In Einstein's 1907 paper,
https://einsteinpapers.press.princeton.edu/vol2-trans/319 , he gives his
exponential gravitational time dilation equation. Actually, he was
working a special relativity problem (with accelerations and no
gravitation) because he knew how to do that, and he was hoping the
result would give him (via the equivalence principle) some help in his
search for a gravitational theory. So Einstein's equation was actually
a time dilation equation for accelerating clocks that are separated by a
fixed distance. According to Einstein, for a pair of accelerating
clocks separated by the distance "L" in the direction of the
acceleration, the leading clock tics faster than the trailing clock by
the factor

R = exp(A*L),

where "A" is the acceleration. In the iterations described below, I
will limit myself to the case where "A" is constant during the
acceleration. I will show that the exponential time dilation equation
is incorrect. (I suspect that Einstein, and also physicists who came
along later, didn't ever notice that the exponential equation is
incorrect, because they never used it in the nonlinear range where its
argument is large ... they only used it for very small arguments, where
it is very nearly linear.)

Suppose that the two clocks are initially inertial (unaccelerated) at
time "t" = 0, and both read zero at that instant. Then, for t > 0, both
clocks undergo a constant acceleration "A" (as determined by
accelerometers attached to each of them, which control a rocket attached
to each of them, so as to achieve the specified acceleration). Their
separation remains constant at "L" during the acceleration. In all of
my calculations below, I chose L = 7.520. Let "tau" be the duration of
the acceleration. Therefore the reading on the leading clock, when the
trailing clock reads "tau", is given by

AC = tau * R = tau * exp(A*L) .

I first take the case where the duration "tau" of the acceleration is
equal to 1.0. I choose the magnitude of the acceleration "A" to be such
that the velocity of the two clocks, after accelerating for a duration
"tau" = 1.0, is 0.8660. The product of the constant "A" and the
duration "tau" gives the rapidity "theta", which is related to the
velocity by the equation

v = tanh(theta),

where tanh() is the hyperbolic tangent function, and is equal to

tanh(theta) = { [exp(theta) - exp(-theta)] / [exp(theta) +
exp(-theta)] } .

So, for v = 0.8660, the rapidity theta = 1.3170. Since

theta = A * tau,

A = theta / tau = 1.3170 / 1.0 = 1.3170.

So for the first case with tau = 1.0, we get

R = exp(A*L) = exp( 1.3170 * 7.520 ) = exp(9.90384) = 20007.

The reading on the leading clock, at the end of the acceleration at tau
= 1.0, is

AC = tau * R = tau * exp(A*L) = 1.0 * 20007 = 20007 = 2.0 *
10^4,

where 10^4 is just 10 raised to the 4th power.

So for the first calculation (with tau = 1.0), we have that at the end
of the acceleration, the leading clock reads

AC = 2.0 * 10^4.

For the second case, we increase the acceleration "A" by a factor of 10,
and decrease the duration "tau" by the factor 10 (which still, as
required, results in the same speed change as in the first case). So we
now have

A = 13.170

and

tau = 0.1.

So

R = exp(A*L) = exp( 13.170 * 7.520 ) = exp(99.0384) = 1.028
* 10^43 ,

and

AC = tau * R = 0.1 * R = 1.028 * 10^42.

Note that when we increased the acceleration by a factor of 10, the
reading of the leading clock didn't increase by a factor of 10, it
increased by ten raised to a power that increased by a factor of about ten.

For the third case, we increase the acceleration "A" again by a factor
of 10, and again decrease the duration "tau" by the factor 10 (which
still, as required, results in the same speed change as in the first and
second case). So we now have

A = 131.70

and

tau = 0.01.

So

R = exp(A*L) = exp( 131.70 * 7.520 ) = exp(990.38) = 1.31 *
10^430 ,

and

AC = tau * R = 0.01 * R = 1.31 * 10^428.

Note that, again, when we increased the acceleration by a factor of 10,
the reading of the leading clock didn't increase by a factor of 10, it
increased by ten raised to a power that increased by a factor of roughly
ten.

So we get the following table:

tau AC

1.0 2.0 * 10^4

0.1 1.0 * 10^42

0.01 1.0 * 10^428

Clearly, this iteration is NOT approaching a finite value for the
leading clock's reading, as tau goes to zero. The leading clock's
reading is clearly diverging as tau goes to zero. I.e., the leading
clock's reading goes to infinity as tau goes to zero.

