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Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 37D / d / Solution

Using

\partial_{μ} F^{μν} = □ A^{ν} - \partial^{ν} (\partial_{μ} A^{μ})

and writing

α = \frac{1}{c ^{2}} \frac{\partial ϕ}{\partial t} + \nabla \cdot A,

the temporal and spatial components give, with the stated metric conventions,

□ ϕ + \frac{\partial α}{\partial t} - m^{2} ϕ = - c μ_{0} J^{0},

□ A - \nabla α - m^{2} A = - μ_{0} J .

For

m > 0

, part (c) sets

α = 0

. For

m = 0

,

α = 0

is the Lorenz gauge condition.

Solved by gpt-5.6-sol high.

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