Talk:Electromagnetic radiation/Archive 1

Archive 1

Comments

Answer to Robert van Hoff: Dear Sir I think your answeres are in our references below :Dr Bo Thelin,btn602003@yahoo.se

A new spectral line intensity formula for optical emission spectroscopy was presented by Dr:s Bo Thelin (experimental physicist) and Sten Yngström (theoretical physicist) at the beginning of the 1980:s. This formula has shown very good agreement between experiments (Ref 1) and the new theory (Ref 2). These references are summaries of earlier papers.

I = K \ left (frac { v^2 }{ c^2 }\ right ) \ e^left (\ frac { -J}{ k \ T \ right )} \ \ left (\ e^left ( \ frac { h \ v }{ k \ T \right - \ 1)^-1

K includes transition rates and element concentrations I= spectral line intensity, v= frequency , J=ionization energy and T=temperature

Many independent experimental methods strongly support the new formula, while the standard intensity formula with the Boltzmann term (upper energy level), deviates very much from experiments. Ref 1 Yngström,S. and Thelin,B. Applied Spectroscopy, 44, 1566, (1990) Ref 2 Yngström,S. International Journal of Theoretical Physics,Vol.33, No 7,(1994) —Preceding unsigned comment added by 79.138.180.147 (talk) 13:50, 11 March 2008 (UTC)

Ratio of ${\frac {E_{0}}{H_{0}}}$?

The article says at the end$E_{0}=c_{0}B_{0}$ while the book "Fundamentals of Photonics" says ${\frac {E_{0}}{B_{0}}}=c*\mu ={\sqrt {\frac {\mu }{\epsilon }}}$ So is there something wrong or did I miss a point?

Continuum of Frequencies in the Spectrum?

I cannot find a reference on whether frequencies of electromagnetic radition constitute a perfectly smooth continuum with all possible frequencies represented, or are frequencies 'clustered' with some frequencies in the spectrum completely absent?--Robert van der Hoff 03:18, 3 May 2007 (UTC)

Frequencies are continuous, in general, for example in a black-body spectrum. For simple systems like isolated atoms, the spectrum is discrete. Dicklyon 03:44, 3 May 2007 (UTC)

Thank you! While the spectrum is continuous, it is not infite. What factors determine its upper and lower limit? I can't find a reference for simple minds like mine explaining why EM waves are self-propagating, and why at the speed of light and not any other speed, or why they just don't stand still? --Robert van der Hoff 03:49, 3 May 2007 (UTC)

See black body for the equation that determines the spectrum of black-body light source. It is nonzero at all energies (or frequencies, or wavelengths), but at high enough frequencies (short enough wavelenghts) is an exponentially decreasing function of energy, so gets pretty close to zero, at energy proportional to absolute temperature. The propagation follows from Maxwell's equation; the speed of propagation is determined by a couple of the constants (the electric and magnetic constants of free space, I forget what they're called). That's how the speed of light and the properties known to electrical theorists turned out to be related through Maxwell's equations. It's not so easy to explain in layman's terms, but I'm sure you can find good descriptions some place, or maybe someone else can try to describe it better here. Dicklyon 01:29, 4 May 2007 (UTC)

Your explanation is much appreciated--Robert van der Hoff 09:29, 4 May 2007 (UTC)

Merge Light and Electromagnetic Radiation??

Merging the topic electromagnetic radiation with the topic light would make no sense since many forms of electromagnetic radiation are not IR, UV or visible light. Most people would consider light to be a seperate though related topic which should be listed and accessed seperately.

Electromagnetic radiation within a specific frequency range happens to be visible to the human eye, and is called light. —Preceding unsigned comment added by 84.86.91.251 (talk) 13:20, 26 April 2008 (UTC)

Electromagnetic radiation

The term electromagnetic radiation is also used as a synonym for electromagnetic waves in general, even if they are not radiating or travelling in free space. This sense includes, for example, light travelling through an optical fiber, and electrical energy travelling within a coaxial cable.

I removed the above paragraph as I find it misleading. Electromagnetic waves are always "radiating", to radiate is to move in "rays" and that is how electromagnetic waves propagate. Also the sentence about electrical energy is just plain wrong, it is not correct to describe the movement of electrons in a cable of any type as an electromagnetic wave. --Jpowell 23:25, 18 June 2006 (UTC)

Gamma rays

How it is possible that some gamma rays have longer wave length than some x-rays? They would be called x-rays then, wouldn't they? Or is there some definition of gamma rays which does not refer to wavelength? --AxelBoldt

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My understanding is that at least originally, gamma-ray was the name given to the photons generated from nuclear decay. X-rays on the other hand were generated by electronic transitions involving highly energetic inner electrons. Therefore the distinction between gamma-ray and x-ray is related to the radiation source rather than the radiation wavelength. Generally, nuclear transitions are much more energetic than electronic transitions, so most gamm-rays are more energetic than x-rays. However, there are a few low-energy nuclear transitions (eg. the 14.4 keV nuclear transition of Fe-57) that produce gamma-rays that are less energetic than some of the higher energy x-rays.

--Matt Stoker

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The 'conflict' with some gamma rays having longer wave length than some X-rays arises becauses we use the terms (gamma ray and X-ray) for both i) certain parts of the electromagnetic radiation spectrum and ii) electromagnetic radiation from certain processes. --Css

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Maybe that should be clarified in the article? —Preceding unsigned comment added by 128.205.204.56 (talk) 15:58, 2 April 2009 (UTC)

Cause of electromagnetic spectrums

The recently added section titled "What causes electromagnetic spectrums" would probably be more appropriate on a page about Spectroscopy. In my opinion the electromagnetic radiation page should be constrained more to a discussion of the properties of the radiation itself, perhaps with references to spectroscopy and other uses of the radiation. Also, the new section needs some work, since emission and absorption of quanta are not only associated with electronic transitions, but are also associated with rotational, vibrational, and nuclear transitions. Also the section titled "Temperature" has some problems, since the continuous spectrum is not due to doppler broadening of atomic emissions, but is more likely due to vibrational emissions. --Matt Stoker

It have moved my section to spectroscopy - I agree this seems more relevant. I was also dubious of the doppler effect being the cause of continuous spectrum but it was the only cause I could find. Thanks for advice. -- sodium

Similar to Electromagnetic spectrum

This article is very similiar in topic to Electromagnetic spectrum -- The Anome

should these be mereged? JDR

I think there's a place for both, as this one could address radiation in general - not just the frequency ranges of radiation. It would need a lot of work though. --Laura Scudder 22:13, 31 Mar 2005 (UTC)

This article is referenced about 360 times, so there is a clear demand for a catchall article on electromagnetic radiation, but I find the current version unsatisfactory. With this in mind I created a To do list that's mostly about article structure to get it going. Laura Scudder 00:12, 1 Apr 2005 (UTC)

the spector of lifht is tinier the dust the pin head it travels at interveni due to 1/20th nano dot perinch second causing raditon dust the like to form jesus.........ariel micheal hanial —Preceding unsigned comment added by 67.101.39.125 (talk) 08:17, 22 March 2010 (UTC)

X-rays vs gamma rays

By definition the difference between X-rays and gamma rays is that X-rays are produced by electrons releasing energy in the form of photons when they change energy levels while gamma rays are released by the nucleus as part of the process of radioactive decay. X-rays can have ridiculously high eV, even overlapping the wavelengths of more normal gamma rays but are still characterized as X-rays if they originate from electrons. -- Alex.tan 07:11, 16 Sep 2003 (UTC)

Moving charge

Thought I'd address the accelerating versus moving charge. Not all moving charge creates radiation. For instance, an infinite wire carrying constant current is an arrangement with moving charges. However, there is only a constant magnetic field - no electric field - so no power is radiated. Therefore, moving charges do not always create radiation. Accelerating charges do always create radiation. I think the author was explicitly thinking of Bremsstrahlung and synchrotron radiation. -- Laura Scudder 22:13, 31 Mar 2005 (UTC) what do u call a spiritual charge? supernova? what is super and nova about that? —Preceding unsigned comment added by 67.101.39.125 (talk) 08:21, 22 March 2010 (UTC)

wavelets

The "wavelet" article link referred in 1.1 points talks about the algorithm, not the physics aception. I don't know what should be done about this since it's the only article on wavelets. I guess a disambiguation page on wavelets could be added, and then a stub about "wavelets" in the sense needed here. I'm not really sure about how to do this since I haven't edited the Wikipedia in a while, so i'm just posting it here.

Yes, this is unfortunate. I suggest that the present wavelet article should be renamed wavelet analysis, and wavelet should be used for the principle in physics. --Heron 16:24, 3 September 2005 (UTC)

"Wavelets" must be removed here. A quick google search on "wavelets" will confirm for anyone what that word means nowadays, and what it will likely continue to mean for a long time to come.

Electromagnet Shield

I live near high voltage electrical cable. How can I make something simple - not to expensive - to shield my monitor from electromagnet field? -Hace--

Turn your room into a Faraday Cage. (CHF 05:47, 28 October 2005 (UTC))

A sheet of ferrous metal (like the metal used for repairing car bodies) attached to a wall or floor between the monitor and the source of em interference may reduce the effects on the monitor. Due regard must be given to not blocking ventilation of the monitor, not shorting out electrical conductors, avoiding sharp edges etc etc. Images on a monitors may do a "hula dance" in the presence of a 60 Hz field, as when a transformer or AC power cable is nearby. A nearby DC magnetic field, as from subway power cables,or a bank of storage batteries may cause the image to shift laterally.Edison 21:56, 6 June 2006 (UTC)

Klystrons and ionising radiation

Just a stupid question. I didnt realise that klystrons gave off ionising radiation. depends on the voltage on the anode I suppose. Does any one have any details?. ie what sort of radiation is given off: X rays or what?--Light current 18:56, 11 September 2005 (UTC)

Natural and man made

Natural and man made should be distinguished soon. The term radiation, is commonly used for this type of energy, although it has a broader meaning, Also called electromagnetic energy or simply radiation. Scott 14:05, 16 October 2005 (UTC)

Derivation

I added a derivation from Maxwell's equations. I feel this is one of the most important derivations in all of physics. (CHF 06:38, 28 October 2005 (UTC))

Excellent work! This gives the article more authority. --Heron 16:04, 28 October 2005 (UTC)

Broken code (Internet Explorer 6 and Firefox 1.5.0.2); repetitive "Failed to parse (Can't write to or create math output directory)" notifications with varying parameters. Not sure if anyone else sees it, I barely know Wiki so its impossible for me to fix it unfortunately (if its a global bug.) --Phopojijo 17:56, 19 April 2006.

I changed a minus sign in equation (6), small error

Maybe you put out a refference on why you ddid it, okay ...!Special+Utilizator+$ (talk) 04:19, 3 March 2011 (UTC)

Where is the energy stored?

Since the electric wave and the magnetic wave are in phase, they both become zero at the same time. At these times, where is the energy of the wave stored?--Light current 03:00, 26 December 2005 (UTC)

Why, then the energy is in places where the electric and magnetic field are not zero. There's always the same number of peaks and nodes in the wave at all times. — Laura Scudder ☎ 17:31, 26 December 2005 (UTC)

Yes, good answer! I suppose if the wave is travelling (which it is) then the wave packet always contains energy also travelling. I suppose I must have been thinking of a zero length wave packet which of course can contain no energy--Light current 19:31, 26 December 2005 (UTC)

Electromagnetic energy travelling at the speed of light in free space is actually not stored energy - it is radiating energy. Energy is stored when it is confined with no velocity. - Marlowgs 13:59, 21 March 2006 (UTC)

In general I don't think that the electric and magnetic fields are ever zero at the same time. The the relationship between the two is that the magnetic field is the derivative of the electrive field with respect to time (and vice versa according to Maxwell's Equations?) which means that they are never in phase: when the electric field is at its peak the magnetic field is at zero and vice versa. Therefore one is always nonzero and "storing" the energy. Someone please correct me if I'm wrong on this.

