# 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. ### Neddy BateValued Senior Member

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That is not only true for a perpetually-inertial observer and her helpers. It is also true for ANY inertial observer who is stationary with respect to those clocks. ANY different answer will contradict those direct measurements.

I agree that it would be absurd to contend that. It seems to me, in light of the above requirement that he cannot "contradict those direct measurements," that he has to use the CMIF method. Any other answer will have him contradicting those direct measurements. I don't understand why you don't see that.

3. ### SsssssssRegistered Senior Member

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Is that me? I have eight Ss.

I think that the general point you are making is that there isn't a reciprocal clock synchronisation method for arbitrarily accelerating clocks and I agree because I think you need a symmetry that generates planes perpendicular to both worldlines and those are fairly limited. However there are ways to establish non-inertial rest frames that don't have the multiple/missing coverage problems that the naive MCIF method has.
My guess would be that Gull doesn't care about what you are asking enough to work it out because he knows simultaneity is arbitrary and a plot of proper time versus coordinate time which is what you want isn't particularly exciting. For what its worth I agree with your analysis and would add that the differently sloped region corresponds to the time that the inertial observer is outside both future and past lightcones of the turnaround.
You mean you want the age A assigns to B at some proper time $\tau$ on A's worldline to be greater than at $\tau-\delta\tau$ and not to depend on the acceleration history of A after $\tau$?
I don't see what's missing from what I wrote in the post you quoted where I showed that the planes perpendicular to the four velocity of an accelerating observer cross at that distance which is just the Minkowski space equivalent of showing that perpendiculars to a circular path cross at the center. You can look up Rindler coordinates if you want which are a coordinate system based on that fact but I don't know if they're in Taylor and Wheeler because I don't have my copy here.
Age decreases of any sort imply your simultaneity planes cross somewhere and that's always a problem. Age jumps imply that your coordinate system doesn't cover some part of spacetime which is also a problem as I said in my last reply to Neddy Bate and in either case you need to do something about it. The major difference is that positive age jumps disappear if you disallow instantaneous acceleration while double covered regions remain so "negative aging" is a problem while positive jumps are an artefact of an idealisation of acceleration.
The only absolutely true answer to that is that he sees a bit of blueshift in the middle of his redshift. There is no answer to what your age correspondence diagram (or proper time versus coordinate time) diagram shows unless you first define your simultaneity convention.
That's not right unless you add a requirement that the one way speed of light is isotropic which is an extremely requirement to the point where any othet choice pretty much has to be bloody mindedness but it is not obligatory to assume that and any other assumption tips and curves your simultaneity planes and makes maths unnecessarily difficult but has no measurable effects.
No there isn't any more than there is for an inertial observer although in both cases for eternally inertial or accelerating observers there is a set of simultaneity planes picked put by symmetry but again you are not required to use them and they don't apply if you ever change acceleration.
That's because it's the wrong thing to be looking at since the uniformly accelerating observers drop below Rindler horizons and are causally disconnected until there is an acceleration change. You really need to look at Rindler coordinates.

5. ### SsssssssRegistered Senior Member

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I agree and that's perfectly sensible but I don't think it's what Mike_Fontenot means by "the MCIF method" which is exactly the stitching method. You can always work in any inertial frame and that's a completely valid thing to do but what Mike_Fontenot wants is a non-inertial rest frame for the traveller with some other restrictions that I haven't figured out yet and he cannot do that by stitching inertial frames which is all I was pointing out.

A way to look at the differences is that you had an example where the traveller had $\gamma=2$ and said the stay at home was 10 just before turn around and 70 just after. My question is asked the day after turnaround. How old was the stay at home one year ago ny the traveller's clock? If I understand your positions you (Neddy Bate) would say 69.5 because you recalculate everything forever when you switch frames and you might add that you've changed frames recently snd would have said 9.5 if I'd asked you a couple of days earlier. Mike_Fontenot would say 9.5 but is struggling to find a general coordinate system to justify that.

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7. ### Neddy BateValued Senior Member

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Yes, I would agree 100% with this. Thank you.

8. ### Neddy BateValued Senior Member

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I agree that a portion of Gull's ACD starts to change slope before the turnaround, and it is a disqualifier for me as well. However, I don't think that has anything to do with causality. I think they compile the data well after the fact, and then fill in the data points later.

The reason I think this is because they say they rely on the "radar" method. I think that means they are willing to wait until much later when the radar signal has made a complete round-trip, and then retro-actively calculate the times from that.

I also think they might be allowing signals which were sent during the outbound leg of the journey, and subsequently received during the inbound journey, to be calculated together without correction for the fact that the turnaround happened in the meantime.

