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In a charge-free and current-free vacuum, Maxwell equations giveTake the curl of Faraday's law and use the curl of the curl identity $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$. The divergence term vanishes by Gauss's law, while the time derivative of the magnetic curl supplies $-{\mu }_0{ϵ}_0{\partial }_t^2\mathbf{E}$. ThusThis is the transverse electromagnetic wave equation.

For the oscillatory magnetic field, Faraday's law gives $\mathbf{B}=(1/\omega )\mathbf{k}\times \mathbf{E}$. Indeed its time derivative equals $-\nabla \times \mathbf{E}$, and the dispersion relation makes it satisfy the remaining Maxwell equations. ThusFor $\omega =ck$, its amplitude is $E_0/c$. The magnetic field, electric field and direction of propagation are mutually perpendicular. A constant background magnetic field could be added; the expression above is the wave-associated field.

In a source-free vacuum, Maxwell equations areTaking a curl of the last two equations and using $\nabla \times (\nabla \times \mathbf{A})=\nabla (\nabla \cdot \mathbf{A})-{\nabla }^2\mathbf{A}$ gives the electromagnetic wave equationsSubstitution of the complex plane wave amplitudes into Maxwell equations, then taking real parts, givesFor a nonzero propagating wave these imply the dispersion relation $\boxed{{\omega }^2=c^2|\mathbf{k}{|}^2}$. Conversely choose any nonzero transverse complex $\mathbf{e}$ and set $\mathbf{b}=\mathbf{k}\times \mathbf{e}/\omega$ to obtain such a wave. The static uniform-field sector at $\omega =|\mathbf{k}|=0$ is separate from propagation.

At a perfect conductor, the oscillatory boundary conditions are $\mathbf{n}\times \mathbf{E}=0$ and $\mathbf{n}\cdot \mathbf{B}=0$. Surface charge and surface current allow nonzero normal electric and tangential magnetic fields. The magnetic condition here concerns the time-dependent wave field; a perfect conductor can retain a pre-existing static magnetic field.

Let $R=I-2\mathbf{n}{\mathbf{n}}^T$ be the orthogonal reflection across the boundary plane, so $\det R=-1$. The proposed reflected amplitudes are ${\mathbf{k}}^{\prime }=R\mathbf{k}$, ${\mathbf{e}}^{\prime }=-R\mathbf{e}$, and ${\mathbf{b}}^{\prime }=R\mathbf{b}$. For an orthogonal reflection, $(R\mathbf{a})\times (R\mathbf{d})=-R(\mathbf{a}\times \mathbf{d})$. HenceAlso $|{\mathbf{k}}^{\prime }|=|\mathbf{k}|$, so the reflected wave obeys the same dispersion relation. At $\mathbf{n}\cdot \mathbf{x}=0$, the incident and reflected phases coincide because ${\mathbf{k}}^{\prime }-\mathbf{k}$ is parallel to $\mathbf{n}$. Their summed amplitudes satisfyThe first has zero tangential part and the second zero normal part, proving both boundary conditions. Finally ${\mathbf{k}}^{\prime }\cdot \mathbf{n}=-\mathbf{k}\cdot \mathbf{n}>0$, as required for reflection into the vacuum.

Use natural units $\hslash =c=1$ and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ throughout. An Abelian gauge transformation is $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$. For a sufficiently regular $\chi$, commuting the partial derivatives gives $F_{\mu \nu }\mapsto F_{\mu \nu }$, so the Maxwell Lagrangian is invariant pointwise.

Vary the action with respect to the electromagnetic four-potential, using variations of compact support or vanishing on the boundary. The antisymmetry of the electromagnetic field tensor givesAfter integration by parts, $\delta S=\int d^{4}x({\partial }_{\mu }{F}^{\mu \nu })\delta A_{\nu }$. The principle of stationary action therefore gives the source-free Maxwell equations,In the Lorenz gauge, ${\partial }_{\mu }{A}^{\mu }=0$, this becomes $\square A_{\nu }=0$ with $\square ={\partial }_{\mu }{\partial }^{\mu }$. The PDF's name “Lorentz gauge” denotes the usual Lorenz gauge, named after Lorenz. Residual gauge transformations preserve it when $\square \chi =0$; this condition is not itself a complete removal of the gauge freedom.

An electromagnetic field tensor obeys the homogeneous Maxwell equations $dF=0$. Together with its invariance under the Killing vector field $V$, Cartan's magic formula gives $d({\iota }_{V}F)={\mathcal{L}}_{V}F-{\iota }_{V}dF=0$. The local potential of a closed differential one-form construction, the one-form Poincare lemma, therefore supplies a scalar field $\Phi$ on a sufficiently small contractible neighborhood withFor an explicit local proof, write ${\iota }_{V}F={\alpha }_i(x)d{x}^i$ on a coordinate ball centered at zero and take $\Phi (x)={\int }_0^1{x}^i{\alpha }_i(sx)ds$. Since $d\alpha =0$, differentiating under the integral gives ${\partial }_j\Phi ={\int }_0^1[{\alpha }_j(sx)+s{x}^i{\partial }_i{\alpha }_j(sx)]ds={\alpha }_j(x)$. There need not be a global potential when this closed one-form has nonzero periods. The homogeneous Maxwell equation is essential; an arbitrary invariant two-form would not suffice.

