# Movement of repelled particles in a ball

EDIT:

Given a system of $$N\geq 3$$ charged point particles in $$\mathbb{R}^3$$ of the same charge which interact according to Coulomb law (thus they repell one from each other). Is it possible that the system remains in a fixed ball all the time? (For $$N=2$$ this is impossible, and this is what I expect in general.)

More precisely, denote $$m_1,\dots, m_N>0$$ the masses of the particles. Assume that the $$i$$th particle acts on $$j$$th one with the force

$$\vec F_{ij}=\frac{k_ee_ie_j}{|\vec x_j-\vec x_i|^3}\cdot (\vec x_j-\vec x_i),$$ where $$k_e>0$$ is a constant, $$e_i$$ is a charge of $$i$$th particle such that $$e_ie_j>0$$, $$\vec x_i$$ is the location of the $$i$$th particle. The equations of motions are $$m_j\frac{d^2 x_j}{dt^2}=\sum_{i\ne j}\vec F_{ij}, \mbox{ where } j=1,\dots,N.\,\,\,(1)$$

The question is whether there is a solution such that for some $$R$$ one has $$||\vec x_i(t)||0, \, i=1,\dots, N.$$

ADDED: I expect that this is impossible. In fact I expect that not only for Coulomb law, but still in greater generality. Assume that the equations (1) are satisfied when the force $$\vec F_{ij}=\vec F_{ij}(x_i,x_j)$$ has the same direction as the vector $$\vec x_j-\vec x_i$$. Assume moreover that if all points are in a fixed ball of the radius $$R$$ then for some constant $$\varepsilon >0$$ such that $$||\vec F_{ij}||>\varepsilon.$$ Is there a solution of (1) such that all the point are in the ball of radius $$R$$ for all $$t>0$$?

• Why should one expect this to be possible? At least for $N=2$ (Coulomb scattering) it looks impossible. – gmvh Mar 25 '20 at 8:06
• Nitpick: For $N=1$, it's possible. – Michael Engelhardt Mar 25 '20 at 14:37
• @SteveHuntsman Those are, unfortunately, for an attractive force; the repulsive case is very different. – Steven Stadnicki Mar 25 '20 at 17:29
• An idle thought, I don't know if it can pan out but it's a possible angle: consider particles by their distance from the origin. Then if at any time $t_0$ the particle furthest from the origin is outward-moving (i.e., its velocity has positive dot product with the vector from the origin to its position), that will be true for all times $t\gt t_0$. It may be possible to show that in that case velocity of the most distant body is $\Omega(1/r)$, in which case the distance from the origin would have to grow as at least $\Omega(\log r)$; then all that's left is showing that it's true at some point. – Steven Stadnicki Mar 25 '20 at 19:19
• Let $r(t)$ be the distance of the farthest particle from the origin. It is piecewise analytic. When $r$ is not analytic, a particle has just passed another particle and it is easy to see that $r'$ doesn't decrease. When $r$ is analytic, the net force on an extremal particle is directed outwards and bounded below by some fixed $\epsilon$ (thanks to a short calculation with the $1/r^2$-law and the assumption that $r<R$), so $r''>\epsilon$. These two conditions show $r\to\infty$. This argument breaks down when the force is weaker than a $1/r$-law or isn't analytic. – MTyson Mar 25 '20 at 19:22

• Very nice. We can also check how the $N=1$ case (cf. nitpick to OP) escapes this: Then, the virial theorem reduces to $0=0$ (the particle can, and must, remain at rest) and there is no contradiction. As soon as $N>1$, the potential energy is strictly positive, and the contradiction arises. – Michael Engelhardt Mar 26 '20 at 14:03
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.
• I am not sure why in the second case that at time $t_n$ the radial acceleration is less than $\epsilon$? May I get more details? – makt Mar 26 '20 at 8:28
• One can even choose points $t_n$ with radial acceleration $\leq 0$. The particle is either oscillating towards $\partial B$, then $(t_n)$ can be chosen to be a subsequence of the locally closest points, in which accelerations are $\leq 0$, or the particle is creeping towards $\partial B$ with oscillating non-negative radial velocity, then the $t_n$ can be chosen to be points of locally maximal radial velocity, again with accelerations $\leq 0$, or the radial velocity is creeping towards zero, and again from some time on radial acceleration is $\leq 0$. – Karl Fabian Mar 26 '20 at 11:38