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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    389
    Note that Fontenot's equation is a function ONLY of the two velocities v_1 and v_2. In particular, it is NOT a function of the separation of the twins at the turnaround. That is a bit surprising, since we know that the EFFECT of a given velocity change on the difference in the ages of the twins at the reunion depends LINEARLY on their separation at the turnaround. But Fontenot's simultaneity method gets the correct result at the reunion, because the duration of the center section of the age correspondence diagram (ACD) DOES depend linearly on their separation at the turnaround. I.e., the slope of the center section doesn't depend on the separation at the turnaround, but the width of the center section DOES.

    For details, see https://sites.google.com/site/cadoequation/cado-reference-frame .
     
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  3. Halc Registered Senior Member

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    1) That's very limiting.
    2) It isn't clear how to do that.
    Bob does two distant instant accelerations instead of just one. Maybe they're years apart, maybe not. For the second velocity change, what value do I plug into the equation? v_2 is obviously the new velocity going home (leg 3). What value do I use for v_1? The velocity in leg 1, or in leg 2? Maybe that answer is buried in there somewhere.

    There is no such ACD in the web page. You mention in there that you've done it, but you don't show any diagram or any of the work. There can be no v_1 in such an integrated case, so it isn't clear how the equation as written is applicable.

    I HAVEN'T thought about how to handle two and three spatial dimensions with my method. Years ago, I DID generalize my CADO equation (which simplifies computing CMIF simultaneity) to two and three spatial dimensions. But I suspect it would be much more difficult for my method. It's not a high priority for me.

    Nonsense. None of them are non-causal. I've pointed this out in multiple prior posts and you've not responded to any of them.
    OK, admittedly I drop the claim on D&G, but only because I don't understand it. I've not seen it clearly explained anywhere, including their own paper.
     
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  5. Mike_Fontenot Registered Senior Member

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    If the two instantaneous velocity changes are separated by only an infinitesimal time, then the first change has no effect and can be ignored. That's true in both CMIF simultaneity and in Fontenot simultaneity. I.e., v_1 in my equation is the velocity before any change has happened, and v_2 is the velocity after the ignorable velocity in between.

    If the two velocities are separated by a finite time, you just handle the two changes independently. Suppose v_a is the velocity change when the twins are born, v_b is the initial velocity change at some distance later, and v_c is the velocity change at a still greater distance. Then you get the first slope from the equation using v_1 = v_a and v_2 = v_b, and the second slope using v_1 = v_b and v_2 = v_c.

    I didn't put the ACD for the finite acceleration case on the webpage. I have little interest in the finite acceleration case, because it's so little different from the instantaneous velocity change case. That's one of the nice features of my method, compared to the CMIF method ... the instantaneous velocity change case is all you need in practice.

    And to answer your second question, my equation doesn't apply to the finite acceleration case at all. There IS no equation in the finite acceleration case ... it's just a numerical integration.

    I asked Steve Gull, many years ago, if he realized that his method is non-causal. His reply was "Yes, it is blatantly non-causal". I've described his ACD in my webpage. The slope changes from 1/gamma_1 to a slope greater than 1 well before the turnaround is reached. Since we can't know for sure that the traveler actually DOES change his velocity, the slope shouldn't change before velocity actually changes, for a causal method.

    Minguzzi's method is also non-causal, because his ACD has a slope of 1.0 on the entire outbound leg. Since we can't know if the traveler will actually change his velocity, the slope should be 1/gamma on the outbound leg, for a causal method. So Minguzzi's method is non-causal.
     
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  7. Halc Registered Senior Member

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    The two cases use completely different methods. Numeric integration sounds like you're using CMIF for continuous acceleration and F-method for the instant case. That's hardly 'so little different'. The former can have Alice's age going backwards for instance, so the curves are not similar. The latter is only useful for thought experiments, not anything practical where infinite power is not plausible.

