An Inconsistency Between the Gravitational Time Dilation Equation and the Twin Paradox

Discussion in 'Physics & Math' started by Mike_Fontenot, Sep 26, 2021.

  1. Mike_Fontenot Registered Senior Member

    Messages:
    360
    Obviously, the distant twin (she) doesn't suddenly feel like she's getting younger. At each instant in her life, her brain is in a state that is different from all of her other brain states. Nothing can change those states. But the accelerating observer DOES conclude that she instantaneously gets younger when he instantaneously changes his velocity in the direction away from her. And so, FOR HIM, she ACTUALLY does get younger. She doesn't agree. And all other perpetually-inertial observers, who are momentarily co-located with him when he instantaneously changes his velocity, disagree with him about her getting younger when he instantaneously changes his velocity, but they also all disagree AMONG THEMSELVES about what her current age is when the accelerating observer changes his velocity. And FOR EACH OF THEM, she ACTUALLY has the current age they compute, ACCORDING TO THEM. That's just the way special relativity IS ... different observers disagree, and they each think they are right, and none of them is wrong!

    What is really new, though, in my latest results, is the fact that the accelerating observer can assemble an array of clocks (and attending "helper friends" (HF's)), which give him a "NOW" that extends throughout all space (analogous to what Einstein did for inertial observers). And THAT guarantees that the accelerating observer's conclusions about the home twin's age are fully MEANINGFUL to him. His conclusions agree with the CMIF simultaneity method, which means that the CMIF simultaneity method is the only correct simultaneity method.
     
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  3. Mike_Fontenot Registered Senior Member

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    360
    I've written it all up, in the viXra repository:

    https://vixra.org/abs/2201.0015

    It's definitely not a short read, but it's all there. It's all free ... viXra doesn't charge anything for it.
     
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  5. Neddy Bate Valued Senior Member

    Messages:
    2,365
    That is all correct, but her brain states have nothing to do with it. SR is very clear that no clock ever goes backwards in its own reference frame.

    But as you said, Einstein already gave those perpetually-inertial observers the ability to assemble their own arrays of clocks (and attending "helper friends" (HF's)), which give each of them a "NOW" that extends throughout all space. So the home twin's age already has meaning to all of them. So, if you now give the accelerating observer a clock array, he does not get any more information from it than what he could have gotten from simply using the co-moving inertial frame (CMIF).

    Speaking of this new clock array you have given to the traveler, you never did answer my question in post #37: "Using the traveler's ONE array of clocks, what is the time on his array clock which is located next to his twin sister at the moment when he says that she changes from 40 to 10 years old?" Please let me know, as I am interested to know how one array of clocks handles this. -- Edited to add: I would also like to know, using the traveler's ONE array of clocks, what is the time on his array clock which is located next to himself at the moment when he says that she changes from 40 to 10 years old.
     
    Last edited: Apr 7, 2022
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  7. Mike_Fontenot Registered Senior Member

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    360
    I haven't worked out your particular case, but you can find the equations for the rate ratio R of the two clocks on page 8 of the pdf. Unfortunately, I never figured out how to get my Mac laptop's Pages program to number the pages, but when I bring up the pdf on my Mac, it shows the page number on the left. (It's in Section 3.) Or you can just count the pages.

    The rate ratio R (which says how much faster the leading clock (the HF's clock) is ticking, compared to the accelerating observer (the "AO") (who's opinion we want about the home twin's age), is

    R(t) = 1 + L A sech_sqrd( A t ).

    That equation is for the case where the AO is accelerating TOWARD the home twin. When he is accelerating AWAY FROM her, the plus sign changes to a minus sign.

    And the "Age Change" (AC) is the integral, from zero to tau, of R(t) dt. The result turns out to be analytically integrable: it's just

    AC(tau) = tau + L tanh( A tau ).

    (And the plus sign changes to a minus sign when he is accelerating away from her).

    For instantaneous velocity changes (where tau goes to zero and A goes to infinity, such that (A tau) = theta (the rapidity)), it's just

    AC = L tanh(theta)
    = L v.

    (And the above is preceded by a minus sign when he is instantaneously changing his velocity in the direction away from her).

    (Those sign changes can be be obtained automatically if you take "A" to be positive for accelerations TOWARD her, and negative for accelerations in the direction AWAY FROM her).

    I just recently recognized the above equation as a special case of the delta_CADO equation that I derived )and published) more than 20 years ago. The delta_CADO equation is given in the background material of my paper, in Section 6b. That's the equation I usually use to determine the CMIF simultaneity, because it is easier and quicker than drawing the lines of simultaneity (LOS's) on a Minkowski diagram. So all this proves that the array of clocks DOES give the same answer as the CMIF method.

