) pre-Galilean prejudice.
Den 25.10.2023 21:44, skrev Lou:
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.
I see. No gravity, no horizontal force.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.
I see. With gravity, a 1600kph horizontal force.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 doHard to make up your mind? :-D
with the horizontal 1600 kph force.
A horizontal force due to earthsYes. Your wrong explanation is stated over and over.
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.
| On Monday, 23 October 2023 at 22:04:42 UTC+1, Paul B. Andersen wrote:
In the ECI frame, the rocket is moving eastwards at 1600 km/h
before it is launched, when NO HORIZONTAL FORCE is acting on it.
So when the rocket fires, it is obviously best to accelerate eastwards, where you start with the HORIZONTAL speed 1600 km/h eastwards.
The 1600 km/h speed doesn't "counter gravity’s downward force",
it only give the rocket an initial horizontal 1600 km/h speed
in the ECI-frame.
(Horizontal speed is tangential, that is perpendicular to radius
vector from Earth's centre. Vertical is parallel to radius vector.)
Have you never seen the liftoff of a rocket on TV?
Before liftoff, the rocket is vertical and moving sideways
with the speed 1600 km/h in the ECI frame along with the ground.
When the rocket fires, the thrust of the rocket engine will
overcome gravity’s downward force, the rocket will lift off
and accelerate vertically. The vertical speed will increase
while the sideways horizontal speed will remain the same.
Some time after liftoff the rocket will veer off eastwards
and then the tangential speed will increase from the initial
1600km/h.
(I don't know why I bother to explain what you seem not
to be able to understand.)
Are you feeling Ok?
Well .. my stomach hurts a bit. Laughed too much!
Have you got echophelia?
No, I have got echophobia.
So please don't state again what you have stated repeatedly.
____________________________________
See the simulation of satellites:
https://paulba.no/Satellites.html
Choose the scenario "Rockets in opposite directions"
Two rockets are launched from the ground with the same
accelerations, but one (green A) is going eastwards,
and the other (red B) is going westwards.
The first stage for both rockets is:
accelerated at 40 m/s² vertically for 200 seconds
The second stage is accelerated at 30 m/s² for 300 seconds,
Green A is accelerating 120⁰ to the east of vertical.
Red B is accelerating 120⁰ to the west of vertical.
If you don't want to run the simulation,
screenshots of a run can be found here:
https://paulba.no/temp/Satellites_run.pdf
--
Paul
https://paulba.no/
On Thursday, 26 October 2023 at 20:48:55 UTC+1, Paul B. Andersen wrote:
Poor desperate Pauls lies get bigger and bigger with each lie.
So far you have failed to supply any evidence to refute my
claim that NASA uses eastward rotation of earth to assist its rockets in achieving escape velocity.
| On Monday, 23 October 2023 at 22:04:42 UTC+1, Paul B. Andersen wrote:
In the ECI frame, the rocket is moving eastwards at 1600 km/h
before it is launched, when NO HORIZONTAL FORCE is acting on it.
So when the rocket fires, it is obviously best to accelerate eastwards, >> |> where you start with the HORIZONTAL speed 1600 km/h eastwards.
Best to accelerate eastwards 😂🤣😂💩💩
Hilarious.
I can just hear the NASA Godard flight Center conversation at liftoff:
Flight control to Bob: “Hey Bob..which direction do you want today. East or west?”
Bob: “”Ummm..let me see,... I think it’s best today if we do it eastward “
The 1600 km/h speed doesn't "counter gravity’s downward force",
it only give the rocket an initial horizontal 1600 km/h speed
in the ECI-frame.
How could it “only give the rocket an initial horizontal 1600 km/h speed “
if you say there is no horizontal force acting on rocket?
Let me guess...Generally Special Woodland Troll Relativity?
(Horizontal speed is tangential, that is perpendicular to radius
vector from Earth's centre. Vertical is parallel to radius vector.)
Have you never seen the liftoff of a rocket on TV?
Before liftoff, the rocket is vertical and moving sideways
with the speed 1600 km/h in the ECI frame along with the ground.
When the rocket fires, the thrust of the rocket engine will
overcome gravity’s downward force, the rocket will lift off
and accelerate vertically. The vertical speed will increase
while the sideways horizontal speed will remain the same.
Some time after liftoff the rocket will veer off eastwards
Why does it veer off eastward? Magic?
and then the tangential speed will increase from the initial
1600km/h.
____________________________________
See the simulation of satellites:
https://paulba.no/Satellites.html
Choose the scenario "Rockets in opposite directions"
Two rockets are launched from the ground with the same
accelerations, but one (green A) is going eastwards,
and the other (red B) is going westwards.
The first stage for both rockets is:
accelerated at 40 m/s² vertically for 200 seconds
The second stage is accelerated at 30 m/s² for 300 seconds,
Green A is accelerating 120⁰ to the east of vertical.
Red B is accelerating 120⁰ to the west of vertical.
If you don't want to run the simulation,
screenshots of a run can be found here:
https://paulba.no/temp/Satellites_run.pdf
--
Paul
https://paulba.no/
Den 27.10.2023 15:27, skrev Lou:
On Thursday, 26 October 2023 at 20:48:55 UTC+1, Paul B. Andersen wrote:
Poor desperate Pauls lies get bigger and bigger with each lie.The point is that I never disputed the obvious fact that most
So far you have failed to supply any evidence to refute my
claim that NASA uses eastward rotation of earth to assist its rockets in achieving escape velocity.
satellites not in polar orbit are orbiting in the same direction as
the Earth is spinning.
Quite the contrary, I have explained why a below:
And this explanation is correct.
| On Monday, 23 October 2023 at 22:04:42 UTC+1, Paul B. Andersen wrote: >> |> In the ECI frame, the rocket is moving eastwards at 1600 km/h
before it is launched, when NO HORIZONTAL FORCE is acting on it.
So when the rocket fires, it is obviously best to accelerate eastwards,
where you start with the HORIZONTAL speed 1600 km/h eastwards.
Best to accelerate eastwardsAre you drunk?
Hilarious.
I can just hear the NASA Godard flight Center conversation at liftoff: Flight control to Bob: “Hey Bob..which direction do you want today. East or west?”
Bob: “”Ummm..let me see,... I think it’s best today if we do it eastward “
The 1600 km/h speed doesn't "counter gravity’s downward force",
it only give the rocket an initial horizontal 1600 km/h speed
in the ECI-frame.
How could it “only give the rocket an initial horizontal 1600 km/h speed “Can't you read?
if you say there is no horizontal force acting on rocket?
Let me guess...Generally Special Woodland Troll Relativity?
"the rocket is moving eastwards at 1600 km/h before it is launched,
when NO HORIZONTAL FORCE is acting on it."
And of course it will keep moving at 1600 km eastwards immediately
after launch because no horizontal force is acting on it.
Inertia!
How is it possible to fail to understand the obvious?
And this is classical, Newtonian mechanics.
(Horizontal speed is tangential, that is perpendicular to radius
vector from Earth's centre. Vertical is parallel to radius vector.)
Have you never seen the liftoff of a rocket on TV?
Before liftoff, the rocket is vertical and moving sideways
with the speed 1600 km/h in the ECI frame along with the ground.
When the rocket fires, the thrust of the rocket engine will
overcome gravity’s downward force, the rocket will lift off
and accelerate vertically. The vertical speed will increase
while the sideways horizontal speed will remain the same.
Some time after liftoff the rocket will veer off eastwards
Why does it veer off eastward? Magic?Yes. Rockets are steered by magic. Didn't you know?
Got it now?and then the tangential speed will increase from the initial
1600km/h.
Off course not.
--____________________________________
See the simulation of satellites:
https://paulba.no/Satellites.html
Choose the scenario "Rockets in opposite directions"
Two rockets are launched from the ground with the same
accelerations, but one (green A) is going eastwards,
and the other (red B) is going westwards.
The first stage for both rockets is:
accelerated at 40 m/s² vertically for 200 seconds
The second stage is accelerated at 30 m/s² for 300 seconds,
Green A is accelerating 120⁰ to the east of vertical.
Red B is accelerating 120⁰ to the west of vertical.
