Re: Fun problem

Ralph Hartley wrote:
This is a *generic* property of this problem, and does not depend at all on many of the details. It is a consequence of the Poincaré recurrence theorem.

The hole can have any (possibly parameterized) shape, the particle can have internal degrees of freedom (e.g. a rolling ball, an *elastic* ball), or any of the changes you suggest, but *still* if you fix all the parameters but one, then the remaining parameter will have a dense, but measure zero, set of values for which the particle escapes.

That wasn't quite right.

It is true that the set of escaping solutions must be dense, but the in the generic case the particle *always* escapes.

In the original problem there is no way for energy to transfer between the vertical and and horizontal (or other) degrees of freedom, so the particle can never rise above its original level. It must reach the edge exactly at its highest point to escape.

If there is less symmetry, e.g. if the walls are sloped (in either direction), then only the total energy is conserved, and there is more than enough to escape. Sometimes the particle will rise *above* the edge, while moving more slowly in the horizontal direction (or spinning more slowly etc). Then there is a continuous range of positions in which it escapes, which means that eventually it always will (with probability 1).

In general when it escapes it won't be sliding on the surface at its original speed, it will be bouncing with less horizontal velocity.

Igor Khavkine wrote:
Or, one could keep all the idealizations, but
consider another resolver of singularities: quantum mechanics. The
table can be modeled as an impenetrable potential wall, while gravity
can be provided by a linear potential. What do solutions of the single
particle Schroedinger equation with such a potential tell us about the
escape probability?

I'm fairly sure that the quantum version always escapes, even with the perfectly cylindrical hole.

Ralph Hartley

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