You are wrong on this, I'm afraid. Take a look at Electromagnetic wave equation#Solutions to the homogeneous electromagnetic wave equation, particularly at the sinusoidal steady-state case, and you will see that E and H are always in phase. --Heron 20:44, 12 April 2006 (UTC)

I'm less sure about this. See http://www.play-hookey.com/optics/transverse_electromagnetic_wave.html for an example discussion where the author believes the electric and magnetic fields to be out of phase with one another. The link Heron provided above seems to be solving the B and E form of the equations separately, which gives the same answer since they are essentially identical equations (see the beginning of that article). But the point is that each one is independently a solution. You'd need to use Maxwell's eqn to derive the value of the other field. Grj23 (talk) 20:38, 29 April 2009 (UTC)

In fact it's pretty clear, I think. If

$\mathbf {E} (\mathbf {r} ,t)=\mathbf {E} (\mathbf {r} )\cos(\omega t)$ then from Maxwell's equation: $\nabla \times (\mathbf {E} (\mathbf {r} )\cos(\omega t))=-\partial \mathbf {B} /\partial t$ and so $\mathbf {B} =-(\nabla \times \mathbf {E} (\mathbf {r} )/\omega )\sin(\omega t)$ Similarly if you start with $\mathbf {B} =-\mathbf {B} (\mathbf {r} )\sin(\omega t)$ then $-(\nabla \times \mathbf {B} (\mathbf {r} ))\sin(\omega t)=\mu _{0}\epsilon _{0}\partial \mathbf {E} /\partial t$ and so $\mathbf {E} =(\nabla \times \mathbf {B} /\omega \mu _{0}\epsilon _{0})cos(\omega t)$ that is it's all consistent. So they must be out of phase...Grj23 (talk) 11:11, 30 April 2009 (UTC)

Apologies for the multiple edits. I guess the above only makes sense for a standing wave. For a travelling wave you might have for example

$\mathbf {E} (\mathbf {r} ,t)=E_{0}\sin(\omega t-kz)\mathbf {i}$ which gives $\partial \mathbf {B} /\partial t=\nabla \times \mathbf {E} =-kE_{0}\cos(\omega t-kz)\mathbf {j}$ and so $\mathbf {B} (\mathbf {r} ,t)=-k/mE_{0}sin(\omega t-kz)\mathbf {j}$ so Heron was right in the case of a travelling wave, though not in the case of a standing wave.131.111.74.72 (talk) 12:21, 30 April 2009 (UTC)

It wouldn't be right to say the energy is stored "in places where the electric and magnetic field are not zero". The energy is stored in the changing EM field. The energy is in the change. An electron in an otherwise empty universe does not create energy nearby it from an electric field. It is in the change of the field which would have energy. For example, if the lone electron was jiggled by the hand of god (who isn't in the "universe" i.e. closed system), then an EM wave would be emitted.

Also, if the electric field of light is zero, then the magnetic field of the light will be zero as well - they are directly related. Fresheneesz 19:42, 12 April 2006 (UTC)

The electric field and the magnetic field must be 90 degree out of phase ! The emitted wave from an antenna must follow the variation of the electric field and the magnetic field in the antenna. You can see the correct representation right here on wikipedia on Heinrich Hertz , page: http://en.wikipedia.org/wiki/Heinrich_Hertz — Preceding unsigned comment added by 71.185.135.27 (talk) 03:03, 9 February 2012 (UTC)

"Must be"? How funny. Your intuition is just wrong here. The graph in the Hertz article is from an experiment done ~120 years ago, before some of this stuff was well understood. It's not clear to me just what Hertz was measuring, but it does not seem probable that he was directly measuring the electric and magnetic wavefronts in a travelling wave. That is not easy to do. I suggest you are probably misinterpreting that graph.--Srleffler (talk) 05:22, 10 February 2012 (UTC)

Please decide what you do not agree with...The age of these experiments or that i misinterpreted the graph! The graph is obvious, and the math didn't change since then and so our understanding about the electromagnetic phenomena...Your reasoning suck's by the way ! — Preceding unsigned comment added by 71.175.121.199 (talk) 23:16, 11 February 2012 (UTC)

The graph certainly does not show the relative phase of the electric and magnetic components in a travelling electromagnetic wave. Hertz was studying the fields in a standing wave. The electric and magnetic fields in a standing wave are 90° out of phase, unlike in a travelling wave. See for example time 23:00–24:17 of the video at http://academicearth.org/lectures/boundary-conditions-at-perfect-conductors-reflection-and-standing-em-waves, which also confirms that the electric and magnetic fields of a travelling wave are in phase. So, the answer to your question is that you misinterpreted the graph, mistaking data taken on a standing wave for a representation of a travelling wave.--Srleffler (talk) 05:53, 12 February 2012 (UTC)

An important point mentioned in the video I linked to above: if the electric and magnetic fields were 90° out of phase in a travelling electromagnetic wave, it would be impossible for that wave to carry energy, since the time average of the Poynting vector would be zero.--Srleffler (talk) 06:04, 12 February 2012 (UTC)

German version has a very nice pic

Would be great if there was a pic like that here too. The great German pic cannot be copied because it has many German words on the it. Andries 23:17, 3 January 2006 (UTC)

• http://de.wikipedia.org/wiki/Elektromagnetische_Welle#Spektrum (here)
• http://de.wikipedia.org/wiki/Elektromagnetisches_Spektrum (here)
• http://es.wikipedia.org/wiki/Espectro_electromagnético#Espectro_visible (here)
• http://es.wikipedia.org/wiki/Espectro_electromagnético
• http://en.wikipedia.org/wiki/Electromagnetic_spectrum#Types_of_radiation
• http://en.wikipedia.org/wiki/Electromagnetic_spectrum
• http://en.wikipedia.org/wiki/Template:Electromagnetic_spectrum ... Ok! deWP, esWP & enWP Ok? --PLA y Grande Covián (talk) 13:58, 30 January 2009 (UTC)

$e=mc231tothirdpowerof500000nanoseconds?$

Electromagnetic waves

How is it that electricity be considered electromagnetic wave? Electricity fails one of the fundamental properties of electromagnetic waves; that is, all electromagnetic waves travels through vacuum at the speed of light. So even though photons travelling through an optic fibre can be considered an electromagnetic wave since photons travel through vacuum at the speed of light. What is the justification for electricity travelling through a wire? How would you even apply Maxwell's equations to electrons? The first paragraph of this article seems to be in error.

Electricity itself is not necessarily an electromagnetic wave. However, electrical enegergy in a coaxial cable might be an electromagnetic wave, or might become one, especially if it is oscilating fast enough and if there is an antenna attached to the coaxial cable. See Radio frequency. --ssd 13:08, 12 January 2006 (UTC)

1. All electromagnetic waves do not travel through vacuum at the speed of light. They can travel through any medium except a superconductor, and except in a vacuum they travel slower than c.
2. All practical electrical energy transmission, whether in a vacuum or in wires, is by electromagnetic waves. If you relied solely on the movement of electrons in wires, then electrical energy would travel at about walking speed. In fact, wires just confine the electromagnetic wave to a narrow region of space, but it's still a wave. Even DC is a wave: when you close a DC circuit, the wave travels from the switch to the load until the whole circuit reaches equilibrium, and then the wave is absorbed by the load and disappears.
3. Maxwell's equations apply to electrons just as well as to anything else, but you'd need a very powerful computer to do the calculations.

--Heron 21:33, 21 March 2006 (UTC)

das ist verry güt der ist das klassa simmer im das schula und im die house das ist nict güt

Direction of propagation

From what (little) I understand of EM waves it seems as though they are always propagated in ALL directions; the front of the wave is a sphere in 3-space whose radius increases (at the speed of light in a vacuum) with time, hence EM RADIATion. I think it is for this reason that lasers are difficult to make as they confine the direction of propagation largely to a single spatial direction. This property of EM radiation would seem to completely destroy any conception of light as a particle (the photon) as the photon would "exist" everywhere in this growing sphere simultaneously. Rather it seems a photon should be thought of as the smallest burst of this spherical wave at a given frequency (according to quantum theory the energy of a single photon depends on frequency). -Kyp4

Lasers don't do what you say. They only produce a parallel beam when you put a collimating lens in front of them [1]. --Heron 18:30, 29 March 2006 (UTC)

• That link is also partially incorrect, in fact. For a discrete beam, there's no way to make a truly parallel, absolutely collimated beam - diffraction will always spread the beam eventually. --Bob Mellish 18:45, 29 March 2006 (UTC)

This is something I find confusing too. Light is classically thought of as propogating outward in a sphere, as Kyp4 says, but a laser obviously doesn't do this. Is the article trying to say that the lens bends the partial sphere of light into a flat surface? If you have a point source emiting one photon at a time, what does the magnetic field around the source look like? Fresheneesz 19:35, 12 April 2006 (UTC)

The magnetic aspect of the em radiation we perceive is 90 degrees to the electro aspect. My question is: Is the magnetic aspect 90 degrees clockwise or anticlockwise to the electro aspect? -Gonegahgah (talk) 00:14, 27 October 2008 (UTC)

This is purely a matter of convention as to which way electric fields go an which way magnetic ones go. Martin Hogbin (talk) 20:07, 27 October 2008 (UTC)

Thanks Martin for your reply but what do you mean? Have they tested it? How do you test it? Were there results? -122.111.79.43 (talk) 22:11, 27 October 2008 (UTC) -Gonegahgah (talk) 22:16, 27 October 2008 (UTC)

Induction vs Hertzian Wave

There was a lot of confusion, even among top scientists, about whether various 19th century electrical experimenters working to develop wireless telegraphy or wireless telephony were producing anything beyond induction when they generated electrical transmissions of signals from an electrical arc or a simple coil of wire, and were able to produce sparks from a crude antenna nearby, or to receive audible voice frequency signals in a receiving coil connected to a telephone receiver. For instance, Edison used the term "Etheric Force" in the 1870's to describe his ability to detect a small arc between two carbon points with two metal squares as antenna, when a buzzer was arcing nearby. Leading scientists dismissed it as mere induction, ignoring Maxwell's equations, but the IEEE website now describes it as Hertzian wave transmission and detection. Hertz used a similar transmitter and detector, but did a rigorous demonstration of how the rf signals exhibited the properties predicted from Maxwell. Sources in radio history say that with induction the energy is localized, but Hertzian Waves travel a longer distance. A better "bright-line" distinction is needed. Various inventors like Preece and Stubblefield achieved wireless electrical transmission of voice signals in the late 19th century by induction. Does audio frequency current sent from a transmitting coil to a receiving coil technically constitute electromagnetic radiation, when at audio frequencies?Some help in the article for distinguishing what is and is not electromagnetic radiation would be helpful.Edison 22:14, 6 June 2006 (UTC)

Perhaps a test could be the following: for a coil of wire used as transmitter and another for receiver, each being a a certain diameter, with a certain number of turns of wire, at an audio frequency of say 1 khz, how rapidly does the induced voltage decrease as the colis are moved apart (their centers remaining coaxial). One falloff curve would be characteristic of electromagnetic waves, and a more rapid decrease would be characteristic of simple induction. Formulae I have seen for the mutual induction between two such coils indicate a very rapid falloff in mutual coupling when the antennas act as simply two windings of an air-core transformer.Edison 15:56, 13 June 2006 (UTC)

Yes, that might work. For more on the rates of attenuation with distance for the different types of field, see Near and far field.

