= Transversality and tensor conventions
Write a nonzero <vacuum> <plane electromagnetic wave> with <wavevector> $\mathbf k=(\omega/c)\mathbf s$, where $\mathbf s$ is a unit propagation direction. Its propagation <Lorentz four-vector> is $K^\mu=(\omega/c,\mathbf k)$, or, up to a scalar normalization in any one frame, $n^\mu=(1,\mathbf s)$. It is null because $K_\mu K^\mu=0$.
Choose coordinates $x^\mu=(ct,\mathbf x)$ and <Minkowski metric> $\eta_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$. With the supplied three-dimensional boost formulas, a consistent convention assigns the printed array to upper-index components,
$$
F^{0i}=-E_i/c,\qquad F^{ij}=-\epsilon_{ijk}B_k.
$$
The <index conventions for the electromagnetic field tensor> then require $F_{0i}=+E_i/c$ and $F_{ij}=F^{ij}$ when both indices are lowered. The printed lower-index label therefore cannot be kept together with its electric signs and a forward propagation vector in this convention.
This is not merely an arbitrary overall sign. For a valid forward wave with $\mathbf s=\widehat{\mathbf z}$, $\mathbf E=E_0\widehat{\mathbf x}$ and $\mathbf B=(E_0/c)\widehat{\mathbf y}$, multiplying the printed array, unchanged, by $n^\mu=(1,0,0,1)$ gives an $x$ component $-2E_0/c$, rather than zero. One consistent repair is to retain the printed array as $F^{\mu\nu}$ and contract with $n_\mu=(1,-\mathbf s)$. Equivalently, lower both tensor indices before forming the contraction asked for in the question.
Using that corrected lower-index tensor gives
$$
n^\mu F_{\mu\nu}
=\left(-\frac{\mathbf s\cdot\mathbf E}{c},\frac{\mathbf E}{c}+\mathbf s\times\mathbf B\right).
$$
To establish both electric and magnetic transversality, include the <dual electromagnetic field tensor>. Choose $\widetilde F^{0i}=-B_i$ and $\widetilde F^{ij}=\epsilon_{ijk}E_k/c$; lowering both indices then gives
$$
n^\mu\widetilde F_{\mu\nu}
=\left(-\mathbf s\cdot\mathbf B,\mathbf B-\frac{\mathbf s\times\mathbf E}{c}\right).
$$
The source-free plane-wave <Maxwell equations> are precisely the pair of covariant conditions
$$
\boxed{n^\mu F_{\mu\nu}=0,\qquad n^\mu\widetilde F_{\mu\nu}=0.}
$$
Their temporal components give $\mathbf k\cdot\mathbf E=0$ and $\mathbf k\cdot\mathbf B=0$. Their spatial components give the associated amplitude and orientation relations, $\mathbf E/c=-\mathbf s\times\mathbf B$ and $\mathbf B=\mathbf s\times\mathbf E/c$.
To see the <Lorentz covariance> explicitly, let $V_\nu=K^\mu F_{\mu\nu}$. It is a <covector>, so if $V_\nu=0$ in one <inertial frame>, its <Lorentz transformation> is also zero in every other <inertial frame>. The same is true of $K^\mu\widetilde F_{\mu\nu}$. Since the transformed wave still has $K'^0>0$ and $|\mathbf k'|=\omega'/c$, their temporal components imply
$$
\boxed{\mathbf k'\cdot\mathbf E'=\mathbf k'\cdot\mathbf B'=0.}
$$
Here $K^\mu$, rather than a vector renormalized to have temporal component one in every frame, is the actual <Lorentz four-vector>; multiplying a zero contraction by a nonzero normalization does not affect the argument.
Finally, the two scalar perpendicularity statements alone are not equivalent to one tensor contraction. They do not enforce the amplitude relation between $\mathbf E$ and $\mathbf B$. Conversely $n^\mu F_{\mu\nu}=0$ alone cannot exclude a longitudinal <magnetic field>: adding $B_\parallel\mathbf s$ leaves its cross product unchanged. For a genuine source-free <plane electromagnetic wave>, the remaining <Maxwell equations> supply that missing constraint, and both tensor contractions vanish. This qualification and the corrected index placement are necessary for the intended invariance argument.
Back to article page
