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On the Lighter Side => New Theories => Topic started by: MikeFontenot on 13/06/2021 15:14:09

Title: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 13/06/2021 15:14:09

According to the CMIF (Co-Moving-Inertial Frames) simultaneity method, an observer (he) who accelerates in the direction away from a distant person (she) will conclude that she rapidly gets YOUNGER during his acceleration. But I think I may have found a counterexample that shows that doesn't happen.

It is well-known that two stationary clocks at different positions in a gravitational field will run at different rates. The clock that is closer to the source of the gravitational field will run slower than the clock that is farther from the source of the field.

Because of the equivalence principle, it is also true that if two clocks that are separated by a fixed distance "d" ly are both accelerated with a constant equal acceleration of "A" ly/y/y, the trailing clock runs slower than the leading clock, by the factor exp(Ad).

So consider the following scenario:

At some instant, the perpetually-inertial "home twin" (she) is 20 years old, and is holding a display that always shows her current age. Facing her is the "helper friend" (the "HF") of an observer (he) who is "d" ly away to her right. Both the HF and he are also 20 years old, and are stationary wrt her at that instant. Like her, he and the HF are each holding a display that always shows their current ages.

Now, suppose that he and his helper then both start accelerating at a constant "A" ly/y/y toward the right. He knows that his helper friend (the HF) is then ageing at a constant rate that is slower than his own rate of ageing, by the factor exp(Ad).

An instant later, his display shows the time 20 + epsilon_1, where epsilon_1 is a very small positive number. He knows that HF's display shows the time 20 + epsilon_2, where epsilon_2 = epsilon_1 / exp(Ad).
She can still see HF's display (because HF has only moved an infinitesimal distance away from her, to her right). She will see that HF's display reads 20 + epsilon_1 / exp(Ad). And likewise, HF can still see her display. What does HF see on her display? Does HF see that she is now slightly younger than 20? No! It would clearly be absurd for someone essentially co-located with her to see her get younger. HF would see her display reporting that she was some very small amount epsilon_3 OLDER that she was at the instant before the acceleration. HF then sends a message to him, telling him that she was 20 + epsilon_3 right then. When he receives that message, he then knows that her current age, when he was 20 + epsilon_1, was 20 + epsilon_3. So he KNOWS that she didn't get younger when he accelerated away from her. That contradicts what CMIF simultaneity says.

Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 14/06/2021 18:57:47
[...]
What does HF see on her display? Does HF see that she is now slightly younger than 20?
[...]

Since my argument makes use of very small (unspecified) quantities,
it could be argued that time delays due to the speed of light might also need to be taken into account when describing what the HF sees on her display. But I think any such concerns can be addressed by pointing out that the separation "d" between him and her can be made arbitrarily large, and CMIF simultaneity says that the amount of negative ageing that occurs is proportional to their separation. Since the errors involved due to the finite speed of light between her and the HF are independent of the distance "d", those errors become negligible for sufficiently large "d".

 
Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 07/08/2021 18:02:43

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

Michael Leon Fontenot


According to the CMIF (Co-Moving-Inertial Frames) simultaneity method, an observer (he) who accelerates in the direction away from a distant person (she) will conclude that she rapidly gets YOUNGER during his acceleration. But I think I may have found a counterexample that shows that such an accelerating observer does NOT conclude that.

It is well-known that two stationary clocks at different positions in a gravitational field will run at different rates. The clock that is closer to the source of the gravitational field will run slower than the clock that is farther from the source of the field.

Because of the equivalence principle, it is also true that if two clocks that are separated by a fixed distance "d" ly are both accelerated with a constant equal acceleration of "A" ly/y/y, the trailing clock runs slower than the leading clock, by the factor exp(Ad).

So consider the following scenario:

At some instant, the perpetually-inertial "home twin" (she) is 20 years old, and is holding a display that always shows her current age. Facing her is the "helper friend" (the "HF") of an observer (he) who is "d" ly away to her right. Both the HF and he are also 20 years old, and are stationary wrt her at that instant. Like her, he and the HF are each holding a display that always shows their current ages.

Now, suppose that he and his helper then both start accelerating at a constant "A" ly/y/y toward the right. He knows that his helper friend (the HF) is then ageing at a constant rate that is slower than his own rate of ageing, by the factor exp(Ad).

An instant later, his display shows the time 20 + epsilon_1, where epsilon_1 is a very small positive number. He knows that HF's display shows the time 20 + epsilon_2, where
epsilon_2 = epsilon_1 / exp(Ad). Epsilon_2 is less than epsilon_1, but is also positive. She can still see HF's display (because HF has only moved an infinitesimal distance away from her, to her right). She will see that HF's display reads 20 + epsilon_1 / exp(Ad). And likewise, HF can still see her display. What does HF see on her display?
Does HF see that she is now slightly younger than 20? No! It would clearly be absurd for someone essentially co-located with her to see her get younger. What HF would see her display reporting is that she was now some very small amount epsilon_3 OLDER that she was at the instant before the acceleration. HF then sends a message to him, telling him that she was 20 + epsilon_3 right then. When he receives that message, he then knows that her current age, when he was 20 + epsilon_1, was 20 + epsilon_3, which is greater than 20. So he KNOWS that she didn't get younger when he accelerated away from her. That contradicts what CMIF simultaneity says.

In the above, I asked

"What does HF see on her display?".

And I answered

"HF would see her display reporting that she was some very small amount epsilon_3 OLDER that she was at the instant before the acceleration."

Since the above argument makes use of very small (unspecified) quantities, it could be argued that time delays due to the speed of light might also need to be taken into account when describing what the HF sees on her display.

But I think any such concerns can be alleviated by pointing out that the separation "d" between him and her can be made arbitrarily large, and CMIF simultaneity says that the amount of negative ageing that occurs is proportional to their separation. Since the errors involved due to the finite speed of light between her and the HF are independent of the distance "d", those errors become negligible compared to the change in her displayed age seen by the HF, for sufficiently large "d".

