 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-312/4/ii/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

df / d η = 0

. At first order, directional deflections act only on the isotropic background and hence give no angular contribution. Differentiating

f = \overset{ˉ}{f} - Θ d \overset{ˉ}{f} / d ln ϵ

along the unperturbed trajectory gives

0 = \frac{d f ˉ}{d l n ϵ} [\frac{d l n ϵ}{d η} - (\partial_{η} + e \cdot \nabla) Θ] .

Using the tensor redshift of the comoving energy,

\dot{Θ} + e \cdot \nablaΘ = - \frac{1}{2} \dot{h}_{ij} e^{i} e^{j} .

The method of characteristics follows the backward ray

x (η^{'}) = x - (η - η^{'}) e

. With zero initial perturbation, the line-of-sight solution is

Θ (η, x, e) = - \frac{1}{2} \int_{0}^{η} d η^{'} \dot{h}_{ij} (η^{'}, x - (η - η^{'}) e) e^{i} e^{j} .

To this order the frame and coordinate direction components may be identified inside a quantity already proportional to

h

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