If you don't want to run the simulation,
screenshots of a run can be found here:
https://paulba.no/temp/Satellites_run.pdf
--
Paul
https://paulba.no/
Paul
https://paulba.no/
Fact is that classical theory predicts that the force responsible for an objects
rotation (acceleration) around an axis will decrease the natural resonant frequency of caesium atoms (ie its “tick rate”)...the greater the force the
atom experiences.
A prediction confirmed by various experiments including Hafael Keating.
Who confirmed that the faster the rotational velocity of the caesium atoms relative to the ECI frame...the greater the force on the atoms and the slower
the observed tick rate will be.
So it’s up to you relativists to disprove this successful prediction made by classical theory.
On Friday, 27 October 2023 at 19:47:53 UTC+1, Paul B. Andersen wrote:
Den 27.10.2023 15:27, skrev Lou:
On Thursday, 26 October 2023 at 20:48:55 UTC+1, Paul B. Andersen wrote:
Poor desperate Pauls lies get bigger and bigger with each lie.
So far you have failed to supply any evidence to refute my
claim that NASA uses eastward rotation of earth to assist its rockets in >>> achieving escape velocity.
The point is that I never disputed the obvious fact that most
satellites not in polar orbit are orbiting in the same direction as
the Earth is spinning.
Quite the contrary, I have explained why a below:
And this explanation is correct.
| On Monday, 23 October 2023 at 22:04:42 UTC+1, Paul B. Andersen wrote: >>>> |> In the ECI frame, the rocket is moving eastwards at 1600 km/h
before it is launched, when NO HORIZONTAL FORCE is acting on it.
So when the rocket fires, it is obviously best to accelerate eastwards, >>>> |> where you start with the HORIZONTAL speed 1600 km/h eastwards.
Have you never seen the liftoff of a rocket on TV?
Before liftoff, the rocket is vertical and moving sideways
with the speed 1600 km/h in the ECI frame along with the ground.
When the rocket fires, the thrust of the rocket engine will
overcome gravity’s downward force, the rocket will lift off
and accelerate vertically. The vertical speed will increase
while the sideways horizontal speed will remain the same.
Some time after liftoff the rocket will veer off eastwards
You didn’t explain anything except confirm what I already told you.
Which is that NASA uses the force of the 1600 kph eastward rotation of
the earth to assist its rockets in achieving escape velocity to get into orbit.
So far none of you relativists has been able to disprove this fact.
And repeating it and pretending by repeating a fact disproves it is
a fact. Is typical of a relativist like yourself who understands little of physics.
On October 27, Lou wrote:
Fact is that classical theory predicts that the force responsible for an objects'classical theory' refers to a pre-quantum, or pre-relativity era,
rotation (acceleration) around an axis will decrease the natural resonant frequency of caesium atoms (ie its “tick rate”)...the greater the force the
atom experiences.
A prediction confirmed by various experiments including Hafael Keating. Who confirmed that the faster the rotational velocity of the caesium atoms relative to the ECI frame...the greater the force on the atoms and the slower
the observed tick rate will be.
So it’s up to you relativists to disprove this successful prediction made
by classical theory.
depending on context. That would be 1905 or 1920.
At that time, little was known about atoms, and nothing about
atomic resonances. Classical theory says nothing about such
resonances. Those "predictions" are nought but your imagination.
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!!!!!
Lift a bicycle wheel up off the ground. Spin it with your free hand.
Then touch that hand to the freely spinning wheel. According
to Paul you should feel no tangental force act on your hand.
Yet something forced your hand tangentally away from
the spinning wheel when you touched it!!!
Where did that tangental force come from if you and Paul say it
doesn’t exist?
On November, Lou wrote:
Lift a bicycle wheel up off the ground. Spin it with your free hand.Momentum. Newton's third. Elasticity of materials.
Then touch that hand to the freely spinning wheel. According
to Paul you should feel no tangental force act on your hand.
Yet something forced your hand tangentally away from
the spinning wheel when you touched it!!!
Where did that tangental force come from if you and Paul say it
doesn’t exist?
If a motor is attached to the wheel axis, then your argument
has some content. A spinning automotive tire, in gear, has
momentum and tangential velocity, and also force (torque),
to maintain speed against resistance. The system of rocket
and eastern launch doesn't work that way.
If the wheel hub disengages from the axle, and the tire rolls away,
the torque disappears. No force.
You have flunked a quiz that Newton could solve, 400 years ago.
But the earth also spins. And that spin is not produced by earths gravity. And that spin exerts an additional force on the caesium atoms that is from a tangental
force caused not by gravity but by the tangental acceleration of the clock in the ECI frame.
On 11/1/2023 6:42 PM, Lou wrote:
But the earth also spins. And that spin is not produced by earths gravity. And that spin exerts an additional force on the caesium atoms that is from a tangentalCesium atoms in an atomic clock are in freefall. That means there is no force on them whatsoever, from any source.
force caused not by gravity but by the tangental acceleration of the clock in
the ECI frame.
Besides, if there were a force, it would go as GM/r^2, which doesn't
match GR's inverse r relationship.
Basically what you just said is that the effects of the force of
gravity at different altitudes for a classical model have to be
calculated by incorrectly using r^2 to make sure the prediction is
false.
But for GR it’s OK to correctly model the force of gravity at
different alttudes using just r.
On Thursday, 2 November 2023 at 04:15:44 UTC, Volney wrote:
On 11/1/2023 6:42 PM, Lou wrote:
But the earth also spins. And that spin is not produced by earths gravity. >>> And that spin exerts an additional force on the caesium atoms that is from a tangentalCesium atoms in an atomic clock are in freefall. That means there is no
force caused not by gravity but by the tangental acceleration of the clock in
the ECI frame.
force on them whatsoever, from any source.
This is an answer that is a ‘have it two ways’ answer only a relativist would have the nerve to make: “There’s a force of gravity in free fall when calculating for the (imaginary) “time dilation” effects using GR .
But no force of gravity in free fall for a classical model.”
You probably don’t even realise that your claims are contradictory.
Anyways If there was no force of gravity acting on them, there would be no free fall. That’s why they are falling. The force of gravity is pushing them.
Besides, if there were a force, it would go as GM/r^2, which doesn't
match GR's inverse r relationship.
Basically what you just said is that the effects of the force of gravity
at different altitudes for a classical model have to be calculated by incorrectly using r^2
to make sure the prediction is false.
But for GR it’s OK to correctly model the force of gravity at different alttudes using just r.
You also ignores the fact that you were previously unable to supply
any calculation to refute the fact that the area of the gravity shadow
of earth at different altitudes falls off with r. Not r^2.
I provided you with all the data and calculations. You were unable to refute it.
So you didn’t answer and pretended that by not supplying any calculations to back up your false claims about r^2 you could avoid having to admit you failed.
A classic relativist tactic. When a relativist gets their claims proven wrong, they change the subject and come back a few weeks later with the
same false fact free claim again.
On 11/2/23 6:37 AM, Lou wrote:
Basically what you just said is that the effects of the force ofNo. One must use 1/r^2 to model the gravitational force at different altitudes (values of r) to make sure THE MODEL AGREES WITH MEASUREMENTS.
gravity at different altitudes for a classical model have to be
calculated by incorrectly using r^2 to make sure the prediction is
false.
But for GR it’s OK to correctly model the force of gravity atMore nonsense -- GR has no "gravitational force".
different alttudes using just r.
Lift a bicycle wheel up off the ground. Spin it with your free
hand.
Then touch that hand to the freely spinning wheel. According
to Paul you should feel no tangental force act on your hand.
Because he says there is no tangental force in a spinning
mass.
Den 01.11.2023 22:49, skrev Lou:
Lift a bicycle wheel up off the ground. Spin it with your freeLou, I am responding to your nonsense only because
hand.
Then touch that hand to the freely spinning wheel. According
to Paul you should feel no tangental force act on your hand.
I resent that you claim I have said something I never did.
Of course there is a tangential force on your hand when
you touch the wheel. It's called friction.
The heat that's burning your hand comes from the rotational
energy stored in the spinning wheel. The wheel will slow down.
Because he says there is no tangental force in a spinningThere is no tangential force acting on the rubber in
mass.
the wheel when you don't touch it. And the tangential force
that is acting on the rubber when you touch it is in
the opposite direction of the velocity, and in the opposite
direction of the force that is acting on your hand.
