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Show that c^2(t^2) - x^2 - y^2 - z^2 is invariant under a change of frame

  1. Oct 20, 2012 #1
    [itex][/itex]1. The problem statement, all variables and given/known data

    Show that the quantity

    [itex]T = c^2(Δt)^2 - (Δx)^2 - (Δy)^2 - (Δz)^2[/itex]

    is invariant under a change of frame

    2. Relevant equations

    Lorentz transformations

    [itex]Δx' = \gamma(Δx - vΔt)[/itex]

    [itex]Δt' = \gamma(Δt - vΔx/c^2)[/itex]

    [itex]Δy' = Δy[/itex]

    [itex]Δz' = Δz[/itex]

    3. The attempt at a solution

    I know that the way to do this is to substitute the Lorentz transformations into the original invariant and then do some algebra, but I end up with the following:

    [itex]T' = c^2(Δt')^2 - (Δx')^2 - (Δy')^2 - (Δz')^2[/itex]

    [itex]=c^2(\gamma(Δt - vΔx/c^2))^2 - (\gamma(Δx - vΔt))^2 - (Δy)^2 - (Δz)^2[/itex]

    After doing some algebra I get

    [itex]T' = \gamma^2c^2Δt^2 + \gamma^2v^2Δx^2/c^2 - \gamma^2Δx^2 - \gamma^2v^2Δt^2 - (Δy)^2 - (Δz)^2[/itex]

    I also try to express gamma in terms of

    [itex]\gamma=\frac{1}{\sqrt{1-v^2/c^2}}[/itex]

    But the equation gets much worse without any apparent progress. I only see it stated in textbooks and websites that plugging in the formula leads to the answer, but no actual steps in between. I'm really confused here and would greatly appreciate help.
     
    Last edited: Oct 20, 2012
  2. jcsd
  3. Oct 20, 2012 #2
    Try grouping all the Δx2 and Δy2 terms, see what you get then.
     
  4. Oct 20, 2012 #3
    Ahh, so the gamma terms cancel out the v and c squared terms. Thanks for that, I've been trying to get it for ages!
     
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