There is another fundamental difference between induction and radiation. With induction, energy is transferred into the EM field during one half of the AC cycle and returned to the source during the other half. If a receiver is placed nearby, it picks up some of the energy which therefore does not return to the source. Thus, you can tell whether or not energy is being transferred by measuring the net amount of power flowing out of the source. With radiation, on the other hand, the energy flow is all one way: out of the source. None of it ever comes back, regardless of the presence or absence of a receiver.

A third way of looking at it is through Maxwell's equations. When you solve them for a radio wave, you get expressions for the E and H fields that are in phase, allowing them to radiate. In induction, they are out of phase, so they can't radiate. (For more on this, see Radiation from an Antenna by Peter Dodd.) --Heron 21:18, 13 June 2006 (UTC)

electromagnetic radiation and light

The first line of this article reads "Electromagnetic radiation (also more informally called light)" This must be corrected since light is only a specific frequency range. You could just as easily say "also more informally called radio waves" or "cosmic rays."

I agree. I think whoever wrote that was trying to make the point that some other parts of the spectrum near to visible light, like UV and IR, are sometimes called light, even though they're not visible - as in "ultraviolet light" and "infrared light". They then forgot that there are other types, like the ones you mentioned, that are never called light. The problem is that the boundaries between 'definitely light', 'sort of light-ish' and 'definitely not light' are subjective. Perhaps the criterion is that the radiation should behave like light under the conditions of whatever experiment you are doing at the time. Unfortunately, the American Heritage^® dic·tion·ar·ies says, in definition no. 2, "EM radiation of any frequency". [2] --Heron 18:36, 18 June 2006 (UTC)

P.S. The OED, in subsense (f) of sense (d) of light, n., includes a quote from Maxwell himself: "...light itself including radiant heat, (and other radiations if any), is..." Perhaps Maxwell was the first and last person to use the word in this way, and all subsequent dictionaries have blindly copied him. He didn't know much about the EM spectrum at that time - not even about radio. --Heron 18:43, 18 June 2006 (UTC)

In my experience, most physicists have no problem using light for EM radiation outside the visible (especially with regards to non-visible laser light). — Laura Scudder ☎ 19:04, 18 June 2006 (UTC)

Even though light can be defined with electromagnetic waves of any wavelength, I still think starting the article with that statement just comes off as uneducated. The most important thing is that people who are not physicists who read this article understand light to be visible light and nothing else. Keep in mind that those of us who understand the different wavelengths of light already know that EM radiation is light so it will not matter if that is taken out, but for those who don't know, that statement is just confusing. -(guy who started this)

I tried to avoid the whole "what wavelengths are exactly light?" discussion by just changing it to "sometimes informally called light". I don't know if that addresses the above complaints, but I just thought that delving into the semantic debate in the lead was something to be avoided. — Laura Scudder ☎ 03:52, 20 June 2006 (UTC)

I just avoided the problem even more by deleting the statement. I left in the word light, but only as part of a list. I think the best place to discuss the "what is light?" question is under Light. People who want to know about "informal" usage are more likely to go to that article first. --Heron 20:22, 22 July 2006 (UTC)

I have a question why can't we see other forms of radiation? What is necessary for our eye's to be able to see all forms of electromagnetic radiation/light? I think it's because only a range of radio waves and visible light can penetrate the atmosphere, the others don't, fortunately. It begs the question why beings with eyes cannot 'see' radio waves as well. I don't know the answer.--Robert van der Hoff 00:51, 8 May 2007 (UTC)

Mainly because we have set detector cells in our retinas (rods and cones on the outer retina, and photosensitive melanopsin ganglion cells on the inner retina) that are designed to "detect" visible light. Some animals can see in infra-red I believe as well. Birds and bees can detect the earth's static magnetic field, and use both it and visible light for navigation. Clearly bits of us are just designed for set purposes ;) Topazg 16:57, 29 June 2007 (UTC)

The Null Effect of magnetic or electric fields on EM radiation/light

The opening sentence: "Electromagnetic radiation, sometimes informally called light, is generally described as a self-propagating wave in space with electric and magnetic components" is misleading because light or EM rad is never deflected or affected in any way by a weak or strong static electric field or a permanent magnetic field located in a vacuum or space.

Shall we change this sentence. bvcrist Bvcrist 20:52, 10 July 2006 (UTC)

We can't change the sentence, because it's correct. We could, however, explain that light is not influenced by static fields in a vacuum, as you say. --Heron 19:45, 11 July 2006 (UTC)

Mass and frequency are equivalent, a photon carries a change of frequency between the individual charged particles.

--79.67.155.20 (talk) 11:02, 5 May 2008 (UTC)

All of the EM spectrum is light?

When did physicists start to make this assertion? At the start of the article we have "In some technical contexts the entire range is referred to as just 'light'." The given reference is not a good one. It points to a website of an organisation that most pepople have never heard of, and if the assertion is made, it's not easy to find. Any physicists out there? Please come up with a much better reference than this. Maybe we should put it as needing a citation. In any case, physicists can't just change the definition of light. That's down to people who write dictionaries. I think most dictionaries give a definition of light as being something that is visible - to a natural eye. Are physicists wrong (what a question - are they ever) ? Arcturus 21:16, 20 October 2006 (UTC)

To define light as only the part of the spectrum visible to human eyes might be common in general use and in dictionaries, but its at least very common in Physics to also include nearby parts of the spectrum (ultraviolet, infrared) that are affected by the same phenomena, or that produce the same effects, or can be analyzed with the same mathematical tools. For example, "infrared light" can be focussed by lenses, detected by semiconductor photodetectors, and analyzed by ray tracing very much like visible light.

I don't think its very common for anyone to refer to AM radio frequencies, for example, as light, though there may be certain contexts where its useful to do that. For that reason, an optics text might very well mention that the whole spectrum can be considered as "light" at least in some situations.

To me, the term "light" is the part of the spectrum that was historically analyzed by ray optics or geometric optics. This developed long before the rest of the spectrum was even discovered: Modern telescopes were invented in the early 1600's, but the rest of the spectrum wasn't known until Heinrich Hertz's discovery in 1888.

Finally, any modern description of "light" or "optics" must also include the full electromagnetic analysis to describe things like diffraction. This makes the boundary between light and microwave and rf radiation very blurry. But the need to see light as part of the electromagnetic spectrum developed historically long after the word "light" was defined.

I agree the article should not claim that all of the spectrum is "light" (otherwise, we should merge Light and Electromagnetic radiation). But it is definitely correct to include some of the spectrum beyond the visible spectrum.

-- The Photon 22:09, 21 October 2006 (UTC)

Agreed, UV and IR are commonly referred to as light. Should we then clarify this in the article and adjust the text so that it doesn't imply that the whole EM spectrum is regarded as light? Arcturus 12:34, 26 October 2006 (UTC)

I think the usage of the word light is appropiate. I think the name of the area on the EM spectrum referred to as 'visible light' should change. Just as the other areas of the EM spectrum have technical names, that area should be given a better technical name, as well. My reason being, all wavelengths on the spectrum are comprised of photons. When we hear the word photon, the words photo and camera come to mind. A camera captures light and photo film is light sensitive. I'll leave the name change up to scientists.Eevathediva (talk) 07:21, 23 April 2008 (UTC)

Explanation for a Grunt Lost in Space

Just what exactly is an EM level? How is it measured in regard to human physiology? If one turns 90 degrees and faces into the the four different directions: North, South, East, or West, or even up or down would a person's EM level vary? If so why? Please reply. Be gentle and kind. It may not look that way, but I am very fragile. User:Kazuba 31 Jan 2007

Misleading mention of AM and FM in diagram

The way that "AM" and "FM" appear in the wavelength diagram and the fact that it is otherwise devoid of specific mentions of applications for various bands of wavelength (e.g. microwave oven, television) could easily fool an uninformed reader into thinking that AM and FM are synonymous with specific parts of the radio wave spectrum rather than correlating with them. --AceMyth 03:27, 16 February 2007 (UTC)

Merge

This merge request should be removed. Light = Radiation, but Radiation != Light.

Arrgh. Forgot to sign. Real programmers know that 1 and 1 makes 10. E9 07:23, 3 April 2007 (UTC)

My concerns looking at electromagnetic radiation and light are that it is not clear that light is talking about visual light, while electromagnetic radiation is talking about electromagnetic radiation. Feel free to take the tag if the definitions are defined distinctly.100110100 07:30, 3 April 2007 (UTC)

Error on diagram

The label $10^{2}$ appears twice on the wavelength axis of the diagram. Peter Harriman 06:29, 21 May 2007 (UTC)

You are right; there's a minus missing. On my Firefox (Mac 1.5.0.11) the diagram completely fails to show up at all, and I just see a white box. Dicklyon 15:05, 21 May 2007 (UTC)

I attempted to fix it, but my SVG-fu is weak, doing so seemed to break the image completely. (Actually, the ways of the server-side SVG renderer are mysterious, and seem to result in a white box at least half the time, even when files will show correctly when rendered locally.) --Bob Mellish 15:42, 21 May 2007 (UTC)

Thanks to Sakurambo, the creator, who has fixed it! Peter Harriman 20:07, 21 May 2007 (UTC)

Electrosmog

Hi, I'm trying to move the info from the electrosmog article to this, so that we can get rid of that rather poor effort. Unfortunately, other than the definition of the term there is nothing much useful left. The supposed effects on human health are covered elsewhere, and the natural world effects are a media non-story. I put the definition in the intro, but maybe that's too high. I don't think it warrants it's own subsection, but I'll leave that up to you guys and gals. 128.243.220.41 15:05, 28 June 2007 (UTC)

Having spent quite a bit of time trying to get some general agreement to have electrosmog as a redirect only and get any relevant information concisely written on the electromagnetic radiation page, I would like to ask first that it stays here instead of having to have another article for it, and secondly that it doesn't have its own heading, as it is not deserving of one. I actually think having it at the top (perhaps with suitable intro: "Recent coverage in the media has referred to electromagnetic radiation as "Electrosmog".....") and then the text that we currently have would be the best way of having it in the article. Topazg 16:51, 29 June 2007 (UTC)

I think what's written is supremely neutral, but as the term IS used is a negative sense and pretty much always in relation to possible health concerns, shouldn't there be a link to said concerns? 129.215.141.101 13:56, 1 August 2007 (UTC)

To whoever keeps putting it in there, please note that the word electrosmog is not a portmanteau word. Electromagnetic is shortened to the standard prefix electro-. Smog is not shortened at all. There is no blending of words, and electrosmog is no more a portmanteau than electromagnetic. Sbacle 18:09, 6 September 2007 (UTC)

Accelerated electrons

Could any of you Experts out there put up something about the quantum explanation of why an accelerated electron radiates?

I know the classical maths indicates that it happens, though not everyone seems to agree - Feynman among them I believe; he says it is rate of change of acceleration apparently - , but no one seems to have a quantum explanation.

I thought I was on to it when Stumble introduced me to Hydrino, but contributors there seem to agree in nothing but disagreeing with Mills.

So come on someone, and sort out all those struggling engineers who can't see how an antenna works! And while you are at it, how does a photon know if it is part of the induction field or the radiated field?

