New Arguments In A Possible Proof That Negative Ageing Doesn't Occur In Special Relativity

Discussion in 'Physics & Math' started by Mike_Fontenot, Aug 7, 2021.

1. phytiRegistered Senior Member

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625
Mike;
Only if the aos is assumed to continuously have an independent existence, which it does not. Referring to fig.1 in the previous post, the aos is only determined after receiving the return signal.

SR does not mandate what the observer concludes. They are free to choose an inertial frame or a pseudo rest frame. That is the essence of the 'relativity principle'.
Also, the term 'perpetual' is redundant, it doesn't provide any useful information.

Poor choice of words. She appeared to age more slowly. Converging or diverging, the closing speed of the gap between A and B observers produces doppler shifts, perceptions that indicate changing clock rates, NOT aging.

system of synchronized clocks.
from OTEOMB, A. Einstein, 1905, par.1:
"But it is not possible without further assumption to compare, in respect of time, an event at A with an event at B. We have so far defined only an A time'' and a B time.'' We have not defined a common time'' for A and B, for the latter cannot be defined at all unless we establish by definition that the time'' required by light to travel from A to B equals the time'' it requires to travel from B to A.
... It is essential to have time defined by means of stationary clocks in the stationary system,"

How far can a local system be extended? Not much. The system is very hypothetical and logistically impossible. All considerations of 'time' resulted in the coordinate transformations which bridge the gap between local and distant 'time'.

If you don't understand the basics, you are just spinning your wheels.

There is no instant knowledge.

3. Mike_FontenotRegistered Senior Member

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What I agreed to was that "ACCORDING TO HER, the entire array is synchronized." (That is in your post #45.)

And I also said:

The established fact is that she concludes that her array of clocks are synchronized. That doesn't require that anyone but her (and her perpetually-inertial helpers permanently co-located with those clocks) has to agree that her clocks are synchronized. Other perpetually-inertial observers who are moving with respect to her don't agree that her clocks are synchronized. And the accelerating observer, or an observer who has accelerated recently, doesn't necessarily agree with her either, even if he is stationary with respect to her for some period of time.

5. Q-reeusBannedValued Senior Member

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4,695
Looking back over #55 it now seems clear nonsimultaneity is not the only concern. Constant proper separation top vs bottom requires a given proper acceleration at the bottom to be greater relative to that measured at the top owing to time dilation at bottom relative to at top. Fine. Does that invalidate exponential metric derivation? The Appendix A derivation given by Robertson posits a fixed value g as measured by an on-board accelerometer. The final exponential form (37) there will be accurate provided an appropriate average proper value for g is used. As the spatial separation shrinks it's clear variation of proper g over that interval becomes negligible and disappears in the differential limit of a redshift spatial gradient.
The same exponential metric final form given by Alley linked to in #52 derives from the differential form of invariant interval, its intensive character obviating any need to consider an average proper value.
What comes out of this is the need to properly consider and define what 'uniform g field' or 'uniform acceleration' really refers to and where and when it applies or not.

7. Neddy BateValued Senior Member

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2,323
Mike,

Let's assume you are correct that her array of clocks can somehow be synchronised only for her and her friends, but not synchronised for the traveler, even after he has stopped moving relative to them. The scenario is that, according to her coordinates, he starts at (x,t)=(0,0), instantaneously achieves a constant speed v=0.866c relative to her, and travels until he reaches (x,t)=(34.64,40.00) where he instantaneously achieves a speed of v=0.000 relative to her (i.e. he stops). Notice the equation x=0.866*40.00=34.64 holds true, as it should (x=vt).

After he stops, the traveler knows he is located at x=34.64 because it had been marked on the ground as a sort of "finish line" for his planned journey. He also knows one of her array clocks displays t=40.00 because he is now standing still right next to it. If he were to use the CMIF method, he would say she is 40 years old. In that case, the equation x=0.866*40.00=34.64 still holds true, as it should (x=vt).

But he does not use the CMIF method, he uses your method instead. So he says she is 10 years old, because that was how old he calculated her to be just before he instantaneously stopped. So now if we use x=vt again, we calculate that he should be located at x=0.866*10.00=8.66. Agreed?? If he had stopped when she was 10 years old, he would have only reached x=8.66 but somehow he made it to x=34.64. Doesn't that prove your method fails? If not, please explain.

In your method, what is the relationship between t=10.00 and x=34.64? Is it that he says his speed was v=34.64/10.00=3.464c? Or does he have his own units of length? Or what?