    More important, the array of clocks proves that the CMIF simultaneity method is the ONLY correct simultaneity method. And it does that WITHOUT requiring the assumption that the accelerating observer must always agree with the perpetually-inertial observer who is momentarily stationary and co-located with him. He DOES agree with that perpetually-inertial observer, but we no longer have to make that assumption.
     
  8. Neddy Bate Valued Senior Member

    Messages:
    2,365
    That particular case is as follows:

    Twins (he & she) are born together at t=0.000. Right then the traveler (he) jumps on a train going v=+0.866c. When he is 20 years old, he reckons she is 10. Right then, he jumps off the train so now v=0.000c and he reckons she is 40. He quickly jumps back on that same train and again reckons she is 10. My questions is very simple, how does the one clock array that you have given him tell him that she changed from 40 to 10 in that last acceleration?

    Well, that is what I want to see for myself. You say you have given him a helper and a clock array, so I would like to know how he can use it to determine her age changes. If you would like, I can give the you my helper clock array times, and show you how the traveler uses them.
     
  9. Mike_Fontenot Registered Senior Member

    Messages:
    360
    I've told you enough that you can do it yourself. I've already shown you that the equation for AC in my paper is just a special case of my delta_CADO equation. And my delta_CADO equation gets exactly the same result you get when you draw a Minkowski diagram, with lines of simultaneity. You've actually already done that Minkowski analysis for your example, so you're done.
     
  10. Neddy Bate Valued Senior Member

    Messages:
    2,365
    But when I do it my way (the Minkowski & CMIF way), the traveler uses multiple clock arrays. You are saying you can do it with only one. Do the times on your clock array change when he accelerates? Are your array clocks even synchronised? If not, how can draw any conclusions from it? Those are the kinds of things I am trying to find out from you now, and if you would tell me your clock readings, I might be able to understand better.

    But since you don't seem interested in that, I will show you how he does it using my method:

    He uses one clock array when his velocity is v=+0.866c which tells him that her clock showing t=10 is simultaneous with his own nearby clock showing t'=20. He verifies this by looking at a photo taken by his helper which shows her celebrating her 10th birthday while the helper's clock shows t'=20. Knowing that he and his helper's clocks are synchronised, he concludes she is 10 when his nearby clock shows 20.

    Then, after he jumps off the train, and his velocity is v=0.000c relative to his sister, he uses a different clock array which tells him that her clock showing t=40 is simultaneous with his NEW nearby clock showing 40. Once again, this can be verified by his NEW helper who can take a photo of her celebrating her 40th birthday while the NEW helper's clock shows 40. Knowing that he and his helper's clocks are synchronised, he concludes she is 40 when his NEW local clock shows 40.

    Then once he jumps back on the train, he goes back to using the first clock array, and concludes she is 10 when his nearby clock shows 20.
     
    Last edited: Apr 8, 2022
  11. Mike_Fontenot Registered Senior Member

    Messages:
    360
    It doesn't sound like you've read any of my paper, or even any of my statements about my paper.

    Yes, the accelerating observer (the "AO") has only a single array of clocks. He is always stationary with respect to them (i.e., all of the distances between him and each of the clocks are constant. The clocks definitely do NOT stay synchronized. But the relationship between their readings at any time can be CALCULATED by the AO, and that effectively creates a "NOW" moment for him that extends throughout all space.

    Immediately before the acceleration starts, the clocks all ARE set to zero. But after that, for as long as the (constant) acceleration lasts, each clock tics at a rate R times the rate of the AO's clock. I call "R" the "rate ratio". I've told you before that it is

    R(t) = 1 + L A sech_sqrd( A t ),

    for all positive "t" until the constant acceleration ends. "L" is the fixed distance between the clock of interest and the AO.

    (The sech function (and also the square of the sech function) starts off equal to 1.0 when t is zero, and initially has a slope near (but slightly less than) 0.0, but then curves downward, and eventually approaches zero (from above) at infinity.)

    The reading on the clock, at some time tau, which I call the"age change" (AC),
    is the integral, from zero to tau, of R(t) dt. Fortunately, that turns out to be analytically integrable: it's just

    AC(tau) = tau + L tanh( A tau ).

    (And the plus sign changes to a minus sign when he is accelerating away from her).

    (The tanh function is zero at the origin, and rises initially at 45 degrees, but then gradually reduces that slope until it assymptotically approaches the horizontal line at 1.0. )

    For instantaneous velocity changes (where tau goes to zero and A goes to infinity, such that (A tau) = theta (the rapidity)), AC is just

    AC = L tanh(theta)
    = L v.

    THAT is the equation that corresponds to my delta_CADO equation, for the case where the velocity before the instantaneous velocity change is zero.

