Basic Special Relativity Question

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

  1. Russ_Watters Not a Trump supporter... Valued Senior Member

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    I don't see why there is a need to bother bringing SR into such a situation at all except to make it just a little more complicated - nor why this is a confusing issue. Signal delay causes observations of far away events to happen after nearby events even though to an observer in between they happen at the same time. So what? That's a feature of Galilean Relativity, not SR and not even a little bit profound/confusing. Bringing SR into it just changes the timing calculations a little, it doesn't change the qualitative result.

    What am I missing about why this is at all interesting?
     
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  3. Tach Banned Banned

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    Post 8. You made some basic algebraic mistakes. Once you figure what the mistakes are, we can continue.
     
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  5. Pete It's not rocket surgery Moderator

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    Please point out the mistakes you found, so I can correct them.
     
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  7. Tach Banned Banned

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    Basic algebra would give the result \(tan(\theta)=-\frac{uV\gamma'}{c^2}\).
    You got the wrong sign and you got the wrong fraction. I do not know why you are so astute in pointing out other's mistakes while you seem totally incapable of detecting your own. This is a recurring problem with your derivation.
     
  8. Pete It's not rocket surgery Moderator

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    Sorry, I don't see how you get that result.
     
  9. Tach Banned Banned

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    One way would be by doing fraction simplification correctly. Try calculating \(\frac{y^"_B-y^"_A}{x^"_B-x^"_A}\) from your definitions.
    Fednis48 managed to figure out the fraction correctly but he still didn't get the sign right.
    Neddy Bate managed nothing.
     
  10. Fednis48 Registered Senior Member

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    You misrepresent me. After seeing your criticism, I thought I agreed with your fraction, but not your sign. After closer inspection, I realized I didn't agree with any part of your answer, and I think Pete is entirely right. We started this whole thread just so you could point out errors. Could you please be more specific than "the algebra is wrong somewhere"?

    Edit: Since you seem averse to precision, let's break it down with line numbers:

    \(\begin{align} 1.&y_A''(t'') &= -u\gamma'(t'' + \frac{Vx_A''}{c^2})\\ 2.& &= \frac{-ut''}{\gamma'} \end{align}\)
    \(\begin{align} 3.&y_B''(t'') &= -u\gamma'(t'' + \frac{Vx_B''}{c^2})\\ 4.& &= \frac{-ut''}{\gamma'} + \frac{uV\gamma'}{\gamma c^2} \end{align}\)
    \(\begin{align} 5.&\tan(\theta'') &= \frac{y_B''(t'') - y_A''(t'')}{x_B''(t'')-x_A''(t'')} \\ 6.& &=(\frac{-ut''}{\gamma'} + \frac{uV\gamma'}{\gamma c^2}- \frac{-ut''}{\gamma'})/(\frac{1}{\gamma'} - Vt''-(-Vt'')) 7.& &= \frac{uV\gamma' / \gamma c^2}{1 / \gamma'} \\ 8.& &= \frac{uV\gamma'^2}{\gamma c^2} \end{align}\)​

    By number, which step(s) are wrong? (Sorry about the ugly alignment; I'm not super-great at Tex.)
     
  11. Tach Banned Banned

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    LOL

    So, you can't calculate the fraction either. Let's try to further dumb it down for you: what is \(y^"_B-y^"_A\)?
     
  12. Pete It's not rocket surgery Moderator

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    \(\begin{align} y_B''(t'') &= \frac{-ut''}{\gamma'} + \frac{uV\gamma'}{\gamma c^2} \\ y_A''(t'') &= \frac{-ut''}{\gamma'} \\ y_B''(t'') - y_A''(t'') &= \frac{uV\gamma'}{\gamma c^2} \end{align}\)​
    Looks OK...

    \(\begin{align} x_B''(t'') &= \frac{1}{\gamma'} - Vt'' \\ x_A''(t'') &= -Vt'' \\ x_B''(t'') - x_A''(t'') &= \frac{1}{\gamma'} \end{align}\)​
    Looks OK...

