Solving Homework: Sigma w/ 2PI from Radial Potential?

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SUMMARY

The discussion centers on the origin of the factor of 2π in the function for sigma when solving a homework problem related to scattering from a radial potential. The contributor initially derived sigma without the 2π by changing the variable to t = cos(theta) and integrating between -1 and 1 using m = n in the Legendre polynomial. It is confirmed that the 2π arises from integrating over the phi dependence, necessitating its inclusion in front of the integral sign in the final equation.

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david_clint
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Homework Statement


Hi. For the question given (attachment) where does the factor of 2PI in the function for sigma come from in the first step because I don't get it in my working for the first step.

Homework Equations


see attachment

The Attempt at a Solution


I got sigma (without the 2PI) by changing variable to t =cos theta, thus integrating between minus and plus 1and using m=n in the legendre polynomial,

so I am wondering if the 2PI came from integrating over the phi dependence as we are dealing with scattering from a radial potential?

I also need prodding in the right direction for the next step!
 

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Yes, the 2pi comes from integrating over phi, and so there should be a 2pi in front of the integral sign in the last equation.
 
Avodyne said:
Yes, the 2pi comes from integrating over phi, and so there should be a 2pi in front of the integral sign in the last equation.

Thanks:)
 

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