A New Simultaneity Method for Accelerated Observers in Special Relativity

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

  1. Neddy Bate Valued Senior Member

    Messages:
    2,112
    There is no single time of the turnaround, that is an idealization. In reality there are only velocities which are positive, zero, or negative. You are using the positive velocity to get one line of simultaneity, and then you fail to use the negative velocity to get a completely different line of simultaneity. You just claim it is "obvious" that you can do that, even though it is not obvious at all.

    No you haven't. All you've clearly done is make a mistake.

    No. I don't see any violation in the principle of causality in using the correct equations of SR or drawing correct Minkowski diagrams. If you want to show a violation of causality you need to show a "cause" occurring at a later time than its "effect". But first you need to assume the following is true, which you have not even done:

    STEP 1 ASSUME THIS IS TRUE:
    The calculation you show for the amount of her aging during pulse transit, 26.67 - 16.91 = 9.76, is only valid for the outbound leg of the trip, not the inbound leg. The reason is because the line of simultaneity you used to get 26.67 is not valid in the reference frame of the inbound leg. The line of simultaneity for the inbound leg would point to 53.33 and so your calculation for the amount of her aging during pulse transit for AFTER THE TURNAROUND should be 53.33 - 16.91 = 36.42.

    STEP 2 SHOW THAT IT RESULTS IN A CAUSE OCCURING AT A LATER TIME THAN ITS EFFECT:
    You fill in this part, please. If you can show a real violation of causality here, I will personally write to the Nobel Committee and recommend you. But you can't because the above (step 1) is standard SR and has stood for over 100 years.
     
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  3. phyti Registered Senior Member

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    Mike;
    In the graphic of post 51,
    compare signal transit times to object E,
    and assigned times by observer.
    (send, receive, assign t)
    B straight (0, 6, 3)
    C straight (3, 12, 7.5)

    The path BC with instant reversal is only possible in cartoon physics.
    The path BC with B and C exchanging data at reversal is possible and would verify the time for the 2-segment path is less than the A 1-segment path.

    As shown in the D&G paper, the path with a transition curve from B to C, is not an inertial path. There is no instantaneous aos for each position on the curve. If there was what could you do in an instant?

    For some perspective, assume you change direction every 2.5 seconds to approximate the curve. That allows you to make a measurement over a distance of 1.25 lightsec, which is the distance to the moon from earth surface. That limits long distance measurements. To resynch your clocks requires time.
     
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  5. Neddy Bate Valued Senior Member

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    Mike,

    First of all I would like to say that I do like your idea of drawing Minkowski diagrams with the time axis horizontal, and the space axis vertical. I like it because it makes the slope of the worldline equal to the velocity, and of course whether the velocity is positive or negative is very important.

    That brings up an important point, which I think has been missing from this conversation so far. With a hypothetical instantaneous turnaround point, the slope at the peak of the diagram is undefined. That is why I always say either "before the turnaround" or "after the turnaround" because we cannot say much about the actual turnaround itself, because of the undefined slope.

    But I think you are quite focused on determining the simultaneity line of the actual midpoint of the journey, not before, and not after. For that, there really should be a radius where those two worldlines meet, instead of just an intersection of two lines. That way we do not have an undefined slope at the midpoint. So, I drew a diagram showing that:

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    Now hopefully you can see why I say that "before the turnaround" the simultaneity line points to 22.67, and "after the turnaround" the simultaneity line points to 53.33. But the shift is not instantaneous now that we have a continuously defined slope. It sweeps over from pointing leftward, to pointing straight down, to pointing rightward.

    So, if you are really concerned with determining the simultaneity line of the actual midpoint of the journey, not before, and not after, then that simultaneity line points to her age being 40.
     
    Last edited: Jan 16, 2020
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  7. Halc Registered Senior Member

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    Agree. This is carried over from another forum in which I post as noAxioms.

    Section 7 contains errors, as pointed out by Neddy I see.

    I mean the twin scenario as described in a non-absolute interpretation (not Neo-Lorentz for example) by somebody not offended by non-parallel lines of simultaneity from different frames crossing each other.
    I notice Dolby considers that a problem to be resolved. I didn't look at Gull or Minguzzi, but I assume similar issues with crossing lines.

    Those guys seem to suffer from the same problem as do you: asserting that there is a current age for the twin not present. Her age simultaneous with any given event in the traveler's worldline is in fact a frame dependent thing and has nothing to do with what the traveler happens to be doing. He doesn't cause her to age by a bunch of years by turning around. She's still young in the outbound frame and always was much older in the return frame, and all the guy does at the turnaround point is change the frame in which he is stationary from the former to the latter.[/QUOTE]
     
  8. Mike_Fontenot Registered Senior Member

    Messages:
    251
    I have discovered an error in my proof that the CMIF simultaneity method is incorrect. Near the end of Section 7, I wrote:

    "So when he asks, at the instant "T+", "What is her current age right now?", he must still conclude
    that she aged by 9.76 years during the complete transit of her message. And the message said she
    was 16.91 years old when she transmitted the message. So therefore he must conclude at the
    instant "T+" that she is still 26.67 years old."