--Boletusedulis 22:35, 24 July 2007 (UTC)

Direction of propogation

Just wondering here, if the direction of propogation is ExB as mentioned at the end of the article, should the direction of propogation for the wave in the first image be from right to left and not left to right, as is currently the case? EZG 14:10, 1 August 2007 (UTC)

Yes, you are correct. — Laura Scudder ☎ 15:19, 1 August 2007 (UTC)

Effect of Rare Earth Magnet on EM Radiation: None !!!

If EM radiation is the propagation (or self-propagation) of EM fields, then why does a strong magnetic field have no effect on the direction of EM radiation? Surely the EM fields of light are affected by strong magnetic or strong electric fields, yet no one has published such effects. Strange isn't it.

In my world, light is generated by EM fields, but it is does not have electric or magnetic properties as it moves through space. When light hits a material, the EM fields of the material are able to interact with light, but that does not prove that light has and possesses EM fields as an inherent property. Light, at all wavelengths, does not have any EM properties. It is purely energy, which is a form of matter or something else, that no-one yet recognizes or understands.

Vince Crist Jan 7, 2008 —Preceding unsigned comment added by 64.165.113.102 (talk) 08:39, 7 January 2008 (UTC)

Strong electromagnetic fields do affect light, but they have to be stronger than the field of your magnets, and the mechanism involves virtual charged particles. Try searching for "photon-photon scattering" online. Melchoir (talk) 09:12, 7 January 2008 (UTC)

Send me the paper and I'll shut up. Otherwise you are simply a Catholic Pope trying to shut up Galielo. Vince —Preceding unsigned comment added by 64.165.113.102 (talk) 09:23, 7 January 2008 (UTC)

Sweet, I get to become a world leader, and all I have to do is ignore you? Thanks! Melchoir (talk) 09:25, 7 January 2008 (UTC)

Well HotDog, where's my paper, where's the link? Don't be shy or lazy. Prove to me that you have some proof to back up your claim. Com'on send the paper or send the web-site link. Don't be a PrimaDonna like all the wanna-be Physics Grad students and their inability to learn by or think for themselves. As you surely noticed, I don't hide behind a monniker as do nearly all Wiki-editors and would-be maintainers of modern day misconceptions. My name is Vince Crist and my e-mail is: bvcrist@xpsdata.com. Want a phone number? —Preceding unsigned comment added by 64.165.113.102 (talk) 19:44, 7 January 2008 (UTC)

Vince - although I appreciate your enthusiasm, this is not the place for this kind of discussion. This is intended for discussion of the article. If you would like to discuss theories and ask questions, there are more appropriate places. Good luck with your quest. PhySusie (talk) 20:53, 7 January 2008 (UTC)

Susie - I am discussing the article by addressing the adjectival description that has no basis in fact or proof of any sort that is published anywhere. Since there is no proof as per Wiki laws for the verifiablility of that adjective, why is it written here. Everyone just accepts it with no proof. Show me the original proof or some recent proof is all I ask.

Maxwell died before Hertz produced his result. Hertz never proved that his microwave radiation had either magnetic or electric fields as it propagated through the room to his receiver. Hertz did use as large metal mesh to collect some of the radiation, but he did not bias the mesh in any way except to rotate it by 90 deg. Who in history decided to define light as being and having electric and magnetic fields? I'd love to know. If you have an article, book or recent test that shows that light has either or both electric and magnetic field properties as it moves through the air, I'd love to read it.

I've been meaning to do another little, potentially very interesting, experiment. Run a small DC circuit with a micro or pico-ampmeter and place a rare earth magnet next to one of the wires to see if there is any effect on the current. If there is a drop, does it level match expectation? If you keep the magnet in place, and have a thin wire with a high current, you might be able to melt the plastic on the wire or heat up a bare metal wire. Care to try? Vince Crist Bvcrist (talk) 08:01, 16 January 2008 (UTC)

"Surely the EM fields of light are affected by strong magnetic or strong electric fields" Visible light is caused by photons. Photons are also responsible for magnetism. A rare earth magnet just isn't strong enough. —Preceding unsigned comment added by 69.18.178.18 (talk) 08:30, 23 April 2008 (UTC)

OK, I found this little discussion quite intriguing and it got me thinking... I'm not convinced that a non-varying magnetic field would have any effect on the propagation of a photon due to the superposition principle. But I read here that people are doubting the integrity of Maxwell's equations; unless I have mis-interpreted it, it asks here who decided light consisted of electric and magnetic fields and that photons have no EM properties themselves. Well this was shocking to read, if one were to read most University-level texts on waves and the derivation of em wave is a frequent occurrence (one such excellent text, in my opinion, is Optics 4th E.d (HECHT, E.) - that I mention later in the thread). EM waves are the solution to the wave equation consequential to Maxwell's equations. It describes how disturbances in em fields propagate in space-time. The interaction of light and matter depends on the fact that em-radiation has both electric and magnetic components and that dispersion occurs from the ability for the induced dipoles in a dielectric substance to keep up with this oscillating field. --Ukberry (talk) 06:52, 1 May 2008 (UTC)

Ambient electromagnetic radiation

This is a tough request, but it would really help this article if someone could find a chart of the ambient electromagnetic radiation over the entire spectrum in some ordinary location (the middle of Central Park on a sunny day, for example). I mention this because the statement in the text

"Natural sources produce EM radiation across the spectrum, and our technology can also manipulate a broad range of wavelengths."

doesn't sound right to me. My guess is that sunlight would be a huge peak even on a log graph, which would help explain why that sliver of spectrum is so important, perhaps rivalled by some of the IR, and that even in a big city the total radio energy would be pretty tiny in comparison, while many of the more exotic frequencies would be practically empty. But I could be wrong - I can't recall ever having seen such a helpful comparison. 70.15.116.59 (talk) 00:58, 15 January 2008 (UTC)

The statement is correct. Sunlight contains wavelengths across the electromagnetic spectrum, not just visible light. Although our atmosphere shields us from much of it, more than just the visible part makes it to the surface of our planet. I am not aware of any measurement of the different parts of the spectrum at a particular location.PhySusie (talk) 17:42, 15 January 2008 (UTC)

Well, Planck's law has a relatively sharp peak, and the sun really is yellow, not just within the visual spectrum but in the absolute sense. I think the yellow peak should stand out plainly over all the rest, while as for the radio transmissions of the sun... you can see a 100-watt radio transmitter a lot further away than you can see a 100-watt light bulb. (The atmospheric window also imposes some of those limitations) 70.15.116.59 (talk) 03:24, 16 January 2008 (UTC)

Yes - em radiation from the sun peaks in the yellow part of the visible spectrum - though what you mean by "not just with the visual spectrum but in the absolute sense" is not clear. And I think I did state that the atmosphere blocks certain parts of the spectrum. I'm not sure what you mean by seeing a 100W radio transmitter better than a 100W light bulb - but that is besides the point. The original statement is still correct. The sun is not the only natural source of em radiation - and natural sources do emit across the spectrum regardless of their intensity. And our technology can manipulate across the spectrum as well. Both sentences are correct. I don't see any reason to change it. PhySusie (talk) 15:52, 16 January 2008 (UTC)

A clarification

Are there any particles emitted, when radiation is created? I was always under the impression that light = stream of photons, but the article does not mention it anywhere. Thus I wonder what does actually radiation consist of? If it consists of "nothing" and is just a wave (=the light is not transferred by any particle) wouldnt it mean that vacuum in fact consists of some "invisble" bricks that interact with each other? (e.g. when you put 5 billard balls in a line and hit the first one, the 4 other balls will move too). Agameofchess (talk) 19:49, 10 April 2008 (UTC)

Light is transfered by particles. Photons. All radiation is particles. All the various forces also have corresponding particles, except maybe gravity. —Preceding unsigned comment added by 69.18.178.18 (talk) 08:32, 23 April 2008 (UTC)

Actually the question is too deep for such a flip answer. Light can not be completely described as either particles or as waves. And no, there are no interacting "bricks" in space; that's the old luminiferous aether theory that was displaced by Einstein. Read about photon and wave–particle duality. Dicklyon (talk) 00:13, 12 May 2008 (UTC)

Serious Diagram Error

The wave diagram just after "Properties" has a serious error. Magnetic and electric fields are 90° out of phase and in quadrature. Both do not go to zero at the same time.Trojancowboy (talk) 21:07, 14 April 2008 (UTC)

Trojancowboy i9s not alone. Not an error. The first of 20,000 Google hits for "Transverse Electromagnetic Wave", www.play-hookey.com/optics/transverse_electromagnetic_wave.html makes the same mistake, and sticks to it although I tried to correct him years ago. This error is pervasive, and is part of the unrecognised crisis, part of which I discuss at www.electromagnetism.demon.co.uk/17136.htm . - Ivor Catt (talk) 21:11, 26 July 2008 (UTC) Ivor Catt

Several Errors

Agree with above comment - The magnetic and electrical components of an electromagentic wave are out of phase with each other by 90deg. To answer the question regarding energy and oscillation it is measured by the RMS value.

Please alter reference to 'Light' when refering to that range of frequencies that the human eye can detect. Light = Electromagnetic Radiation. Those frequencies which the human eye can see are more correctly termed 'Visible Light'.

Alternating electrical currents are not a flow of electrons. Please refer to electrostatics for more information.

Electical currents (individualy) do not produce electromagnetic radiation and neither do magnetic fields: Electrical currents induce a magnetic field and vice versa. Your explanation of electrical current traveling down a wire producing electromagnetic radiation is very missleading. It is the resitive effect of the wire that produces the EM radiation. —Preceding unsigned comment added by 62.6.149.17 (talk) 14:30, 16 April 2008 (UTC)

COMMENT - Disagree with the comment and response to comment

The electric and magnetic fields are in fact in phase in EM light. They are out of phase in AC circuitry. Eugene Hecht has the best book on Optics I've read through my academic career in Physics and when referring to electromagnetic radiation zone he states with mathematical analysis:

"In this zone, a fixed wavelength has been established; E and B are transverse, mutually perpendicular, and in phase" [1]

A more useful thought experiment I thought about when asking myself what would happen if they were out of phase setting up normal modes. If they were out of phase, the nodes of the electric field would be at different locations to the magnetic ones; so you would be able to couple energy from these points, so interference patterns would not occur. So if the components are out of phase in EM radiation, they would behave like EM waves in a circuit (which do not exhibit interference in the same way).

The reason I disagree with the diagram is that, although the direction of propagation is not explicitly noted (it is assumed to travel with increasing distance) and base vectors are not defined (assume E field on the +y axis and M field on the +x axis with propagation down the +z axis), the light would not be travelling in this direction on the diagram.

The reason I disgree with:

Electical currents (individualy) do not produce electromagnetic radiation and neither do magnetic fields

is that this is the basis of telecommunications and optoelectronic effects. Radio and TV use this effect all the time.

[1] - HECHT, E. Optics 4th Ed. (2002)

--Ukberry (talk) 14:25, 27 April 2008 (UTC)

Ukberry is correct (I've just checked in Hecht myself). The waves are in phase. Now why in the world did I have in mind that they were 90 degrees out of phase?Headbomb (talk • contribs) 17:49, 27 April 2008 (UTC)

All, see the Maxwell's equation. Faraday's law and Ampere's circuital law are responsible for EM waves propagation. They state (approximately) that curl of Electric field is equal to the rate of change of Magnetic field and vice-versa. curl is the rotation vector (which is itself defined in terms of differentials). Hence E and B will be in phase!! Ahirwav (talk) 13:08, 2 May 2008 (UTC)

Ahirwav, are you implying that if ANY two fields are related by the time derivative of the other, then then their solution to a wave equation (where space and time are considered) must be in phase??

