- #1

- 325

- 3

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 [tex]\theta[/tex] 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,

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

[/tex]

my attempted solution:

so far i've worked out that

**q**

^{2}= (

**p**

_{i}-

**p**

_{f})

^{2}=

**p**

_{i}

^{2}-

**p**

_{f}

^{2}- 2

**p**

_{i}.

**p**

_{f}

ie. q

^{2}= 2p

^{2}- 2p

^{2}cos([tex]\theta[/tex])

= 2p

^{2}(1-cos([tex]\theta[/tex]))

since the magnitudes of pi and pf are equal.

then I got 1 - cos([tex]\theta[/tex]) = 2sin

^{2}([tex]\theta/2[/tex])

by using cos(A+B) = cos(A)cos(B) - sin(A)sin(B) where A = B = [tex]\theta/2[/tex]

then I looked at Fermi's Golden Rule:

[tex]\frac{d\sigma}{d\Omega}[/tex] = ([tex]2\pi[/tex]/[tex]\hbar[/tex])|

**M**

_{fi}|

^{2}D

_{f}

where the matrix element |

**M**

_{fi}| = [tex]\frac{4g^2\pi\hbar}{q^2 + (mc)^2}[/tex]

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/q

^{4}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.