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Conservation of four momentum

  1. Oct 11, 2011 #1
    Hi physics people,

    This is a past (3rd year university level) exam question, so I hope it's ok that I didn't post this in the homework section even if it's set out like a homework question.

    The Question:
    Suppose we are observing the collision

    Anti-electron-neutrino + electron ---> W-minus boson

    What energy has the Anti-Electron-Neutrino, to produce the W particle?

    Relavant equations and data:
    mass of electron = 0.511 MeV/c^2
    mass of W-minus boson = 80.403 GeV/c^2

    Four-momentum equation
    [tex]S = -(\overrightarrow{{p_{e}}}+\overrightarrow{{p_{\nu }}})^{2} = -(cp_{e}+cp_{\nu })^{2}+(E_{e}+E_{\nu})^{2}[/tex]

    Thoughts so far:
    My understanding is that the four-momentum is always conserved so I have to equate the initial four-momentum to the final four-momentum using the assumption that the electron and the W boson have no kinetic energy.

    [tex]S_{initial} = -c^{2}p_{\nu }^{2} + E_{\nu }^{2}[/tex]
    [tex]S_{final} = ((m_{W^{-}}) c^{2})^{2}[/tex]

    to the initial four-momentum becomes zero since
    cp = E

    So I'm not sure how to carry on, the energy of the neutrino can't be zero!?

    Thanks to anyone who helps
    Last edited: Oct 12, 2011
  2. jcsd
  3. Oct 12, 2011 #2
    You'll find that it isn't possible for both the electron and the W to be at rest.
  4. Oct 12, 2011 #3
    Is there not a frame where they would both be at rest?. That seems to be at odds with the "no preferred rest frame" postulate.
  5. Oct 12, 2011 #4


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    Gold Member

    I shoot a green pea at billiard ball and it sticks, moving the billiard ball slightly. Is there *one* frame where both the initial billiard ball and the composite (billiard+pea) are at rest? No. The existence of such a frame would mean you have a frame where momentum isn't conserved. The described situation is equivalent.
  6. Oct 12, 2011 #5


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    I hope its ok to just show how to do this as it isn't a homework question. I just use conservation of momentum in the rest frame of the W.
    p_{\nu} +p_{e} = p_W[/tex]
    (E_{\nu},E_{\nu} \hat{z}) + (E_{e}, p_e \hat{z}) = (E_W, \vec{p}_W)
    Now in this frame we have:

    (E_{\nu}, E_{\nu} \hat{z}) + (E_{e}, -E_{\nu} \hat{z}) = (M_W, 0)

    [tex]m_e^2 = E^2_{e} - E^2_{\nu}
    from the momentum squared of the electron in this frame, we have [itex]E_{e} = \sqrt{m_e^2 + E^2_{\nu}}[/itex]

    (E_{\nu}, E_{\nu} \hat{z}) + (\sqrt{m_e^2 + E^2_{\nu}}, -E_{\nu} \hat{z}) = (M_W, 0)

    Looking at just the energies:

    [tex]E_{\nu} + \sqrt{m_e^2 + E^2_{\nu}} = M_W[/tex]

    Giving us

    [tex]E_{\nu} = \frac{M_W^2 -m_e^2}{2 M_W}[/tex]

    This is all assuming all particles are on-shell.
  7. Oct 12, 2011 #6
    Nice analogy, thanks.
  8. Oct 12, 2011 #7
    Thanks very much, this helps with a lot of other things as well, that aren't so greatly explained in my notes. Cheers!
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