If the CMIF simultaneity method is incorrect, as the above proof contends, what is the alternative? I am aware of only a single alternative that (like the CMIF method) obeys the principle of causality. I describe that alternative in my paper, “A New Simultaneity Method for Accelerated Observers in Special Relativity”, on the viXra repository:

http://viXra.org/abs/2106.0133

I've put the above possible proof that there is no negative ageing (and thus that the CMIF simultaneity method can't be correct) on the viXra repository:

http://viXra.org/abs/2106.0142

There is another argument that shows that the HF ("Helper Friend") can't conclude that the home twin (she) is less than 20 years old when the HF is 20 + epsilon_2. We can require that she transmits NO light messages to him when she is 20 years old or younger. Suppose the HF receives a light message from her when he is 20 + epsilon_2 years old. By the requirement, she must have been older than 20 years old when she sent that message. When the HF receives that message, he knows that she must be older than when she sent the message, so she must definitely be older than 20 years old when the HF is 20+epsilon_2. Therefore, she did NOT get younger, according to him, when he accelerated away from her.

A still simpler argument is that, if the HF ever concluded that she got younger when he accelerated away from her, he would be concluding that she was less than 20 years old at that instant of his acceleration. But the HF was co-located with her when she was less than 20, and he couldn't be two places a that same instant.

It seems to me that, once the distant accelerating observer has a way to set up an array of clocks (with attending observers) that he can use to define his concept of "NOW" (analogous to how Einstein did it for perpetually-inertial observers), it becomes impossible for the home twin to age negatively, according to the distant accelerating observer. It's true that those clocks aren't synchronized as they are in the perpetually-inertial case, but they don't have to be, since the distant accelerating observer knows exactly how the rates of those clocks compare to his own clock.

I suspect that the same type of argument can be used to show that the essentially instantaneous (positive) ageing of the home twin (according to the traveler who instantaneously changes his velocity) also cannot occur. If these arguments are correct, then the commonly-used CMIF (Co-Moving Inertial Frames) simultaneity method can't be correct.
Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: Halc on 08/08/2021 01:27:02
Topic has been merged because
1) It violates posting guidelines for the main sections of the forum
2) It’s the same as this prior topic with which it has been merged

A Possible Proof That Negative Ageing Doesn't Occur In Special Relativity
SR does not conclude or assert any particular preferred coordinate system, and thus SR would not make any frame-independent statement such as “Negative aging doesn’t occur”.

SR does not bother discussing something like the CMIF coordinate system, which is a coordinate system based on a time-like worldline, because such a coordinate system does not seem useful for any useful purpose for making any predictions.

Quote
It is well-known that two stationary clocks at different positions in a gravitational field will run at different rates. The clock that is closer to the source of the gravitational field will run slower than the clock that is farther from the source of the field.

Because of the equivalence principle, it is also true that if two clocks that are separated by a fixed distance "d" ly are both accelerated with a constant equal acceleration of "A" ly/y/y, the trailing clock runs slower than the leading clock, by the factor exp(Ad).
Equivalence principle is applied incorrectly in multiple ways. In gravity, the thing further down experiences greater proper acceleration if it is to maintain the same proper distance from the thing higher up.
If you have say two observers (Ted at the head, higher up, Bob at the bottom) both with identical proper acceleration, moving at the same velocity at some time T, then in the CMIF of either observer’s worldline, the proper separation between the two after local time T will increase, which is why the string breaks in Bell’s spaceship scenario. You should know these basics.

If the acceleration is 1G and the separation 1 LY at T, then relative to Ted’s CMIF of Ted, Bob ages backwards.

The rest of the post seems to rest on these invalid assertions, so it ends here. The CMIF method isn’t ‘incorrect’ as you conclude, but merely useless.

From the OP:
Because of the equivalence principle, it is also true that if two clocks that are separated by a fixed distance "d" ly are both accelerated with a constant equal acceleration of "A" ly/y/y,
Under the equivalence principle, neither coordinate nor proper acceleration is identical between the two clocks. The upper.lead clock undergoes lower acceleration (of both kinds) than the lower/trailing clock.

Quote
Now, suppose that he and his helper then both start accelerating at a constant "A" ly/y/y toward the right.
But here you're giving them identical acceleration of one kind. You fail to specify if this is constant proper or coordinate acceleration, but the two points of measurement in the former example (with the equivalence principle) will maintain constant proper separtion, and will not do so with what you're doing here. Your conclusions seem to be based on the two situations being equivalent.

What you're essentially doing with the distant guy constant proper acceleration is to set up Rindler coordinates based on his acceleration. Under that, 'she' will not age negatively if she exists within his Rindler frame, and she will age negatively in those coordinates if not.

If the acceleration is constant coordinate acceleration, then the Rinder coordinates do not apply, and everybody's ages is a function of the chosen coordinate system and not a function of anything he is doing.
Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: puppypower on 08/08/2021 16:01:30
[Long rambling off-topic post removed]
Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 15/08/2021 15:32:57

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

Michael Leon Fontenot


According to the CMIF (Co-Moving-Inertial Frames) simultaneity method, an observer (he) who accelerates in the direction away from a distant person (she) will conclude that she rapidly gets YOUNGER during his acceleration. But I think I may have found a counterexample that shows that such an accelerating observer does NOT conclude that.
[...]

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.

Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: Bored chemist on 15/08/2021 22:51:44
A Possible Proof That Negative Ageing Doesn't Occur In Special Relativity
Had anyone said that it did?
Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: Halc on 15/08/2021 23:36:30
Had anyone said that it did?
Andromeda 'paradox' is one of the more classic examples.
I am walking one way, and simultaneously (relative to my immediate inertial frame), the Andromeda race has launched a fleet against Earth.  Now I turn around and walk the other way, and simultaneously (relative to my new immediate inertial frame), the Andromeda race is still in the process of making the decision to launch the fleet a month from now. By my act of turning around, I have 'caused' the Andromeda race to age (relative to me) negative one month.
These sorts of RoS things seem to keep Mike up at night, and all his threads are about ways to resolve this problem that isn't actually a problem.
Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 16/08/2021 15:37:32
Andromeda 'paradox' is one of the more classic examples.
I am walking one way, and simultaneously (relative to my immediate inertial frame), the Andromeda race has launched a fleet against Earth.  Now I turn around and walk the other way, and simultaneously (relative to my new immediate inertial frame), the Andromeda race is still in the process of making the decision to launch the fleet a month from now. By my act of turning around, I have 'caused' the Andromeda race to age (relative to me) negative one month.

That was a very good response.

