Movement of repelled particles in a ball

If all the particles remained in a bounded domain, the virial theorem would apply. In the case of a radial inverse square power law, it states that twice the asymptotic time average of kinetic energy of the system equals minus the asymptotic time average of its potential energy. However, while the kinetic energy is always nonnegative, the potential energy for repulsive coulomb forces is positive, contradiction.

Let $B$ be the smallest ball such that all $N$ particles remain inside $B$ for all $t\geq0$. Either the trajectory of one of the particles intersects $\partial B$ at some finite time $t_0$, or there is one particle and a sequence $(t_n)_{n\in\mathbb{N}}$ with $\lim \limits_{n \to \infty} t_n ~=~\infty$ such that the particle position at $t_n$ has distance $<1/n$ from $\partial B$, and no other particle at $t_n$ is closer to $\partial B$.

In the first case the radial velocity of the particle at $t_0$ is zero, and therefore the radial component of its acceleration must be $\leq 0$, in contradiction to the fact that the radial component of all forces is positive.

In the second case for each $\epsilon>0$ one can find a time $t_n$ such that the radial acceleration of the particle is less than $\epsilon$. But the radial component of the force from the other particles has a global lower bound because they cannot get arbitrarily close to the particle due to global energy conservation, but have to stay inside the sphere. Choosing a sufficiently small $\epsilon>0$ therefore leads to a contradiction.