- Could I have calculated the integral in a different order than the one in Griffiths?
- Could one still calculate the result analytically if more particles were produced in the decay?
- Is there a faster way to obtain the result?

The starting point of the calculation is Fermi's golden rule \begin{equation}\label{eq:20160730a} d\Gamma = |\mathcal{M}|^2 \frac{1}{2 m}\ D^4 p_2\ D^4 p_3\ D^4 p_4\ (2 \pi)^4 \delta(p_1 - p_2 - p_3 -p_4) \end{equation} with $m$ the mass of the muon, $p_2$, $p_3$, $p_4$ the momenta of the electron anti neutrino, muon neutrino and electron, $D^4 p = \dfrac{d^4 p}{(2 \pi)^3} \delta(p^2) (E >0)$ the Lorentz invariant phase space differential for a particle with mass zero and $|\mathcal{M}|^2 = 2 \left( \dfrac{g_w}{M_W}\right)^2 (p_1 p_2) ( p_3 p_4) $. I calculate in the center of mass frame, then $p_1 p_2 = m E_2$ and $p_3 p_4 = \dfrac{m^2}{2} - m E_2$

The essence of the set up of my calculation is to split the decay as $1 \to 2 + q$ and $q \to 3 + 4$. Thus instead of a 3-particle decay, I will calculate the product of two 2-particle decays and then integrate over $q$. More concretely, I multiply \eqref{eq:20160730a} with \begin{equation*} 1 = \int\frac{d^4 q}{(2 \pi)^4} (2 \pi)^4 \delta^4 (q - p_3 - p_4)\ 1_{q_0 > 0} \int d(w^2) \delta( q^2 - w^2) \end{equation*}

I integrate over $p_3$ and $p_4$ and use the well known formula for 2-body decay \begin{equation*} \int D^4 p_3 \ D^4 p_4\ (2 \pi)^4 \delta^4(q - p_3 -p_4) = \frac{1}{8 \pi} \end{equation*} This gives \begin{equation*} d\Gamma = |\mathcal{M}|^2 \frac{1}{2 m}\ D^4 p_2 \frac{1}{8 \pi} \ (2 \pi)^4 \delta(p_1 - p_2 - q) \frac{d^4 q}{(2 \pi)^4} d(w^2) \delta( q^2 - w^2)\ 1_{q_0 > 0} \end{equation*} This is the formula for the decay rate of the process $1 \to 2 + q$. I again use the formula for the 2-body phase space, then \begin{equation*} d\Gamma = \frac{1}{2 \pi} \left( \frac{1}{ 8\pi} \frac{|\vec{p}|}{m^2}\right) |\mathcal{M}|^2 \frac{1}{8 \pi} d(w^2) \ 1_{ m \ge w} \end{equation*} Because particle 2 is massless $|\vec{p}| = E_2$. It is also easy to see that $E_2 = \dfrac{m^2 - w^2}{2 m}$. The final integration over $w^2$ is straightforward and gives \begin{equation*} \Gamma = \frac{1}{12} \frac{1}{(8 \pi)^3} \left( \frac{g_w}{M_W} \right)^4 m^5 \end{equation*} which is the correct formula, see Griffiths equation 10.36

Although my calculation is perhaps shorter than in Griffiths, Griffiths also obtains the spectrum $\frac{d\Gamma}{dE_2}$ along the way. I do not obtain this because my calculation steps are larger. Another nice derivation of the result can be found in exercise 11.3 in Srednicki [8]. All in all I think nothing special is going on in the calculation. There are many different ways to attack the integrals, and the integrals can also be calculated if the mass of the electron is not ignored. I also think similar methods can be used if there are more particles in the decay.

If I have more time I plan to see if using the spinor helicity formalism leads to a quicker calculation. It seems that not only amplitudes can be calculated quicker with this formalism [3], but also cross sections [4].

Muons are very interesting particles. They are heavy cousins of electrons, and one could repeat parts of chemistry with muons replacing electrons. Muons can be used in scanners [5]. They were once proposed to help building fusion reactors [6]. There is also some mystery about the magnetic moment of the muon: the calculation of the value disagrees with experiments [7]. There is also a proposal to build a muon collider, which has some advantages over the already used electron colliders.

__Referenties__

[1] Introduction to Elementary Particles, David Griffiths

[2] Notes on Phase Space, Hitoshi Murayama

[3] Scattering Amplitudes in Gauge Theory and Gravity, Elvang and Huang

[4] arXiv:0905.2715

[5] Muon tomography

[6] Muon-catalyzed fusion

[7] Magnetic dipole moment of muon

[8] Quantum Field Theory, Srednicki

## No comments:

## Post a Comment