= Solution
Let starred quantities denote the <capacitor>'s rest <inertial frame>, $u_0=\epsilon_0E^2/2$, $\beta=v/c$, 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 transformation>
$$
L=\begin{pmatrix}\gamma&\gamma\beta&0&0\\\gamma\beta&\gamma&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix},\qquad
F=LF_*L^T,\qquad T=LT_*L^T.
$$
For a purely electric field, the spatial block of the <electromagnetic stress-energy tensor> is
$$
T_*^{ij}=u_0\delta_{ij}-\epsilon_0E_iE_j.
$$
This 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.
\b[Plates parallel to $xy$.] Choose $\mathbf E_*=E\hat{\mathbf z}$ and $\mathbf B_*=0$. The <electromagnetic field tensor> and <electromagnetic stress-energy tensor> are
$$
F_*^{\mu\nu}=\begin{pmatrix}
0&0&0&E/c\\0&0&0&0\\0&0&0&0\\-E/c&0&0&0
\end{pmatrix},\qquad
T_*^{\mu\nu}=u_0\operatorname{diag}(1,1,1,-1).
$$
The 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> gives
$$
\boxed{T^{\mu\nu}=u_0\begin{pmatrix}
\gamma^2(1+\beta^2)&2\gamma^2\beta&0&0\\
2\gamma^2\beta&\gamma^2(1+\beta^2)&0&0\\
0&0&1&0\\0&0&0&-1
\end{pmatrix}.}
$$
In particular, the laboratory <Poynting vector> is $\mathbf S=2u_0\gamma^2v\hat{\mathbf x}$, 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 is
$$
S_x-uv=u_0v.
$$
The extra term is $vT_*^{xx}$, 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.
\b[Plates parallel to $yz$.] Now $\mathbf E_*=E\hat{\mathbf x}$. The rest <tensors> are
$$
F_*^{\mu\nu}=\begin{pmatrix}
0&E/c&0&0\\-E/c&0&0&0\\0&0&0&0\\0&0&0&0
\end{pmatrix},\qquad
T_*^{\mu\nu}=u_0\operatorname{diag}(1,-1,1,1).
$$
The <electric field> is parallel to the <Lorentz boost>, so $F^{01}=E/c$ remains unchanged and $\mathbf B=0$. The cancellation between $T_*^{00}$ and $T_*^{11}$ gives
$$
\boxed{T^{\mu\nu}=u_0\operatorname{diag}(1,-1,1,1),\qquad
u=u_0,\qquad \mathbf S=0.}
$$
There 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: $S_x=\gamma^2v(u_0+T_*^{xx})=0$. 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 <capacitor fringing field>[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 $U_0=u_0V_0$. 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 <1/b>.
Back to article page
