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Lorentz transform for momentum

  1. Nov 21, 2008 #1
    1. The problem statement, all variables and given/known data

    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!)

    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[tex]^{}2[/tex])]

    3. 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.
  2. jcsd
  3. Nov 21, 2008 #2


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    Hi insynC! :smile:

    (have a gamma: γ)

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

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

    Px = γ(u)ux , Px' = γ(v)vx
  4. Nov 21, 2008 #3
    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.
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