What is meant by a longitudinal electromagnetic wave?

I reckon this is most likely a rather obtuse question, but I was wondering what would be meant by a longitudinal electromagnetic wave? Questions such as this one refer to there being 'no longitudinal electromagnetic waves', without formally defining what this would be.

My issue is that in the case of electromagnetism, we have 4 fields (?) that could be varying: D, E, H and B. I would think that formally, a longitudinal wave would be one for which the direction of the varying field is in the direction of the wave propagation given by the wavevector k. Maxwell's equations only exclude D and B having a component parallel to k (where there is no free charge present), but clearly in non-isotropic media, E will necessarily therefore have a component parallel to k (and a non-isotropic magnetisable medium ill hae a component of H parallel to k).

I have never heard of EM waves in ordinary media being referred to as longitudinal (at least in part). In my lecture notes, the realisation of longitudinal waves was only mentioned briefly in the context of a vaninishing dielectric constant (plasma waves) and when there is a free charge present so that D is not divergenceless...

• I think, usually people consider EM waves in vacuum where D=E und H=B and both are perpendicular to k (that is, transversal) such that there is no component along k (longitudinal). – Photon Dec 28 '18 at 10:02
• @Photon Thanks for your reply! Though clearly my lecture notes consider non-vacuum cases, (and Wikipedia also mentions longitudinal plasma waves) ad yet the ordinary media case isn't covered. – Meep Dec 28 '18 at 10:29
• I see. Well, this case is covered by the answer by Vladimir Kalitvianski below. – Photon Dec 28 '18 at 10:38

A wave is longitudinal or transverse depending on thr direction of the field with respect to the propagation vector $$\mathbf{k}$$. If the field is parallel, then it is longitudinal, and if it is perpendicular then it is transverse.

In the Fourier representation $$\nabla \rightarrow i \mathbf{k}$$. This means Gauss' Law becomes

$$\nabla \cdot \mathbf{E} = 4\pi \rho \rightarrow i \mathbf{k} \cdot \mathbf{E} = 4\pi \rho$$

Since $$\rho$$ is exactly zero in vacuum, this means that $$i \mathbf{k} \cdot \mathbf{E} =0$$ so the electric field is perpendicular to the k-vector and the wave must be transverse. However in a medium, the charge density $$\rho$$ is non-zero and so the electric field can have a component parallel to the k-vector ($$i \mathbf{k} \cdot \mathbf{E} \neq 0$$), giving rise to longitudinal waves.

If you take a charge and start to make it oscillate, next to the charge there will be a longitudinal retarded EMF. Generally, a near field is not transversal, but a superposition of a transversal and a longitudinal waves.

Look at the general solution $$\vec{E}$$ for a radiating charge: it consists of two distinct terms, one being proportional to $$1/R^2$$ (the near field) and another one proportional to $$1/R$$ (radiated).

• Thank you for your reply. Do you mean to say that a wave should be characterised as transverse or longitudinal based on whether the E field vector at that point is parallel or perpendicular to the k vector at that point? – Meep Dec 28 '18 at 13:25
• Vector $\vec{k}$ is a variable in the Fourier transform of the field $\vec{E}(\vec{r})$. Normally there are many "directions" in a Fourier spectrum. So one may define "transversality" of a harmonic $\vec{E}_{\vec{k}}$ with respect to $\vec{k}$ or transversality of the total field $\vec{E}(\vec{r})$ with respect to the vector $\vec{R}$. – Vladimir Kalitvianski Dec 28 '18 at 17:38
• @,21joanna12 yes that's the definition of longitudinal and transverse. – KF Gauss May 19 at 7:52

An example of a longitudinal electromagnetic mode is a surface plasmon. This requires a corresponding longitudinal charge-density wave, which is how you get this to work with Gauss’ Law. Surface plasmons can be excited at any metal-dielectric interface, or in a conducting material like graphene.