Solution

A gauge transformation is

A_{μ} ⟼ A_{μ} + \partial_{μ} χ .

Because partial derivatives commute, the added contribution to

F_{μν}

\partial_{μ} \partial_{ν} χ - \partial_{ν} \partial_{μ} χ = 0,

Define

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

Since

\partial_{0} = c^{- 1} \partial_{t}

and

A_{0} = - Φ/ c

F_{0 i} = - \frac{E _{i}}{c}, F_{i 0} = \frac{E _{i}}{c}, F_{ij} = ϵ_{ijk} B_{k} .

The identity

\partial_{λ} F_{μν} + \partial_{μ} F_{ν λ} + \partial_{ν} F_{λ μ} = 0

follows directly from

F = d A

. Its spatial and mixed components are

\nabla \cdot B = 0, \nabla \times E = - \partial_{t} B .

For the other two equations define the four-current

j^{μ} = (c ρ, J) .

Then

\partial_{ν} F^{0 ν} = μ_{0} j^{0}

gives

\nabla \cdot E = μ_{0} c^{2} ρ = \frac{ρ}{ϵ _{0}},

and

\partial_{ν} F^{i ν} = μ_{0} j^{i}

gives

\nabla \times B - \frac{1}{c ^{2}} \partial_{t} E = μ_{0} J .

Using

F^{ρ σ} F_{ρ σ} = 2 (B^{2} - \frac{E ^{2}}{c ^{2}}),

one finds

T^{00} = \frac{1}{2 μ _{0}} (B^{2} + \frac{E ^{2}}{c ^{2}}) = \frac{1}{2} (\frac{B ^{2}}{μ _{0}} + ϵ_{0} E^{2}) .

This is the electromagnetic energy density.

For a null vector, the trace term in

T^{μν} k_{μ} k_{ν}

vanishes. Put

q^{ρ} = k_{μ} F^{μ ρ} .

Antisymmetry gives

q \cdot k = 0

. A vector orthogonal to a null vector has nonnegative Minkowski norm, so

T^{μν} k_{μ} k_{ν} = \frac{1}{μ _{0}} q^{ρ} q_{ρ} \geq 0.

Explicitly, in a frame with

k^{μ} = κ (1, 1, 0, 0)

q^{2} = κ^{2} [(B_{3} - \frac{E _{2}}{c})^{2} + (B_{2} + \frac{E _{3}}{c})^{2}] .

Hence the null energy condition for the electromagnetic field is strict whenever the contraction is nonzero:

T^{μν} k_{μ} k_{ν} > 0 .

Solved by gpt-5.6-sol high.

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