# Talk:Electromagnetic radiation

Active discussions
WikiProject Physics (Rated C-class, Top-importance)
This article is within the scope of WikiProject Physics, a collaborative effort to improve the coverage of Physics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
C  This article has been rated as C-Class on the project's quality scale.
Top  This article has been rated as Top-importance on the project's importance scale.
WikiProject Electrical engineering (Rated C-class, Top-importance)
This article is within the scope of WikiProject Electrical engineering, a collaborative effort to improve the coverage of Electrical engineering on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
C  This article has been rated as C-Class on the project's quality scale.
Top  This article has been rated as Top-importance on the project's importance scale.
 To-do list for Electromagnetic radiation: edit · history · watch · refresh · Updated 2007-04-05 Implement a structure following Wikipedia:WikiProject Science guidelines: Introduction accessible to the general public and citing related topics What is electromagnetic radiation? Types/sources of electromagnetic radiation Properties of electromagnetic radiation (power, energy, momentum, polarization) Re-read first paragraph under heading Properties History History of understanding safety around radioactivity edit the first sentence for vandalism Priority 2
Wikipedia Version 1.0 Editorial Team / v0.5 / Vital (Rated Start-class)
This article has been reviewed by the Version 1.0 Editorial Team.
Start  This article has been rated as Start-Class on the quality scale.
???  This article has not yet received a rating on the importance scale.
Additional information:
This article is within of subsequent release version of Natural sciences.
This article has been selected for Version 0.5 and subsequent release versions of Wikipedia.
This article is a vital article.

## Suggestions for EMR additions

Let's say you have a geosynchronous Satellite at 2,000 km above the earth, and its emitting some type of directional EMF radiation towards a target on the earth. I'm interested in learning the attenuation with respect to EMF frequency, and also size of how spread out the surface area is that will receive this EMF on the ground (blooming area). I specifically want to know if its possible to get any microwaves, IR, ultaviolet, xray's, gamma rays, through the atmosphere and to know what surface area of the earth will receive this energy, one city block or instance or a pin hole for each emf frequency. — Preceding unsigned comment added by Violationofairspace (talkcontribs) 07:21, 31 December 2013 (UTC)

## Magic quantum effects?

I do not understand the last edits from today. [1] When I said that photons can be created at will, that doesn't mean by power of thought alone-- the phrase in English certainly covers more than that. What do you mean that "photons are not really particles"? When a positron meets an electron two of something are produced (not three or four or one). What do YOU call them? I also said that there is only one way EMR can be created, and that is by acceleration of charged particles (true). This was changed to be "in classical physics." No, that is wrong. In any type of physics, you must accelerate a charge in some way to create EMR, or a photon. There are no exceptions (if you disagree, provide one). Quantum physics may be needed to explain the line spectra of (say) nuclear gamma emission (just as it is needed to explain the line spectra of atomic emission!) but the fact of the emission itself is not a quantum process per se, only the quantization of it is. IOW, quantum physics is necessary to explain the quantization, not the process itself. The emission and absorption of gamma rays is heavily dependent on the electric dipole and quadrupole moment of nuclei as well as the charge contributions of electrons that penetate them [2] and this means that gamma rays are always (no exceptions) produced and absorbed by rearrangment of nuclear charges (which are certainly accelerated in the process). When positrons are annihilated and when neutral pions decay, again charged particles and antiparticles are pulled into each other (that means accelerated), before they disappear into photons. Same process. In QED, real photons (which are necessary for EMR) couple to charges. They do not couple to non-accelerated charges, which instead proceed in straight world lines, with no transfer of momentum. SBHarris 06:51, 19 January 2012 (UTC)

Regarding "at will": sorry if my attempt at humour made my meaning less clear. My objection to "at will" is that as far as I can tell, it doesn't mean anything here. EMR can be produced or destroyed by many different processes, and energy is always conserved. What does it mean to say that it can be produced or destroyed at will?? This seems to imply an arbitrariness to whether EMR is created or destroyed. If you had some specific meaning in mind that the current wording doesn't capture, let me know and perhaps we can find a clearer way to express it.
You seem to be taking a classical point of view, thinking of quantum mechanics as a perturbation on top of classical physics. While that's often a fine analytical method for solving problems, it's not really accurate. The world is quantum mechanical. Classical mechanics is an approximation, valid in certain limits. Nuclear and atomic emission are both fundamentally quantum processes. An electron in an atom does not have well-defined velocity. When the atom absorbs a photon, the electron changes to a state with higher energy, and still does not have well-defined velocity. The change is instantaneous; there is no well-defined change in velocity with respect to time. Describing this as "acceleration" is not, from a quantum point of view, a very useful description. I found your statement "The creation of EMR happens by only one known mechanism..." much, much too strong, and misleading.
I didn't like the wording "Photons are particles, but their properties become more 'particle-like'..." Light is not really a particle or a wave. It is what it is; some aspects of light's behaviour are conveniently described using a particle model, while others are conveniently described using a wave model. Quantum mechanics requires neither model; one can predict the results of measurements without presuming that light is either a particle or a wave. It's better just to say "Photons' properties become more 'particle-like'...". I don't object in general to calling photons particles, anymore than I would object to calling light a wave. It's just that in this specific context saying "photons are particles" added nothing useful to the sentence.--Srleffler (talk) 05:15, 20 January 2012 (UTC)

## Wave model chart is wrong

The Electric and the Magnetic components of the wave should have phase difference of pi/2. Then the wave can carry energy in the direction of propagation, represented by the Poisson's vector, rotating in the plane, perpendicular to the direction of propagation. bspasov@yahoo.com — Preceding unsigned comment added by 98.154.17.2 (talk) 04:32, 13 March 2012 (UTC)

You are mistaken. The electric and magnetic components are in phase. If they were not, the time average of the Poynting vector would be zero, and no net energy could be carried by the wave. This has been discussed before.--Srleffler (talk) 04:40, 13 March 2012 (UTC)

It is YOU who are mistaken! Consult the Maxwell equations! rotE=−B The rotation of the E field is maximal at the zero (or inflexion) point. (And vice versa.) It is horrible to distribute such a poisoning stupidity in the Wiki. (after a due consideration which make it even worse)

On the other hand: I found some 30 pictures related to the topics in the google and EVERY ONE OF THEM WAS WRONG. So you are not alone with your misconception.