So when we use this method to determine by how much the home twin's
(her) age increases when the traveling twin (he) instantaneously changes
his velocity by 0.866 when he reverses course at the turnaround, it
tells us that the home twin gets INFINITELY older, which is not true.
From the time dilation equation for an inertial observer (which the
home twin IS), we KNOW that both she and he have a finite age at their
reunion. Therefore the exponential gravitational time dilation equation
CAN'T be correct.

On Sunday, February 26, 2023 at 12:28:58 PM UTC-8, Mike Fontenot wrote:
> In Einstein's 1907 paper,

The calendar reads 2023. The correct answer (based on the Schwarzchild solution) has been known since 1916. Why do you keep beating up the poor strawman, Mike?

On Sunday, February 26, 2023 at 12:28:58 PM UTC-8, Mike Fontenot wrote:
> ...a time dilation equation for accelerating clocks that are separated by a
> fixed distance.

If the clocks are separated by a fixed proper distance (meaning the distance in terms of their co-moving system of inertial coordinates), then they must be undergoing Born rigid motion, which implies that the proper accelerations of the clocks are not equal. Conversely, if you stipulate that the proper accelerations are equal, then the distance between the clocks (in terms of their co-moving inertial coordinates) can't be constant. On the other hand, the coordinate distance in terms of the original coordinates would be constant. To derive the general result, applicable to all magnitudes of acceleration and distances, it is necessary to specify each of these things..

Note that, in Einstein's 1907 survey article, Section 18, he side-stepped the need to answer these questions by restricting to sufficiently small accelerations such that terms of second or higher power in the acceleration can be neglected. What you seem to be trying tyo do is consider arbitrarily large accelerations, so you can't neglect the higher order terms (and note that this is distinct from the higher order terms due to increasing the distance).

Once you specify precisely which case you are considering, I'll be happy to show you how to derive the exact answer.

On 2/26/23 2:53 PM, Trevor Lange wrote:
> On Sunday, February 26, 2023 at 12:28:58 PM UTC-8, Mike Fontenot wrote:
>> ...a time dilation equation for accelerating clocks that are separated by a
>> fixed distance.
>
> If the clocks are separated by a fixed proper distance (meaning the distance in terms of their co-moving system of inertial coordinates), ...

I intentionally make NO use of ANY inertial coordinates, other than at
the instant before the (constant) acceleration starts. I have no
interest at all in what any momentarily co-moving inertial observers
conclude about what's happening while the clocks are accelerating.
Before the acceleration starts, the clocks are both synchronized to
zero, and the separation between the clocks is set with yardsticks to
the desired distance "L". The only coordinates being used are the
coordinates of the accelerating frame. Time is measured by the separated
clocks in that frame. Distance is measured by the yardsticks between the
clocks, with each yardstick having a rocket attached to it, with an
accelerometer that controls the rocket so as to keep the yardsticks
accelerating along with the clocks. And each clock also has a rocket
and accelerometer attached to it. Every piece of matter is separately
accelerated in that manner ... there are no stresses or strains due to
the acceleration.

Einstein certainly believed that the distance between the clocks was
constant during the acceleration ... otherwise, he would have had to
specify how "L " varies with time in his equation, AND HE DIDN'T DO THAT!

On Sunday, February 26, 2023 at 3:10:30 PM UTC-8, Mike Fontenot wrote:
> >> ...a time dilation equation for accelerating clocks that are separated by a
> >> fixed distance.
> >
> > If the clocks are separated by a fixed proper distance (meaning the distance
> > in terms of their co-moving system of inertial coordinates), then the proper
> > accelerations of the clocks are not equal. Conversely, if you stipulate that the
> > proper accelerations are equal, then the distance between the clocks (in terms
> > of their co-moving inertial coordinates) can't be constant.
>
> Before the acceleration starts, the clocks are both synchronized to
> zero, and the separation between the clocks is set with yardsticks to
> the desired distance "L".

Yes, but thereafter if you are subjecting the clocks to equal constant proper acceleration, the proper distance between them will change. Conversely, if you want to maintain constant proper distance, they cannot have equal proper accelerations. So, you need to specify which of these cases you are talking about, i.e., constant equal proper accelerations or constant proper distance.

> Einstein certainly believed that the distance between the clocks was
> constant during the acceleration ... otherwise, he would have had to
> specify how "L " varies with time in his equation, AND HE DIDN'T DO THAT!

Again, as Einstein clearly stated at the beginning of section 18, he is considering only the case of sufficiently small accelerations so that the higher order terms (the effect associated with Born rigid motion) are negligible. But you are intertested in the case of large accelerations, so you need to account for those effects, as explained above. You need to specify if you are talking about constant equal proper accelerations, or constant proper distance.

On 2/26/23 4:25 PM, Trevor Lange wrote:
> On Sunday, February 26, 2023 at 3:10:30 PM UTC-8, Mike Fontenot wrote:
(I, Mike Fontenot said):
>> Before the acceleration starts, the clocks are both synchronized to
>> zero, and the separation between the clocks is set with yardsticks to
>> the desired distance "L".

(Trevor Lange said):
>
> Yes, but thereafter if you are subjecting the clocks to equal constant proper acceleration, the proper distance between them will change. Conversely, if you want to maintain constant proper distance, they cannot have equal proper accelerations. So, you need to specify which of these cases you are talking about, i.e., constant equal proper accelerations or constant proper distance.

If the clocks undergo the same PROPER acceleration, then the PROPER
separation between them will be constant. It is INERTIAL observers who
will conclude otherwise, and I'm not talking about, and I'm not
interested in, the conclusions of any inertial observers about the
accelerating clocks. I'm only interested in the conclusions of the
people who are located with, and accelerating with, the clocks. That's
what "proper acceleration" and "proper separation" MEAN.

There is another reason that we know the separation of the clocks
(according to the people co-located with the clocks) is constant, when
the accelerations of the two clocks are the same. We can make use of
the equivalence principle. The gravitational scenario, which is
equivalent to the acceleration scenario that I have described, is this:

Two clocks are mounted on a vertical pole, a constant distance "L"
apart. The surface of the ground that the pole is mounted in is
PERFECTLY flat (it's a very large plane), and it is horizontal,
perpendicular to the pole. Both the horizontal extent and the depth of
the ground are very large, creating a very strong and completely uniform
gravitational field at the pole, of magnitude "g" (numerically equal to
the acceleration "A" in the special relativity scenario). I.e., each
clock experiences exactly the same constant gravitational field strength
"g", and the separation of the two clocks is perfectly and permanently
constant at "L". The upper clock tics faster than the lower clock by
the constant factor exp(L*g), just like the leading clock tics faster
than the trailing clock by the constant factor exp(L*A) in the special
relativity scenario. And according to the equivalence principle, the
separation "L" is constant in each scenario.

On Monday, February 27, 2023 at 10:02:01 AM UTC-8, Mike Fontenot wrote:
>
> If the clocks undergo the same PROPER acceleration, then the PROPER
> separation between them will be constant.
No, read the "Bell paradox"

On 2/27/23 12:01 PM, Mike Fontenot wrote:
> If the clocks undergo the same PROPER acceleration, then the PROPER
> separation between them will be constant.

This is just plain false. Constant proper separation implies Born rigid
motion, which requires varying proper accelerations along the direction
of acceleration -- see the Bell spaceship paradox.

> [...]

Your invocation of the equivalence principle is flawed, as you violate
its locality condition.

> [... rest ignored, based on the above errors]

Tom Roberts

On 2/27/23 11:10 AM, Tom Roberts wrote:
> On 2/27/23 12:01 PM, Mike Fontenot wrote:
>> If the clocks undergo the same PROPER acceleration, then the PROPER
>> separation between them will be constant.
>
> This is just plain false. Constant proper separation implies Born rigid
> motion, which requires varying proper accelerations along the direction
> of acceleration -- see the Bell spaceship paradox.

Here's what Wiki says about that:

"Bell's spaceship paradox is a thought experiment in special relativity.
It was designed by E. Dewan and M. Beran in 1959[1] and became more
widely known when J. S. Bell included a modified version.[2] A delicate
thread hangs between two spaceships. They start accelerating
simultaneously and equally as measured in the inertial frame S, thus
having the same velocity at all times as viewed from S."

Here's the important part of what I quoted above, with the "all capital
letters" made by me:

"They start accelerating simultaneously and equally AS MEASURED IN THE
INERTIAL FRAME "S" thus having the same velocity at all times AS VIEWED
FROM S."

