A New Simultaneity Method for Accelerated Observers in Special Relativity

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

  1. phyti Registered Senior Member

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    471
    Mike;

    SR defines an inertial frame in motion as equivalent to a pseudo rest frame. When a light signal is sent outbound to reflect from a distant object and return, the observer expects the transit times out and back to be equal. That’s the basis for establishing the aos/los. You are not in agreement with the definitions.

    The relative speed is .6c with a gamma = 1/.8.
    left:
    This is the distortion from mixing frames, as in B2 interpreting light signals sent by B1. B2 moving toward A after reversal, intercepts the B1 signals earlier than B1 would have, thus perceiving A as closer. The A speed profile appears to change twice in returning to B.

    right:
    The clock rate correspondence of A to B, resulting from the left graphic. It's incorrect since A was an inertial frame and its clock rate was constant, B1 and B2 were both inertial frames, and their clock rates were constant. It shows a nonsensical speed profile with the A clock running 2x the rate of the B clock, while moving at the same speed!

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  3. Neddy Bate Valued Senior Member

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    But I think you agreed in your post #148 that there is no discontinuity in her age in the CMIF method either, provided that the turnaround takes a finite amount of time, (an arbitrarily small amount of time lets us still use our other calculations with sufficient accuracy). The only reason we get a discontinuity in her age is because you specified that the velocity must go instantaneously from v=+0.577c to v=-0.577c without passing through the velocities that are in between those two values. That is not a real acceleration, in real life, that is looking at two different inertial frames. I think what you are really concerned with is her age getting younger, not that there are any discontinuities, because there aren't.
     
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  5. Neddy Bate Valued Senior Member

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    Mike,

    Here's something you might want to think about. For now let's just let him stay at v=0.000 indefinitely after your point labeled T. From that moment on, let him watch her clock with his powerful telescope. There is no relative movement between them, so he does not have to do anything difficult.

    All he has to do is watch her clock through his telescope. He currently sees it displaying 16.91 and its tick rate is exactly the same rate that his own clock is now ticking. Using your method, he predicts that her current age is 26.67 at that moment, and that her clock actually starts ticking faster than his own at that time. So all he has to do is wait and see.

    Using the CMIF method, he would have predicted that her current age is 40.00 and that her clock would never tick faster than his own. So all he has to do is wait and see which of the two different predictions turns out to be correct.

    After 9.76 years he finally sees her clock displaying 26.67 and he is expecting to see its tick rate increase. But it never does. It continues to tick at the same rate as his own. He can still pretend whatever he wants to believe is true, (just put another ad hoc explanation here), but the evidence does not support it. The evidence (and all logic of what inertial means) supports the CMIF method. The end.
     
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  7. Mike_Fontenot Registered Senior Member

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    OK, I finally finished my analysis of the Dolby and Gull answer to Neddy's question "How old is she when he is 32.66 (according to him)"? (Note that, since there are no discontinuities in the D&G ACD, I don't have to be specific in the above question about whether I'm talking about infinitesimally BEFORE he changes velocity, or infinitesimally AFTER he changes velocity: her age doesn't change by a finite amount in that infinitesimal time interval.) The answer I gave in my previous post (33.3 years old) wasn't correct. The correct answer is "at the velocity change, her age is about 32 years old". The reason I say "ABOUT 32 years old" is because I can only give an upper and lower bound for her age at the velocity change. Her age is less than 32.66 and greater than 31.63. The reason that I can only give a range for her age is that the D&G method works "backwards": you specify an exact age for HER, and then COMPUTE using their algorithm the corresponding age for HIM (according to him). So, in order to determine her exact age, for some specified age for him, you have to do an infinite iteration.

    The above problem wouldn't occur if the center section of the D&G ACD were a straight line. It IS a straight line in the standard scenario with v1 = 0.57735 and v2 = -0.57735. But I discovered yesterday that for the scenario where v1 = 0.57735 and v2 = 0.0, the center section of the D&B ACD ISN'T quite straight: the average slope over the whole center section is 1.318, but the average slope over a lower portion of the center section is 1.232.