(István Dániel, physicist, Hungary) — Preceding unsigned comment added by Dániel I fiz (talkcontribs) 18:26, 15 April 2012 (UTC)

Sigh. The curl (also called the rotation or vorticity) of a moving wave is 90 degrees out of phase with its simple amplitude. See those max E and max B points, at the crests of their waves? There is zero rot or curl there at those points, which is to say that vorticity is zero on the x axis under the crest of an E or B wave. The max rot or curl (vorticity) actually occurs where the amplitude is zero (the node), where it's changing the fastest in space (along the axis if propagation). Vorticity there is maximal. You must be able to visualize this if you can visualize curl or vorticity. Remember that the capping surface for Stoke's law for an EM plane wave can't be orthogonal to the direction of the wave, since all curls in that direction are always zero (which is why a loop antenna can't "see" an EM wave directed exactly "through" it.). Instead, your capping surface, or loop antenna, must be directed so it lines up with all wire in the same direction as the E field, so that the magnetic flux lines of the polarized wave go exactly through the loop area orthogonally, as the wave passes. Then, it's easy to see that when a crest of magnetic field amplitude is directly in the center of such a loop, there's no change in flux momentarily according to Maxwell, and thus no EMF around the loop. This, at a time when the E field at the loop is largest! But E and induced EMF are 90 degrees out of phase in time, in the far field. They only work like Faraday induction (in phase) in the near-field, where current and B are strongest at times when E (due to minimal charge separation in the antenna) is smallest. (And vice versa, as when charge separation in the antenna is largest and E is largest in the near field, current is smallest and B is smallest in th near-field)
If you don't believe it, write out the 3 two-term vector cross products for curl E and B for the components of the E and B fields in 3D and free-space (no currents or charges) and set the components of dEx/dx and dBx/dx in the direction of x propagation, to zero (since of course they are, or can be in the absense of static sources) and then pick a linear polarization so that you're only looking at By and Ez for propagation direction x. Thus, for pure linear polarization set Bz = 0 and Ey = 0. Then the remaining x spacial derivatives of By and Ez per Maxwell are equal to the time derivatives of each other, on the other side of the equation, per Maxwell. That's Maxwell's equations for a plane wave in free space. Set c = 1. Answer is below.
Of course ALSO (easier to visualize), the rate of change of traveling wave with time at any point, is ALSO 90 degrees out of phase with the amplitude. This is easier to see right off, as clearly dB/dt and dE/dt must be zero at the crest of the wave, since for an instant there, the wave is not changing in either time or space. And also dE/dt and dB/dt are maximal at nodes where E and B are zero.
Since both rot and time derivative of fields are out of phase with the amplitude by the same amount, it turns out the simple E and B amplitudes are exactly in phase with each other! If rotE = -dB/dt and rotB = -dE/dt (in units where c = 1), then it turns out that dEz/dx = dBy/dt, and dBy/dx = dEz/dt. You can see that derivatives with regard to x (x = axis of propagation) are zero at the top of sine waves and cosine waves on axis x. If the wave is moving, you can see that the time-derivative of the same waves must momentarily be zero at the crests also. So, basically, Ez = By at any time t, and any place x. And there you are. SBHarris 23:29, 15 April 2012 (UTC)
• Later: Okay, here's the simple version, noting first that Maxwell, like all good relativistically correct equations, treats space and time on equal footing. The formula for an E field of EMR is something like E = A sin(argument), like E = A sin (kx-wt). Okay. Maxwell's exquations for free space say :${\displaystyle \nabla \times \mathbf {E} =-dB/dt}$ . The left side reduces to simple first-order derivatives of E with respect to space, and the right side is a first order derivative with respect to time. Thus, both derivatives of such a wave function result in the same 90 degree sin -> cosine phase change, or the reverse, and the phase stays the same for E and B on the other side. The same happens for the other curl B equation. Voila. A real difficulty here is that the "natural language" statement of the two relevant free-space Maxwell equations often goes something like this: a change in the E field results in a B field. Not quite-- in fact that's sort of wrong and misleading and results in problems like this. More correctly, a spacial change in the E field results in a time change in the B field and vice versa. But both are first derivatives, so whatever B is produced from E, has the same phase as E, and vice versa. SBHarris 01:49, 16 April 2012 (UTC)
• Yes. The rate of change of E in space equals -1 times the rate of change of B in time. This clearly forbids a 90° phase difference, since the rate of change of E at a maximum is zero, and the rate of change of B where it crosses zero is not zero, but rather maximum. --Srleffler (talk) 02:19, 16 April 2012 (UTC)

OK. You are perfectly right. I made 2 errors at the same time. First I forgot about the time derivative. And I also thought something wrong about the shape of the standing waves. Sorry about this. (DI) — Preceding unsigned comment added by 193.224.139.4 (talk) 08:37, 16 April 2012 (UTC)

Without the shift of 90° there is no propagation. -- 2003:45:E814:6340:40F6:C9DA:5F2D:B124 (talk) 18:40, 30 October 2014 (UTC)
No, that's quite wrong, and one reason this article exists is to destroy such misconceptions. For generations students have been told that when light propagates, the changing B field "creates" the E field, and vice versa. As though when energy goes into the E field it goes out of the B field, so it's always stored somewhere. WRONG! It may be intuitive, but it's completely and utterly in error. (By the way, the same thing happens in gravitational waves-- there is a point in the nodes of any wave where space and time is not disturbed at all-- and is utterly flat. One wonders where the energy is? Well, and EMR wave is like that, also).

In any case, the E and B fields in an EMR wave (far-field part) are totally independent of each other. One is not "creating" the other out in space. They have both been created at the same time, at the source, by acceleration of a charge, but after that, they travel next to each other like a pair of cufflinks or socks. One doesn't generate the other.

A lot of the confusion here comes from the fact that the source-free Maxwell equations (the two that are just for fields) tend to be looked on as somehow CAUSAL, as though one term on one side of a field equation is responsible for CAUSING the terms on the other side. And that is because we use the equations in near-field situations where they do describe near-field phenomena like induction, which do look causal because they connected with charges which CAUSE fields.

But the E from Maxwell's equations has 3 components when you look at the fields of an oscillating dipole (as we do with EMR) and two of these are caused by charge (static charge and moving charge) in a way that decrease rapidly with distance. The one associated with moving charge (current) is inductive--- but EMR is not an inductive phenomena. The B component responsible for EMR (EM radiation) is NOT the inductive one from simple current, but the radiative one that is caused by how fast the current changes (accelerated charge). So radiated E is not like the E induced inside a transformer, and is NOT caused by a changing B field through a imaginary loop. And so on. Studying Faraday induction gives you a feeling for near fields (as in a transformer where E and B are indeed out of phase), but gives a completely wrong intuition for distances longer than lambda/2pi where induction E and B because less important. And the components caused by charge acceleration, where E and B are in-phase, and not caused by an induction-like process, and this new process start to dominate at distance. That is far-field or EMR, and is what this article is about. See here SBHarris 23:39, 30 October 2014 (UTC)

Once more: Without the shift of 90° there is no propagation. Neglecting the shift it is impossible to detect one of the two posible directions using a switch and a LC circuit in a amateur receiver for the so called ″Fuchsjagd″ [3]. Look: User_talk:Una_Smith#Amateur_radio_direction_finding -- Wefo (talk) 13:31, 31 October 2014 (UTC)
The problem is verry verry simple: The picture in literature is simply wrong, but the WP relaing on literature is unable to handle such a fact. So the are many people who know ist better, but this is not accepted. Therefore I have been deleted by enemies in the German WP. And I wanted this because I dont want to be responsable for wrong information in the WP. Everything clear? -- Wefo (talk) 17:41, 31 October 2014 (UTC)
It is imposible to dissipate real power in the vacuum. Therefore the apparent power has to be zero, what means, a delay of 90° between E and H is nessary. -- Wefo (talk) 21:26, 31 October 2014 (UTC)
The "Leitungsgleichungen" [4] describe the analog propagation by means of electrical engeneering [5] -- Wefo (talk) 01:59, 1 November 2014 (UTC)

