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Lorentz gamma factor in a decay of arbitrary leptons

  1. May 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that the Lorentz gamma factor in the decay

    [tex]K^+ \rightarrow l^+ \nu_l[/tex]

    can be written as

    [tex]\gamma = \frac{m_K^2 + m_l^2}{2m_K m_l}[/tex]

    where nu is either e or mu.


    3. The attempt at a solution

    I'm stuck on a part of the proof. I think I understand most of it, except for one crucial part:

    So I'm starting with the invariant and listing out the energies of the particles involved:

    [tex]E^2 - \textbf{p}^2c^2 = m^2c^4[/tex]

    Energy of the Kaon (at rest initially)
    [tex]E_k = m_K c^2[/tex]

    Energy of subsequent decay particles, ignoring the leftover mass that is the UC(bar). I don't really know why, but it seems like its pertinent to ignore that for now. So for the arbitrary lepton:
    [tex]E_l = c (m_l^2 c^2 + \textbf{p}_l^2)^{1/2}[/tex]

    and the arbitrary associated nutrino we have:
    [tex]E_{\nu} = |\textbf{p}_{\nu}|c = |\textbf{p}_l|c[/tex]

    Now we know that the rest energy of the Kaon is going to be equal to the energies of the lepton plus neutrino (easy stuff)

    [tex]m_K c^2 = c (m_l^2 c^2 + \textbf{p}_l^2)^{1/2} + |\textbf{p}_l|c[/tex]

    Now subtracting the momentum term to the other side, dividing by c and squaring:

    [tex](m_K c - |\textbf{p}_l|)^2= m_l^2 c^2 + \textbf{p}_l^2 [/tex]

    and solving for the momentum gives:

    [tex]|\textbf{p}_l| = \frac{m_K^2 - m_l^2}{2 m_K}c[/tex]

    Here's where I get into trouble. I know from a problem in Griffiths that for this kind of decay, velocity of the lepton is given by:

    [tex]\frac{v}{c} = \frac{m_K^2 - m_l^2}{m_K^2 + m_l^2}[/tex]

    and that the lorentz factor:

    [tex]\gamma = \frac{1}{(1- \left( \frac{v}{c} \right )^2)^{1/2}}[/tex]

    when you slap in that v, becomes:

    [tex]\gamma = \frac{m_K^2 + m_l^2}{\left( (m_K^2 +m_l^2)^2 - (m_K^2 - m_l^2)^2 \right)^{1/2}}[/tex]

    which reduces to:

    [tex]\gamma = \frac{m_K^2 + m_l^2}{2m_K m_l}[/tex]

    which is what I set out to prove.

    I don't get this step:

    [tex]|\textbf{p}_K| = \frac{m_K^2 - m_l^2}{2m_K} c \rightarrow \frac{m_K^2 - m_l^2}{m_k^2 + m_l^2}c[/tex]

    Doing that seems like an easy way to solve the problem, but ignores the fact that the kaon mass is NOT half of the kaon squared plus the lepton squared (or is it?). Plus there's the issue of the units.
     
  2. jcsd
  3. May 3, 2012 #2
    I am not really sure why you are mixing results from two different problems. If you solve the problem correctly then you should get the final velocity directly; you don't need to find it out from somewhere else. Now, in this case it seems unnecessary. The total energy of the lepton is [itex]E^2 = p^2 c^2 + m^2 c^4 = \gamma^2 m^2 c^4[/itex], so once you have correctly obtained the momentum it should be fairly straightforward to determine the Lorentz factor.
     
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