_________________________________________________________________

Abstract:

I have previously shown that the exponential version of the
gravitational time dilation (GTD) equation (first given by Einstein in
1907) is incorrect, because it is inconsistent with the outcome of the
twin paradox. I then gave a corrected version of the GDT equation which
IS consistent with the outcome of the twin paradox, and which is also
consistent with the co-moving-inertial-frames (CMIF) simultaneity
method. In this brief paper, I describe an experimental test of my GTD
equation that might be feasible to conduct.

___________________________________________________________________________

Section 1. Einstein’s Exponential GTD Equation

In Einstein’s 1907 paper [
https://einsteinpapers.press.princeton.edu/vol2-trans/319 ], Einstein
stated that the GTD equation is

R(g) = exp(g L),

where “g” is the force per unit mass exerted by the uniform
gravitational field, and “L” is the constant distance (in the direction
of the field) between two stationary clocks. The quantity “R” is the
ratio of the tic rates of the two clocks: the clock which is higher in
the field (farther from the source of the field) will run “R” times
faster than the other clock.

According to the Equivalence Principle, we can then also say that when
there are no gravitational fields (i.e., in a Special Relativity
scenario), two clocks which are initially unaccelerated, and which are
separated by a constant distance “L”, and which are then simultaneously
accelerated with an acceleration “A” (in the direction of their
separation), then the rate ratio “R” is

R(A) = exp(A L).

The leading clock runs “R” times faster than the trailing clock. Note
that for constant “A” and “L”, the rate ratio “R” DOES NOT VARY WITH TIME.

I showed in https://vixra.org/abs/2109.0076 that the above exponential
equation is inconsistent with the outcome of the twin paradox.
Specifically, if the traveling twin (he) changes his velocity
instantaneously at his turnaround, the exponential equation says that
the home twin (she) will be INFINITELY old after his turnaround, and
when the twins are reunited. That isn’t true: both of their ages are
finite at the reunion. Thus the exponential equation for “R” is
incorrect. (And the exponential equation is also incorrect for the case
where the turnaround is not instantaneous, but is just very quick.)

Section 2. My GTD Equation

The corrected rate ratio equation is R(A) = [ 1 + L A sech_sqrd(theta) ],

where sech_sqrd( ) is the square of sech( ), the hyperbolic secant. (The
hyperbolic secant is the reciprocal of the hyperbolic cosine, cosh,
which is more likely to be available in tables). Cosh(theta) can also be
calculated using the equation

cosh(theta) = { exp^theta + exp^(-theta) } / 2.

The CRC Standard Mathematical Tables book (14th Edition) show plots of
the hyperbolic functions on page 527.

The quantity “theta” in the above equation for R(A) is theta(t) = A t,

where “t” specifies how long the constant acceleration “A” has been
going on, since it abruptly started from zero acceleration at time zero.
So we really should write the “R” equation as

R(A,t) = [ 1 + L A sech_sqrd{ theta(t) } ].

A plot of R(A,t) versus “t” is given in section 4 of
https://vixra.org/abs/2201.0015 for the case A = 1 ls/s/s (about 40 g’s)
and L = 7.52 ls.

The R(A,t) equation produces quite different qualitative results than
those produced by the exponential R(A) equation. The exponential R(A)
equation says that if the constant acceleration “A” goes on for an
essentially infinite time, the rate ratio NEVER CHANGES ... i.e., the
leading clock keeps ticking faster than the trailing clock by the same
ratio, forever. And for large “A”, that constant ratio is HUGE! In
contrast, the new R(A,t) equation says that, as “t” goes to infinity,
R(A,t) approaches 1.0. I.e., the two clocks eventually tic at
essentially the same rate. That is quite a qualitative difference, which
might be observable experimentally.

The above results are derived and explained more thoroughly in
https://vixra.org/abs/2201.0015 and in https://vixra.org/abs/2206.0133 .

