What is the Angular Dependence of Electron Scattering?

[jeebs]

hi,
here's the problem:

imagine a heavy nucleus and an electron approaching it with some momentum p_i.
it scatters elastically off the nucleus at some angle $$\theta$$ and momentum p_f. the nucleus can be considered to remain at rest, since it is so massive.

I have to start with the definition of vector q = p_i - p_f, get an expression for q^2 (the square of the magnitude of the vector q), and show that the angular dependence of the scattering is given by the Rutherford formula,

$$\frac{d\sigma}{d\Omega}$$ $$\propto 1/(sin^4(\theta/2))$$

my attempted solution:

so far I've worked out that q² = (p_i - p_f)² = p_i² - p_f² - 2p_i.p_f
ie. q² = 2p² - 2p²cos($$\theta$$)
= 2p²(1-cos($$\theta$$))
since the magnitudes of pi and pf are equal.

then I got 1 - cos($$\theta$$) = 2sin²($$\theta/2$$)
by using cos(A+B) = cos(A)cos(B) - sin(A)sin(B) where A = B = $$\theta/2$$

then I looked at Fermi's Golden Rule:

$$\frac{d\sigma}{d\Omega}$$ = ($$2\pi$$/$$\hbar$$)|M_fi|²D_f

where the matrix element |M_fi| = $$\frac{4g^2\pi\hbar}{q^2 + (mc)^2}$$
which is a result that is given to me in a previous part of the problem.

clearly when used in Fermi's Golden Rule, the square of the matrix element gives 3 terms, one of which is the 1/q⁴ term I am looking for, but also 2 other terms. Is there a way for me to get rid of the 2 extra terms or am I barking up the wrong tree here?

thanks in advance.

 

[jeebs]

actually, ignore this question, i figured it out. the mass m = 0. I'm an idiot. can't even find a way to delete the question.