A New Simultaneity Method for Accelerated Observers in Special Relativity

Discussion in 'Alternative Theories' started by Mike_Fontenot, Dec 26, 2019.

  1. Mike_Fontenot Registered Senior Member

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    That's the ASSUMPTION that DEFINES the CMIF simultaneity method. But it is not true in the Dolby&Gull method, or in the Minguizzi method, or in my method (which I'm going the christen as "the Fontenot method").

    That was the reason for it being called a paradox. People thought that the traveler SHOULD be able to use the time dilation equation for the entire trip, because they thought that surely that one instant of acceleration couldn't make a difference. But we know that SOMETHING has to make a difference, because they have to agree about who's older when they are reunited (because they are standing right there looking at each other). He may conclude she is ageing more slowly than he is during some parts of the trip, but he DOES find her to be the older at the reunion, so she HAS to be ageing faster than him SOMETIME during the trip (according to him AND to her).
     
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  3. Mike_Fontenot Registered Senior Member

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    If anyone is trying to find an error in my proof, be aware that I've improved the wording of the proof, so be sure to use the latest version of my webpage ... I've just updated it. If I've made an error in the proof, it shouldn't be hard to localize.

    Also, be aware that the proof does NOT make use of my new simultaneity method (which I'm calling "Fontenot's method" or "Fontenot's simultaneity method"). I show that the CMIF simultaneity method is incorrect purely from first principles, combined with the fact that on the outbound leg (up to but not including the instantaneous velocity change) the traveler (he) is equivalent to an inertial observer (because he has never accelerated), and we only need the time-dilation equation to determine his conclusion about his twin's current age on that leg.
     
    Last edited: Jan 20, 2020
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  5. Halc Registered Senior Member

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    Where is this updated proof? What does it prove?
    I looked at section 7 of the cadoequation/cado-reference-frame doc and it still makes all the same mistakes, somehow concluding:
    This is completely wrong. The message transit time in the new frame is massively more than it was in the outgoing frame. It has much greater distance to travel in that frame, so her age is much greater than it was before the velocity change.

    Maybe you've fixed some other document, so you should include a link when you post an update like that.
     
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  7. Neddy Bate Valued Senior Member

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    True, and the reason "the traveler must always adopt the perpetually-inertial observer's conclusion with whom he is stationary at that instant as to the current age of the home twin" is because you cannot have different people stationary with respect to one another disagreeing on her current age. She can only have one age, according to all of the people who are stationary with respect to each other, otherwise it results in nonsense.

    Worse that that, it results in all of the perpetually-inertial people saying, "The speed of light is the constant c in our inertial reference frame," and the traveler saying, "No it isn't, not for me anyway, I'm different than you." It is nonsense, AND it violates the postulate of the constancy of the speed of light in all non-accelerating reference frames, from which SR is derived. No legitimate result in SR can violate the postulate on which it is built.

    You are assuming the legitimacy of that line of simultaneity as soon as you start your proof by drawing a Minkowski diagram, which has specific rules for the angles and directions of the lines of simultaneity. Just as you assumed the legitimacy of the first line of simultaneity that points to 26.67 on the same grounds.

    You seem to think that it is still up for debate as to how SR resolves the twin "paradox" but it isn't. If Dolby and Gull and others are saying it is, and choosing to use whatever lines of simultaneity they choose, then they are making the same type of mistake also.
     
  8. Neddy Bate Valued Senior Member

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    Mike,

    I respect your desire to keep the deceleration instantaneous, and keep everything as simple as possible. So, instead of having the traveling twin instantaneously turn around and go back toward his sister at v=-0.557c, let's look at the even simpler case where the traveling twin instantaneously changes his velocity to v=0.000c. He simply instantaneously stops. Here is the Minkowski diagram:

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    Now, let's assume for a moment that you are correct, that in the instant his velocity is v=0.000c that he continues to claim that her current age is 26.67 and that the amount she aged while her image pulse was in transit was 26.67-16.91=9.76 years. Now that he is stationary with respect to her, he knows the distance between them is 23.09 light years, which means he is claiming that the speed of that light pulse was 23.09/9.76=2.365 light years per year. So, we have a violation of the constancy of the speed of light, which was one of the premises on which SR was built. So, you have clearly made a mistake.