Quote
These sorts of RoS things seem to keep Mike up at night, and all his threads are about ways to resolve this problem that isn't actually a problem.

I personally don't see it as a problem at all ... I prefer the CMIF simultaneity method to all other alternatives.  But the problem that I see is that many physicists DON'T accept the CMIF method, or at least the part of CMIF that says that the distant person can get younger, according to the accelerating observer.  Also, my recent proof seems to show (contradicting the CMIF method) that the distant person doesn't ever get younger (according to the accelerating person), and I'm still having trouble finding the error in that proof, if there is an error.
Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: Halc on 17/08/2021 00:54:15
But the problem that I see is that many physicists DON'T accept the CMIF method
Citation needed. From what I see, most physicists simply find no utility in a coordinate system based on a worldline with haphazard motion. Such a coordinate system does not foliate spacetime and thus doesn't assign unique coordinates to most events.
But they very much find utility in coordinate systems based on say continuous linear acceleration, despite having most of the same anomalies mentioned above.

Quote
the part of CMIF that says that the distant person can get younger
Depends on what you mean by getting younger.

A(bstract): If it is simply an abstract assignment of arbitrary 'current times' to various points on a worldline, there's nothing preventing this abstract action. There's no contradiction to say that it's Monday for me at your-now, but to assign the time of <your-now plus five minutes> to what is last Friday to me. It's just your abstract assignment of numbers to my worldline which concerns me not one hoot.

M(etaphysical): It is a very different thing to make a metaphysical statement that M1) I have an actual age, and M2) that this metaphysical designation of my current age is a function of non-local (or even local) events. I don't think you're making this (very idealistic) assertion, but your wording is open to interpretation. You calling it some kind of 'law of the land' suggests the latter interpretation. Your statement of "Time has to move to the future" suggests that you assert at least M1, in which case your CMIF method is completely wrong. The actual age of the distant person is whatever it actually is and only the one preferred coordinate that corresponds to reality, and any alternate 'according to X' coordinate system is simply wrong by definition.
Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 17/08/2021 20:01:09

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.


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.

Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 23/08/2021 18:59:58

I've had a new thought that is seeming "air tight" to me:

If some person (the "HF") is momentarily co-located with the home twin (she), the HF cannot possibly witness her age to instantaneously change at that instant, either positively of negatively. That would be an absurdity.

The distant accelerating observer (the "AO") is able to say that the HF always shares his (the AO's) notion of "NOW". Therefore the AO must also conclude that the home twin's age didn't instantaneously change (either positively or negatively) at that instant.

If the above argument is correct, then the CMIF simultaneity method can't be correct.
Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 25/08/2021 19:44:05

I've had a new thought that is seeming "air tight" to me:

If some person (the "HF") is momentarily co-located with the home twin (she), the HF cannot possibly witness her age to instantaneously change at that instant, either positively of negatively. That would be an absurdity.

The distant accelerating observer (the "AO") is able to say that the HF always shares his (the AO's) notion of "NOW". Therefore the AO must also conclude that the home twin's age didn't instantaneously change (either positively or negatively) at that instant.

If the above argument is correct, then the CMIF simultaneity method can't be correct.

The way the accelerating observer (the "AO") defines his "NOW" instant at distant locations comes directly from the gravitational time dilation equation, via the equivalence principle. It says that a "helper friend" (HF) who always is accelerating exactly as the AO is accelerating, with acceleration A, will age at a rate that is a fixed known ratio of the AO's rate. The given HF and the AO are always a constant distance "d" apart. If the chosen HF is BEHIND the AO (compared to the direction of the acceleration), that HF will age SLOWER than the AO by the factor exp(A d). To keep things as simple as possible, we can always let all of the HF's and the AO's ages be the same, immediately before they all start accelerating. Then the ratio of the age of the "behind" HF's age to the AO's age is just 1/exp(A d). And if, instead, another HF is AHEAD of the AO (compared to the direction of the acceleration), then the ratio of that "ahead" HF's age to the AO's age is just exp(A d). (Of course, different "behind" HF's will have different distances "d" to the AO, and likewise for the "ahead" HF's.) So, at some instant T in the AO's life, he computes that the original "behind" HF's current age is T/exp(A d). Or, alternatively, he computes that the "ahead" HF's current age is T exp(A d) The way he SELECTS the HF from among all possible HF's (both ahead and behind him) is such that the chosen HF is momentarily co-located with the home twin (her) at the instant the AO wants to know her current age.

So, if all of the above is correct, that allows the AO to construct an array of (effectively) synchronized clocks and helper observers attached to him, similar to what a perpetually-inertial observer can do, that can put an observer momentarily co-located with the distant twin (her) at the instant in the AO's life when he wants to know her current age. And in both the perpetually-inertial and the accelerated cases, it would be ABSURD for that momentarily co-located observer to observe a large and abrupt change in her age at that instant.
Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 27/08/2021 18:13:59
Because of the equivalence principle, it is also true that if two clocks that are separated by a fixed distance "d" ly are both accelerated with a constant equal acceleration of "A" ly/y/y, the trailing clock runs slower than the leading clock, by the factor exp(Ad).
You made the same assertion in June, the OP of this topic.
This makes no mention of gravity, so the equivalence principle is irrelevant.
I START with the "gravitational time dilation" equation.  Here's a link to a Wiki article on it:

 https://en.wikipedia.org/wiki/Gravitational_time_dilation

For a CONSTANT, UNIFORM gravitational field, the equation given in the Wiki article reduces to

 T_d(h) = exp(G h),

where T_d is the ratio of the tic rates of the two clocks, G is the strength of the gravitational force, and h is the constant separation of the two clocks.  The clock which is closest to the source of the gravitational field will tic slower than the other clock by the factor T_d.  (I have assumed that the units are such that the speed of light "c" is equal to 1, so for simplicity I don't include "c" in the equation.  I usually use units of years for time and lightyears for distance.

When the equivalence principle is applied to this GR equation, we get the SR equation

  T_d(h) = exp(A h),

where "A" is the constant acceleration that the two clocks are undergoing.  The clock in the rear tics at the slower rate.