Section 3. A Proposed Experimental Test of My GTD Equation

Charged particles can be accelerated to speeds that are a large fraction
of the speed of light, by exposing them to very large electric fields.
So we can start with a stationary pair of them, separated by the
distance “L”, and then switch on a very strong uniform electric field
(in the direction of their separation) that will accelerate both of them
at the same rate, and maintain their separation at “L”.

But how can we use each particle as a “clock”? We might be able to
accomplish that by using UNSTABLE charged particles ... particles that
have a known average lifetime (before they decay into uncharged
particles that won’t accelerate in the electric field). That way, if the
leading particle is ageing faster than the trailing particle, the
leading particle will (on average) decay quicker than the trailing
particle, which might be observable. That might allow an experimental
way to verify or falsify my GDT equation.

(The above content is also available at https://vixra.org/abs/2210.0072 )

Mike Fontenot <mlfasf@comcast.net> wrote:

> _________________________________________________________________
>
> Abstract:
>
> I have previously shown that the exponential version of the
> gravitational time dilation (GTD) equation (first given by Einstein in
> 1907) is incorrect, because it is inconsistent with the outcome of the
> twin paradox. I then gave a corrected version of the GDT equation which
> IS consistent with the outcome of the twin paradox, and which is also
> consistent with the co-moving-inertial-frames (CMIF) simultaneity
> method. In this brief paper, I describe an experimental test of my GTD
> equation that might be feasible to conduct.
>
> ___________________________________________________________________________
>
> Section 1. Einstein's Exponential GTD Equation
>
> In Einstein's 1907 paper [
> https://einsteinpapers.press.princeton.edu/vol2-trans/319 ], Einstein
> stated that the GTD equation is
>
> R(g) = exp(g L),
>
> where "g" is the force per unit mass exerted by the uniform
> gravitational field, and "L" is the constant distance (in the direction
> of the field) between two stationary clocks. The quantity "R" is the
> ratio of the tic rates of the two clocks: the clock which is higher in
> the field (farther from the source of the field) will run "R" times
> faster than the other clock.

If you go on to the next page you see that Einstein gives the correct
general result in terms of the Newtonian potential \Phi.
(in the Newtonian approximation)
For the special case of a homogeneous field
you get \Delta(\Phi) = g\Delta(h),
with g the local acceleration of gravity.

There is nothing to correct.
In general the total observed value for the twin paradox
is a combination of the special and the general relativistic effects.
This has been thoroughly confirmed both with clocks on Earth
and with clocks in space, on sats and on interplanetary probes.
If you predict anything else it stands falsified,

Jan

On 10/22/22 2:46 PM, J. J. Lodder wrote:
>
> There is nothing to correct.
> In general the total observed value for the twin paradox
> is a combination of the special and the general relativistic effects.

No, the outcome of the twin paradox is determined purely by the time
dilation equation for an inertial observer ... it is strictly a special
relativity result ... no general relativity is involved. An inertial
observer (she) will conclude that anyone who is moving at a speed "v"
relative to her will be ageing "gamma" times slower than she is. So
that tells her how old the traveling twin (he) will be when they are
reunited. For example, for "v" = 0.866 ly/y, "gamma" is equal to 2.0.
So if the traveler is always traveling at a speed of 0.866 ly/y, she
will be twice as old as he is at their reunion. In contrast, the
exponential time dilation equation for an accelerating observer (making
an essentially instantaneous velocity change), says that her age is
INFINITE when they are reunited (because she ages by an infinite amount
during his instantaneous velocity change, according to him). That
contradicts the time dilation equation for an inertial observer, which
is sacrosanct, so the time dilation equation for an accelerating
observer must be incorrect.

> This has been thoroughly confirmed both with clocks on Earth
> and with clocks in space, on sats and on interplanetary probes.
> If you predict anything else it stands falsified,
>

All of those experiments are for very weak gravitational fields and/or
small accelerations ... they don't test the exponential equation in it's
highly nonlinear regions, only in its linear region near the origin.
The problem with the exponential equation is in its highly nonlinear region.