--Ukberry (talk) 13:54, 2 May 2008 (UTC)

A possible resolution of this problem is to say that the electric and magnetic components are 180 degrees out of phase, noting that the polarity of the fields (N/S, +/-) is purely conventional.86.132.189.39 (talk) 20:47, 27 July 2008 (UTC)

Serious phase errors!

That there are widespread and serious errors with regard to Maxwell's equations and the creation of EM waves can be seen by the above discussions. It is also seen in the widespread error that Electric and Magnetic fields create each other. THEY DO NOT! Also the Electric and Magnetic fields ARE IN PHASE as was indicated above. I corrected this error in the article which was quickly returned to its erroneous state (as expected). I provided proof of my assertion with reference to the actual mathematics of the causality of Maxwell's equations and the conclusion that the equations that everyone believes show that E and M fields create each other are NOT in fact causal. This calculation and the arguments it is based upon, was done by O.D. Jefimenko for which I provided the reference. This same conclusion was also arrived at by Panofsky for whom I did not provide a reference. Indeed the 90 degree phase shift concept arises because of the mistaken belief that E and M fields "create each other". Clearly there is some determination here to insure that these widespread errors remain included in this article. The mathematics and arguments of Jefimenko and Panofsky do not appear open to debate and absolutely none of the widely accepted physics and E&M textbooks show a 90 degree field shift. (And neither do the illustrations in this article) I don't think that proven erroneous concepts such as the E and M creation story or the 90 degree phase shift error have any place in a quality article. I don't think that a trick of saying that these fields are "out of phase" (which is obviously NOT 90 degrees)by cleverly assigning conventions represents an honest approach to this matter either. —Preceding unsigned comment added by 69.218.219.89 (talk) 20:31, 17 June 2009 (UTC)

Why are you talking about a 90 degree phase shift between E and B? Of course, everyone knows that E and B are in phase in a linearly-polarized EM plane-wave. There's no controversy about this. Is this error in the article somewhere? I don't see it... --Steve (talk) 21:43, 17 June 2009 (UTC)

Mentioned it for two reasons. One is that in these comments above someone claimed that there was a 90 degree lag. This is a common misconception due to the lag in found in resonant electrical circuits. But there is also another important point. Electric and magnetic field clearly are occurring simultaneously! The principle of causality clearly states that all events MUST be preceded by a cause. That means the cause must be prior in time. Simultaneous event CANNOT "cause each other! The article plainly states that the electric and magnetic fields are causing each other. This is obviously a totally incorrect statement although it is also a widespread error due to the false assumption that Maxwell's equations as usually expressed represent causal relations. They do not (for the most part). This error has been clearly shown by Jefimenko as well as Panofsky whose publications provide ready reference for this assertion. OK? —Preceding unsigned comment added by 69.218.235.59 (talk) 19:17, 18 June 2009 (UTC)

Oh, OK. I wouldn't worry about the above comments that there's a 90 degree lag. You can find all sorts of nonsense on a wikipedia talk page, it doesn't mean it's a "common misconception".

When I kick a ball, the ball accelerates. One might say that the force of my foot causes the acceleration of the ball. But the force and the acceleration are at the same time, F=m*a. Right?

Say I have a capacitor, which I repeatedly charge and discharge with some frequency. Most textbooks would say that the charges on the capacitor plates create the electric field within the capacitor, and that changing electric field causes a magnetic field within the capacitor. Jefimenko, I guess, would say that the charges on the capacitor plates create both the electric and the magnetic fields within the capacitor. I don't think one of these points of view is correct and one is incorrect. I think they're different ways to think about the same event.

If you intend to create a changing electric field, you will also create a magnetic field, whether you intended to or not. I think that's what people mean by "a changing electric field causes a magnetic field". Obviously no one is saying that the electric field has a force of will and intentionally creates the magnetic field, or that the electric field needs to have been changing for 10 milliseconds before the magnetic field finally gets a chance to start responding. And it can also be the case that the magnetic field can cause a changing electric field.

Maybe a better example is a resistor: Sometimes it makes sense to say that the voltage across a resistor causes a current to flow, sometimes it makes sense to say that the current across a resistor causes a voltage drop, and in reality the current and voltage are simultaneous. These aren't contradictory. Do you agree? --Steve (talk) 22:33, 18 June 2009 (UTC)

No, both views of a capacitor are NOT correct. If one accepts the law of causality, then clearly the charges create the electric and magnetic fields simultaneously which then travel at the speed of light from those charges to where they are going. Since they are simultaneous it's impossible for them to "cause" each other without violating causality!

The whole problem here (and it IS a common error) is to look at Maxwell's equations in the usual form which obviously seem to show that a changing electric field dD/dt seems to be "causing" H and and a changing magnetic field dB/dt seems to be causing E. It seems so obvious! But the problem is these equations represent equalities NOT a causal relationship! The quantities can be EQUAL but they do not "cause" each other! Indeed, both electric and magnetic fields are caused by charges and their changes. They are created simultaneously and thence proceed out into space at the speed of light. Maxwell himself seemed aware of this problem using the term "measured by" rather than "caused by" in certain non-causal situations. (see Maxwell section 531)

Of course STATIC charges can create an electric field without a magnetic field and a constant current can create a magnetic field without an electric field, but these are not the cases involved with EM radiation where both electric and magnetic fields are changing with time.

Now look at the section in question:

"According to Maxwell's equations, a time-varying electric field generates a magnetic field and vice versa. Therefore, as an oscillating electric field generates an oscillating magnetic field, the magnetic field in turn generates an oscillating electric field, and so on. These oscillating fields together form an electromagnetic wave"

It CLEARLY implies that we are going to use the above erroneous conclusion and implies that the way electromagnetic waves keep going through space is sort of like the way energy exchanges back and forth between a capacitor and Coil in an LC circuit. Yes, they don't actually SAY that but the implication is there. And of course the difference is that in the LC electric and magnetic energy DO have a phase shift between them whereas in EM waves both are IN phase. That makes all the difference in the world! One might argue about the semantics of the word "generate" vs "create" but to me they both seem to imply the same thing. Namely that electric and magnetic fields are "creating" each other, which they do not do.

Hence a correct statement would be that a "source" of time-varying charges and/or currents (a flow of charge) simultaneously creates electric and magnetic fields which travel through space together at the speed of light to form an electromagnetic wave. The statement as it now stands is simply incorrect.

To consider the resistor example, to put a voltage (potential) on the resistor means to bring charges from infinity to the ends of the resistor. Once those charges stop moving they create an electric field in the resistor. Note that the Electric Field proceeds from the charges down the resistor at the speed of light! That electric field then creates a force upon the static charges within the resistor causing them to accelerate. As this occurs current begins to flow. Hence to say that a voltage "causes" a current in a resistor is correct! They do NOT happen simultaneously. To say that the current "causes" a voltage drop in the resistor would not be true and we see the voltage happens before the current does. However, in a practical case we do know that there is a law (Ohm's law) which relates voltage and current in a resistor. But as above Ohm's law ONLY represents a causal relationship if it is written in the correct form. We would also observe in this case that once the current is flowing in the resistor, it creates a magnetic field which proceeds away from the resistor at the speed of light. So in this case we see that electric and magnetic fields are NOT created together and in fact one could say that the electric field DOES indeed create the magnetic field (but not vice versa). But in this case to be correct we would like Maxwell have to say that the voltage drop in a resistor is "measured by" the current in it, rather than "created" by it!

The problem here is that because things happen really fast there is the tendency to assume they are actually at the same time. In a practical case that would be essentially true! But when one starts insisting that this or that CAUSES something else, one has to be very careful to figure out what comes first, because simultaneous things cannot "cause each other." —Preceding unsigned comment added by 69.218.211.181 (talk) 17:28, 6 July 2009 (UTC)

So you would agree that there's cause and effect in an LC circuit? What's causing what? As I'm sure you know, the voltage across the inductor and the voltage across the capacitor are in phase... --Steve (talk) 18:21, 6 July 2009 (UTC)

Old Revision Reverted (Opening Paragraph)

I was confused with the edit at first. Need to talk about this. So I don't get the word phenomenon was removed, this clarifies that light is "an occurrence of fact" (from Dictionary Definition), where the fact lies in the physics of the system.

Then the verb "perceive" was removed and replaced with sense; however the rewording could imply that the eye could perceive other phenomena. The eye certainly does perceive, defined as: "to become aware of (something) through the senses". Thus, we get a more concise description.

I must also clarify that a disturbance does not require a physical "medium", the disturbance with EM radiation is a disturbance of the EM field that is present everywhere in space; thus, I have reverted the edit.

I also want to see if everyone is happy with the diagram with the text and remove the dispute of factual accuracy; although I think that a new diagram would be better (would do it myself, but I'm not so good at graphics). And the sections on wave and particle models, I want to have a link between the two - why we observe both models.

Can I have some feedback please

Dictionary Definitions from: Collins Online Dictionary

--Ukberry (talk) 00:17, 12 May 2008 (UTC)

"Eighty-one octaves"

So does that mean that there is a minimum or maximum energy level (frequency) for EM? I was under the impression that there were theoretically no limits.

Stonemason89 (talk) 23:39, 13 July 2008 (UTC)

There may be some theoretical limitations to the frequency of EM radiation but the reference cited does not seem to give any and I suspect that the maximum possible range is more than 81 octaves. I suggest that the statement is removed until it can be better justified.Martin Hogbin (talk) 17:04, 14 July 2008 (UTC)

A little research has shown that cosmic gamma rays up to 100 GeV have been detected. I see no problem with EM radiation with uHz frequencies which gives us a range of a little over 81 octaves. A fundamental limit on the range might be set by the number of Planck lengths in the observable universe, which is estimated at 2.7 × 10⁶¹ - over 200 octaves. I will therefore remove the statement until someone has value which can be verified.Martin Hogbin (talk) 09:43, 16 July 2008 (UTC)

I have just noticed that the 81 octave statement cites Isaac Asimov's Book of Facts. I suspect that this is the range that been observed rather than the maximum theoretically possible. I will still remove the statement pending clarification.Martin Hogbin (talk) 09:48, 16 July 2008 (UTC)

Properties Diagram Is For t = constant?

Is the Properties diagram meant to portray a snapshot of a wave at a fixed time or is it meant to imply that you could stand at a point on the distance axis where the E and M components are zero and never detect the wave no matter how long you waited? If it is a representation of a function like sin(A*distance + B*time) then I think something should be added to the description of the diagram to say that time is constant.

Tashiro (talk) 17:39, 19 January 2009 (UTC)

Informal RfC for 'Speed of light' article

There is a dispute concerning the wording of the 'Light as electromagnetic radiation' section of the 'Speed of light' article. Editors are requested to give their opinions on the 'Speed of light' talk page. We decided to ask on related article talk pages rather than go for a full RfC so that we would get editors with a knowledge of and interest in the subject. Martin Hogbin (talk) 18:28, 6 February 2009 (UTC)

Why is this article called Electromagnetic radiation, not Electromagnetic wave?

