Why no longitudinal electromagnetic waves?

Question

According to wikipedia and other sources, there are no longitudinal electromagnetic waves in free space. I'm wondering why not.

Consider an oscillating charged particle as a source of EM waves. Say its position is given by $x(t) = sin(t)$. It is clear that at any point on the $x$ axis, the magnetic field is zero. But there is still a time-varing electric field (more or less sinusoidal in intensity, with a "DC offset" from zero), whose variations propagate at the speed of light. This sounds pretty wave-like to me. Why isn't it? Is there perhaps a reason that it can't transmit energy?

A very similar question has already been asked, but it used a "rope" analogy, and I feel that the answers overlooked the point that I'm making.

Answer 1

http://en.wikipedia.org/wiki/Longitudinal_wave#Electromagnetic has a good summary of the situation. There are no longitudinal solutions of the Maxwell equations in a vacuum, but you can get such solutions in a plasma.

Answer 2

I think this is partly a question of vocabulary, and partly a reflection of the fact that the longitudinal Coulomb oscillations you describe fall off so rapidly with distance. (Basically $1/r^2$ instead of $1/r$.) Therefore they are usually called "near field effects" and are totally dominated by the transverse "waves" after a distance of only a very few wavelengths. Nevertheless, they do exist, even in a vacuum, and they do extend to infinity, just very, very weakly.

Answer 3

I don't know if this really qualifies as an answer, but if I read your question rightly I think you might find this quote interesting:

"The original forms of quantum mechanics ... [quantized] ... the electromagnetic field ... by Fourier transformation, as a superposition of plane waves having transverse, longitudinal, and timelike polarizations ... The combination of longitudinal and timelike oscillators was shown to provide the (instantaneous) Coulomb interaction of the particles, while the transverse oscillators were equivalent to photons." [1]

[1] Laurie M. Brown, Feynman's Thesis, pp. xi-xii. World Scientific (2005), paperback edition.

Answer 4

Once you get far enough away from a radiating source, your field will look approximately like a plane wave.

If you look at a plane wave, where $\vec{E}(\vec{x},t)=\vec{E}_0(\vec{k}\cdot\vec{x}-\omega t)$ and $\vec{B}(\vec{x},t)=\vec{B}_0(\vec{k}\cdot\vec{x}-\omega t)$ (for fixed functions of a single variable $\vec{E}_0$, $\vec{B}_0$), you will find that satisfying Maxwell's equations in empty space requires that $\vec{k}\cdot\vec{E}_0=\vec{k}\cdot\vec{B}_0=0$. That is, the electric and magnetic fields must be perpendicular to the direction of propagation.

Why? Because variation along the direction of propagation would lead to a non-zero divergence in $\vec{E}$ or $\vec{B}$, which is strictly forbidden. Unless, of course, you have non-zero charge density, in which case $\vec{E}$ can have a corresponding divergence. This is why longitudinal waves are possible in plasmas.

Answer 5

Longitudinal electromagnetic fields are required to satisfy Maxwells divE = 0 + rho_free. They always exist even in vacuum. Plane wave approximation does not hold very well outside of a few (very limited) conditions.