    To understand why the D&G method is "backwards", you have to know what their algorithm IS ... and I don't think many people have figured out what that algorithm is (mostly because they don't work out a simple example to illustrate their method). What you do is, you PICK a specific age for her, locate that point on the horizontal axis of the Minkowski diagram, and then construct two pulses intersecting that point. One of the pulses is received by her at that point (and transmitted earlier by him), and the other pulse is transmitted by her, and received later by him. Denote his age when he transmitted the first pulse as "t_s". Denote his age when he receives the second pulse as "t_r". Then average those two ages (sum them, and divide by 2). The resulting number is the his age when she is the specific age that we started out by arbitrarily choosing.

    When I discovered that the D&G method doesn't give a straight line for the center section of the ACD diagram for the scenario with v2 = 0.0, I wondered whether MY method might also not have a straight center section for that scenario either. THAT'S what I'm going to investigate now.
     
  8. Neddy Bate Valued Senior Member

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    2,114
    I followed your instructions, but did not get the same answer that you got. I chose a point on the horizontal axis where she is 32.0 years old, let her send a light pulse to him at that time, and he receives it when he is 40.5 years old, on the return leg of his trip. Earlier, when he was 16.5 years old, he sent her a light pulse and she received it when she was 32.0y ears old. But the average is (40.5+16.5)/2=28.5 which is not the turnaround point. What went wrong?

    Oh, never mind, I see now. You were using the scenario where he stays at v=0.000 indefinitely, rather than do a return trip. In that case the numbers are (16.5+47.8)/2=32.2 which is close enough to 32.66 I guess.

    But please see my post #183. If he thinks she is 32 when she is really 40, he will expect her clock to tick fast at some point. He can watch it with a powerful telescope, and see that it never ticks faster than one second per second. He is at rest in her inertial frame, after all, and proper clocks don't tick faster than one second per second in inertial frames. So your method and this one are both proven wrong by simply observing that clocks don't do what these methods predict.
     
    Last edited: Feb 27, 2020
  9. Mike_Fontenot Registered Senior Member

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    251
    I forgot that I had already done that, when I posted the Minkowski diagram and the ACD for my method back in post #169, for the scenario where v1 = 0.57735 and v2 = 0.0 for a while, and then eventually switching to v3 = -0.57735. The point R4 in that ACD is the point where he finally agrees with the PIO who has been riding along with him ever since he changed his velocity to 0.0. The average slope of the ACD from there down to the point T is 1.577. And the average slope of the ACD from the point R1 to the point T is ALSO 1.577. And the average slope of the ACD from the point R0 to the point T is ALSO 1.577. So the entire segment of the ACD between the velocity change to 0.0 and the point R4 (when he agrees with the PIO) IS a straight line. My method doesn't suffer from the nonlinearity of the Dolby and Gull method. And my method doesn't "work backwards" like the D&G method does: in my method, you can arbitrarily specify an age for him, and then directly determine her current age then (according to him) ... that's usually what we want to do. With the D&G method, you have to arbitrarily specify HER age, and use their method to determine his current age then (according to him). And perhaps most important, the D&G method is non-causal, whereas mine is causal.
     
  10. Mike_Fontenot Registered Senior Member

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    251
    No, he is NOT. For a while after he accelerates in a way that makes her conclude that their relative velocity is zero, he disagrees with her about a lot of things. He disagrees about the correspondence between their ages. He also disagrees about their relative velocity. (The quantities v1, v2, and v3 in the specification of the scenario are their relative velocities ACCORDING TO HER.) Look at Section 11 of my webpage

    https://sites.google.com/site/cadoequation/cado-reference-frame .

    And he also disagrees with her contention that the velocity of a light pulse is always the constant "c".