Newton's third law.
The only force that is acting on the rubber when you don't touch
the wheel is the centripetal force mediated through the spokes,
giving the rubber the centripetal acceleration v²/r, where v is
the tangential speed of the wheel, and r is the radius.
------------------
A word about Force and acceleration:
The Newtonian gravitational force acting on a mass m is:
F = GMm/R² where:
G is the gravitational constant,
M is the mass of Earth
R is the distance from the mass m to the centre of the Earth.
The gravitational acceleration of a mass m is:
g = F/m = GM/R²
note that the gravitational acceleration is independent
of the mass. (Galileo and the Leaning Tower of Pisa, remember?)
On 11/2/2023 7:37 AM, Lou wrote:
On Thursday, 2 November 2023 at 04:15:44 UTC, Volney wrote:
On 11/1/2023 6:42 PM, Lou wrote:
But the earth also spins. And that spin is not produced by earths gravity.Cesium atoms in an atomic clock are in freefall. That means there is no >> force on them whatsoever, from any source.
And that spin exerts an additional force on the caesium atoms that is from a tangental
force caused not by gravity but by the tangental acceleration of the clock in
the ECI frame.
This is an answer that is a ‘have it two ways’ answer only a relativistNo, there is no force! The GR effects are inversely proportional to r,
would have the nerve to make: “There’s a force of gravity in free fall when calculating for the (imaginary) “time dilation” effects using GR .
not r^2,
so GMm/r doesn't even have dimensionality of force, while
GMm/r^2 does. You seem to think everything is a force, including the tangential velocity of a launched rocket. Which is, obviously, a
velocity, not a force.
But no force of gravity in free fall for a classical model.”A free falling object doesn't experience any force.
And this is no double standard, since GMm/r is not even a force, while GMm/r^2 is.
You probably don’t even realise that your claims are contradictory.Since GMm/r isn't even a force, no contradiction.
Anyways If there was no force of gravity acting on them, there would be no free fall. That’s why they are falling. The force of gravity is pushing them.In free fall, there is no effect of gravity, pretty much by definition. That's why astronauts float around the space station (orbits are freefall).
Besides, if there were a force, it would go as GM/r^2, which doesn't
match GR's inverse r relationship.
Basically what you just said is that the effects of the force of gravity at different altitudes for a classical model have to be calculated by incorrectly using r^2That is not incorrect, that's just plain Newtonian physics!
to make sure the prediction is false.Newton wanted to help prove relativity?
But for GR it’s OK to correctly model the force of gravity at different alttudes using just r.Once again, it is not even a force!
You also ignores the fact that you were previously unable to supplyFirst, the shadow is irrelevant for this discussion. Second, the shadow
any calculation to refute the fact that the area of the gravity shadow
of earth at different altitudes falls off with r. Not r^2.
area is not proportional to either 1/r nor 1/r^2. At large distances,
where the sine of the angle of the edge of the shadowing object can be approximated as 1 (meaning a very small angle) it is approximately proportional to 1/r^2. For close distances once cannot make that
assumption, the function is more complex, involving the angle of the
center of the object-observer-edge of the object. This is not
proportional to 1/r (or 1/r^2).
I provided you with all the data and calculations. You were unable to refute it.I'm not about to look up the near distance function for you. For large distances it is close to 1/r^2 as the source is pointlike (subtended
So you didn’t answer and pretended that by not supplying any calculations
to back up your false claims about r^2 you could avoid having to admit you failed.
angle almost 0).
A classic relativist tactic. When a relativist gets their claims proven wrong, they change the subject and come back a few weeks later with the same false fact free claim again.The size of a shadow isn't even relevant here! We're discussing the
effects of gravitational force and potential, not shadows.
On 11/2/23 6:37 AM, Lou wrote:
Basically what you just said is that the effects of the force ofNo. One must use 1/r^2 to model the gravitational force at different altitudes (values of r) to make sure THE MODEL AGREES WITH MEASUREMENTS.
gravity at different altitudes for a classical model have to be
calculated by incorrectly using r^2 to make sure the prediction is
false.
But for GR it’s OK to correctly model the force of gravity atMore nonsense -- GR has no "gravitational force". In the Newtonian limit
different alttudes using just r.
of GR, there is a gravitational force that goes as 1/r^2 (as in
Newtonian gravitation).
Your fundamental problem is that you just do not know what "force"
means. Ditto for just about every technical word you use. You REALLY
need to learn basic physics before attempting to write about it.
Den 01.11.2023 22:49, skrev Lou:
Lift a bicycle wheel up off the ground. Spin it with your freeLou, I am responding to your nonsense only because
hand.
Then touch that hand to the freely spinning wheel. According
to Paul you should feel no tangental force act on your hand.
I resent that you claim I have said something I never did.
Of course there is a tangential force on your hand when
you touch the wheel. It's called friction.
The heat that's burning your hand comes from the rotational
energy stored in the spinning wheel. The wheel will slow down.
Because he says there is no tangental force in a spinningThere is no tangential force acting on the rubber in
mass.
the wheel when you don't touch it. And the tangential force
that is acting on the rubber when you touch it is in
the opposite direction of the velocity, and in the opposite
direction of the force that is acting on your hand.
Newton's third law.
The only force that is acting on the rubber when you don't touch
the wheel is the centripetal force mediated through the spokes,
giving the rubber the centripetal acceleration v²/r, where v is
the tangential speed of the wheel, and r is the radius.
------------------
A word about Force and acceleration:
The Newtonian gravitational force acting on a mass m is:
F = GMm/R² where:
G is the gravitational constant,
M is the mass of Earth
R is the distance from the mass m to the centre of the Earth.
The gravitational acceleration of a mass m is:
g = F/m = GM/R²
note that the gravitational acceleration is independent
of the mass. (Galileo and the Leaning Tower of Pisa, remember?)
According to Newton, g is the acceleration of an object
in free fall, and the acceleration is downwards.
According to GR, g = GM/R² is the proper acceleration of a body
when the body is stationary on the ground. It is acting upwards. (Equivalence principle.)
Note that an accelerometer on the ground shows that the acceleration
is upwards, and an accelerometer i free fall shows no acceleration.
An accelerometer measures proper acceleration, and proves
the equivalence principle.
But you shouldn't bother with relativity.
Learn Newtonian physics first!
-------------------------
Say whatever you want about this, but don't try to
paraphrase me. Quote me literally, or not at all!
--
Paul
https://paulba.no/
OOn Friday, 3 November 2023 at 04:47:40 UTC, Tom Roberts wrote:
On 11/2/23 6:37 AM, Lou wrote:
Basically what you just said is that the effects of the force ofNo. One must use 1/r^2 to model the gravitational force at different
gravity at different altitudes for a classical model have to be
calculated by incorrectly using r^2 to make sure the prediction is
false.
altitudes (values of r) to make sure THE MODEL AGREES WITH MEASUREMENTS.
But the measurements of gravitational time dilation at different
altitudes is consistent with 1/r.
Not r^2
Volney can confirm this.
But for GR it’s OK to correctly model the force of gravity atMore nonsense -- GR has no "gravitational force". In the Newtonian limit
different alttudes using just r.
of GR, there is a gravitational force that goes as 1/r^2 (as in
Newtonian gravitation).
Not what I hear. GR uses r, gravitational potential, to model
the so called ‘time dilation’ effects of gravity on in tick rates of atomic clocks at different altitudes.
Your fundamental problem is that you just do not know what "force"
means. Ditto for just about every technical word you use. You REALLY
need to learn basic physics before attempting to write about it.
Your fundamental problem is you don’t understand what force and gravity are.
On Friday, 3 November 2023 at 04:52:55 UTC, Volney wrote:
On 11/2/2023 7:37 AM, Lou wrote:
On Thursday, 2 November 2023 at 04:15:44 UTC, Volney wrote:not r^2,
On 11/1/2023 6:42 PM, Lou wrote:
But the earth also spins. And that spin is not produced by earths gravity.Cesium atoms in an atomic clock are in freefall. That means there is no >>>> force on them whatsoever, from any source.
And that spin exerts an additional force on the caesium atoms that is from a tangental
force caused not by gravity but by the tangental acceleration of the clock in
the ECI frame.