    Consider Bob accelerating away at 1G indefinitely. Alice is going to age backwards in his frame. We see this now with distant objects that are continuously accelerating away from us due to dark energy. These distant objects have negative ages in our inertial frame, which is why inertial frames cannot be used to describe distant objects. Instead, cosmologists use something very similar to Minguzzi's method (using a reference event rather than a reference frame). They don't call it Minguzzi's method because Minguuzzi isn't consistent about the method (use it all cases, not just from one PoV), plus Minguzzi selects an event that doesn't foliate all spacetime.


    OK, I didn't see that in the document, maybe only for lack of finding it, so thanks.
    We have triplets Alice, Bob, and Charlie. Bob and Charlie take the ship out 10 years to planet Zog to retrieve the famed jewel of Zog. Zog is moving toward's Earth at -0.8c, but rather than waiting for it to pass nearby and let somebody else get the jewel, they're going to go out to it and pick it up first.
    The jewel would be damaged by the infinite power of the ship turning around, so the plan is to jettison Charlie near the planet where he matches its speed by the amazing ball pit they've got. He spends 20 minutes retrieving the jewel, and then leaps up as the returning ship scoops him up with a butterfly net, boosting his return speed by a paltry 0.066c. Bob stays on the ship and turns around after a minute or so. They are once again bound for home, but how old is Alice?
    They agree she's 5 years aged at first (a few minutes discrepancy maybe), but disagree as to her new rate of aging.

    S = (1 / gamma_2) + gamma_2 * (1 - v_2) * (v_1 - v_2)

    Bob computes S = .5 + 2 * 1.866 * 1.732 = 6.964
    Charlie computes S = .5 + 2 * 1.866 * .066 = 0.746

    Hmm... OK, that's two problems. One, they're with each other but in complete disagreement about the rate of her aging. Second problem is that Charlie has her aging too slowly, less than 1 in fact. She's not going to be her actual age in his frame until long after Charlie has passed Earth by.

    Did I do that wrong?

    OK, you're using non-causal to describe a situation where a calculation is based on actions that have yet to happen, where my take was an effect outside the future light cone of its cause.

    This assessment doesn't hold water. His method is never based on future intended actions. It is always computed for here and now. I must defend this because cosmologists use an objective variant of this method, the comoving reference frame, which is often asserted as the absolute frame for those absolutists among you.
     
  8. Neddy Bate Valued Senior Member

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    2,409
    Mike, I have a question for you.

    In section 4 you wrote, "The turnaround is an event, so everyone must agree about his age at the turnaround." Does that mean everyone, including people who may or may not have accelerated in the recent past? Or does that only apply to perpetually inertial people? I think you will say it applies to everyone.

    Based on that, please consider that for the outbound half of the journey, the traveler can be on a very long train which extends far enough so that the stay-home twin is always co-located with part of it. Likewise, for the inbound half of the journey, the traveler can be on a very long train which extends far enough so that the stay-home twin is always co-located with part of it. In fact, all the traveler has to do to at the turnaround point is jump off one train and onto the other. (You can add a ball pit, or a butterfly net if you want.)

    We can also fill the outbound train with an infinite number of synchronised clocks everywhere. Of course those clocks are synchronised in the outbound train's own reference frame. We can also fill the inbound train with an infinite number of synchronised clocks everywhere. Of course those clocks are synchronised in the inbound train's own reference frame.

    Now, it would be an 'event' when one of those train's synchronised clocks happened to be co-located with the stay-home twin. So I think you would agree that everyone would have to agree on her age at each of those events.

    So, when she is t=0.00 years old and she is co-located with an outbound train clock which reads t'=0.00 that is an event. And when she is t=26.67 years old and she is co-located with an outbound train clock which reads t'=32.66 that is an event. And when she is t=53.33 years old and she is co-located with an inbound train clock which reads t''=32.66 that is an event. And when she is t=80.00 years old and she is co-located with an inbound train clock which reads t''=65.32 that is an event.

    Note that if all of those t' and t'' values correspond to the traveler's age, then your age correspondence diagram should reflect those ages corresponding to each other. But it doesn't, because you are letting one person disagree with the times of events that everyone else must agree with.

    Or maybe you do let the traveler agree with the times of those events, but he just doesn't agree that they correspond to his age. That would mean that he disagrees that the clocks on the inbound train are synchronised, even while he is riding on that train. (Because his age is always the same as that of the train-clock that is co-located with himself.*) But then, after some time, he finally does agree that the clocks on the inbound train are synchronised, even though nothing has happened to the clocks, and even though nothing has happened to his location relative to them.