Einstein was the first one, I believe, to derive these results, in 1907.
Here is a reference:

 https://einsteinpapers.press.princeton.edu/vol2-trans/315

Start with the bottom of that first page, labeled part V.
Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 27/08/2021 23:50:48

I've had a new thought that is seeming "air tight" to me:

If some person (the "HF") is momentarily co-located with the home twin (she), the HF cannot possibly witness her age to instantaneously change at that instant, either positively of negatively. That would be an absurdity.

The distant accelerating observer (the "AO") is able to say that the HF always shares his (the AO's) notion of "NOW". Therefore the AO must also conclude that the home twin's age didn't instantaneously change (either positively or negatively) at that instant.

If the above argument is correct, then the CMIF simultaneity method can't be correct.

I now see that the above argument is NOT correct. The fact that it is not correct DOESN'T prove that she never abruptly gets older, however. It may be true that she doesn't abruptly get older, but the above argument doesn't prove that.

I think the only way to prove that she doesn't abruptly get older, or to prove the opposite, is to make use of two DIFFERENT HF's. Immediately before the speed change, she is co-located with HF1. Immediately after the abrupt speed change, she is co-located with HF2. If she DOES get a lot older because of the speed change, HF2 will be far from HF1. If HF2 is close to Hf1, then she DIDN'T get much older. Which one of those outcomes actually happens isn't trivial to determine ... I think it can be done, but I think it will be complicated. It will require, among other things, looking at it from her perspective, and that is complicated by the fact that, according to her, the distances between the various HF's isn't constant. (Those distances ARE constant, according to the HF's and the AO.)

So, I've got a lot of work to do. But I HAVEN'T been able to find a flaw in my original proof that she can't abruptly get younger, according to the AO. So as of now, I think that proof is valid. If so, the CMIF simultaneity method is incorrect. And I also believe that if she can't abruptly get younger, then she can't abruptly get older either ... they either must both be true, or else they must both be false.
Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: Halc on 28/08/2021 01:14:28
The fact that it is not correct DOESN'T prove that she never abruptly gets older
Couple of issues with that wording.
1) I agree that only if the AO's velocity is discontinuous can her age relative to his worldline also be discontinuous (older, younger, whatever. Distance matters not so long as it is nonzero). Since it requires infinite force for AO's velocity to be discontinuous, it isn't going to happen.
2) Your conclusion " she never abruptly gets older" is worded as if it was some kind of metaphysical fact instead of a relation to something. So say "she never abruptly gets older relative to AO's worldline, or AO's CMIF as you seem to prefer naming it. A statement of simultaneity without a frame reference is either an absolute statement or is meaningless, and given your posts, I'm never quite sure if you're not some kind of solipsistic absolutist.

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I think the only way to prove that she doesn't abruptly get older
Technically, everybody is always 'getting older', so I think you need to be more precise in what you mean by 'abruptly'. It's not like we suddenly start doing it when we were not getting older before.
And again, lack of frame reference makes it sound like solipsistic absolutism.
Quote
Immediately before the speed change, she is co-located with HF1. Immediately after the abrupt speed change, she is co-located with HF2.
Ouch! Meaningless statements without a frame reference. You're talking I think about immediately after AO's speed change, meaning simultaneous at a distance with that speed change event. Simultaneity at a distance is meaningless without a frame reference. You have to learn to be clear about things, else we're all just guessing.

Quote
But I HAVEN'T been able to find a flaw in my original proof that she can't abruptly get younger, according to the AO.
Yay! A frame reference! She can get younger relative to AO, but not abruptly (instantly) due to my reasoning in point 1 above. She might lose 50 years in one of AO's seconds, but that's not abruptly.
Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 28/08/2021 14:47:47

The fact that it is not correct DOESN'T prove that she never abruptly gets older


Couple of issues with that wording.
[...]
2) Your conclusion " she never abruptly gets older" is worded as if it was some kind of metaphysical fact instead of a relation to something. So say "she never abruptly gets older relative to AO's worldline, or AO's CMIF as you seem to prefer naming it.

You are correct that I should have made it clear that I was referring to the conclusion of the AO.  I probably failed to do that because in my previous post that I was quoting, I had said: "Therefore the AO must also conclude that the home twin's age didn't instantaneously change (either positively or negatively) at that instant."  But I should have repeated that in my new post.  Sorry.

Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 01/09/2021 18:18:54

Instantaneous Velocity Changes in the Equivalence Principle Version of the Gravitational Time Dilation Equation

When using the CMIF simultaneity method, the analysis is GREATLY simplified by using instantaneous velocity changes, rather than finite accelerations that last for a finite amount of time.  So I decided to try using instantaneous velocity changes in the Equivalence-Principle Version of the Gravitational Time Dilation equation (the "EPVGTD" equation).  The result (assuming I haven't made a mistake somewhere) is unexpected and disturbing.  My analysis found that the age change of the HF, produced by an instantaneous velocity change by the AO and the HF, from zero to 0.866 lightseconds/second (ls/s), directed toward the home twin (her), is INFINITE!

I'll describe my analysis, and perhaps someone can find an error somewhere.

Before the instantaneous velocity change, the AO (he), HF, and the home twin (she) are all mutually stationary.  She and the HF are initially co-located, and the AO (he) is "d" lightseconds away from her and the HF.

I start by considering a constant acceleration "A" ls/s/s that lasts for a very short but finite time of "tau" seconds.  That acceleration over tau seconds causes the rapidity, theta, (which starts at zero) to increase to

  theta = A tau ls/s,

and so we get the following relationship:

  A = theta / tau.

We will need the above relationship shortly.

(Rapidity has a one-to-one relationship to velocity.  Velocity of any object that has mass can never be equal to or greater than the velocity of light in magnitude, but rapidity can vary from -infinity to +infinity.)

We want the velocity, beta, to be 0.866 ls/s after the acceleration.  Rapidity, theta, is related to velocity, beta, by the equation

  theta = arctanh (beta) = (1/2) ln [ (1 + beta) / (1 - beta) ].

("arctanh" just means the inverse of the hyperbolic tangent function.)

So velocity = 0.866 corresponds to a rapidity of about 1.317 ls/s.

The "EPVGTD" equation says that the acceleration A will cause the HF to age faster than the AO by the factor exp(A d), where d is the constant separation between the AO and the HF.

