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Relativity- CM and Lab frame energy

  1. Dec 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi, I've been doing some relativity past paper questions and I totally got stuck on 2 of them.
    Both regarding an elastic collision of 2 electrons at relativistic speeds.

    First question: I am to show that it is true that the total energy of the electron which is in motion in the lab frame can be given by the following formula:
    [tex]E _{1}= \frac{2E* ^{2}-m ^{2} c ^{4} }{mc ^{2} } [/tex]

    Where E* is the energy of one of the electrons in the CM frame.
    E is energy in lab frame, m is mass, c is speed of light

    Second question: According to an observer in CM frame scattering angle is 90 degrees.
    Whais is the scattering angle in Lab frame (in terms of E* and m)

    2. The attempt at a solution

    I tried doing it in various ways and I can't get an answer.

    My 2 best bets are noticing that:
    [tex]E _{1}= mc ^{2} + \frac{2p ^{*2} _{1}}{m }[/tex]
    where p* is the momentum of an electron in CM frame, but what next??

    Or I try this way, don't even know if its valid:
    (total energy in Lab frame) squared=(total energy in CM frame) squared
    So for lab frame total energy is
    [tex] E _{1} +mc ^{2} [/tex]
    and for CM frame i take 2E*
    I can then expand any of these E's taking
    [tex] E = \sqrt{m ^{2} c ^{4}+(pc) ^{2} } [/tex]
    This gets me really close to the answer but I have an additional term i can't get rid of: [tex] \frac{p _{1} ^{2}c ^{2} }{2} [/tex], which is the momentum of moving particle in lab frame times c^2

    For second question I don't even know what to start with
  2. jcsd
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