Classical scattering off an ellipsoid

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Homework Help Overview

The discussion revolves around the scattering of particles off the surface of an ellipsoid defined by the equation x^2 + y^2 + z^2/f^2 = R^2. Participants are tasked with finding the differential cross-section for this scenario.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to express the impact parameter as a function of the scattering angle and is uncertain about the differentiation process. Some participants question the ability to differentiate the function and suggest using the chain rule. Others express confusion regarding the integration process related to the differential cross-section.

Discussion Status

Participants are exploring various approaches to differentiate the impact parameter and calculate the differential cross-section. There is a discussion about a potential discrepancy between the original poster's result and the book's answer, indicating an ongoing examination of the problem without a clear consensus.

Contextual Notes

There is mention of integrating from 0 to pi/2 and multiplying by 4, but the purpose of this integration is unclear to some participants. The original poster also notes that they are working with the differential cross-section rather than a total cross-section.

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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}.
 
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why can't 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?
 
How did you integrate? Try integrating from 0 to pi/2 and multiplying by 4.
 
Dr Transport said:
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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