Note that the argument in the exponential exp(A d) can be separated like this:

  exp(A d) = [exp(d)] sup A,

where "sup A" means "raise the quantity exp(d) to the power "A" ".  The rationale for doing that is because the quantity exp(d) won't change as we make the acceleration greater and greater, and the duration of the acceleration shorter and shorter.  That will make the production of the table below easier.

The change in the age of the HF, caused by an acceleration "A" that lasts "tau" seconds is just

  tau [exp(d)] sup A,

because [exp(d)] sup A is the constant rate at which the HF is ageing, during the acceleration, and tau is how long that rate lasts.

But we earlier found that A = theta / tau, so we get

  tau [exp(d)] sup {theta / tau}

for the change in the age of the HF due to the short acceleration.  So we have an expression for the change in the age of the HF that is a function of only the single variable tau ... all other quantities in the equation (d and theta) are fixed.  We can now use that equation to create a table that shows the change in the age of the HF, as a function of the duration of the acceleration (while keeping the area under the acceleration curve constant).

In order to make the table as easy to produce as possible, I chose the arbitrary value of the distance "d" to be such that

  exp(d theta) = 20000.

Therefore we need

  ln[ exp (d theta) ] = d theta = ln (20000) = 9.903,

and since theta = 1.317, d = 7.52 lightseconds.

If we were creating this table for the CMIF simultaneity method, we would find that as the duration of the acceleration decreases (with a corresponding increase in the magnitude of the acceleration, so that the product remains the same), the amount of ageing by the HF approaches a finite limit.  I.e., in CMIF, eventually it makes essentially no difference in the age of the HF when we halve the duration of the acceleration, and make the acceleration twice as great.

But here is what I got for the EPVGTD simultaneity method:

(in the table, "10sup4" means "10 raised to the 4th power".)

  tau     |    (tau) (2000)sup(1/tau)

____________________________

  1.0     |     2x10sup4 = 20000

  0.5     |     2x10sup8

  0.4     |     2.26x10sup10

  0.3     |     6.3x10sup13

  0.2     |     0.64x10sup21

  0.1     |     1.02x10sup42

  0.01   |     1.27x10sup428

  0.001 |     ? (My calculator overflowed at 10sup500)

Clearly, for the EPVGTD simultaneity method, the HF's age goes to infinity as the acceleration interval goes to zero.  That seems like an absurd answer to me.  And it is radically different from what happens with CMIF simultaneity, where the HF's age quickly approaches a finite limit as tau goes to zero.

Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: Halc on 01/09/2021 19:40:33
The result (assuming I haven't made a mistake somewhere) is unexpected and disturbing.
If the result is nonsense, maybe that assumption is best not made. I always assume the mistake is with me if I get inconsistent results.

Quote
My analysis found that the age change of the HF, produced by an instantaneous velocity change by the AO and the HF, from zero to 0.866 lightseconds/second (ls/s), directed toward the home twin (her), is INFINITE!
The equation you're using gives the rate of age change relative to AO, but you are interpreting that as an age change, not a rate of age change. Simple as that. The difference in velocity (.866 in this case) is irrelevant since the same answer is computed from any abrupt positive change in velocity.

If you insist on using this formula for an abrupt change in velocity, do it as a limit as the time to change velocity by .866c approaches zero. But don't plug in zero to something that just yields NaN.

I stand unconvinced that it will yield a correct value since the formula is a local one being used in a non-local way.
Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 02/09/2021 00:22:01

The equation you're using gives the rate of age change relative to AO, but you are interpreting that as an age change, not a rate of age change.


No, I DID compute the HF's total age change during the acceleration.  If you look back at my post, you'll see that I said:

"The "EPVGTD" equation says that the acceleration A will cause the HF to age faster than the AO by the factor exp(A d), where d is the constant separation between the AO and the HF."

So there, I WAS talking about the rate.  But then I said:

"The change in the age of the HF, caused by an acceleration "A" that lasts "tau" seconds is just

  tau [exp(d)] sup A"

So there I AM talking about the total age change of the HF:  I multiplied the rate by the amount of time "tau" that he was ageing at that rate.

Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: Halc on 02/09/2021 02:46:15
If you look back at my post, you'll see that I said:

"The "EPVGTD" equation says that the acceleration A will cause the HF to age faster than the AO by the factor exp(A d), where d is the constant separation between the AO and the HF."

So there, I WAS talking about the rate.  But then I said:

"The change in the age of the HF, caused by an acceleration "A" that lasts "tau" seconds is just

  tau [exp(d)] sup A"
Which is zero times NaN which is NaN, not infinity. tau is zero for an instantaneous velocity change.

And my 2nd criticism also still stands.
Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 03/09/2021 20:34:54

Instantaneous Velocity Changes in the Equivalence Principle Version of the Gravitational Time Dilation Equation - Revised Model (the LGTD Model)

_______________________________________________________

I repeated my previous analysis of the instantaneous increase in the home person's (her) age (according to the accelerating person, AO, him), according to the Equivalence Principle Version of the Gravitational Time Dilation Equation, (the "EPVGTD" equation), and replaced it with the new equation, which I'll call the "Linearized Gravitational Time Dilation Equation", (the "LGTD" equation).  I simply replace the exponential exp(A d) with the quantity (1 + A d).  (This is the same approximation that Einstein used in his 1907 paper).  In what follows below, I'll repeat each affected calculation that I made in my last post, and show the revised calculation.

[...]

[Previous]:

The "EPVGTD" equation says that the acceleration A will cause the HF to age faster than the AO by the factor exp(A d), where d is the constant separation between the AO and the HF.

[Revised]:

The "LGTD" equation says that the acceleration A will cause the HF to age faster than the AO by the factor (1 + A d), where d is the constant separation between the AO and the HF.

(Both of the above are for the case where the AO accelerates TOWARD the unaccelerated person (her).)

[...]

[Previous]:

The change in the age of the HF, caused by an acceleration "A" that lasts "tau" seconds is just

  tau [exp(d)] sup A,

because [exp(d)] sup A is the constant rate at which the HF is ageing, during the acceleration, and tau is how long that rate lasts.

[Revised]:

The change in the age of the HF, caused by an acceleration "A" that lasts "tau" seconds is just

 tau (1 + A d),

because (1 + A d) is the constant rate at which the HF is ageing, during the acceleration, and tau is how long that rate lasts.


