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Spacetime Invariance and Lorentz Equations

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data
    So, I am working on a question that requires me to prove that s^2 = s'^2 from the Lorentz equations. It seemed like it'd be trivial... and then I ended up here a few hours later, not willing to waste any more time.

    2. Relevant equations
    By definition: s^2 = x^2 - (ct)^2 & s'^2 = x'^2 - (ct')^2
    And the Lorentz equations are x' = y(x - vt) & t' = y(t - vx/(c^2) ) --> y = gamma for the lazy man

    3. The attempt at a solution
    So I followed this line of algebra:
    Start with: s'^2 = x'^2 - (ct')^2
    Sub in Lorentz Equations: = [y(x - vt)]^2 - c^2*[y(t - vx/(c^2) )]^2
    Factor out y and Expand Brackets: = y^2 { [x^2 - 2xvt + (vt)^2] - c^2*[t^2 -2xvt/c^2 + ((vx)^2)/(c^4))] }
    Multiply in -c^2 over on the right: = y^2 [x^2 - 2xvt + (vt)^2 -(ct)^2 + 2xvt - (v^2*x^2)/(c^2)]
    Eliminate the 2xvt terms: = y^2 [x^2 + (vt)^2 -(ct)^2 - (v^2*x^2)/(c^2)]
    Now, t = x/c, so by applying this to the rightmost term: = y^2 [x^2 + (vt)^2 -(ct)^2 - (vt)^2]
    Eliminate the (vt)^2 terms: = y^2 [x^2 - (ct)^2]
    And recall that s^2 = x^2 - (ct)^2: Therefore: ==> s'^2 = y^2[s^2]

    There's my dilemma. This would mean spacetime is NOT invariant, since it depends on gamma. I'm not quite prepared to call the people who invented this liars, or to say I'm better at math than them... but I thought my algebra was pretty good and it led me to this. So... what's the problem?
     
    Last edited: Sep 19, 2009
  2. jcsd
  3. Sep 20, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi LUphysics! Welcome to PF! :smile:

    (have a gamma: γ :wink:)

    Sorry, but this is almost unreadable :redface:

    can you type it again, using the X2 tag just above the Reply box? :smile:
     
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