This is an answer that is a ‘have it two ways’ answer only a relativist >>> would have the nerve to make: “There’s a force of gravity in free fall >>> when calculating for the (imaginary) “time dilation” effects using GR . >> No, there is no force! The GR effects are inversely proportional to r,
Exactly. So too are the effects of gravity in a classical model inversely proportional to r. Not r^2
That’s how a classical model also successfully predicts tick rates of caesium atoms at different altitudes.
so GMm/r doesn't even have dimensionality of force, while
GMm/r^2 does. You seem to think everything is a force, including the
tangential velocity of a launched rocket. Which is, obviously, a
velocity, not a force.
But no force of gravity in free fall for a classical model.”A free falling object doesn't experience any force.
And this is no double standard, since GMm/r is not even a force, while
GMm/r^2 is.
Pure nonsense. Those are arbritrary formulas. Not forces. Made up
to try to model the force of gravity.
And r....models the force of gravity
on tick rates of atoms at different altitudes for both SR and classical.
Not r^2.
You probably don’t even realise that your claims are contradictory.
Since GMm/r isn't even a force, no contradiction.
Completely ignoring the fact Einstein used r to succesfully model the force of gravity on various things in GR.
I provided you with all the data and calculations. You were unable to refute it.
So you didn’t answer and pretended that by not supplying any calculations >>> to back up your false claims about r^2 you could avoid having to admit you >>> failed.
I'm not about to look up the near distance function for you. For large
distances it is close to 1/r^2 as the source is pointlike (subtended
angle almost 0).
In other words...you know I’m right. So you don’t try to prove I’m wrong.
A classic relativist tactic. When a relativist gets their claims proven
wrong, they change the subject and come back a few weeks later with the
same false fact free claim again.
The size of a shadow isn't even relevant here! We're discussing the
effects of gravitational force and potential, not shadows.
There you go. Cant refute the fact that the area of the gravity shadow in a classical
model falls off at r
Thought you wouldn’t be able to answer.
------------------
A word about Force and acceleration:
The Newtonian gravitational force acting on a mass m is:
F = GMm/R² where:
G is the gravitational constant,
M is the mass of Earth
R is the distance from the mass m to the centre of the Earth.
The gravitational acceleration of a mass m is:
g = F/m = GM/R²
note that the gravitational acceleration is independent
of the mass. (Galileo and the Leaning Tower of Pisa, remember?)
According to Newton, g is the acceleration of an object
in free fall, and the acceleration is downwards.
On 11/3/2023 12:24 PM, Lou wrote:
On Friday, 3 November 2023 at 04:52:55 UTC, Volney wrote:
On 11/2/2023 7:37 AM, Lou wrote:
On Thursday, 2 November 2023 at 04:15:44 UTC, Volney wrote:No, there is no force! The GR effects are inversely proportional to r,
On 11/1/2023 6:42 PM, Lou wrote:
But the earth also spins. And that spin is not produced by earths gravity.Cesium atoms in an atomic clock are in freefall. That means there is no >>>> force on them whatsoever, from any source.
And that spin exerts an additional force on the caesium atoms that is from a tangental
force caused not by gravity but by the tangental acceleration of the clock in
the ECI frame.
This is an answer that is a ‘have it two ways’ answer only a relativist
would have the nerve to make: “There’s a force of gravity in free fall
when calculating for the (imaginary) “time dilation” effects using GR .
not r^2,
Exactly. So too are the effects of gravity in a classical model inversely proportional to r. Not r^2Once again, Newtonian gravity is GMm/r^2. NOT /r. Classical physics has
time the same everywhere, no such thing as time dilation, so no time dilation caused by any potential GMm/r. You are grasping at straws.
That’s how a classical model also successfully predicts tick rates of caesium atoms at different altitudes.WHAT "classical model"? You can't even get the units correct!
so GMm/r doesn't even have dimensionality of force, while
GMm/r^2 does. You seem to think everything is a force, including the
tangential velocity of a launched rocket. Which is, obviously, a
velocity, not a force.
But no force of gravity in free fall for a classical model.”A free falling object doesn't experience any force.
And this is no double standard, since GMm/r is not even a force, while
GMm/r^2 is.
Pure nonsense. Those are arbritrary formulas. Not forces. Made upAre you saying that Newton just made up his GMm/r^2?
to try to model the force of gravity.
And r....models the force of gravityClassical physics doesn't even have time dilation! What classical
on tick rates of atoms at different altitudes for both SR and classical. Not r^2.
formula can model something it doesn't even have?
You probably don’t even realise that your claims are contradictory.
Since GMm/r isn't even a force, no contradiction.
Completely ignoring the fact Einstein used r to succesfully model the forceNo, he did not.
of gravity on various things in GR.
I provided you with all the data and calculations. You were unable to refute it.
So you didn’t answer and pretended that by not supplying any calculations
to back up your false claims about r^2 you could avoid having to admit you
failed.
I'm not about to look up the near distance function for you. For large
distances it is close to 1/r^2 as the source is pointlike (subtended
angle almost 0).
In other words...you know I’m right. So you don’t try to prove I’m wrong.Nope. Again, first the shadow is irrelevant to the discussion. Second,
it is up to YOU to prove your claim (with a derived formula or reference
to one) that shadows go as 1/r.
A classic relativist tactic. When a relativist gets their claims proven >>> wrong, they change the subject and come back a few weeks later with the >>> same false fact free claim again.
The size of a shadow isn't even relevant here! We're discussing the
effects of gravitational force and potential, not shadows.
There you go. Cant refute the fact that the area of the gravity shadow in a classicalWhy would I answer since it up to YOU to prove your claim? In addition,
model falls off at r
Thought you wouldn’t be able to answer.
once again, it's irrelevant.
On Friday, 3 November 2023 at 17:33:37 UTC, Volney wrote:
On 11/3/2023 12:24 PM, Lou wrote:
On Friday, 3 November 2023 at 04:52:55 UTC, Volney wrote:
On 11/2/2023 7:37 AM, Lou wrote:
On Thursday, 2 November 2023 at 04:15:44 UTC, Volney wrote:
On 11/1/2023 6:42 PM, Lou wrote:
A classic relativist tactic. When a relativist gets their claims proven >>>>> wrong, they change the subject and come back a few weeks later with the >>>>> same false fact free claim again.
The size of a shadow isn't even relevant here! We're discussing the
effects of gravitational force and potential, not shadows.
And notice for a classical model area of gravity shadow/altitude is how the force of gravity
is calculated. And it is inversely proportional.
Not some BS relativists fantasy of r^2 .
Notice GR itself calculates the effect of gravity on atomic clock tick
rates using inverse proportional.
Relativists try to con everyone by pretending gravitational potential
has nothing to do with gravity!!
Ignoring the fact that to calculate “time dilation” all GR reference says it is *gravity* which slows down or speeds up the imaginary photons which leads to the imaginary “time dilation”
Hypocrisy supreme from relativists. Saying gravity causes time dilation.
And then in the same breath saying it doesn’t💩.
There you go. Cant refute the fact that the area of the gravity shadow in a classicalWhy would I answer since it up to YOU to prove your claim? In addition,
model falls off at r
Thought you wouldn’t be able to answer.
once again, it's irrelevant.
I already have proved my claim.
If one measures the angle subtended by the earth
to an observer at R, 2R , 3R etc....The area of the shadow falls off with r
Not r ^2 as you falsely claim
You are so dishonest that as usual for a relativist you make claims about a classical model that you know are patently false. Here below are percentages based on angle subtended by earth to observer at various radius distances
for an observer. Notice the relationship is...Inverse proportional.
r=100%
2r=33% 1/2
3r =21% 1/3
4r=16% 1/4
5r=13% 1/5
On 11/5/2023 5:16 AM, Lou wrote:
On Friday, 3 November 2023 at 17:33:37 UTC, Volney wrote:
On 11/3/2023 12:24 PM, Lou wrote:
On Friday, 3 November 2023 at 04:52:55 UTC, Volney wrote:
On 11/2/2023 7:37 AM, Lou wrote:
On Thursday, 2 November 2023 at 04:15:44 UTC, Volney wrote:
On 11/1/2023 6:42 PM, Lou wrote:
A classic relativist tactic. When a relativist gets their claims proven
wrong, they change the subject and come back a few weeks later with the
same false fact free claim again.
