Compute decay rate of muon according to given equation

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SUMMARY

The discussion focuses on calculating the decay rate of a muon, which decays into an electron, an electron neutrino, and a muon neutrino, represented by the equation ## \mu \rightarrow e \nu_\mu \nu_e##. The decay rate is derived from the matrix element ## |\mathcal{M}|^2 = G^2_F (m^2-2mE)mE##, leading to the conclusion that the decay rate is given by $$ \Gamma = \frac{G^2_F m^5 }{192 \pi ^3}$$. Participants suggest utilizing spherical symmetry and the three-momentum delta function to simplify the integral calculations involved in deriving the decay rate.

PREREQUISITES
  • Understanding of particle decay processes in quantum field theory.
  • Familiarity with the concepts of matrix elements and decay rates.
  • Knowledge of energy-momentum conservation principles.
  • Proficiency in performing integrals over phase space in particle physics.
NEXT STEPS
  • Study the derivation of decay rates in "Quantum Field Theory and Standard Model" by Matthew D. Schwartz.
  • Learn about spherical symmetry in integrals and its applications in particle physics.
  • Explore the use of delta functions in energy-momentum conservation in decay processes.
  • Investigate techniques for simplifying three-dimensional integrals in quantum mechanics.
USEFUL FOR

This discussion is beneficial for graduate students in physics, particularly those specializing in particle physics and quantum field theory, as well as researchers working on muon decay processes and related calculations.

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Homework Statement


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 $$ \Gamma = \frac{G^2_F m^5 }{192 \pi ^3}$$

Homework 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.

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
 
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Did you consider the spherical symmetry of the problem? That rshould simplify the three-dimensional integral.
 
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 d^3p of the muon three-momenta in the form p^2 dp d \Omega. 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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