On Saturday, 22 October 2022 at 22:46:57 UTC+2, J. J. Lodder wrote:

> There is nothing to correct.
> In general the total observed value for the twin paradox
> is a combination of the special and the general relativistic effects.
> This has been thoroughly confirmed both with clocks on Earth
> and with clocks in space, on sats and on interplanetary probes.
> If you predict anything else it stands falsified,

And in the meantime in the real world, forbidden by your
bunch of idiots GPS and TAI keep measuring t'=t in
forbidden by your bunch of idiots old seconds.

Mike Fontenot <mlfasf@comcast.net> wrote:

> On 10/22/22 2:46 PM, J. J. Lodder wrote:
> >
> > There is nothing to correct.
> > In general the total observed value for the twin paradox
> > is a combination of the special and the general relativistic effects.
>
> No, the outcome of the twin paradox is determined purely by the time
> dilation equation for an inertial observer ... it is strictly a special
> relativity result ... no general relativity is involved.

That's your definition, nobody else agrees with it.
In Hafele and Keating for example the clock differences upon return
depend on both special and general relativity.
[-]

> > This has been thoroughly confirmed both with clocks on Earth
> > and with clocks in space, on sats and on interplanetary probes.
> > If you predict anything else it stands falsified,
> >
>
> All of those experiments are for very weak gravitational fields and/or
> small accelerations ... they don't test the exponential equation in it's
> highly nonlinear regions, only in its linear region near the origin.
> The problem with the exponential equation is in its highly nonlinear region.

Your point is pointless.
The exponential equation is derived in the Newtonian approximation,
so for \Delta(\Phi/c^2) << 1

If the exponential differs significantly from its linearisation
the derivation that led to it has become invalid,

Jan

On Sunday, 23 October 2022 at 15:37:46 UTC+2, J. J. Lodder wrote:
> Mike Fontenot <mlf...@comcast.net> wrote:
>
> > On 10/22/22 2:46 PM, J. J. Lodder wrote:
> > >
> > > There is nothing to correct.
> > > In general the total observed value for the twin paradox
> > > is a combination of the special and the general relativistic effects.
> >
> > No, the outcome of the twin paradox is determined purely by the time
> > dilation equation for an inertial observer ... it is strictly a special
> > relativity result ... no general relativity is involved.
> That's your definition, nobody else agrees with it.
> In Hafele and Keating for example the clock differences upon return
> depend on both special and general relativity.

And in the meantime in the real world, however, forbidden
And iby your bunch of idiots GPS and TAI keep measuring
t'=t in forbidden by your bunch of idiots old seconds.

On 10/22/22 4:30 PM, Mike Fontenot wrote:
> the outcome of the twin paradox is determined purely by the time
> dilation equation for an inertial observer ... it is strictly a
> special relativity result ... no general relativity is involved.

So you are considering a twin scenario in flat spacetime. Presuming that
the stay-at-home twin is moving inertially, we can use the coordinates
of their inertial frame to compute the elapsed proper time of each twin:
T_traveler = \Integral dt sqrt(1-v^2/c^2)
T_stayathome = \Integral dt
where t is the time coordinate of that frame, v is the speed of the
traveling twin relative to that frame (a function of t), and the
integrals are taken over each twin's path expressed in that inertial
frame (between their parting and their reunion).

> An inertial observer (she) will conclude that anyone who is moving at
> a speed "v" relative to her will be ageing "gamma" times slower than
> she is. So that tells her how old the traveling twin (he) will be
> when they are reunited. For example, for "v" = 0.866 ly/y, "gamma"
> is equal to 2.0. So if the traveler is always traveling at a speed of
> 0.866 ly/y, she will be twice as old as he is at their reunion.

Yes. That is just putting specific numbers into the equations above.

> In contrast, the exponential time dilation equation for an
> accelerating observer (making an essentially instantaneous velocity
> change), says that her age is INFINITE when they are reunited
> (because she ages by an infinite amount during his instantaneous
> velocity change, according to him).