So bottom line: If you use the curl equations you will deduce the incorrect result that the electric and magnetic fields are out of phase. So the curl equations should not be used in deriving the EM wave equations. — Preceding unsigned comment added by 74.110.125.244 (talk) 13:53, 30 September 2020 (UTC)

## Biological effects

The content of the first paragraph of the "Biological effects" section has been disputed. There is a discussion at Wikipedia:Dispute resolution noticeboard#Electromagnetic radiation. Alternate proposed text for the section can be seen in this diff, and this one. Help would be appreciated in locating reliable sources that indicate what the current consensus is in the scientific community.--Srleffler (talk) 04:20, 7 April 2012 (UTC)

## Reference with dead link

I have reverted variants of this edit twice, because of the edit summary "drop unconventional statement backed by dead link". The validity of a reference is not affected by a link becoming dead. The statement is supported by a citation. If you want to remove it, let's discuss it here first.--Srleffler (talk) 06:00, 21 June 2012 (UTC)

The statement that observing light being absorbed in discrete quantities is not by itself evidence of the quantization of light is not really controversial. Demonstrating that the light itself was quantized, not merely just the interaction with matter, won Einstein a Nobel prize. The leap from the observation to the conclusion is not a trivial one.

Perhaps we can rephrase the statement to make it clearer?--Srleffler (talk) 06:12, 21 June 2012 (UTC)

My apologies -- wasn't familiar with the guidelines in this case. My thinking was that the reference needed to be a legitimate one, and in this case a dead link didn't seemed to disqualify it. I realize now that the reference can be valid even if the link is dead. It does seem like the link should be dropped, however. modify 06:30, 21 June 2012 (UTC)
I looked up the guidelines on this to make sure I'm right. From Wikipedia:Link rot: "Do not delete factual information solely because the URL to the source does not work any longer. WP:Verifiability does not require that all information be supported by a working link, nor does it require the source to be published online." Also from that page, the dead link should remain, because having the URL may aid future editors in locating the original source. It's also possible that the link may become active again later—not unlikely in this case since it looks like a site configuration error has taken out most of this research group's online presentations.--Srleffler (talk) 06:56, 21 June 2012 (UTC)
Sounds good. modify 07:18, 21 June 2012 (UTC)
The phenomenon of the quantization of light is not in question for me. It's the thesis that it's matter, per se, that is causing the quantum behavior that seems novel to me, and for that reason, possibly suspect. My knowledge of the subject is not adequate to confirm or challenge this point, however, so I have no issue with the reversion. modify 06:37, 21 June 2012 (UTC)
I see. The statement is not very clear. The fact that one observes that absorption of light is quantized is not by itself evidence that light is quantized; it is only evidence that the absorption process operates in discrete steps. If all one knew was that light is absorbed in discrete quanta, the simplest theory would be that the quantum nature of matter causes this phenomenon. In fact, it is also true that light itself is quantized; the issue is just that observing that absorption is quantized is not by itself sufficient to prove this.--Srleffler (talk) 06:56, 21 June 2012 (UTC)
I like that clarification. From your explanation I gather that the statement is not about the nature of light or matter, it's about what can be concluded from the particular experiment in question. That makes sense. modify 07:18, 21 June 2012 (UTC)
And as a historical note, Einstein's explanation of the photoelectric effect didn't really put the nail in the coffin, either, since it could as well be explained as just another example of quantum absoption by matter (in this case an ionizing atom) while the light itself continued as a field which was available as a bank from which you could take any withdrawal so long as you had the right sized "bag". Even the dim light of the photoelectric experiment contained far too large an intensity to see single "photon" events in real time, so you're free to blame which ever side (or both sides) of the transaction for the discrete nature. Einstein didn't convince Planck. What did convince Planck (and the world) was the Compton effect, which is pretty hard to explain without single photons interacting with single electrons. In the Compton effect, an electron can absorb any energy it likes, and does, but is limited not by its own quantum nature, but by conservation of momentum for ingoing and outgoing photon, as well as itself. So, two particles. And end of controversy. SBHarris 18:13, 21 June 2012 (UTC)

## E and B fields in/out phase

In traveling waves in-phase ("real"), but in standing waves out-of-phase ("quadrature").[6] Also in near field (or in the antenna) the electric and magnetic field components are 90 degrees out of phase for traveling waves[7]. (see also Near and far field or Near-field electromagnetic ranging or experimental [8] or for sound waves [9]). — Preceding unsigned comment added by 195.113.87.138 (talk) 09:51, 7 January 2013 (UTC)

Sure, but this is an article on EM radiation, not the EM field, which has its own article. The facts you mention are in the article on EMF and perhaps need more emphasis. But here, EMR already *means* far field and traveling wave. So in-phase it must be. In order to radiate it must be in phase for B and E. SBHarris 18:36, 7 January 2013 (UTC)

## Dr. Richard Feynman, Nobel Laureate

Please study at least the first three lectures on Quantum Electrodynamics (for non-physicists): Richard Feynman - The Douglas Robb Memorial Lectures on Quantum Electrodynamics

Maxwell's Equation should be interpreted as the Probability Amplitude of finding a single photon at point B that goes from A to B or A-C-B. collinear with C somewhere between A and B. If there is more than one path the photon can travel, then Maxwell's Equation is of no use in determining the Probability Amplitude of finding a non-collinear solution to the probability that you will find a photon at point B. You need to use Quantum Electrodynamics equations which use complex vectors and the frequency of the photon. A photon is a particle not a wave. In all of the experiments where multiple photons might be interpreted as a wave, the initial question needs to be examined. The Probability Amplitude in a double-slit experiment where you cannot determine which slit a photon passed through, will result in a pattern that seems to be caused by wave interference, but is actually the sum of the Probability Amplitudes of all possible paths the photon can travel. When you contrive a way to determine which slit a photon passes through, perhaps by placing a photomultiplier at one slit, you change the question from the probability of a photon leaving the single-photon laser source and arriving at the wall to that probability plus the probability that the photomultiplier at that slit goes off. In that case, you there will not be an interference pattern, but one that shows the same probability amplitude when plotting location x vs. time t with varying x, that is, a flat line.

That which is referred to in the article as a wave is a stream of photons. What exactly happens when both the electric and magnetic fields are zero in the case of in-phase fields? What force exists that can generate an electric and magnetic field from pi to 2pi in the sine wave analogy? It is the collapsing electric field that induces the increasing magnetic field. The electric and magnetic fields have to be 90-degrees out of phase in order for the fields to interact in such a way as to keep the pattern going along the direction of propagation. Using a sine function for one and a cosine for the other makes the equations of force balance. I forgot Bessel functions, Laplace transforms decades ago, but you need to place the fields in such a manner that one field going from 1 to 0 generates the other other field from 0 to 1. There is also the probability that the photon will twist as it propagates, that is, the electric and magnetic fields will rotate along the direction of propagation. That explains polarization, why you can take two polarizing lens and by rotating one, vary the amplitude of the stream of photons from light to dark.