    Instead, let's assume for a moment that Minkowski is correct, that in the instant his velocity is v=0.000c that he claims that her current age is 40.00 and that the amount she aged while her image pulse was in transit was 40.00-16.91=23.09 years. Now that he is stationary with respect to her, he knows the distance between them is 23.09 light years, which means he is claiming that the speed of that light pulse was 23.09/23.09=1.000 light year per year. So, we see that when his line of simultaneity shifts to point to t=40, it preserves the constancy of the speed of light, as it must in SR. Note also that no other line of simultaneity can be chosen and still satisfy the requirement that c=1.000 light year per year.
     
  9. Mike_Fontenot Registered Senior Member

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    Special relativity is based on the assumption that the speed of light will be measured to be the same for all perpetually-inertial observers. The traveler who has recently accelerated is not a perpetually-inertial observer.

    But what you SHOULD be doing is trying to identify a specific error in my proof of Fontenot's theorem. I've laid the proof out in a sequence of simple statements. Tell me which statement in that proof is incorrect.

    (Note that my proof DOESN'T assume anything about my simultaneity method.)
     
    Last edited: Jan 21, 2020
  10. Neddy Bate Valued Senior Member

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    Could you produce a citation for anyplace in SR where it is specified that the speed of light can have two different values according to two people who are at rest in the same perpetually inertial reference frame? Of course not.

    Consider a clock at rest in the stay-home twin's frame, located at x=23.09, and synchronized to her clock by Einstein's clock synch method. Such a clock (and all such sychronised clocks at rest throughout her whole reference frame) will all be displaying t=40.00 simultaneously in the home twin's frame right after the traveler stops. He can simply look at that clock which is right next to him and see that it displays 40 years. Everyone at rest in that frame can tell him that his sister must be 40 years old because of it. But you want him to stand there and say, "No that only means she is 40 years old according to you guys, but according to me she is 26.67 years old." Even as he is standing there looking at the synchronised clock right in front of him displaying 40 years. It is complete nonsense.

    I have already told you, it is a mistake to say that you can use the same line of simultaneity for both before and after the instantaneous velocity change. You cannot do that with a Minkowski diagram, which is a graphical representation of certain equations (the Lorentz transforms). You seem to think SR does not have fixed equations, but rather is flexible enough to allow whatever you choose. It isn't.

    Also, naming it "Fontenot's theorem," after yourself, is one of the signs of a crank.
     
    Last edited: Jan 21, 2020
  11. Mike_Fontenot Registered Senior Member

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    You obviously haven't been able to identify a specific error in my proof.
     
  12. Halc Registered Senior Member

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    Again, where is this proof? What is the thing postulated that the proof demonstrates? The latter question I suppose might be answered by the first.

    My prior post commented on one section 7, which still contains an invalid proof that CMIF is wrong, so I don't think that's the thing you updated.
     
  13. Neddy Bate Valued Senior Member

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    LOL, whatever you say. Using the line of simultaneity from the moving frame, when the traveler is at rest in the home frame, is a very specific error.

    But you are convinced that the traveler can decelerate to v=0.000c , and know that the distance between him and her is 23.09 light years, and still conclude that the amount of time that light pulse was in transit from her to him was 9.76 years, even though it traveled a distance of 23.09 light years making the speed of light 23.09/9.76=2.365 light years per year. And you think that is allowable under SR. Amazing.

    Furthermore, you think you have proved that if the traveler decelerates to v=0.000c , and knows that the distance between him and her is 23.09 light years, THAT IT WOULD BE INCORRECT for him to conclude that the amount of time that light pulse was in transit from her to him was 23.09 years, even though that would make the speed of light 23.09/23.09=1.000 light years per year in the stay-home twin's inertial frame. You actually think that would be incorrect under SR. Absolutely amazing.

    Well at least I tried to help you. I will bow out now, and let you enjoy your nonsense. Best wishes with "Fontenot's Theorem".
     
  14. Mike_Fontenot Registered Senior Member

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    Quote the exact statement that I make in my proof that you maintain is incorrect.
     
