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Spacetime Inverval Invariance using Lorentz Transformations

  1. Aug 27, 2011 #1
    1. The problem statement, all variables and given/known data
    Prove that the spacetime interval
    -(ct)[itex]^{2}[/itex] + x[itex]^{2}[/itex] + y[itex]^{2}[/itex] + z[itex]^{2}[/itex]
    is invariant.

    2. Relevant equations
    Lorentz transformations
    [itex]\Delta[/itex][itex]x' = \gamma(\Delta[/itex][itex]x-u\Delta[/itex][itex]t)[/itex]
    [itex]\Delta[/itex][itex]y' = \Delta[/itex][itex]y[/itex]
    [itex]\Delta[/itex][itex]z' = \Delta[/itex][itex]z[/itex]
    [itex]\Delta[/itex][itex]t' = \gamma(\Delta[/itex][itex]t-u\Delta[/itex][itex]x/c^{2})[/itex]

    3. The attempt at a solution
    I have tried to prove that [itex]\Delta S = \Delta S'[/itex]
    So first I said that [itex]\Delta S' = - \Delta (ct')^{2} + \Delta (x')^{2} + \Delta (y')^{2} + \Delta (z')^{2}[/itex]

    And inserted all the Lorentz Transformations above into the above formula.

    I end up simplyfying it to get

    [itex]\gamma^{2} (x^{2} + u^{2}t^{2} - c^{2}t^{2} - \frac{u^{2}x^{2}}{c^{2}}) + y^{2} + z^{2}[/itex]

    How does this equal [itex]S = - \Delta (ct)^{2} \Delta (x)^{2} + \Delta (y)^{2} + \Delta (z)^{2}[/itex] ? I can't see a way to get rid of the extra terms to get this simple function.

    Any help would be really really great!
  2. jcsd
  3. Aug 27, 2011 #2


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    If you collect the terms, you'll see that the coefficient of x2 is [itex]\gamma^2(1-u^2/c^2)[/itex]. Use the definition of [itex]\gamma[/itex] to simplify that.
  4. Aug 27, 2011 #3
    Oh of course! How did I miss that?

    Thank you so much I got it out now!
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