NONSENSE! That "infinite" change occurs over zero time, which makes the
result be indeterminate.

In general, attempting to analyze instantaneous changes is fraught with
mathematical difficulties, and you have just shown that you don't know
how to deal with them, and don't even know there is a problem in
attempting to do that. See the exercise below for one approach.

> That contradicts the time dilation equation for an inertial observer,
> which is sacrosanct, so the time dilation equation for an
> accelerating observer must be incorrect.

No. Your "analysis" is incorrect.

> The problem with the exponential equation is in its highly nonlinear
> region.

The problem YOU have is attempting to deal with infinity multiplied by
zero, yielding nonsense.

Instead of instantaneous accelerations, set up the problem in a
physically realizable manner: let the traveler
1. start at rest in the above inertial frame adjacent to
the stay-at-home twin
2. accelerate in the +x direction with constant proper
acceleration A for proper time T1
3. coast inertially for proper time T2
4. accelerate in the -x direction with constant proper
acceleration -A for proper time 2*T1
5. coast inertially for proper time T2
6. accelerate in the +x direction with constant proper
acceleration A for proper time T1
7 end up at rest in the inertial frame adjacent to the
stay-at-home twin

It is straightforward but tedious to sum up the elapsed proper times of
both twins. You are free to consider those periods of acceleration as
being in a (uniform) gravitational field -- of course you will obtain
EXACTLY the same values as the integrals above.

[Simplification: let T1 be short, A be large, and T2
be long. Then during steps 2 and 6 the twins are
approximately co-located so their elapsed proper
times during those steps are approximately equal.
The entire difference is due to the inertial steps
3 and 5, and the turn-around step 4 in which the
stay-at-home twin is located very far above the
traveling twin in the "gravitational field".]

Exercise for the reader: calculate that simplification,
and then take the limit (T1->0, A->infinity) while
holding A*T1 constant. Hint: this is one simple way
to deal with the instantaneous accelerations that
Mike wants to consider.

It is easy to prove the results are the same, without the tedious
calculations, because one is actually integrating the metric tensor over
each path -- tensors and paths are invariant.

Earlier in this thread you claimed:
> I have previously shown that the exponential version of the
> gravitational time dilation (GTD) equation (first given by Einstein
> in 1907) is incorrect, because it is inconsistent with the outcome of
> the twin paradox.

If the above is a description of that, your claim is manifestly false.

