# 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. ### Mike_FontenotRegistered Senior Member

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I DID check it, and it worked perfectly. I just checked it again, and it still works fine for me. Maybe whether it works or not depends on which country you are located in. In the version I referenced, here is a quote of the most important part, at the 24:46 point:

"Amazingly, the alien's time slice has swept back through 200 years of Earth's history, and now includes events that we consider part of the distant past, like Beethoven finishing his 5th symphony."

3. ### Mike_FontenotRegistered Senior Member

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Brian Greene DOESN'T agree that when the MCIF lines of simultaneity (LOS's) cross, they become "invalid". I don't agree with that either. It smacks of a "rule" created purely to eliminate results that one doesn't like.

5. ### Mike_FontenotRegistered Senior Member

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Brian Greene has some recent videos in which he still expresses exactly the SAME point of view as in the NOVA TV program. And he has provided two versions of those new videos: one without the math, for the general public, and the other with the math, for serious physics students.

7. ### Mike_FontenotRegistered Senior Member

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In general relativity, one-to-one charts are needed to be able to piece together different coordinate systems where they meet in spacetime. But in special relativity, a given accelerating observer needs only a single coordinate system, because it covers all of spacetime for him. All he requires in a coordinate system is that, for each instant in his life, it tells him the current age and position of all objects, and the CMIF system does that. There is no requirement that his coordinate system be one-to-one.

8. ### SsssssssRegistered Senior Member

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Print out a map of the center and eastern half of your town and another map of the center and western half of your town and lay the two maps side by side and you'll find that the mapped distance from the east to the west of the town center is double what it should be and distances in the town center are ambiguous because you can make mistakes and measure distances from one copy of point A to the other copy of point B. Anyone with any sense wouldn't lay the two partial maps out that way they'd cut the overlap off one map because you'd have to be dumb as a bag of hammers to deliberately set up a map with two copies of the same region on it. Is not having two copies of the same region on the same map a "rule" created purely to eliminate results that one doesn't like or is it just common sense?

As soon as simultaneity planes cross you have created a map of spacetime with two copies of the event where they cross. I'd say it's common sense you don't want that exactly like the map example and formally it is expressed as the map between coordinate space and real space being both one to one and onto i.e. a diffeomorphism. This is textbook GR stuff and you can look in chapter 2 of Sean Carroll's online GR notes for free if you want to confirm it so I very very much doubt that Greene believes non-diffeomorphic maps are fine whatever he chooses to gloss over in popularisations.

9. ### SsssssssRegistered Senior Member

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You are mistaking your obsession for what people in general want coordinates for which includes writing equations of motion and measuring distances and intervals and specifying values of fields and all of that is made much harder and more error prone if you don't require a diffeomorphism between coordinates and real spacetime.

If you have two coordinates for one event how do you guarantee that you assigned the same magnetic field value to both versions and why would you want to have to do that instead of just picking a diffeomorphism and letting the maths take care of it?

Last edited: Aug 9, 2021
10. ### Mike_FontenotRegistered Senior Member

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The CMIF simultaneity method, and my recent possible proof that the CMIF method is invalid, concern SPECIAL relativity, not GENERAL relativity.

I stand by what I said in the two quotes you gave in your last two posts.

11. ### Neddy BateValued Senior Member

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Yes, so using v=+0.866c, if the traveling twin (he) is 20 years old, he would say his stay-home twin sister (she) is 10 years old, and if at that point he jumps off the train he was riding so v=0.000c now, he will have to re-calculate and say that she is 40 years old. That is because all of her grid clocks are stationary on the ground and they are all synchronised to her age, and they are all displaying 40 years at that time, simultaneously. He has no choice but to agree that she must be 40 when he is stationary with respect to those grid clocks, because in that case they are synchronised and all displaying 40 at the same time.

Of course if he then jumps back on the train so v=+0.866c again (in the same direction as before), then he must recalculate again and say she is 10 years old again. He would not want to keep referring to those other clocks that he is no longer stationary with respect to, because when he is moving with respect to them they are not synchronised. Instead he would have to refer to the clocks on the train which are synchronised to each other, and all displaying his age 20 simultaeously in the train frame. Then he would have to figure out her age based on its relationship to the train clocks, and he figures she must be 10. And someday he will find proof that he was correct, because there can be a photograph of her celebrating her 10th birthday while a train clock displaying 20 happens to be passing her by.