The size of a shadow isn't even relevant here! We're discussing the >>>> effects of gravitational force and potential, not shadows.
And notice for a classical model area of gravity shadow/altitude is how the force of gravityNewton was a "relativist"? Remember, F=GMm/r^2 is from Newton.
is calculated. And it is inversely proportional.
Not some BS relativists fantasy of r^2 .
Notice GR itself calculates the effect of gravity on atomic clock tick rates using inverse proportional.That is the effect of the POTENTIAL, not a force!
Relativists try to con everyone by pretending gravitational potentialNo, it is not a force. GMm/r^2 is a force (Newton model).
has nothing to do with gravity!!
Ignoring the fact that to calculate “time dilation” all GR reference saysPhotons always travel at c.
it is *gravity* which slows down or speeds up the imaginary photons which leads to the imaginary “time dilation”
Gravitational FORCE does not cause time dilation. Time dilation is not proportional to the FORCE (1/r^2).
On 11/5/2023 5:16 AM, Lou wrote:
On Friday, 3 November 2023 at 17:33:37 UTC, Volney wrote:
On 11/3/2023 12:24 PM, Lou wrote:
On Friday, 3 November 2023 at 04:52:55 UTC, Volney wrote:
On 11/2/2023 7:37 AM, Lou wrote:
On Thursday, 2 November 2023 at 04:15:44 UTC, Volney wrote:
On 11/1/2023 6:42 PM, Lou wrote:
A classic relativist tactic. When a relativist gets their claims proven
wrong, they change the subject and come back a few weeks later with the
same false fact free claim again.
The size of a shadow isn't even relevant here! We're discussing the >>>> effects of gravitational force and potential, not shadows.
And notice for a classical model area of gravity shadow/altitude is how the force of gravityNewton was a "relativist"? Remember, F=GMm/r^2 is from Newton.
is calculated. And it is inversely proportional.
Not some BS relativists fantasy of r^2 .
Notice GR itself calculates the effect of gravity on atomic clock tick rates using inverse proportional.That is the effect of the POTENTIAL, not a force!
Relativists try to con everyone by pretending gravitational potentialNo, it is not a force. GMm/r^2 is a force (Newton model).
has nothing to do with gravity!!
Ignoring the fact that to calculate “time dilation” all GR reference saysPhotons always travel at c.
it is *gravity* which slows down or speeds up the imaginary photons which leads to the imaginary “time dilation”
Hypocrisy supreme from relativists. Saying gravity causes time dilation. And then in the same breath saying it doesn’t💩.Gravitational FORCE does not cause time dilation. Time dilation is not proportional to the FORCE (1/r^2).
There you go. Cant refute the fact that the area of the gravity shadow in a classicalWhy would I answer since it up to YOU to prove your claim? In addition, >> once again, it's irrelevant.
model falls off at r
Thought you wouldn’t be able to answer.
I already have proved my claim.Assertions are not proofs.
If one measures the angle subtended by the earthAssertion. Where is the formula which shows how the shadow falls off
to an observer at R, 2R , 3R etc....The area of the shadow falls off with r
with distance? You have not provided it, so you haven't proven your claim.
Not r ^2 as you falsely claimR^2 is NEWTON's claim!
You are so dishonest that as usual for a relativist you make claims about aThose numbers are not inversely proportional.
classical model that you know are patently false. Here below are percentages
based on angle subtended by earth to observer at various radius distances for an observer. Notice the relationship is...Inverse proportional.
r=100%
2r=33% 1/2
3r =21% 1/3
4r=16% 1/4
5r=13% 1/5
WHERE IS YOUR FORMULA?
Notice GR itself calculates the effect of gravity on atomic clock tick
rates using inverse proportional.
Den 05.11.2023 11:16, skrev Lou:
Notice GR itself calculates the effect of gravity on atomic clock tick rates using inverse proportional.According to GR, gravitation doesn't affect the tick rate of a clock.
The clock always run at its proper rate.
But an observer in a gravitational field, say on the Earth,
will observe that a clock higher up seems to run faster
than his own clock, because he can observe the clock in his
telescope, and see that it advances faster than his own clock.
And this is no illusion.
Den 05.11.2023 11:16, skrev Lou:
Notice GR itself calculates the effect of gravity on atomic clock tick rates using inverse proportional.According to GR, gravitation doesn't affect the tick rate of a clock.
The clock always run at its proper rate.
But an observer in a gravitational field, say on the Earth,
will observe that a clock higher up seems to run faster
than his own clock, because he can observe the clock in his
telescope, and see that it advances faster than his own clock.
And this is no illusion.
Consider the following scenario:
On the North pole we have a clock, a sender transmitting
a radio signal with frequency f₀, and a radio receiver
with a frequency counter.
Hoovering above the North pole at altitude h we have
a spaceship with a rocket engine. It has equal instrumentation
as on the ground. The two clocks are equal, and the transmitted
frequencies are both f₀.
According to GR, what are the consequences of the gravitation?
#1:
A person sitting on a chair at the ground will feel that the chair
exerts a force on him, F₁ = m₁⋅GM/R² where m₁ is his mass, G is
the gravitational constant, M is the mass of the Earth and R is
the radius of the Earth. This force will give him an upwards
proper acceleration g₁ = GM/R²
#2:
To stay at the altitude h, the rocket engine must exert a force
on the spaceship, F₂ = m₂⋅GM/(R+h)² where m₂ is the mass of the spaceship. This force will give the spaceship an upwards proper
acceleration g₂ = GM/(R+h)².
#3:
The observer on the ground will receive the frequency f₁ from
the spaceship, where:
f₁ = √(1 + (GM/c²)⋅(1/R - 1/(R+h)))⋅f₀
Since f₁ > f₀ the observer on the ground will say that
the clock in the spaceship appears to 'tick faster' than his own clock.
#4:
The observer in the spaceship will receive the frequency f₂ from
the ground, where:
f₂ = √(1 + (GM/c²)⋅(1/(R+h)-1/R))⋅f₀
Since f₂ < f₀ the observer in the spaceship will say that
the clock on the ground appears to 'tick slower' than his own clock.
-----------------
Let's put in numbers.
f₀ = 1GHz, h = 20309 km (GPS height)
f₁ = 1000000000.52867 Hz
f₂ = 999999999.47133 Hz
The ground clock would measure a solar day to be 86400 seconds,
while the clock in the spaceship would measure a solar day
to be 86400.0000456771 seconds.
Get this:
According to NM and GR _the force on a stationary body in the ECI
frame is inversely proportional to the square of the distance to
the centre of the Earth_.
And remember, GR's predictions are thoroughly confirmed.
On Sunday, 5 November 2023 at 21:37:17 UTC, Paul B. Andersen wrote:
Den 05.11.2023 11:16, skrev Lou:
According to GR, gravitation doesn't affect the tick rate of a clock.
Notice GR itself calculates the effect of gravity on atomic clock tick
rates using inverse proportional.
The clock always run at its proper rate.
But an observer in a gravitational field, say on the Earth,
will observe that a clock higher up seems to run faster
than his own clock, because he can observe the clock in his
telescope, and see that it advances faster than his own clock.
And this is no illusion.
Consider the following scenario:
On the North pole we have a clock, a sender transmitting
a radio signal with frequency f₀, and a radio receiver
with a frequency counter.
Hoovering above the North pole at altitude h we have
a spaceship with a rocket engine. It has equal instrumentation
as on the ground. The two clocks are equal, and the transmitted
frequencies are both f₀.
According to GR, what are the consequences of the gravitation?
#1:
A person sitting on a chair at the ground will feel that the chair
exerts a force on him, F₁ = m₁⋅GM/R² where m₁ is his mass, G is
the gravitational constant, M is the mass of the Earth and R is
the radius of the Earth. This force will give him an upwards
proper acceleration g₁ = GM/R²
#2:
To stay at the altitude h, the rocket engine must exert a force
on the spaceship, F₂ = m₂⋅GM/(R+h)² where m₂ is the mass of the
spaceship. This force will give the spaceship an upwards proper
acceleration g₂ = GM/(R+h)².