Dr. Feynman explains it much better than I can. There is also a transcript of his QED lectures. The other Maxwell's Equations define electrostatic functions, the force on an imaginary charge placed in arbitrary locations of electrostatic and magnetic fields. A practical use is the precise aiming of a stream of electrons upon the phosphors of a cathode ray tube.

Hpfeil (talk) 15:52, 1 December 2013 (UTC)

This is an article mostly on the field theory of EMR. The QM version is covered in other articles but may need additional treatment here. It is important only when hv is large compared with beam energy-- something that happens only at high frequencies or low beam powers.

There is no mechanism in the far field where one field generates the other-- that is myth taught to hapless generations, but is true only near the source. Instead, far from the source, both fields are in phase and caused by the same processes at the source. Everywhere in the far field E = cB. That means E and B functions are in phase. That means dE/dx from curl = dB/dx = c dB/dt. That's already simple free space plane wave Maxwell's equation for motion along z in time. These are both first derivatives of E and B functions, so they need to be in phase. People talk about Maxwell as though dB/dt generated E, which would suggest out of phase. But that's not what the the equation says. It says dB/dt is proportional to curl E which means dE/dx , a first order space derivative. With two first order derivatives of functions equated that puts their two satisfying functions IN phase. OK? SBHarris 17:59, 1 December 2013 (UTC)

## Using EMR and "light" synonymously

This article sometimes refers to "EMR" and in many cases simply refers to the properties of "light". One of the many things people are attempting to learn when they come to a page like this is "is all EMR considered light?"

If the article is going to use the terms synonymously, it should explicitly call that out. If the terms cannot be considered synonymous, please use EMR when appropriate, and call out usage of "light" as specific examples.

Going to be honest here - came to the article while trying to establish the above, and I can't from this article, or the article on "light". TheRealJoeWiki (talk) 22:54, 8 June 2014 (UTC)

I would have considered only EMR that is in the visible spectrum to be "light". A common definition of light is that it illuminates or makes things more visible to our eyes. Using them interchangably would diverge from that definition when the wider term has already been established. I see the article on light now mentions physicists can use the term more variously. Spur (talk) 02:55, 5 September 2014 (UTC)

They do. Although your average physicist will proudly and without prejudice proclaim that all EMR is just different kinds of light, in the event, physicists tend to call UV and IR "light" but when it gets to X-rays and microwaves, they tend to talk photons and radiation. At the same time, it's really not true that "light" is just the stuff we see by. Just about everybody (not just physicists) considers UV light (blacklight) and infrared light, to be invisible light.

I'll stick in something about this in the lede, and see if somebody boldly reverts me. SBHarris 04:40, 5 September 2014 (UTC)

I understand your POV, but sticking light in the lede sentence as a synonym for EMR:

Electromagnetic radiation (EM radiation, EMR, or light) is a form of energy released by electromagnetic processes. Colloquially light often refers exclusively to visible light, or collectively to visible, infrared and ultraviolet light.

is going to be really confusing for the nontechnical readers. How about:

Electromagnetic radiation (EM radiation or EMR) is a form of energy released by electromagnetic processes. Light is a form of EMR, and all forms of electromagnetic radiation are often referred to colloquially as light in physics.

--ChetvornoTALK 01:49, 17 January 2015 (UTC)

The problem with that is, that it's wrong. Physicists really don't generally refer to radio waves or gamma rays as "light." They all understand them as EMR, and the same as light but different frequency, but if they started talking about them as "light" they would be misunderstood and misunderstand each other. "I'm going to use the radio telescope at so-and-so" "You could have just said the telescope" "Well, they do have a light telescope and I don't want to use that." "But radio waves are a kind of light." "Don't be pedantic."

In a similar way, LIDAR doesn't include RADAR. Even for a physicist. SBHarris 03:00, 17 January 2015 (UTC)

I thought your point in your previous post was that physicists did refer to it that way. Anyway, the current lead is much better. I don't know who stuck light in the lead, but it was pretty confusing for general readers. --ChetvornoTALK 14:43, 17 January 2015 (UTC)
When it's important to distinguish different frequency ranges, as when discussing particular telescopes, it would be confusing to use "light" and "EMR" synonymously. But when talking broadly about all EMR, physicists often do just say light, as in "light can act like particle or a wave", or use it synonymously with "photons". Perhaps it's a difference between theorists and experimentalists? Either way, I think the lead should make some mention of the generic use of light. "Sometimes, especially theoretical physics, the term 'light' is used to refer to any EMR", maybe. Easy Secrets (talk) 21:15, 18 January 2015 (UTC)
Thanks for a great article/graphics; I'm starting to get the picture. -- Charles Edwin Shipp (talk) 15:18, 21 April 2015 (UTC)

## Wifi bridge

Shouldn't we mention wifi bridges here as well ? Especially when using digital TV at home (in which case the wifi is often constantly on all day, using the latest WiFi version (ac), transmitting huge amounts of data, between decoder and wireless router), I think it may have an effect on human health. 15:57, 13 May 2015 (UTC)

I suggest that the passage: " Quanta of EM waves are called photons, which are massless, but they are still affected by gravity " is replaced by; " Quanta of EM waves are called photons, that have no rest mass but have a relativistic mass given by m=E/c2 and thus are affected by gravity. The energy E is given by E= hv, wher h is Plancks constant and v the frequency in radians/second."

Best regards, Roland Eisenträger rolande@techni.no141.0.71.79 (talk) 22:52, 6 January 2016 (UTC)

## Induction ??

I do have an odd problem. My wife has the problem , that is she is at a check out at the store, the cash register will freeze and has to be re-boot The same thing is at an ATM machine and also if she is close to a computer. Could that be an electronic circuit / radiation ?? Maybe you could give me some sugestions or get me in a direction where to look for.

My E-Mail address is Vpetervantol@aol.com — Preceding unsigned comment added by 172.164.35.23 (talk) 22:07, 10 February 2016 (UTC)

## External links modified

Hello fellow Wikipedians,

I have just modified one external link on Electromagnetic radiation. Please take a moment to review my edit. If you have any questions, or need the bot to ignore the links, or the page altogether, please visit this simple FaQ for additional information. I made the following changes:

When you have finished reviewing my changes, please set the checked parameter below to true or failed to let others know (documentation at {{Sourcecheck}}).

As of February 2018, "External links modified" talk page sections are no longer generated or monitored by InternetArchiveBot. No special action is required regarding these talk page notices, other than regular verification using the archive tool instructions below. Editors have permission to delete these "External links modified" talk page sections if they want to de-clutter talk pages, but see the RfC before doing mass systematic removals. This message is updated dynamically through the template {{sourcecheck}} (last update: 15 July 2018).

• If you have discovered URLs which were erroneously considered dead by the bot, you can report them with this tool.
• If you found an error with any archives or the URLs themselves, you can fix them with this tool.

