# Muon decay correction

by Catria
Tags: correction, decay, muon
 P: 43 In my class notes, the muon decay width is: $\Gamma_\mu=\frac{G^2_F m^5_\mu}{192\pi^3}$ Yet, in Wikipedia (take that for what it's worth) it says that, once the corrections have been applied, the muon decay width takes the form: $\Gamma_\mu=\frac{G^2_F m^5_\mu}{192\pi^3}I\left(\frac{m^2_e}{m^2_\mu}\right)$ where $I(x)=1-8x-12x^2 ln(x)+8x^3+x^4$ is a correction factor. I wonder how does one arrive at that correction factor. If it is possible to get enough steps so that someone could understand its origin, as well as understanding its derivation, it would be very appreciated.
 Sci Advisor Thanks P: 3,850 Catria, The muon decay rate is calculated from evaluation of a Feynman diagram -- simple in principle but complicated in practice. Here's a paper describing in general terms how one does it. (Note especially Eq 44.) The calculation for muon decay involves a three-particle final state, and the energies and angles of all three decay products must be integrated over to get the total decay rate. It simplifies a lot if you neglect the mass of the electron, giving the answer from your class notes. The "corrected" version in Wikipedia is the full result.
 Sci Advisor P: 5,935 Note that (me/mμ)² << 1, so that I(x)≈I(0)=1.
P: 81

## Muon decay correction

Like bill says, as you probably know from your post, you calculate the amplitude at tree level. Then one can improve the calculation by adding the correction factor.

Of course one first calculates the largest correction first.

To do this, you have to extend the Feynman diagram literature to loop level and perform integrals over loop momentum.

In decays, these are regularized by the massive exchange particle. Anyway, it is these loop integrals that give you these ratios and logarithms.
 Mentor P: 15,571 These are not loop corrections. These are phase space corrections - due to the fact that the electron has a tiny mass. These have the wrong dimensionality to be loop corrections.
 P: 81 Oops. I should have looked at the link before replying.

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