Use proper time for the massive particle, let $u$ be its tangent, and assume $m\ne 0$. The Killing equation makes $u^a{u}^b{\nabla }_a{V}_b=0$. The Lorentz force equation then impliesConsequently the local conserved quantity isThe sign follows from contracting $F$ in its first argument, exactly as in the source. Adding a constant to $\Phi$ merely shifts the conserved quantity. This is a charged-particle conserved quantity from a Killing symmetry.

The Maxwell equations with vacuum permittivity and permeability give $\rho ={ϵ}_0\nabla \cdot E$ and $J={\mu }_0^{-1}\nabla \times B-{ϵ}_0{\partial }_{t}E$. The current identity is the Ampère-Maxwell equation, and Faraday's law gives ${\partial }_{t}B=-\nabla \times E$. With $g={ϵ}_{0}E\times B$,The identity $(\nabla \times B)\times B=(B\cdot \nabla )B-\frac{1}{2}\nabla B^2$, and its electric analogue, turn this into the divergence of the Maxwell stress tensor $T$. Thus the tensor in the question is its negative:The vector $g$ is electromagnetic momentum density, and ${\sigma }_{ij}$ is the outward flux of its $i$th component across a surface normal in the $j$ direction. In static conditions the total force on matter is $F_i=-\oint {\sigma }_{ij}{n}_{j}dS$; generally subtract the time derivative of the enclosed electromagnetic field momentum as well.

For the two infinite sheets, superposition and the electromagnetic field jumps give $E=(\sigma /{ϵ}_0)e_x$, $B={\mu }_{0}K{e}_y$ between them and zero outside. Substitution yields exactlyA thin pillbox enclosing the $x=0$ sheet has only its $+x$ face in the interior electromagnetic field. ThereforeDirectly, omit the sheet's self-electromagnetic field: the other sheet supplies $E_{\mathrm{other}}=\sigma e_x/(2{ϵ}_0)$ and $B_{\mathrm{other}}={\mu }_{0}K{e}_y/2$. The Lorentz force per area is $\sigma E_{\mathrm{other}}+K{e}_z\times B_{\mathrm{other}}$, giving the same result. The electric contribution is attractive and the magnetic contribution repulsive.

If $K=\sigma v$ for a single moving charge population, use ${\mu }_0{ϵ}_0=c^{-2}$ to obtain $F/A={\sigma }^2(1-v^2/c^2)e_x/(2{ϵ}_0)$. For physical $|v|<c$ this is attractive, so motion of these charges cannot make the force repulsive. Independently imposed currents need not satisfy that restriction on $K/\sigma$.

For a time-independent current, the Maxwell equations reduce to magnetostatics:Write $\mathbf{B}=\nabla \times \mathbf{A}$ using a vector potential and choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. The vector identity for the curl of a curl gives the Poisson equationThe free-space Green function therefore givesTaking the curl with respect to $\mathbf{r}$, moving it under the integral, and using $\nabla (1/|\mathbf{r}-{\mathbf{r}}^{\prime }|)=-(\mathbf{r}-{\mathbf{r}}^{\prime })/|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3$ gives the Biot-Savart law

Antisymmetry gives $-F^{\rho \mu }{\partial }_{\rho }{A}^{\nu }=F^{\mu \rho }{\partial }_{\rho }{A}^{\nu }$. Combining this with the canonical stress-energy tensor removes the potential in favour of the electromagnetic field tensor:This is the electromagnetic stress-energy tensor in the $(+,-,-,-)$ convention. The product $F^{\mu \rho }{F}^{\nu }{}_{\rho }$ is symmetric in $\mu ,\nu$, since the contracted metric is symmetric. Its expression only in $F$ makes it gauge-invariant. These two properties hold without using the equations of motion.

For conservation, the source-free Maxwell equations are ${\partial }_{\rho }{F}^{\rho \mu }=0$. Hence the given correction equalson shell. This is an antisymmetric superpotential improvement: $B^{\rho \mu \nu }=-F^{\rho \mu }{A}^{\nu }$ is antisymmetric in $\rho ,\mu$, so ${\partial }_{\mu }{\partial }_{\rho }{B}^{\rho \mu \nu }=0$. Conservation of the canonical stress-energy tensor therefore implies ${\partial }_{\mu }{\Theta }^{\mu \nu }=0$. Equivalently, direct differentiation gives ${\partial }_{\mu }{\Theta }^{\mu \nu }=-F^{\nu }{}_{\rho }{\partial }_{\mu }{F}^{\mu \rho }$ after the homogeneous Maxwell equations cancel the field-strength derivative terms.