    (Note that the rapidity "theta" can get arbitrarily large when tau gets arbitrarily large, but the velocity "v" is limited by the tanh function to a maximum of infinitesimally less that 1.0, the speed of light.)

    (The right-hand-side of above equation is preceded by a minus sign when he is instantaneously changing his velocity in the direction away from her).

    (Those sign changes can be be obtained automatically if you take "A" to be positive for accelerations TOWARD her, and negative for accelerations in the direction AWAY FROM her).
     
  12. Neddy Bate Valued Senior Member

    Messages:
    2,365
    Mike, in the time it took you to you typed that, you could have calculated the times I asked for, and given me the answers directly.

    So, in the example I gave you, the time on the train-clock next to the traveler is t'=20 when he wants to know his sister's age. And you say the time on his distant helper's clock (located where the twin sister is located) is not 20, but it only differs by some known value. Since that value is known, we can compensate for that amount, and convert it as if both clocks display t'=20 simultaneously, just like the train-clocks. Thus your single clock array at this point has been converted to match the Einstein-synchronised clocks on the train.

    The sister's time here is t=10, so what I was wondering was if your clock array would give you that time directly. But if so, you don't know it, or don't want type t=10 for me for some reason.

    The next step is for the traveler to jump off the train. When he does that, all of the times on your single clock array change to some new non-synchronised values which esseentialy makes it a new array of non-synchronised clocks. Once again, the times can be compensated for by adding or subtracting the known amount they are off by, and now your clock array at this point has been converted to match the Einstein-synchronised clocks on the ground.

    The sister's time here is t=40, so what I was wondering was if your clock array would give you that time directly. But if so, you don't know it, or don't want type t=40 for me for some reason. When the traveler jumps back on the train, The sister's time is t=10 again, so what I was wondering was if your clock array would give you that time directly.

    At least that is as much sense as I can make of it. So I would say that you are still essentially using multiple arrays of clocks, because your clocks all change completely when the traveler accelerates. Anyway, none of this seems useful to me, because it seems you cannot easily give all of the numbers shown on all of the clocks, the way I did in post #47.
     
    Last edited: Apr 8, 2022
  13. Write4U Valued Senior Member

    Messages:
    18,613
    Question:
    Rather than comparing time frames inside the universe, is it not true that there is a universal NOW if measured against the "beginning of universal time", some 13.8 billion years ago.

    How can we make a statement of "duration" if we have set t = 0 as a beginning of universal time?

    Is the universe 13.77 billion years "old" ? If yes, then there must also be a universal time duration from then to "now", no?

    Universal time cannot be relative to Universal time, it must always be the same.
    It can be relative to our POV "inside" the universe, but that is not the question.
     
  14. Neddy Bate Valued Senior Member

    Messages:
    2,365
    You would probably need someone who understands GR to answer that, so that's not me. But in SR at least, where things are moving linearly relative to one another, there is no universal now. The best you can do is set up an array of synchronised clocks, which works for the reference frame in which it is stationary. But for someone moving relative to that array, they will say those clocks are not in synch, but they can set up their own clock array. That would work for them, but again, not for others moving relative to it.
     
  15. Write4U Valued Senior Member

    Messages:
    18,613
    I understand the SR perspective, but then you are talking from a POV inside the universe.
    An observer (clock) outside the Universe would see a single expanding singularity with a single increase of duration of existence, no?
    The universe is an expanding singularity, no ? How can a singularity be differently relative to itself?

    I just realized that may have to be a separate thread? If so I am not trying to hijack the thread.
     
    Last edited: Apr 8, 2022
  16. phyti Registered Senior Member

    Messages:
    627
    Mike;

    In SR, simultaneity is subjective for the observer.
    Example:
    She S (black) is at rest and He H (green) is moving away at .4.
    He passes S1, one of a system of synchronized clocks, which reads 1.79.
    H has sent a blue light pulse (Ht=.98) to S and receives an image (or time encoded signal) (St=1.50) at (Ht=2.29). By definition, H assigns his corresponding time (Ht=(2.29+.98)/2=1.64). His axis of simultaneity is parallel to his x axis.
    Simultaneous events for S are not simultaneous for H, if there is relative motion of S to H.

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  17. Motor Daddy Valued Senior Member

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    5,425
    Oh good, he knows the velocity of the train. Now all he needs to do is measure distance and he can know the time.

    So he travels .866 x 299,792,458 meters every second. So he travels 259,620,268.628 meters per second. So if he jumps on a train traveling at .866c, at t=0, and jumps off the train 259,620,268.628 meters from where he jumped on, then he traveled for exactly 1 second, and his sister is 1 second older than she was when he jumped on. At that point they are both 1 second old.

    You will not distort distance and time because your clock messes up. A clock is not time, it is a measuring device that MEASURES time.

