 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-3/37d/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Put

ρ = r^{2} + a^{2}, ϖ = ρ sin θ, z = r cos θ .

Since

x^{2} + y^{2} = ϖ^{2}

, the Euclidean spatial line element is

d x^{2} + d y^{2} + d z^{2} = d ϖ^{2} + ϖ^{2} d ϕ^{2} + d z^{2} .

Now

d ϖ = \frac{r}{ρ} sin θ d r + ρ cos θ d θ, d z = cos θ d r - r sin θ d θ .

The cross terms cancel. With

Σ = r^{2} + a^{2} cos^{2} θ, Δ = r^{2} + a^{2},

d x^{2} + d y^{2} + d z^{2} = \frac{Σ}{Δ} d r^{2} + Σ d θ^{2} + Δ sin^{2} θ d ϕ^{2} .

Adding the time coordinate gives

d s^{2} = - d t^{2} + \frac{r ^{2} + a ^{2} cos ^{2} θ}{r ^{2} + a ^{2}} d r^{2} + (r^{2} + a^{2} cos^{2} θ) d θ^{2} + (r^{2} + a^{2}) sin^{2} θ d ϕ^{2} .

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