Is relativity of simultaneity measurable?

Discussion in 'Physics & Math' started by Pete, May 8, 2013.

  1. Tach Banned Banned

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    No, they are not, you just need to use the reciprocal Lorentz transforms to prove that your claim is false.

    No, they don't , see above.
     
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  3. Pete It's not rocket surgery Moderator

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    Didn't I? Probably because it's trivial in principle (although complex in practice, particularly if you want femtosecond precision.)
    Try a photocell and light source on each clock, with an aligned mirror on the clock it passes. The photocell is activated when the passing mirror reflects the light source.
     
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  5. Pete It's not rocket surgery Moderator

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    You are wrong. We specifically set up the experiment to make the passing events simultaneous in \(S\); ie in \(S\), both the rail clocks and the ground clocks are \(L\) units apart.

    The rail clocks are \(L\gamma\) units apart in \(S'\), and \(L\) units apart in \(S\).
    The ground clocks are \(L/\gamma\) units apart in \(S'\), and \(L\) units apart in \(S\).
    This is basic length contraction.

    Yes, they do. In \(S\), the A clocks line up at the same time as the B clocks. Your equations:
    \(t_A=t_B=t_0\)
     
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  7. Tach Banned Banned

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    Yes, you get this type of error every time you use the dumbed down approach of length contraction. You need to use the complete Lorentz transforms and you will realize your mistake.
     
  8. Tach Banned Banned

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    No, you didn't. Because you couldn't.

    This is nonsense.
     
  9. Pete It's not rocket surgery Moderator

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    What error?
    The rail clocks are \(L\gamma\) units apart in \(S'\), and \(L\) units apart in \(S\).
    The ground clocks are \(L/\gamma\) units apart in \(S'\), and \(L\) units apart in \(S\).

    I'm using your maths, Tach:
    \(t_B - t_A = 0\)
    \(t'_B - t'_A = Lv\gamma/c^2\)

    The events happen at the same time in \(S\), as measured by \(S\) clocks.
    They happen at different times in \(S'\), as measured by \(S'\) clocks.
    We've measured relative simultaneity.

    No, it's pretty straightforward.
     
  10. Neddy Bate Valued Senior Member

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  11. Tach Banned Banned

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    Like I said, when you use the dumbed down approach of length contraction, you get errors in return.
    Using the full Lorentz transforms, you get the correct answers:

    In \(S'\):

    \(x'_A=0\)
    \(x'_B=L\)
    \(x'_B-x'_A=L\)


    In \(S\):

    \(x_A=\gamma(x'_A-vt'_A)=-\gamma v t'_A\)
    \(x_B=\gamma(x'_B-vt'_B)=\gamma L - \gamma v t'_B\)
    \(x_B-x_A=\gamma L -\gamma v (t'_B-t'_A)=\gamma L-\gamma v \gamma Lv / c^2=\frac{L}{\gamma}\)


    So, obviously, the pairs of clocks don't line up in either frame, exactly as I told you. No matter how you twist and turn, your setup is invalid and you will never be able to run a "Ros detection" test, despite your disrespect of the mainstream physics community.
     
  12. Pete It's not rocket surgery Moderator

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    Wrong starting condition.
    Start with
    \(x'_B=L\gamma\)
     
  13. Tach Banned Banned

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    This only make your explanation even more invalid: when the clocks at the A endpoint coincide, the B endpoints are at respectively \(L\gamma\) and \(\frac{L}{\gamma}\). You will never be able to come up with a workable experimental setup capable of measuring RoS. Never.
     
    Last edited: May 13, 2013
  14. OnlyMe Valued Senior Member

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    Tach, are you saying that there is no experimental set up that with today's technology RoS can be measured?
     
  15. Tach Banned Banned

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    The opposite, the technology is available , as I demonstrated to Pete. What Pete cannot do is to put together a coherent theory for the experiment. He simply fumbles from one theoretical mistake to another, as shown above.
     
  16. OnlyMe Valued Senior Member

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    If I understand your objection, it is because the train is moving relativistically.., so from either frame—the train's or the embankment's—the other is length contracted and their respective points A and B do not line up when their midpoints M and M' line up. Correct?
     
  17. Tach Banned Banned

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    ...and the misalignment happens no matter how Pete tries to fiddle the spatial separations of the clocks.
     
  18. Neddy Bate Valued Senior Member

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    ...and you are saying that the only way to measure RoS would be if the ends could somehow align simultaneously in both frames, at which time you would measure what? Lack of RoS?
     
  19. Tach Banned Banned

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    I can see you are still struggling with the setup.
     
  20. Neddy Bate Valued Senior Member

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    Actually I'm just pointing out that if there were no misalignment in either frame, then there would be no RoS to measure. But that's okay, keep digging.
     
  21. OnlyMe Valued Senior Member

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    I think what Tach is saying is that since A and B do not line up in either frame any measurements made in the two frames could not be compared directly.

    I'm not sure that is really necessisary though. It seems if you can establish that the flashes occur simultaneously in one frame and sequentially in the other, you have proof of RoS. Einstein's hypothetical goes further than that but it was a teaching tool. You don't really need to know anything about what parts of the train were aligned with the embankment points A and B, or even that the flashes were simutaneous in the embankment frame, or even simutaneous at all. All you need to know is that the event(s) were measured to occurr with a different timing when measured from the two frames.

    All Pete needed was the embankment frame, a rest frame for the simultaneous measurement of two events, and an observer in motion relative to the embankment frame. One observer measures the events as simutaneous (whether they are in fact or not) and the other measures them as sequential = RoS.
     
  22. Neddy Bate Valued Senior Member

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    That sounds a lot like your earlier post:


    Do you understand why Pete said your earlier post was not RoS? If so, can you tell me what you have changed in your newer post that resolves the problem?
     
  23. Pete It's not rocket surgery Moderator

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    This is the situation in \(S'\).
    Now check in \(S\):
    When the clocks at the A endpoint coincide, the B endpoints are at respectively \(L\) and \(L\).

    Tach, this is really simple - the experiment is arranged so that the clocks line up simultaneously in the lab frame.
    Is that impossible?
     

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