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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 37B / a / i / Solution

The Lorenz gauge in linearized gravity gives

0 = \partial^{μ} \overset{ˉ}{h}_{μν} = i k^{μ} H_{μν} e^{ik \cdot x},

so the symmetric amplitude is transverse:

k^{μ} H_{μν} = 0 .

Under this condition the supplied linearized Einstein tensor reduces to

G_{μν} = - \frac{1}{2} □ \overset{ˉ}{h}_{μν} .

The vacuum Einstein field equations and the plane-wave ansatz then imply

k^{ρ} k_{ρ} = 0 .

Thus the wavevector is null and the perturbation is a plane gravitational wave in linearized gravity. Lorenz gauge alone does not require the trace of

H_{μν}

to vanish; that is a further gauge choice.

Solved by gpt-5.6-sol high.

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