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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.

In Einstein’s 1907 paper, Einstein stated that the GTD equation is R(g) = exp(g L),

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).

Note that for constant “A” and “L”, the rate ratio “R” DOES NOT VARY WITH TIME.

I showed in [in another paper] 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.

Section 3. A Proposed Experimental Test of My GTD Equation...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.

In my paper, I said:[...]A plot of R(A,t) versus “t” is given in section 4 for the case A = 1 ls/s/s (about 40 g’s) and L = 7.52 ls.[...]

Quote from: MikeFontenot on 26/10/2022 21:35:18 the number of seconds in a year, which, if I've done the arithmetic correctly, is about a factor of 16 MILLION greater.31.5 million actually, or almost exactly πe7 by coincidence.

the number of seconds in a year, which, if I've done the arithmetic correctly, is about a factor of 16 MILLION greater.

Your equations are not always wrong, but your representation of Einstein's theories are almost always wrong.[...]

Thank you for responding. It is a bit more clear now as to what you're trying to convey.Quote from: MikeFontenot on 27/10/2022 18:08:28In his 1907 paper, Einstein said that when two people (each with his own clock) are initially stationary at time zero on the clocks, and not accelerating, and separated by "L" lightseconds (ls), and who then simultaneously start accelerating (according to their accelerometers) at a constant "A" ls/s/s (in the direction of their constant separation), then the leading clock will tic faster than the trailing clock by the factor R(A) = exp(L A) .Einstein did not say this. There's several inaccuracies here.A) You left out specification of coordinate systems / frame-references.If we use frame S (the inertial frame relative to which the clocks are synced and the clocks are both initially stationary), then the clocks experience identical acceleration relative to S and must remain synced in S by symmetry.If we use frame ∑, the accelerating reference frame of the rear clock, then L is not constant, the front clock is not stationary in ∑, but it does indeed run slower than the stationary clock at the origin of ∑. Also, there is more than one kind of accelerating coordinate system, and Einstein is not using the one you think he is.B) Einstein's equation (a related but different one) referenced an entirely different scenario than the Bell's spaceship one which you are describing. This is something I pointed out above, but you're not big on reading comments so you're persisting in this inappropriate scenario. QuoteIn each of the cases, the acceleration "A" and the duration "tau" of the accelerationThere is no duration of acceleration in Einstein's scenario. It never quits. Just so you know. Einstein describes a particular time interval of that acceleration.Anyway, duration as measured in which clock? Again, the answer is worldline dependent, but I'm guessing tau on each of the accelerating clocks, not as measured by S where the acceleration wouldn't be constant.Quote... are chosen such that the product "A tau" is constant for all cases at 1.317 ls/s. That results in all cases in the velocity of the two clocks, at the end of the acceleration, being 0.866 ls/s.A funny way to say that you want both clocks accelerated to a rapidity of 1.317c, which indeed is a speed relative to S of √3/2 where the Lorentz factor is 2. Nice round number are easy.QuoteFor that first case, compute the reading on the leading clock when the trailing clock reads time "tau".Frame dependent question which has an ambiguous answer. Both simultaneously read 'tau' relative to S.Judging by your subsequent text, I'm guessing you didn't want it relative to S, but rather relative to the immediate inertial frame S' of the rear clock at time tau. If you want that, you should explicitly say that. Neither of Einstein's equations yield results relative to S', so that guess of mine is also off topic.QuoteThat reading will be given by tau R = tau exp(L A).No. That is Einstein's formula for a different situation (and different coordinate system) than the one you are describing, where both clocks maintain a constant proper separation throughout, effectively putting the clocks at opposite ends of a rigid continuously accelerating thing. You're describing Bell's scenario where both clocks undergo identical proper acceleration, and then only temporarily.QuoteWhat do you get for the answer to the first case?I admittedly didn't get one since the scenario is complicated, but it is larger than the one you get with your inappropriately applied formula.QuoteNow [after three different insanely high g's], look at your results. Does that sequence of readings on the leading clock appear to be approaching a finite limit as tau goes to zero? Or does it appear to be diverging to infinity as tau goes to zero?OK, this seems to get to the crux of the matter. The two formulas give the same answer for small values of L, but yours is not small, so the equation (30) is inappropriate for the coordinate system Einstein chooses to use. This describes it in more detail:https://en.wikipedia.org/wiki/Rindler_coordinates#Variants_of_transformation_formulasEinstein is using his light-signal sync convention, but with an accelerated frame, which yields equation (2i): dt = e^{αx}dt_{0} where α is proper acceleration and dt_{0} is your tau. This is Einstein's choice of coordinate system, which is essentially Lass (or radar) coordinates which do indeed tend towards infinity as acceleration increases to really high values. His equation (30): dt = (1 + αx)dt_{0} is seen for the Kottler–Møller coordinates shown in (2c). Einstein is not using Kottler–Møller coordinates, but the two coordinate systems are differ only in second order effects and are similar for reasonable values for α and x, so he chooses to use the simpler formula in his paper.The third equation (the Rindler one, 2f) is similar, but without the "1 +" part..So yes, the equation you quote does indeed tend to infinity as acceleration increases, but that's correct for Lass coordinates, and equation (30) works as an approximation of Lass coordinates for unexceptional values such as those one might see in a real situation. Your mistake seems to be an assumption that simultaneity is being determined by frame S' (as described

In his 1907 paper, Einstein said that when two people (each with his own clock) are initially stationary at time zero on the clocks, and not accelerating, and separated by "L" lightseconds (ls), and who then simultaneously start accelerating (according to their accelerometers) at a constant "A" ls/s/s (in the direction of their constant separation), then the leading clock will tic faster than the trailing clock by the factor R(A) = exp(L A) .

