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Momentum - Energy Question

  1. Dec 15, 2007 #1
    So, I am reading about a compton scattering problem, and I don't understand part of the derivation of a formula. I will explain my confusion.

    If a gamma photon with energy [tex]E_{\gamma}[/tex], undergoes compton scattering with an electron which is at rest, how does one arrive at the following expression?

    [tex]E^{'}_{\gamma}=\frac{E_{\gamma}}{1+(2E_{\gamma}/m_{o}c^{2})}[/tex]

    So far it says we start with the conservation of energy and momentum:
    [tex]E_{\gamma}=E^{'}_{\gamma}+E_{e} \ \ (eqn 1)[/tex]
    [tex]\frac{E_{\gamma}}c=P_{e}-\frac{E^{'}_{\gamma}}{c}\ \ (eqn 2)[/tex]

    From eqn 2 we get:
    [tex]E_{\gamma}+E^{'}_{\gamma}=p_{e}c=\sqrt{(E_{e}+m_{o}c^{2})^{2}-(m_{o}c^{2})^{2}}[/tex]

    This is where I am confused. I don't understand where the term inside of the radical comes from. Any ideas?
     
  2. jcsd
  3. Dec 15, 2007 #2

    jtbell

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    Staff: Mentor

    The [itex]E_e[/itex] in the square root on the right is the electron's kinetic energy, which most books call [itex]K_e[/itex]. [itex]E_e[/itex] usually means the total energy:

    [tex]E_e = K_e + m_0 c^2[/tex]

    Using this notation, the relationship between energy, momentum and mass is

    [tex]E^2 = (pc)^2 + (m_0 c^2)^2[/tex]

    so

    [tex]p_e c = \sqrt {E_e^2 - (m_0 c^2)^2} = \sqrt {(K_e + m_0 c^2)^2 - (m_0 c^2)^2}[/tex]
     
    Last edited: Dec 15, 2007
  4. Dec 17, 2007 #3
    hi all,

    a similar post for me. pls help me.
    i need to calculate the energy lost by a solid sphere (~ 0.5 mm dia) on hitting the solid wall obliquely.
    i know the tangetial and normal velocities and the restitution coefficients. what other parameters do i need to know ??

    help with the expression or a reference to articles in the web would be of great help.

    thanks you.
     
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