    That last sentence is incorrect. ALL he can conclude from that argument is that it is still true that she was 26.67 years old (according to him) at the instant "T-". I did NOT prove that she was still 26.67 years old at the instant "T+", which I was claiming.

    In order to complete the proof correctly, I need to consider another pulse. She transmitted the first pulse when she was 16.91 years old. Call that the instant in her life "TAU". She transmits the additional pulse at the instant "TAU+", which is only infinitesimally later than the instant "TAU". The first pulse was received by him at the instant "T+". The additional pulse is received by him at the instant "T++", which in only infinitesimally different than the instant "T+". So when he receives that additional pulse, he knows that her age only changed by an infinitesimal amount during his instantaneous velocity change. CMIF simultaneity says that she aged by a large finite amount during his instantaneous velocity change. Therefore the CMIF simultaneity method is incorrect.
     
  9. Neddy Bate Valued Senior Member

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    2,112
    That doesn't fix your error, because you used a line of simultaneity based on v= +0.577c to obtain the value 26.67 and that is not valid for the exact midpoint of the trip where the velocity is either v=0.000c (reality) or undefined (for an instantaneous turnaround). The slope of the worldline is the velocity. What do you consider to be the slope of the worldline at the midpoint of the trip?
     
  10. Mike_Fontenot Registered Senior Member

    Messages:
    251
    The slope is undefined at that exact point, cause the slope is discontinuous there. And that causes me no problem at all. The slope of the two legs of his worldline are well-defined at all other points on the diagram.
     
  11. Neddy Bate Valued Senior Member

    Messages:
    2,112
    Okay, so if the undefined slope is not a problem for your "proof", then your "proof" should also work for a case where there is no point on the worldline with an undefined slope. I posted a Minkowski Diagram for such a case in post #63.

    If you let the stay-home twin send three light pulses, you can have one intersect the worldline just before the radius (the last moment where the slope is still v= +0.577c), another intersect at the exact midpoint of the worldline (the moment where the slope is v= 0.000c), and another intersect the worldline just after the radius (the first moment where the slope is v= -0.577c). In the interest of keeping the pulses as close together as possible, you can make the radius arbitrarily small, but the slope still must be defined everywhere on the worldline.

    So, can you show us what your "proof" would look like using that situation?
     
  12. Mike_Fontenot Registered Senior Member

    Messages:
    251
    The instantaneous velocity change scenario is the simplest and easiest scenario to analyze, and it gives me everything I need for the proof that the CMIF simultaneity is incorrect. Using a finite acceleration scenario just unnecessarily complicates things, and I'm not at all interested in doing that.

    Just FYI, though, I HAVE extended my simultaneity method to handle finite accelerations. It requires a numerical integration, which I carried out with a C program that I wrote. (I doubt that a closed-form analytical solution exists). For a finite acceleration of 1g that last a few years, the resulting age correspondence diagram (ACD) is almost indistinguishable from the ACD for the corresponding instantaneous velocity change scenario. The only case where the finite solution was interesting was for the case where a low-power rocket is used for the entire trip, producing 0.2g, and reversing the rockets direction twice during the trip (first in order to produce the turnaround, and then to produce the slowdown before the reunion).
     
  13. Halc Registered Senior Member

    Messages:
    227
    This seems to be a major factor in this strange way your method frames things.
    At the instant T-, when he's asked what his speed is, he replies 'stationary', and a short time and intense acceleration later at T+, when he's asked what his speed is, he still replies 'stationary.' This seems to be a contradiction that his acceleration has done nothing to his momentum. It seems to violate what the word stationary means.

    No you don't. The same pulse is going to get there at T+, so a different one is not. You need to recompute the transit time of that same pulse in the new frame.

    BTW, I also have no problem with the instantaneous turnaround. It is effectively 'short enough not to require accounting in our computations'.
     
    Last edited: Jan 18, 2020
  14. Neddy Bate Valued Senior Member

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    2,112
    It is actually very easy. No matter how small you make the radius in the wordline, there will still be a slope of v=0.000c at the exact midpoint of the trip. So, at that point, using a simultaneity line based on any other velocity other than v=0.000c will be flat out incorrect.
     
  15. Neddy Bate Valued Senior Member

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    2,112
    The instantaneous turnaround is meant to simplify things. But Mike is using it to complicate things.
     
