Particle physics: calculating the phase space factor for pion to muon decay

[ncs22]

Show that the phase space factor \rho \propto p^2 dp/dE for the decay \pi\rightarrow \mu + \upsilon is

\rho \propto \frac{({m_\pi}^2 - {m_\mu}^2)^2}{{m_\pi}^3}E_\mu

where E is the total energy.I can show that p^2 = ({m_\pi}^2 - {m_\mu}^2)^2/4{m_\pi}^2

but then I get stuck, I don't know how to evaluate dp/dE and I'm not sure what p here is referring to i.e. which particle and in which frame. I worked out the above expression for p² taking it to be the energy for the muon (or neutrino) in the center of mass frame.

 

[turin]

I'm not sure what exactly is meant by phase-space factor, but I would assume that a Dirac delta (for 4-momentum conservation) has already been factored out, which means that you can treat the momentum as the final-state momentum of the particle of your choice. Usually, you choose the visible one (e.g. the muon).

 

[ncs22]

Thanks for the reply.

ok cool at least that means the first bit is probably right :-)

The phase space factor is the number of final states per initial energy, for example the term in Fermi's Golden rule usually denoted by a \rho. Yes I know you can write the phase space as a product of integrals over every particles momentum in which case a delta function has to be introduced to account for the fact that not every momentum is independent.

When I find the answer I will post it here.

 

[Spammed]

I understand I can't post solutions to homework questions until the OP has solved the problem or the thread has been dormant a long time. However, it has been over 15 years and I just solved this same problem, so I would like to finally put this thread to rest.

To start, note that $p$ here is the magnitude of the muon's momentum $\mathbf{p}$ in the COM frame. As this is a two-body decay, $\mathbf{p_\nu} = -\mathbf{p}$ by conservation of momentum. As stated in the original post, $E$ is the total energy in the COM frame, so $E = m_\pi$ (we set $c=1$ throughout).

The 4-momenta of the particles in the COM frame can be written as $$(p_\pi)^\lambda = (m_\pi, \mathbf{0})$$ $$(p_\mu)^\lambda = (E_\mu, \mathbf{p})$$ $$(p_\nu)^\lambda = (p, -\mathbf{p})$$ where we assume the neutrino is massless. By conservation of 4-momentum, $$(p_\pi)^\lambda - (p_\nu)^\lambda = (p_\mu)^\lambda$$ By squaring both sides (recalling the Minkowski metric) we get $$-m_\pi^2+2m_\pi p = -m_\mu^2$$ $$\implies\quad p^2=\frac{(m_\pi^2-m_\mu^2)^2}{4m_\pi^2} \tag{1}$$ Now we consider $\frac{dp}{dE}$. Equating the zeroth component of the 4-momenta we find $$m_\pi=E_\mu+p=\sqrt{m_\mu^2+p^2} + p$$ From earlier we know $m_\pi=E$, so we get $$E=\sqrt{m_\mu^2+p^2} + p$$ Differentiating both sides wrt $E$, we find $$1=\frac{1}{2}(m_\mu^2+p^2)^{-\frac{1}{2}}\cdot2p \frac{dp}{dE}+\frac{dp}{dE}$$ $$\implies \quad 1=\frac{dp}{dE} [1+\frac{p}{(m_\mu^2+p^2)^{\frac{1}{2}}}]=\frac{dp}{dE} [1+\frac{p}{E_\mu}]$$ $$\implies \quad \frac{dp}{dE} = [1+\frac{p}{E_\mu}]^{-1} = \frac{E_\mu}{E_\mu+p}$$ $$\implies\quad \frac{dp}{dE} = \frac{E_\mu}{m_\pi}\tag{2}$$ Combining equations $1$ and $2$ and ignoring the constant gives the required answer.