How to Show Lorentz's Identity with Relative Speed?

[wtronic]

Homework Statement

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'

Homework Equations

show that

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

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.

 

[siddharth]

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.

 

[wtronic]

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

 

[wtronic]

this the actual equation

[text]1+u_{x}single-quoteV/c^2=\sqrt(1-usingle-quote^2/c^2)*\sqrt(1-V^2/c^2)/\sqrt(1-u^2/c^2)[/text]

 

[wtronic]

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