[Previous]:

But we earlier found that A = theta / tau, so we get

  tau [exp(d)] sup {theta / tau}

[Revised]:

But we earlier found that A = theta / tau, so we get

 tau (1 + [ ( theta d ) / tau ]  =  tau + (theta d)

[...]

It is still true that d = 7.52 lightseconds and theta = 1.317.

Therefore the revised result is that the change in HF's age during the acceleration is equal to

  tau + ( theta d ) = tau + (1.317)(7.52) = tau + 9.904.

So, in the revised model, as tau approaches zero (to give an instantaneous velocity change), the change in the HF's age during the speed change approaches 9.904 seconds from above.  So the HF's age increased by a finite amount, unlike the infinite increase that the EPVGTD equation gave.

Before the instantaneous velocity change, the AO, the HF, and the home twin (she) were all the same age.  She and the HF were co-located.  So after the instantaneous speed change, the AO hasn't aged at all, but the HF is 9.904 seconds older than he was before the speed change, according to the AO.  And since she and the HF have been colocated during the instantaneous speed change, they couldn't have ever differed in age during the speed change ... it would be absurd for either of them to see the other have an age different from their own age at any instant.  So after the instantaneous speed change, the AO must conclude that she and the HF both instantaneously got 9.904 seconds older than they were immediately before the speed change.

By comparison, the CMIF simultaneity method says that the AO will conclude that her age instantaneously increases by 6.51 seconds, so the LGTD and CMIF don't agree.

Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 05/09/2021 14:14:30

Wait. The HF spends 9.9 proper seconds (years, units, whatever) accelerating to .866c and you assert that he's still co-located with her after that?  Try again.


The speed change is instantaneous.  Neither she nor the HF move during the instantaneous speed change.  And she and the HF are colocated immediately before the instantaneous speed change.


Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 05/09/2021 20:56:49

I just repeated my previous analysis of instantaneous velocity changes in the "linearized" (LGTD) version of the equivalence principle version of the gravitational time dilation equation, but for the case where the instantaneous velocity change is AWAY FROM the home twin (her). The result is exactly like the previous result, except that she instantaneously gets YOUNGER, not older. (This contradicts my previous possible proof that negative ageing doesn't occur.)

Below, I'll repeat the previous calculations, and show the changes.

[Previous]:

I simply replace the exponential exp(A d) with the quantity (1 + A d).

[New]:

I simply replace the exponential exp(-A d) with the quantity (1 - A d).

[Previous]:

The "LGTD" equation says that the acceleration A will cause the HF to age FASTER than the AO by the factor (1 + A d), where d is the constant separation between the AO and the HF.

(The above is for the case where the AO accelerates TOWARD the unaccelerated person (her).)

[New]:

The "LGTD" equation says that the acceleration A will cause the HF to age SLOWER than the AO by the factor (1 - A d), where d is the constant separation between the AO and the HF.

(The above is for the case where the AO accelerates AWAY FROM the unaccelerated person (her).)

[...]

[Previous]:

The change in the age of the HF, caused by an acceleration "A" that lasts "tau" seconds is just

tau (1 + A d),

because (1 + A d) is the constant rate at which the HF is ageing, during the acceleration, and tau is how long that rate lasts.

[New]:

The change in the age of the HF, caused by an acceleration "A" that lasts "tau" seconds is just

tau (1 - A d),

because (1 - A d) is the constant rate at which the HF is ageing, during the acceleration, and tau is how long that rate lasts.


[Previous]:

But we earlier found that A = theta / tau, so we get

tau (1 + [ ( theta d ) / tau ] ) = tau + (theta d)

[New]:

But we earlier found that A = theta / tau, so we get

tau (1 - [ ( theta d ) / tau ] ) = tau - (theta d)

[both Previous and New]:

It is still true that d = 7.52 lightseconds and theta = 1.317.

[Previous]:

Therefore the revised result is that the change in HF's age during the acceleration is equal to

tau + ( theta d ) = tau + (1.317)(7.52) = tau + 9.904.

[New]:

Therefore the revised result is that the change in HF's age during the acceleration is equal to

tau - ( theta d ) = tau - (1.317)(7.52) = tau - 9.904.

[Previous]

So, in the revised model, as tau approaches zero (to give an instantaneous velocity change), the change in the HF's age during the speed change approaches 9.904 seconds from above. So with an instantaneous velocity change, the HF's age INCREASED instantaneously by a finite amount.

[New]

So, in the revised model, as tau approaches zero (to give an instantaneous velocity change), the change in the HF's age during the speed change approaches -9.904 seconds from above. So with an instantaneous velocity change, the HF's age DECREASED instantaneously by a finite amount.

[Previous]:

Before the instantaneous velocity change, the AO, the HF, and the home twin (she) were all the same age. She and the HF were co-located. So after the instantaneous speed change, the AO hasn't aged at all, but the HF is 9.904 seconds OLDER than he was before the speed change, according to the AO. And since she and the HF have been colocated during the instantaneous speed change, they couldn't have ever differed in age during the speed change ... it would be absurd for either of them to see the other have an age different from their own age at any instant. So after the instantaneous speed change, the AO must conclude that she and the HF both instantaneously got 9.904 seconds OLDER than they were immediately before the speed change.

By comparison, the CMIF simultaneity method says that the AO will conclude that her age instantaneously increases by 6.51 seconds, so the LGTD and CMIF don't agree.

[New]:

Before the instantaneous velocity change, the AO, the HF, and the home twin (she) were all the same age. She and the HF were co-located. So after the instantaneous speed change, the AO hasn't aged at all, but the HF is 9.904 seconds YOUNGER than he was before the speed change, according to the AO. And since she and the HF have been colocated during the instantaneous speed change, they couldn't have ever differed in age during the speed change ... it would be absurd for either of them to see the other have an age different from their own age at any instant. So after the instantaneous speed change, the AO must conclude that she and the HF both instantaneously got 9.904 seconds YOUNGER than they were immediately before the speed change.

By comparison, the CMIF simultaneity method says that the AO will conclude that her age instantaneously decreases by 6.51 seconds, so the LGTD and CMIF don't agree.

Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 07/09/2021 19:10:13

So, what to make of all these different and contradictory results?

