Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 357 / 2 / c / Solution

Only the transverse metric components are needed. At fixed

(u, r, θ)

,

\partial_{ϕ} x = - y, \partial_{ϕ} y = x .

Contracting this vector with the

x, y

block of the metric makes all angular factors combine to

ρ^{2}

:

g_{ϕϕ} = κ^{2} ρ^{2} + κλ ρ^{2} E = r^{2} sin^{2} θ (1 + \frac{λ}{κ} E) .

Substitution into the preceding news formula gives

\partial_{u} c = - \frac{λ}{2 κ} r \to \infty lim (r \partial_{u} E) .

Therefore the requested constants are

p = - \frac{λ}{2 κ}, q = 1 .

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