    You may think that's odd, but it is actually much LESS odd than what happens at the velocity changes in the CMIF method. In the CMIF method, he concludes that their separation INSTANTANEOUSLY increases by a large amount during the velocity change from 0.57735 to zero, which implies an INFINITE relative velocity between them. And he will likewise say that a light pulse near her is traveling INFINITELY fast during his velocity change.
     
  11. Neddy Bate Valued Senior Member

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    2,114
    You mean according to your method. In SR, he is at rest in her frame, and concludes their relative velocity is v=0.000c (which it is), and that the speed of light is c (which is one of the premises from which the rate of time dilation is derived in the first place), all of which you discard when you make your own version.

    Because of the premises of SR, we know he must say her clock ticks at a rate of 1/gamma for the entire return leg of the trip, which you also discard when you make your own version.

    The "instantaneous" aspect of that is because you neglect to consider the velocities between 0.577c and zero. And you cannot conclude any greater velocity than 0.577 just because the distance increases, because that distance increasing is a result of deceleration! A speedometer attached to him would report a decreasing velocity until he reaches v=0.000. The rest is just length contraction becoming less contracted.
     
    Last edited: Feb 27, 2020
  12. Mike_Fontenot Registered Senior Member

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    251
    After I wrote the above, I went to my webpage and re-read what I had written there about HIS conclusion about their relative velocity. I was surprised that I hadn't given ANY information about how I got the result that I stated there (that he concludes that their relative velocity is equal to -1.0, the speed of light). I looked through my LARGE stack of notes that I've accumulated over the last few months, but didn't spot my analysis that gave that result of 1.o for their relative velocity (according to him). So I tried to do it from scratch for the scenario where v1 is 0.57735 and v2 = 0.0. I got that he agreed with her: that their relative velocity is zero (averaged over the period where he normally disagrees with the PIO). (I later found that that analysis was erroneous). I was perplexed ... why did I earlier get the v = -1.0 result?

    So this morning I looked more carefully through my notes, and finally found the spot where I had gotten the result v = -1.0. And I reviewed the procedure I had used to get that result. I fairly quickly spotted a careless mistake I had made. So, I decided to work out the correct answer for a large number of scenarios.

    To do such a large number of scenarios in a reasonable amount of time, I decided to do it graphically, drawn roughly to scale, so I could just use a centimeter scale to get quick (but rough) answers. It was still three or four hours of work. What I found was quite surprising. He does NOT agree with her for the v1 = 0.57735 and v2 = 0.0 case. He concludes that their relative velocity is about 0.18, not zero. And oddly, he DOES agree with her for the scenario v1 = 0.866 and v2 = -0.866: he gets -0.866! And he probably agrees with her for the v1 = 0.57735 and v2 = -0.57735 case: my rough calculations said that he says their relative velocity is about -0.57.

    I've got to check over these calculations tomorrow, and then do the most important scenarios accurately (instead of doing the quick scaled graphics approach). If the results check out, I'll expand section 10 in my webpage so that it shows how the analysis is done ... it's a bit complicated, and easy to get wrong.
     
    Last edited: Feb 29, 2020
  13. Mike_Fontenot Registered Senior Member

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    251
    I redid the v1 = 0.57735 and v2 = 0.0 scenario again, doing it analytically (with high precision) instead of the "scaled-graphical" approach I used yesterday. Their relative velocity, according to him, is 0.18347. That result is their relative velocity, according to him, AVERAGED over the "disagreement interval", between when he changed his velocity, until he receives the pulse from her that she transmitted when SHE says he changed his velocity. Early in that interval, their relative velocity was greater than 0.18347. And late in that interval, there relative velocity was very close to zero.

    I also did an analysis to determine what he says the average velocity of a light pulse leaving him when he changes his velocity relative to her to zero, until it reaches the point in his life when he begins to agree with her. That average velocity is 1.0.