I am curious about this. Matthew 百家姓之四 Discussion 討論 04:33, 23 March 2009 (UTC)

Merge with light

Making a distinction between EM radiation and light seems a bit POVish to me. The only purpose seems to be to make a distinction between light and visible light. But there already is an article on visible light. Yes there are a number of cultural elements in the light article that don't really belong in a scientific article about light but these could be merged with visible light. Serendi^podous 08:13, 18 May 2009 (UTC)

I will say that, in my opinion, rather than Electromagnetic Radiation, Light, Visible Spectrum, and Electromagnetic Spectrum, we should have Electromagnetic Radiation, Visible Light, and Electromagnetic Spectrum. (That is, merge "Light" with "Visible spectrum".) As a physicist, I feel that, broadly speaking, using "light" for "electromagnetic radiation" is sometimes appropriate and sometimes not. If the situation pertains to "almost-visible" (near-IR and near-UV) radiation which interacts with matter much like visible light and can be manipulated with similar optics, it is usually acceptable to call it "light". Also, when discussing phenomena that are highly applicable to visible light (e.g. Rayleigh scattering or refraction), it is justifiable. This may include those applicable to any electromagnetic radiation, such as the "Doppler shift of light", "polarization of light", etc. However, in the context of physical phenomena occurring only far from the visible portion of the electromagnetic spectrum, it sounds highly odd to use the word "light" for the electromagnetic radiation involved. An example would be Compton scattering of X-rays and gamma rays: we simply do not say "Compton scattering of light", as the phenomenon is not observed for visible (or almost-visible) light. (But yet physicists do get away with (ab-?)using the prefix photo- as in photonuclear reaction or pion photoproduction even though they are talking about a reaction induced by gamma rays!) So I would say it's more of a style issue when to use "light" for "electromagnetic radiation". 69.140.12.180 (talk) 21:19, 26 May 2009 (UTC)Nightvid

Another peculiar case is the Advanced Light Source and Advanced Photon Source, both of which produce mostly X-Rays. :-) Anyway, I agree with 69.140.12.180: Merge visible spectrum and light into a new article: visible light, about the past and present understanding of visible light. That's the common-sense, near-universal definition of the term "light". Even 800nm photons are normally called "infrared radiation", not "infrared light". Of course, much of the current light article would move here instead. --Steve (talk) 07:31, 27 May 2009 (UTC)

Agree. Merge light into visible light, with "light" redirecting there. Jheald (talk) 10:54, 27 May 2009 (UTC)

Disagree. The Visible spectrum article is about the spectrum, not about light; it can logically stay separate from the light article, which covers many aspects of light unrelated to spectrum. Dicklyon (talk) 15:43, 27 May 2009 (UTC)

On second thought, I agree with Dicklyon: "visible spectrum" is worthy of a separate, narrowly-focused article, about the different wavelengths from 400nm to 700nm, and their relation to color and other physical phenomena. How about just making visible light redirect to light instead of redirecting to visible spectrum? (Plus, moving some content from visible spectrum to light.) --Steve (talk) 19:41, 27 May 2009 (UTC)

Oppose - The current division seems broadly correct: Light (a less technical article, using the common or garden term and focussed on the part of the spectrum that we can see) and Electromagnetic radiation (more heavy-duty physics content, discusses things which are common across the whole spectrum). Perhaps these articles could be augmented with content from Visible spectrum and Electromagnetic spectrum, and I agree with Steve's suggestion that Visible light should redirect to Light, but that's separate to this merger proposal! Djr32 (talk) 22:07, 27 May 2009 (UTC)

I would support merging Visible spectrum with light, and merging electromagnetic spectrum with electromagnetic radiation. Both article pairs seem to have a large overlap. I would oppose merging light/vis. spectrum with EM spectrum/EM radiation. I think visible light, as the stuff we see, has an important distinction from the rest of the spectrum. It also gives a place for a slightly more physics-light description - I do not think this page needs to go into QED. The EM radiation page can then go into more depth, and of course cover the full range. The light page would link there for more details. GyroMagician (talk) 10:27, 3 June 2009 (UTC)

Given the lack of enthusiasm shown for the original merge proposal, I have removed the merge tags. There were a number of other suggestions above for articles which could be merged, which could be followed up on. Djr32 (talk) 17:41, 6 June 2009 (UTC)

There seems to be consensus that a merge is necessary, but no consensus on what that merge should be. I would prefer it if such a merge were conducted by someone with a working knowledge of the subject. Serendi^podous 19:10, 6 June 2009 (UTC)

Oppose - Electromagnetic radiation includes radio waves and microwave energy, which are NOT considered light. Light is a form of electromagnetic radiation but NOT all electromagnetic radiation is light. --96.225.33.226 (talk) 21:53, 28 June 2009 (UTC)

EM wave quantized

The article states that "energy of an EM wave is quantized." This clearly is not true, although it is a common misperception. What is true is that energy is added to and removed from the EM field in quanta because this (the conversion process) involves matter which gives up and accepts energy in discrete "quantized" amounts. No citation is given for the EM wave itself being quantized. Maxwell's equations for EM waves do not (and can not) predict any quantization. blackcloak (talk) 19:19, 28 July 2009 (UTC)

This is a common correct perception. :-) I added a ref.

I read the section starting on pg 511 of the reference. To me the logic appears circular. The cited experiment does not confirm the quantization of the EM field; it merely confirms conservation of energy in the conversion from the EM field to the various energy components of ejected electrons. Stated another way, an analysis of the remaining energy of an electron emitted from a surface of a material with a work function in no way says anything about the quantization of (i.e. particle nature of) the EM field itself; it does say something about one (frequency) component of what might be a very complex (many frequencies and associated intensities in the) EM field that happened to interact with the (detector's) surface. To make this any clearer, we'll probably have to start exchanging descriptions of specific thought experiments. Maybe there is something I'm missing. blackcloak (talk) 06:31, 29 July 2009 (UTC)

Maxwell's equations for EM waves do not predict any quantization, but Maxwell's equations do not describe our universe. That's why we figured out the more accurate theory of quantum electrodynamics. This more-accurate theory does predict that monochromatic light has quantized energy, independent of any light-matter interaction. :-) --Steve (talk) 04:30, 29 July 2009 (UTC)

The only reason for mentioning Maxwell's equation in this context is that the article covers the subject in some detail. I did not go back to check if your caution about limitations of Maxwell's equation is in the text. blackcloak (talk) 06:31, 29 July 2009 (UTC)

I have no intention of discussing thought experiments. You seem to believe that the quantization comes from the matter and not the light. That was also Planck's belief, but then Einstein figured it out and physics moved on, and that was 100 years ago. The understanding of light in modern physics is something that's been understood for decades and withstood extremely stringent and specific experimental tests...see Precision tests of QED for example. This goes way beyond the photoelectric effect. :-)

Your reference was to a section in a book dealing with the photoelectric effect. That is what I was responding to. Now you send me elsewhere. I assume you've learned a simple dismissal is not going to work. Your response was a reincantation of dogma, not a response to content. If you think you actually understand an experiment that shows the EM field is itself quantized, perhaps you'd like to take a crack at explaining it. No wiki article I've read has attempted this. blackcloak (talk) 04:05, 30 July 2009 (UTC)

I don't think the article as written implies that Maxwell's equations are exact laws of the universe. But if you disagree, by all means we can make it clearer. :-)

Anyway, regardless of your assessment, the quantization of light energy is universally accepted by physicists, which means it's good enough for wikipedia. See WP:RS. :-) --Steve (talk) 07:43, 29 July 2009 (UTC)

No argument here. That is, when it comes to the part about wikipedia. blackcloak (talk) 04:05, 30 July 2009 (UTC)

The ideas of light being quantized and light being a wave are both accepted, but they don't mix very well. That's "duality". I changed that statement after a source was added, since the source didn't support the idea of the wave being quantized. Dicklyon (talk) 14:33, 29 July 2009 (UTC)

They mix perfectly well as described by quantum electrodynamics. But your edit is fine. --Steve (talk) 15:03, 29 July 2009 (UTC)

Should I be assuming that you think your benediction should be good enough for me? blackcloak (talk) 04:05, 30 July 2009 (UTC)

Someone who appreciates consistency? And then does something about it? Am I to believe my eyes? blackcloak (talk) 04:05, 30 July 2009 (UTC)

One does what one can. I personally think the idea of the EM "field" being "quantized" is misleading, or misleadingly stated, frequently. I'm pretty familiar with QED (though I'm not really a physicist), and I don't think it says anything to suggest that the "wave" is quantized; rather, it talks about wave functions as amplitudes, and quantized observables, which is fine. In any case, adding a source to a contested statement is always good, as it makes it more clear how to fix the statement to agree with the source. Dicklyon (talk) 04:39, 30 July 2009 (UTC)

The original statement -- that the energy of the wave is quantized -- was correct; consult any textbook on quantum field theory. The new version is almost meaningless. But you are right that the original needs a better reference than the one that was provided. Jim E. Black (talk) 07:42, 21 October 2009 (UTC)

Gravity

I am sure if any one says "black hole" the next thing that is thought of is how gravity effects EMR. But, there are many properties of EMR which are related to gravity. For example, pair production, gravitational red/blue shift of EMR, bending of light near a star. There may be more, but I will let the good people who can write clear articles write them. I just hope someone will make a section on these gravitational properties of EMR. —Preceding unsigned comment added by 76.186.80.92 (talk) 13:45, 13 August 2009 (UTC)

"EM radiation carries energy and momentum that may be imparted to matter with which it interacts."

The last line of the intro. paragraph:

"EM radiation carries energy and momentum that may be imparted to matter with which it interacts."

...needs some background information and explanation. It seems like it belongs in the middle of a paragraph in the middle of a chapter in a great book on the subject of Electromagnetic Radiation! 207.199.220.113 (talk) 16:45, 13 September 2009 (UTC)

Question

Is there any type of energy RESISTANT to Electro magnetic radiation? —Preceding unsigned comment added by 67.176.66.143 (talk) 00:26, 12 October 2009 (UTC)

Mass Question

... highly confusing. EM article includes visible light; the visible light article defines light as "packets" of EM or "photons"; the photon article defines photons as massless. Back to his EM article: "EM radiation carries energy and momentum", since momentum = mass*velocity - I take issue with *something*. I dont know what yet, but definitely something :D (71.58.235.30 (talk) 13:41, 18 November 2009 (UTC))

Momentum = mass*velocity only holds for massive bodies, and only in the framework of Newtonian mechanics. Formally, momentum is the Noether charge of translational invariance (see relativistic momentum). This notion can also be applied to quantum fields and massless particles. — Tobias Bergemann (talk) 17:42, 18 November 2009 (UTC)

Experiments Section or New Page

There should be a page or a section involving experiments the electromagnetic radiation. Please people look up.

Electric vs Magnetic Phasing

The opening para. says "Electromagnetic radiation....consists of electric and magnetic field components which oscillate in phase." WHAT! This suggests they are in phase with each other. Aren't they actually 90° out of phase with each other?'

Electric field strength is proportional to the rate of change of the magnetic field. SO MAX magnetic field rate of change is at it's ZERO crossing = MAX Electric field. THEREFORE, 90° out of phase?. Please confirm and/or rewrite this paragraph! If I'm correct the diagrams in article are wrong as well.

• "ubiquitous"? Why the'big' words? Not needed. Makes it harder to understand. --220.101.28.25 (talk) 20:40, 24 November 2009 (UTC)

The spatial and temporal oscillations of the E and B fields are IN PHASE, but the orientation of their field vectors is at right angles. 155.198.198.134 (talk) 14:51, 7 October 2010 (UTC)

Your error is in stating that electric field strength is proportional to the rate of change of magnetic field. That is not what Maxwell's equations say. For a plane wave, the derivative of the electric field with respect to distance along the optical axis is equal to the rate of change of magnetic field in time. This means that when the rate of change of the magnetic field (in time) is maximum, the rate of change of the electric field (in space) is also maximum. --Srleffler (talk) 06:08, 12 February 2011 (UTC)

I think that theres somthing wrong with the derivation!!

