1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Lorentz's identity, relative speed

  1. Dec 1, 2007 #1
    1. The problem statement, all variables and given/known data
    start from:
    x = [x' + vt']/sqrt[1 - v^2/c^2]
    ct = [v/cx' +ct']/sqrt[1 - v^2/c^2]
    y = y'
    z = z'

    2. Relevant equations
    show that

    ( 1 - [tex]\frac{u^{2}}{c^{2}}[/tex])(1+[tex]\frac{vux'^{2}}{c^{2}}[/tex]) = ( 1 - [tex]\frac{v^{2}}{c^{2}}[/tex])(1-[tex]\frac{u'^{2}}{c^{2}}[/tex])

    3. The attempt at a solution

    ok, I have spent many hours on this crappy thing. We have no book in class so.....
    I derived the lorentz transformation for ux, uy, and uz... as well as u'x', u'y', u'z'.... then i computed the velocities in each fram using u = sqrt[ ux^2 + uy^2 + uz^2] and the same for u'. Nevertheless I end up in some mess of algebraic letters that get me nowhere close to the answer. I just need some sort of hit as to how to approach this problem.

    thansk for any hints.
    Last edited: Dec 1, 2007
  2. jcsd
  3. Dec 2, 2007 #2


    User Avatar
    Homework Helper
    Gold Member

    What are u,v,u' ? Is -v the velocity of unprimed system wrt the primed system? Then what's u? I suggest that you post the entire problem as it was given. That way, there's no scope for confusion.
  4. Dec 2, 2007 #3
    yeah, I know it is confusing..... but that is the whole problem... exactly as it was given to us.
    For what understand it is like this

    u = speed of particle 1 in S frame of reference
    u = sqrt[ux^2 + uy^2 + uz^2]
    u' = speed of same particle after a lorentz transformation in the S' frame of reference
    u' = sqrt[u'x^2 + u'y^2 + u'z^2]

    now, v would be the speed of one reference with respect to the other. I assume it is the v that carries over from the gamma sqrt[1-v^2/c^2] from the lorentz transformation.

    sorry about my notation, but I can't understand how to use latex yet.

    thanks for the reply
  5. Dec 2, 2007 #4
    this the actual equation

    Last edited: Dec 2, 2007
  6. Dec 2, 2007 #5
    never mind guys, i found the answer.... i will post the stepwise solution when i get a chance to write it on latex or scan it
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook