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Question on Lorentz space-time transformations

  1. Jan 27, 2009 #1
    Hey, not strictly homework but this is probably the best place for it, I wonder if you guys can help me out with a past paper question I've been pondering:

    Two events occur at the same place in an inertial reference frame S, but are
    separated in time by 3 seconds. In a different inertial frame S' they are separated in
    time by 4 seconds.
    (a) What is the distance between the two events·as measured in S'? [4]
    (b) What is the speed of S relative to S'? [4]


    Now, I've tried rearranging the Lorentz transformations for the reference frame velocity in order to equate them, and sub in the numbers given above, but I still have trouble because the Lorentz factor (gamma) sticks around. It wouldn't be too hard to do the second part of the question first, and get the distance from that, but I'd like to know the "proper" method.
     
    Last edited: Jan 27, 2009
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  3. Jan 27, 2009 #2

    Fredrik

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    You can take the coordinates of the two events in frame S to be x=(0,0) and y=(3,0). A Lorentz transformation takes the origin to itself, so x'=(0,0), and according to your specifications we must have y'=(4,d), with d to be determined, and this is how you determine it:

    [tex]\begin{pmatrix}4\\ d\end{pmatrix}=\gamma\begin{pmatrix}1 & v\\ v & 1\end{pmatrix}\begin{pmatrix}3\\ 0\end{pmatrix}[/tex]

    The same equation gives you v, which is the velocity of S in S'. Note that I'm using units such that c=1.
     
  4. Jan 27, 2009 #3
    Hmm, thanks for your response, using column vectors with time as an extra co-ordinate was something I hadn't considered.

    The equation you quote appears to be similar to a set that I have derived, which is encouraging, I think I'm on the right track. I'm still not sure how to obtain d from the bottom row though.
     
  5. Jan 27, 2009 #4

    Fredrik

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    That part is extremely easy. Do you know how to multiply matrices? Do you see what the bottom row is?

    Edit: I meant that it's extremely easy if you have already obtained [itex]\gamma[/itex] from the first row and calculated v from it. That part is slightly harder, but only slightly.

    If you're suprised by the fact that I used "column vectors with time as an extra coordinate", I have to ask if you really know what a Lorentz transformation is. What else would a Lorentz transformation act on?
     
    Last edited: Jan 27, 2009
  6. Jan 28, 2009 #5

    turin

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    I wonder if you are making a pun. Anyway, the proper way to do it is the "proper" way to do it. You should know that

    [tex]\Delta{}t^2-\Delta{}x^2=\Delta{}t'^2-\Delta{}x'^2[/tex]

    is a Lorentz invariant. So, you can use the information in [itex]S[/itex] to calculate it, and then use this value in [itex]S'[/itex] to calculate [itex]\Delta{}x'[/itex] for part (a). Then, for part (b) simply use

    [tex]v'=\frac{\Delta{}x'}{\Delta{}t'}[/tex]
     
  7. Feb 15, 2009 #6
    @Frederik: I didn't say it surprised me, I just said it was a method of representing the problem I hadn't considered. Now that I re-read what you've said there, I understand perfectly how to solve it, I was simply misreading your equation, my bad.

    @Turin: No, no pun intended, but well spotted!#

    Thanks for the help, both of you. :smile:
     
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