# $\tau \rightarrow \nu_{\tau} e^- \bar{\nu}_e$ decay. Where is my mistake?

I want to calculate $$\tau \rightarrow \nu_{\tau} e^- \bar{\nu}_e$$ decay rate $$\Gamma$$, with the effective four fermion interaction to leading order.
$$\mathcal{L}_{eff} = -\frac{G_F}{\sqrt{2}} [e^+\gamma^{\mu} (1-\gamma_5) \nu_e][\bar{\nu}_{\tau} \gamma_{\mu} (1-\gamma_5) \tau]$$ I take masses of outcoming particles to zero. I get
$$|\mathcal{M}|^2 = 64 G_F^2~p_{\tau} \cdot p_3~p_1\cdot p_2$$ where $$\tau(p_{\tau}), \nu_{\tau}(p_1), e^-(p_2), \bar{\nu}_e(p_3)$$. I am relatively confident that this is correct. We have: $$d\Gamma = \frac{1}{2E_{\tau}} |\mathcal{M}|^2 d\Pi_{LIPS} \\ d\Pi_{LIPS} = \frac{d^3p_1}{2E_1(2\pi)^3} \frac{d^3p_2}{2E_2(2\pi)^3} \frac{d^3p_3}{2E_3(2\pi)^3} (2\pi)^4 \delta^4(p_{\tau}-p_1-p_2-p_3)$$

I go on with inserting some 1's to reduce the 3 body phase space integral to two 2 body phase space integrals. Define $$s= (p_1+p_2)^2 = 2p_1\cdot p_2$$ $$\Gamma \stackrel{(p_1+p_2)^2 > 0}{=} \frac{1}{2E_{\tau}} \int \frac{d^3p_1}{2E_1(2\pi)^3} \int \frac{d^3p_2}{2E_2(2\pi)^3} \int \frac{d^3p_3}{2E_3(2\pi)^3} \int d^4p \int_0^{\infty} ds (2\pi)^4 \delta(p^2-s) \delta^4(p-p_1-p_2) \delta^4(p_{\tau}-p-p_3)|\mathcal{M}|^2 \\ \stackrel{E := p^0 = \sqrt{\vec{p}^2+s}}{=} \frac{1}{2E_{\tau}} \int_0^{\infty} ds \int \frac{d^3p_1}{2E_1(2\pi)^3} \int \frac{d^3p_2}{2E_2(2\pi)^3} \int \frac{d^3p_3}{2E_3(2\pi)^3} \int \frac{d^3p}{2E(2\pi)^3} (2\pi)^7 \delta^4(p-p_1-p_2) \delta^4(p_{\tau}-p-p_3)|\mathcal{M}|^2 \\ = \frac{1}{4\pi E_{\tau}} \int d\Pi_{3} \int d\Pi_{12} |\mathcal{M}|^2$$ with $$d\Pi_{12} := \frac{d^3p_1}{2E_1(2\pi)^3} \frac{d^3p_2}{2E_2(2\pi)^3} (2\pi)^4 \delta^4(p-p_1-p_2)\\ d\Pi_3 := \frac{d^3p}{2E(2\pi)^3} \frac{d^3p_3}{2E_3(2\pi)^3} (2\pi)^4 \delta^4(p_{\tau}-p-p_3)$$ Since these quantities are Lorentzinvariant I can calculate $$\Pi_{12} = \frac{1}{8\pi}$$ in the frame where $$\vec{p} = 0$$.
In the tauon rest frame $$p_{\tau} = (m_{\tau}, 0)$$ I then get \begin{aligned}\Gamma &= \frac{1}{4\pi m_{\tau}} \frac{1}{8\pi} \int_0^{\infty} ds \int \frac{d^3p}{2E(2\pi)^3}\int \frac{d^3p_3}{2E_3(2\pi)^3} (2\pi)^4 \delta^4(p_{\tau}-p-p_3) 32 G_F^2 E_3 m_{\tau} s \\ &\stackrel{E_3 = |\vec{p}|}{=}\frac{G_F^2}{16\pi^4} \int_0^{\infty} ds \int d^3p ~\frac{s}{E}\delta(m_{\tau} - E - E_3) \\&= \frac{G_F^2}{4\pi^3} \int_0^{\infty} ds \int_0^{\infty} dE~sE ~\delta(m_{\tau} - E - \sqrt{E^2-s}) \\ &= \frac{G_F^2}{8\pi^3} \int_0^{\infty} dE~(m_{\tau}^2-2m_{\tau} E) E(m_{\tau}-E) \theta(m_{\tau}-2 E) \\&= \frac{G_F^2}{8\pi^3} \int_0^{m_{\tau}/2} dE (m_{\tau}^2-2m_{\tau} E) E (m_{\tau}-E) \\&= \frac{G_F^2 m_{\tau}^5}{256\pi^3}\end{aligned} However the result should be $$\frac{G_F^2 m_{\tau}^5}{192\pi^3}$$, so I am a factor $$4/3$$ off, but I can't find my mistake.

• Your formula for $|{\cal M}|^2$ is wrong. You have dropped off terms involving the spin of the τ and that of the e , which do marvels in modifying the actual angular distribution: a very distinctive angular distribution! Dirty hint: compare to the decay of the μ. – Cosmas Zachos Mar 20 at 22:13