# Classical scattering off an ellipsoid

## 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.

## 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 $$\frac{d \sigma}{d \Omega}$$.

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## Answers and Replies

malawi_glenn
Science Advisor
Homework Helper
why cant you take the derivatite of $s(\theta )$? It is just the chain rule.

How stupid of me!

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

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

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

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

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Dr Transport
Science Advisor
Gold Member
Brain fart.........