Momentum density of states for pion decay.

[pondzo]

Homework Statement

I am trying to calculate the ratio of the density of states factor, $\rho(p)$, for the two decays:
$$\pi^+\rightarrow e^++\nu_e~~$$ and $$\pi^+\rightarrow \mu^++\nu_{\mu}~~$$

Homework Equations

$\rho(p)~dp=\frac{V}{(2\pi\hbar)^3}p^2~dp~d\Omega$

Which is the number of states with momentum between $p$ and $dp$ and lie within a small solid angle $d\Omega$. $V$ is an arbitrary volume to which we confine the system.

Also, $\rho_{Total}=\rho_1(p_1)\rho_2(p_2)...\rho_n(p_n)$

The Attempt at a Solution

Using the above equation:

$\rho_n(p_n)=\frac{V}{(2\pi\hbar)^3}p_n^2~d\Omega$

The ratio R should be:

$R=\frac{p^2(e^+)p^2(\nu_e)}{p^2(\mu^+)p^2(\nu_{\mu})}$

The only way I can think to proceed is:

$M_{\pi^+}^2=(P_e+P_{\mu_e})^2~~\text{ where } P_x \text{ is the 4-momentum of particle } x$
After assuming the mass of the decay products is negligible when compared to its momentum, and that the angle between the two products is 180 degrees, I arrive at: $p^2(e)p^2(\mu_e)=\frac{1}{16}M_{\pi^+}^4$.
But I will just get the same expression for the second decay, so I feel I am doing it wrong. Any suggestions?

 

[mfb]

If you neglect masses, the decays are completely identical.
The muon mass is not small compared to the pion mass.

 

[pondzo]

Let the sub script one mean electron and the subscript two mean electron neutrino, then:

$M_{\pi^+}^2=(P_1+P_2)^2)$
$~~~~~~~=E_1^2+E_2^2+2E_1E_2-p_1^2-p_2^2-2\vec{p_1}\cdot\vec{p_2}$
Using $E_i^2=m_i^2+p_i^2$
$~~~~~~~=m_1^2+m_2^2+2p_1p_2+2\sqrt{(m_1^2+p_1^2)(m_2^2+p_2^2)}$ assumes angle between $p_1$ and $p_2$ is 180 degrees
Where can I go from here? Maybe I need to make some sort of approximation.. Do I use the fact that $\vec{p_1}=-\vec{p_2}$ in the pions rest frame ?

 

[mfb]

I would certainly work in the pion rest frame, yes. You can approximate the neutrino masses with 0, and if you don't care about the amplitude of the decay process I guess the same works for the electron.