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Proving x^2-c^2*t^2 invariance

  1. Jan 20, 2010 #1
    How do you prove x2-c2t2 is invariant under the lorentz transformations given that;
    78195e8f63116bf11b2bbef574fbcc25.png
     
  2. jcsd
  3. Jan 20, 2010 #2
    Ive tried the obvious replacing x and t with x' and t' but i still cant get it to drop out :(
     
  4. Jan 20, 2010 #3

    George Jones

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    In x'^2 - ct'^2, replace x' and t' be the expressions that you gave in the original post, and factor x^2 and t^2 out of terms in which they occur.
     
  5. Jan 20, 2010 #4

    Mentz114

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    Try again. I just worked it out and found that c2t2-x2=c2t'2-x'2.
     
  6. Jan 20, 2010 #5

    jtbell

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    You need to prove that

    [tex](x^{\prime})^2 - c^2 (t^{\prime})^2 = x^2 - c^2 t^2[/tex]

    After substituting the Lorentz transformation on the left side, everything should eventually cancel out.
     
  7. Jan 20, 2010 #6
    Thanks everyone! I didn't see that gamma could be taken out of the equation and cancelled. It works now :D
     
  8. Jan 20, 2010 #7
    What you've done is a special case, in the general case, x^2-(ct)^2 is a constant and gamma does not cancel out, the calc is a bit rigorous then, you need to differentiate and substitute v
     
  9. Jan 20, 2010 #8
    Ok, I'l definitely keep that in mind! Thanks.
     
  10. Jan 20, 2010 #9
    Oops, mistake.No differentiation.The invariance yields directly by substituting LT.
    I differentiated x^2-(ct)^2=k and substituted v , but that's again a special case when v is the velocity of the object in the unprimed frame
     
  11. Jan 20, 2010 #10
    Mmm, ok. I think I might be doing that next semester at uni.
     
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