The scenario I've described says NOTHING about the conclusions of any
inertial people, unlike what's being described in Bell's spaceship
paradox. I talk about PROPER acceleration and PROPER separation, which
refers only to the conclusions of the accelerating people themselves
(based on accelerometers attached to the clocks and to the accelerating
people). Your (and Bell's) scenario is entirely different from mine.

On Monday, February 27, 2023 at 11:09:57 AM UTC-8, Mike Fontenot wrote:
> The scenario I've described says NOTHING about the conclusions of any
> inertial people, unlike what's being described in Bell's spaceship
> paradox. I talk about PROPER acceleration and PROPER separation, which
> refers only to the conclusions of the accelerating people themselves
> (based on accelerometers attached to the clocks and to the accelerating
> people). Your (and Bell's) scenario is entirely different from mine.
Crank

Proper {acceleration, time, speed, velocity, etc} are frame independent. That is, they are the same for all observers.

On 2/27/23 1:09 PM, Mike Fontenot wrote:
> On 2/27/23 11:10 AM, Tom Roberts wrote:
>> On 2/27/23 12:01 PM, Mike Fontenot wrote:
>>> If the clocks undergo the same PROPER acceleration, then the
>>> PROPER separation between them will be constant.
>> This is just plain false. Constant proper separation implies Born
>> rigid motion, which requires varying proper accelerations along
>> the direction of acceleration -- see the Bell spaceship paradox.
>
> Here's what Wiki says about that: [...] "They start accelerating
> simultaneously and equally AS MEASURED IN THE INERTIAL FRAME "S"
> thus having the same velocity at all times AS VIEWED FROM S."

Sure. They also have a constant separation, measured simultaneously in S
-- but that is not their PROPER separation, because they are not at rest
in S.

> [...] I talk about PROPER acceleration and PROPER separation, [...]

No, you don't. You don't seem to know what those words actually mean.

Tom Roberts

On 2/27/23 4:43 PM, Tom Roberts wrote:
> On 2/27/23 1:09 PM, Mike Fontenot wrote:
>>
>> Here's what Wiki says about that: [...] "They start accelerating
>> simultaneously and equally AS MEASURED IN THE INERTIAL FRAME "S"
>> thus having the same velocity at all times AS VIEWED FROM S."
>
> Sure. They also have a constant separation, measured simultaneously in S

You've still misunderstood what I'm saying.

First of all, as stated above, the Wiki article gives the case where the
inertial observers conclude that the two accelerating rockets accelerate
in such a way that they always have the same velocity at any instant
(according to the inertial observers). That's a different scenario from
my scenario.

The information I've provided concerns what two humans who are
co-located with the two accelerating clocks conclude about their
acceleration and their separation. YOU are talking about what inertial
observers conclude about the accelerating observers. Those two groups
of people don't agree ... everybody knows that.

The accelerating observers conclude that they are each accelerating at a
the same constant rate (as shown by accelerometers attached to them),
and that they have a constant separation (as shown by yardsticks laid
out between them).

The inertial observers conclude that the accelerating observers are NOT
accelerating at the same rate, and that the separation of the
accelerating observers is NOT constant.

All I care about is what the people who are accelerating conclude about
their acceleration and about their separation. I don't care at all
about what the inertial observers conclude about those two things.

On Monday, February 27, 2023 at 5:12:06 PM UTC-8, Mike Fontenot wrote:
> On 2/27/23 4:43 PM, Tom Roberts wrote:
> > On 2/27/23 1:09 PM, Mike Fontenot wrote:
> >>
> >> Here's what Wiki says about that: [...] "They start accelerating
> >> simultaneously and equally AS MEASURED IN THE INERTIAL FRAME "S"
> >> thus having the same velocity at all times AS VIEWED FROM S."
> >
> > Sure. They also have a constant separation, measured simultaneously in S
> You've still misunderstood what I'm saying.
>
> First of all, as stated above, the Wiki article gives the case where the
> inertial observers conclude that the two accelerating rockets accelerate
> in such a way that they always have the same velocity at any instant
> (according to the inertial observers). That's a different scenario from
> my scenario.
>
> The information I've provided concerns what two humans who are
> co-located with the two accelerating clocks conclude about their
> acceleration and their separation. YOU are talking about what inertial
> observers conclude about the accelerating observers. Those two groups
> of people don't agree ... everybody knows that.
>
> The accelerating observers conclude that they are each accelerating at a
> the same constant rate (as shown by accelerometers attached to them),
> and that they have a constant separation (as shown by yardsticks laid
> out between them).
>
> The inertial observers conclude that the accelerating observers are NOT
> accelerating at the same rate, and that the separation of the
> accelerating observers is NOT constant.
>
> All I care about is what the people who are accelerating conclude about
> their acceleration and about their separation. I don't care at all
> about what the inertial observers conclude about those two things.
Crank

All observers agree on the value of proper acceleration.