Cheers.—InternetArchiveBot 07:15, 22 December 2016 (UTC)

## model of running waves wrong

You should think of a swinging wave, transversal swinging. Here we see just a "swimming" - like water waves. It is a longitudinal wave You are showing, not transversal. Ok. this way it would be if there are many photons building that collective wave.. But if it have to be for just one, then the wave must be a standing wave with swingung transversally thouse wave forms of sinuses..

## Introduction has way too much information

I think the introduction is way too long, and incomprehensible for general readers. Intros are supposed to be a maximum of 4 paragraphs long (MOS:LEAD), and this one is six. I would suggest getting rid of the 4th paragraph, which is this indigestible infodump of unnecessary and unsourced technical gobbledygook:

"In the domain of classical electrodynamics, EMR satisfies the following self-evident physical laws. EMR carries energy. Therefore, due to the energy conservation law, every EMR must have a source of electric charges that supplied its energy. EMR propagates in the vacuum without losing energy. For this reason, its energy current must abide by the inverse square law. Energy current of an electromagnetic wave is described by the Poynting vector, which is proportional to the vector product of the wave’s electric and magnetic fields. Hence, the inverse square law proves that the strength of the EMR electric field and that of its magnetic field must decrease like 1/R, where R denotes the distance from the source. The energy of electromagnetic radiation and the laws of Special Relativity prove that electromagnetic fields propagate at a finite speed, and in the vacuum, it is the speed of light. Therefore, radiation fields measured at a given space-time point were produced by electric charges of the source at an earlier time (called the retarded time). Mathematically, electromagnetic fields are obtained as spatial and temporal derivatives of 4-potentials called the Lienard-Wiechert 4-potentials. These 4-potentials depend on the retarded position and the retarded velocity of the charges at the source. Referring to the distance from the source, these 4-potentials decrease like 1/R. It follows that a derivative of the 4-potentials with respect to a spatial coordinate, decreases like 1/R2. This result proves that radiation fields are obtained only from a time-derivative of the 4-potentials. Since the 4-potentials depend on the velocity of the electric charges, one finds that due to the time-derivative, an acceleration of the charged particles at the source is a necessary condition for radiation. Furthermore, electromagnetic fields satisfy the wave equation. Therefore, the actual radiation emitted from a source is determined by the interference of the fields that are associated with the accelerating charges at the source. (A common misconception states that charge-acceleration is a sufficient condition for radiation. This is not true. For example, take the electric current that flows along a circular conductor which is connected to a battery. The ring itself is motionless. However, charges of the electric current accelerate towards the ring’s center. Moreover, the system is time-independent, and for this reason it transfers no electromagnetic energy to the environment. Hence, this system contains accelerating charged particles, but it emits no radiation. This is an example where a destructive interference cancels the entire radiation.)"

The introduction is supposed to be a summary of the info in the body of the article, not a separate derivation or proof. The derivation of the Helmholtz wave equation and mention of the inverse square law and Lienard-Wiechert potentials is already in the body of the article, and any of the rest that is worth saving should be moved there. ----ChetvornoTALK 21:13, 6 April 2018 (UTC)

Done [10] --ChetvornoTALK 22:37, 9 May 2018 (UTC)

## Left-hand coordinate system used in lede figure

An IP editor has felt the need to introduce a qualification [11] which I don't believe to be correct. His/Her confusion is apparently related to the use a left-hand coordinate system, but this does not affect the EM wave being illustrated. I suggest that the figure be corrected somehow, though I don't see this as urgent. Thoughts? Attic Salt (talk) 17:16, 11 September 2018 (UTC)

Done. I agree the IP editor's comment in the caption was wrong; although the coordinates were left-handed, the relation of E, B, and propagation direction v in the wave was right-handed as it should be. Corrected the image to show a right-hand coordinate system, and removed the remarks about it in the caption. --ChetvornoTALK 09:11, 12 September 2018 (UTC)
The figure is still in left-hand coordinates for (x,y,z). Contrary to the caption, propagation is shown in the x direction, not the stated z direction. So we still have a problem. IMO, the figure should be adjusted. Attic Salt (talk) 20:54, 11 September 2018 (UTC)
The figure has been adjusted. The labels on the coordinate axes have been exchanged so the horizontal axis, along which the wave is propagating, is now labeled "z". Are you sure your browser is showing you the updated image? Flush your cache. --ChetvornoTALK 21:09, 11 September 2018 (UTC)
Okay, that did it. Thank you. Attic Salt (talk) 21:19, 11 September 2018 (UTC)
I did not perceive any change in the pic and made a new suggestion. BTW, vacuum is homogeneous, isotropic, ..., maybe unbounded is required, but for un-physical plane waves??? ;) Purgy (talk) 07:20, 12 September 2018 (UTC)
I am sorry, my cash was more resistant than I thought of (I thought I had flushed it). Nevertheless, I suggested some tweaks, because I consider "planes in a direction" as confusing. Purgy (talk) 10:24, 12 September 2018 (UTC)

## Recent edit - electromagnetic waves do not travel slower in materials?

User:Attic Salt recently changed a sentence in the introduction from

"...electromagnetic waves.... propagate at the speed of light through a vacuum."

to

"...electromagnetic waves.... propagate at the speed of light through a vacuum or through a material medium."

This is contradicted by just about every physics, optics, and electromagnetics textbook; electromagnetic waves (group velocity) travel slower than the speed of light in materials: [12], [13], [14], [15]. In addition, the Speed of light article does not support his change: "The speed at which light propagates through transparent materials, such as glass or air, is less than c.", contradicting his claim that it does. This change should be reverted. --ChetvornoTALK 16:47, 27 September 2018 (UTC)