#3:
The observer on the ground will receive the frequency f₁ from
the spaceship, where:
f₁ = √(1 + (GM/c²)⋅(1/R - 1/(R+h)))⋅f₀
Since f₁ > f₀ the observer on the ground will say that
the clock in the spaceship appears to 'tick faster' than his own clock.
#4:
The observer in the spaceship will receive the frequency f₂ from
the ground, where:
f₂ = √(1 + (GM/c²)⋅(1/(R+h)-1/R))⋅f₀
Since f₂ < f₀ the observer in the spaceship will say that
the clock on the ground appears to 'tick slower' than his own clock.
-----------------
Let's put in numbers.
f₀ = 1GHz, h = 20309 km (GPS height)
f₁ = 1000000000.52867 Hz
f₂ = 999999999.47133 Hz
The ground clock would measure a solar day to be 86400 seconds,
while the clock in the spaceship would measure a solar day
to be 86400.0000456771 seconds.
You are the expert on GR. I’m just trying to show how the ‘apparent’ change in tick rates can be explained classically as *actual* real tick rate changes due to different strengths of gravity at different altitudes on
atoms using r, not r^2.
(GR also seems to use r, as per #3 below)
It seems #1 and #2 on your above list are concerned with how
much energy it takes to hold the rocket up at a certain altitude so that’s a seperate issue I believe. Although I remind you below that the reason
why the rocket needs to exert a force on the rocket is because of
the gravitional pull of earth below it.
However your #3 seems to be directly concerned with why the tick rates
change with altitude under GR. Or appear to change.
It seems that formula 3 doesn’t use r^2. But rather r.
Isnt that the case?
If so then it doesnt matter why the grounds tick rate appears slower to
the spaceship observer, but rather.....HOW you calculate this apparent slowing down of the tick rate.
And so ultimately it seems you still say GRAVITY is the reason for the apparent
slowing up of the tick rate for the spaceship observer. After all if the spaceship and the ground observer were not subject to earths gravity then there would be no apparent change in the tick rates between the two.
And the formula #3 you use is essentially saying the relationship between the different ‘apparent’ tick rates and altitude is based on r. Not r^2.
So you can’t use r to model the effects of gravity on the frequency of
your photons going from surface to spaceship and then say a classical
model *can’t* use r to model the same observation.
After all both appear the same. The actual classical *real physical” change in tick
rates looks exactly the same as a relativistic “apparent” change in tick rates.
On Tuesday, 7 November 2023 at 22:28:48 UTC+1, Paul B. Andersen wrote:
Weight can go away not gravity force.Get this:Learn your Shit, poor idiot. No force according to it.
According to NM and GR _the force on a stationary body in the ECI
frame is inversely proportional to the square of the distance to
the centre of the Earth_.
Get this.What predictions?
And remember, GR's predictions are thoroughly confirmed.
While in the meantime in the real world, forbidden byClocks have time as well as space...
it "improper" clocks keep measuring t'=t, just like all
serious clocks always did.
Den 06.11.2023 11:58, skrev Lou:
On Sunday, 5 November 2023 at 21:37:17 UTC, Paul B. Andersen wrote:
Den 05.11.2023 11:16, skrev Lou:
According to GR, gravitation doesn't affect the tick rate of a clock.
Notice GR itself calculates the effect of gravity on atomic clock tick >>> rates using inverse proportional.
The clock always run at its proper rate.
But an observer in a gravitational field, say on the Earth,
will observe that a clock higher up seems to run faster
than his own clock, because he can observe the clock in his
telescope, and see that it advances faster than his own clock.
And this is no illusion.
Consider the following scenario:
On the North pole we have a clock, a sender transmitting
a radio signal with frequency f₀, and a radio receiver
with a frequency counter.
Hoovering above the North pole at altitude h we have
a spaceship with a rocket engine. It has equal instrumentation
as on the ground. The two clocks are equal, and the transmitted
frequencies are both f₀.
According to GR, what are the consequences of the gravitation?
#1:
A person sitting on a chair at the ground will feel that the chair
exerts a force on him, F₁ = m₁⋅GM/R² where m₁ is his mass, G is >> the gravitational constant, M is the mass of the Earth and R is
the radius of the Earth. This force will give him an upwards
proper acceleration g₁ = GM/R²
#2:
To stay at the altitude h, the rocket engine must exert a force
on the spaceship, F₂ = m₂⋅GM/(R+h)² where m₂ is the mass of the >> spaceship. This force will give the spaceship an upwards proper
acceleration g₂ = GM/(R+h)².
#3:
The observer on the ground will receive the frequency f₁ from
the spaceship, where:
f₁ = √(1 + (GM/c²)⋅(1/R - 1/(R+h)))⋅f₀
Since f₁ > f₀ the observer on the ground will say that
the clock in the spaceship appears to 'tick faster' than his own clock. >>
#4:
The observer in the spaceship will receive the frequency f₂ from
the ground, where:
f₂ = √(1 + (GM/c²)⋅(1/(R+h)-1/R))⋅f₀
Since f₂ < f₀ the observer in the spaceship will say that
the clock on the ground appears to 'tick slower' than his own clock.
-----------------
Let's put in numbers.
f₀ = 1GHz, h = 20309 km (GPS height)
f₁ = 1000000000.52867 Hz
f₂ = 999999999.47133 Hz
The ground clock would measure a solar day to be 86400 seconds,
while the clock in the spaceship would measure a solar day
to be 86400.0000456771 seconds.
You are the expert on GR. I’m just trying to show how the ‘apparent’ change in tick rates can be explained classically as *actual* real tick rateNonsense.
changes due to different strengths of gravity at different altitudes on atoms using r, not r^2.
(GR also seems to use r, as per #3 below)
It seems #1 and #2 on your above list are concerned with howI never mentioned energy.
much energy it takes to hold the rocket up at a certain altitude so that’s
a seperate issue I believe. Although I remind you below that the reason why the rocket needs to exert a force on the rocket is because of
the gravitional pull of earth below it.
I only pointed out that the gravitational
force that acts on objects that are stationary in the non rotating
Earth centred frame (ECI) is: F = GMm/r²
Get this:
According to NM and GR _the force on a stationary body in the ECI
frame is inversely proportional to the square of the distance to
the centre of the Earth_.
Stop making a fool of yourself by claiming otherwise.
However your #3 seems to be directly concerned with why the tick rates change with altitude under GR. Or appear to change.No.
It seems that formula 3 doesn’t use r^2. But rather r.
Isnt that the case?
It is meaningless to talk about "tick rate" of a clock
without having something to compare it with.
So let me repeat what I showed you above.
Note:
The "tick rate" f compared to the clock on the ground,
means that if the clock emits a frequency 1 Hz, then
the frequency f will be received at the ground.
(Gravitational blue shift, Gravitational Doppler shift)
Below is "rate" the 'tick rate' relative to a clock on the ground.
"r" is the distance from the centre of the Earth.
"F/F₀" is the gravitational force on the clock relative to
the force on a clock at the ground.
r rate F/Fo
---------------------------------
10R 1.000000000626 0.0100
9R 1.000000000618 0.0123
8R 1.000000000608 0.0156
7R 1.000000000596 0.0204
6R 1.000000000579 0.0278
5R 1.000000000556 0.0400
4R 1.000000000522 0.0625
3R 1.000000000464 0.1111
2R 1.000000000348 0.2500
1R 1.000000000000 1.0000
R is the radius of the Earth
GPS r = 4.12R
Geostationary r = 6.6R
Moon r = 60R
In case your newsreader has cluttered up the table,
you can find it Here:
https://paulba.no/temp/ClockRate.pdf
Note that F/Fo is inversely proportional to r².
At 10R the force is one hundred part of the force on 1R.
But the rate at 10R is only very slightly higher.
It is nothing near inversely proportional to the force,
and it nothing near inversely proportional to r.
And remember, GR's predictions are thoroughly confirmed.
If so then it doesnt matter why the grounds tick rate appears slower to the spaceship observer, but rather.....HOW you calculate this apparent slowing down of the tick rate.Of course gravity is the cause of the gravitational blue shift (gravitational clock dilation).
And so ultimately it seems you still say GRAVITY is the reason for the apparent
slowing up of the tick rate for the spaceship observer. After all if the spaceship and the ground observer were not subject to earths gravity then there would be no apparent change in the tick rates between the two.