On February 27, Tom Roberts wrote:
>>>> If the clocks undergo the same PROPER acceleration, then the
>>>> PROPER separation between them will be constant.
>
>>> This is just plain false. Constant proper separation implies Born
>>> rigid motion, which requires varying proper accelerations along
>>> the direction of acceleration
>
>> [...] I talk about PROPER acceleration and PROPER separation, [...]

Given two rockets with identical constant accelerations, as per this
example. The lead rocket is at higher grav. potential. Is there a
length contraction effect at work here? i.e. concerning the separation
between the ships.

Gravitational length contraction?

--
Rich

On Monday, February 27, 2023 at 10:02:01 AM UTC-8, Mike Fontenot wrote:
> > If you are subjecting the clocks to equal constant proper acceleration, the proper distance between them will change. Conversely, if you want to maintain constant proper distance, they cannot have equal proper accelerations. So, you need to specify which of these cases you are talking about, i.e.., constant equal proper accelerations or constant proper distance.
>
> If the clocks undergo the same PROPER acceleration, then the PROPER
> separation between them will be constant.

That is incorrect. Learn about hyperbolic motion (constant proper acceleration) and Born rigid motion.

> Two clocks are mounted on a vertical pole, a constant distance "L"
> apart. The surface of the ground that the pole is mounted in is
> PERFECTLY flat (it's a very large plane), and it is horizontal, perpendicular
> to the pole... creating a very strong and completely uniform
> gravitational field at the pole...

No, what you are trying to describe is the gravitational field of an infinite flat wall, which in Newtonian gravity produces a perfectly uniform gravitational acceleration to all distances, but in general relativity it does not result in a homogeneous gravitational field in the sense of uniform gravitational acceleration. This is a well-studied subject. No, this does not contradict the equivalence principle, because that principle refers to differential relations in the infinitessimal region around any event, not over extended regions. Before you even think about starting to thing about general relativity, I suggest you first study the special relativity aspects, i.e., hyperbolic motion and Born rigid motion.

On Tuesday, 28 February 2023 at 00:43:31 UTC+1, Tom Roberts wrote:
> On 2/27/23 1:09 PM, Mike Fontenot wrote:
> > On 2/27/23 11:10 AM, Tom Roberts wrote:
> >> On 2/27/23 12:01 PM, Mike Fontenot wrote:
> >>> If the clocks undergo the same PROPER acceleration, then the
> >>> PROPER separation between them will be constant.
> >> This is just plain false. Constant proper separation implies Born
> >> rigid motion, which requires varying proper accelerations along
> >> the direction of acceleration -- see the Bell spaceship paradox.
> >
> > Here's what Wiki says about that: [...] "They start accelerating
> > simultaneously and equally AS MEASURED IN THE INERTIAL FRAME "S"
> > thus having the same velocity at all times AS VIEWED FROM S."
> Sure. They also have a constant separation, measured simultaneously in S
> -- but that is not their PROPER separation, because they are not at rest
> in S.

In the meantime in the real world, however, IMPROPER
clocks measure IMPROPER t'=t.

On 2/27/23 9:16 PM, Trevor Lange wrote:
> On Monday, February 27, 2023 at 10:02:01 AM UTC-8, Mike Fontenot wrote:
>> If the clocks undergo the same PROPER acceleration, then the PROPER
>> separation between them will be constant.
>
> That is incorrect. Learn about hyperbolic motion (constant proper acceleration) and Born rigid motion.
>

Hyperbolic motion and Born rigid motion concern the perspective of
INERTIAL OBSERVERS describing accelerating objects, NOT the perspective
of the observers who are themselves accelerating. My results concern
only in the latter.