    1 year is the Earth traveling 1 lap around the Sun. If the Earth makes 17 laps, then 17 years have elapsed for EVERYONE in the universe, regardless of their velocity.
     
  18. Neddy Bate Valued Senior Member

    Messages:
    2,365
    Welcome back MD! Long time no see.

    You are just ignoring the effects of SR (special relativity), but that is just how you roll, I guess. You could have mastered SR in the time you've been gone, and then come back and impressed everyone by showing us. Then you could still claim it was wrong if you wanted to, but at least you'd know what you were claiming was wrong.

    Oh well, congratulations on quitting smoking cold turkey, that is very impressive. I know a few people who quit, but they did so by smoking nicotine-free cigarettes instead. None of them were ever able to do it cold turkey.
     
  19. Motor Daddy Valued Senior Member

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    5,425
    That was 23 years ago.

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    I notice you didn't respond with any substance. Same ol' Neddy Bate, ignore the details when it doesn't fit your theory.

    How old is Sissy when Bro' travels 259,620,268.628 meters at .866c?
     
  20. Neddy Bate Valued Senior Member

    Messages:
    2,365
    Well yeah I ignored the details, because there is not much sense in me telling you what SR says, when you are not even considering SR in your analysis. Do you really want me to try to teach you SR?

    Your question does not specify whose clock measures 1 second, or who measures the distance. Bro and Sis agree on the relative speed being 0.866c while they are in relative motion, but each would use their own clocks and their own meter sticks. So your scenario could be interpreted to mean SHE measures 1 second of time and 1 lightsecond of distance, or it can be interpreted to mean HE measures 1 second of time and 1 lightsecond of distance.
     
    Last edited: Apr 25, 2022
  21. Motor Daddy Valued Senior Member

    Messages:
    5,425
    I never mentioned measuring time or clocks.

    YOU claimed the velocity was .866c.
    That necessarily means 259,620,268.628 meters every second. Or put another way, if they start together at t=0, and he rides the train to the 259,620,268.628 meter mark on the tracks, that is 1 second of travel, and necessarily means 1 second has elapsed for everyone in the universe, and that they both aged 1 second. No clock required because YOU know the velocity of .866c.

    BTW, How do you know the velocity is .866c at t=0? Did you measure the velocity in the past, prior to t=0? How do you know the velocity is still .866c after t=0?
     
  22. Neddy Bate Valued Senior Member

    Messages:
    2,365
    259,620,268.628 meters is 0.866 lightseconds. I prefer to use lightseconds rather than meters.

    Okay, let there be a nice red mark on the tracks that is 0.866 lightseconds away from where the twins are born. Let that distance be measured in the reference frame in which the tracks are stationary. That is Sissy's own reference frame for the whole thought experiment, because she is the "stay-home twin".

    First let's look at what Sissy says:
    Sissy says 1 second of time has elapsed on her own clock, and she also says that Bro must be arriving at that red mark on the tracks because his speed away from her is 0.866 lightseconds per second. But Sissy is smart (even though she was just born) and she already understands SR, so she also says that only 0.5 seconds has elapsed on Bro's clock, because the 1/gamma factor for 0.866c is 0.5. That is time dilation.

    Now let's look at what Bro says:
    Bro arrives at the red mark at the time his own clock displays 0.5, just as SR predicts. Knowing the relative speed between them is 0.866c, he figures the distance between the red mark and Sissy must be 0.5*0.866=0.433 lightseconds. That is length-contraction. While he is moving relative to the tracks, he says the tracks are length-contracted by the 1/gamma factor of 0.5.

    So we see that there are at least two different effects going on in SR, time dilation, and length contraction. Those are the easy things to understand in SR. There is actually one more effect called relativity of simultaneity, which is trickier to understand.

    It is a thought experiment. The train is moving at a constant 0.866c in this thought experiment, because it is stipulated to do so.
     
  23. Motor Daddy Valued Senior Member

    Messages:
    5,425
    Sissy knows that 1 second has elapsed, because the train traveled to the red mark at a velocity of .866c.
    Bro knows that being at the red mark at a velocity of .866c means 1 second has elapsed. But he also knows his clock changed rate compared to a standard second. His clock no longer keeps time accurately.

    He can verify this by watching the Earth make 1 lap around the Sun, while he rides the train. His clock will not show 1 year has elapsed while riding the train, but he knows that 1 lap of Earth around the Sun has occurred, and that is 1 year. He throws his clock out the window and calls Sissy to get an accurate time check, which she reveals to him that 1 year has elapsed, which is substantiated by the Earth completing 1 lap around the Sun. Lesson learned for Bro, never trust a clock on a train moving at a velocity of .866c!
     

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