  15. Neddy Bate Valued Senior Member

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    Okay, I read your updated page, and this is the part I think is wrong:

    "Call the time when she transmitted that additional message "TAU+", which is only infinitesimally later than the instant "TAU". So when he receives that additional pulse, he knows that her ageing during that message's transit is only infinitesimally different than her ageing during the transit of the first message."

    How do you know that is true? You are assuming the very thing you are claiming to have proven. You are correct that her age as viewed in the image pulses will only be infinitesimally different from each other, but that is not the same thing as saying that the line of simultaneity that he is supposed to use can only change infinitesimally.

    The turnaround point has an undefined slope, so there can be infinitely many lines of simultaneity there. You are assuming that the only line of simultaneity there is the one pointing to her age being 26.67, and ignoring that there is also a line of simultaneity there which points to her age being 53.33. You even claim to understand that such a line is in fact the correct line for the perpetually inertial frame which is traveling toward her at v=-0.577c, so why can't it be the correct line for the traveler after he changes frames? If it is, (and it is), then it is clearly not true that "he knows that her ageing during that message's transit is only infinitesimally different than her ageing during the transit of the first message."

    In other words, you are not even allowing for the possibility that he could use a different line of simultaneity after he changes velocity.

    If all of these other lines of simultaneity exist at the undefined point, then how do you know that the only correct one to use for T+ is the one from T- regardless of how he changes his velocity? The answer is that you don't know that. You are assuming it, and then claiming to have proven it.
     
    Last edited: Jan 22, 2020
  16. Neddy Bate Valued Senior Member

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    Mike, please consider my simpler version where he instantaneously changes from v=0.577c to v=0.000c. In that version, the distance between him and her goes from being length-contracted, to not being length contracted, which is clearly a difference you should think about.

    Also, before he stops, he considers the image pulse to have originated from a time in the past when when they were even closer together because of the relative motion.

    Since the distance her light pulse travels (at constant speed c both before and after the stopping, contrary to your claims) changes from small to large, so must the time elapsed during transit change from small to large.
     
  17. Mike_Fontenot Registered Senior Member

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    What part of that quoted statement do you think is untruthful? And what EXACTLY do you think is unjustified in the statement itself?
     
  18. Halc Registered Senior Member

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    I quoted the same statement in post 83 and bolded the part that is wrong.
    In the new frame (TAU+) the distance to the event where the message was sent was much larger, and hence the message had a much larger transit time. That makes her current age much older, whereas you conclude it's infinitesimally different.
     
  19. Neddy Bate Valued Senior Member

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    Mike,

    I re-wrote your passage in a way that it makes only justified claims, no unjustified ones:

    Call the time when she transmitted that additional message pulse "TAU+", which is only infinitesimally later than the instant "TAU". So, when he receives that additional message pulse, he knows that her age as shown in the second message pulse is only infinitesimally different than her age as shown in the first message pulse. However, this does not tell us how much he considers her to have aged while the image pulses were in transit.

    In order to determine that, we need to know what age she is, according to him, at the two times he receives the two message pulses. This requires that we consider lines of simultaneity on our diagram, but which ones? The CMIF method would have us use two different lines of simultaneity, but I am determined to prove that wrong, so I can't do that...

    And I'll let you take it from there...
     
  20. Mike_Fontenot Registered Senior Member

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    The only LOS that is needed is the one I used for the FIRST message (which is the LOS of the perpetually-inertial observer on the outbound leg).

    Sketch that first pulse on the Minkowski diagram, but make a comment on the sketch that he receives it at the instant "T-" in his life, an infinitesimal time before the velocity change when he is "T" years old. So we know that her current age (according to him) when he receives that first pulse is infinitesimally less than what the LOS analysis told us.

    Now, ask yourself how you're going to plot the additional pulse on the Minkowski diagram. You know it has the same slope as the first pulse (45 degrees), so the two pulses are obviously parallel on the diagram. You also know that the two pulses are infinitesimally close together where they start on the horizontal axis. So the two lines are infinitesimally close together on the diagram. That means the two pulses will be indistinguishable on the diagram ... they will just look like one pulse. The amount of her ageing while those two pulses are in transit CAN'T differ by more than an infinitesimal amount. And her age when those two pulses were transmitted don't differ by more than an infinitesimal amount, so her age when those two pulses are received can't differ by more that an infinitesimal amount.

