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Proof for momentum Lorentz Transformation

  1. Feb 21, 2016 #1
  2. jcsd
  3. Feb 22, 2016 #2

    TeethWhitener

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    Which part of this step don't you understand? Can you show exactly where you get stuck?
     
  4. Feb 23, 2016 #3
    I tried substituting the Lorentz transformation for u'
    ux' = (ux-v)/(1-uxv/c2)
    into the LHS
    and simplifying it from there, but I couldn't derive the expression on the RHS :(
     
  5. Feb 23, 2016 #4

    PeroK

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    Yes, it's quite difficult algebra to derive this identity. You have three gamma factors and a velocity transformation formula to manage. You just have to keep working at it.

    There is a quicker way using the properties of 4-vectors and the proper time of the particle if you are comfortable with those?

    Actually, I've just noticed a very quick way simply using the transformation of the time coordinates ##t## and ##t'##.

    Let me know if you are interested in the quick way!
     
    Last edited: Feb 23, 2016
  6. Feb 23, 2016 #5

    TeethWhitener

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    Maybe this is a trivial point, but you have to use the Lorentz transformed expressions for ##u'_y## and ##u'_z## as well as ##u'_x##. Start with ##u'^2=u'^2_x+u'^2_y+u'^2_z## in the LHS. As PeroK says, there's a lot of messy algebra to work through, but it should work out.

    Well, I'm certainly interested if OP isn't. :biggrin:
     
  7. Feb 23, 2016 #6

    PeroK

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    The relationship between the proper time of the particle and the two coordinate times is:
    ##\frac{dt}{d\tau} = \gamma, \frac{dt'}{d\tau} = \gamma '## (1)

    Differentiate the transformation for time wrt proper time:

    ##t' = \gamma_V(t - Vx/c^2)##

    ##\frac{dt'}{d\tau} = \gamma_V(\gamma - V \gamma v_x/c^2) = \gamma_V\gamma (1- Vv_x/c^2) ##

    Then, using (1) we have:

    ##\gamma ' = \gamma_V\gamma (1- Vv_x/c^2)##

    In fact, you can derive the energy-momentum transformation much more easily by differentiating the components of position wrt proper time of the particle.
     
  8. Feb 23, 2016 #7

    TeethWhitener

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    Wow, that is pretty.
     
  9. Feb 23, 2016 #8

    robphy

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    With rapidities (i.e. trigonometry), [itex]\theta'=\theta-\theta_V[/itex],
    [itex]\begin{align*}\cosh\theta'
    &=\cosh(\theta-\theta_V)\\
    &=\cosh\theta\cosh\theta_V-\sinh\theta\sinh\theta_V\\
    &=\cosh\theta\cosh\theta_V(1-\tanh\theta\tanh\theta_V)\\
    \gamma_{u'}
    &=\gamma_{u}\gamma_{V}(1-uV)
    \end{align*}[/itex]
     
  10. Feb 27, 2016 #9
    Sflr I was caught up with some stuff this week :/ I just figured out the algebraic manipulation after substitution of u' ; I realised that I had to observe the final expression more closely instead of blindly manipulating the algebra. Thanks so much for your help! :)
     
  11. Feb 27, 2016 #10
    Yup, I finally worked through the algebra. Thanks! :)
     
  12. Feb 27, 2016 #11
    Woah that's sweet XD Thanks for the alternative solution!
     
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