There are four conserved currents, one for each fixed $\nu$, with charges $P^{\nu }=\int d^{3}x{\Theta }^{0\nu }$. The Belinfante-Rosenfeld stress-energy tensor differs from the canonical tensor only by boundary contributions to these charges when the fields decay suitably.

Finally, in four spacetime dimensions,Thus all requested properties are

For a translation with the sign used in the paper, the field variation at fixed coordinates is $\delta A_{\rho }=\beta {ϵ}^{\nu }{\partial }_{\nu }{A}_{\rho }$, and $\delta \mathcal{L}={\partial }_{\mu }(\beta {ϵ}^{\mu }\mathcal{L})$. Differentiating the Maxwell Lagrangian with respect to the field gradient givesThus Noether's theorem yields $j^{\mu }={ϵ}_{\nu }{T}^{\mu \nu }$, where the canonical stress-energy tensor of the Maxwell field isIt is conserved on the source-free Maxwell equations, but it is not generally symmetric. Indeed,need not vanish. For example, take $A_0=a{x}^1$, $A_2=b{x}^0$, and the other components zero, with nonzero constants $a,b$. This produces constant field strength and satisfies the source-free equations, but $T^{12}=0$ and $T^{21}=ab$.

Nor is it gauge-invariant: while $F$ and $\mathcal{L}$ are unchanged, a gauge transformation giveswhich is generally nonzero. These shortcomings motivate the improvement in part (vi); they do not contradict conservation of the canonical stress-energy tensor.

In vacuum the Maxwell equations areSubstitution of the complex plane electromagnetic wave amplitudes into these equations givesThus the electric field must be transverse to $k$; taking the positive-frequency branch gives $\omega =c|k|$.

For $E_0=E(1,i,0)$, $k=k\hat{z}$, and $\phi =kz-\omega t$, the real fields areThe electric vector rotates with constant magnitude, so this is circular polarization. Its Poynting vector isThe Poynting vector is the electromagnetic energy flux: its direction is the direction of propagation and its magnitude is the power crossing unit area.

Let $\mathbf{\sigma }$ be the stated Maxwell stress tensor. Taking its divergence and usinggivesThe Maxwell equations in vacuum with sources areSubstitution, together with antisymmetry of the cross product, yieldsTherefore the required vector is the electromagnetic momentum densityandThis is local conservation of electromagnetic momentum. The tensor component ${\sigma }_{ij}$ is the flux of $i$-momentum in the $j$ direction in the convention of the question, $\mathbf{g}$ is field momentum per unit volume, and $\rho \mathbf{E}+\mathbf{J}\times \mathbf{B}$ is the Lorentz force density transferred to matter.

In a vacuum with no charge or current, the Maxwell equations areFor the proposed plane electromagnetic wave, $\nabla \cdot B=0$ givesThe Ampère-Maxwell equation givesand substituting this into Faraday's law gives the dispersion relationThus both fields have transverse polarization relative to $k$, and the corresponding real electric field is

For incidence in the positive $x$-direction, choose the incident fieldsThe perfect conductor requires the tangential electric field to vanish at $x=0$. The reflected wave therefore hasThis is normal reflection of an electromagnetic wave from a perfect conductor. The magnetic field in $x\le 0$ isso its tangential value at the surface is

The Maxwell equation $\nabla \times \mathbf{E}=-\partial \mathbf{B}/\partial t$, integrated over a fixed spanning surface $S$, gives by Stokes theoremThus the electromotive force and magnetic flux obey Faraday's law

Differentiate $\mathbf{g}={ϵ}_0\mathbf{E}\times \mathbf{B}$, use the vacuum Maxwell equations, and use $\nabla \cdot \mathbf{E}=\nabla \cdot \mathbf{B}=0$. Componentwise vector identities then givewhereThis is the negative of the conventional Maxwell stress tensor; with the sign used here, its divergence is the outward momentum flux.

The Lorenz gauge is ${\partial }_{\mu }{A}^{\mu }=0$. Substituting the electromagnetic four-potential into the covariant Maxwell equations then removes the mixed derivative and givesEquivalently,

In SI units the Maxwell equations in free space are

The trace of the electromagnetic stress-energy tensor in four spacetime dimensions isFor its covariant divergence, the source-free Maxwell equations eliminate the derivative of the first factor. Contracting the Bianchi identity ${\nabla }_{[a}{F}_{bc]}=0$ with $F^{ac}$ giveswhich cancels the derivative of the trace term. HenceFor $J^a=T^{ab}{V}_b$, stress-energy conservation and symmetry of $T^{ab}$ now imply

Differentiating $\mathbf{g}={ϵ}_0\mathbf{E}\times \mathbf{B}$, substituting Maxwell equations, and using the stated vector identity gives the Maxwell stress tensor conservation lawMoving the divergence to the left identifies