Classical scattering off an ellipsoid

[ehrenfest]

Homework Statement

Particles of scattered off the surface of an ellipsoid given by x^+y^2+z^2/f^2 = R^2, where f and R are constants. Find the differential cross-section.

Homework Equations

The Attempt at a Solution

Let s be the impact parameter. I can find s as a function of the scattering angle, theta, but I am not sure where to go from there. I get:
$$s(\theta) = \frac{R}{\sqrt{f^2 tan^2(\theta/2)+1}}$$
When I did this for a sphere instead of an ellipse, that expression for s was much simpler and could be easily differentiated. Then I just used the formula:
$$\frac{d\sigma}{d\theta}=\frac{ s(\theta)}{\sin \theta} \left|{\frac{ds(\theta)}{d\theta}\right|$$

But, I have no idea how to differentiate this w.r.t theta. Is there another way to get $$\frac{d \sigma}{d \Omega}$$.

 

[malawi_glenn]

why can't you take the derivatite of $s(\theta )$? It is just the chain rule.

 

[ehrenfest]

How stupid of me!

$$\left| \frac{ds}{d\theta} \right| = \frac{R f^2\tan(\theta/2)\sec^2 (\theta/2)}{2 (f^2 \tan^2(\theta/2)+1)^{3/2}}$$

 

[ehrenfest]

For the differential scattering cross-section, I get:

$$\frac{d\sigma}{d\Omega} = f^2 R^2/4 \frac{1}{(f^2+(1-f^2)\cos(\theta/2))^2}$$

My book says that I am missing a factor of 4 and that I need to add one to the denominator before squaring. Who is right: me or my book?

 

[Dr Transport]

How did you integrate? Try integrating from 0 to pi/2 and multiplying by 4.

 

[ehrenfest]

How did you integrate? Try integrating from 0 to pi/2 and multiplying by 4.

I didn't integrate. This is the differential cross-section.

What would I integrate from 0 to pi/2? can you elaborate?

EDIT: I just used the formula

$$\frac{d\sigma}{d\theta}=\frac{ s(\theta)}{\sin \theta} \left|{\frac{ds(\theta)}{d\theta}\right|$$

 

[Dr Transport]

Brain fart...