Tom Roberts

On Wednesday, October 26, 2022 at 8:50:30 PM UTC-7, Tom Roberts wrote:
> On 10/22/22 4:30 PM, Mike Fontenot wrote:
> > the outcome of the twin paradox is determined purely by the time
> > dilation equation for an inertial observer ... it is strictly a
> > special relativity result ... no general relativity is involved.
> So you are considering a twin scenario in flat spacetime. Presuming that
> the stay-at-home twin is moving inertially, we can use the coordinates
> of their inertial frame to compute the elapsed proper time of each twin:
> T_traveler = \Integral dt sqrt(1-v^2/c^2)
> T_stayathome = \Integral dt
> where t is the time coordinate of that frame, v is the speed of the
> traveling twin relative to that frame (a function of t), and the
> integrals are taken over each twin's path expressed in that inertial
> frame (between their parting and their reunion).
> > An inertial observer (she) will conclude that anyone who is moving at
> > a speed "v" relative to her will be ageing "gamma" times slower than
> > she is. So that tells her how old the traveling twin (he) will be
> > when they are reunited. For example, for "v" = 0.866 ly/y, "gamma"
> > is equal to 2.0. So if the traveler is always traveling at a speed of
> > 0.866 ly/y, she will be twice as old as he is at their reunion.
> Yes. That is just putting specific numbers into the equations above.
> > In contrast, the exponential time dilation equation for an
> > accelerating observer (making an essentially instantaneous velocity
> > change), says that her age is INFINITE when they are reunited
> > (because she ages by an infinite amount during his instantaneous
> > velocity change, according to him).
> NONSENSE! That "infinite" change occurs over zero time, which makes the
> result be indeterminate.
>
> In general, attempting to analyze instantaneous changes is fraught with
> mathematical difficulties, and you have just shown that you don't know
> how to deal with them, and don't even know there is a problem in
> attempting to do that. See the exercise below for one approach.
> > That contradicts the time dilation equation for an inertial observer,
> > which is sacrosanct, so the time dilation equation for an
> > accelerating observer must be incorrect.
> No. Your "analysis" is incorrect.
> > The problem with the exponential equation is in its highly nonlinear
> > region.
> The problem YOU have is attempting to deal with infinity multiplied by
> zero, yielding nonsense.
>
> Instead of instantaneous accelerations, set up the problem in a
> physically realizable manner: let the traveler
> 1. start at rest in the above inertial frame adjacent to
> the stay-at-home twin
> 2. accelerate in the +x direction with constant proper
> acceleration A for proper time T1
> 3. coast inertially for proper time T2
> 4. accelerate in the -x direction with constant proper
> acceleration -A for proper time 2*T1
> 5. coast inertially for proper time T2
> 6. accelerate in the +x direction with constant proper
> acceleration A for proper time T1
> 7 end up at rest in the inertial frame adjacent to the
> stay-at-home twin
>
> It is straightforward but tedious to sum up the elapsed proper times of
> both twins. You are free to consider those periods of acceleration as
> being in a (uniform) gravitational field -- of course you will obtain
> EXACTLY the same values as the integrals above.
>
> [Simplification: let T1 be short, A be large, and T2
> be long. Then during steps 2 and 6 the twins are
> approximately co-located so their elapsed proper
> times during those steps are approximately equal.
> The entire difference is due to the inertial steps
> 3 and 5, and the turn-around step 4 in which the
> stay-at-home twin is located very far above the
> traveling twin in the "gravitational field".]
>
> Exercise for the reader: calculate that simplification,
> and then take the limit (T1->0, A->infinity) while
> holding A*T1 constant. Hint: this is one simple way
> to deal with the instantaneous accelerations that
> Mike wants to consider.
>
> It is easy to prove the results are the same, without the tedious
> calculations, because one is actually integrating the metric tensor over
> each path -- tensors and paths are invariant.
>
> Earlier in this thread you claimed:
> > I have previously shown that the exponential version of the
> > gravitational time dilation (GTD) equation (first given by Einstein
> > in 1907) is incorrect, because it is inconsistent with the outcome of
> > the twin paradox.
>
> If the above is a description of that, your claim is manifestly false.
>
> Tom Roberts

( https://groups.google.com/g/sci.physics.relativity/c/Vzr4XFUJIO4/m/Lqnzi_QLCAAJ )

"You can easily convince yourself that the product of these
transformations is not a pure rotation, there is in general
a net boost left over."

But, in 1.2 there isn't "a pure rotation
lines up the coordinate frame again".

