I am stuck working on a problem that involves finding the Lagrangian for a free particle constrained to move on the surface of a disk of radius $a$. The particle collides elastically with the edge of the disk. I have tried the standard method of multipliers, but that doesn't work. Is there a way to convert this non-holonomic constraint into a holonomic one?


1 Answer 1


Hmm, cool problem. At each collision with the wall, $v_{r}=\dot{r}$ flips sign but $\dot{\theta}$ stays the same, and thus so does the angular momentum $L$. No idea if this works, but it seems worth a shot - I was inspired by quantum mechanics. You basically have a particle in a well. So \begin{align} L &= T-V\\ &= \frac{1}{2}mv^{2} - V\\ &= \frac{1}{2}m(\dot{r}^{2}+r^{2}\dot{\theta}^{2}) - V(r) \end{align} where $V(r) = \begin{cases} 0 &\mbox{if } r<a \\ \infty & \mbox{if } r\geq a \end{cases}$. You could possibly write $$V=V_{0}\Theta(r-a)$$ and then send $V_{0}\rightarrow\infty$, where $\Theta(x)$ is the Heaviside step function.

  • $\begingroup$ Are you sure about the Heaviside step function version? $\endgroup$
    – alarge
    Nov 26, 2014 at 8:57
  • $\begingroup$ Sure about it in the sense...? Oh you mean having the $\theta(r)$ term. As $V_{0}\rightarrow\infty$ it's not a problem, as it enforces the $r\geq 0$ condition. I've found it useful in electrostatics where it helps me remember the Dirac delta from the divergence of $\frac{\hat{r}}{r^2}$. But I see your point, I'll edit to include the more traditional version. $\endgroup$
    – G. Paily
    Nov 26, 2014 at 16:42
  • $\begingroup$ Sorry, I wasn't very clear, was I. I meant that should it not be just $V = V_0\Theta(r-a)$? $\endgroup$
    – alarge
    Nov 26, 2014 at 16:45
  • $\begingroup$ Ha, I completely missed that! So used to stuff being in solid spheres, I forgot which part was zero! Thanks, fixed, finally. $\endgroup$
    – G. Paily
    Nov 26, 2014 at 16:49
  • $\begingroup$ I am trying to do something similar with KKT conditions. I wanted to see if this would work instead of the potential trick you did. physics.stackexchange.com/questions/328129/… $\endgroup$
    – user92177
    Apr 21, 2017 at 23:38

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