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 $$||\vec x_i(t)||<R \mbox{ for all } t>0, \, i=1,\dots, N.$$

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 $t>0$?

  • $\begingroup$ Why should one expect this to be possible? At least for $N=2$ (Coulomb scattering) it looks impossible. $\endgroup$ – gmvh Mar 25 '20 at 8:06
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    $\begingroup$ Nitpick: For $N=1$, it's possible. $\endgroup$ – Michael Engelhardt Mar 25 '20 at 14:37
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    $\begingroup$ @SteveHuntsman Those are, unfortunately, for an attractive force; the repulsive case is very different. $\endgroup$ – Steven Stadnicki Mar 25 '20 at 17:29
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    $\begingroup$ 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. $\endgroup$ – Steven Stadnicki Mar 25 '20 at 19:19
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    $\begingroup$ 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. $\endgroup$ – MTyson Mar 25 '20 at 19:22

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.

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    $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ 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? $\endgroup$ – makt Mar 26 '20 at 8:28
  • $\begingroup$ 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 $\leq 0$. $\endgroup$ – Karl Fabian Mar 26 '20 at 11:38

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