 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ii/paper-1/37c/a/solution

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

With

\partial_{μ} = (\frac{1}{c} \frac{\partial}{\partial t}, \nabla),

F^{μν} = \partial^{μ} A^{ν} - \partial^{ν} A^{μ} .

x^{' μ} = Λ^{μ}_{ν} x^{ν}

, the four-potential transforms as a four-vector and therefore

F^{' μν} (x^{'}) = Λ^{μ}_{ρ} Λ^{ν}_{σ} F^{ρ σ} (x) .

Thus $F$ is an antisymmetric rank-two tensor under Lorentz transformations.

 New to topics? Read the docs here!