On Wednesday, October 26, 2022 at 8:50:30 PM UTC-7, Tom Roberts wrote:
> On 10/22/22 4:30 PM, Mike Fontenot wrote:
> > the outcome of the twin paradox is determined purely by the time
> > dilation equation for an inertial observer ... it is strictly a
> > special relativity result ... no general relativity is involved.
> So you are considering a twin scenario in flat spacetime. Presuming that
> the stay-at-home twin is moving inertially, we can use the coordinates
> of their inertial frame to compute the elapsed proper time of each twin:
> T_traveler = \Integral dt sqrt(1-v^2/c^2)
> T_stayathome = \Integral dt
> where t is the time coordinate of that frame, v is the speed of the
> traveling twin relative to that frame (a function of t), and the
> integrals are taken over each twin's path expressed in that inertial
> frame (between their parting and their reunion).
> > An inertial observer (she) will conclude that anyone who is moving at
> > a speed "v" relative to her will be ageing "gamma" times slower than
> > she is. So that tells her how old the traveling twin (he) will be
> > when they are reunited. For example, for "v" = 0.866 ly/y, "gamma"
> > is equal to 2.0. So if the traveler is always traveling at a speed of
> > 0.866 ly/y, she will be twice as old as he is at their reunion.
> Yes. That is just putting specific numbers into the equations above.
> > In contrast, the exponential time dilation equation for an
> > accelerating observer (making an essentially instantaneous velocity
> > change), says that her age is INFINITE when they are reunited
> > (because she ages by an infinite amount during his instantaneous
> > velocity change, according to him).
> NONSENSE! That "infinite" change occurs over zero time, which makes the
> result be indeterminate.
>
> In general, attempting to analyze instantaneous changes is fraught with
> mathematical difficulties, and you have just shown that you don't know
> how to deal with them, and don't even know there is a problem in
> attempting to do that. See the exercise below for one approach.
> > That contradicts the time dilation equation for an inertial observer,
> > which is sacrosanct, so the time dilation equation for an
> > accelerating observer must be incorrect.
> No. Your "analysis" is incorrect.
> > The problem with the exponential equation is in its highly nonlinear
> > region.
> The problem YOU have is attempting to deal with infinity multiplied by
> zero, yielding nonsense.
>
> Instead of instantaneous accelerations, set up the problem in a
> physically realizable manner: let the traveler
> 1. start at rest in the above inertial frame adjacent to
> the stay-at-home twin
> 2. accelerate in the +x direction with constant proper
> acceleration A for proper time T1
> 3. coast inertially for proper time T2
> 4. accelerate in the -x direction with constant proper
> acceleration -A for proper time 2*T1
> 5. coast inertially for proper time T2
> 6. accelerate in the +x direction with constant proper
> acceleration A for proper time T1
> 7 end up at rest in the inertial frame adjacent to the
> stay-at-home twin
>
> It is straightforward but tedious to sum up the elapsed proper times of
> both twins. You are free to consider those periods of acceleration as
> being in a (uniform) gravitational field -- of course you will obtain
> EXACTLY the same values as the integrals above.
>
> [Simplification: let T1 be short, A be large, and T2
> be long. Then during steps 2 and 6 the twins are
> approximately co-located so their elapsed proper
> times during those steps are approximately equal.
> The entire difference is due to the inertial steps
> 3 and 5, and the turn-around step 4 in which the
> stay-at-home twin is located very far above the
> traveling twin in the "gravitational field".]
>
> Exercise for the reader: calculate that simplification,
> and then take the limit (T1->0, A->infinity) while
> holding A*T1 constant. Hint: this is one simple way
> to deal with the instantaneous accelerations that
> Mike wants to consider.
>
> It is easy to prove the results are the same, without the tedious
> calculations, because one is actually integrating the metric tensor over
> each path -- tensors and paths are invariant.
>
> Earlier in this thread you claimed:
> > I have previously shown that the exponential version of the
> > gravitational time dilation (GTD) equation (first given by Einstein
> > in 1907) is incorrect, because it is inconsistent with the outcome of
> > the twin paradox.
>
> If the above is a description of that, your claim is manifestly false.
>
> Tom Roberts

Oh, and thanks, Dr. Roberts, you're about the most respected opinion here.

On Thursday, 27 October 2022 at 05:50:30 UTC+2, Tom Roberts wrote:
> On 10/22/22 4:30 PM, Mike Fontenot wrote:
> > the outcome of the twin paradox is determined purely by the time
> > dilation equation for an inertial observer ... it is strictly a
> > special relativity result ... no general relativity is involved.
> So you are considering a twin scenario in flat spacetime. Presuming that
> the stay-at-home twin is moving inertially, we can use the coordinates
> of their inertial frame to compute the elapsed proper time of each twin:
> T_traveler = \Integral dt sqrt(1-v^2/c^2)
> T_stayathome = \Integral dt
> where t is the time coordinate of that frame, v is the speed of the
> traveling twin relative to that frame (a function of t), and the
> integrals are taken over each twin's path expressed in that inertial
> frame (between their parting and their reunion).

