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Midframe Lemma?

  1. Aug 31, 2007 #1
    I came across this term for the first time less than a day ago in Rindler's book Relativity (2 Ed. paperback p40). It's the assertion "that 'between' any two inertial frames S and S' there exists an inertial frame S'' relative to which S and S' have equal and opposite velocities." I am not able to understand his two-line "proof." I tried searching Google and Wikipedia for more on this but nothing turned up (:surprised). I was wondering if someone here knows about this lemma and could explain it to me, or provide a link to some site that explains it.

    Thanks.
     
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  3. Aug 31, 2007 #2

    George Jones

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    It might first help think about things from a non-relativistic, Galilean point of view.
     
  4. Aug 31, 2007 #3

    robphy

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    One can also consider a Euclidean analogue.
     
  5. Aug 31, 2007 #4
    So, basically, S moves with [itex]\vec{v_1}[/itex], S' moves with [itex]\vec{v_2}[/itex] and S'' moves with [itex]\frac{\vec{v_1} + \vec{v_2}}{2}[/itex]? (All of that with respect to me, of course.) I was never thinking "'SR'ically" or in terms of Minkowskian geometry, though.
     
  6. Aug 31, 2007 #5

    JesseM

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    No, because of relativity's velocity addition formula it wouldn't work so simply. For example, if in the S'' frame S is moving at 0.6c to the left and S' is moving at 0.6c to the right, then if in your frame S'' is moving at 0.8c to the right, in your frame S' is moving at speed (0.8c + 0.6c)/(1 + 0.8*0.6) = 0.9459c, while in your frame S is moving at speed (0.8c - 0.6c)/(1 - 0.8*0.6) = 0.3846c. So you can see that the velocity of S'' in your frame, 0.8c, is not equal to [itex]\frac{\vec{v_1} + \vec{v_2}}{2}[/itex] which would work out to (0.3846c + 0.9459c)/2 = 0.6653c.
     
  7. Aug 31, 2007 #6
    That might be true, but I was thinking along the lines of GeorgeJones' advice, i.e. Galilean relativity.

    So, now how can I prove this lemma in terms of SR? Any pointers? (In the book it is mentioned, long before the velocity-composition formulae, in a discussion of the homogeneity (in space and time) and isotropy of inertial frames, and how one can arrive at Einstein's principle of relativity from this assumption/axiom.)
     
  8. Aug 31, 2007 #7

    JesseM

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    A straightforward way to prove it would be to use the velocity addition formula to find a general formula for the velocity v an object should have in your frame such that if two other objects have velocities v1 and v2 in your frame, their speeds will be equal in v's own frame. But this probably isn't the simplest approach. You said the textbook had a simple two-line proof, can you post it and then people here could explain whatever it is you don't understand?
     
  9. Aug 31, 2007 #8
    Figure 2.4 shows three frames - S, S'' and S', in that order. (Three big L's). S" prime has a relatively long arrow pointing to the right. S' has one even longer pointing in the same direction. They also have some "experiments" drawn inside, but that has to do with the next paragraph.
     
  10. Aug 31, 2007 #9

    George Jones

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    Let: V be the velocity of S' with respect to S; u be the velocity of S with respect to S''; v be the velocity of S' with respect to S''.

    Consider a sequence of frames S'' that start at S'' = S, and that vary continuously to S'' = S'.

    If S'' = S, then u = 0 and v = V.

    If S'' = S', then u = -V and v = 0.

    So, the *magnitude* of u varies continuously from 0 to V at the same that the magnitude of v varies continously from V to 0. Somewhere, they have to cross.
     
  11. Aug 31, 2007 #10
    Ah...now I can at least start to wrap my head around it. :) Thanks.
     
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