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Source bundle pages 70–73; the cover identifies Trinity Term 2021 despite the attachment filename. Use coordinates $x^{\mu }=(ct,x,y,z)$, metric signature $(-,+,+,+)$, and $A^{\mu }=(\varphi /c,\mathbf{A})$. For the electromagnetic field tensor use $F^{0i}=E_i/c$ and $F^{ij}={ϵ}_{ijk}{B}_k$. Prescribed external fields are treated classically, neglecting radiation reaction in the trajectories; radiation losses are computed separately where requested.

Use the electromagnetic field tensor convention stated above, and lower indices with the Minkowski metric $g=\mathrm{diag}(-1,1,1,1)$. Lowering a single time index changes its sign. Summing both orders of each antisymmetric index pair gives the Lorentz scalarFor $T^{00}$, $F_j{}^0=F^{j0}=-E_j/c$ and $g^{00}=-1$. Direct substitution in the printed electromagnetic stress-energy tensor yieldsFor a spatial index $i$, $g^{0i}=0$ and $F_j{}^i=F^{ji}=-{ϵ}_{ijk}{B}_k$. HenceThus the full top row isHere $u_{\mathrm{EM}}$ is the electromagnetic energy density and $\mathbf{S}$ the Poynting vector, using ${ϵ}_0{\mu }_0{c}^2=1$. By the symmetry $T^{0i}=T^{i0}$, these components describe both energy flux divided by $c$ and $c$ times the electromagnetic momentum density $\mathbf{g}=\mathbf{S}/c^2$. Keeping the factor $c$ is essential because the time coordinate is $ct$.

Let starred quantities denote the capacitor's rest inertial frame, $u_0={ϵ}_0{E}^2/2$, $\beta =v/c$, and $\gamma =(1-{\beta }^2{)}^{-1/2}$. Work in the uniform interplate region of an ideal parallel-plate capacitor, neglecting its fringing field of a capacitor. To transform from that rest inertial frame to a laboratory in which the capacitor moves along $+x$, use the positive-sign Lorentz transformationFor a purely electric field, the spatial block of the electromagnetic stress-energy tensor isThis is the negative of the Maxwell stress tensor: it has $-u_0$ along the electric field and $+u_0$ perpendicular to it. The stress contribution to energy transport in a moving capacitor is essential to the Lorentz transformation of the energy density and flux.

Plates parallel to $xy$. Choose ${\mathbf{E}}_{\ast }=E\hat{\mathbf{z}}$ and ${\mathbf{B}}_{\ast }=0$. The electromagnetic field tensor and electromagnetic stress-energy tensor areThe transformed nonzero independent electromagnetic field tensor components are $F^{03}=\gamma E/c$ and $F^{13}=\gamma \beta E/c$. Thus $\mathbf{E}=\gamma E\hat{\mathbf{z}}$ and $\mathbf{B}=-\gamma vE\hat{\mathbf{y}}/c^2$, consistent with $\mathbf{B}=\mathbf{v}\times \mathbf{E}/c^2$. Transforming the electromagnetic stress-energy tensor givesIn particular, the laboratory Poynting vector is $\mathbf{S}=2u_0{\gamma }^{2}v\hat{\mathbf{x}}$, and the electromagnetic energy density is $u=u_0{\gamma }^2(1+{\beta }^2)$. The ratio $S_x/u=2v/(1+{\beta }^2)$ is subluminal for $v<c$, but differs from the capacitor's velocity. The difference from a simple convective flux isThe extra term is $v{T}_{\ast }^{xx}$, the rest-frame stress multiplied by the velocity. This shows why treating stored field energy as an unstressed parcel moving rigidly with the plates misses part of the energy transport.

Plates parallel to $yz$. Now ${\mathbf{E}}_{\ast }=E\hat{\mathbf{x}}$. The rest tensors areThe electric field is parallel to the Lorentz boost, so $F^{01}=E/c$ remains unchanged and $\mathbf{B}=0$. The cancellation between $T_{\ast }^{00}$ and $T_{\ast }^{11}$ givesThere is no Poynting flux in the ideal interior in this orientation, although the region containing the electric field moves. The longitudinal negative stress contribution cancels the convective term: $S_x={\gamma }^{2}v(u_0+T_{\ast }^{xx})=0$. Field energy is removed at the receding boundary and supplied at the advancing boundary by work on the charged plates; it need not flow through the interior to accompany the moving region. At edges, omitted fringing fields provide additional local structure.