Let's look carefully at what happened. The slope of the traveler's line of simultaneity changed twice. First it pointed to her being 10, then it pointed to her being 40, and then it pointed to her being 10 again. You can choose to call that "negative aging", but we know she does not age negatively in her own fame, or in any inertial reference frame, so the actual meaning of that statement is highly questionable.

There is no problem with CMIF simultaneity. It never produces any absurd result such as your method does. Namely, in the example above using v=+0.866c:

If the traveler is 20 years old and says his twin sister is 10 years old, and if at that point he jumps off the train so v=0.000c, your method would have him still saying that she is 10 years old (although aging faster now) instead of simply recalculating her age to be 40.

This means that you have one inertial person (him) claiming she is 10, even though he is stationary in an inertial frame where all of the frame clocks synchronised to her age are all simultaneously displaying 40 years, including one that might be located right next to him.

You want physicists to believe that SR was never clear about whether or not she is actually 40 years old in her own frame when all of those frame clocks synchronised to her age are simultaneously displaying 40. You think maybe SR was saying that she can be both 10 and 40 at the same time, in her own frame, which is absurd. You are attempting to make a mockery of SR and the physicists who understand it.

Last edited: Aug 9, 2021
12. ### SsssssssRegistered Senior Member

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Special relativity is a special case of general relativity and something that is a restriction on what you can do in the general theory is also a restriction on a special case of it.

Chaining momentarily comoving inertial frames works ok on an open tube around your male observer but I infer that your female observer and "helper friend" are outside that tube and not covered by those coordinates so you obviously get rubbish like time coordinates that reverse sense with respect to proper times when you try to use them where they don't work. That doesn't mean any actual clocks reverse so nobody reverse ages and the only reason you might think they would is if you try to use a broken coordinate system and assume it means something real. None of that ever needed proving.
Here you seem not to be constructing a single coordinate system from slices of multiple inertial frames but instead switching between global inertial frames which is fine but not what Mike_Fontenot is trying to do AFAICT. I think he's trying to build a single non-inertial rest frame that covers all of spacetime for an accelerated observer which can be done in several ways but not by chaining slices of inertial frames because you get loads of overlaps and reversed time coordinates. You can kind of do it (thinking about it you probably need to think about matching derivatives but you can use the spatial planes) as long as you keep proper acceleration below $\alpha$ and stay within $c^2/\alpha$ of the male observer's worldline but beyond that you need a better method.

13. ### Mike_FontenotRegistered Senior Member

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In my last post to you, I asked you this:

"But my question for you, Neddy, is this: Do YOU subscribe to CMIF simultaneity? I always thought you did, but I think lately, at least, you've been in the "no negative ageing" camp ... i.e., I think you hold that the accelerating observer (he) will never conclude that the distant person (she) is abruptly getting YOUNGER." If so, that would mean that you AREN'T a solid CMIF believer, because CMIF clearly does says that .

You didn't directly respond to my question, but what you say below seems to say that you DO believe that, according to the accelerating observer (he), that the distant perpetually-inertial "home-twin" (she) gets YOUNGER when he accelerates in the direction AWAY from her. So you ARE solidly in the CMIF camp.

Yes. He says that the distant person got older when he accelerated in the direction TOWARD her. That is solid CMIF.

Yes. He says that the distant person got YOUNGER when he accelerated in the the direction AWAY from her. That is solid CMIF.

Yes, he concludes that she got 30 years younger during his last acceleration, so her ageing was negative during that acceleration.

Of course we know that.

If my new proof is correct (that negative ageing doesn't occur when he accelerates in a direction AWAY from the distant person), then the CMIF method ISN'T correct. Maybe my proof has a flaw in it somewhere, but I haven't been able yet to find any flaw. Feel free to point out any flaw that you see in it.

14. ### phytiRegistered Senior Member

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Mike;

