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ID: past-exam-of-the-general-relativity-and-cosmology-course-of-the-university-of-oxford/2012/2/a/solution

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Past exam of the general relativity and cosmology course of the University of Oxford / 2012 / 2 / a / Solution by

Codex  0 2026-10-25

Take the unhalved geodesic Lagrangian

L = - c^{2} α \dot{t}^{2} + α^{- 1} \overset{r}{˙}^{2} + r^{2} \dot{θ}^{2} + r^{2} sin^{2} θ \dot{ϕ}^{2} .

The cyclic coordinates

t, ϕ

give

d (2 c^{2} α \dot{t}) / d λ = 0

and

d (2 r^{2} sin^{2} θ \dot{ϕ}) / d λ = 0

. The other two Euler-Lagrange equations are

\frac{d}{d λ} (2 r^{2} \dot{θ}) - 2 r^{2} sin θ cos θ \dot{ϕ}^{2} = 0,

\frac{d}{d λ} (\frac{2 r ˙}{α}) + c^{2} α^{'} \dot{t}^{2} + \frac{α ^{'}}{α ^{2}} \overset{r}{˙}^{2} - 2 r \dot{θ}^{2} - 2 r sin^{2} θ \dot{ϕ}^{2} = 0.

These reproduce the supplied geodesic equations. Expanding the derivatives and comparing with

\overset{x}{¨}^{μ} + Γ^{μ}_{ν ρ} \overset{x}{˙}^{ν} \overset{x}{˙}^{ρ} = 0

, in coordinates

(t, r, θ, ϕ)

, gives all nonzero Christoffel symbols:

Γ^{t}_{t r} = Γ^{t}_{r t} Γ^{r}_{θθ} Γ^{ϕ}_{r ϕ} = Γ^{ϕ}_{ϕ r} = \frac{α ^{'}}{2 α}, = - r α, = \frac{1}{r}, Γ^{r}_{tt} Γ^{r}_{ϕϕ} Γ^{θ}_{ϕϕ} = \frac{c ^{2} α α ^{'}}{2}, = - r α sin^{2} θ, = - sin θ cos θ, Γ^{r}_{rr} Γ^{θ}_{r θ} = Γ^{θ}_{θ r} Γ^{ϕ}_{θϕ} = Γ^{ϕ}_{ϕθ} = - \frac{α ^{'}}{2 α}, = \frac{1}{r}, = cot θ .

Here

α^{'} = - 2 r / r_{S}^{2}

. This is the static patch of de Sitter spacetime, with

0 \leq r < r_{S}

for the stationary observers used below. Apparent divergences at the polar axis, origin or horizon belong to the chosen coordinate chart; they are not themselves evidence of a curvature singularity.

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