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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.

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Technically, they would be considered electrostatic waves as $\mathbf{k} \times \mathbf{E} = 0$, i.e., the phase velocity is parallel to the oscillation of the electric field (which are linearly polarized acoustic waves as John Rennie suggested). – honeste_vivere Oct 19 '15 at 21:20

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.

share|cite|improve this answer 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.

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Then, can EM waves be longitudinal in plasma? – Evil Angel Apr 13 '14 at 5:03
Yes, but they're really sound waves in a charged gas not EM waves. – John Rennie Apr 13 '14 at 5:51
I am a layman, so I apologize for a possible dumb question, but these undistorted progressive waves do not count as longitudinal EM waves? Maybe solitons? Thanks in advance. – Gustavo Jul 10 at 19:17

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.

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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.

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Transverse waves are not obligatory propagating. Consider a moving uniformly charge. Its electric field has longitudinal and transverse components, but nothing is a radiation. – Vladimir Kalitvianski Mar 11 '12 at 18:23

Is this not related to the fact that the massless photon cannot have a longitudinal mode? It would have to satisfy,

$$ k_\mu \epsilon^\mu = -\vec k \cdot \vec \epsilon = 0 $$ If it were longitudinal, $\vec k = \vec \epsilon\times|\vec k|$ so that $\vec k \cdot \vec \epsilon=|\vec k|\ne0$.

Notice that if the photon were massive we would be allowed its rest frame in which $\vec k =0$, but it isn't, so we're not.

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If you look at a light wave as a rotating x and y axis which propagates forward in the z direction, the equation which might result takes the appearance of a screw [or helix]. The equation of the wave is not only a function of time, but also in z.

$$y = A e^{i( Bz + wt )}, \quad i=\sqrt{-1}$$

Note an equation of a helix which is:

$$X = A \sin Bz, \quad y = A \cos Bz, \quad z = z$$

It seems that the helix is formed by rotating the polarization of the light wave at an angular velocity. This seems like the description of a " longitudinal " wave. I hope this will help.

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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.

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Because you are looking in the wrong parts of science, one long forgotten and never pursued. You might research Marconi and Tesla, both of which where using longitudinal electromagnetic waves in their transmission devices. Tesla was not concerned with wireless signal transmission, but wireless "power" transmission.

You won't find longitudinal electromagnetic waves outside of the Tesla and Marconi era, which modern science does not bother to investigate any longer.

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Simply wrong. Longitudinal waves can be shown to not work in free propagation but they are used regularly in wave guides. – dmckee Jul 5 '13 at 21:40

protected by Qmechanic Apr 13 '14 at 9:31

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