Does a vertically or horizontally polarized electromagnetic wavefront have a thickness ?

Or is it really thin and just the height or width of it's wavelength ?

enter image description here

  • $\begingroup$ I'm not quite sure what you mean with "thickness"? in an X, Y, Z coordinate system, lets say your wave propagates in Z-direction. So, along which of the three axes do you measure "thickness"? $\endgroup$ Oct 29, 2018 at 10:35
  • $\begingroup$ The obvious assumption is that thickness would be the measurement perpendicular to the linear polarization. This, again, assumes that polarization is purely linear, which it rarely is, and rarely states that way. $\endgroup$
    – user10489
    Oct 29, 2018 at 11:28
  • $\begingroup$ A wavefront is a surface created by joining all the points in space which correspond to (say) the crest of the wave. This surface moves at the speed of light. It's a *concept" not a physical thing, it makes no sense to talk about its thickness. Its just a way of tagging one wave so we can think about how it moves and changes shape. As a counter-example, imagine setting the trigger on a 'scope to 50%, seeing it trigger on a sine wave, and asking how thick the trigger point is. $\endgroup$
    – tomnexus
    Oct 30, 2018 at 4:24

2 Answers 2


That image illustrates the electric field just along one line for "clarity". The line is shown in dark grey, running perpendicular to the antenna, from the lower-left to upper-right of the image. I suppose the author thought trying to show an animated electric field in three dimensional space when mapped onto a two-dimensional image would just be too much information to be clear.

The length of the green arrows should not be interpreted as any kind of "width". Rather, the length of the arrow is proportional to the electric field intensity at the single point at the base of the arrow. This kind of visual analogy is a common way to represent a vector field.

Here's another visualization of an electric field around an oscillating dipole (source: https://gfycat.com/MelodicThornyHairstreak):

enter image description here

Here we can see the oscillating electric field exists all around the dipole in three dimensions.

In this animation it also appears that the field changes everywhere at once, which implies that the distance visible from one side of the image to the other is much less than one wavelength. If we could "zoom out", we would see the field doesn't actually change everywhere at once, but that changes begin at the origin and then propagate out at the speed of light.


In antenna theory, a plane wave is the spherical propagating wave that, after a short distance, appears essentially as a flat plane wave to the observer's antenna. This is due to the rapid expansion of the spherical wave and the relatively small size of a receiving antenna compared to the sphere. This can be likened to the sensation that the earth is flat from an individual's point of observation.

The "thickness" of the plane wave is determined by how long the originating transmitter continues to transmit. As an example, imagine sending a single dit in morse code at 13 WPM. At this speed, a dit lasts approximately 100 milliseconds. Since an RF signal travels at 300,000 kilometers/second in free space, the "thickness" of this plane wave is 30,000 kilometers. If nothing else is sent from the transmitter, the plane wave will have passed the receiving antenna in ~100 milliseconds.

In deference to comments made to this post, a harmonic plane wave is technically a flat surface (i.e. two dimensions) of identical phase from the emitting device. A true harmonic plane wave does not exist as it would require an infinite amount of energy to generate it. So we often take the liberty of analyzing EM energy as a plane wave when in fact it is not. It is in this same context of inexactitude that I consistently referenced thickness in quotes since in truth this would comprise an infinite number of non-existent two dimensional plane waves. That would be quite pedantic but not necessarily helpful.

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    $\begingroup$ To say a dit has a particular thickness is fine in a sense, but to say a wavefront has such a thickness is an entirely different thing. A wavefront is a locus of points with identical phase: to say this locus has a thickness would imply the transmitter's phase remained constant for some amount of time, which isn't how it works. $\endgroup$ Oct 31, 2018 at 1:03
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    $\begingroup$ Interesting interpretation, Glenn. By thickness, I (mistakenly?) thought the questioner was asking about the wavefront's dimensions perpendicular to the direction of propagation, i.e., in proportion to the cross-section of the antenna element(s). $\endgroup$
    – Brian K1LI
    Oct 31, 2018 at 1:21
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    $\begingroup$ Note that Glenn said "...to the observer's antenna". $\endgroup$
    – Mike Waters
    Oct 31, 2018 at 1:41

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