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The middle frame lemma

  1. Feb 10, 2009 #1
    Hi,

    I'm currently reading "Introduction to SR", by Wolfgang Rindler (2nd Ed, 1991 Oxford Science Publications).
    Now I have troubles with Problem Ex I No. 7.
    Since this problem is not homework and also a general question on SR, I post it here.
    (If I should post other problems from that book in the homework section anyway, I will do that.)

    The middle frame lemma is given as follows:

    Given two inertial frames (IF) S and S', there exists a third IF S''' relative to which S and S''
    have opposite velocities of the same magnitude. We can picture it 'in the middle' between S and S'.

    Now the question:

    Prove that at any instant there is just one plane in S on which the clocks
    of S agree with the clocks of S', and that this plane moves with velocity

    [tex]\frac{c^2}{v}(1-\frac{1}{\gamma})[/tex]

    How is this plane related to the frame S''
    of the 'middle frame' lemma ?

    My idea:

    From the Lorentz transformation (LT) of

    [tex]t^{'}=\gamma(t-\frac{vx}{c^2})[/tex]

    I get,

    [tex]t=t^{'}[/tex]:

    [tex]t=\frac{v}{c^2}( 1-\frac{1}{\gamma} )^{-1} x[/tex]

    in S and hence this is a plane through

    [tex]x=\frac{c^2}{v}(1-\frac{1}{\gamma})t[/tex]

    . One now finds

    [tex]v_x=\frac{c^2}{v}(1-\frac{1}{\gamma})[/tex]

    for the velocity of such a plane.

    I now see that this gives

    [tex]t^{'}=\frac{v}{c^2} x \frac{\gamma}{\gamma-1} [/tex]

    in S', so

    [tex]v_x^{'}=-\frac{c^2}{v}(1-\frac{1}{\gamma})[/tex].

    So this planes move in the opposite direction in S'.

    But how are they connected with S'''. I would say, that this are the x'''-axes ?
    But what is then the t'''-axis ?

    Any comments are welcome, thanks.
     
    Last edited: Feb 10, 2009
  2. jcsd
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