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Proof Minkowski metric is invariant under Lorentz transformation

  1. Aug 1, 2013 #1
    Ok, this should be an easy one but it's driving me nuts. When we take the Lorentz transformations and apply them to x2-c2t2 we get the exact same expression in another frame. I can do this math easily by letting c=1 and have seen others do it by letting c=1 but I have never seen anyone actually do it with the c's in there. It doesn't sound like it should be that hard but I just can't get it to work. Can anyone go through it?
     
  2. jcsd
  3. Aug 1, 2013 #2
  4. Aug 1, 2013 #3

    WannabeNewton

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    It's always the arithmetic isn't it :wink:? Glad it worked out!
     
  5. Aug 1, 2013 #4

    Bill_K

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    The neatest way:

    x' = γ(x - vt)
    t' = γ(t - vx/c2)

    Take linear combinations:

    (1) x' - ct' = γ(x - ct - v(t - x/c)) = γ(1 + v/c)(x - ct)
    (2) x' + ct' = γ(x + ct - v(t + x/c)) = γ(1 - v/c)(x + ct)

    Multiplying (1) and (2) together, all the leading factors cancel and we get

    x'2 - c2t'2 = x2 - c2t2

    (Note that x ± ct are the eigenvectors of the transformation. Plus the factors in front are just the Doppler shifts.)
     
    Last edited: Aug 1, 2013
  6. Aug 1, 2013 #5
    Wow, I just realized I had simply forgotten to distribute the c squared to everything originally. Talk about dumb!
     
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