    * Note: The outbound train's clocks happen to be set so that they all display t'=0.00 in that train's frame at the time when the traveler first instantaneously accelerated, and the inbound train's clocks happen to be set so that they all display t''=32.66 in that train's frame at the time when the traveler did his second instantaneous acceleration (jumping from one train to the other).
     
    Last edited: Feb 19, 2020
  9. Mike_Fontenot Registered Senior Member

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    389
    I'll respond to Halc and Neddy as soon as I finish the analysis of the multiple velocity changes scenario. I worked on it most of the day yesterday. Almost done (I think). The analysis requires using the material in my webpage that defines my method, the heart of which is in Section 8, entitled "Pulses Partly in Both Halves of the Minkowski Diagram". Take a look at that section:
    https://sites.google.com/site/cadoequation/cado-reference-frame
     
  10. phyti Registered Senior Member

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    679
    Mike;

    Read section 8. It doesn't take 5 pages to explain the simple version of the 'twins'.

    An observer must make measurements to get factual data.

    Your data 44.9 and 38.6 don't fit the scenario.

    B1 and B2 are coincident at the reversal, which is one event.

    You can still use the coordinate transforms, for v=.6c, and B(x, t) = B(0,32).

    x' = (x-vt)/.8 = (0-.6(32))/.8 = 24

    t' = (t-vx)/.8 = (32-.6(0))/.8 = 40

    A(x, t) = A(24, 40).

    An answer in 3 lines!

    If only you would study the basics of SR theory.
     
  11. Mike_Fontenot Registered Senior Member

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    389
    The short answer is, "The traveler (he) can, after he changes his velocity, use my equation to determine how fast his home twin (she) is ageing (compared to his own rate of ageing) IF he currently agrees (about her current age) with the perpetually-inertial observer (PIO) riding along with him then. If he DOESN'T agree with the PIO, he CAN'T use my equation, and he must use the Minkowski diagram analysis to determine her relative rate of ageing then.

    I'll elaborate as soon as possible.
     
  12. Mike_Fontenot Registered Senior Member

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    389
    I'm going to try to upload two jpegs: the ACD and the Minkowski diagram, for the case of two separated velocity changes, starting from v1 = 0.57735 at the birth of the twins, then changing to v2 = 0.0 when he is 32.66 years old, and then later changing to v3 = -0.57735. The three different ages for him on that last velocity change are 36.89, 44.21, and 55.75 years old. I've marked on the ACD where he begins to agree with the perpetually-inertial observer, for each of those three scenarios ... they are the final low-slope lines of slope 1/gamma = 0.817.

    Please Register or Log in to view the hidden image!


    And here is the Minkowski diagram:

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    In the Minkowski diagram above, I failed to write the amount of her ageing during the upper portion of the L0 pulse: it is 4.227 years. That is what the PIO (perpetually-inertial observer) AFTER the velocity change calculates. The PIO BEFORE the velocity change determined that her ageing during the lower portion of the pulse (up to the point P0) is 7.974 years. So the traveler concludes that her aging during the entire pulse is 7.974 + 4.227 = 12.201 years. And she was 21.133 years old when she transmitted the pulse. So he concludes that she was 21.133 + 12.201 = 33.334 years old when he received her pulse. He was 36.887 years old then. The fact that he ADDS the amounts of her ageing during the two portions of the pulse (as determined by the two PIO's), to determine her current age when he receives her pulse, is the HEART of the definition of my simultaneity method ... everything follows from that.
     
  13. Halc Registered Senior Member

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    227
    The first change to v2=0 initiates a slope of 1.577 for Alice's age. This is consistent with the S equation you give.
    All three scenarios are identical except for the duration they linger at the turnaround point. That duration is not part of the S equation, and each scenario has Bob accelerating to the same v3=-.577c, so the S equation (based on this new speed -.577c, its gamma, and the old speed of 0) should yield the same slope for all three scenarios, which is 1.6333, but I see the ACD diagram shows different slopes for each of the three cases, none of them 1.6. It's like you're using new rules, but you haven't told us what they are.

