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_Fontenot Registered Senior Member

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    Of course he doesn't. He isn't co-located with her at that instant. It's co-located observation that I'm talking about.

    Until now, the only way we knew how to get a "helper observer" ("HO") momentarily co-located with the home twin (her), with the HO sharing the same "now" instant with a distant observer, him, (whose opinion of her current age we are interested in), was for the case where he is a perpetually-inertial observer himself. And of course in that case, we know from the time-dilation result that he will always say that she is ageing gamma times slower than he is, and so the HO will just witness her slower ageing by the factor gamma.

    But the use of the equivalence principle on the gravitational time dilation equation allows us for the first time to arrange for an HO to be co-located with her, in the case where he is an ACCELERATING observer ("AO") who shares the same "NOW" moment with the HO. If her age were suddenly changing, according to him, when he accelerates, then the HO would be able to directly witness that. But it would be absurd for anyone co-located with her to witness any sudden age change by her. Therefore it doesn't happen.

    It is the ability, for the first time, to construct an array of clocks (and associated helper observers) for an accelerating observer, that is so exciting.
     
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  3. Neddy Bate Valued Senior Member

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    But SR never said there would be any 'drastic age change' at her location. The drastic age change requires there to be distance between her and her brother when her brother accelerates, and even then the drastic age change is something he says happens from his distant location, but she says never happens at her location. You are trying to disprove it by assuming that if it happens at his location, then it must also happen at her location, which is not how it works at all.
     
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  5. Neddy Bate Valued Senior Member

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    If you truly had an HO and AO sharing the same "now" in the moment BEFORE the 20 year old traveler jumps off the train, they both would agree that their own time is t'=20 and that her time is t=10. The HO could take a photo of her, and she would be 10 years old.

    Likewise, if you truly had an HO and AO sharing the same "now" in the moment AFTER the 20 year old traveler jumps off the train, they both would agree that her time is t=40. That HO could take a photo of her, and she would be 40 years old.

    You can do this easily by letting acceleration equal zero, and using the premise that as long as there is no relative velocity between the HO and the AO, then they share the same "NOW". So before he jumps off the train, he uses an HO on the train who photographs her at 10 years old serving as proof for examination later. And after he jumps off the train, he uses an HO on the ground who photographs her at 40 years old serving as proof fore examination later.

    She ages 30 years in between the two photographs, but there is also 30 years of time between the photographs in her frame, so she does not have any sudden age change. Instead it is his changing from one HO to the other HO that causes him to say that her age changed suddenly. He can quickly switch back and forth from using one HO to the other HO as many times as he wants to, by jumping off and on and off the train over and over.
     
    Last edited: Aug 26, 2021
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  7. Neddy Bate Valued Senior Member

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    Here is the sequence of events as they happen chronologically according to HER (times t are her clock times):
    (t=0) Male and female twins are born, and the boy jumps on a very long passing train.
    (t=10) An HO on the train takes a picture of her as he is passing close by her, and she is celebrating her 10th birthday in the picture.
    (t=40) A different HO on the ground takes a picture of her, and she is celebrating her 40th birthday in the picture. The traveler also jumps off the train at this time.

    Here is the sequence of events as they happen chronologically according to HIM (times t' are his clock times):
    (t'=0) Male and female twins are born, and the boy jumps on a very long passing train.
    (t'=20) An HO on the train takes a picture of her as he is passing close by her, and she is celebrating her 10th birthday in the picture. The traveler jumps off the train at this time, and the HO on the ground takes a picture of her celebrating her 40th birthday.
     
  8. Mike_Fontenot Registered Senior Member

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    I've read your arguments in your last three posts, and I think I understand what you are saying. I'll continue to give them more thought.
     
  9. phyti Registered Senior Member

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    S8;

    G&B
    "This formula simply says that an observer can assign a time
    to a distant event by sending a light signal to the event and back, and averaging the (proper) times of sending and receiving."

    So it's just reworking Einstein's definition.

    I don't see any simultaneity planes perpendicular to world lines except for the ref. frame. Space is not full of imaginary lines as you and G&B depict.
    No CMIF is required for clock synchronization, only light signals, and it applies to the local ref. frame. If it were to apply to distant locations, it becomes a universal simultaneity, which is impossible with finite light speed. Which local frame would be the 'chosen' one?

    I use the spacetime graphic (Minkowski spacetime diagram) since it is a geometric representation of the coordinate transformations, and it works.

    I used the same curve as in the G&B paper.
    In the graphic, the lower green aos doesn't apply since the path doesn't remain inertial moving forward, and the upper green aos doesn't apply since the path doesn't remain inertial moving backward. The middle one (2-4) qualifies, since the outbound and inbound transit times are equal, and the inertial and accelerated portions are equal.
    The object M only has one reflection for each time value.
    Without any calculation, a clock will integrate the time for any speed profile, so all that is needed is to compare clocks at reunion.
    The 'twin' scenario has received much more attention than it deserves.
     
  10. phyti Registered Senior Member

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    Neddy;

    I don't think the psychedelic patterns in the G&B paper are necessary.
    Using your example will show the original Minkowski spacetime diagram is sufficient.
    The red line replaces the hyperbola, since there are only 2 clocks being compared.
    If we were predicting the results for a successful motion as shown, we could use the same v for all of B's motion and calculate t'=.5t.
    If we follow the motion as it unfolds, with no prior knowledge, then we compare the perception of A and B.
    A:
    The signal sent at t=5.4 returns at t=74.6. A assigns t'=20 to t=40.
    B:
    The signal sent at t'=1.4 returns at t'=20. B assigns t=5.4 to t'=10.7.
    The signal sent at t'=5.4 returns at t'=21.3. B assigns t=10 to t'=12.
    The green x' is replaced with the magenta x', distortion from a discontinuous profile.
    The signal sent at t'=10.7 returns at t'=29.3. B assigns t=40 to t'=20.

