 Solution

ID: past-exam-of-the-electromagnetism-course-of-the-university-of-oxford/2023/3/a/solution

ρ

J

c = (ϵ_{0} μ_{0})^{- 1/2}

\nabla \cdot E \nabla \times E = ρ / ϵ_{0}, = - \partial_{t} B, \nabla \cdot B \nabla \times B = 0, = μ_{0} J + c^{- 2} \partial_{t} E .

E = - \nabla ϕ - \partial_{t} A, B = \nabla \times A .

\nabla \cdot A + c^{- 2} \partial_{t} ϕ = 0.

Substituting into Gauss's law and using the Lorenz gauge condition gives

\frac{ρ}{ϵ _{0}} = - \nabla^{2} ϕ - \partial_{t} (\nabla \cdot A) = - \nabla^{2} ϕ + c^{- 2} \partial_{t}^{2} ϕ .

For the Ampère-Maxwell equation, the double-curl identity gives

\nabla (\nabla \cdot A) - \nabla^{2} A = μ_{0} J - c^{- 2} \nabla \partial_{t} ϕ - c^{- 2} \partial_{t}^{2} A .

□ = \nabla^{2} - c^{- 2} \partial_{t}^{2}

, the remaining equations are

□ ϕ = - ρ / ϵ_{0}, □ A = - μ_{0} J .

Conversely, if the electric potential and magnetic vector potential satisfy these wave equations and the Lorenz gauge condition, the same substitutions in reverse give Gauss's law and the Ampère-Maxwell equation. The other two Maxwell equations follow identically from the definitions, proving equivalence within the Lorenz gauge.

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