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Special relativity - 2 meter sticks mooving along each other

  1. Jun 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Two sticks (both with the proper length ##\scriptsize 1m##) travel one toward another along their lengths. In the proper system of one stick they measure time ##\scriptsize 12.5ns## between two events:

    1. right ends are aligned (this happens first)
    2. left ends are aligned (this happens last)

    Question:
    What is the relative speed ##\scriptsize u## between the sticks?

    2. Relevant equations
    • Lorentz transformations
    • Lenght contraction

    3. The attempt at a solution
    I first draw the picture:
    mjSP0.png

    If i use the Lorentz transformation and try to calculate the relative speed ##\scriptsize u## i get speed that is supposed to be greater than the speed of light:

    \begin{aligned}
    \Delta x' &= \gamma (\Delta x - u \Delta t) \xleftarrow{\text{Is this ok? I don't think my system is in the standard configuration}}\\
    \frac{1}{\gamma}\Delta x &= \gamma (\Delta x - u \Delta t)\xleftarrow{\text{$\Delta x$ is equal to the proper length (in $xy$ the botom stick is standing still)}}\\
    \tfrac{1}{\gamma^2}\Delta x &= \Delta x - u \Delta t\\
    \left( 1 - \tfrac{u^2}{c^2}\right)\Delta x &= \Delta x - u \Delta t\\
    \Delta x - \tfrac{u^2}{c^2}\Delta x &= \Delta x - u \Delta t\\
    \tfrac{u^2}{c^2}\Delta x &= u \Delta t\\
    \tfrac{u}{c^2}\Delta x &= \Delta t\\
    u &= \tfrac{\Delta t}{\Delta x} c^2\\
    u &= \tfrac{12.5\cdot 10^{-9}s}{1m} 2.99\cdot 10^{8}\tfrac{m}{s} c\\
    u &= 3.74 c
    \end{aligned}

    In the solutions it says i should get ##\scriptsize u=0.5 c##, but i get ##\scriptsize u=3.74c##. I used the Lorentz transformation found on Wikipedia. These are for the standard configuration, but in my case is it the standard configuration?
     
  2. jcsd
  3. Jun 25, 2013 #2
    Try instead taking on the frame of one of the meter sticks

    Also, the step where you find the speed u seems a little suspect to me. Remember we're dealing with relativity!

    And yes this is a fairly standard configuration so, if used correctly, the lorentz transforms apply
     
    Last edited: Jun 25, 2013
  4. Jun 25, 2013 #3

    Doc Al

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    Staff: Mentor

    The standard configuration has the primed frame moving in the +x direction, so you'll have to reverse the sign of the velocity.

    Your second equation is incorrect. Why are you using the length contraction formula? Since both observers see the events happen at the ends of their meter sticks, what must they both see as the distance between those events? Careful with the sign of Δx.
     
  5. Jun 26, 2013 #4
    Thanks for pointing that out. I wasn't sure about this and now i am.

    I thought that ##\scriptsize \Delta x## in lorentz transformation ##\scriptsize \Delta x' = \gamma(\Delta x - u\Delta t)## is a length of a stick in coordinate system ##xy##, but it is not! It is the distance between events! and i should incorporate symbol ##\ell## for proper length to avoid confusion. This explains my mistakes right? What do you think?
     
  6. Jun 26, 2013 #5

    Doc Al

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    I think you've got the right idea. ##\scriptsize \Delta x## is the distance between the two events according to the unprimed frame; ##\scriptsize \Delta x'## is the distance between events in the primed frame.

    Of course, since the events take place at the ends of the rods, you already know the distances between the events in each frame.
     
  7. Jun 27, 2013 #6
    As reckoned from the S frame of reference, the meter stick in the S' frame of reference is length contracted, to a length of 1/γ meters. So, when the right end of the stick of the S' frame of reference is at x = 1 meter (as reckoned in the S frame of reference), its left end, as reckoned from the S frame of reference is at x =1-1/γ meters. This is the additional distance it has to travel before its left end coincides with the left end x = 0 of the meter stick in the S frame of reference. The amount of time it takes for the left end of the S' meter stick to reach the left end of the S meter stick is thus (1-1/γ)/v seconds, where v is in meters/sec. So,

    [tex](1-\sqrt{1-(\frac{v}{c}})^2)=v × 12.5×10^{-9}[/tex]

    or

    [tex](1-\sqrt{1-(\frac{v}{c}})^2)=3.75\frac{v}{c}[/tex]
     
  8. Jun 27, 2013 #7

    Doc Al

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    That's a perfectly OK way to solve the problem, and entirely equivalent to applying the Lorentz transformation as the OP attempted. (Of course!)
     
  9. Jun 27, 2013 #8
    Thanks Doc Al. But there is a problem here. There is no acceptable solution to this equation. 12.5 ns is too long a time interval for the problem to have a solution. Try solving the equation, seeing what you get, and substituting the result back in.

    Chet
     
  10. Jun 27, 2013 #9

    Doc Al

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    Hmmm.... Unless I made an error, I solved it the other day and got the expected answer of 0.5c. I'll check again.

    Update: I had no problem finding a solution.
     
  11. Jun 27, 2013 #10

    Doc Al

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    I would think that any time interval would have a solution. The longer the time interval, the slower the speed.

    Give it another try.
     
  12. Jun 27, 2013 #11

    TSny

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    I think Chestermiller is correct. There is no solution for this problem as stated. Squaring both sides of an equation can lead to an "extraneous solution".
     
  13. Jun 27, 2013 #12

    Doc Al

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    I think you might be right! Interesting. (I'll have to think this over when I've gotten some sleep.)

    Edit: I think it has a solution; I just had the wrong sign for Δx'. (The time order is reversed.)
     
    Last edited: Jun 27, 2013
  14. Jun 27, 2013 #13
    I got 0.5c also, but when I substituted it into the equation, only the negative square root gave a match (and this does not satisfy the physics).

    Chet
     
  15. Jun 27, 2013 #14

    TSny

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    I think the answer would be 0.5c if the events were defined when the ends of the sticks are as shown.
     

    Attached Files:

  16. Jun 29, 2013 #15

    Doc Al

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    Well, this statement of mine is nonsense. :uhh: :redface:

    Chestermiller and TSny are correct: The problem, as stated, has no solution. (The data are physically impossible.)

    I suspect that the problem was meant to be stated as TSny gave in his last post, where the time given is for the second stick to completely pass by the first stick. The events are as described by TSny.

    Note that the only difference when solving the corrected problem (as per TSny) using the LT is the sign of Δx'. Worth doing as an exercise.
     
    Last edited: Jun 29, 2013
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