In each of the cases, the acceleration "A" and the duration "tau" of the acceleration

... are chosen such that the product "A tau" is constant for all cases at 1.317 ls/s. That results in all cases in the velocity of the two clocks, at the end of the acceleration, being 0.866 ls/s.

For that first case, compute the reading on the leading clock when the trailing clock reads time "tau".

That reading will be given by tau R = tau exp(L A).

What do you get for the answer to the first case?

Now [after three different insanely high g's], look at your results. Does that sequence of readings on the leading clock appear to be approaching a finite limit as tau goes to zero? Or does it appear to be diverging to infinity as tau goes to zero?

[...]

You have misunderstood almost everything I've said. Way too many mistakes to respond to separately.

The ONLY reference frame I'm using is the frame in which the two observers (and their clocks) are stationary (during the entire experiment).

I don't care at all what any perpetually-inertial observer anywhere thinks about anything.

The frame I'm using is EXACTLY what Einstein was using.

At the instant (t = 0) when the two observers fire their rockets, they are the same age (say, zero years old), and their clocks are synchronized at zero.

Each of the two observers agree that their separation is constant (at L lightseconds) during the ENTIRE experiment.

All you need to do (for each of the set of experiments) is use the exponential equation to calculate what the leading clock reads at the end of the constant acceleration "A"

Imagine that the two clocks are connected by a steel beam of length "L". And suppose that all along that beam are attached rockets and accelerometers that insure that each part of the beam is accelerating with acceleration "A", just like the clocks (with their rockets and accelerometers) are. So the beam can't expand or contract in length ... it has the length "L" always.

Method 2: Just use the gravitational scenario that is equivalent to the acceleration scenario, as given by the equivalence principle.

In that gravitational scenario, there is a gravitational field that does NOT vary in strength along the direction of the field ... the gravitational field is always and everywhere of strength "g" = "A"

No. All of the rockets are instructed to maintain an acceleration of exactly "A"

Quote from: MikeFontenot on 30/10/2022 16:51:54No. All of the rockets are instructed to maintain an acceleration of exactly "A" ls/s/s. From whose point of view?

No. All of the rockets are instructed to maintain an acceleration of exactly "A" ls/s/s.

The acceleration varies along the length of a rigid beam.

From whose point of view?

No. All of the rockets are instructed to maintain an acceleration of exactly "A" ls/s/s at their location on the beam, as shown on their accelerometers.

So the acceleration IS exactly "A" ls/s/s everywhere along the beam, and also at both clocks (which also have their own accelerometers, and their own rockets which precisely maintain a constant "A" ls/s/s).

No. By the equivalence principle, the gravitational scenario that is equivalent to the accelerational scenario

Most of your (Halc's) statements, about the accelerations and the distances that differ from my statements, differ because you are using the conclusions about the scenario ACCORDING TO SOME PERPETUALLY-INERTIAL OBSERVER.

If the endpoints of a body are accelerated with constant proper accelerations in rectilinear direction, then the leading endpoint must have a lower proper acceleration in order to leave the proper length constant so that Born rigidity is satisfied....However, if a different acceleration profile is chosen by which the endpoints of the body are simultaneously accelerated with same proper acceleration as seen in the external inertial frame, its Born rigidity will be broken.

No it doesn't. The "Method 1" example I gave you in an earlier post clearly produces constant acceleration, and a constant separation "L" of the accelerating observers (and their clocks). I'll repeat it here

[...]I am using mostly inertial frames here since your observers are all inertial all the time except for one second each.[...]

It's a bit like your experiment needs you to move faster than light. You are relying on something that can't be done.

Quote from: pzkpfw on 01/11/2022 20:29:02It's a bit like your experiment needs you to move faster than light. You are relying on something that can't be done.There isn't any faster than light motion in my scenario, or in my proposed experimental test.

Quote from: Halc on 01/11/2022 00:44:02your observers are all inertial all the time except for one second each.Your statement (in red) is incorrect. NONE of my observers are inertial, except at the instant before they turn on their rockets. They are accelerating with a constant acceleration, as indicated by accelerometers attached to them.

your observers are all inertial all the time except for one second each.

In each of the cases, the acceleration "A" and the duration "tau" of the acceleration are chosen such that the product "A tau" is constant for all cases at 1.317 ls/s. That results in all cases in the velocity of the two clocks, at the end of the acceleration, being 0.866 ls/s.

Perhaps I read your scenario wrong. I was under the impression the acceleration ceased for both ships after one second,[...]