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 Bate Valued Senior Member

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    Yes, so the OP cannot have both constant spacing in their own frame and the same proper acceleration for both.

    Very nice diagrams, by the way. How in the world did you draw the one with curved lines?
     
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  3. Ssssssss Registered Senior Member

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    Yup.
    Thanks. All the graphs are drawn by a bit of python that you can give a list of velocities and durations and it figures out the locations of the acceleration events between the velocities and the equations of their lightcones and then it takes a list of times and initialises simultaneity planes and figures out where they intersect the light cones and how they bend there and draws them with matplotlib. In the curvy line one the worldline is actually 30-odd straight segments with 30-odd lightcones from the joins so it's not really curvy strictly speaking it's just a more complicated version of the one before with most of the lightcones not plotted. You could actually write code to handle arbitrary curved worldlines properly but you'd need to use a numerical integrator and solve more complex intercept equations and you wouldn't be able to see the difference.
     
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  5. Ssssssss Registered Senior Member

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    Sorry I got this the wrong way round I should have said an observer on the left hand side crosses more simultaneity planes than a similar observer on the right in the same lapse of their proper times so the one on the left ages the same in a greater lapse of coordinate time which means they age more slowly as seen by the accelerating observer if he uses this system.
     
    Last edited: Aug 24, 2021
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  7. Neddy Bate Valued Senior Member

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    Ssssssss,

    Using your method, can you tell me the home twin's age, according to the traveling twin, for my earlier scenario? It is the one where v=0.866c and the traveler stops (rapid deceleration to v=0.000c) when he is 20 years old. By the MCIF method he would say that she is 20/2=10 years old just before he decelerates, but then after he stops he would say she is 20*2=40 years old. Is it the same for yours?

    And then what about if he quickly accelerates back to v=0.866c at that point? By the MCIF method he would say that she is 20*2=40 years old just before he accelerates, but then after he quickly reaches speed he would say she is 20/2=10 years old again. Is it the same for yours?
     
  8. phyti Registered Senior Member

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    625
    forum;

    From "On Radar Time And The Twin Paradox", Dolby & Gull, 2008
    "Although this period of acceleration can indeed fix the gap between G and H, it cannot resolve the more serious problem (mentioned also in Marder[7] and in Misner et al.[8]) which occurs to Barbara’s left.
    Here her hypersurfaces of simultaneity are overlapping, and she assigns three times to every event!"

    The last line is false.
    The graphic shows an object M at rest relative to Alex. Barb sends a signal at event 1.
    She can't assign a corresponding time for event 2 until the return signal at event 3. Because of symmetry she assigns her clock event 4 to event 2.
    The local time is determined by current time signals, not planes of simultaneity. The local observer is present for emission and detection. The question is when does the distant reflection event occur. A second signal before event 4 shows the closer agreement with the typical SR clock synch convention.

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  9. dart_ship Registered Member

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    Ssssssss what display shows age?
     
  10. dart_ship Registered Member

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    Ssssssss birth certificate?
     
  11. dart_ship Registered Member

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    is a circle with a cross below a woman and a circle with an arrow top right a man?
     
  12. Ssssssss Registered Senior Member

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    Not quite. In my system the traveller only uses inertial coordinates inside the future lightcone of an acceleration event and until the time that frame calls the next acceleration event so in this case this system agrees with the MCIF method on the outbound leg but then there is no age jump and the stay at home twin ages rapidly until she is 75 which is when she enters the future lightcone of the traveller's stop and enters the next region where inertial frames are used. So in the MCIF method I think the home twin's age rises smoothly to 10 in the first 20 years of the traveller's time then jumps to age 40 then ages smoothly at one year per year of traveller's time but in my method the twin's age rises smoothly to 10 in the first 20 years then smoothly to 75 in the next 35 years then runs at one year per year the same as the MCIF method thereafter. Here's a Minkowski diagram with the traveller's planes at five year intervals shown and the stay at home added in blue and a few of her lines of simultaneity in dashed lines where you can see the traveller's 20 year simultaneity line crosses the blue line at 10 years and that the lightcone bounding the final inertial frame planes crosses the blue line at 75 years.

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    No her age would always increase. In this case where he stops for a bit her aging speeds up until he starts moving again which is when her aging rate reduces again but won't quite agree with the MCIF method because the temporary stop introduces a kink in the simultaneity planes that the MCIF simultaneity planes don't have and you can see the kink between the two lightcones in the Minkowski diagram below.

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  13. Ssssssss Registered Senior Member

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    It's not false because they're talking about the MCIF method where simultaneity planes are perpendicular to the worldline and not about the radar method that you appear to be constructing and in fact they are constructing the radar coordinate method specifically to avoid the issue.
     
  14. Ssssssss Registered Senior Member

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    None of the graphs I've posted show age except for the green worldline where you can calculate age at any event by counting the simultaneity planes the worldline crosses because they are drawn at one year intervals of the traveller's proper time but this method won't work for other observers although you can always draw an arbitrary worldline and integrate \(\sqrt{dt^2-dx^2}\) along it.
     
  15. dart_ship Registered Member

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    did you mean CMIF Ssssssss?
     
  16. Ssssssss Registered Senior Member

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    Momentarily Comoving Inertial Frame.
     
  17. QuarkHead Remedial Math Student Valued Senior Member

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    "Frames" (aka coordinate systems) do not move, not even momentarily.
     
  18. Ssssssss Registered Senior Member

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    True but you can define the inertial frame in which an arbitrarily moving object is momentarily at rest and that is called the MCIF.
     
    Last edited: Aug 25, 2021
  19. QuarkHead Remedial Math Student Valued Senior Member

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    Objects do not move in their rest frames, by definition, and likewise conversely
     
  20. Ssssssss Registered Senior Member

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    True.
     
  21. Ssssssss Registered Senior Member

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    Ah I see I made a typo and put "rest" in twice because I was going to say what I said differently so I'll correct that.
     
  22. Mike_Fontenot Registered Senior Member

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    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.
     
  23. Neddy Bate Valued Senior Member

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    Are you not even aware by now that no one ever observes any abrupt change in anyone's age? Even in the case where my 20 year old traveler calculates that his sister is now 10 years old, and then jumps off the train and calculates that she is now 40 years old, HE DOES NOT OBSERVE ANY ABRUPT CHANGE IN HER AGE.

    Using his telescope, he sees the same image of her before and after his instantaneous jump from v=0.866c to v=0.000c. So you can't disprove the age jump by talking about how no one observes an age jump. That "observation" is not part of the model in the first place.
     

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