Einstein (in his 1907 paper) said that the leading clock of a pair of
clocks accelerating at a constant rate "A" will tic faster than the
trailing clock by the factor exp(L*A), where the separation of the
clocks is "L". Einstein clearly believed (correctly) that the
separation "L" is constant (according to human attendants co-located
with the clocks). Otherwise, he would have needed to specify how "L"
varies with time, in order for the equation to be of any use. He didn't
do that.

On 2/27/23 7:12 PM, Mike Fontenot wrote:
> [...] The accelerating observers conclude that they are each
> accelerating at a the same constant rate (as shown by accelerometers
> attached to them), and that they have a constant separation (as shown
> by yardsticks laid out between them).

This is inconsistent with SR. As you are discussing a gedanken, you MUST
use the predictions/descriptions of SR.

In SR, two observers separated in space and accelerating along the line
between them can either:
* have equal proper accelerations (in which case the proper
distance between them is increasing)
OR
* have constant proper separation (in which case the
front observer must have a smaller proper acceleration
than the rear observer)
It is not possible to have both, as you assume.

For equal proper accelerations, this _IS_ the Bell spaceship paradox
(the string between the spaceships breaks because the proper distance
between them is increasing). That is, for equal proper accelerations in
your scenario, gaps will appear between the yardsticks and/or the
observers; the gaps will increase over time.

[BTW in this case, equal accelerations relative to the
initial inertial frame S is equivalent to equal proper
accelerations (though actual values will differ).]

This is all well known, and has been since Bell popularized his
spaceship paradox.

> [... further errors based on the above]

Tom Roberts

On 2/28/23 1:24 PM, Tom Roberts wrote:
> On 2/27/23 7:12 PM, Mike Fontenot wrote:
>> [...] The accelerating observers conclude that they are each
>> accelerating at a the same constant rate (as shown by accelerometers controlling rockets
>> attached to them), and that they have a constant separation (as shown
>> by yardsticks laid out between them, with the yardsticks also having attached accelerometers and rockets).
>
> In SR, two observers separated in space and accelerating along the line
> between them can either:
>     * have equal proper accelerations (in which case the proper
>       distance between them is increasing)
> OR
>     * have constant proper separation (in which case the
>       front observer must have a smaller proper acceleration
>       than the rear observer)
>

That may be the conclusion of INERTIAL observers observing accelerating
clocks. It is NOT the conclusion of observers who are undergoing the
acceleration.

On Tuesday, February 28, 2023 at 11:49:43 PM UTC+1, Mike Fontenot wrote:
> On 2/28/23 1:24 PM, Tom Roberts wrote:
> > On 2/27/23 7:12 PM, Mike Fontenot wrote:
> >> [...] The accelerating observers conclude that they are each
> >> accelerating at a the same constant rate (as shown by accelerometers controlling rockets
> >> attached to them), and that they have a constant separation (as shown
> >> by yardsticks laid out between them, with the yardsticks also having attached accelerometers and rockets).
> >
> > In SR, two observers separated in space and accelerating along the line
> > between them can either:
> > * have equal proper accelerations (in which case the proper
> > distance between them is increasing)
> > OR
> > * have constant proper separation (in which case the
> > front observer must have a smaller proper acceleration
> > than the rear observer)
> >
> That may be the conclusion of INERTIAL observers observing accelerating
> clocks. It is NOT the conclusion of observers who are undergoing the
> acceleration.

It is the conclusion of observers who are undergoing the acceleration.
What makes you think it's not?

--
Jan

On 2/28/23 4:07 PM, JanPB wrote:
>
> It is the conclusion of observers who are undergoing the acceleration.
> What makes you think it's not?
>

Because it is the conclusion of inertial observers, and inertial
observers don't agree with accelerating observers.

On Tuesday, February 28, 2023 at 9:59:36 AM UTC-8, Mike Fontenot wrote:
> >> If the clocks undergo the same PROPER acceleration, then the PROPER
> >> separation between them will be constant.
> >
> > That is incorrect. Learn about hyperbolic motion (constant proper acceleration)
> > and Born rigid motion.
> >
> Hyperbolic motion and Born rigid motion concern the perspective...