For a finite idealized interplate rest volume $V_0$, length contraction gives $V=V_0/\gamma$. The field-only integrated laboratory energies would consequently be $U_{xy}=\gamma (1+{\beta }^2)U_0$ and $U_{yz}=U_0/\gamma$, with $U_0=u_0{V}_0$. Their different transformation laws are not paradoxical: the field alone is not a closed system. Include the plates and the supports holding them apart, whose mechanical stresses balance the Maxwell stress tensor. The total closed-system four-momentum then has the ordinary Lorentz transformation law. This is the same distinction between a subsystem and a conserved whole that appears in b.

Use natural units and the Minkowski metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, with $A^{\mu }=(\Phi ,\mathbf{A})$ and $A_{\mu }=(\Phi ,-\mathbf{A})$. Then $F_{0i}=E_i$, $F^{i0}=E_i$, and $F^{ij}=-{ϵ}_{ijk}{B}_k$. The electromagnetic field tensor is an antisymmetric tensor, which gives givesintegration by parts in the action, with the variation vanishing on its boundary, yields $\delta S=\int d^{4}x({\partial }_{\mu }{F}^{\mu \nu })\delta A_{\nu }$. Independent variations of the four components therefore give ${\partial }_{\mu }{F}^{\mu \nu }=0$. The time component is $\nabla \cdot \mathbf{E}=0$; the spatial components are $\nabla \times \mathbf{B}-{\partial }_t\mathbf{E}=0$. These are the two source-free Maxwell equations obtained dynamically. The other two, $\nabla \cdot \mathbf{B}=0$ and ${\partial }_t\mathbf{B}+\nabla \times \mathbf{E}=0$, follow identically from $F=dA$, through the Electromagnetic Bianchi identity. They are not additional Euler-Lagrange field equations.^[ref]

Use Minkowski metric $g_{\mu \nu }=\mathrm{diag}(-1,1,1,1)$ and units with the speed of light equal to one. Thus $u^{\mu }{u}_{\mu }=-1$, and $h_{\mu \nu }=g_{\mu \nu }+u_{\mu }{u}_{\nu }$ is the spatial projector of a timelike congruence. Keep the medium velocity fixed when varying the electromagnetic four-potential. Since the electromagnetic field tensor is $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$, its variation givesBoth tensors are antisymmetric. Integration by parts, with variations vanishing at the boundary, therefore givesThe Euler-Lagrange field equation is ${\partial }_{\mu }(F^{\mu \nu }-M^{\mu \nu })=0$. Renaming the free index and using antisymmetry givesThese are the inhomogeneous Maxwell equations with the bound charge and bound current encoded by the polarization-magnetization tensor.

Introduce the electromagnetic electromagnetic field tensor $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$ and the charged vector field strength ${\mathcal{G}}_{\mu \nu }=D_{\mu }{B}_{\nu }-D_{\nu }{B}_{\mu }$, with $D_{\mu }={\partial }_{\mu }+ie{A}_{\mu }$. Minimal coupling givesUnder $B_{\mu }^{\prime }=e^{i\alpha (x)}{B}_{\mu }$ and $A_{\mu }^{\prime }=A_{\mu }-{\partial }_{\mu }\Lambda$,Consequently the required relation isThen ${\mathcal{G}}_{\mu \nu }^{\prime }=e^{i\alpha }{\mathcal{G}}_{\mu \nu }$, its complex conjugate transforms oppositely, the mass bilinear is invariant, and $F_{\mu \nu }^{\prime }=F_{\mu \nu }$. Each term in the total Lagrangian density is therefore invariant under the electromagnetic U(1) gauge symmetry. This phase gauge symmetry does not reinstate the broken gradient-shift gauge symmetry of the massive field itself.