Basic Special Relativity Question

Discussion in 'Physics & Math' started by Fednis48, Apr 22, 2013.

  1. Pete It's not rocket surgery Moderator

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    The Lorentz transform tells us what would be measured in \(S''\).

    Posting [post=3067875]maths[/post] does:
    Your equations are a subset of my equations:
    \( \begin{align} \large R''(k) &= \left(\gamma'(t' - \frac{Vk}{c^2}), \ \gamma'(k - Vt'), \ -ut' \right) \ \normalsize \left{0<k<L \\ -\frac{H}{u}<t'<0\right} \\ &= \left(t'', \ \frac{k}{\gamma'} - Vt'', \ -u(\frac{t''}{\gamma'} + \frac{Vk}{c^2}) \right) \ \normalsize \left{0<k<L \\ -\gamma'(\frac{H}{u} - \frac{Vk}{c^2}) < t'' < -Vk\gamma'\right} \\ \large R''(k) &= \left(\gamma'(t' - \frac{Vk}{c^2}), \ \gamma'(k - Vt'), \ 0 \right) \ \normalsize \left{0<k<L \\ t'>=0\right} &= \left(t'', \ \frac{k}{\gamma'} - Vt'', \ 0) \right) \ \normalsize \left{0<k<L \\ t'' >= \frac{-Vk\gamma'}{c^2}\right} \\ \end{align} \)​

    They predict that the rod bends in \(S''\) and not in \(S'\), and that's OK - it does not imply any absolute physical contradiction.

    Length contracted in the lab reference frame. The proper length is obviously unchanged. Length contraction can't crush a brittle rod, just like the transient bending of the rod in \(S''\) can't break the rod.
    In that sense, the sense used by Gron and Johanessen, they're both not 'physical effects.'
    This semantic sidetrack is done. If you want to discuss the physicality of length contraction, open another thread.
     
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  3. Tach Banned Banned

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    So, measure it. I challenge you to put together a valid experimental setup. You can't.


    Measure the timestamps \(t"=\frac{-Vk\gamma'}{c^2}\) in the following:

    \( \begin{align} \large R''(k) &= \left(\gamma'(t' - \frac{Vk}{c^2}), \ \gamma'(k - Vt'), \ -ut' \right) \ \normalsize \left{0<k<L \\ -\frac{H}{u}<t'<0\right} \\ &= \left(t'', \ \frac{k}{\gamma'} - Vt'', \ -u(\frac{t''}{\gamma'} + \frac{Vk}{c^2}) \right) \ \normalsize \left{0<k<L \\ -\gamma'(\frac{H}{u} - \frac{Vk}{c^2}) < t'' < -Vk\gamma'\right} \\ \large R''(k) &= \left(\gamma'(t' - \frac{Vk}{c^2}), \ \gamma'(k - Vt'), \ 0 \right) \ \normalsize \left{0<k<L \\ t'>=0\right} &= \left(t'', \ \frac{k}{\gamma'} - Vt'', \ 0) \right) \ \normalsize \left{0<k<L \\ t'' >= \frac{-Vk\gamma'}{c^2}\right} \\ \end{align} \)​

    No, they don't. You keep "wishing" that the "rod bends", there is no measurement that would confirm that.

    Neither is the rod bent in the train car frame.

    but it flattens the ions and it packs them tighter, a measurable effect

    Nice try, the length contraction effect is measurable in the lab frame. Now try that with the "rod bending" in the lab frame. Time for you to stop playing games, Pete.
     
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  5. Undefined Banned Banned

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    Is this guy even more stupid than previously imagined? First he says this:
    Then in the same post he says this:


    Which depend on animations!

    No actual "proof" of anything from Tach's links, just more demonstrations of his stupid double standard on what is "proof" from others and what is "proof" from himself!

    Is this guy the site clown or something? No one could be so stupid to make such a self contradicting post like that for real, could they?
     
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  7. Tach Banned Banned

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    Nope, they don't, they depend on actual measurements. You need to try understanding the links, not jump at soundbites like "animations". The people conducting the experiments really measured the effects and generated the animations in order to illustrate the measured effects, so people could understand what went on in the experiment they already conducted.
     
  8. Undefined Banned Banned

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    Are you for real? Measuring "effects" and "interpreting" these effects are two different things. The "effects" only show "collision". The "results" are then "interpreted via theoretical framework of assumptions which are used to "model" via "simulations". No where along does any "proof" of SR theoretical length contraction arise that is independent of "simulations" of "interpretations" of "assumptions" etc etc.

    You need to go to a school for scientists and learn the difference before you try to "correct" other people again.
     
  9. Tach Banned Banned

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    Did I say any different? The scientists that put together the two websites (that, incidentally, agree with each other), first developed the theoretical foundations, then did the experimental measurements, and only after that they produced the explanations that include the animations. This is how science is conducted.

