# Lorentz transform for momentum

1. Nov 21, 2008

### insynC

1. The problem statement, all variables and given/known data

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

Px' = $$\gamma$$ (Px - vE/(c$$^{}2$$))

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

2. Relevant equations

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

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

3. The attempt at a solution

Applying this, I used the fact Px = $$\gamma$$ 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 / [ $$\gamma$$ - $$\gamma$$ ux v /(c$$^{}2$$) ]

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.

2. Nov 21, 2008

### tiny-tim

Hi insynC!

(have a gamma: γ)

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

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

Px = γ(u)ux , Px' = γ(v)vx

3. Nov 21, 2008

### insynC

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

I have to now introduce γ(ux') and γ(ux), 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.