  16. Mike_Fontenot Registered Senior Member

    Messages:
    251
    No, he replies "+0.57735 ly/y wrt her".

    No, he replies "-0.57735 ly/y wrt her".
     
  17. Halc Registered Senior Member

    Messages:
    227
    You've not been using 'wrt her' in any of your assertions. In that frame, her age doesn't change at all when he turns around. You're being inconsistent with which frame you're referencing.
     
  18. Neddy Bate Valued Senior Member

    Messages:
    2,112
    Mike, just to be clear, are you saying that it is too complicated for you to consider an extremely small radius at the midpoint of the worldline? I find that odd.

    It seems to me that you would just say that with a small enough radius, the situation would be essentially identical to the instantaneous turnaround scenario, except instead of the midpoint of the worldline having an undefined slope, the midpoint would have a slope of zero. Why is that too complicated for you? I'll let you make the radius as small as you want, as long as it is not zero radius.
     
    Last edited: Jan 18, 2020
  19. Mike_Fontenot Registered Senior Member

    Messages:
    251
    Just FYI, your drawing of the finite acceleration scenario is wrong. The finite acceleration should START at the instant "T". It shouldn't start before "T" and be halfway through the acceleration at "T". The slope midway through the acceleration is of no special importance to me.

    But for the purpose of proving that the CMIF simultaneity is incorrect, I'm not at all interested in the finite acceleration case. I have proved that, in the standard twin paradox scenario with an instantaneous velocity change, the CMIF simultaneity is incorrect. If you want to identify a specific error in THAT proof, for THAT scenario, then do it. Otherwise, I'm not interested.

    One other comment: I'm not sure you are aware that the fundamental ASSUMPTION that DEFINES the CMIF simultaneity method is that at each instant in his life, 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. When you tell me that I am improperly ignoring the line of simultaneity that intersects the traveler's worldline shortly after the instantaneous velocity change, my reply is that if I accepted the legitimacy of that LOS, I would be assuming that CMIF simultaneity is true at the beginning of my proof that CMIF is incorrect! Thanks, but no thanks.
     
  20. phyti Registered Senior Member

    Messages:
    471
    Mike;

    This is to show the difference between assuming and knowing.

    At the reversal event, B and C see the same image of At=16.
    B assigns At=16 to Bt=20. Using that ratio of .8, he calculates At=25.6 when Bt=32, assuming her speed hasn't changed.
    For C to assign a time for At=16, via his aos (green/purple), he would have to send a signal at
    (38.4/.4=96)t before At=16, or At=-80. Ct=120(.8)=96 when he arrives at the reversal. If t is lyr, there would be no funding.
    Let C assume the role of B at reversal with an exchange of data. C can verify the A clock reading at At=40/Ct=36.8.

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  21. phyti Registered Senior Member

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    Mike;

    Al moving in inertial frame for time interval A, which extends into the past.
    Al decelerates to inertial frame for time interval B.
    Al decelerates to inertial frame for time interval C.
    Al decelerates to inertial frame for time interval D.
    Al decelerates to inertial frame for time interval E, which extends into the future.

    For each short time interval Al re-synchs his clocks (blue/orange rectangles) and establishes his aos (green/purple). I.e., they don't exist until he makes (imagines) them. They do not extend into the universe without limit, as commonly stated. Making a measurement using light is the same method as synchronizing local clocks. During the short time intervals while decelerating, he can't make any long distance (lyr) measurements.

    The reality, there is no instant knowledge of the state of a distant system, since it requires some form of information transmitted at light speed. All knowledge is historical, to a degree dependent on distance, from nano seconds to a million lyr.

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  22. phyti Registered Senior Member

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    Mike;
    Clocks c1 and c2 static .
    c2 signal path is ABC, aos is purple.
    c2 accelerates toward c1, path is ABD, aos is blue.
    c2 accelerates away from c1, path is ABF, aos is green.
    colors are reversed on chromebook?
     
  23. phyti Registered Senior Member

    Messages:
    471
    Mike;

    questions:

    section 1.
    "But the ‘traveling’ twin (whom I’ll refer to throughout as ‘he’) is not perpetually inertial, and so he is not always entitled to use the time-dilation equation."

    [and in the real world, no one is always in an inertial frame (of reference), but for practical purposes, a frame with no significant acceleration can approximate an inertial frame. It's done every day on earth.]

    "He might believe that he can use the time dilation equation whenever he is not accelerating,"

    [true]

    "and he might believe that nothing happens to their respective ages during his instantaneous turnaround."

    [She is not affected by object motion at a distance. (Action at a distance was obsolete when light speed was measured as finite.) He is affected by his own motion.]

    "If so, he will conclude that she will be the younger at the reunion, which we know is incorrect."

    [How does he conclude that?]
     

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