= Solution
The <electromagnetic four-potential> is $A^\mu=(\Phi/c,\mathbf A)$, and the vacuum <four-wavevector> is $K^\mu=(\omega/c,0,0,k)$ with $k=\omega/c>0$. Define $\psi=K_\mu X^\mu=\omega t-kz$ and $C=(\omega k)^{-1/2}$. The given transverse potential is $A_y=-C\sin\psi$. Its physical fields are
$$
\boxed{E_y=C\omega\cos\psi,\qquad B_x=-Ck\cos\psi,}
$$
so the <Poynting vector> points along positive $z$. For our signature, $\partial^\mu=(c^{-1}\partial_t,-\nabla)$, $F^{0i}=-E_i/c$, and $F^{ij}=-\epsilon_{ijk}B_k$. Therefore
$$
\boxed{F^{\mu\nu}=C\cos\psi
\begin{pmatrix}
0&0&-\omega/c&0\\
0&0&0&0\\
\omega/c&0&0&k\\
0&0&-k&0
\end{pmatrix}.}
$$
The square-root prefactor is simply the source's chosen wave normalization; an overall dimensional amplitude could multiply every field.
The <four-current> is $J^\mu=(c\rho,\mathbf j)$, with $J_\mu=(c\rho,-\mathbf j)$. <Charge conservation> takes the equivalent forms
$$
\boxed{\partial_t\rho+\nabla\cdot\mathbf j=0,\qquad
\partial_\mu J^\mu=\partial^\mu J_\mu=0.}
$$
Lowering both indices of $F$ changes the sign of its time-space entries: $F_{0i}=E_i/c$, while $F_{ij}=-\epsilon_{ijk}B_k$. The temporal component of $\partial^\mu F_{\mu\nu}$ is $\nabla\cdot\mathbf E/c=\mu_0c\rho$, by <Gauss's law>. Its spatial components are $c^{-2}\partial_t\mathbf E-\nabla\times\mathbf B=-\mu_0\mathbf j$, by the <Ampère-Maxwell equation>. Hence
$$
\boxed{\partial^\mu F_{\mu\nu}=\mu_0J_\nu.}
$$
For the displayed free wave the current vanishes because $k=\omega/c$; the component derivation also proves the sourced identity. The two homogeneous <Maxwell equations> follow separately from expressing $F$ in terms of a potential.
\b[Force-sign qualification.] With these definitions the physical <Lorentz force density> is $f_\mu^{\mathrm{phys}}=F_{\mu\nu}J^\nu$, whereas the source prints the reversed contraction $J^\nu F_{\nu\mu}=-f_\mu^{\mathrm{phys}}$. Consequently
$$
\boxed{f_0^{\mathrm{phys}}=\frac{\mathbf j\cdot\mathbf E}{c},\qquad
f_0^{\mathrm{printed}}=-\frac{\mathbf j\cdot\mathbf E}{c}.}
$$
The physical time component is the power transferred to matter per volume divided by $c$, as in <Poynting's theorem>. For example parallel positive $j_x$ and $E_x$ do positive work, but the literal printed contraction has negative time component. This sign cannot be repaired while simultaneously retaining the source's tensor definition and the Maxwell identity above; reverse the force contraction to describe the physical force.
Back to article page
