- #1
pondzo
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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?