The "EPVGTD Equation" (the one with the exponential), says that, if the AO (he) instantaneously changes his velocity in the direction TOWARD the home time (her), she instantaneously gets INFINITELY older, according to him.  That's nonsense, because it gives incorrect ages for the twins when they are reunited.

On the other hand, when he instantaneously changes his velocity in the direction AWAY from her, the EPVGTD equation says that her age doesn't change instantaneously.  While it's not certain that that result itself is incorrect, it seems to result in an inconsistency at the reunion. The EPVGTD equation says that, with zero acceleration, he and all the HF's age at the same rate.  That seems to require that on the outbound and inbound legs, his conclusion about the correspondence between his and her ages must be the same.  And clearly, on the OUTBOUND leg, he MUST say she is ageing SLOWER than he is, by the factor gamma.  So he must say that, on the INBOUND leg, she is ageing slower by the factor gamma.   But in that case, their conclusions about the correspondence between their ages at the reunion won't be consistent: she says she is the OLDER, but he says she is the YOUNGER. So his conclusions won't match her conclusions at the reunion, which is impossible since they are colocated then and they MUST agree about the correspondence between their ages then.

So much for the EPVGTD equation. What about the LGTD equation? The linearized equation (the LGTD equation) gives results that are qualitatively similar to the CMIF simultaneity method: her age instantaneously changes, according to him, during his instantaneous velocity change (instantaneously increasing when his momentarily infinite acceleration is TOWARD her, and instantaneously decreasing when his momentarily infinite acceleration is AWAY FROM her).  But the AMOUNT of the instantaneous change is greater than CMIF says it should be.  It is interesting that the amount of the instantaneous age changes would be exactly the same for CMIF and LGTD if the linearized equation multiplied the distance "d" by the velocity "v", rather than by the rapidity "theta".  But, in determining the velocity effect obtained by integrating the acceleration "A", it IS necessary to use the rapidity "theta", not the velocity "v", as the variable of integration. (Taylor and Wheeler go over this in detail).

WHY does the EPVGTD equation fail so miserably in this example? Isn't the GTD equation a well-established result in general relativity?  And the equivalence principle is certainly well-established.  Is the GTD equation WRONG?

And WHY goes the LGTD work better than the EPVGTD, at least qualitatively?  The LGTD should be a justified approximation of the EPVGTD only when the argument (A d) is small, and an infinite "A" (even though it lasts only an infinitesimal time) certainly isn't small!  The LGTD equation shouldn't give results that are even qualitatively correct, but it does. Why?

Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: Halc on 08/09/2021 01:57:44
The answers to your problems have already been posted, and perhaps you should read them rather than dismiss them. I will repeat them, attempting to flesh it out for additional clarity.

The HF spends 9.9 proper seconds (years, units, whatever) accelerating to .866c and you assert that he's still co-located with her after that?  Try again.
The speed change is instantaneous.
Mostly meaningless statement. OK, AO's speed change is instantaneous relative to AO or to the UO (unaccelerated observer, or 'she'), but not relative to HF.
A statement of simultaneity without a frame reference is either an absolute statement or is meaningless
Your statement lacked a reference, and what is going on would become much more clear to you if you would include them. A meaningful statement would be that the speed change is instantaneous relative to AO. It is not instantaneous relative to HF for instance since by definition HF and AO are always stationary in their proper frame, and HF takes t=9.9 to complete the acceleration.

Quote
Neither she nor the HF move during the instantaneous speed change.
Again no references, so again meaningless. To reword:
Relative to AO's ARF (accelerated reference frame), HF does not move because his proper separation from AO would change if he did move. Likewise AO does not move relative to HF's ARF (the same ARF). Relative to that ARF, the UO very much does move since she moves to a point halfway (3.76) between AO and HF and ages 6.51 in the process. So HF and UO do not remain colocated.
Relative to HF there is t=9.9 of constant proper acceleration away from AO and UO after which AO is still 7.52 behind and UO is 3.76 behind.
Relative to UO's inertial frame, UO is stationary and nobody goes anywhere during the AO's instantaneous acceleration event.  But HF commences acceleration away from AO, so over time he departs company from UO.

So things are quite different depending on the point of view. AO and HF always share the same ARF but different potentials in it, so their time does not remain synced.

So, what to make of all these different and contradictory results?
The problems in this post are mostly using correct equations in inappropriate situations.

Quote
The "EPVGTD Equation" (the one with the exponential), says that, if the AO (he) instantaneously changes his velocity in the direction TOWARD the home time (her), she instantaneously gets INFINITELY older, according to him.
No, it says undefined, not infinite. This is correct since the inputs are acceleration and duration instead of a celerity change. The former do not define a unique change in celerity.

Quote
The linearized equation (the LGTD equation) gives results that are qualitatively similar to the CMIF simultaneity method: her age instantaneously changes, according to him, during his instantaneous velocity change
This is what I mean by using it inappropriately. The LGTD equation is for comparing two locations at constant proper separation in an accelerated reference frame, and she (UO) does not do this, so the equation simply does not apply. It only works for HF who does maintain this constant proper separation per you specification. He’s at the front end of a rigid ship with AO at the rear, accelerating past UO.  After t=6.51 of UO, she sees the midpoint of the ship go by at .866c and the abrupt cessation of acceleration of the thing.

UO is inertial, and thus the equation appropriate for her (relative to AO) is the CMIF maths just like we do with the twins scenario.

Quote
WHY does the EPVGTD equation fail so miserably in this example?
It was spot on actually, but only applicable to what HF does relative to AO.
Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 26/09/2021 15:23:59
The complete title should be:
  "An Inconsistency Between the Gravitational Time Dilation Equation and the Twin Paradox"

_______________________________________________

I've pulled together the material I've posted recently on the "A possible proof" thread about the gravitational time dilation equation being inconsistent with the outcome at the reunion of the twin paradox, and I've put it on the viXra repository:

http://viXra.org/abs/2109.0076?ref=12745236

Here is the title and abstract:

"An Inconsistency Between the Gravitational Time Dilation Equation and the Twin Paradox"

Abstract:

It is shown in this monograph that the Gravitational Time Dilation Equation, together with the well-known Equivalence Principle relating gravitation and acceleration, produce results that contradict the required outcome at the reunion of the twins in the famous twin ‘paradox’. The Equivalence Principle Version of the Gravitational Time Dilation Equation (the “EPVGTD” equation) produces results that say that, when the traveling twin (he) instantaneously changes his velocity, in the direction TOWARD the distant home twin (her), that he will conclude that her age instantaneously becomes INFINITE. It is well known that, according to her, at their reunion, she will be older than him, but both of their ages will be FINITE. The twins clearly MUST be in agreement about their respective ages at the reunion, because they are co-located there.