Last edited: Aug 22, 2021
9. ### SsssssssRegistered Senior Member

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Yes but that's true of any method because you can't know how old anyone is until you see their birthday party through your telescope and you are predicting their age until then because you can't know that the other person didn't get kidnapped by the Illuminati and stuffed into another fast spaceship specifically to screw around with your calculations so Mike_Fontenot's age correspondence diagram is either compiled after the fact at the same time Dolby and Gull are compiling their coordinate system or it's a prediction and subject to change as new information comes to light. If the female observer is one light year away her age at any moment is fixed up to one year ago and can never be changed due to Illuminati attack or anything else but both Mike_Fontenot and Dolby/Gull might have to update the parts after that and the only real difference is that Dolby and Gull might need to update their charts due to Illuminati kidnapping or due to their own acceleration but Mike_Fontenot wants a method that only needs to be revised in the case of Illuminati kidnapping.
Yes because the point is to construct a non inertial coordinate system where the radar set is at rest so by definition there is no correction for changes in the radar set's state of motion because Dolby and Gull chose to define it as always at rest. If you corrected for its acceleration then you'd be implicitly treating inertial observers in whose frame the radar set was initially stationary as the state of rest and you'd just recover their standard SR inertial frame by a devious route.

10. ### SsssssssRegistered Senior Member

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Stated more generally "her" age in "his" past lightcone is fixed in any reasonable system but at any time after she leaves his past lightcone her age is a prediction and subject to revision and Dolby and Gull and I don't mind revising our predictions due to our own actions after we've made them but you (Neddy and Mike) do.

11. ### Mike_FontenotRegistered Senior Member

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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.

12. ### phytiRegistered Senior Member

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Mike;

Here is a graphic for the A and B frames. It is a history of events for each. You can make your own ACD. B is the primed frame.
A-frame:
Since one observer cannot follow the discontinuous B speed profile, there are two, B1 and B2.
B1 records the A history from t=0 to t=2 using the 'radar' method with optical frequencies, for his inertial interval.
B2 records the A history from t=8 to t=10 using the 'radar' method with optical frequencies, for his inertial interval.
Neither one can use the 'radar' method for the interval t=2 to t=8. As a team, B1 could send signals to A, and B2 receive them.
B-frame:
Combining the B1, B2, and team histories, the drawing is symmetrical. The A-clock runs at .8 from t'=0 to t'=2.5, and t'=5.5 to t'=8. The A-clock runs faster from t'=2.5 to t'=5.5, at a rate of 6/3.

1. The 'radar' method is not new. It is the same method used for measurement in SR.
2. Clocks separated by large distances run independently of each other.
3. The current age of a distant person is unknowable as explained in Einstein's paper, covering simultaneity. You have to make a measurement via light, to get a roundtrip time that can be divided by 2, per the SR clock synch convention. That only gives an historical time, not a 'current' time. The answer is relative to the observer.

In the D&G paper, a gradual transition from outbound to inbound would cure the time jump/lost time. The multiple times for a single event is nonsense due to interpreting the aos/simult.planes as literal things. Another instance of making simple things overly complicated.

As to a network of synchronized clocks, how did they get there, and how did they get synchronized?
Quartz crystal time keeping devices originated about 1940.
Einstein couldn't have had such a system around 1900.

A. Einstein, 1905 paper:
"Thus with the help of certain IMAGINARY physical experiments we have settled what is to be understood by synchronous stationary clocks located at different places,"

In the A-frame, t'=.8t. When At=2, x=1.2.
A sends a signal at t=.8 which returns at t=3.2.
A calculates t'=1.6 is synchronous with t=2, but doesn't know until t=3.2.
Her knowledge is historical.

13. ### SsssssssRegistered Senior Member

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No for about the fourth time observers with the same proper acceleration do not share a notion of now unless they are colocated or that proper acceleration is zero for the same reason that non concentric circles do not share a radius. You should look up Rindler coordinates and Bell's spaceship paradox or just draw two identical hyperbolas and try to find lines that are Minkowski perpendicular to both (spoiler: there is exactly one which is the line through the two hyperbolas' centers but you need every line to do it).

14. ### SsssssssRegistered Senior Member

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Mike_Fontenot I think you are looking for a definition of "now" that

works for an observer with arbitrarily time varying acceleration
AND
may depend on the past acceleration history but not the future acceleration history of that observer
AND
never gives negative aging
AND
never gives age jumps even with infinite accelerations

Do I have that right?

15. ### Mike_FontenotRegistered Senior Member

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622
Yes.

Yes.

No, I was not looking for that. I actually like the CMIF method more than any other simultaneity method. But the argument I gave in my most recent post seems correct to me, and if it IS correct, I have no choice but to accept that the observer who instantaneously changes his velocity must conclude that the distant home twin's current age never changes instantaneously, either positively or negatively.

16. ### Mike_FontenotRegistered Senior Member

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The clocks of the "AO" and all his helpers are not synchronized (as they would be in a perpetually-inertial frame), but the relationship between the tic rates of the AO's clock and each of the helper's clocks IS known to him and his helpers. So the relationship between the AO's age and each of his helpers ages is KNOWN to the AO and to each of the helpers. That establishes de facto synchronization.

17. ### phytiRegistered Senior Member

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Mike;

U is the universal ref. frame containing Al and clocks at events e1 and e2.
Al accelerates to the right at -t1 until t0 (= now).
Common practice adds the green x', and states a clock at e1 is simultaneous with t0.
Al cannot know that since his signal sent before -t1 hasn't returned. The same is true for e2 and all events on x'.
The simultaneity convention does not magically provide times for distant locations.
Even if Al had remained at rest (magenta) the return signals have not returned.

