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Squaring the energy

  1. Feb 14, 2010 #1
    Hey, I'm stuck on part of the calculation for the compton wavelength. Basically, my notes go:

    Energy of Electron

    Before collision: E0 = mec2

    After collision: Ee = [tex]\sqrt{P_{e}^{2}c^{2} + m_{e}^{2}c^{4}}[/tex]

    Pe = momentum of electron after collision
    Ee = energy of electron after collision
    me = mass of electron

    I can't see why this follows as the energy after the collision. Any ideas?

  2. jcsd
  3. Feb 14, 2010 #2


    Staff: Mentor

    That is the formula for the total energy of the electron. The first term is the kinetic energy which is zero before the collision, and the second term is the mass-energy or rest-energy which is unchanged, i.e. the first formula is derived from the second for the special case of the electron at rest.
  4. Feb 14, 2010 #3

    Doc Al

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    That second equation is true for any particle and is often written like this:

    E^2 = m^2c^4 + p^2c^2

    The 'Before collision' version is just a special case where P = 0, since the electron starts out at rest.

    (Edit: DaleSpam beat me to it.)
  5. Feb 14, 2010 #4
    Ah sorry maybe I should clarify my question. What i'm really wondering is where the formula for E2 comes from. If it's a general equation, where can I find a derivation or explanation of it?

  6. Feb 14, 2010 #5


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    Any book on special relativity.
  7. Feb 14, 2010 #6


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    Are you acquainted with the usual formulas for relativistic energy and momentum?

    [tex]E = \frac {m_0 c^2} { \sqrt {1 - v^2 / c^2}}[/tex]

    [tex]p = \frac {m_0 v} { \sqrt {1 - v^2 / c^2}}[/tex]

    Solve one equation for v, substitute into the other equation, and simplify.
  8. Feb 14, 2010 #7
    Apparently I'm not! Thanks for the help everyone, that's cleared it up.
  9. Feb 14, 2010 #8

    Vanadium 50

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    You must love algebra!

    It's simpler and faster to calculate E2-(cp)2. Then divide out the denominator and you're done.
  10. Feb 14, 2010 #9


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    Yeah, since we actually know where we want to end up, that's the easiest way to do it. I've been working out too many solutions for homework problems where you don't know the answer in advance. :yuck:
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