    I also don't see our concerns addressed. For starters, if Bob lingers only a day before starting for home, the slope of 1.6333 isn't going to get Alice's age up to snuff by the time he meets her, putting him in an embarrassing situation where he tells her to her face she's some other age than what she says. OK, she'll take that as a compliment, but still.
    Second concern: The method is a function not of what is going on now, but partly of one's history, hence you can have two observers in each other's presence but differing in their history having an agreement as to Alice's current age.
    This second concern is echo'd by Neddy's scenario with the long train full of clocks which the PIO passenger says are synchronized, but Bob (who just got on the train) says are not. Again, without any acceleration of train or observers or effort to send sync signals to the clocks, suddenly Bob agrees they're in sync when they were not a minute ago.

    I don't see how your last post addresses any of that. The last post didn't seem to correspond to the short story posted just prior:

    Your method, using the S equation, is never going to agree with any PIO, so you're saying the equation can never be used? It can only be used if there is no other human nearby??? The rules seem to be getting more and more weird.
     
    Last edited: Feb 22, 2020
  14. Write4U Valued Senior Member

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    18,732
    And this is an example of serious science?
     
  15. Neddy Bate Valued Senior Member

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    No, just a strange tangent to it.
     
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  16. phyti Registered Senior Member

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    679
    Mike;

    Throughout history, people have traveled over the surface of the earth. It would be a natural inclination to represent distance as a horizontal plane, i.e. x and y directions. Time or any additional variable would be vertical on a graph. This eliminates readers having to learn a new method, when a standard already exists.

    From SR, B knows his coordinates as B(0, 32) at the reversal event. He then applies the coordinate/Lorentz transformations to calculate A's coordinates for that event and gets A(24, 40). That's it.
    The aos (los) is established when you make a measurement using light, as described in the clock synch procedure. It is thus meaningful and relative to an inertial ref. frame, which implies the return signal is only useful to the sender. In this simple, symmetrical case, there is an exception.

    The experiment is reduced to 3 computer-clock systems, A, B1, and B2, . Since a single system cannot physically follow the profile with the abrupt reversal, B2 assumes the role of B1 at the reversal event, which includes exchanging all data and synchronizing clocks.
    Initially A and B1 are synchronized. B2 is set to a random time, and is at a distance approaching A. B1 periodically sends a signal with time Bt1 to A. A reflects that signal with the time At. B2 can record receiving it at Bt2, after the reversal. When (Bt2-Bt1)/2 = B32 (time of reversal), then At was simultaneous with B32. This will work only when outbound and inbound speeds are equal.
    Using the transformations above, B is not obtaining A's age instantly as it may appear. The transformations are conditional, based on constant uniform/inertial motion. There is an assumption that a 2-way measurement will be made at the current speed. If there is a change in speed or direction, the aos must be re-established.
     
  17. Neddy Bate Valued Senior Member

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    2,409
    I agree that when Mike makes the readers of his paper draw the Minkowski diagram in that way, he asks a bit much. But I actually like having the time axis horizontal, because then the slope of the worldline is the velocity. Also, we tend to think of time as a timeline, and a horizontal timeline makes more sense than a vertical timeline.

    I have no idea where you got your numbers. Mike is using v=0.57735 which is gamma = γ = 1 / √(1 - v²/c²) = 1.224745. The traveler's transformations of the coordinates are from (x', t') = (-18.85, 32.66) to (x, t) = (0.00, 26.67) via the inverse Lorentz transformation, as follows:

    x = γ(x' + vt')
    Substitute:
    γ = 1.224745
    x' = 23.09/-γ = -vt' = -18.85
    v = 0.57735
    t' = 32.66
    Result:
    x = γ(x' + vt') = 0.00

    t = γ(t' + (vx' / c²))
    Substitute:
    γ = 1.224745
    t' = 32.66
    v = 0.57735
    Result:
    t = γ(t' + (vx' / c²)) = 26.67