In the meantime in the real world, of course, forbidden
by your bunch of idiots GPS and TAI keep measuring t'=t
in forbidden by your bunch of idiots old seconds.

On 2022-10-27 05:46:45 +0000, Maciej Wozniak said:

> On Thursday, 27 October 2022 at 05:50:30 UTC+2, Tom Roberts wrote:
>> On 10/22/22 4:30 PM, Mike Fontenot wrote:
>>> the outcome of the twin paradox is determined purely by the time
>>> dilation equation for an inertial observer ... it is strictly a
>>> special relativity result ... no general relativity is involved.
>> So you are considering a twin scenario in flat spacetime. Presuming that
>> the stay-at-home twin is moving inertially, we can use the coordinates
>> of their inertial frame to compute the elapsed proper time of each twin:
>> T_traveler = \Integral dt sqrt(1-v^2/c^2)
>> T_stayathome = \Integral dt
>> where t is the time coordinate of that frame, v is the speed of the
>> traveling twin relative to that frame (a function of t), and the
>> integrals are taken over each twin's path expressed in that inertial
>> frame (between their parting and their reunion).
>
> In the meantime in the real world, of course, forbidden
> by your bunch of idiots GPS and TAI keep measuring t'=t
> in forbidden by your bunch of idiots old seconds.

530

--
Athel -- French and British, living mainly in England until 1987.

On Saturday, October 22, 2022 at 4:30:08 PM UTC-5, Mike_Fontenot wrote:

> All of those experiments are for very weak gravitational fields and/or
> small accelerations ... they don't test the exponential equation in it's
> highly nonlinear regions, only in its linear region near the origin.
> The problem with the exponential equation is in its highly nonlinear region.

Does your equation even predict correctly for weak fields?
The elongated mark for ISS is hypothetical, but the data point for
DSAC is firmly established, as are the geosynchronous, GPS
and GP-A data points.
https://en.wikipedia.org/wiki/Time_dilation#/media/File:Time_Dilation_vs_Orbital_Height.png
Any disagreement with any of these, and your theory is dead.

Show your work.

Depicts the time dilation as a function of orbital height relative to a stationary observer on Earth. It is common to consider the total dilation as being due to two distinct effects: Kinematic time dilation (customarily referred to as the Special Relativity effect) accounts for slowed time in orbit (relative to the observer on Earth) depending on the orbital velocity associated to a specific orbital height. Gravitational time dilation (customarily referred to as the General Relativity effect) accounts for accelerated time (relative to the observer on Earth) due to the distance to the Earth's gravitational center. Although these sources of time dilation are regularly referred to as "SR time dilation" and "GR time dilation", this usage is incorrect, because general relativity accounts for both effects. The point for "GPS" has been firmly confirmed by decades of measurement, both by satellites of the U.S. Global Positioning System as well as by the satellites of the Russian GLONASS and European Galileo systems in similar orbits. Geosynchronous time dilation has been firmly confirmed by the Chinese Beidou and the Indian Regional Satellite Systems, which use geostationary satellites to improve accuracy above their respective countries. Also included are the measurement at peak altitude of the 1976 Gravity Probe A experiment, which measured time dilation effects throughout most of its nearly vertical trajectory, and 2021 results from the Deep Space Atomic Clock mission. Time dilation for the ISS (which has orbited at various altitudes, hence the elongated mark) has not yet been confirmed by actual measurement, but this should change with the launching of the Atomic Clock Ensemble in Space (ACES) mission.

There is a thread with lots of back and forth between me and Halc (the
moderator) on this issue here:

https://www.thenakedscientists.com/forum/index.php?topic=85706.0

On Thursday, November 3, 2022 at 11:15:23 AM UTC-7, Mike_Fontenot wrote:
> There is a thread with lots of back and forth between me and Halc (the
> moderator) on this issue here:
>
> https://www.thenakedscientists.com/forum/index.php?topic=85706.0
Effectively, they tell you are full of shit.