Depends on what "speed of light means". It is slower in a medium than in a vacuum, yes. I am concerned that this article has tended to focus on EM in a vacuum. Attic Salt (talk) 17:14, 27 September 2018 (UTC)
There is no ambiguity in physics about what the "speed of light" means; it is a universal constant equal to exactly 299,792,458 metres per second. There might be some confusion among readers about the self-contradictory nature of the fact that light does not travel at the "speed of light" in materials. But that is what the wikilink on the phrase is for; they can go to the article and find out. But your new sentence is simply incorrect. --ChetvornoTALK 17:39, 27 September 2018 (UTC)
There was never(?) a surprise about waves traveling through a medium, but through vacuum(!), that was the story. So there is good reason to focus on vacuum. Furthermore, I think that EM-waves propagate in a medium in a way more complicated manner (susceptibility, magnetizability, phase, attenuation, ...) than in vacuum. The caption of the pic should retain the vacuum property for previously asked about reasons (isotropy, ...). I see no chance to understand wave propagation in media without a thorough understanding of the vacuum case. I also assume that "speed of light" without any further attribute is generally understood as "in vacuum". Describing propagation with only most simple boundary conditions is already very complicated, and the (overstretched) near/far-field approximations (derived above a half-plane for vacuum) lose their importance already in air, with increasing frequency. I don't have anything relevant to say outside the focus on vacuum. Purgy (talk) 18:37, 27 September 2018 (UTC).
I put in "vacuum" into the figure caption myself, some several edits ago, but the more general case of lossless (electrically insulating) material is also relevant for the figure, as the E and M waves are in phase with each other. Note that the article does, at several place, discuss EM propagation through media. I recognise that it is not a simple subject, but some alerting to the reader that EM propagates through vacuum and through media seems like an important issue. 19:17, 27 September 2018 (UTC)
Looking for a proper revision, I took out the statement "in a material medium" from the relevant sentence in the lede, but I tried to make it clear that the propagation speed in a vacuum is pretty special, and to emphasise this, I noted that this speed is denoted "c". Now that we mention this, please note that a few other symbols are already defined in the lede, Planck's constant, frequency, etc. So also defining c seems sensible. Anyway, I hope this helps. Attic Salt (talk) 21:31, 27 September 2018 (UTC)
As stated in my last edit summary, I am unsure whether EM-wave propagation should be called depending on any medium. At the core EM-waves always and everywhere propagate according to the same rules, it's just that they themselves create sources for "new" waves, creating an overall impression of wavy interactions, propagating differently. I think that an interpretation, focused on "media dependency", is a misleading pedagogic vehicle. Purgy (talk) 07:16, 28 September 2018 (UTC)
Purgy Purgatorio, the subject of EM propagation through media is very important, and is not at all a "misleading pedagogic vehicle", as you suggest. An entire volume in the series by Landau and Lifschitz is devoted to the subject: "Electrodynamics of Continuous Media": [16]. I also note that with this edit: [17], Chetvorno removed the reference to propagation in a vacuum from the caption to the figure in the lede, saying in his edit summary that "electric and magnetic fields are always in phase in an electromagnetic wave". This is incorrect. In a lossy medium, the impedance relationship between E and M is complex, and where the "quasi-static" approximation applies their waves are out of phase by pi/4. This is discussed beginning on page 300 of Landau and Lifshitz. It is also discussed in Chapter 11-6 of Panofsky and Phillips. Note that these subjects are of importance to people working with subjects as diverse as EM interaction with metals to magnetotellurics. Interested readers might alo consult the following video on the subject of EM propagation in lossy dielectrics is here: [18], with nice visualisations starting at about 29:30. Anyway, I hope that helps. Presently, the caption to the lede figure, focussing on E and M waves that are in phase is not so much wrong as it is just not descriptive enough. That is not how an EM wave looks in a lossy material. Sincerely, Attic Salt (talk) 13:26, 28 September 2018 (UTC)
I agree with Attic Salt that this "slower than c" propagation (of either phase velocity or group velocity) is not misleading if presented properly. First, make it clear that light needs no medium to propagate, and travels at c in a vacuum. Second, make it clear that where EM waves interact with matter, that interaction slows their propagation; you don't need to explain that by "new" waves, though that's one approach. Dicklyon (talk) 14:31, 28 September 2018 (UTC)

I apologize for my exaggerating wording. I did not intent to deny existence and importance of an elaborate and useful theory of EM-waves propagating through media, and neither the factual difference in the speed of propagation, as manifest, e.g., in the refraction coefficient. My goal was to put emphasis on the native propagation properties of EM-waves (not confined to sinusoidals or linear polarization!), as expressed in the homogeneous Maxwell equations, without any medium and at a fundamental speed, that remain true, even when additional circumstances are in effect -like e.g. media. However, I am not sure whether the (as known to me!) pedagogy about EM-fields in media were not, at least sometimes (?"quasi-static"-regime?), seducing to regionally useful, but in the whole picture wrong tracks. Certainly, a proper presentation is always fine, and, yes, the linearly excited new waves is just my simplistic pet metaphor on this, not suited to do advanced calculation with. Generally, I am not into fighting for or against any content. Purgy (talk) 07:57, 29 September 2018 (UTC)

Purgy Purgatorio, don't worry, I wasn't taking any of this personally. I've also given this some thought and I will look for ways to fix some of the text that are correct but not overly distracting regarding "media" and "speed". I recognise that the article is presently focussed on vacuum propagation, and I'm not trying to rewrite the whole thing. Thank you. Attic Salt (talk) 12:37, 29 September 2018 (UTC)
Okay, I reworked the figure caption. To me (at least) this is clear, but I know that is only my view of my writing. Attic Salt (talk) 13:00, 29 September 2018 (UTC)

## Syntax

This is a common misuse of syntax among scientists (I am a retired Physician and biochemist). Physical phenomena do not "Obey" laws, nor to they behave in order to "Obey " or "Satisfy" laws. To say that physical phenomena can obey or satisfy implies that they have a conciousness, which is nonsense. Physical phenomena can be described by laws, that is all.Historygypsy (talk) 23:48, 10 November 2018 (UTC)

It's an idiom. Saying that physical phenomena "obey" laws is an extremely common idiom accepted in scientific writing and no one is mislead by it to think that they have a "consciousness". Wikipedia is not the vocabulary police. --ChetvornoTALK 23:43, 19 January 2019 (UTC)

## Was anyone wondering if you can smell electromagnetic waves??

I just removed the following paragraph:

Soundwaves are not electromagnetic radiation. At the lower end of the electromagnetic spectrum, about 20 Hz to about 20 kHz, are frequencies that might be considered in the audio range. However, electromagnetic waves cannot be directly perceived by human ears. Sound waves are instead the oscillating compression of molecules. To be heard, electromagnetic radiation must be converted to pressure waves of the fluid in which the ear is located (whether the fluid is air, water or something else).

It seems silly and unnecessary to me. - Aleck, Smart (talk) 13:50, 27 May 2019 (UTC)

Indeed. Excellent edit, this. Soundwaves are not electromagnetic radiation. Mexican waves aren't either. No need for the article to rub it in  . Thanks. - DVdm (talk) 13:42, 27 May 2019 (UTC)

## Radio waves are not electromagnetic

Nobody has proven that radio waves have an electric component. Hertz has not proved that radio waves are electromagnetic. His wire-loop detector detected the waves coming from his apparatus through electromagnetic induction. Hertz had no way to discern if the waves he was receiving in his experiments had an electric component and a magnetic component. The reception of radio waves is through electromagnetic induction in the receiving antenna, and their emission is due to the high-frequency currents in the emiting antenna producing oscillating magnetic fields around it that propagate away from it as magnetic waves only. To be able to say that radio waves are electromagnetic, you have to cite experiments that show clearly the presence of the electric component and that of the magnetic component in the radio wave. Please add citations of such experiments, if you know them. (Ionel DINU)--(talk) 10:08, 15 April 2020 (UTC)