/media/math/2/0/9/20913029e6919606f6debc3e687d97c3.png

This equation is currect but the next one after it:

/media/math/a/a/3/aa320df3ec6536d867f3433b6e328905.png

does not follow from the first one! the derivative of f in the first equation is with respect to position and B's derivative is with respect to time! so you cannot integrate the right side of the first equation with respect to time and get the second formula! Am I wrong?

Tone issue

I tagged this article because the tone in the 'Derivation' section is not approriate (e.g. "Let's consider ..." and "... with what corresponding magnetic field?"). -Roger (talk) 00:27, 3 April 2010 (UTC)

Speed of propagation =

First sentence in the second paragraph states, "Depending on the circumstances, electromagnetic radiation may behave as a wave or as particles"

I believe this is outdated. As I understand it, it is particles of radiation being propagated in a wave. —Preceding unsigned comment added by 99.72.119.145 (talk) 03:21, 3 February 2011 (UTC)

The article is correct. Light is neither a wave nor a particle, but behaves like both in different circumstances. --Srleffler (talk) 05:34, 12 February 2011 (UTC)

E & B Fields are shown with improper phase

For the EM wave of a dipole emitter, like the one shown on this page, the E & B fields are actually out of phase. For reference, see the current/voltage relationship for an LC oscillator. If one is a sine wave, the other will be (+/-) the cosine wave. — Preceding unsigned comment added by McGinnis (talk • contribs) 03:11, 26 April 2011 (UTC)

The picture in the article is correct. You're mistaken. Dauto (talk) 19:00, 7 September 2011 (UTC)

book to add

Albert Einstein didn’t prove that the photon exists. He did theorize the quantum effect and received the Nobel Prize in 1921 for the photoelectric effect. Arthur Compton did the experiment to prove that the photon exists in 1923. In 1927 is when he received the Nobel Prize for proving the existence of the photon. Read the Nobel Prize paper in which Arthur Compton mentions for the first time the photon as a particle and the experiment that proves it. The book X-rays and electrons An outline of recent X-ray theory By Arthur H. Compton Ph. D. Copyright 1926 By D. Van Nostrand Company This book includes papers from 1923 on and uses the word photoelectrons for the one particle and electrons for the other. He shortened the word photoelectrons to photon by eliminating electro. The new particles thus became the photons from a shortening of photoelectrons. Read the book. The term’s used for electrons from the many papers are: • photoelectrons • recoil electrons • beta rays The term’s used for photons from the many papers are: • x-rays • x-ray quantum • light • light darts • electromagnetic waves • radiation • radiation quanta • radiation quantum • quantum He used photoelectrons a lot for the electrons but for the photons there was no common term. 68.171.143.254 (talk) 12:22, 7 September 2011 (UTC)

Image problem

The diagram of the EM spectrum used in this article and many others seems to have some problems. As noted on its talk page, the visible spectrum does not appear to be placed accurately on the broader EM spectrum. The color rendering is also not good: 555 nm appears to be yellow, while it is in fact green. --Srleffler (talk) 13:56, 15 October 2011 (UTC)

jarring image

This image (/media/wikipedia/commons/thumb/4/4c/Electromagneticwave3D.gif/220px-Electromagneticwave3D.gif) is incredibly distracting. 74.132.249.206 (talk) 19:23, 23 November 2011 (UTC)

Of the first five, I found this to be the most informative and useful image. 171.64.68.161 (talk) 19:57, 14 December 2011 (UTC)

Archive 1

Assessment comment

The comment(s) below were originally left at Talk:Electromagnetic radiation/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

It is a bit confusing between radio like FM and AM witch seems to be something between Microvawe and longue wave and the generale definition of radio frequency witch includes microwave as well. the Electromagnetic spectrum (http://en.wikipedia.org/wiki/Electromagnetic_radiation) seems to have big categories like : -Gama rays 1-10pm -X rays 10pm-10nm -UV rays 10nm-380nm -Visible Light 380nm-780nm -Infra Red 750nm-1mm -Microwave 1 mm-30 cm -Radio Wave 1 mm-1m /!\ it does overlap :) -Long radio wave 1m- shall we put microvawe inside Radio Wave or find an other way to call the wave between 30cm and 1m ?

Last edited at 13:24, 25 September 2006 (UTC). Substituted at 14:44, 1 May 2016 (UTC)

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 (talk • contribs) 07:21, 31 December 2013 (UTC)

Magic quantum effects?

I do not understand the last edits from today. [3] 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 [4] 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 (talk • contribs) 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 :$\nabla \times \mathbf {E} =-dB/dt$. 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″ [5]. 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" [6] describe the analog propagation by means of electrical engeneering [7] -- 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").[8] Also in near field (or in the antenna) the electric and magnetic field components are 90 degrees out of phase for traveling waves[9]. (see also Near and far field or Near-field electromagnetic ranging or experimental [10] or for sound waves [11]). — 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.

--Chetvorno^TALK 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. --Chetvorno^TALK 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)

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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. ----Chetvorno^TALK 21:13, 6 April 2018 (UTC)

Done [12] --Chetvorno^TALK 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 [13] 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. --Chetvorno^TALK 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. --Chetvorno^TALK 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: [14], [15], [16], [17]. 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. --Chetvorno^TALK 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. --Chetvorno^TALK 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": [18]. I also note that with this edit: [19], 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: [20], 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. --Chetvorno^TALK 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. --Chetvorno^TALK 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 (talk • contribs) 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? --Chetvorno^TALK 01:54, 1 July 2020 (UTC)

@Chetvorno: 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. --Chetvorno^TALK 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. --Chetvorno^TALK 02:18, 3 July 2020 (UTC)

@Chetvorno: 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. --Chetvorno^TALK 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)

@Chetvorno: 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. --Chetvorno^TALK 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. --Chetvorno^TALK 19:42, 5 July 2020 (UTC)

First sentence and lead structure

An edit has been proposed and contested that added mention of photons to the first sentence. The format, with parentheticals, is awkward. Any description of the dual wave/particle nature of light contained in a single sentence is going to be incomplete, but what if we just added the word "quantized" piped to photon so that the first sentence read, "In physics, electromagnetic radiation (EMR) consists of quantized waves of the electromagnetic (EM) field, propagating through space, carrying electromagnetic radiant energy. It includes radio waves, microwaves, infrared, (visible) light, ultraviolet, X-rays, and gamma rays. All of these waves form part of the electromagnetic spectrum." That approach is terse but at least introduces the concept right away. I also suggest removing the sources from the first sentence, since this information is sourced in the body.

We don't currently mention photons until the third paragraph of the lead, and even then we say "...an alternate way of viewing EMR is that it consists of photons..." which isn't terribly accurate and we only mention after two paragraphs on the classical model. I think the lead should be rewritten to focus on the contemporary understanding of EM, with minimal focus on the classical model. VQuakr (talk) 16:44, 7 October 2021 (UTC)

Light is quantized. Radiation, being light, is quantized. The interaction of radiation with matter is quantized. But I am unsure that it is accurate to say that the wave is quantized.Constant314 (talk) 17:11, 7 October 2021 (UTC)

Fair point. What about moving it to "...quantized electromagnetic radiant energy" instead? VQuakr (talk) 17:58, 7 October 2021 (UTC)

I'm fine with VQuakr's initial wording: "...electromagnetic radiation (EMR) consists of quantized waves of the electromagnetic (EM) field...". It's true that EM waves quantum nature only manifests when it interacts with matter, but that is too confusing a point to make in the lead paragraph. I feel "quantized electromagnetic radiant energy" is too nonspecific for general readers. I like how the introduction describes the wave and particle viewpoints in separate paragraphs, introduced by the sentence: "...an alternate way of viewing EMR is that it consists of photons..." (fair disclosure: I wrote it). But I agree that some mention of the quantized nature should be in the lead. --Chetvorno^TALK 19:20, 7 October 2021 (UTC)

I am OK with photons and quantized energy, but unsure of quantized wave. Constant314 (talk) 19:24, 7 October 2021 (UTC)

Whatever we say, it should be directly supported by a source we cite there. The problem before was that the cited source (or at least the parts quoted from it) talked about electromagnetic radiation only in terms of wave. If we want to add some words about a quantized view of electromagnetic radiation, let's make sure that what we say is backed up by a cited source. Personally, I would say that what's quantized is the interaction of radiation with matter, and that photons are how we describe those interactions (between things at zero interval in spacetime, as Gilbert Newton Lewis observed). But I know this is not likely to be the view in most sources, so I don't put it that way. Dicklyon (talk) 02:35, 8 October 2021 (UTC)

Disagree, there's no need to cite the first sentence as long as it summarizes cited information in the article body. Cites in the lead for non-BLP material are not desirable. VQuakr (talk) 17:15, 8 October 2021 (UTC)

Perhaps so. But presently there's ref 1 on the opening sentence, so that would have to be removed (and used later) if material that it doesn't support is added into that sentence. Dicklyon (talk) 02:50, 9 October 2021 (UTC)

Looking at refs 6, 7, and 8 cited in the lead for the QM point of view, I note that 6 doesn't mention "electromagnetic radiation", 7 associates it only with waves (except in one place), and 8 associates it only with waves, though mentioning quantum transitions. To me, these treatments make good sense, treating "radiation" as primarily a wave effect, and bringing in quanta at the emission and absorption events, which is the only place they matter. Photons don't exist in a vacuum -- only in the interactions with matter (as I said before, I realize you probably won't find that pov in a source, but if you look at how big a photon is, it never has finite extent in space, due to relativity; it extends from emitter to absorber, which are at zero interval in spacetime). So if you want to talk about radiation as photons, you at least need a source that clearly does so, as opposed to the cited sources that associate radiation with waves (as is necessary for looking at how radiation propagates). Dicklyon (talk) 03:06, 9 October 2021 (UTC)

Photons are weird and I agree that we should stick closely to what reliable sources say about them.Constant314 (talk) 03:14, 9 October 2021 (UTC)

Some refs are flaky, too. Ref 7 says "PHOTON: A particle of electromagnetic radiation carrying a specific amount of energy, measured in electron volts (eV)." But as we all know, the energy of a photon is not "specific", but rather entirely dependent on what inertial frame it's measured in. A monochromatic source emits photons of a certain energy (in the frame of that source), which can be absorbed by an absorber in a different frame, with a relative velocity, as photos of some other specific energy. So the energy per photon is not a property of the photons in the radiation so much as of the interaction of radiation with matter; see Redshift. Dicklyon (talk) 03:31, 9 October 2021 (UTC)

That same source also says "RADIATION: The transfer of energy by means of electromagnetic waves, which require no physical medium (for example, water or air) for the transfer. Earth receives the Sun's energy, via the electromagnetic spectrum, by means of radiation." This is more like how radiation is usually regarded. Dicklyon (talk) 03:48, 9 October 2021 (UTC)

Looking for books that mention "particle of electromagnetic radiation", I do find quite a few. This one seems good, in discussing Compton's experiments and 5 ways that QM rescued classical physics in the early 20th century; all 5 are about the interaction of light with matter. Dicklyon (talk) 05:21, 9 October 2021 (UTC)

Is the first diagram wrong?

