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Homework Help: Decay rate of a particle into three particles

  1. Apr 30, 2017 #1
    1. The problem statement, all variables and given/known data
    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,
    [tex]|\mathcal{M}|^2 = 32G_F^2(m^2-2mE)mE[/tex]
    being [itex]E[/itex] the electron antineutrino energy and [itex]m[/itex] the muon mass. Find the decay rate.

    2. Relevant 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
    [tex]\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}[/tex]
    being [itex] \delta^4(\Sigma_p) = \delta^4 ( p_1^\mu - (p_2^\mu + p_3^\mu + p_4^\mu))[/itex] a four-momentum conserving delta function.

    3. 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 [itex]\mathbf{p}_1=0[/itex] and [itex]E_1 = m[/itex]. With this I've rewritten the delta function as [itex]\delta^4(\Sigma_p)=\delta(m-E_2-E_3-E_4)\delta^4(-\mathbf{p}_2-\mathbf{p}_3-\mathbf{p}_4[/itex]. Thus we get

    [tex]\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[/tex]

    now since the problem tells to ignore the masses of the electron and the neutrinos we can assume that [itex]E_i=|\mathbf{p}_i|[/itex] for [itex]i=2,3,4[/itex]. Thus we simplify further to

    [tex]\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[/tex]

    now we integrate over [itex]d^3p_4[/itex] using the delta. This will compute every [itex]\mathbf{p}_4[/itex] in [itex]-\mathbf{p}_2 - \mathbf{p}_3[/itex]. Thus we find that

    [tex]\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[/tex]

    Now we use spherical coordinates. Considering [itex]\mathbf{p}_2[/itex] fixed we arrange the axis of [itex]\mathbf{p}_3[/itex] so that [itex]\mathbf{p}_2[/itex] lies along the [itex]z[/itex]-axis. With this, writing [itex]k_i = |\mathbf{p}_i| [/itex] we have [tex]|\mathbf{p}_2+\mathbf{p}_3| = \sqrt{k_2+k_3 +2k_2k_3\cos \theta_3}[/tex] which turns the integral into the horrible form

    [tex]\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[/tex]

    where we can integrate over the [itex]\phi[/itex] variables and the [itex]\theta_2[/itex] and simplify to

    [tex]\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[/tex]

    now is where I'm stuck. My better idea was to change variables defining [itex]k = k_2+k_3+2k_2k_3\cos\theta_3[/itex]. That is, the energy on the center of mass frame. This would make the delta simplify to [itex]\delta(m-k)[/itex] 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?
  2. jcsd
  3. Apr 30, 2017 #2


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    2017 Award

    Staff: Mentor

    The decay just has two non-trivial degrees of freedom. Isolate them and all other integrals are trivial (you integrate over a constant).
    Typically one would choose the invariant mass of two particles for two different pairs, but as the electron neutrino mass appears in your matrix element, it might be better to choose this as one variable.
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