Phase space factor for a two-body decay

Question

I was trying to understand how pion decays to muons and not electrons because of helicity suppression. So I was trying to figure out the ratio of the decay widths.

PDG review for kinematics (subsection 49.4.2) (PDF) reports that for a two-body decay we have $$ d\Gamma = \frac{1}{32 \pi^2} |\mathcal{M}|^2 \frac{|\mathbf{p}_1|}{M^2} d\Omega\tag{49.18} $$ with $$ E_1 = \frac{M^2-m_2^2-m_1^2}{2M} \quad |\mathbf{p}_1| = \frac{1}{2M} \sqrt{\lambda(M^2,m_1^2,m_2^2)} $$ and $\lambda$ defined as $$ \lambda(\alpha,\beta,\gamma) = \alpha^2+\beta^2+\gamma^2 -2\alpha\beta-2\alpha\gamma-2\beta\gamma. $$

Now if we label the neutrino as the second particle, and consider it massless, we get $$ E_1 = \frac{M^2+m_1^2}{2M} $$ $$ |\mathbf{p}_1| = \frac{1}{2M} \sqrt{M^4+m_1^4-2M^2m_1^2} = \frac{1}{2M} \sqrt{(M^2-m_1^2)^2} = \frac{1}{2M} (M^2-m_1^2) $$

and I'm guessing that, except for a few constants, I could call "phase space factor" all that is not $|\mathcal{M}|^2$, and the total width is $$ \Gamma \propto |\mathcal{M}|^2 \frac{|\mathbf{p}_1|}{M^2} \propto \frac{M^2-m_1^2}{M^2}. $$

However this to me seems incompatible with what I read on Perkins chapter 7.10, that says that the phase-space factor for pion decay (neglecting neutrino mass) is

$$ p^2 \frac{dp}{dE_0} = \frac{(M^2+m_1^2)(M^2-m_1^2)^2}{4M^2} $$ derived from $$ E_0 = M = |\mathbf{p}_1| + \sqrt{|\mathbf{p}_1| ^2+m^2}. $$

The derivation steps I from perkins seem to make sense, however I was trying to approach this both ways. My guess is that I am not interpreting the formula by PDG correctly.

I am aware I did not compute the matrix element but I will just take it for granted as of now, I was trying to focus only on the phase-space part.

Answer 1

Starting from the decay rate given by $$ \Gamma(\pi \rightarrow \mu+\nu) = \frac{1}{32\pi} |\mathcal{M}|^2 \frac{|\mathbf{p}_\mu|}{m_\pi^2} \, \text{d}\Omega, $$ where I changed to notation to match the particles. Assuming a massless neutrino we get $$ |\mathbf{p}_\mu| = \frac{1}{2m_\pi} \sqrt{\lambda(m_\pi^2,m_\mu^2,m_\nu^2)}=\frac{1}{2m_\pi}\sqrt{m_\pi^4+m_\mu^4-2m_\pi^2m_\mu^2}=\frac{1}{2m_\pi}\sqrt{(m_\pi^2-m_\mu^2)^2}=\frac{m_\pi^2-m_\mu^2}{2m_\pi}. $$ This leads to the following proportionality $$ \Gamma(\pi \rightarrow \mu+\nu) \propto \frac{|\mathbf{p}_\mu|}{m_\pi^2} = \frac{m_\pi^2-m_\mu^2}{2m_\pi^3}. $$ I could not find the equation you referenced in Peskin but following the calculations in [1] the rate for pion decay turns out to be $$ \Gamma(\pi \rightarrow \mu+\nu) = \frac{G_{\text{wk}}F_\pi^2m_\mu^2(m^2_\pi-m_\mu^2)^2}{16\pi m_\pi^3}, $$ and to answer your question take the ratio of the half-widths of the pion–electron and the pion–muon decay reactions

$$ \frac{\Gamma(\pi\rightarrow e+\nu)}{\Gamma(\pi\rightarrow \mu+\nu)} = \frac{m_e^2(m_\pi^2-m_e^2)^2}{m_\mu^2(m_\pi^2-m_\mu^2)^2}, $$ which presumably is a small number which is the helicity suppression you are talking about.

[1] The Quantum Theory of Fields Vol II, Steven Weinberg