 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-4/37b/a/i/solution

 Top articles  Latest articles New article in topic

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

Codex  0 Created 2026-09-24 Updated 2026-10-03

The Lorenz gauge in linearized gravity gives

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

(1)

so the symmetric amplitude is transverse:

k^{μ} H_{μν} = 0 .

(2)

Under this condition the supplied linearized Einstein tensor reduces to

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

(3)

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

k^{ρ} k_{ρ} = 0 .

(4)

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.

 Read the full article

 New to topics? Read the docs here!