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  11. phyti Registered Senior Member

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    Mike;

    How will you construct the array of clocks?

    There was a TV commercial recently of a teenage boy with a bat and pail of balls.
    He would toss a ball into the air and try to hit it, yelling 'the greatest hitter in the world'.
    After many failures, he changed his slogan to 'the greatest pitcher in the world'.
     
  12. Ssssssss Registered Senior Member

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    You seem to be quoting from Dolby and Gull which would be D&G no?
    You are missing the point because the introduction is talking about MCIFs which the paper says misrepresent simultaneity and are the result of the k-calculus being "applied wrongly" and the quote you've just given is after that part and is where they are talking about the correct application of k-calculus or radar coordinates. So Dolby and Gull are correct when they say that the wrong-headed MCIF method forces Barbara "to conclude that the distant planets swept backwards and forwards in time whenever she went dancing". The whole point of the paper is that this correct statement about MCIFs suggests that MCIFs are inappropriate for distant simultaneity and that they/Bondi have a better approach.
    All of the pale grey dotted lines in figure 1 and figure 2 are MCIF simultaneity planes that are perpendicular to the worldline.
    Space or at least the MCIF definition of space is those imaginary lines in spacetime. Other better definitions are available.
    Yep although there are limits to that for eternally accelerating observers.
    I have no idea what you are saying here.
    So do Dolby and Gull and in fact every single diagram in the paper is a Minkowski diagram.
    Which curve? Which graphic? Which green aos? And what's an aos anyway? I have lost track entirely of what you are talking about.
    They are just surfaces of simultaneity by their definition drawn on a Minkowski diagram.
     
  13. Ssssssss Registered Senior Member

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    Sure but Neddy and Mike don't like Dolby and Gull's coordinates because the shape of the simultaneity plane associated with "now" on my worldline depends on acceleration phases in my future which doesn't actually matter because it's all fixed in my past lightcone but seems to bother them hence my revised approach.
     
  14. Ssssssss Registered Senior Member

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    Yes Mike_Fontenot if two people are colocated then any simultaneity convention will do and everyone will agree because they are at the same event so always get the same time coordinate. You only get apparent age changes if you are a remote observer inferring stuff you haven't seen yet and by the time any signal has got to you all the differences will iron out and you'll agree on the ages of everyone you can see and there will never be jumps forward or backward only rate changes when one or other of you accelerates.
     
  15. Ssssssss Registered Senior Member

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    This is wrong.
    This is wrong.
    This is wrong.
    This is wrong.
    This is wrong.
    This is wrong.
    This is wrong.
    Um....
    Yeah so why do you think it needs proving?
     
  16. Mike_Fontenot Registered Senior Member

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    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.
     
  17. Ssssssss Registered Senior Member

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    No one gets abruptly older or younger but this has no bearing on whether or not CMIF is correct because coordinate time has nothing to do with aging just what some people say about aging.
     
  18. phyti Registered Senior Member

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    S8;

    On the left, U observes A synchronizing his clocks c1 and c2.
    U does not agree they are synchronized.
    The green aos (axis of simultaneity, since motion is confined to x direction) or x' axis is only there after using light signals. It is a fictitious mathematical tool per the SR clock synch convention, and graphically used to indicate a relation between the two clocks.
    On the right, A perceives his clocks as synchronized in an orthogonal coordinate system.
    The simultaneity references are there as a result of measurement.

    My argument in the Dolby & Gull paper is their interpretation of fig.2, that the overlapping aos implies three different times assigned to the same event.
    As the G&D twins gif shows, there is only one signal for any event to the left.

    A system of synchronized clocks extended throughout space, as Mike and Neddy prefer, is not possible. It would also be considered a universal simultaneity, which was eliminated with Relativity.

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  19. phyti Registered Senior Member

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    Mike;

    Each observer receives a sphere of images from the surrounding world, represented as a circle converging to observer A at t0. The last image from B was at -t1.
    If B independently (thus ?) changed velocity at -t1, A would not be aware of it until some time between t2 and t3.
    If A1 was synchronized to A, B's motion would alter his perception of the clock rate for A1, and consequently that for A.
    B's motion alters his perception of the world, but not the distant processes.
    When SLAC accelerates particles (in a straight line) to near light speed, astronomers do not detect any contraction of the universe in that direction.
    In all cases, A's knowledge of the B-clock and its position is historical.
    There is no instant knowledge.

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  20. Mike_Fontenot Registered Senior Member

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    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.
     
  21. Neddy Bate Valued Senior Member

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    Infinite proper acceleration is not a physical possibility in our universe. It would not surprise me if the pseudo-gravitational potential is undefined for an infinite proper acceleration.
     
  22. Neddy Bate Valued Senior Member

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    Even the velocity is undefined at the turnaround point, if you assume it goes from +0.866c to -0.866c in a zero time interval. Infinite proper acceleration is not a physical thing which can be taken literally. It should be interpreted as going from +0.866c to 0.000c to -0.866c in a very small time interval.
     
  23. Ssssssss Registered Senior Member

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    It's defined but only for things in the observer's Rindler wedge which has its point where the turnaround happens so it doesn't include any region where the MCIF method has planes sweeping backwards and Mike_Fontenot is still using the wrong potential for observers with a constant distance relation anyway.
     

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