- #1
leo.
- 90
- 4
Homework Statement
Consider the process of decay of a muon into one electron, one electron antineutrino and one muon neutrino using the Fermi theory. Assume the matrix element is, ignoring the electron's and the two neturino's masses,
|\mathcal{M}|^2 = 32G_F^2(m^2-2mE)mE
being E the electron antineutrino energy and m the muon mass. Find the decay rate.
Homework Equations
I believe the most relevant equation would be the equation for the decay rate, particularized for the case of three particles in the final state
\Gamma = \dfrac{1}{2E_1} \int (2\pi)^4 \delta^4(\Sigma_p) |\mathcal{M}|^2 \dfrac{d^3 p_2}{(2\pi)^3}\dfrac{1}{2E_2}\dfrac{d^3 p_3}{(2\pi)^3}\dfrac{1}{2E_3}\dfrac{d^3 p_4}{(2\pi)^3}\dfrac{1}{2E_4}
being \delta^4(\Sigma_p) = \delta^4 ( p_1^\mu - (p_2^\mu + p_3^\mu + p_4^\mu)) a four-momentum conserving delta function.
The Attempt at a Solution
Here I'm labeling (1) the initial muon, (2) the electron antineutrino, (3) the electron and (4) the muon neutrino.
My attempt was to work in the center of mass frame, so that \mathbf{p}_1=0 and E_1 = m. With this I've rewritten the delta function as \delta^4(\Sigma_p)=\delta(m-E_2-E_3-E_4)\delta^4(-\mathbf{p}_2-\mathbf{p}_3-\mathbf{p}_4. Thus we get
\Gamma = \dfrac{1}{2m} \dfrac{(2\pi)^4}{8(2\pi)^9}\int \dfrac{|\mathcal{M}^2|}{E_2E_3E_4} \delta(m-E_2-E_3-E_4)\delta^3(-\mathbf{p}_2-\mathbf{p}_3-\mathbf{p}_4)d^3 p_2 d^3p_3 d^3p_4
now since the problem tells to ignore the masses of the electron and the neutrinos we can assume that E_i=|\mathbf{p}_i| for i=2,3,4. Thus we simplify further to
\Gamma = \dfrac{1}{16m} \dfrac{1}{(2\pi)^5}\int \dfrac{|\mathcal{M}^2|}{|\mathbf{p}_2| |\mathbf{p}_3| |\mathbf{p}_4|} \delta(m-|\mathbf{p}_2|-|\mathbf{p}_3|-|\mathbf{p}_4|)\delta^3(-\mathbf{p}_2-\mathbf{p}_3-\mathbf{p}_4)d^3 p_2 d^3p_3 d^3p_4
now we integrate over d^3p_4 using the delta. This will compute every \mathbf{p}_4 in -\mathbf{p}_2 - \mathbf{p}_3. Thus we find that
\Gamma = \dfrac{1}{16m} \dfrac{1}{(2\pi)^5}\int \dfrac{|\mathcal{M}^2|}{|\mathbf{p}_2| |\mathbf{p}_3| |\mathbf{p}_2+\mathbf{p}_3|} \delta(m-|\mathbf{p}_2|-|\mathbf{p}_3|-|\mathbf{p}_2+\mathbf{p}_3|)d^3 p_2 d^3p_3
Now we use spherical coordinates. Considering \mathbf{p}_2 fixed we arrange the axis of \mathbf{p}_3 so that \mathbf{p}_2 lies along the z-axis. With this, writing k_i = |\mathbf{p}_i| we have |\mathbf{p}_2+\mathbf{p}_3| = \sqrt{k_2+k_3 +2k_2k_3\cos \theta_3} which turns the integral into the horrible form
\Gamma = \dfrac{1}{16m} \dfrac{1}{(2\pi)^5} \int_0^{2\pi}\int_0^{2\pi} \int_0^\pi \int_0^\pi \int_0^\infty \int_0^\infty \dfrac{|\mathcal{M}^2|\delta(m-k_2-k_3-\sqrt{k_2+k_3+2k_2 k_3\cos\theta_3})}{k_2 k_3 \sqrt{k_2+k_3+2k_2k_3\cos\theta_3}} k_2k_3\sin\theta_2\sin\theta_3 dk_2 dk_3 d\theta_2d\theta_3 d\phi_2d\phi_3
where we can integrate over the \phi variables and the \theta_2 and simplify to
\Gamma = \dfrac{1}{8m} \dfrac{1}{(2\pi)^3} \int_0^\pi \int_0^\infty \int_0^\infty \dfrac{|\mathcal{M}^2|\delta(m-k_2-k_3-\sqrt{k_2+k_3+2k_2 k_3\cos\theta_3})}{ \sqrt{k_2+k_3+2k_2k_3\cos\theta_3}}\sin\theta_3 dk_2 dk_3 d\theta_3
now is where I'm stuck. My better idea was to change variables defining k = k_2+k_3+2k_2k_3\cos\theta_3. That is, the energy on the center of mass frame. This would make the delta simplify to \delta(m-k) but I didn't get much further.
So is my approach correct, or I made some mistake on how to tackle this? And how do I proceed?