    All of the above is from the moment before the turnaround. In the moment after the turnaround, the x' reverses sign, as follows:

    x = γ(x' + vt')
    Substitute:
    γ = 1.224745
    x' = 23.09/γ = 18.85
    v = 0.57735
    t' = 32.66
    Result:
    x = γ(x' + vt') = 46.18

    t = γ(t' + (vx' / c²))
    Substitute:
    γ = 1.224745
    t' = 32.66
    v = 0.57735
    Result:
    t = γ(t' + (vx' / c²)) = 53.33
     
    Last edited: Feb 24, 2020
  18. Neddy Bate Valued Senior Member

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    2,409
    Using the stay-home twin's coordinate system, she is always located at the origin (x=0) of her own coordinate system. At the time she is born, imagine there is a train whose own origin is located at x=23.09*2=46.18 on her x axis. Imagine that train is moving toward her at v=0.57735c. After 40 years her time, the origin of that train will be located at x=46.18-(vt)=23.09 which is exactly where and when the traveler does his turnaround. After a total of 80 years her time, the origin of that train will be located at x=46.18-(vt)=0.00 which is exactly where and when the two twins are reunited at the end of the journey.

    That is why the inbound leg of the journey has her located at x=46.18 for the whole inbound journey. Because at t'=0.000 in that train's coordinate system, the origin of the train was co-located with x=0.000 which is not where the stay-home twin is located, because the origins of those two systems only get farther apart with time. She gets closer with time. She will only be co-located with the origin of that train's coordinate system at the end of the inbound journey. So she has to be 46.18 light years away from her own origin, on the positive x axis, according to the inbound train. That point gets closer to the origin of the inbound train, as required.
     
  19. Neddy Bate Valued Senior Member

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    You are correct that the traveler says the stay-home twin's age is 40 at the turnaround point, provided no one is insisting on using hypothetically instantaneous velocity changes, and never allowing for the velocity to go to v=0.000c between v=+0.577c and v=-0.577c. In any realistic turnaround, the velocity would have to go to v=-0.000 at some time between v=+0.577c and v=-0.577c.
     
  20. Mike_Fontenot Registered Senior Member

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    He says her age at the turnaround [at the instant that his velocity first becomes zero] is 40, IF you are using the CMIF simultaneity method. If you are using Dolby and Gull's simultaneity method, or Minguzzi's simultaneity method, or my simultaneity method, then she is 26.67 years old then. That is the point labeled T in the ACD I posted above, for the revised scenario where he continues with v = 0 for a while before finally changing to -0.57735.
     
  21. Neddy Bate Valued Senior Member

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    If that is the case, then that is why all methods other than CMIF are wrong. She is indisputably 40 in her own reference frame at the turnaround point of her brother's trip, and everyone stationary with respect to her reference frame must agree, because simultaneity is only reference-frame-dependent, not person-dependent. When the traveler's velocity is v=0.000c with respect to her, he is stationary with respect to her frame, and should conclude her age is 40 at that time. I understand that you (and perhaps D, G & M) disagree, but that just means that the you think it is part of SR to make simultaneity something other than reference-frame-dependent, which it is not.

    I remember awhile back, you said you had posted a proof that CMIF was the only correct simultaneity in SR, but then you said that you discovered a flaw in your proof. Was that "flaw" by any chance the fact that simultaneity in SR is never anything other than reference-frame-dependent?
     
    Last edited: Feb 25, 2020
  22. Mike_Fontenot Registered Senior Member

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    In my statement above, the text I've highlighted in red is incorrect. The red text is true for my method, but in both the Dolby and Gull method, and in Minguzzi's method, her age is greater than 26.67 years old at T. In Minguzzi's method, she is 32.66 years old at T. I don't know how old she is at T in the D&G method, because I've never worked out this revised scenario for the D&G method. I'll see if I can determine that. But what I DO know is that in all three of those methods (the non-CMIF methods), there is no DISCONTINUITY in the ACD at T ... i.e., her age immediately before T is the same as her age immediately after T.
     
  23. Mike_Fontenot Registered Senior Member

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    389
    I THINK her age at T in the Dolby and Gull method is 33.3 years old. But I want to confirm that.
     

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