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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-varying 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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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.

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  • $\begingroup$ What decreases at $1/r^2$ and $1/r$? The field or the potential? The question asks about fields but your answer seems to imply potentials or plane waves. If the static electric field decreases as $1/r^2$ then how could a time varying point source field not decrease similarly? $\endgroup$ Commented Jul 8, 2022 at 11:15
  • $\begingroup$ I do not see clearly how this answer answers the question. $\endgroup$
    – Davius
    Commented Jul 10, 2022 at 13:39
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    $\begingroup$ @Davius this answer says, in other words, that longitudinal waves do exist in the sense electric field component in direction of propagation may oscillate, and a pattern of this component along the line of propagation is travelling with speed of light. But this pattern propagation is usually not considered proper wave in theoretical physics, because it does not obey the wave equation; its intensity falls off too fast with distance from the source. These "waves" are traditionally understood as time variation of conservative field connected to the source, not as travelling waves. $\endgroup$ Commented Aug 22, 2022 at 21:48
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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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  • $\begingroup$ You cannot use plane wave approximation to analyze this situation as presented in the question. That applies to a plane charge distribution. $\endgroup$ Commented Jul 8, 2022 at 11:20
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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.

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    $\begingroup$ Then, can EM waves be longitudinal in plasma? $\endgroup$ Commented Apr 13, 2014 at 5:03
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    $\begingroup$ Yes, but they're really sound waves in a charged gas not EM waves. $\endgroup$ Commented Apr 13, 2014 at 5:51
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    $\begingroup$ 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? arxiv.org/pdf/hep-th/9606171v4.pdf Thanks in advance. $\endgroup$
    – Gustavo
    Commented Jul 10, 2016 at 19:17
  • $\begingroup$ The summary just refers to Griffiths' book en.wikipedia.org/wiki/Introduction_to_Electrodynamics . Can you mention or quote the argumentation? $\endgroup$ Commented Jul 8, 2022 at 12:33
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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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One of Maxwell's equation is Gauss' law: $$ \nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0} $$ which integrated over a sphere gives Coulomb's law: $$ \vec E(r) = \frac{Q}{4\pi \varepsilon_0} \frac{\hat r}{r^2} $$ By varying the distance to the charge, as the question suggests: $$ r = r_0+ k \sin(\omega t) r_0 = r_0 (1+k \sin(\omega t)) $$ the field oscillates as: $$ \vec E(r) = \frac{Q}{4\pi \varepsilon_0} \frac{\hat r}{(r_0(1+ k \sin(\omega t) ))^2} $$

So indeed this field oscillates with the oscillating remote charge. Does this field radiate? It is maybe not according to the most common definitions of a radiation field, which use plane wave approximation, but it might transport energy if a charge at the field point at $r$ oscillates with a similar frequency. The phase difference determines whether the field point receives or transmits energy to the source point charge. If the phase difference is $0$ or $\pi$ there is no radiation. There is a maximum of energy exchange at phase difference $\pi/2$ and $3\pi/2$.

There are real applications of this, often with spherical antennas. The spherical antenna prevents transverse electric fields and magnetic fields. Nicola Tesla used such antennas. Prof. Dr-Ing. Konstantin Meyl uses them too. I have personally witnessed Meyl operating a boat being powered by $20$ or $40$ MHz longitudinal electric fields (sometimes erroneously mentioned as scalar fields). Nicola Tesla compared the waves with sound, which they do resemble. Sound is a longitudinal vector field in a scalar medium.

How these waves are retarded and what their speeds are, would be better to treat in separate questions but in general pressure speeds are higher than shear speeds. The Coulomb gauge is specified as having no retardation in the electric potential.

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  • $\begingroup$ Konstantin claims also that Coulomb, Lorenz, Heavidise mistakenly gauged Maxwell's equation. in his extended field theory he adds the potential Vortex, in replacement of Maxwell's Vector Potential A (which was mistakenly set to zero but yet does not describe formation of structures). Turns out, apparently, that we started diverging from the truth since Maxwell :) even more so since the gauges applied after his death. $\endgroup$
    – juanmf
    Commented Aug 30, 2022 at 19:36
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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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  • $\begingroup$ Transverse waves are not obligatory propagating. Consider a moving uniformly charge. Its electric field has longitudinal and transverse components, but nothing is a radiation. $\endgroup$ Commented Mar 11, 2012 at 18:23
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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 $z$.

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

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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EM waves are generated by the oscillation of an electron-a point source. Neat the electron is the so called near field and all components of the wave have non-zero value. Away from the point source we have a propagating spherical waves forming the far-field and it is symmetric about the origin. If we take a small section of the advancing wave surface we get a plane wave. Due to symmetry, all force components cancel leaving only the z- component(x-components in some answers)- the one along the increasing radius. This leave only the z components of the electric and magnetic fields varying with time and distance as cos(kz-wt), where k,w are wave number and frequency. The wave amplitude varies as 1/r and not 1/r^2 as in the case of the static fields. As given above, this results in the static field becoming insignificant compared to the EM field forces as the distance from the source is larger and it is neglected. Energy can be carried only by EM field or a wave and can't be carried by a constant static field. That is because Energy=force x distance. That means only a force that causes the distance to change can carry energy. Example you can push hard on a standing car for hours without feeling that tired.. but the moment the car moves and you wanted to keep the force/pressure you find you get tired in no time.. as in the second case you are spending energy due to motion. Regarding longitudinal EM waves in fact they exist. They are given by Poynting vector S=E^H. This is a longitudinal wave but S/c^2 is a momentum wave carrying the energy of the wave S/c per unit volume.

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Longitudinal electromagnetic fields are required to satisfy Maxwell's $\nabla \cdot E = \rho_\text{bound} + \rho_\text{free}$. 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.

https://en.wikipedia.org/wiki/Nikola_Tesla

http://www.capturedlightning.com/frames/Tesla0.html

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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    $\begingroup$ Simply wrong. Longitudinal waves can be shown to not work in free propagation but they are used regularly in wave guides. $\endgroup$ Commented Jul 5, 2013 at 21:40
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Light can have polarization along the k-vector. See circular polarized light.

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    $\begingroup$ circularly polarised light is transverse... $\endgroup$ Commented Oct 22, 2017 at 18:07
  • $\begingroup$ @AccidentalFourierTransform Circularly polarized light has no net polarization over spacetime. Elliptically polarized light has. ole has to be more descriptive though. $\endgroup$ Commented Jul 10, 2022 at 13:22

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