Optimise the N body decay!

For this problem, it is tempting to try and make up from scratch a configuration that will maximise the total energy of $m_1$ . However, it is not trivial to prove that this is in fact the optimal solution. A simple approach I found (most probably not original at all) makes it obvious what configuration that is and quickly gives the maximum energy. Throughout this solution I will assume $c = 1$ . I use $\left \| . \right \|$ to denote the length of a four-vector and $\left | . \right |$ to denote the length of a spatial vector.

For timelike four-vectors in Minkowski space with sign convention (+, -, -, -), the inverse of the triangle inequality holds:

$\left \| x \right \| + \left \| y \right \|\leq \left \| x + y \right \|$

Using this it's not hard to show that for any set of vectors in Minkowski space $\left \{ x_i \right \}$ we have:

$\sum\limits_{i=2}^{N} \left \| x_i \right \| \leq \left \| \sum\limits_{i=2}^{N} x_i \right \|$

Let this set be the four-momenta of particles $2, 3, ... ,N$ .

By conservation of four-momentum, we have:

$\sum\limits_{i=2}^{N} p_i = P - p_1$ where $P = \left ( M,0,0,0 \right)$ is the four-momentum of the mother particle.

Plugging this into our inequality, and using the fact that $\left \| p_i \right \| = m_i$ , we get:

$\sum\limits_{i=2}^{N} m_i =\sum\limits_{i=2}^{N} \left \| p_i \right \| \leq \left \| P-p_1 \right \| = \sqrt{\left(M-E_1\right)^2 - \left|\vec{p_1}\right|^2} = \sqrt{M^2 - 2ME_1 + E_1^2 - \left|\vec{p_1}\right|^2} = \sqrt{M^2+m_1^2-2ME_1}$

Manipulating the terms, we obtain:

$E_1 \leq \frac{M^2 + m_{1}^2 - \left( \sum\limits_{i=2}^{N} m_i \right )^2}{2M} = \frac{M}{2} = 2m_1$

Putting back the $c$ where it needs to be, we get:

$\gamma = \frac{E_1}{m_{1}c^2} \leq 2$ .

Using the properties of the triangle inequality, we can conclude that equality occurs when all four-vectors after the decay are scalar multiples of each other. This means that all spatial momenta are along one direction, and for particles $2, ..., i , ... , j, ..., N$ we have:

$\beta_i = \frac{\left|\vec{p_i}\right|}{Ei} = \frac{\left|\vec{p_j}\right|}{Ej} = \beta_j$ .

This means that the velocities of all other particles are equal.

To maximise the energy of particle $m_1$ , consider the rest of the particles as one composite particle with no relative motion such that the energy of all the particles is minimised. Since $\sum_{i=2}^{\infty} {m_i} = \frac{M}{4} = m_1$ , the decay is symmetric. Therefore $E_1 = \frac{1}{2}M c^2$ . The general case for $m_i$ can be solved using an identical approach.