My question is based on thinking about a point particle in electromagnetic fields, but the idea should apply to other problems.

The point $\mathbf x_0$ is an equilibrium point of the force field $\mathbf F$ given that,

(i) The force at $\mathbf x_0$ is zero. That is, $\mathbf F(\mathbf x_0)=\mathbf 0$

(ii) There exists a neighborhood of $\mathbf x_0$ where the forces have a component that point towards it. That is, $\exists~ \varepsilon>0$ such that for every $\mathbf x$ that satisfies $0<|\mathbf x_0-\mathbf x|\leq\varepsilon$, the expression $\mathbf F(\mathbf x)\cdot(\mathbf x_0-\mathbf x)>0$ holds.

Is this correct?

  • $\begingroup$ This situation cannot happen with electrostatic field. If the field is electromagnetic, force is not function of position only. What are you trying to do? Definition of stable equilibrium in EM theory? $\endgroup$ – Ján Lalinský Feb 10 '20 at 23:38

Let's start off by what we mean by stable equilibrium.

A point of stable equilibrium of a system is a configuration of the system such that, if we perturb it slightly and let the time evolution make its course, then the configuration will remain in a neighborhood of the stable equilibrium point.

To put it more precisely, using the Hamiltonian formalism, a point $P = (\vec{q}_0, \vec{p}_0)$ in the phase space $\Gamma$ of the system is a stable equilibrium point if and only if for every $r>0$ (and eventually lower than some threshold to account for any different behaviour the potential might have far from $P$) there exists some $s>0$ such that the time evolution of any point initially in the the open ball $B(P, r)$ [where by $B(P, r)$ we mean the ball centered in $P$ with radius $r$] stays in $B(P, s)$ for every time $t$ (I don't think the openness of the ball plays any major role here but I might be wrong don't quote me on that).

In your case you are referring to equilibrium points that are of the type $P=(\vec{q}, \vec{0})$ (i.e. with all vanishing momenta). By comparing the definition with your conditions it appears that if a point in phase space $\Gamma$ satisfies condition (i) and (ii) then it will of course be a stable equilibrium point. The reverse though might not be true: there might be systems with stable equilibrium points that do not satisfy conditions (i) and (ii).

So to wrap it all up, the answer depends on the question. If you are asking whether a point satisfying conditions (i) and (ii) is a stable equilibrium point than the answer is yes.

If, on the other hand you were asking whether every stable equilibrium point satisfies conditions (i) and (ii) then the answer is no.


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