    I'll try to upload a jpeg of the Minkowski diagram I used in yesterday's analysis, and also a jpeg of the Minkowski diagram I used in today's analysis.

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  14. Mike_Fontenot Registered Senior Member

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    I just got several new results that I find very interesting.

    For the scenario when v1 = 0.57735 and v2 = 0.0, I had previously determined that he says their average relative velocity during the "disagreement interval" (DI) is 0.18347 ly/y. And I had previously determined that he agrees with her about the speed of the pulse that spans the DI: it's 1.0 ly/y. (That pulse starts from her when she is 40, and ends when he receives that pulse). But I've just discovered that for a DIFFERENT pulse that she transmits and that he receives, he concludes that it's AVERAGE speed is 2.115 ly/y, NOT 1.0 ly/y. That particular pulse starts when she is 26.6667 years old (which is where his LOS on the outbound leg intersects the horizontal axis), and terminates when he receives it when he is 42.422 years old. So, in general, he will consider different pulses to travel at different speeds, when they are distant from him. But he DOES always conclude that any light pulse that is momentarily co-located with him has the speed 1.0 at that instant.

    For the v1 = 0.57735 and v2 = -0.57735 scenario (the standard twin paradox situation), I determined that he says their relative velocity is -0.57735 during the DI ... i.e., in this case, he agrees with her. But for the pulse that leaves her when she is 26.6667 years old, and is received by him when he is 37.7138, he says its AVERAGE velocity over its complete transit is 3.155 ly/y, NOT 1.0 ly/y. (But again, at the instant that that pulse passes him, he says its speed is 1.0 ly/y).

    The fact that pulses travel at different speeds in my simultaneity method is similar to what happens in the CMIF simultaneity method during the instantaneous turnaround: there, a distant pulse's speed can be infinite, and it can even travel in the opposite direction that it did momentarily before the instantaneous turnaround! Compared to that weirdness, my method is pretty tame.
     
  15. Neddy Bate Valued Senior Member

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    2,114
    Mike,

    For that scenario, after he stops, he sees her clock displaying 16.91 by using a powerful telescope. From that moment on, assuming he remains stationary, he can watch her clock ticking at a rate of one second per second for the rest of their lives. When he sees her clock displaying 26.67 he imagines that her clock has actually begun ticking faster than one second per second, but all of the things that are part of the fontenot theory (speed of light changing, distance changing) happen to work together in such a way that all of the evidence of it is undetectable, correct? Or do you think he actually sees her clock ticking at a different rate at any point? Just curious...
     
  16. Mike_Fontenot Registered Senior Member

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    251
    I also analyzed ANOTHER pulse in that scenario: for the pulse that leaves her when she is 40 years old, and is received by him when he is 44.6, he says the average pulse speed is 2.49 ly/y.

    And in the v1 = 0.57735 and v2 = 0.0 case, for the pulse that begins when she is 26.6667, he says its average speed
    is 2.115 ly/y.

    So different distant pulses, even in the same scenario, generally have different average speeds (according to him).
     
  17. Mike_Fontenot Registered Senior Member

    Messages:
    251
    I assume you're talking about the case where v1 = 0.57735 and v2 = 0.0. Right?

    No, that's not true in my method. As I've shown, in my method, after he changes velocity, immediately and for a while he says that distant pulses generally travel a speeds different from 1.0 ly/y, and he says that their speed generally isn't constant as his age increases. So he WON'T see her clock is ticking at a constant rate of one second of her time per one second of his time (or at ANY constant ratio).
     
  18. Neddy Bate Valued Senior Member

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    2,114
    Yes.

    Yes, I understand that you say that light traveling from her to him is not traveling at c (according to the now-stopped traveler), but you also say that her clock is not ticking at one second per second (according to the now-stopped traveler), so don't the two effects cancel so that he sees (with his eyes + a telescope) that her clock is ticking at a rate of one second per second constantly?

