Moore's technique suggested the following QUESTION: If it can be shown that some N-body (or other Hamiltonian system) periodic solution is an action minimum, does that prove a stability claim? And by minimum I mean strict local minimum within some space of small-perturbations. Actually Moore's numerical technique I would think only shows it is an action min within the space of periodic solutions (and maybe even that was only shown nonrigorously and within some subspace, but it is probably possible to show it rigorously using second variations and some bounding argument about high frequencies). BEGINNINGS OF AN ANSWER: I'm a little confused because the space we are dealing with in variational problems is infinite dimensional, but I'm used to finite-dimensional. In the finite-dim'l case, certainly action-min implies stability, because 1. every Newtonian mechanics solution is an action-min (unlike, e.g. Einstein GR where the action is merely stationarized, nobody said minimized...) and 2. it is known (at least in the 3-body case, but I presume in N-body case for any N if we stay away from collisions) that solutions are analytic... and 3. there cannot be another min arbitrarily close to your min for the same reason analytic functions cannot have unboundedly many zeros within a neighborhood of a point; zeros always must be isolated points (any "analytic" function disobeying this would have an essential singularity there, not analytic) and this is also true for analytic functions of any finite number of variables. QED. BUT that reasoning does not work in the infinite dimensional case, because the following entire analytic function of infinity variables x1,x2,x3,... SUM F(x[n]*n)/n^9 where F(x) is a polynomial with two mins, one at x=0 and the other at x=1, has (x1,x2,x3...)=(0,0,0,...) as a non-isolated min. MY USUAL NASTY COUNTEREXAMPLE: is a point mass moving along a geodesic on a compact hyperbolic manifold (boundaryless surface of constant negative curvature). As I mentioned before, all such solutions are unstable. There are plenty of closed orbits (closed geodesics). These are not *strict* action mins since the action is the same for every solution with same kinetic energy. But given any particular closed geodesic, you could modify the problem by adding a smooth potential energy function which was 0 on that geodesic and slightly positive off it (say proportional to the squared distance to the geodesic, to be concrete) , in which case that orbit would then become a strict action min, indeed the unique global action min. So would this orbit then become stable to infinitesimally small perturbations? I believe the answer is NO, it still (if the point has enough mass and kinetic energy and the manifold's |curvature| is large enough, versus the magnitude of the potential energy) is unstable. So if I'm right about that (and I think I am) that answers my original question in the negative. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)