18. ### Neddy BateValued Senior Member

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You were the one who did not even accept the idea that two inertial people with no relative motion between them can say that they have the same notion of "NOW", unless both of them were perpetually inertial. Now you claim the above nonsense? You are going in the wrong direction.

Let me know when you decide to accept that two inertial people with no relative motion between them can say that they have the same notion of "NOW" regardless of past acceleration history, and then we can talk. After all, it's just the special case of your above statement, with acceleration equal to zero.

Last edited: Aug 23, 2021
19. ### Neddy BateValued Senior Member

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2,540
I had mentioned Bell's spaceship paradox to Mike before also. In the reference frame of one of the spaceships, the distance between the spaceships does not remain constant either. His reply was that Bell does not apply to his case, because that distance remains constant in his case. But wouldn't that necessitate that they cannot have identical proper accelerations?

20. ### SsssssssRegistered Senior Member

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Distance remains constant if he continues to use the initial inertial frame of the ships ironically enough but not if he uses any other frame which is what I think he wants to do. My impression is that Einstein had not completely grasped this in 1907 although he'd clearly figured ut out by 1915 but it was certainly far from common knowledge among physicists as late as the 1960s when Bell promulgated his paradox.

21. ### Neddy BateValued Senior Member

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This is one of the reasons I mentioned earlier that the equivalence principle can only apply locally. I doubt that the distance would keep increasing like that for the equivalent case in a gravitational field, would it?

22. ### SsssssssRegistered Senior Member

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But that's true of any observer on a known course and there's nothing special about accelerating observers and the natural synchronisation (flat planes Minkowski perpendicular to the worldline) of an accelerating observer is singular at the center of the hyperbola.

Here's how I'd do this. You want a perpetually inertial observer to use an Einstein inertial frame so if we consider an observer who undergoes a single impulsive acceleration we must use an Einstein inertial frame at least until the acceleration event with "until" defined using that frame otherwise you'd need to retcon your age history due to the acceleration. Here's a Minkowski diagram with the accelerated worldline drawn in green and the simultaneity planes so far defined shown in blue and with the last line heavier to show it's a cutoff.

You want things to look more or less like an inertial frame again after the acceleration but we don't want planes to cross so that we avoid negative aging so why not steal an idea from Dolby and Gull and use new Einstein frame planes just in the future lightcone of the acceleration event? The Minkowski diagram below adds the lightcone in yellow and the new velocity's Einstein simultaneity planes inside the lightcone.

We don't have simultaneity planes in the parts to the left and right of the lightcone but we know we don't want the planes we define there to meet the old planes so they must be at least asymptotically parallel to the old planes so we might as well just make them parallel and we don't want any age jumps so we need them to be a continuous extension of the planes inside the lightcone. That gives us the diagram below where I've added the simultaneity planes in the gap and you can see an inertial observer on the left would age more rapidly in that region and an observer on the right more slowly.

That's it I think although you need to define distance which I think you could do by measuring proper distance in the planes. You can handle multiple acceleration events by nesting their lightcones and drawing Einstein frame simultaneity planes inside each one and simply extending them parallel to the planes of the previous period until they meet the next lightcone and repeating which looks like this for a 0.6c twin paradox where the lightcones are in yellow and the boundary of each inertial frame region is marked in heavier blue.

You can also handle smooth arbitrary acceleration by making a polygonal approximation to the worldline and considering the limit as the number of segments goes to infinity and figuring out the integrals. Here's a smoothish acceleration (actually it's a 32 line polygonal approximation because I haven't done the integrals) twin paradox where you can see the distortion of the simultaneity planes is restricted to the region between the future lightcones of the start and end of the acceleration except for some spacing variation.

And here's your jinking course with two sharp reversals in quick succession.

I think this system does everything you want. You can see that simultaneity planes only change direction because if the lightcones of past events and not the past lightcones of future events so you would only have to revise your age predictions if the other person turns out to have done something unexpected and it's mostly based off the current inertial frame with a twist and it is possible to write a formal definition of the simultaneity planes in terms of the integral curves of a vector field defined from lightcones and your observer's four velocity.

Last edited: Aug 23, 2021
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23. ### SsssssssRegistered Senior Member

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It isn't easy to get a uniform gravitational field in general relativity surprisingly. There's a paper by Jones and MuĂ±oz I think that I can dig out if you want to look where they derive two ways of getting one and there's either a non zero cosmological constant or some other complexity that I forget needed. So there isn't a simple real gravity equivalent of a long stack of Bell's spaceships accelerating at the same proper acceleration so you are correct that the distance doesn't keep growing but I would guess that neither would it for Bell's spaceships if they were in a universe with a carefully tuned large cosmological constant.

The simple case where distance doesn't change for the ships is a stack of spaceships with increasing proper acceleration "down" the stack. Their worldlines are hyperbolas with a common center and spaceships accelerating like that get closer to each other as seen in their initial inertial frame but remain a constant distance apart in their own MCIFs because the changing distance seen from the inertial frame is exactly the same as the length contraction.

Last edited: Aug 23, 2021