Last edited: Aug 11, 2021
8. Mike_FontenotRegistered Senior Member

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359
I remember when we originally talked about my simultaneity method (about a year ago), I concluded that after he changes velocity, and during the period of time when he disagrees with the co-located and co-stationary PIO (the "disagreement interval", "DI"), that he says the speed of a distant light pulse isn't always equal to "c", and that their relative velocity isn't always what she thinks it is. That may be what's going on here.

If my recent proof is correct, I think I MAY be able to use it to determine what the correct simultaneity method actually is. The proof says it's not CMIF, but it may not be my existing alternative simultaneity method either. I've got a lot of work to do, and THAT'S what I'm going to be spending my time on for quite a while.

9. Neddy BateValued Senior Member

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2,323
Yes, I remember that. And I think you were also saying that if he can disagree with her about whether or not her clock array is synchronised in her perpetually inertial rest frame, then he can also disagree with the distance 34.64. So maybe he will insist that he is really located at x=8.66 despite all evidence to the contrary, just as he insists that she is 10 years old despite all evidence to the contrary. Maybe he can also insist that he is the real winner of the 2020 US election.

Also, please keep in mind that there is obviously nothing wrong with the traveler in this case simply agreeing with her and her friends and simply agreeing that she is 40 years old, because that is the only answer that fulfills the simple equation t=x/v=34.64/0.866=40.00 as the scenario gives us x=34.64 and v=0.866c. So obviously there can never be any valid proof that he is not allowed to do that, since it is obviously a correct answer.

10. phytiRegistered Senior Member

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625
Mike;
Refer to the gif:
Blue lines are light signals.
No need for a redundant helper.
Ann will receive the latest image of ε on the Bart clock.
Bart will receive the latest image of ε on the Ann clock.
Bart will be a distance of 20+.5at^2, with t=ε.
Neither can assign a corresponding time to the reflection event until it returns.

You keep describing events as if light speed is instantaneous.

11. SsssssssRegistered Senior Member

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I don't understand your problem. You can do the equivalence principle calculation in a uniform gravity field or a non uniform one you just get different values of the potential function and kinetic energy but it should work out either way. Uniform gravity fields are really hard to make in GR and require a carefully tuned cosmological constant IIRC so it shouldn't be particularly surprising that one doesn't drop out of the boost symmetry of Minkowski spacetime and it's irrelevant in practice anyway because the fractional difference in acceleration between two objects $h$ apart is $gh/c^2$ which is like one part in $10^{16}$ for a real elevator.

12. SsssssssRegistered Senior Member

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302
There isn't a single correct simultaneity method but there are loads that keep their definition of now monotonic in the proper time of an arbitrary worldline like just picking an inertial frame and sticking with it whether you are at rest in it or not or radar coordinates.

13. SsssssssRegistered Senior Member

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I've been trying to make that point too but it doesn't seem to stick.

14. Neddy BateValued Senior Member

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Mike has convinced himself that a person who has accelerated in the recent past is allowed to have completely different laws of physics than a perpetually inertial person. For example, let two inertial people be standing together, stationary with respect to one another. Let one be perpetually inertial, and let the other have accelerated in the recent past.

The perpetually inertial person says, "That distant light-clock is ticking at a constant rate of 1 second per second, because it is stationary with respect to us, and the speed of light is constant." Mike thinks the other person can say, "Only for you, because you have been perpetually inertial. I accelerated recently, so for me, that distant light-clock is ticking at a non-constant rate that is faster than 1 second per second, and the speed of light is not constant." He also thinks that idea is just standard SR, not something that he misunderstood.

I think he learned SR by studying the twin paradox, and he believes the lesson to be learned there is that accelerated people are allowed a completely different reality than perpetually inertial ones. (The stay-home twin is assumed to have been perpetually inertial, as far as he can tell.)

15. SsssssssRegistered Senior Member

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That's using non inertial coordinates like radar coordinates which is just as fine as always using one set of inertial coordinates or always using your rest inertial frame because there is no unique answer to what time it is at a distant point. Nothing measurable changes if you change your simultaneity planes so as long as they don't cross or become null or timelike you can make them as weird a shape as you want but you can't make global non inertial frames out of cut up bits of inertial frames and I don't know why he thinks he needs to prove it.

16. Neddy BateValued Senior Member

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2,323
Strangely enough, he wants to prove that an inertial person who has accelerated in the recent past would be wrong to use the simultaneity plane of his rest frame. So, in the example you just replied to where the perpetually inertial person says,"That distant light-clock is ticking at a constant rate of 1 second per second, because it is stationary with respect to us, and the speed of light is constant," Mike is trying to prove that the other inertial person would be wrong to say the same thing, (even though they are standing next to each other, and stationary with respect to one another).

