Special theory of relativity > Conservation of mass and momentum > Problem 14.5.21

Statement

$14.5.21$. In what range of energies do the kinetic energies of the electron and neutrino lie when a $\mu^-$-meson decays:
$$
\mu^- \to e^- + \nu + \bar{\nu}
$$

Solution

Neutrinos are considered massless particles, so their kinetic energy equals their total energy.
It is easy to show that the decay is possible with zero kinetic energies of the electron or neutrino (for example, when the other two particles fly apart in opposite directions with equal momenta). Let us find the maximum possible energy.

In the muon rest frame, the total momentum is zero. The electron energy is maximized when the electron acquires the maximum momentum, i.e., when both neutrinos fly in the same direction opposite to the electron. The momenta are collinear, so let $E = E_{\nu} + E_{\bar{\nu}}$, $P = p_{\nu} + p_{\bar{\nu}} = \frac{E}{c}$.
Conservation laws:
$$
m_{\mu^-}c^2 = E + E_e \qquad p_e - P = 0 \tag{1}
$$
Invariant for the electron:
$$
m_e^2 c^2 = \frac{E_e^2}{c^2} - p_e^2 \tag{2}
$$
Substituting (2) into (1):
$$
E^2 = E_e^2 - m_e^2 c^4 = (m_{\mu^-}c^2 - E_e)^2 \tag{3}
$$
$$
E_e = c^2 \frac{m_{\mu^-}^2 + m_e^2}{2m_{\mu^-}} \tag{4}
$$
$$
E_{eK} = E_e - m_e c^2 = c^2 \frac{m_{\mu^-}^2 + m_e^2 - 2m_{\mu^-}m_e}{2m_{\mu^-}} = c^2 \frac{(m_{\mu^-} - m_e)^2}{2m_{\mu^-}} \tag{5}
$$

Now similarly, let a neutrino fly in one direction, and the electron and antineutrino in the opposite direction.

Conservation laws and invariant:
$$
m_{\mu^-}c^2 = E_{\nu} + E_{\bar{\nu}} + E_e \qquad p_e + \frac{E_{\bar{\nu}}}{c} = \frac{E_{\nu}}{c} \tag{6}
$$
$$
m_e^2 c^2 = \frac{E_e^2}{c^2} - p_e^2 \tag{7}
$$
$$
(E_{\nu} - E_{\bar{\nu}})^2 = E_e^2 - m_e^2 c^4 \tag{8}
$$
$$
(E_{\nu} - E_{\bar{\nu}})^2 = (m_{\mu^-}c^2 - E_{\nu} - E_{\bar{\nu}})^2 - m_e^2 c^4 \tag{9}
$$

Rewrite (9) in notation $x = E_{\nu}$, $y = E_{\bar{\nu}}$, $M = m_{\mu^-}c^2$, $m = m_e c^2$:
$$
(y - x)^2 = (M - x - y)^2 - m^2.
$$
Expanding the brackets, for example using the difference of squares, we obtain
$$
(2x - M)(2y - M) = m^2.
$$
$$
y = \frac{1}{2} \left( \frac{m^2}{2x - M} + M \right)
$$

Examine monotonicity:
\[
\frac{dy}{dx} = -\frac{m^2}{(2x - M)^2} < 0.
\]
Thus the function strictly decreases as $x$ increases. Therefore, the maximum value of $x = E_{\nu}$ is achieved at the smallest possible $x$. The smallest possible $x = E_{\bar{\nu}}$ is zero (energy cannot be negative). Substituting $x = 0$:
$$
E_{\nu} = \frac{c^2}{2} \left( m_{\mu^-} - \frac{m_e^2}{m_{\mu^-}} \right)
$$
It appears the author has a typo in the answer.

Answer

$$
E_{\nu}\in \left[0;\frac{c^2}{2}\left(m_{\mu^-}-\frac{m_e^2}{m_{\mu^-}}\right)\right]
$$
$$
E_{eK}\in \left[0;\frac{c^2(m_{\mu^-}-m_e)^2}{2m_{\mu^-}}\right]
$$