The gravitational force is inversely proportional to r².
The gravitational clock dilation ("clock rate") increases
with r, but it is neither proportional nor inversely proportional
to either r or r². Why should it be?
And the formula #3 you use is essentially saying the relationship between the
different ‘apparent’ tick rates and altitude is based on r. Not r^2. So you can’t use r to model the effects of gravity on the frequency of your photons going from surface to spaceship and then say a classical model *can’t* use r to model the same observation.
After all both appear the same. The actual classical *real physical” change in tick
rates looks exactly the same as a relativistic “apparent” change in tick rates.
On Tuesday, 7 November 2023 at 21:28:48 UTC, Paul B. Andersen wrote:
The gravitational force is inversely proportional to r².
Except for when GR calculates tick rate changes. Then you
use just r.
Den 08.11.2023 16:46, skrev Lou:
On Tuesday, 7 November 2023 at 21:28:48 UTC, Paul B. Andersen wrote:Indisputable!
The gravitational force is inversely proportional to r².
Except for when GR calculates tick rate changes. Then youSo stop disputing!
use just r.
On Wednesday, 8 November 2023 at 17:56:52 UTC, Paul B. Andersen wrote:
Den 08.11.2023 16:46, skrev Lou:
On Tuesday, 7 November 2023 at 21:28:48 UTC, Paul B. Andersen wrote:
The gravitational force is inversely proportional to r².
Indisputable!
Then why isn’t r^2 in your formula for calculating the effect of gravity
on caesium clock tick rates?
f₂ = √(1 + (GM/c²)⋅(1/(R+h)-1/R))⋅f₀
It’s not surprising the predictions made by GR that *you* cited in your earlier post of increases in tick rates with altitude are *proportional to r*.
After all, you used r to calculate them. And then tried to pretend
you used r^2.
Except for when GR calculates tick rate changes. Then you
use just r.
So stop disputing!
Finally!
I’m glad you admit that GR, like classical theory, assumes the force
of gravity is proportional to r to predict tick rates.
Den 08.11.2023 16:46, skrev Lou:
On Tuesday, 7 November 2023 at 21:28:48 UTC, Paul B. Andersen wrote:Indisputable!
The gravitational force is inversely proportional to r².
On Wednesday, 8 November 2023 at 17:56:52 UTC, Paul B. Andersen
wrote:
Den 08.11.2023 16:46, skrev Lou:
On Tuesday, 7 November 2023 at 21:28:48 UTC, Paul B. AndersenIndisputable!
wrote:
The gravitational force is inversely proportional to r².
Then why isn’t r^2 in your formula for calculating the effect of
gravity on caesium clock tick rates?
[...] And then tried to pretend you used r^2.
I’m glad you admit that GR, like classical theory, assumes the force
of gravity is proportional to r to predict tick rates.
On Wednesday, 8 November 2023 at 17:56:52 UTC, Paul B. Andersen wrote:
Den 08.11.2023 16:46, skrev Lou:
On Tuesday, 7 November 2023 at 21:28:48 UTC, Paul B. Andersen wrote:Indisputable!
The gravitational force is inversely proportional to r².
Then why isn’t r^2 in your formula for calculating the effect of gravity
on caesium clock tick rates?
f₂ = √(1 + (GM/c²)⋅(1/(R+h)-1/R))⋅f₀
Den 08.11.2023 19:24, skrev Lou:
I am beginning to suspect that you are raving mad.
Enough, now.
On 11/8/23 12:24 PM, Lou wrote:
On Wednesday, 8 November 2023 at 17:56:52 UTC, Paul B. AndersenWell, in weak fields such as near earth.
wrote:
Den 08.11.2023 16:46, skrev Lou:
On Tuesday, 7 November 2023 at 21:28:48 UTC, Paul B. AndersenIndisputable!
wrote:
The gravitational force is inversely proportional to r².
This is HIGHLY DISPUTABLE in strong field regions (were "gravitational force" simply does not make sense, and r might not make sense, either).
Then why isn’t r^2 in your formula for calculating the effect ofBecause clock tick rates don't depend on gravitation at all.
gravity on caesium clock tick rates?
But in the weak field near earth, when clock tick rates are MEASURED
from a different location, the value measured depends on the difference
in GRAVITATIONAL POTENTIAL between the clock and the measuring
instrument [#]. In the weak field near earth, gravitational potential is proportional to 1/r (r measured from the center of the earth; this
neglects minor effects).
[#] Presuming the measurement is mediated by EM signals,
and the clock and instrument are at rest in a static field.
[...] And then tried to pretend you used r^2.
Nobody ever "pretended" like that, YOU misread. Repeatedly. Because YOU
do not understand the difference between potential and force.
I’m glad you admit that GR, like classical theory, assumes the forceNo! Nobody ever said that (except you). YOU have repeatedly misread.
of gravity is proportional to r to predict tick rates.
In the weak field near earth, measurements of clock tick rates depend on
the difference in GRAVITATIONAL POTENTIAL between clock and measuring instrument; it is proportional to 1/r. Gravitational FORCE is
proportional to the gradient of the gravitational potential, i.e. proportional to 1/r^2.
You REALLY need to learn basic physics before attempting to write about it. Tom Roberts
On Wednesday, 8 November 2023 at 23:09:59 UTC, Tom Roberts wrote:
On 11/8/23 12:24 PM, Lou wrote:
On Wednesday, 8 November 2023 at 17:56:52 UTC, Paul B. AndersenWell, in weak fields such as near earth.
wrote:
Den 08.11.2023 16:46, skrev Lou:
On Tuesday, 7 November 2023 at 21:28:48 UTC, Paul B. AndersenIndisputable!
wrote:
The gravitational force is inversely proportional to r².
This is HIGHLY DISPUTABLE in strong field regions (were "gravitational
force" simply does not make sense, and r might not make sense, either).
Then why isn’t r^2 in your formula for calculating the effect ofBecause clock tick rates don't depend on gravitation at all.
gravity on caesium clock tick rates?
But in the weak field near earth, when clock tick rates are MEASURED
from a different location, the value measured depends on the difference
in GRAVITATIONAL POTENTIAL between the clock and the measuring
instrument [#]. In the weak field near earth, gravitational potential is
proportional to 1/r (r measured from the center of the earth; this
neglects minor effects).
[#] Presuming the measurement is mediated by EM signals,
and the clock and instrument are at rest in a static field.
[...] And then tried to pretend you used r^2.
Nobody ever "pretended" like that, YOU misread. Repeatedly. Because YOU
do not understand the difference between potential and force.
I’m glad you admit that GR, like classical theory, assumes the forceNo! Nobody ever said that (except you). YOU have repeatedly misread.
of gravity is proportional to r to predict tick rates.
In the weak field near earth, measurements of clock tick rates depend on
the difference in GRAVITATIONAL POTENTIAL between clock and measuring
instrument; it is proportional to 1/r. Gravitational FORCE is
proportional to the gradient of the gravitational potential, i.e.
proportional to 1/r^2.
You REALLY need to learn basic physics before attempting to write about it. >> Tom Roberts
You and the others need to understand that your pretence that “gravitational potential” has nothing to do with gravity is self
delusion on a grand scale.
If Gravitational potential doesn’t have anything to do with gravity,
then why does ‘gravity’ appear in the term?
And if gravity or the force of gravity have nothing to do with
Gravitational potential and the observed tick rates of atoms at
different altitudes, then what fundamental force is responsible
for these observations?
Wake up Tom. GP is a fluid term. Used by various theorists for
different reasons.
Laplace himself considered GP could be used to
calculate *the force of gravity* And Newton’s take of
‘mass generating a scalar field GM/r’ sounds like his “scalar field” is just another term for: the force of gravity changing relative to r.
Anyways regardless of your fantasy that GP has nothing to do with gravity in GR
the fact remains that a classical model of gravity which has gravitational strength at different altitudes dictated by the area of the earth’s shadow as
seen from the observer...is proportional to r. Not r^2. Which means
that the different real tick rates of clocks at different radius can be accurately
modelled classically using r.
You and the others need to understand that your pretence that “gravitational potential” has nothing to do with gravity is self
delusion on a grand scale.
