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  • In free space, where there are no charges or current, it follows from $\nabla\cdot\vec{E}=\nabla\cdot\vec{B}=0$ that the electromagnetic waves are transverse i.e. $\vec{k}\cdot\vec{E}=\vec{k}\cdot\vec{B}=0.$ For a reference, see here.

  • Now consider radiation from a source but far away from it. J.D. Jackson's treatment of radiation from a localized oscillating source, reveals that far away from the source (in particular, in the radiation zone), where Eq. 9.19 applies, both $\vec{E}$ and $\vec{B}$ are perpendicular $\hat{n}$ (which, I think, can be taken to be the direction of $\vec k$).

Thus, both in free space as well as far away from a source, electromagnetic fields are transverse. Also, if we think about it, for electromagnetic radiation to be present in free space, it must be generated by a source somewhere, maybe by a source very far away. Doesn't that mean what we refer to as free space is implicitly the radiation zone?

Solidification
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1 Answers1

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In the first case, only plane/spherical wave (the simplest kinds of EM radiation) has $\vec{k}$ perpendicular to fields $\vec{E},\vec{B}$. In general, EM radiation has no single $\vec{k}$, e.g. stationary EM wave inside reflective cavity does not have single $\vec{k}$, and it makes no sense to say such general wave is transverse. At the same time, a general EM radiation may be expressed as sum of many plane waves each with many different $\vec{k}$ though.

In the second case of radiation far from source, it is implicitly assumed that the radiation field is inspected in vacuum (free space). If it is inspected in matter, then E,B need not be perpendicular to k.

Ján Lalinský
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