No, an object is undergoing hyperbolic motion just if a co-moving accelerometer shows a constant value. Born rigid motion of an object consists of motion such that the object remains unstressed, meaning the atoms and molecules maintain their same intrinsic relations to each other. For example, you can place your two clocks at the ends of a solid rod, and subject that rod to Born rigid motion, so the intrinsic length of the rod remains the same. (Here "intrinsic" means what you thoughtlessly presume to be the valid measure.) If you subject one end of the rod to constant proper acceleration A, then the other end of the Born rigid rod must have constant proper acceleration less than A.

> Einstein (in his 1907 paper) said that the leading clock of a pair of
> clocks accelerating at a constant rate "A" will tic faster than the
> trailing clock by the factor exp(L*A)...

Again, in this paper Einstein restricts his considerations to the case of arbitrarily small accelerations, not valid for large accelerations, so it is silly for you to be grandly disproving his statements by invoking large accelerations. Moreover, that little off-hand comment (based on the fact that for a row of uniformly-spaced accelerating clocks, each of which ticks faster than the one behind it by a factor k, their rates will be in proportion to 1, k, k^2, k^3, ...) has no relevance and actually is not valid for arbitrary acceleration. But it is valid in the low-acceleration limit that Einstein was discussing.

Mike Fontenot wrote:

> In Einstein's 1907 paper,
> https://einsteinpapers.press.princeton.edu/vol2-trans/319 , he gives his
> exponential gravitational time dilation equation. […]

Even if you are correct that Einstein had made a mistake there:

Who cares about a “gravitational time dilation” that Einstein formulated *a
decade* before he even postulated general relativity (in 1915/1916), even
before the Schwarzschild metric was found (in 1916)?

The measurable gravitational time dilation can be *easily* derived from the
Schwarzschild metric, which is a solution of the Einstein field equation:

ds² = (1 − rₛ/r) c² dt² − 1/(1 − rₛ/r) dr² − r² (dθ² + sin³θ dφ²),

dr = dθ = dφ = 0

⇒ ds² = c² dτ² = (1 − rₛ/r) c² dt²,

⇔ dτ = dt √(1 − rₛ/r) ,

⇔ Δτ = ∫_W dτ = ∫_{t₁}^{t₂} dt √(1 − rₛ/r)
= Δt √(1 − rₛ/r)
= Δt √(1 − 2 G M/(r c²)).

See <https://qr.ae/prH4Xu> or <https://www.quora.com/What-affects-gravitational-time-dilation-large-masses-or-high-speeds-relative-to-those-masses/answer/Thomas-Lahn> for details.

PointedEars
--
Q: What happens when electrons lose their energy?
A: They get Bohr'ed.

(from: WolframAlpha)

On Tuesday, 28 February 2023 at 21:24:37 UTC+1, Tom Roberts wrote:
> On 2/27/23 7:12 PM, Mike Fontenot wrote:
> > [...] The accelerating observers conclude that they are each
> > accelerating at a the same constant rate (as shown by accelerometers
> > attached to them), and that they have a constant separation (as shown
> > by yardsticks laid out between them).
> This is inconsistent with SR. As you are discussing a gedanken, you MUST
> use the predictions/descriptions of SR.

You can't think differently than our idiot guru has taught!!!
You're FORCED!!!!!!

On Wednesday, 1 March 2023 at 00:07:12 UTC+1, JanPB wrote:
> On Tuesday, February 28, 2023 at 11:49:43 PM UTC+1, Mike Fontenot wrote:
> > On 2/28/23 1:24 PM, Tom Roberts wrote:
> > > On 2/27/23 7:12 PM, Mike Fontenot wrote:
> > >> [...] The accelerating observers conclude that they are each
> > >> accelerating at a the same constant rate (as shown by accelerometers controlling rockets
> > >> attached to them), and that they have a constant separation (as shown
> > >> by yardsticks laid out between them, with the yardsticks also having attached accelerometers and rockets).
> > >
> > > In SR, two observers separated in space and accelerating along the line
> > > between them can either:
> > > * have equal proper accelerations (in which case the proper
> > > distance between them is increasing)
> > > OR
> > > * have constant proper separation (in which case the
> > > front observer must have a smaller proper acceleration
> > > than the rear observer)
> > >
> > That may be the conclusion of INERTIAL observers observing accelerating
> > clocks. It is NOT the conclusion of observers who are undergoing the
> > acceleration.
> It is the conclusion of observers who are undergoing the acceleration.

No, poor halfbrain, it is the conclusion of you and your fellow idiots.