Classical scattering off an ellipsoid

  • Thread starter ehrenfest
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  • #1
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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:
[tex]s(\theta) = \frac{R}{\sqrt{f^2 tan^2(\theta/2)+1}} [/tex]
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:
[tex] \frac{d\sigma}{d\theta}=\frac{ s(\theta)}{\sin \theta} \left|{\frac{ds(\theta)}{d\theta}\right| [/tex]

But, I have no idea how to differentiate this w.r.t theta. Is there another way to get [tex]\frac{d \sigma}{d \Omega}[/tex].
 
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Answers and Replies

  • #2
malawi_glenn
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why cant you take the derivatite of [itex] s(\theta ) [/itex]? It is just the chain rule.
 
  • #3
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How stupid of me!

[tex] \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}} [/tex]
 
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  • #4
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For the differential scattering cross-section, I get:

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

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?
 
  • #5
Dr Transport
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How did you integrate??? Try integrating from 0 to pi/2 and multiplying by 4.
 
  • #6
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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

[tex] \frac{d\sigma}{d\theta}=\frac{ s(\theta)}{\sin \theta} \left|{\frac{ds(\theta)}{d\theta}\right| [/tex]
 
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  • #7
Dr Transport
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Brain fart.........
 

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