17. SsssssssRegistered Senior Member

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If so then that's obviously going to fail because a coordinate system you must or must not use would violate general covariance. From this thread I think he's just looking for a coordinate system that is a rest frame for an arbitrarily accelerating observer and covers all of spacetime which is why I keep mentioning radar coordinates which do that but I also think that there's some history to this topic that I'm not aware of.

18. phytiRegistered Senior Member

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625
Mike;

You didn't specify a distance for d, so I chose 20.
If d=10, the image would have been 10+ε.
If d=5, the image would have been 15+ε.
If d=0, the image would have been 20+ε.
I.e. Ann sees Bart the same age, only if they are in the same place.

19. Mike_FontenotRegistered Senior Member

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359
It's now looking like to me that Einstein's 1907 method of defining an array of clocks for an accelerating observer agrees completely with the CMIF method. If so, that means that CMIF is the law of the land, no further assumptions required. I can also now see what the error in my proof (that negative ageing doesn't occur) was. Details to follow.

20. SsssssssRegistered Senior Member

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A sequence of flat planes orthogonal to the worldline of a particle under constant proper acceleration defines the Rindler wedge while Einstein appears to be considering a uniform gravitational field which is a very different thing so I expect you are wrong.
Can you explain why you think this proof is necessary? Negative aging clearly doesn't happen in reality from our own experience and it clearly doesn't happen in a relativistic model because proper time is an affine parameter on a clock's worldline and so the only place it appears to happen is in maps of spacetime where you let your spatial planes cross where it's obviously a consequence of extending your coordinates through a coordinate singularity so just a consequence of careless map drawing not reality.

Last edited: Aug 16, 2021
21. Mike_FontenotRegistered Senior Member

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359
That was apparently wishful thinking on my part. I still can't find an error in my proof that negative ageing doesn't occur ... the error that I THOUGHT I saw didn't "pan out".

Then, I did an analysis which approximates instantaneous (and large) velocity changes (in both directions). When the acceleration was directed TOWARD the distant person (she), the analysis said she got much older during the acceleration ... consistent with the CMIF simultaneity method. But when the acceleration was directed AWAY from the distant person (she), the analysis said her age changed very little, and she didn't get younger ... which is NOT consistent with the CMIF simultaneity method. In fact, the those two results, taken together, are clearly inconsistent with the standard twin "paradox" scenario: suppose he keeps instantaneously changing his velocity, first accelerating toward her, then immediately away from her, over and over. According to my results using the array of clocks (and "helper friends", HF's), each of those accelerations TOWARD her produces a large increase in her age (according to him), but each of those accelerations AWAY from her produces NO decrease in her age, and very little change in her age at all. So he can make her be as old as he wants to, by continuing to do those paired accelerations, over and over. So when he quits doing all those paired accelerations, and then accelerates toward her one last time, and then coasts back to their reunion (with the usual slow ageing by her, according to him), his computation of her age at his arrival won't be what is required by the twin "paradox" result. (They HAVE to agree with each other about their respective ages at the reunion). So this analysis produces an absurd and inconsistent result, and it can't be correct. But Einstein's gravitational time dilation equation had been well-tested, so how could it produce an erroneous result when its acceleration equivalent (using the equivalence principle) is used? Is it possibly because the gravitational time dilation equation is not intended to be used with gravity being switched on and off, and therefore we can't use it to describe what happens when the observer switches his acceleration on and off?

Or maybe I've just made a mistake in my analysis somewhere. I'll give a few facts about my analysis that will allow anyone who wishes, to check my work. I use an acceleration of +-1 ls/s/s (equivalent to about 30.5 million g's), and the acceleration lasts only 1.317 seconds. That results in a velocity of +-0.866, which gives the convenient gamma = 2.0. When computing distances traveled by various observers, I integrate the velocity v with respect to t, the accelerating person's time. The relation between velocity v and "rapidity", theta, is v = tanh(theta). Theta is simply related to acceleration A and time t: theta = A t. (Rapidity is a non-linear equivalent of velocity: rapidity can vary from -infinity to +infinity, whereas velocity can't exceed the speed of light in magnitude.) The integral of tanh is equal to ln(cosh). But you can get a rough idea of the magnitude of the results by just integrating theta rather than v to get approximate distances.

22. SsssssssRegistered Senior Member

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302
Your twins started colocated? Then you won't get negative aging effects because your simultaneity planes don't cross but when the distance grows to light second you'll start to see it.

23. Mike_FontenotRegistered Senior Member

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No, at the beginning of the scenario, right before he accelerates, they are separated by the distance "d" lightyears. It is unspecified, but it is non-zero. Regardless of how big you make "d", her age doesn't decrease (according to him) when he accelerates away from her, but her age suddenly increases by a large amount (according to him) when he accelerates toward her.