    NO second LOS is needed in the above reasoning.
     
  21. Neddy Bate Valued Senior Member

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    Sorry, but your conclusion just doesn't follow from your premise. I understand your reasoning, but it just doesn't hold up to scrutiny.

    Imagine that instead of the traveling twin instantly changing velocity, you have two perpetually inertial travelers passing each other at the midpoint of the diagram, one with v=+0.577c and the other with v=-0.577c. The Minkowski diagram for that case has all of the same properties as the one for the twin at the turning point with the undefined slope. And you have already agreed that the correct LOS for the outgoing traveler in that case would be the LOS that points to 26.67, and the correct LOS for the incoming traveler points to 53.33. That is two different LOS lines which are both valid at one point on the diagram.

    Now add her message pulses to the diagram, and then test your claim that "The amount of her ageing while those two pulses are in transit CAN'T differ by more than an infinitesimal amount." You should be able to see that it does not hold up.

    The geometry on the diagram is the same, so if your reasoning were sound, it should apply in this case as well, but it doesn't. There is no reason to assume the LOS that points to 26.67 is the only valid LOS at the turnaround point (with the undefined slope), and after the turnaround, the LOS points to 53.33 without any doubt. Minkowski knew what he was doing, and the follies of Dolby and Gull et al did not change the fact that there are set rules for the slopes and directions of LOS on Minkowski diagrams.

    Two different calculations reveal two different amounts of her aging while the messages were in transit:
    26.67 - 16.91 = 9.76 (before)
    53.33 - 16.91 = 36.42 (after)
     
    Last edited: Jan 23, 2020
  22. Neddy Bate Valued Senior Member

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    Mike, I have a thought experiment regarding your "theorem" which I think you would be wise to consider. But I don't want to explain the whole thing only to have you tell me that you are not interested in it. So, if you and I can first agree on certain aspects of your theorem, I will then reveal the thought experiment afterward.

    We would need to use the simpler scenario, where the traveling twin instantly changes velocity from v=+0.577c to v=0.000c. Everything else in the scenario could stay the same, but I would expect you to actually consider certain things about times when his velocity is v=0.000c, not just the point with the undefined slope. Are you willing to explore that a little bit? If not, I understand, but you will be missing out on something I think you would find interesting.
     
  23. phyti Registered Senior Member

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    Mike;
    The missing time or 'time jump' for the overly simplified triangle is due to the unreal speed profile, with its discontinuity. There is no transition from the outbound segment to the inbound segment.
    This can be corrected with a segment of accelerated motion for B as shown on the left.
    This restores the aos (green) for the entire trip, but with distortion.
    A sends a time coded signal to B at regular intervals of 4t.

    B accelerates toward A from B24 to B40.
    On the right, B sends light signals at regular intervals, to measure the distance from A.
    The return times vary due to relative motion of source and receiver, and is not an indication of aging. It's doppler effects, known before relativity was conceived.
    As B accelerates, he can interpret the effects on himself and A's acceleration toward him as a g-field per the equivalence principle.
    Note on the left, his acceleration shortens the distance A-B from 24 to 21,
    thus on the right, B's measurement is less than 21. His perception of the A path is flattened somewhat because he was accelerating toward his return signals, reducing the round trip time. That does not occur for inertial motion.
    Note on the left, B cannot assign a Bt for A40 until B44. All information is historical (you said it in section 4).
    The descriptions are not symmetrical. B experiences acceleration while A does not.
    B's motion does not affect A or her clock, but it does affect his perception of A and her clock.
    B receives 3 signals outbound, 5 signals accelerating, and 12 signals inbound, or 20 signals for an interval of 80 during his interval of 64. The verification occurs when the clocks are rejoined. No calculations are needed since clocks are automatic and accurate accumulators of time.

    The inertial observer only has a simulated axis of simultaneity, established for his speed in space (even though he doesn't know what that is). It was never intended for use in an accelerated frame.

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