Solution for differential equation

PhysicsRock
Greetings,
in one of the exercise sheets we were given by our Prof, we were supposed to draw the trajectory of a patricle that moves toward a bounded spherical potential that satisfies

##
V(\vec{r}) = \begin{cases}
V_0 & | \vec{r} | \leq a \\
0 & else \\
\end{cases}
##

for ##E_\text{particle} < V_0##. I drew a circular path within the potential boundary, but apparently that is wrong. A friend of mine brought to my attention that the particle couldn't enter the potential at all, if it's Energy is lower than the potential energy. I wanted to take a look at this analytically and derived the differential equation

##
m \ddot{\vec{r}} = V_0 \left[ \delta(r - a) - \delta(r + a) \right] \frac{\vec{r}}{r}
##

with ##r = | \vec{r} |##. Intuitively, it makes sense to me as there is no Force on the particle unless it's right at the boundaries where the potential changes, thus satisfying ##\vec{F} = - \nabla V##.

To come to my question, has anyone got any idea how this could be solved, if it is even possible to solve an equation of that form?

Homework Helper
Gold Member
You have written a three-dimensional equation for what is essentially a one-dimensional problem because the potential depends on ##r## only. For that reason, I think it is safe to assume that angular momentum is conserved, write the standard energy conservation equations, one for each region, used in central force problems and see where it takes you. You will have to figure out how to quantify the "kick" that the particle receives at the boundary. I have not solved this problem, but that would be my approach.

Your friend is correct. If the particle has energy less than ##V_0##, it will not penetrate (classically) the potential, i.e. it will bounce off with no energy loss. This sounds like scattering off a hard sphere to me.