 Solution

ID: past-exam-of-the-mphil-in-nuclear-energy-course-of-the-university-of-cambridge/2021/paper-ne1/3/c/i/solution

This part is a radial differential-equation identity. Expanding the spherical derivative gives

ϕ^{''} + \frac{2}{r} ϕ^{'} + B^{2} ϕ = 0.

Put

ψ = r ϕ

. Since

ψ^{''} = r ϕ^{''} + 2 ϕ^{'}

r

yields

ψ^{''} + B^{2} ψ = 0

. Its real solution is

ψ = A sin (B r) + C cos (B r), ϕ_{cf} = A \frac{sin ( B r )}{r} + C \frac{cos ( B r )}{r} .

For a regular field at the centre, the cosine-over-radius contribution must vanish. The sine-over-radius term is regular because

sin (B r) / r \to B

r \to 0

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