Lorentz transform for momentum

[insynC]

Homework Statement

Derive the Lorentz transformation for the x component of momentum, i.e.

Px' = [tex]\gamma[/tex] (Px - vE/(c[tex]^{}2[/tex]))

I've used Px = x component of momentum (not very good with latex, sorry!)

Homework Equations

I thought the best place to start was the Lorentz transformation for velocity (which was given):

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

The Attempt at a Solution

Applying this, I used the fact Px = [tex]\gamma[/tex] m0 ux - where m0 is rest mass - and then fiddled around with it.

I was able to almost get the answer, except on the RHS I got what is required multiplied by a factor of:

1 / [ [tex]\gamma[/tex] - [tex]\gamma[/tex] ux v /(c[tex]^{}2[/tex]) ]

Unfortunately I couldn't show this was equal to 1 and am not even convinced it is. Was the approach I took the easiest way to the answer? I've tried it again and got the same problem, so maybe there is a better way to tackle it.

 

[tiny-tim]

Applying this, I used the fact Px = [tex]\gamma[/tex] m0 ux - where m0 is rest mass - and then fiddled around with it.

Hi insynC!

(have a gamma: γ)

(and use the X₂ tag just above the reply field: P_x )

I'm not sure that you used γ's dependence on speed:

P_x = γ(u)u_x , P_x' = γ(v)v_x

 

[insynC]

Good point, I hadn't taken that into consideration.

I have to now introduce γ(u_x') and γ(u_x), but ultimately the answer has a γ(v) in it. I'm struggling to remove the excess γ's introduced to convert the velocity transform into the momentum transform.