    After all, a perpetually inertial person standing right next to him can look through a telescope and see her rate as one second per second, so the stopped traveler should see the same thing. Think about it and let me know.
     
    Last edited: Mar 3, 2020
  19. Mike_Fontenot Registered Senior Member

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    251
    Of course he and the PIO observe the same signal coming from her. But they DON'T come to the same conclusion about her actual tic rate.

    The PIO says that her actual tic RATE is the SAME as the observed tic rate, but the current reading on her clock has a later time than the time the PIO just observed (and that time difference is constant).

    The traveler, who recently accelerated, says that the elapsed time between the receipt of each clock image, and the transmission of that image, is VARYING from one image to the next. So he knows that even though the received tic rate is constant, the tic rate of the home twin's clock is NOT constant.
     
  20. Neddy Bate Valued Senior Member

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    2,114
    Yes, that is what I thought, thank you. So the varying rate of her clock (according to the now-stopped traveler) and the varying speed of light (according to the now-stopped traveler) and I suppose also the varying distance between them (according to the now-stopped traveler) are all related in a way that makes it so that he SEES (with his eyes and telescope) her clock ticking at a rate of one second per second.

    In other words, the visual evidence does not tell us whether one or the other is correct.
     
    Last edited: Mar 3, 2020
  21. Neddy Bate Valued Senior Member

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    2,114
    Mike,

    Before I approach the logic issue that I raised in the other thread, I would like to verify that the following things are true in your method (NOTE: I am still using the simple case where v1 = 0.577c and v2 = 0.000c):

    1. That if the traveler uses your method, he uses CMIF on the outbound leg of his journey (until he changes velocity to v=0.000c)
    2. That your method has all perpetually inertial frames using CMIF

    If so, then from #1 we can deduce the following: Just before stopping, the traveler and the outbound perpetually inertial frame (OPIF) both agree that she is 26.67 years old.

    And if so, then from #2 we can deduce the following: Just after stopping, the traveler and the stay-home perpetually inertial frame (SHPIF) disagree on her age. The SHPIF says that she is 40, whereas the now-stopped traveler (NST) says she is 26.67.

    Is this all correct?
     
  22. Mike_Fontenot Registered Senior Member

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    251
    I wouldn't associate what happens on the outbound leg with the CMIF method at all. I would just say that on the outbound leg, the traveler's perspective is identical to a perpetually-inertial observer's (PIO's) perspective. There are two things that keep him from actually being a PIO. First, he accelerated right after he was born. Second, he is PLANNING to accelerate in the future.

    His acceleration at the beginning has no effect on her current age in the future (according to him, and to her). So that acceleration doesn't keep him from agreeing with a PIO riding along with him on the outbound leg. (And, as I said in my paper, if that bothers anybody, we can replace him by a true PIO, whose mother just happened to be passing the home "twin's" mother when they were both born.)

    His acceleration at the point "T" (when he changes his velocity to zero, according to the home twin) can have no effect on ANYTHING that happens BEFORE the point T. That is because of the principle of causality, which I believe special relativity assumes. That principle states that effects cannot precede causes.

    So, because of the above two arguments, we conclude that he can be treated as a PIO on the outbound leg. In particular, he is entitled to use the time-dilation equation (TDE). So he concludes that her current age is his current age, divided by the gamma factor (1.2247), on the outbound leg.

    Again, I don't see the relevance of the CMIF method in that sentence. PIO's are PIO's. Period. They have no need or interest in knowing anything about the CMIF method. PIO's can always use the TDE to get the current age of any other PIO anywhere. And, if necessary, a PIO can always make use of his LOS on a Minkowski diagram.

    The CMIF method is DEFINED by stating that an observer who sometimes accelerates (he) always agrees, about the current age of the home twin, with the PIO who is currently co-located and co-stationary with him, whenever he is not currently accelerating.

    That's certainly true of the traveler: it's just what the TDE says. I'm not certain what you mean by the OPIF ... maybe the PIO who is riding along with him? If so, yes, he and the PIO agree at that point.

    Again, I didn't understand your #2. As I've already said, the only thing that matters on the outbound leg is that he is entitled to use the TDE. He is equivalent to a PIO on the outbound leg.

    I think your SHPIF is just HER, the home twin. HE and SHE do indeed disagree about the correspondence between their ages on the outbound leg. In the CMIF method, they agree about their relative ages, immediately after the velocity change and forever thereafter. In my method, they continue to disagree after the velocity change for a while (which can last years), but eventually, they will agree. I call the interval of his ages, after his velocity change, where he disagrees with her, the "disagreement interval" (the "DI"). It is determined, on the Minkowski diagram, by drawing a pulse emitted by her immediately after the velocity change, and received by him. The DI starts when that pulse is emitted by her, and ends when that pulse is received by him. The magnitude of their disagreement in my method is largest immediately after his velocity change, but continuously declines after that, and approaches zero as the end of the DI is approached.

    I'd word it this way:

    "When he is 32.66 immediately BEFORE the velocity change, she says she is 40 then, but he says she is 26.67 then". Both the CMIF method and my method agree on that. And I'd add that, immediately AFTER the velocity change, in my method he STILL says she is currently 26.67, but in the CMIF method he now says she is 40.
    So in the CMIF method, her current age instantaneously changes during the instantaneous velocity change, but in my method, her age doesn't instantaneously change during the instantaneous velocity change.
     
  23. Neddy Bate Valued Senior Member

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    2,114
    Okay, I think I understand better now. In your method, the outbound leg is actually nothing like CMIF at all.

    For example:
    With CMIF, at the moment before the twin accelerates away from his sister, he would say that a clock synched to hers and located at x=23.09 would display t=0.00, but then at the moment after the twin accelerates away from his sister, he would say that same clock displays t=13.33. Then, still using CMIF, he could use the time dilation equation on that clock as well, provided he added 13.33 to the value he gets from the time dilation equation. That is why, when he arrives at that clock and his own clock says t'=32.66, he can calculate what that clock says as t=(32.66/1.2247)+13.33=40.00. Because with CMIF, the traveler says that all of the clocks at rest in the stay-home twin's frame tick at the same time-dilated rate, which is 1/1.2247 the rate of his own clock. They don't all display the same time at the beginning of his journey, but if he knows the time that any one of them displays at the moment after he accelerates away from his sister, then he can use the time dilation equation for any one of them, provided he adds in the time it displayed at the beginning of his journey, (just after he accelerates away from his sister).

    With your method, at the moment before the twin accelerates away from his sister, he would say that a clock synched to hers and located at x=23.09 would display t=0.00, also. But then at the moment after the twin accelerates away from his sister, he would think that same clock still displays t=0.00 but that it starts ticking at some strange rate. Then, still using your method, he could not use the time dilation equation on that clock. When he arrives at that clock, his own clock says t'=32.66 and that clock says t=40.00 just as with CMIF. But other than that, just about everything else about that clock is different (according to him) using your method compared to CMIF. With your method, the traveler says that all of the clocks at rest in the stay-home twin's frame tick at different rates, and about the only one that ticks at a rate of 1/1.2247 the rate of his own clock is his sister's clock. So he can't use the time dilation equation for any other clocks at rest in her frame except her own clock, even though he thinks the time they all display at the moment after he accelerates away from his sister is t=0.00. (I don't know, maybe your method allows him to start using the time dilation equation after he passes very close by a stay-home-frame clock, by adding the time he saw on it when he was momentarily co-located with it? Seems like it should, but I don't know, you tell me.)

    That is quite a big difference. I was thinking your method used the CMIF sometimes, and didn't use the CMIF other times, which would be arbitrary. But your method never uses the CMIF. It is completely different.
     
    Last edited: Mar 4, 2020

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