Others have asked in general about cases in which the action integral is not minimized, but my question is specific: Can we show that the conventional action integral is always maximized for a system at stable equilibrium?

My reasoning is as follows. The conventional action integral is


At static equilibrium (whether stable or unstable), we have that the kinetic energy is identically zero and the potential energy is constant in time. Thus it would seem that, for any static equilibrium configuration, the action reduces to

$$S=-U\Delta t$$

We know that for stable equilibrium configurations, the potential energy is minimized (the principle of minimum potential energy). Since $S=-U\Delta t$, and $\Delta t$ is positive, it would seem that $S$ and $U$ have opposite signs, so that when $U$ is minimized, $S$ is maximized.

Is my logic sound? Or am I oversimplifying?

  1. We assume OP is investigating an action functional of the form $$S~=~\int_{t_i}^{t_f}\! dt ~L, \qquad L~=~T(q,\dot{q})-V(q),$$ with boundary conditions (BCs).

  2. The classical solution(s) (i.e. solutions to the EL equations satisfying the BCs) do not necessarily minimize/maximize the action, cf. e.g. this Phys.SE post.

  3. For generic BCs, a solution will not be static, i.e. a constant solution at a stationary point $q_0$ of the potential $V$. In order to have a static solution, the BCs should be Dirichlet BCs $$ q^j(t_i)~=~q^j_0~=~q^j(t_f), $$ where $q^j_0$ is a stationary points of the potential $V$. Then the stabile solutions are minimum points for the potential $V$. However, these are not necessarily maximum path for the action $S$, cf. pt. 2 and OP's title question.

  4. Finally let us assume that the kinetic term $T$ is absent even off-shell, i.e. the Lagrangian $L=-V$ is static. (Consider e.g. non-relativistic massless point particles.) Then BCs would over-constrain the system, so we should also remove the BCs to be consistent. Then the classical solutions will be static solutions. Then the stabile solutions are minimum (maximum) points for the potential $V$ (Lagrangian $L$), respectively, cf. OP's title question.

  • $\begingroup$ Thanks, @Qmechanic. If I'm understanding you correctly, we cannot exclude the kinetic energy from the action simply on the basis that we are searching for equilibrium configurations. Thus, for example, we cannot prove the principle of stationary potential energy by taking $S=-U\Delta t$ and noting that $\delta S=0$ is then equivalent to $\delta U=0$. So, in general, the answer to my question is 'No.' However, if the system has no mass, then the kinetic energy will be absent regardless, in which case the answer to my question is 'Yes.' Is that an accurate synopsis? $\endgroup$ – John S Aug 8 '20 at 15:55
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$ – Qmechanic Aug 8 '20 at 15:56
  • $\begingroup$ Cool, thanks again @Qmechanic. $\endgroup$ – John S Aug 8 '20 at 16:10
  • $\begingroup$ I am writing this up as a paper. Do you want to be a coauthor? $\endgroup$ – John S Aug 15 '20 at 19:09

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