I think the directions of the first diagram is wrong. It should be, as one field increases (in absolute value, ie including signs + / -, and not just the magnitude), the other decreases. I reference a diagram that show this in Atkin's Physical Chemistry, 8th edition p983. I think this is raised a few headings above. 2A00:23C5:C13C:9F00:4CC8:24FA:C403:BD1B (talk) 01:01, 24 February 2022 (UTC)

The diagram is correct for a traveling plane wave in vacuum. E and B are lock step synchronized, peaking and crossing zero at the same point. However, the fields can get out of phase in non-traveling situations and also when traveling in some materials. Constant314 (talk) 01:20, 24 February 2022 (UTC)

It might be correct for travelling plane waves, but the article is about travelling electromagnetic waves, not any old waves, and they have their own properties. They cross zero at the same points, but one in the reverse direction as shown. 2A00:23C5:C13C:9F00:E8FA:22F2:8266:C2BF (talk) 21:14, 24 February 2022 (UTC)

A traveling plane wave is the most common case and is usually what you find illustrated. The direction of propagation is given by the direction of the cross-product E x B. The direction of the fields need to be as shown to get propagation toward the right. Constant314 (talk) 22:31, 24 February 2022 (UTC)

The vector product is not commutative, and the pair are anti-parallel. If you use a mathematical model, then there is no reason why it is not B x E rather than your quoted E x B. It is therefore important when you use the vector product to model the physics that you remember for EM waves, as the electrical component goes up, the magnetic component comes down. They both peak, it is just one is the top peak, when the other is the bottom peak, and yes they cross over at zeros. 2A00:23C5:C13C:9F00:E8FA:22F2:8266:C2BF (talk) 23:49, 24 February 2022 (UTC)

Not sure what you are saying. It is always E x B, never B x E, that would go the wrong direction. It may not be obvious from the drawing, but E and B are at right angles. E is pointed in the x direction and B is pointed in the y direction. Constant314 (talk) 00:08, 25 February 2022 (UTC)

There are enough people here who have noticed the mistake. For EM waves, when one component is at a maximum, the other component is at its minimum. I have referenced and quoted from a reputable text book. You can always look up the diagram in the quoted reference. If you are not sure what we are talking about, then please look up the diagram from any decent text book. And if your reference differ from what many here have noticed, then please quote the reference. The arrangement in the diagram is when one component is at its maximum, the other component is also at its maximum, which is wrong 2A00:23C5:C13C:9F00:E8FA:22F2:8266:C2BF (talk) 01:55, 25 February 2022 (UTC)

FYI, references for your perusal:

• Griffiths, Introduction to Electrodynamics, third edition, 1999, fig 9.10 on page 379.
• Harrington, Introduction to Electromagnetic Engineering, Dover, 2003, fig. 10-4, page 262.
• Prucell, Electricity and Magnetism, 2011, fig 9.7, page 333.
• Kraus, Electromagnetics, 1984, fig. 10-3, page 381
• Halliday, Resnick & Walker, Fundamentals of Physics, 2003 fig. 34-5, page 805

Constant314 (talk) 02:16, 25 February 2022 (UTC)

If you look at the diagram in the Wiki article "Light" and the following https://ask.learncbse.in/t/draw-electromagnetic-wave-and-lable-its-parts/9917, the diagrams show that the maximum of one component correspond to the minimum of the other component, ie different from the diagram in this article. 2A00:23C5:C13C:9F00:E8FA:22F2:8266:C2BF (talk) 03:21, 25 February 2022 (UTC)

Another website: https://www.doubtnut.com/question-answer-chemistry/draw-the-diagram-of-electromagnetic-wave-161348909 2A00:23C5:C13C:9F00:E8FA:22F2:8266:C2BF (talk) 03:34, 25 February 2022 (UTC)

A simpler book: "Giles Sparrow, Instant Physics ISBN 978-1-78739-981-5" diagram on p69. 2A00:23C5:C13C:9F00:E8FA:22F2:8266:C2BF (talk) 03:39, 25 February 2022 (UTC)

2A00:23C5:C13C:9F00:E8FA:22F2:8266:C2BF, Constant314 is correct, see the animation, right.

Yes, those two websites got it backward. They are not reliable sources. Take a look at poynting vector to see how E x B defined the the direction of propagation. Constant314 (talk) 04:12, 25 February 2022 (UTC)

2A00:23C5:C13C:9F00:E8FA:22F2:8266:C2BF, sorry, I didn't quite understand what your objection was. What you are saying is that the vectors $\mathbf {E}$, $\mathbf {B}$, and the direction of the wave $\mathbf {S}$, are not necessarily related by the right hand rule, they could be in the left hand rule. That is a common graphical error in drawings of electromagnetic waves. No, the order of the vectors in the definition of the Poynting vector is always $\mathbf {S} =\mathbf {E} \times \mathbf {B}$, not $\mathbf {B} \times \mathbf {E}$. From the definition of the cross product, that means in an electromagnetic wave ($\mathbf {E}$, $\mathbf {B}$, direction of motion) are always related by the right hand rule. --Chetvorno^TALK 05:11, 25 February 2022 (UTC)

Please don't take this as a complaint There are 2 schools of thoughts in the discussions here, so it is unlikely that they are both correct. I stand with Atkins. A changing electric field induces a changing magnetic field of the opposite sign, this changing magnetic field then induces an electric field in the opposite direction to the inducing electric field, and therefore reduces it, etc. Boths field then fluctuate sinusoidally, which on the time scale (equivalent to travelled distance) means the peak of one is the trough of the other. This is what is shown in Atkins, and other works I mentioned. The diagram given in the article does not make sense. 2A00:23C5:C13C:9F00:C468:C68D:2787:E989 (talk) 14:31, 25 February 2022 (UTC)

No, there are not two schools of thought, the electric and magnetic fields are always related to the direction of motion by the right hand rule. It's just that the graphic artists who draw diagrams for cheap textbooks sometimes get it wrong. Here are some websites which explain it: How to use right hand rule to find direction of e-m wave, Right hand rule tutorial, Youtube --Chetvorno^TALK 15:44, 25 February 2022 (UTC)

No, there are not two schools of thought. Go with the physicists. What you are missing, I suspect, is that there is an asymmetry in Maxwells equations. One of the equations for curl has a negative sign that the other does not.

$\nabla \times \mathbf {E} =-{\frac {\partial \mathbf {B} }{\partial t}}$

$\nabla \times \mathbf {H} ={\frac {\partial \mathbf {D} }{\partial t}}$

The extra negative sign turns out to be crucial. Without it, you get a self-dampening wave instead of a self-sustaining wave. Constant314 (talk) 16:43, 25 February 2022 (UTC)

2A00:23C5:C13C:9F00:C468:C68D:2787:E989| "...the peak of one is the trough of the other". So you are saying that you think the electric and magnetic fields are 90° out of phase? Like a sine and a cosine wave? --Chetvorno^TALK 18:00, 25 February 2022 (UTC)

Prucell is available online. [21], page 439. Prucell is a Nobel laurite. He did a lot of work on radar in WW2. Constant314 (talk) 18:44, 25 February 2022 (UTC)

I changed the diagram over on light, which was not incorrect, since it did not indicate the direction of propagation. Constant314 (talk) 00:50, 26 February 2022 (UTC)

You have no right to change the diagram in "Light", as you have no agreement to change it. 19:27, 26 February 2022 (UTC) — Preceding unsigned comment added by 2A00:23C5:C13C:9F00:B93E:98BF:6035:AD5 (talk)

I support Constant314's change of the diagram on Light --Chetvorno^TALK 19:58, 26 February 2022 (UTC)

Just watched the indian teacher about the right hand rule for the 3-D Cartesian frame-work. He didn't even answer his own question of which direction the positive z-axis will point. There's nothing wrong with it BTW, but it is much easier to point out that digits 1, 2, 3 of the right hand is congruent to x, y, z (in that order), or you could use the wrist, thumb and 4-fingers model to remember the directions. Why not say the z-axis come out towards the reader? Of course there are no 2 schools of thought. The diagram you put forward cannot be correct. The right hand rule is the same as that for the alternator: As one field increases, the other one falls (taking signs into account, and not just the magnitude). This is what the two partial differential equations mean. This would also agree with Le Chatelier's Principle. Your diagram does not show this. Please examine your diagram. Your diagram does not match the differential equations, and does not agree with Le Chatelier's Principle. It is the artist of your diagram who's got it wrong. As for going with physicists, which physicists are your referring to? There are enough physicists reading this article who have spotted this mistake. And it was physcists that got the direction of flow of particles wrong in a metal wire, and caused problems for students and learners ever since. If we just listen to some physicists, we'll still believe the universe is immersed in aether. Please change the diagram to the correct one and restore the correct one you deleted n the "light" article. 19:23, 26 February 2022 (UTC) — Preceding unsigned comment added by 2A00:23C5:C13C:9F00:B93E:98BF:6035:AD5 (talk)

The physicists are Griffiths, Prucell, Kraus, Halliday, Resnick & Walker. Harrington is an electrical engineer. These reliable sources have all been used as college textbooks in physics or electrical engineering. Do you believe that Prucell, a Nobel laurite that worked extensively in radar got it wrong? You still seem to be ignoring the asymmetry in Maxwell's equations and the definition of the poynting vector. The direction of propagation is given unequivocally by E x B. Any diagram that is inconsistent with that is incorrect. You can see what Prucell wrote here: [22], page 439. Here is is what Feynman, another Nobel laurite, shows [23] in figure 27-2. Constant314 (talk) 19:43, 26 February 2022 (UTC)

There is no one called Prucell. Why don't you write to the publishers and ask whether the authors have proof-read their books in fine detail. Your diagram is wrong for logical reasons (1) the diagram does not match the partial differential equations you quoted, (2) it does not agree with Le Chatelier's Principle, (3) your diagram show both fields rising and falling together, and not as described by the partial differential equations you quote, or Maxwell's equations. Plus the diagram in Atkins' (Oxford prof) does not agree. The problem with some engineers is that they are foggy when it comes to science and maths. As for Nobels, even Einstein is wrong when it come to Quantum Theory. My friend, you do not question other people's diagrams and what they mean enough; it makes you a good parrot, but not so good as a thinker. 2A00:23C5:C13C:9F00:B93E:98BF:6035:AD5 (talk) 20:35, 26 February 2022 (UTC)

Yes, Prucell should have been Purcell. Looks like we are at and end here. The six reliable sources and the consensus is to keep the diagram as is. Constant314 (talk) 21:44, 26 February 2022 (UTC)

Agree, all done. --Chetvorno^TALK 21:49, 26 February 2022 (UTC)

You are both wrong. The correct diagram is as diagram here: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/13%3A_Molecular_Spectroscopy/13.01%3A_The_Electromagnetic_Spectrum 2A00:23C5:C13C:9F00:B93E:98BF:6035:AD5 (talk) 02:57, 27 February 2022 (UTC)

How about this from the same folks [24] Constant314 (talk) 03:07, 27 February 2022 (UTC)

Another correct diagram: https://astronomy.swin.edu.au/cosmos/e/Electromagnetic+Radiation 2A00:23C5:C13C:9F00:B93E:98BF:6035:AD5 (talk) 03:02, 27 February 2022 (UTC)

Another correct diagram: https://physics.stackexchange.com/questions/407619/what-do-electromagnetic-waves-look-like 2A00:23C5:C13C:9F00:B93E:98BF:6035:AD5 (talk) 03:08, 27 February 2022 (UTC)

https://blog.soton.ac.uk/soundwaves/further-concepts/1-mechanical-waves-and-light-waves/ 2A00:23C5:C13C:9F00:89E4:E928:9182:7DE1 (talk) 03:21, 1 March 2022 (UTC)

https://www.dreampointphysics.com/2021/03/definition-of-electromagnetic.html 2A00:23C5:C13C:9F00:89E4:E928:9182:7DE1 (talk) 03:35, 1 March 2022 (UTC)

I'm not looking to restart this discussion, but I have noticed that the diagram in Atkins has been removed in the tenth edition (2014). Constant314 (talk) 20:37, 7 March 2022 (UTC)