= Three-body leptonic tau-decay kinematic bounds
{title2=$m_\ell\leq E_\ell^*\leq(m_\tau^2+m_\ell^2)/(2m_\tau)$}
Combine the two massless <neutrinos> in $\tau^-\to\ell^-\bar\nu_\ell\nu_\tau$ into $q$. In the parent rest frame, <four-momentum conservation> gives
$$
q^2=m_\tau^2+m_\ell^2-2m_\tau E_\ell^*,\qquad
0\leq q^2\leq(m_\tau-m_\ell)^2.
$$
The lower bound follows from the nonnegative <Minkowski norm> of the sum of two future-directed null <four-vectors>; the upper bound follows from $E_\ell^*\geq m_\ell$. Substituting gives the displayed <energy> bounds. They are kinematic: they do not specify the <weak interaction> <decay amplitude> or the <probability density function> inside the interval.
If the charged daughter is also massless, put $x=2E_\ell^*/m_\tau\in[0,1]$. A parent with <Lorentz factor> $\gamma$ and <speed> $\beta$ in <natural units> produces
$$
E_\ell=\gamma\frac{m_\tau x}{2}(1+\beta\cos\theta^*),\qquad
0\leq E_\ell\leq\frac{E_\tau}{2}(1+\beta).
$$
Here $\theta^*$ is the daughter angle relative to the boost direction in the parent rest frame. Two opposite parent <momenta> therefore need not produce two opposite visible daughter <momenta>: the invisible system can recoil differently in each <particle decay>.
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