Scattering angle as a function of impact parameter

[Silviu]

Homework Statement

This is the problem 3.34 from Goldstein, 3rd Edition:

Consider a truncated repulsive Coulomb potential defined as V=k/r, for r>a and V=k/a, for r<=a. For a total energy E>k/a, obtain expressions for the scattering angle $\Theta$ as a function of $s/s_0$, where $s_0$ is the impact parameter for which the periapsis occurs at the point r=a.

Homework Equations

$d^2u/d\theta^2 + u = -m/l^2*dV/du, with u=1/r$
$\Theta(s) = \pi -2\int_{0}^{u_m}(s\cdot du/(\sqrt{1-V(u)/E-s^2u^2}))$, with E - the energy of the particle and $u_m = 1/r_m, r_m$ the distance at the closest approach

The Attempt at a Solution

I tried to calculate $r(\theta)$ outside and inside the sphere, $s_0$ can be calculated assuming that the potential is k/r all the time, as for $s_0$ the particle doesn't go inside the sphere. But I am not sure how to relate $\Theta$ to s, when I have the potential inside the sphere.

 

[mark726]

Hello,

Thank you for sharing this problem from Goldstein's book. The truncated repulsive Coulomb potential is an interesting concept to explore. To obtain the expression for the scattering angle $\Theta$ as a function of $s/s_0$, we first need to understand the dynamics of the particle in this potential.

As you have correctly mentioned, the potential is given by V=k/r outside the sphere (r>a) and V=k/a inside the sphere (r<=a). This means that the particle experiences a repulsive force when it is outside the sphere, and a constant potential when it is inside the sphere. We can use the equations of motion to describe the trajectory of the particle and obtain the expression for the scattering angle.

To start, let us consider the motion of the particle outside the sphere, where the potential is V=k/r. Using the equations of motion, we can write:

$m\ddot{r} = -\frac{k}{r^2}$

Integrating this equation, we get:

$m\dot{r} = -\frac{k}{r} + C$

where C is a constant of integration. We can use the fact that at the periapsis, the particle reaches the distance r=a and has the velocity $\dot{r}=0$. This gives us the value of C as:

$C = \frac{k}{a}$

Substituting this value of C in the above equation, we get:

$m\dot{r} = -\frac{k}{r} + \frac{k}{a}$

We can further simplify this equation by defining the impact parameter s as:

$s = \frac{m}{l}\dot{r}$

where l is the angular momentum of the particle. This gives us:

$\dot{r} = \frac{l}{m}s$

Substituting this in the above equation, we get:

$\frac{ds}{dt} = -\frac{k}{ml}\frac{1}{r^2} + \frac{k}{m}\frac{1}{ar}$

We can further simplify this equation by defining the variable u=1/r. This gives us:

$\frac{ds}{dt} = -\frac{k}{ml}u^2 + \frac{k}{ma}u$

Integrating this equation, we get:

##s = \frac{k}{ml}(\frac