[Physics] Formalism to deal with discontinuous potentials in classical mechanics (hard wall, hard spheres)


It seems to me that Hamiltonian formalism does not suit well for problems involving instantaneous change of momentum, like particle collisions with hard wall or hard sphere gas model. At least I could not apply it straightforwardly to the simplest possible problem of 1D particle hitting a wall:

            │/ wall
  particle  │/
───o────────┼────────────> x

My attempt was quite direct. I took the Hamiltonian to be

$$H = \frac{p^2}{2m} + U(x)$$

with potential $U$ defined as

$$U(x) = \cases{ 0, \; x < 0, \\[.5em] K, \; x > 0, \\[.5em] E, \; x= 0.}$$

where $E$ is the particle energy and $K > E$. Hamiltonian equations should read as

$$\cases{\dot x = \frac{p}{m}, \\[.6em] \dot p = – \frac{d U}{d x}.}$$

It is not hard to integrate the first equation, but my attempts to integrate the second one did not lead to any meaningful result (that's why I do not share them here, it was a complete failure).

So I ask whether it is possible to obtain the solution to the problem by directly integrating Hamiltonian equations in the form above, without relying on general mechanical theorems/principles like energy conservation? Or is such an approach completely unsuitable for the task?

If so, what is the general (and elegant) approach to such systems?

There exist a related question on PSE "Hamiltonian function for classical hard-sphere elastic collision", but the setting is more cumbersome.

Best Answer

Perhaps the simplest and most intuitive approach is to regularize the hard wall potential

$$V_0(x)~=~\left\{ \begin{array}{rcl} 0 &\text{for}& x<0 \cr\cr \infty &\text{for}& x>0\end{array}\right. $$


$$ \lim_{\varepsilon \to 0^+} V_{\varepsilon}(x) ~=~V_0(x).$$

For instance, one could choose the regularized potential as

$$ V_{\varepsilon}(x)~=~\frac{x}{\varepsilon}\theta(x).$$

This corresponds to constant velocity for $x<0$ and constant acceleration for $x>0$. Next write down a continuous solution for the position $x_{\varepsilon}(t)$ as a function of time $t$, say, for given pertinent initial conditions.

Finally, at the end of the calculation, one should remove the regularization $\varepsilon \to 0^+$ again, and check if the limit $$ \lim_{\varepsilon \to 0^+} x_{\varepsilon}(t)$$ makes physical sense.

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