The energy coupled between two metal plates by electrostatic induction and the energy coupled from one loop of wire to another by electromagnetic induction decreases with the sixth power of their separation. If there is no receiving coil within 10 - 100 times the diameter of the coil no energy will leave the coil. For example an operating AC electric motor consumes power from the line, while if you remove the rotor there is nothing to absorb the magnetic field energy of the stator coils and it stops consuming power. In other words the oscillating magnetic induction field around a coil of wire with an alternating current in it does not "radiate" energy to infinity, if there is nothing nearby to couple to it the energy stays in the space near the coil. Whereas the energy density in electromagnetic waves decreases with the square of distance from the source, so if you put a spherical screen around a light bulb or a radio transmitter the screen will intercept the same amount of light power regardless of its diameter. So sources of electromagnetic waves "radiate" away energy regardless of whether there is a "receiver" to receive it, for example your flashlight bulb consumes the same amount of power when it is shining into the empty sky as when it is shining on something, and a radio transmitter consumes the same amount of power regardless of whether it is being received. A dipole antenna like the rabbit ears on your TV receives the electric field component of radio waves, so you prove radio waves have an electric component every time you turn on your TV. A ferrite antenna such as is in your AM radio receives the magnetic component of radio waves, so you prove radio waves have a magnetic component every time you turn on your AM radio. There are of course plenty of statements supporting all these facts in electromagnetics texts, and even high school physics books. We don't have to add citations to support statements that are not in the article. --ChetvornoTALK 04:35, 15 April 2020 (UTC)
You have no experimental proof for your statement that "A dipole antenna like the rabbit ears on your TV receives the electric field component of radio waves". What happens in the TV antenna is that currents are induced in the antenna through electromagnetic induction, which is due to the oscillating magnetic field of the wave. Your mention of the action of a coil proves that you do not know that coils are not used as antennas for transmitting radio waves - they are used only for receiving them because experiments proved that they are directional - loop antennas receive waves coming in the direction of their plane and none at all when the waves fall perpendicular to the loop plane (which is another reason to believe that electromagnetic induction is involved in the reception of radio waves). A one loop antenna or a coil antenna do not emit radio waves -that's why they are called closed oscillating circuits. Only straight wires emit radio waves - and that is because the high frequency current existing in them produce an oscillating magnetic field in the space around them that is pushed away with every reversal of the current. Also you have no experimental proof that electric fields detach from the antenna emitting radio waves. It would just be useful to the average wikipedia reader to know that some of the statements made in the article actually have no experimental proof - they only rest on what textbooks say. (Ionel DINU) — Preceding unsigned comment added by Idnwiki (talkcontribs) 05:38, 15 April 2020 (UTC)--Idnwiki (talk) 10:08, 15 April 2020 (UTC)
Please sign all your talk page messages with four tildes (~~~~) and indent the messages as outlined in wp:THREAD and wp:INDENT — See Help:Using talk pages. Thanks.
This is not the place for discussions about the subject. Here we can only discuss the article — see wp:talkpage guidelines. You might try asking this at the wp:Reference desk/Science. - DVdm (talk) 09:30, 15 April 2020 (UTC)

## Electromagnetic radiation

Okay so every type of electromagnetic radiation is capable of ionizing? ChimeziemMichael (talk) 20:45, 9 June 2020 (UTC)ChimeziemMichael (talk) 20:40, 9 June 2020 (UTC)

No. As discussed in the lead and later on in the article in a few places, only higher energy EMR is ionizing. VQuakr (talk) 21:16, 9 June 2020 (UTC)

## Clarification diagrams needed

Hi, I would like to add to the shown 'plots' of the EM wave, especially the very first image, something like the following: "The length of the arrows represent the strength of the field lines, not their spatial extension. The fields extend infinitely in the planes."

I have noticed that novices, when presented this type of "image" do not realize this and they may interpret the image as some sort of 'beam'[erratum corrected Laura sf (talk) 04:33, 3 July 2020 (UTC)] with the fields beginning and ending at the start and end of the arrows. Without experience in caluculs and vector fields it is not all intuitive that arrow length is an overlaid mathematical 'plot' of the magnitude.

Even better would be to use a double panel image in which the given representation is compared to (mapped onto) the more physical and less mathematical diagram of a 3D slab with the fields again indicated by arrows but with their magnitude plotted as arrow density instead of arrow length. This would make it clear that both are only representations, and also that the fields extend infinitely but with varying magnitude: a true 3D plane wave and not a beam[erratum corrected Laura sf (talk) 04:33, 3 July 2020 (UTC)]. Laura sf (talk) 23:55, 30 June 2020 (UTC)

The 3D diagram sounds helpful, I'm trying to visualize what it would look like. I assume there is no usable one on Commons. Is there an example online of the kind of diagram you mean? --ChetvornoTALK 01:54, 1 July 2020 (UTC)
There is one on Wikimedia that could be acceptable with a well thought-out caption:

the three representations of an EM wave
With this image, the idea would be to clearly state that these are three equivalent representations of the electro (magnetic) wave: the red line, the blue arrows, and the black arrows describe the same thing. (In the first, red line height = field strength; in the second, blue arrow length = field strength; in the third, black arrow density = field strength.) It may be a bit much, but at least once you get this, you are armed for whatever representation of light and fields you're bound to run into. Plus, even though of course none of these are actual "images" of the EM wave, the black one is definitely closer to being an image of the physical reality than the red and blue ones are.
Otherwise, if it'll be used, I'm happy to draw an svg myself with the desired properties. Some similar representations are this from this page, or for a very simple 2D one this one which came from this forum question. Laura sf (talk) 12:00, 2 July 2020 (UTC)
Not done this page is not fully protected, deactivating admin edit request. — xaosflux Talk 13:26, 2 July 2020 (UTC)
@Laura sf and Chetvorno: I am unconvinced this is a useful change for the article. The problem is that there is little to no overlap between the group of readers that do not understand what is shown by the sine wave representation, and the group that will correctly interpret a graphical representation of vector field density. If this is added with commentary, it certainly should not be the first image or in the lead section and should be kept simple - something similar to the one Laura sf linked here for example. VQuakr (talk) 16:47, 2 July 2020 (UTC)
@VQuakr: I agree that very simple 2D one I in the link would be ok for the first image and definitely better than the current one
But I disagree with the assumption on reader groups. Wat is the argument for that? Using an image as a first picture relies on the readers’ intuition. The one that’s there now is not intuitive nor simple. I think a problem lies too in having the magnetic field there as well and plotting it in ‘3D view’: a sine perpendicular on the vertical sine. (I know these representations are commonplace, and they definitely should be included where the maths start) This gives the illusion of representing the wave in 3D which is not at al the case. Laura sf (talk) 22:15, 2 July 2020 (UTC)
@VQuakr: I want to add to the question of what representation the reader will understand, because I think that is my main issue with the "arrows of sinusoidally varying length" representation. (Blue on figure above, and also the current image on the WP page.) What a highschooler would have learnt prior to learning about the contents of this page (EM radiation) is the concept of electric field lines. The problem is that the arrows here represent an electric field (as stated correctly in the caption) and they look exactly like electric field lines, but they are absolutely not electric field lines. Actually, the blue arrows mean nothing. The sine function (red line) conveys all the information. I don't understand how or why this practice of "coloring in" sinusoids by drawing vertical arrows inside emerged. It adds nothing and does not conform to any standardized mathematical or physical representation method (i.e. functions, field lines or vector fields). I've also never seen this practice outside of the context of EM waves. Please correct me if I'm wrong though, I'm curious to know.
It's wouldn't be so bad if it was just an aesthetic frill but in this case, in my experience, it confuses learners on a fundamental topic that already is rather abstract. Laura sf (talk) 01:53, 3 July 2020 (UTC)
@VQuakr: I also disagree with the assumption about readers, however I think the existing drawing should remain the lead drawing and the proposed plane wave drawing should come after it. The existing drawing at the top of the article is of a single ray of radiation along the z-axis; it gives no information about how the electric and magnetic fields are arranged in 3 dimensional space in a beam of radiation like a flashlight beam or the radio waves from a television station. I suspect this is a question most physics students have; I can vaguely remember wondering about it myself when I was learning EM. Once readers have seen the simpler ray drawing, the plane wave drawing will answer questions they have about extended wavefronts. --ChetvornoTALK 02:18, 3 July 2020 (UTC)
@Laura sf: Thanks for all the info. The red and blue arrows are the vectors that depict the magnitude and direction of the magnetic and electric field at the points along the z-axis (at the base of the arrows). The arrows shown just represent sample vectors to give viewers the concept, actually there are two vectors representing the two fields at each point of the z-axis. The sinusoidal lines just show the length of the vectors at each point along the axis. Both arrows are necessary because at any point an electromagnetic wave is composed of two fields, the electric and the magnetic field, which are perpendicular. I got interested and drew a first attempt at a drawing of a plane wave. My drawing shows how the sinusoidal variation of the fields along the direction of motion relates to the plane wave. I think it's better than the one you displayed above, which is misleading in a number of geometric ways. But I'm not trying to preempt you, this was just an experiment. If you want to draw one for the article I'll support yours. --ChetvornoTALK 02:18, 3 July 2020 (UTC)
Your svg drawing is awesome. An important note on your reply:
"The existing drawing at the top of the article is of a single ray of radiation; it gives no information about how the electric and magnetic fields are arranged in 3 dimensional space in a beam of radiation like a flashlight beam or the radio waves from a television station." -> My point is, that the existing drawing you refer to is precisely a 3 dimensional plane wave. I want to make sure we are on the same page here because this is exactly where the confusion lies thanks to images like the one we're discussing.
- A "single ray" = a three-dimensional plane wave with fields extending infinitely in the plane (here x-y) perpendicular to the propagation direction. The ray thus very decidedly is defined in 3 dimensional space. So:The existing drawing gives all information about how the electric and magnetic fields are arranged in 3 dimensional space. They are arranged like you drew in your .svg. The existing drawing and your svg give exactly the same information, but your svg IMO gives it in a clearer way
- You are right that this article does not, and should not, describe "a beam of radiation like a flashlight beam or the radio waves from a television station":
This is a beam of EM radiation. (A beam is a plane wave convoluted with a Gaussian envelope, so that a 'cilinder', by lack of a better word, of propagating radiation results.)
Note that 1 or 2 dimensional EM waves simply do not exist. Not physically, and not mathematically. The equation E=E_o sin(kx)describes the three dimensional EM plane wave. The Maxwell equations that describe the radiation in this article can only be derived in 3D and have only solutions in 3D. The ray optics framework that describes reflection, refraction etc, can only be derived in 3D and has only solutions in 3D. The laws of Snell, Fresnel etc are for a 3D plane wave (= ray). Any other physical manifestation of radiation, i.e. one where the fields do not extend infinitely in the plane perpendicular to the propagation direction, is going to be a more complex modification the 3D plane wave (= ray) discussed on this page, for example the Gaussian beam or a stream of photons. Laura sf (talk) 03:33, 3 July 2020 (UTC)
@Laura sf: Sorry, I didn't mean to be condescending. I misunderstood what you were saying. We're on the same page, I agree absolutely with all you said. Your point is that none of that is conveyed by the lead diagram. From that perspective, not only is the lead diagram misleading, but all the other diagrams in the article also show the fields along a single ray line. The article definitely needs a diagram of a plane wave. However I still think the existing diagram should be the lead, to introduce general readers to the idea of vectors and that an electromagnetic wave consists of two perpendicular coupled vector fields. The more complicated plane wave diagram could be right under it, with a caption explaining that this is what (part of) a beam of (linearly polarized, monochromatic, plane wave) electromagnetic radiation looks like mathematically. I think the plane wave diagram is too much information to hit them with as the first drawing in the article. --ChetvornoTALK 05:13, 3 July 2020 (UTC)
Thanks both of you to taking the time to reply and come up with examples. I know 2D EM waves do not exist, but 2D representations of them do. My main concern here is with overwhelming the lead section; Chetvorno the graphic you made here is great for later in the article. The animation below I think would need some work - it is hard to interpret intensity from those color saturations and there is an alternating black section that should be white. It is easier to visually get intensity from greyscale, but we would need to figure out a way to communicate the polarity. Particularly if you are only showing the electric field component, the isometric projection (vs an orthogonal section) view does not add any information, either. VQuakr (talk) 15:03, 3 July 2020 (UTC)
No worries did not take it as such. I think the diagrams are fine, as long as the reader knows what they're looking at. My original opinion was that the since the aim is to display the solution of Maxwell's equations of electromagnetism, the aim is to display a sinusoidal plane wave electric field and note, to be complete, that it is superposed with a sinusoidal plane wave magnetic field rotated 90 degrees wrt to the first. But from your answer I am reconsidering: the contribution of the magnetic field and how they're tied together deserves equal attention. I'll wait for a bit to see if anyone else wants to read this very long thread and weigh in, but I think I can give it a go with the current image with updated caption + your svg. I also found the following which may be good for the math section:

Evolution of the electric field component of the electromagnetic wave over time. Reddest: maximally positive amplitude (field points upward). White: zero-field. Bluest: maximally negative amplitude (field points downward).
Do advise on wording if you can. Laura sf (talk) 05:53, 3 July 2020 (UTC)
I kind of agree with VQuakr's remarks above. This article is very broad and has to cover a lot of material besides plane waves, and the reader can click links to Plane wave which is where the topic should be explained fully (which also needs better diagrams, by the way). My feeling is that the description of plane waves should be limited to a few sentences in the text, and mostly explained in the captions under the drawings we're adding. I agree the magnetic field should have equal attention. I wouldn't describe a plane electromagnetic wave as a "sinusoidal plane wave electric field superimposed with a sinusoidal plane wave magnetic field rotated 90 degrees"; this will give readers the impression that the two components are independent. One idea for the wording is in the Description section under my plane wave drawing on Commons; something simple like "In a plane wave the electric and magnetic field have a constant direction and magnitude over each plane perpendicular to the direction of motion" If readers want more they can go to Plane wave. --ChetvornoTALK 19:47, 5 July 2020 (UTC)
The proposed multicolored animation of moving planes is kind of abstract and doesn't show the vectors; it's going to require a lot of explanation. As an alternative, we could animate a drawing showing the field vectors like this one you proposed before. --ChetvornoTALK 19:42, 5 July 2020 (UTC)
Return to "Electromagnetic radiation" page.