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Compute decay rate of muon according to given equation

  1. Nov 7, 2015 #1
    1. The problem statement, all variables and given/known data
    A muon decays to an electron, an electron neutrino and a muon neutrino. ## \mu \rightarrow e \ \nu_\mu \ \nu_e## .The matrix element of the process is ## |\mathcal{M}|^2 = G^2_F (m^2-2mE)mE## with ##m## being the mass of the muon and ##E## the energy of the resulting electron neutrino.
    I need to show that decay rate is $$ \Gamma = \frac{G^2_F m^5 }{192 \pi ^3}$$
    2. Relevant equations
    $$ \Gamma = \frac{1}{2 E_1} |\mathcal{M}|^2 d\Pi_{LIPS} $$
    with ##d\Pi_{LIPS}## being an integral over this entitiy
    $$ d\Pi = \prod_{final \ states j} \frac{d^3 p_j}{(2 \pi )^3}\frac{1}{2 E_{p_j}}
    (2 \pi )^4 \delta ^4(\sum p_i^\mu - \sum p_f^\mu)$$
    The delta function shows energy-momentum conservation with ##p_i^\mu## related to initial particles' momenta and ##p_f^\mu## related to final particles' momenta.
    3. The attempt at a solution
    $$d\Pi = \frac{d^3p_e}{(2\pi)^3} \frac{d^3p_{\nu \mu}}{(2\pi)^3} \frac{d^3p_{\nu_e}}{(2\pi)^3} (2 \pi )^4 \delta (E + E_{\nu \mu} + E_e - m) \delta ^3( \vec{p_\mu} - \vec{p_e} - \vec{p_{\nu_{mu}}} - \vec{\nu_e}) \frac{1}{2E_e} \frac{1}{2E_{\nu \mu}} \frac{1}{2E_{\nu _e}}$$
    Integrating over three four-momenta and using delta function over momenta and the fact that we can use a frame in which the initial particle is not moving, I wrote:
    $$d\Pi_{LISP} = \frac{1}{(2 \pi)^5 2^3} \int d^3p_e \int d^3p_{\nu \mu} \delta (E + E_{\nu \mu} + E_e - m) \frac{1}{E_e E_{\nu \mu} E } $$
    This time we have ## \vec{p_e} = -\vec{p_{\nu_{mu}}} - \vec{\nu_e}##.
    Changing the integral variable so that we have delta function of some variable like x we have:
    $$ x= E + E_{\nu \mu} + E_e - m$$
    $$ E_{\nu \mu} = P_{\nu \mu} \quad $$
    $$E_e = \sqrt{p_e^2 + m_e} = \sqrt{|\vec{p_{\nu_{\mu}}} + \vec{\nu_e}| + m_e} = \sqrt{p_{\nu_{\mu}}^2 + p_e^2 -2 p_{\nu_{\mu}} p_e \cos \theta}$$
    $$ \frac{dx}{dp_{\nu \mu}}=1 + \frac{2 p_{\nu \mu}}{E}$$
    Using this new variable in integral we have
    $$d\Pi_{LISP} = \frac{1}{(2 \pi)^5 2^3} \int d^3p_e \int_{E + E_{\nu \mu} + E_e - m}^{\infty} dx \delta(x) \frac{p_{\nu \mu}^2}{E_e E_{\nu \mu} E } (1 + \frac{2 p_{\nu \mu}}{E})$$
    $$d\Pi_{LISP} = \frac{1}{(2 \pi)^5 2^3} \int d^3p_e \frac{p_{\nu \mu}^2}{E_e E_{\nu \mu} E } (1 + \frac{2 p_{\nu \mu}}{E}) \Theta(E + E_{\nu \mu} + E_e - m)$$

    Now I can not reach the result that the problem want ! Please help me!
    This problem is problem number 3 of chapter 5 from the book "Quantum Field Theory and Standard model" by Matthew D. Schwartz
    I really appreciate your help and patience.
    Thankyou
     
  2. jcsd
  3. Nov 8, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    Did you consider the spherical symmetry of the problem? That rshould simplify the three-dimensional integral.
     
  4. Nov 8, 2015 #3

    nrqed

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    Science Advisor
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    Gold Member

    Why don't you use the three-momentum delta function to do the integral over the electron three-momentum (instead of a neutrino three-momentum)? Since this is the most complicated three-momentum (because of the mass), this seems to be the best approach.
    Then rewrite the infinitesimals [itex] d^3p [/itex] of the muon three-momenta in the form [itex] p^2 dp d \Omega [/itex]. The angular integrations ar trivial and you may then convert the p variables to energy variables (which is simple since you consider the massless limit for the neutrinos). You may use the energy delta function to carry out trivially one of these integrals, leaving you a last one that should be easy.
     
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