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When solving the wave equation for electromagnetic waves, $$ \nabla^{2} \mathbf{E}=\mu_{0} \varepsilon_{0} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}}, \quad \nabla^{2} \mathbf{B}=\mu_{0} \varepsilon_{0} \frac{\partial^{2} \mathbf{B}}{\partial t^{2}} $$

assuming solutions of the form $\mathbf{E}=\mathbf{E}_{0} f\left(\hat{\mathbf{k}} \cdot \mathbf{x}-c t\right)$, it arises that the relation between the electric and magnetic fields is $$ \mathbf{B}=\frac{1}{c} \hat{\mathbf{k}} \times \mathbf{E} \tag{1} $$

However, $\mathbf{E}=\mathbf{E}_{0} f\left(\hat{\mathbf{k}} \cdot \mathbf{x}-c t\right)$ is actually a travelling plane wave, since it has the structure $F(\vec{x}, t)=G(\vec{x} \cdot \vec{n}-c t)$.

Is $(1)$ still a valid relation between electric and magnetic fields for non plane waves?

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    $\begingroup$ As I recall, even non traveling plane waves can be expressed as traveling plane waves using Fourier series if f is square integrable , so it should hold under those conditions at least. $\endgroup$
    – R. Romero
    Jun 9, 2021 at 21:36
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    $\begingroup$ The statement is only true in an isotropic medium. In anisotropic medium, the electric field is not perpendicular to wavevector in general. $\endgroup$
    – Kksen
    Aug 2, 2021 at 14:00

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