 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/iii/paper-53/4/i/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

Vary the quadratic Einstein-Hilbert action within the transverse trace-free tensor sector. Integration by parts gives

δ S^{(2)} = - M_{Pl}^{2} \int d η d^{3} x [(a^{2} E_{ij}^{T'})^{'} - a^{2} \nabla^{2} E_{ij}^{T}] δ E^{T ij},

with endpoint terms set to zero. Hence the tensor equation is

E_{ij}^{T''} + 2 H E_{ij}^{T'} - \nabla^{2} E_{ij}^{T} = 0 .

k^{2} ≪ H^{2}

, so its leading equation is

(a^{2} E_{ij}^{T'})^{'} = 0

. Its solutions are

E_{ij}^{T} = C_{ij} + D_{ij} \int^{η} \frac{d η ^{'}}{a ( η ^{'} ) ^{2}} = C_{ij} + D_{ij} \int^{t} \frac{d t ^{'}}{a ( t ^{'} ) ^{3}} .

There is thus a constant tensor mode. During inflation the second mode decays; in exact de Sitter expansion it is proportional to

a^{- 3}

after shifting the constant of integration. Constancy is the leading super-Hubble behaviour, exactly realized for the zero-gradient solution.

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