How to calculate total cross section from differential cross section

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SUMMARY

This discussion focuses on the calculation of total cross section from differential cross section, specifically transitioning from \(\frac{d\sigma}{d\phi}\) to the total cross section through integration. The integral is performed from \(\phi=0\) to \(\phi=2\pi\). A key insight shared is the necessity of multiplying the summation by its complex conjugate to find the absolute value squared of the expression, which simplifies the integration process. The orthonormal properties of the cosine functions are also highlighted as crucial for canceling terms during integration.

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svletana
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I was doing the calculations for this: http://fermi.la.asu.edu/PHY531/cylinder/index.html
But I can't figure out how to go from \frac{d\sigma}{d\phi} to the total cross section. My guess was that you did the integral from \phi=0 to \phi=2\pi, but that's not helping since I can't tell either how they got the absolute value squared inside the sum..

Thanks for anyone who listens :)
 
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If you had to find the absolute value squared of the expression, what would you have to do? Answer: Multiply the summation by its complex conjugate. Do it, integrate and see what you get. Don't forget to use different dummy indices for the summations.
 
kuruman said:
If you had to find the absolute value squared of the expression, what would you have to do? Answer: Multiply the summation by its complex conjugate. Do it, integrate and see what you get. Don't forget to use different dummy indices for the summations.
That helped a lot, thanks! I can see now how with that you can make the terms with cos(nx)*cos(mx) with m=/=n cancel when you integrate from them being orthonormal :) The rest is history, thank you very much again!
 

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