Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 310 / 3

Table of contents
- a 3
  - Solution a
- b 3
  - Solution b
- c 3
  - Solution c
- d 3
  - Solution d
- e 3
  - Solution e
- f 3
  - Solution f

a

Solution

Local energy-momentum conservation is the vanishing covariant divergence of the energy-momentum tensor:

\nabla_{μ} T^{μ}_{ν} = 0 .

Solved by gpt-5.6-sol high.

b

Solution

For a mixed tensor,

\nabla_{μ} T^{μ}_{ν} = \partial_{μ} T^{μ}_{ν} + Γ^{μ}_{μ λ} T^{λ}_{ν} - Γ^{λ}_{μν} T^{μ}_{λ} .

Write

T = \overset{ˉ}{T} + δ T

and

Γ = \overset{ˉ}{Γ} + δ Γ

. The homogeneous background already satisfies

\overset{ˉ}{\nabla}_{μ} \overset{ˉ}{T}^{μ}_{ν} = 0

, and discarding products of two perturbations leaves

δ (\nabla_{μ} T^{μ}_{ν}) = \partial_{μ} δ T^{μ}_{ν} + \overset{ˉ}{Γ}^{μ}_{μ λ} δ T^{λ}_{ν} - \overset{ˉ}{Γ}^{λ}_{μν} δ T^{μ}_{λ} + δ Γ^{μ}_{μ λ} \overset{ˉ}{T}^{λ}_{ν} - δ Γ^{λ}_{μν} \overset{ˉ}{T}^{μ}_{λ} .

This is simply the first variation of the tensor's covariant derivative.

Solved by gpt-5.6-sol high.

c

Solution

For a spatial index

j

, the background Christoffel symbols give

\overset{ˉ}{Γ}^{μ}_{μ 0} = 3 H, - \overset{ˉ}{Γ}^{i}_{0 j} δ T^{0}_{i} = - Hδ T^{0}_{j}, - \overset{ˉ}{Γ}^{0}_{ij} δ T^{i}_{0} = - a^{2} Hδ T^{j}_{0} .

Together with the

3 Hδ T^{0}_{j}

trace term, the first two contributions combine to

2 Hδ T^{0}_{j}

. Varying the Levi-Civita connection and contracting it with

\overset{ˉ}{T}^{μ}_{ν} = diag (- \overset{ρ}{ˉ}, \overset{ˉ}{P}, \overset{ˉ}{P}, \overset{ˉ}{P})

gives

δ Γ^{μ}_{μ j} = \frac{1}{2} \partial_{j} (h^{μ}_{μ}),

and the surviving combination reduces to

- (\overset{ρ}{ˉ} + \overset{ˉ}{P}) (\frac{1}{2} \partial_{j} h_{00} - H h_{j 0}) .

Consequently the spatial momentum equation is

\partial_{0} δ T^{0}_{j} + \partial_{i} δ T^{i}_{j} + 2 Hδ T^{0}_{j} - a^{2} Hδ T^{j}_{0} - (\overset{ρ}{ˉ} + \overset{ˉ}{P}) (\frac{1}{2} \partial_{j} h_{00} - H h_{j 0}) = 0 .

Solved by gpt-5.6-sol high.

d

Solution

In synchronous gauge in cosmology,

h_{0 μ} = 0

, so the explicit metric terms vanish. Symmetry of the covariant energy-momentum tensor and the background metric imply

δ T^{j}_{0} = - a^{- 2} δ^{ji} δ T^{0}_{i} .

Hence the two expansion terms in part (c) combine to

3 Hδ T^{0}_{j}

. Substituting the scalar velocity potential and scalar anisotropic stress definitions gives

\partial_{0} [(\overset{ρ}{ˉ} + \overset{ˉ}{P}) \partial_{j} δ u] + 3 H (\overset{ρ}{ˉ} + \overset{ˉ}{P}) \partial_{j} δ u + \partial_{j} δ P + \partial_{j} \nabla^{2} π^{S} = 0.

Solved by gpt-5.6-sol high.

e

Solution

Every term in part (d) is a spatial gradient. Extracting the longitudinal scalar coefficient, up to a spatially homogeneous function that can be absorbed into the zero mode, yields the Cosmological Euler equation in synchronous gauge:

δ P + \nabla^{2} π^{S} + \partial_{0} [(\overset{ρ}{ˉ} + \overset{ˉ}{P}) δ u] + 3 H (\overset{ρ}{ˉ} + \overset{ˉ}{P}) δ u = 0 .

Solved by gpt-5.6-sol high.

f

Solution

Yes. Synchronous gauge in cosmology uses freely falling time lines and sets the lapse and shift perturbations to zero, so no gravitational-potential force appears explicitly in this component of momentum conservation. Metric perturbations still affect the other field equations, the evolution of matter variables, and the relation between coordinate-dependent variables and gauge-invariant observables; synchronous gauge also retains residual gauge modes.

In Newtonian gauge in cosmology, the time-time metric perturbation is a Newtonian gravitational potential. Its spatial gradient therefore appears explicitly as a force term in the Euler equation. The difference is a coordinate representation of the same covariant conservation law.

Solved by gpt-5.6-sol high.

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