    \(\begin{align} \frac{y_B''(t'') - y_A''(t'')}{x_B''(t'')-x_A''(t'')} &= \frac{uV\gamma'}{\gamma c^2}.\gamma' \\ &= \frac{uV\gamma'^2}{\gamma c^2} \end{align}\)​
    I don't see a mistake.
     
  13. Tach Banned Banned

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    Do you think you can calculate \(y^"_B-y^"_A\) without making basic algebra mistakes?
     
  14. brucep Valued Senior Member

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    Your 'nonzero' predicts' the result of the experiment is frame dependent. Since you and Pete are trying to prove such a paradox exists in reality. Tach gets a nonzero answer but he knows it's nonsense because the physics model of your experiment is contrived. Playing games with the physics is what gets done in these forums. Folks that have a good mathematical understanding of this physics know that your analysis is wrong since it predicts the local proper frame and the remote coordinate frame are both frame dependent. Start with 'events in spacetime' are invariant. This means all experimental measurements made in the local proper frame of the event are invariant while all experimental measurements made from the remote coordinate frame are frame dependent. Both the local proper frame observer and the remote coordinate frame observer agree on all measurements after the appropriate transformations are completed. You chose the local proper frame and you chose the remote coordinate frame. From there you added the RoS nonsense to try and show the rod supports were cut simultaneously in the local proper frame and cut at different times in the remote coordinate frame. What ever measurements you made from a remote coordinate frame recovers the local proper frame measurement if transformed correctly. The only relativistic effect associated with the rod is length contraction. When the rod is released it follows it's natural path in the weak gravitational field. You decided how that path ended. Both ends of the rod hit the floor at the same time in the local proper frame of the event. Any transformation from a remote coordinate frame better recover the local proper frame invariant measurements.
     
    Last edited: Apr 23, 2013
  15. Fednis48 Registered Senior Member

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    More posts while I was editing my own - whoops! But to answer your question,

    \(y^"_B-y^"_A=\frac{-ut''}{\gamma'} + \frac{uV\gamma'}{\gamma c^2}- \frac{-ut''}{\gamma'}=uV\gamma' / \gamma c^2\)
     
  16. Tach Banned Banned

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    I am not done yet with the "three musketeers" (Pete, Fednis48 and Neddy Bate). Once we get past their basic algebraic mistakes, the fun starts

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  17. Tach Banned Banned

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    Nope. Try again. I really narrowed down the expressions such that you cannot continue to make errors. It is fascinating to see you and Pete making the same basic error (and stubbornly repeating it).
     
  18. Pete It's not rocket surgery Moderator

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    Post 29, Tach.
    I don't see the error you describe. I'm beginning to suspect you may be mistaken.
     
  19. Fednis48 Registered Senior Member

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    Nope. The only thing I predict to be frame-dependent is the question: do both ends of the rod hit the ground at the same time? Simultaneity is relative, so this is fine. The results of any well-posed experiment should agree in both reference frames.

    Our thought experiment is that a metal rod is moving straight down, with no acceleration, in a fast-moving train. If you think this is "contrived", I don't know what to tell you.
     
  20. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

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    I'm gonna have the last laugh.

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  21. Pete It's not rocket surgery Moderator

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    Seriously?
     
  22. Tach Banned Banned

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    seriously, try again:

    In \(S''\):
    \(\begin{align} A'' &= \left(x_A''(t''), \ y_A''(t'') \right)\\ &= \left(-Vt'', \ -u\gamma'(t'' + \frac{Vx_A''}{c^2}) \right ) \end{align}\)
    \(\begin{align} B'' &= \left(x_B''(t''), \ y_B''(t'') \right) \\ &= \left(\frac{1}{\gamma'} - Vt'', \ -u\gamma'(t'' + \frac{Vx_B''}{c^2}) \right) \end{align}\)​
     
  23. Fednis48 Registered Senior Member

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    ...you still think that expression is wrong?! Now I really want you to point out the exact line-number of the error as per post 27, because the expression you disagree with is of the form \(a+b-a=b\). It doesn't get more simple than that.
     

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