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Showing that Energy-momentum relation is invariant

  1. Oct 5, 2014 #1
    1. The problem statement, all variables and given/known data

    A particle of mass [itex]m[/itex] is moving in the [itex]+x[/itex]-direction with speed [itex]u[/itex] and has momentum [itex]p[/itex] and energy [itex]E[/itex] in the frame [itex]S[/itex].

    (a) If [itex]S'[/itex] is moving at speed [itex]v[/itex], find the momentum [itex]p'[/itex] and energy [itex]E'[/itex] in the [itex]S'[/itex] frame.
    (b) Note that [itex]E' \neq E[/itex] and [itex]p' \neq p[/itex], but show that [itex](E')^2-(p')^2c^2=E^2-p^2c^2[/itex]

    2. Relevant equations

    [itex]E=p^2c^2+m^2c^4[/itex]
    [itex]E'=\gamma mc^2=\frac{mc^2}{\sqrt{1-\frac{(u')^2}{c^2}}}[/itex]
    [itex]p'=\gamma mu'=\frac{mu'}{\sqrt{1-\frac{(u')^2}{c^2}}}[/itex]
    [itex]u'=\frac{u-v}{1-\frac{uv}{c^2}}[/itex]

    3. The attempt at a solution

    My prof says we need to use (a) to derive (b). It seems like it'd be easier a different way, but whatever. My strategy has been to show that [itex](E')^2-(p')^2c^2=m^2c^4[/itex], which would complete the derivation.

    I tried playing around with some algebra and got [itex](E')^2=\frac{m^2c^4(c^2-uv)^2}{(c^2-u^2)(c^2-v^2)}[/itex] and [itex](p')^2=\frac{m^2(u-v)^2}{(1-\frac{uv}{c^2})^2(c^2-u^2)(c^2-v^2)}[/itex]

    I got a little overwhelmed with the algebra at this point. I'd like to know if this is right so far, at least, before trying anything else.
     
  2. jcsd
  3. Oct 5, 2014 #2

    vela

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    For part (a), you should be able to show that
    \begin{eqnarray*}
    E' &= \gamma(E - \beta pc) \\
    p'c &= \gamma(pc - \beta E)
    \end{eqnarray*} where ##\beta = v/c## and ##\gamma = 1/\sqrt{1-\beta^2}##.

    Your expression for ##E'^2## looks okay. Divide both the numerator and denominator by ##c^4## and take the square root. You should be able to show the first equation above. Just make sure you keep track of the three ##\gamma##'s you have in this problem.
     
    Last edited: Oct 6, 2014
  4. Oct 6, 2014 #3
    Thanks very much for the help. It worked like a charm - the last part of the derivation that I ended up using was pretty neat. My only worry is that my prof will wonder how I thought to express ##E'## and ##cp'## in that way.
     
  5. Oct 6, 2014 #4

    vela

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    Look up the Lorentz transformation. :smile: You might already be familiar with it applied to x and t. It applies to all four-vectors.
     
  6. Oct 6, 2014 #5
    Interesting - we've covered the Lorentz transformation, but not four-vectors. My class gives a brief overview of SR, but maybe I'll see it in future.

    I have another question that I'm editing into the OP now - It's somewhat related, but I didn't want to flood the section with threads.

    EDIT: Woops, guess it's been too long to edit. See you in the next thread XD
     
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