On 11/9/23 4:12 AM, Lou wrote:
You and the others need to understand that your pretence that “gravitational potential” has nothing to do with gravity is self delusion on a grand scale.Why do you make stuff up like this?? -- there is no such "pretence" -- gravitational potential is clearly and obviously an aspect of gravity.
But gravitational potential is not gravitational force [#]; as I said before, gravitational force is minus the gradient of the gravitational potential. Under suitable conditions, the gravitational potential is proportional to 1/r, and the gravitational force is proportional to
1/r^2. THESE ARE DIFFERENT.
[#[ The context here is Newtonian mechanics, not GR.
to model the force of gravity
at any point in its field for a classical model one uses r. Not r^2.
[... more nonsense]
On 11/5/2023 5:16 AM, Lou wrote:
On Friday, 3 November 2023 at 17:33:37 UTC, Volney wrote:
On 11/3/2023 12:24 PM, Lou wrote:
On Friday, 3 November 2023 at 04:52:55 UTC, Volney wrote:
On 11/2/2023 7:37 AM, Lou wrote:
On Thursday, 2 November 2023 at 04:15:44 UTC, Volney wrote:
On 11/1/2023 6:42 PM, Lou wrote:
A classic relativist tactic. When a relativist gets their claims proven
wrong, they change the subject and come back a few weeks later with the
same false fact free claim again.
The size of a shadow isn't even relevant here! We're discussing the >>>> effects of gravitational force and potential, not shadows.
And notice for a classical model area of gravity shadow/altitude is how the force of gravityNewton was a "relativist"? Remember, F=GMm/r^2 is from Newton.
is calculated. And it is inversely proportional.
Not some BS relativists fantasy of r^2 .
Notice GR itself calculates the effect of gravity on atomic clock tick rates using inverse proportional.That is the effect of the POTENTIAL, not a force!
Relativists try to con everyone by pretending gravitational potentialNo, it is not a force. GMm/r^2 is a force (Newton model).
has nothing to do with gravity!!
Ignoring the fact that to calculate “time dilation” all GR reference saysPhotons always travel at c.
it is *gravity* which slows down or speeds up the imaginary photons which leads to the imaginary “time dilation”
Hypocrisy supreme from relativists. Saying gravity causes time dilation. And then in the same breath saying it doesn’t💩.Gravitational FORCE does not cause time dilation. Time dilation is not proportional to the FORCE (1/r^2).
There you go. Cant refute the fact that the area of the gravity shadow in a classicalWhy would I answer since it up to YOU to prove your claim? In addition, >> once again, it's irrelevant.
model falls off at r
Thought you wouldn’t be able to answer.
I already have proved my claim.Assertions are not proofs.
If one measures the angle subtended by the earthAssertion. Where is the formula which shows how the shadow falls off
to an observer at R, 2R , 3R etc....The area of the shadow falls off with r
with distance? You have not provided it, so you haven't proven your claim.
Not r ^2 as you falsely claimR^2 is NEWTON's claim!
You are so dishonest that as usual for a relativist you make claims about aThose numbers are not inversely proportional.
classical model that you know are patently false. Here below are percentages
based on angle subtended by earth to observer at various radius distances for an observer. Notice the relationship is...Inverse proportional.
r=100%
2r=33% 1/2
3r =21% 1/3
4r=16% 1/4
5r=13% 1/5
WHERE IS YOUR FORMULA?
On Friday, 3 November 2023 at 04:52:55 UTC, Volney wrote:
On 11/2/2023 7:37 AM, Lou wrote:
On Thursday, 2 November 2023 at 04:15:44 UTC, Volney wrote:
On 11/1/2023 6:42 PM, Lou wrote:
But the earth also spins. And that spin is not produced by earths gravity.Cesium atoms in an atomic clock are in freefall. That means there is no >> force on them whatsoever, from any source.
And that spin exerts an additional force on the caesium atoms that is from a tangental
force caused not by gravity but by the tangental acceleration of the clock in
the ECI frame.
Exactly. So too are the effects of gravity in a classical model inversely proportional to r. Not r^2This is an answer that is a ‘have it two ways’ answer only a relativistNo, there is no force! The GR effects are inversely proportional to r,
would have the nerve to make: “There’s a force of gravity in free fall
when calculating for the (imaginary) “time dilation” effects using GR .
not r^2,
That’s how a classical model also successfully predicts tick rates of caesium atoms at different altitudes.
so GMm/r doesn't even have dimensionality of force, whilePure nonsense. Those are arbritrary formulas. Not forces. Made up
GMm/r^2 does. You seem to think everything is a force, including the tangential velocity of a launched rocket. Which is, obviously, a
velocity, not a force.
But no force of gravity in free fall for a classical model.”A free falling object doesn't experience any force.
And this is no double standard, since GMm/r is not even a force, while GMm/r^2 is.
to try to model the force of gravity. And r....models the force of gravity on tick rates of atoms at different altitudes for both SR and classical.
Not r^2.
Completely ignoring the fact Einstein used r to succesfully model the force of gravity on various things in GR.You probably don’t even realise that your claims are contradictory.Since GMm/r isn't even a force, no contradiction.
Anyways If there was no force of gravity acting on them, there would be noIn free fall, there is no effect of gravity, pretty much by definition. That's why astronauts float around the space station (orbits are freefall).
free fall. That’s why they are falling. The force of gravity is pushing them.
Besides, if there were a force, it would go as GM/r^2, which doesn't
match GR's inverse r relationship.
In other words...you know I’m right. So you don’t try to prove I’m wrong.Basically what you just said is that the effects of the force of gravity at different altitudes for a classical model have to be calculated by incorrectly using r^2That is not incorrect, that's just plain Newtonian physics!
to make sure the prediction is false.Newton wanted to help prove relativity?
But for GR it’s OK to correctly model the force of gravity at differentOnce again, it is not even a force!
alttudes using just r.
You also ignores the fact that you were previously unable to supplyFirst, the shadow is irrelevant for this discussion. Second, the shadow area is not proportional to either 1/r nor 1/r^2. At large distances, where the sine of the angle of the edge of the shadowing object can be approximated as 1 (meaning a very small angle) it is approximately proportional to 1/r^2. For close distances once cannot make that assumption, the function is more complex, involving the angle of the center of the object-observer-edge of the object. This is not
any calculation to refute the fact that the area of the gravity shadow of earth at different altitudes falls off with r. Not r^2.
proportional to 1/r (or 1/r^2).
I provided you with all the data and calculations. You were unable to refute it.I'm not about to look up the near distance function for you. For large distances it is close to 1/r^2 as the source is pointlike (subtended
So you didn’t answer and pretended that by not supplying any calculations
to back up your false claims about r^2 you could avoid having to admit you
failed.
angle almost 0).
There you go. Cant refute the fact that the area of the gravity shadow in a classicalA classic relativist tactic. When a relativist gets their claims proven wrong, they change the subject and come back a few weeks later with the same false fact free claim again.The size of a shadow isn't even relevant here! We're discussing the effects of gravitational force and potential, not shadows.
model falls off at r
Thought you wouldn’t be able to answer.
On Thursday, 9 November 2023 at 21:05:12 UTC, Tom Roberts wrote:
On 11/9/23 4:12 AM, Lou wrote:
You and the others need to understand that your pretence thatWhy do you make stuff up like this?? -- there is no such "pretence" --
“gravitational potential” has nothing to do with gravity is self
delusion on a grand scale.
gravitational potential is clearly and obviously an aspect of gravity.
But gravitational potential is not gravitational force [#]; as I said
before, gravitational force is minus the gradient of the gravitational
potential. Under suitable conditions, the gravitational potential is
proportional to 1/r, and the gravitational force is proportional to
1/r^2. THESE ARE DIFFERENT.
[#[ The context here is Newtonian mechanics, not GR.
Play word games. But fact is that to model the force of gravity
at any point in its field for a classical model one uses r. Not r^2.
(R^2 is acceleration. Obviously you don’t understand what
‘meters per second ^2’ means.)
And finally...if you weren’t so desperate to ignore the facts...
You would realise that a classical model of gravity relies
on the gravity shadow of earth on any observer at r,2r etc.
And this falls of with r.
A fact you cannot disprove.
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