 Solution

ID: past-exam-of-the-particle-physics-course-of-the-university-of-oxford/2022/4/b/i/solution

For azimuthally symmetric Rutherford scattering, integrate over the solid angle,

d Ω = 2 π sin θ d θ

. Substituting

u = sin (θ /2)

gives

sin θ d θ = 4 u d u

, so

σ = \frac{2 π K}{p ^{2}} \int_{θ_{m i n}}^{θ_{m a x}} \frac{sin θ}{sin ^{4} ( θ /2 )} d θ = \frac{4 π K}{p ^{2}} [csc^{2} \frac{θ _{m i n}}{2} - csc^{2} \frac{θ _{m a x}}{2}] .

The small lower cutoff dominates when

θ_{m a x} ≫ θ_{m i n}

and

θ_{m i n} ≪ 1

. Using

sin (θ_{m i n} /2) ≃ θ_{m i n} /2

, the leading total cross section is

σ ≃ \frac{16 π K}{p ^{2} θ _{m i n}^{2}} .

A nonzero cutoff is essential: the unscreened forward cross section otherwise diverges.

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