 Solution

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

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

Write

m_{ij} = m_{ij}^{(\pm 2)}

and

h = h^{(\pm 2)}

. Since

m_{ij} e^{i} e^{j} = \frac{1}{2} (1 - μ^{2}) e^{\pm 2 i φ}

, the line-of-sight solution is

Θ = - \frac{e ^{i k \cdot x}}{2 2} m_{ij} e^{i} e^{j} \int_{0}^{η} d η^{'} \dot{h} (η^{'}) e^{- ik (η - η^{'}) μ} .

The angular pattern

m_{ij} e^{i} e^{j}

is proportional to

Y_{2, \pm 2}

. Projecting onto this spherical harmonic multiplies its coefficient by

\frac{\int _{- 1}^{1} ( 1 - μ ^{2} ) ^{2} e ^{- i xμ} d μ}{\int _{- 1}^{1} ( 1 - μ ^{2} ) ^{2} d μ} = 15 \frac{j _{2} ( x )}{x ^{2}},

where

x = k (η - η^{'})

and

j_{2}

Θ_{ℓ = 2} = - \frac{15}{2 2} m_{ij} e^{i} e^{j} e^{i k \cdot x} \int_{0}^{η} d η^{'} \dot{h} (η^{'}) \frac{j _{2} ( k ( η - η ^{'} ))}{[ k ( η - η ^{'} ) ] ^{2}} .

\int \frac{d Ω _{e}}{4 π} e^{i} e^{j} e^{a} e^{b} = \frac{1}{15} (δ^{ij} δ^{ab} + δ^{ia} δ^{jb} + δ^{ib} δ^{ja}) .

Because

m

is symmetric and trace-free, contracting this with

m_{ab}

gives

2 m^{ij} /15

. Substitution into the paper's signed neutrino tensor anisotropic stress therefore gives

Π^{ij} = 22 \overset{ρ}{ˉ}_{ν} e^{i k \cdot x} m^{ij} \int_{0}^{η} d η^{'} \dot{h} (η^{'}) K (k (η - η^{'})), K (x) = \frac{j _{2} ( x )}{x ^{2}} .

K (0) = 1/15

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