= Covariant transversality of a vacuum plane wave
{title2=$n^\mu F_{\mu\nu}=n^\mu\widetilde F_{\mu\nu}=0$}
For a source-free <plane electromagnetic wave> in <vacuum>, let $n^\mu=(1,\mathbf s)$ be its null propagation <Lorentz four-vector>. In signature $(+---)$, take $F^{0i}=-E_i/c$, $F^{ij}=-\epsilon_{ijk}B_k$, and define the <dual electromagnetic field tensor> by $\widetilde F^{0i}=-B_i$, $\widetilde F^{ij}=\epsilon_{ijk}E_k/c$. Lower both indices with the <Minkowski metric>. Direct contraction gives
$$
n^\mu F_{\mu\nu}=\left(-\frac{\mathbf s\cdot\mathbf E}{c},\frac{\mathbf E}{c}+\mathbf s\times\mathbf B\right),\qquad
n^\mu\widetilde F_{\mu\nu}=\left(-\mathbf s\cdot\mathbf B,\mathbf B-\frac{\mathbf s\times\mathbf E}{c}\right).
$$
Both contractions vanish exactly when the transverse <plane electromagnetic wave> <Maxwell equations> hold. Being zero <covectors>, they remain zero in every <inertial frame>. Thus the transformed spatial <wavevector> is perpendicular to both transformed fields, even though the separate three-dimensional <dot products> are not themselves <Lorentz scalars> for arbitrary fields.
One contraction alone is insufficient without the other <Maxwell equations>. In particular, adding a <magnetic field> parallel to $\mathbf s$ leaves $n^\mu F_{\mu\nu}$ unchanged but violates $\mathbf s\cdot\mathbf B=0$. Conversely the two scalar perpendicularity conditions alone do not impose the required relation $\mathbf B=\mathbf s\times\mathbf E/c$ between field strengths and directions.
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