fig.1:
B is moving away from A at a constant speed
B needs synchronized clocks in his ref. frame. He uses the method defined in SR to synch c1 and c2 using light, shown as the blue rectangle. A observes c1 set before c2, but also that the c1 time code has a longer return path than that for c2.
The magenta L is the physical t' and x' axis for B (moving parallel to the Ax axis).
The green axis of simultaneity is a fictional mathematical tool for the purpose of calculation. It only exists after a signal is sent and returned to the sender, as shown.
B assumes a pseudo rest frame, and divides the round trip time in half.
If B had accelerated instantly at At=0 to the current speed, the aos would have rotated
instantly to the same angle, pointing to a negative value of t.
fig.2:
A continuous acceleration to the same speed produces a gradual rotation of the aos, with positive increments of t, but at a reduced rate (t1 to t3).
Both are altered perceptions in the mind of B, as a result of his motion.
Fig.1, instant acceleration is not an actual physical motion.
Fig.2, aos requires constant inertial motion.
B could use the equivalence principle to explain the acceleration portion as a g-field, slowing the A-clock rate.
An observer C moving parallel to A with no relative motion would not see any change in the A-clock.
SR is a theory of perception in addition to motion, but Einstein was only focused on the
dynamics/mechanics. Perception is what the observer thinks he is experiencing after analysis of all the sensory input and forming a conclusion. Reality confined to the mind.
Doppler shift is a common example. The source frequency doesn't change yet the receiver/observer thinks it does.

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15. ### Mike_FontenotRegistered Senior Member

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No one is moving away from anyone else at constant speed, in my proof that negative ageing doesn't happen. If you find a flaw in my proof, please describe the flaw.

16. ### Neddy BateValued Senior Member

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I did point out two major flaws in post #4. You can't use gravity in SR, and the equivalence principle only applies locally, not globally.

Last edited: Aug 10, 2021
17. ### Neddy BateValued Senior Member

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If I understand correctly, Mike_Fontenot wants the traveling twin to say that the age of his distant twin sister does not change (no positive aging, no negative aging) when the traveler either stops, or accelerates back toward her. He wants the distant clock times to remain continuous, even for an instantaneous acceleration profile. He is trying to make the people happy who object to the idea of negative aging, even though he himself claims that he has no problem with the concept of negative aging.

18. ### Neddy BateValued Senior Member

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It doesn't matter what anyone's opinion may be, the fact is that there is nothing wrong with the CMIF method of simultaneity.

Okay, so you agree that in the scenario I gave (v=+0.866c), the CMIF is nothing more than the traveler (he) first saying his distant twin (she) is 10 years old (as he is 20 years old), and then him jumping off the train (v=0.000c) and now him saying she is 40 years old, then finally him jumping back on the train (v=+0.866c) and him saying once again that she is 10 years old.

I assume you also agree that in the perpetually inertial reference frame in which she is stationary, when she is 40 years old, all of her synchronised clocks display 40 correct?*

Let's assume your proof is correct. Now, help me explain this to the traveler in the scenario you just agreed with. He has jumped off the train and sees a clock right next to him which displays 40. He knows from the above * important fact that in the perpetually inertial reference frame in which she is stationary, she is 40 years old when that clock displays 40. How shall we go about explaining to the traveler that you have "proof" that it would be "incorrect "for him to say she is 40 years old? Please, this is on the verge of madness.

Last edited: Aug 10, 2021
19. ### Mike_FontenotRegistered Senior Member

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I start out with this well-known result in GR:

"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."

The equivalence principle says that I can remove the constant gravitational field, and replace it with a constant acceleration. By doing that, I've converted a GR problem into an SR problem. And there's nothing "local" about it.

I maintain this is true in special relativity, without any qualification or restriction:

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). There are no limitations on "A" or "d".

20. ### SsssssssRegistered Senior Member

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This is obviously wrong because there's a Rindler horizon at $d=c^2/A$ behind the front observer and the redshift there is infinite. How did you derive it?

21. ### Neddy BateValued Senior Member

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If you look at Bell's Spaceship scenario, you will find that two clocks which start off a distance "d" apart from each other in their mutual rest frame will not maintain a distance of d between each other (as measured by either of their own distance rods) after they have both begun accelerating at the same proper acceleration. That is why the string breaks.

22. ### Mike_FontenotRegistered Senior Member

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If my proof is correct, then there can be no negative ageing when he accelerates in the direction AWAY from (the distant) her. I BELIEVE there can be no sudden positive ageing if there can be no sudden negative ageing ... I don't believe there can be one without the other. Therefore in the instantaneous velocity changes scenarios, there can be no instantaneous age changes (either positive or negative). I.e., the Minkowski diagram cannot have any discontinuities. Therefore the observer who sometimes accelerates can't agree with the perpetually-inertial observer with whom he is currently momentarily co-located and stationary. And he can't agree that her array of clocks, that he has become momentarily stationary with, are synchronized.

23. ### Neddy BateValued Senior Member

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Okay, but if there was no traveler at all, and only the perpetually inertial rest frame of the stay-home twin, you surely must agree that when she is 40 years old, her entire array of clocks displays 40, simultaneously in that reference frame, correct?