How to get inverse Lorentz tranformation from direct Lorentz transformation

  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.

    Thanks!
     
  2. jcsd
  3. JesseM

    JesseM 8,491
    Science Advisor

    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. 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.
    Direct:

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

    Inverse:

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

    Eugene.
     
  5. Thanks guys. It was very clear. Now I get the problem! :)
     
  6. JesseM

    JesseM 8,491
    Science Advisor

    Yes, sorry, I mistyped.
     
  7. 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. 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.

    Pete
     
  9. 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. Why not?
     
  11. 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. 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. Doc Al

    Staff: Mentor

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

    robphy 4,394
    Science Advisor
    Homework Helper
    Gold Member

    If you use rapidity, your Euclidean trigonometric intuition would have guided you.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?