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How to get inverse Lorentz tranformation from direct Lorentz transformation

  1. Oct 1, 2007 #1
    How to get inverse Lorentz tranformation from "direct" Lorentz transformation

    Hello, I am having trouble on deriving the inverse Lorentz transformation from the direct Lorentz transformation. I looked at some threads here and I found in here (https://www.physicsforums.com/showthread.php?t=183057) that all I need to do is to "combine" the equation for x and t and I will get the inverse equation...but I don't really know what does it mean to "combine" the equation...? I also found in textbooks that to get the inverse transformation I just need to solve for x in the direct transformation. However, when I do it...it doesn't give the inverse transform equation. Can anybody give me some help here? I'll greatly appreciate it.

  2. jcsd
  3. Oct 1, 2007 #2


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    It's just algebra, it means solving those two equations (a combined system of equations) for x and t. You have:

    x'=gamma*(x - vt) and t'=gamma*(t - vx/c^2)

    So, with the first one you can do:
    x' = gamma*x - gamma*vt
    x' + gamma*vt = gamma*x
    x'/gamma + vt = x

    And with the second one:
    t' = gamma*t - gamma*vx/c^2
    t' + gamma*vx/c^2 = gamma*t
    t'/gamma + vx/c^2 = t

    Then substitute this expression for t into the earlier equation x = x'/gamma + vt, which gives you:

    x = x'/gamma + v(t'/gamma + vx/c^2) = x'/gamma + vt'/gamma + xv^2/c^2

    and if you subtract xv^2/c^2 from both sides, you get:

    x(1 - v^2/c^2) = x'/gamma + vt'/gamma

    Now since gamma = [tex]\frac{1}{(1 - v^2/c^2)^{1/2}}[/tex] this is the same as:

    [tex]x * (1 - v^2/c^2)^1 = (1 - v^2/c^2)^{1/2} * (x' + vt')[/tex]

    So if you divide both sides by (1 - v^2/c^2) you get:

    [tex]x = (1 - v^2/c^2)^{-1/2} * (x' + vt')[/tex]

    which is just x = gamma*(x' + vt'), the reverse transformation for x in terms of x' and t'. Then you can plug this into t = t'/gamma + vx/c^2 and get the reverse transformation for t in terms of x' and t', which should work out to t = gamma*(t' + vx'/c^2).
    Last edited: Oct 1, 2007
  4. Oct 1, 2007 #3
    I think it should be

    t = gamma*(t' + vx'/c^2)

    The direct and inverse transformations should differ only by the sign of velocity v.

    x'=gamma*(x - vt)
    t'=gamma*(t - vx/c^2)


    x=gamma*(x' + vt')
    t=gamma*(t' + vx'/c^2)

  5. Oct 1, 2007 #4
    Thanks guys. It was very clear. Now I get the problem! :)
  6. Oct 1, 2007 #5


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    Yes, sorry, I mistyped.
  7. Oct 6, 2007 #6
    I've been following this, and I can see how to get to x, but am having trouble with t...I've got up to t = t'/gamma + (gamma*x'v)/c^2 + (gamma*t'v^2)/c^2

    I just can't see where to go from there!
  8. Oct 6, 2007 #7
    That's a lot of work just to say - Switch the sign on the velocity in the Lorentz transformation and you end up with the inverse Lorentz transformation.

  9. Oct 6, 2007 #8
    Yes but then again you can't always do that - I'm working on inverting it mathematically, so we're not allowed to just say that! Unfortunately...!
  10. Oct 6, 2007 #9
    Why not?
  11. Oct 6, 2007 #10
    Because the question I'm working on says "Mathematically invert equations (1) and (2) [ie, x' and t'] to obtain the inverse transformation"

    Which means you can't just look at it from a physics point of view, you have to show it through the method that JesseM said above.
  12. Oct 6, 2007 #11
    Like I said above (sorry for reposting,, feared that it got lost in the much quoting above):

    I've been following this method, and I can see how to get to x, but am having trouble with t...I've got up to t = t'/gamma + (gamma*x'v)/c^2 + (gamma*t'v^2)/c^2

    I can't see how to make it into the inverse Lorentz from there! Have tried rearranging but just can't make it look right...!
  13. Oct 6, 2007 #12

    Doc Al

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    Express v^2/c^2 in terms of gamma.
  14. Oct 6, 2007 #13
    It wouldn't have occurred to me to do that, thankyou! :smile:
  15. Oct 6, 2007 #14


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    If you use rapidity, your Euclidean trigonometric intuition would have guided you.
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