    Why are you so twisted in your knickers? I am just clearing your misconceptions.
     
  10. Undefined Banned Banned

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    But it's not the "proof" you were claiming it as, is it? It's theoretical interpretation of something. Epicycles were theoretical interpretations of something, but they too weren't "proof" of Earth as the center of the universe. Take the "hit" now before you dig yourself deeper, Tach. Admit you have no "proof" of what you claimed to Pete, and just continue the discussion with him in a better frame of mind so that I can improve my naive understandings properly without all the "muddy waters" you stir up. And stop playing games if you really want to be a scientist one day.
     
  11. Tach Banned Banned

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    For the mainstream people, it is a proof. For crackpots, not so much <shrug>
     
  12. Undefined Banned Banned

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    An animation of a simulation of a set of assumptions used to theoretically interpret something is a sufficient "proof" for some people of actual physical effect is as interpreted? Not very "scientific" if you ask me. But then I am naive enough that my mind won't accept or equate such things as "proof" of anything physically true until better support for that claim is made available. Have you any?
     
  13. Neddy Bate Valued Senior Member

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    Tach, I have a sincere question for you. Earlier in the thread you said something about the coordinate times being just labels. If the endpoints of the rod impact the floor simultaneously in both frames (as you claim), then what would be the purpose of "labeling" two different times? Why would relativity require that, rather than just labeling both times the same? I'd like to understand that better.
     
  14. Tach Banned Banned

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    I didn't claim that, you keep trying this cheap trick and I keep correcting you.
    Test for you: What did I say exactly about the impact?
     
  15. Tach Banned Banned

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    That's not what I said, this is your fringe take of what you think I said , try reading again.
     
  16. Undefined Banned Banned

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    It's what your "proof" links amounted to when looked at closely. So by extension it was what you were saying or you wouldn't have offered them to support your claims to Pete. Try reading your linked "proofs" again.
     
  17. Neddy Bate Valued Senior Member

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    I'm not trying any kind of a trick. You said this:

    So now I am asking, if the rod does not hit sequentially, then what is the purpose or meaning of the clock labels which do not match?
     
  18. Tach Banned Banned

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    Much better, you found the actual way I phrased things. The timestamps (labels) \(t"=-\frac{Vk\gamma'}{c^2}\) applied in \(S"\) have no physical meaning, they aren't measurable.
     
  19. Neddy Bate Valued Senior Member

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    What I am asking is why would relativity give us equations to calculate times which are meaningless and non-measurable? Why not just say t''=t' instead? Are the labels good for anything at all?
     
  20. Tach Banned Banned

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    Well, not all mathematical concepts are measurable. A very good example is the amount of RoS, it isn't measurable. Yet, the theory predicts it.

    Because this is a reductionist, incorrect way of thinking, known to be wrong since Einstein discovered RoS.



    Sure, they allow us to organize our thoughts, convey ideas, etc. Unfortunately, they aren't measurable entities.
     
  21. Pete It's not rocket surgery Moderator

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    More games?
    I'm guessing that the point you're not making is that the animation values don't match the equation.
    This is because in the animation, I have k=0 at x'=-3, instead of at x'=0 (The rod is 6 units long, the middle of the rod is at x'=0).
    Here it is again with k=0 at x'=0:

    Please Register or Log in to view the hidden image!


    You can see a slowed-down, collision only animation here: RodCollisionSlow.gif
    And the individual collision frames here: http://sdrv.ms/10nGCzm
    And here are the \(t''\) values for the collision:
    (k, t'')
    (0.00, 0)
    (0.75, -1)
    (1.50, -2)
    (2.25, -3)
    (3.00, -4)
    (3.75, -5)
    (4.50, -6)
    (5.25, -7)
    (6.00, -8)

    Correct

    A measurable effect in the lab frame, just like the rod bending is a measurable effect in the platform frame, according to the Lorentz transform.
     
  22. Neddy Bate Valued Senior Member

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    In Einstein's thought experiment with the train and the two lightning strikes, the observer at the midpoint of the train sees the lighting strikes at two different times. He could measure the time interval between those two strikes, then later when he is at rest with the platform, he could show his measurement to the observer at the midpoint of the platform. The observer at the midpoint of the platform would say, "I saw the two lightning strikes simultaneously". Isn't that an example of measuring RoS?

    But you say the endpoints of the rod impact the floor simultaneously in both frames. That means there is one time when the endpoints impact in the train frame (t') and there is one time when the endpoints impact in the platform frame (t''). It seems to me that you are saying t''=t' are the real times of impact.


    I don't understand how meaningless, unmeasurable times would help with that at all.
     
  23. Pete It's not rocket surgery Moderator

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    That would imply that the train is not length contracted in \(S''\).
    Please demonstrate length contraction in \(S''\) without using \(t''\).
     

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