<amazon sales plug removed>

When my manuscript comes up, you can click on "Look Inside", and on the back cover, there is a small photo of yours truly, impersonating an old codger.

The remaining question that I'd like the answer to, is "Has the gravitational time dilation equation been experimentally confirmed for arguments of the exponential function that are large enough to test the non-linear portion of the exponential curve?"  I.e., is it possible that the "modified linearized Gravitational Time Dilation equation (the MLGTD equation, in Section 7 of my monograph) is the correct GDT equation, and that the exponential version is incorrect?
Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 01/04/2022 16:27:29
Anybody is free to use any coordinate system they want, and if they both choose the same coordinate system, then they'll agree about their respective ages at all times. So for instance, when I'm in the car traveling to Grandma's house, I'm moving relative to Grandma, but both of us probably choose the coordinate system of the ground below us and hence are not in disagreement about our respective ages even while spatially separated.

At any given instant "tau" in the
life of a given INERTIAL observer (he), it's clear that there is just a
single answer to the question "How old is that particular distant person
(she) right now (at the given time "tau" in the life of the inertial
observer): it is what the particular "Helper Friend" (HF) who happens to be momentarily co-located with the distant person (she), says it is, at the instant when he is age "tau". The only way there could be any other
allowable answer is if the synchronization of the clocks isn't valid,
and that is impossible if the velocity of light in that inertial
reference frame is equal to the universal constant "c".

My argument above is that, IF those clocks are synchronized (according to the given observer), then he can't help but conclude that the current age of that distant person IS completely meaningful TO HIM.  And the only way that those clocks AREN'T synchronized according to him, is if the velocity of light in his inertial reference frame ISN'T equal to the universal constant "c".  But the fundamental assumption of special relativity IS that light will be measured in all inertial reference frames to have the value "c".  Therefore, FOR any given inertial observer (he), the current age of a distant person is completely meaningful to him.

But what about a non-inertial observer?  In particular, what about a
given observer who is undergoing a constant acceleration?  What does HE say the current age of a distant person is?  It turns out to be possible for such an accelerating observer to rely on an array of clocks and associated "helper friends" (HF's) to give him the answer.  Unlike in
the inertial case, those clocks DON'T run at the same rate.  But the
ratio of the rates of those clocks can be CALCULATED by the given
observer.  And if he (and the HF's) are initially stationary and
unaccelerated, they can start out with synchronized clocks (and ages).
Then, if they all fire their identical rockets at the same instant, they
can each CALCULATE the current reading of each of the other clocks, at each instant in their lives.  The calculations of each of the HF's all
agree.  So, at any instant in their lives during that acceleration, they
each share the same "NOW" instant with all of the other HF's.  That
means that the given observer (he), at any instant "tau" in his life,
can obtain the current age "T" of some distant person (her), by asking
the HF, who happens to be momentarily co-located with her at that NOW instant, what her age is then.


Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 02/04/2022 16:20:43

In the above post, I said that the given accelerating observer (he) (abbreviated, the "AO"), at each instant of his life, can CALCULATE the current reading on each of the HF's clocks.  What IS that calculation?

Let t = 0  be the reading on his clock at the instant that the constant acceleration "A" begins, and let all the HFs' clocks also read zero at that instant.  Thereafter, he and all of the HFs are accelerating at "A" ls/s/s, and the ratio R of any given HF's clock rate to his (the observer's (he) whose conclusions we are seeking) clock rate is

  R(t)  =  [ 1 +-  L  A  sech^2 (A t) ],

where L is the constant distance between him and the given HF, and sech() is the hyperbolic secant (which is the reciprocal of cosh(), the hyperbolic cosine).  The "^2" after the sech indicates the square of the sech.  The "+-" in the above equation means that the second term is ADDED to 1 for the HF's who are LEADING the accelerating observer, and the second term is SUBTRACTED from 1 for the HF's who are TRAILING the accelerating observer.  For brevity, I'll just take the case where the HF of interest is a leading HF.

The limit of R(t), as "t" goes to zero, is 1 + L A.  The limit of R(t), as "t" goes to infinity, is 1.0  So R(t) starts out at some positive number greater than 1, and then approaches 1.0 as t goes to infinity. So eventually, all the clocks essentially tic at the same rate, but early in the acceleration, the ratio of the tic rates varies significantly with time.

The current reading of the HF's clock (the "Age Change" or "AC"), when the AO's clock reads "tau", is

  AC(tau)  =  integral, from zero to tau, of { R(t) dt }

                 =  tau  +  L tanh( A tau ).

The above result depends on the fact that

  sech^2(u) = d{tanh(u)} / d{u}.

As tau goes to zero, AC goes to zero.  As tau goes to infinity, AC goes to tau + L, which goes to infinity, approaching a slope of 1.0 from above.

So there you have it.  That's the calculation that defines "NOW" for the AO and all of the HF's, and makes simultaneity at a distance a meaningful concept for them.  Simultaneity at a distance is not a choice.

Title: Re: New Possible Proof That Negative Ageing Doesn't Occur In Special Relativity?
Post by: MikeFontenot on 02/04/2022 22:57:55

So there you have it.  That's the calculation that defines "NOW" for the AO and all of the HF's, and makes simultaneity at a distance a meaningful concept for them.  Simultaneity at a distance is not a choice.


But what does the above say about the current age of the home twin (she), according to the traveling twin (he), for each instant in his life on his trip? The answer is that the above equations give the same results as the Co-Moving-Inertial-Frames (CMIF) simultaneity method. That is very fortuitous, because the CMIF method is relatively easy to use. The value of the array of clocks discussed above (which establish a "NOW" moment for the accelerating observer that extends throughout all space) is that they GUARANTEE that the CMIF results are fully meaningful to the traveler, and that the CMIF method is the ONLY correct simultaneity method for him. He has no other choice.