Talk:Photon/Archive 4

This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page.

Dear newcomers,

Hi, and welcome to Photon! There's still lots that can be improved here and we encourage you all to be bold. However, some topics have been debated at length already, and we encourage you to consider the consensus of previous generations of editors. It is not always easy to see why things have been written as they are, or the repercussions of editorial choices

• The term "light" refers to all forms of electromagnetic radiation, for both beauty and brevity.

This is a standard usage among physicists and in most technical contexts (see Wikipedia's Light article). Here at Photon, the general concept of "electromagnetic radiation" is used throughout the article, and it is easier and more illuminating for readers to see a familiar 1-syllable word such as "light" rather than a 10-syllable phrase such as "electromagnetic radiation". So we have chosen to clarify in the header that light is defined as EM radiation to help those readers for whom this definition is not obvious or familiar. Although it might be tempting to replace "light" with "electromagnetic radiation" in the header and throughout the article, it would unfortunately make the article longer and more tiresome to read, less clear without being more precise.

If this is standard usage among physicists then where is the good reference to this? Last time I read this article there was a poor reference to a website where I couldn't find anything about it (though I have to admit I didn't look for long!). This concept seems adsurd to me. You might as well say that the term "radio waves" refers to all forms of electromagnetic radiation. While it might be standard usage in the academic community, it certainly isn't in the world at large. Remember - this article is not for physicists, it's for the man in the street. Arcturus 19:24, 17 October 2006 (UTC)

I didn't know radio waves were light. Funny - I've never SEEN them. "Scientists" taking over the English language - what next?

People with science degrees writing Wikipedia articles perhaps? As a retired science teacher, I'd say using the term light in English to represent all electromagnetic radiation frequencies is perfectly sensible and widely used, not just in academic circles, but to children in school; using "visible" light to mean the frequencies humans can see is also common.--Phil Wardle (talk) 06:51, 14 January 2009 (UTC)

• Oh, now this is an obscenity, people

"All EMS" = "Light" is a Hasty generalization. It is an oversimplification used to help explain a difficult concept to people who cannot or do not want to understand the real truth. It would be inappropriate in a scientific journal, but apparently acceptable in a situation where half-truths are given credence as fact. Saying "All EMS" = "Light" is much like saying the Earth is flat and the center of the universe. Beauty is not being served, and Brevity would seem to have no place in Wikipedia.

One component of science is accuracy. Defining "All EMS" = "Light" is fundamentally inaccurate. As for this usage being "standard", with a circular reference to the wiki, I call for some real data for this "statistic". FYI: Collective EMS emissions; all EMS; radio, light, every electromagnetic phenomenon; are referred to as "Propagated Waves". —Preceding unsigned comment added by 66.150.46.254 (talk) 18:10, 1 December 2009 (UTC)

When I looked up EMS in Wikipedia, because I wondered what the letter S stood for, I found no mention of anything to do with electromagnetic radiation. So I suggest that the letters EMR would be more appropriate.WFPM (talk) 18:59, 16 April 2010 (UTC) Okay, it's already there, so the EMS must be a missprint. Suggest above EMS names be changed to EMR to avoid confusion.WFPM (talk) 17:16, 20 March 2011 (UTC)

Now the Photon is involved with the transfer of a conceptual quantity of EMR called a Quantum, and we're trying to decide whether it's a single entity or an accumulated package of entities. And this wave/particle controversy is derailing a lot of progress in understanding the nature of energy and/or action generation and propagation. But when you consider the conceptual value of the Quantum, which is agreed to be the total amount of action transferred by either conceptual transfer method, you wind up with a concept of an "energy per unit time" times a "time interval" quantity value. And, of course, the argument that whole quantum value can be transferred in one quantity in an instant reduces the conceptual decision to a particle versus package decision. And since we also have to deal with the polarization problem concerning the nature of the photon, it is easier to hypothesize it's having an orientation of spin configuration (like maybe a doughnut?) that can be interfered with or passed by the properties of the propagating medium. So by a process of hypothesis and elimination, were reduced to a concept of a polarizable particle transfer mechanism. And the other "energy wave in the medium" concept does not seem to be capable of explaining the capability of the details of EMR generation to be propogated over long distances while retaining the details, like those noted in the Whirlpool Galaxy.WFPM (talk) 19:54, 16 April 2010 (UTC)

And if you want to retain the multiple radiation particle concept, all you have to do is to is to imagine a small particle that has the polarization properties of the photon and call that appropriately the "radiation particle". And the photon accumulation process would involve the capture of the energy of a number of these particles for use in desired physical and/or chemical activities.

And if you do that you're going to wind up with a concept of a very small radiation particle with a mass of like 10 to the - 47th power grams, and we could call that protomatter and speculate about it's method of successive delivery of energy to a work site as a means of accomplishing such activities as photoelectric emission, chemical activity, and/or refraction.WFPM (talk) 17:57, 20 April 2010 (UTC)

• The photon is massless because it has zero invariant mass.

The unqualified term "massless" refers to invariant mass. No one disputes that photons can contribute to the invariant mass of a system, or to the stress-energy tensor (and, thus, photons gravitate). It's just that we have agreed to reserve "massless" for the intrinsic property of invariant mass, and to use more nuanced wording for those other conceptions of mass. The Talk archives above contain the discussion of this point at length.

For further information, please consult the Talk archives above, the scientific and non-scientific peer reviews and the Featured article discussion. Thanks and good editing! :) Willow 09:48, 3 October 2006 (UTC)

Intro needs a look —Preceding unsigned comment added by Ldussan (talk • contribs) 05:15, 21 April 2010 (UTC)

are governed by quantum mechanics and will exhibit wave-particle duality – they exhibit properties of both waves and particles. For example, a single photon may be refracted by a lens or exhibit wave interference with itself, but also act as a particle giving a definite result when quantitative momentum is measured.

I'm not so sure that "but act as a particle giving a definite result when quantitative momentum is measured" is what we're trying to say. Surely when momentum is measured from any wave it gives a quantitative result. Perhaps "but act as a particle giving repeatable quantized results when energy or momentum is measured over varying areas."Ldussan (talk) 05:12, 21 April 2010 (UTC)

The descriptive explanation of the word "Photon" could be better explained as being the name of an amount of "energy" and "linear momentum" that is associated with it's real and/or theoretical properties. Because the immediate mixed discussion of these properties almost immediately results in a confusion of concepts as to it's identity. And these concepts could be better explained if they were more expressly differentiated from each other, so that those interested in their individual ideas could better pursue their inquiry into the details of the subject matter.WFPM (talk) 18:46, 22 May 2010 (UTC)

How do you have momentum at all without mass? I seem to recall it being intrinsically part of the math for momentum. I suggest considering the photon as energy amount and packet form, increase the energy by a specific amount, get a new frequency. I DO seem to recall some math on THAT one, but can't recall the theory involved right now. It dealt with why one cannot have a transmitter sending a 640 KHz signal at 20GeV, off of the top of my head.Wzrd1 (talk) 05:10, 28 September 2011 (UTC)

Your suggestion to think of the photon as a packet of energy is absolutely apposite, and is a key difference between classical mechanics and quantum mechanics. In classical mechanics, you can divide momentum by mass and get velocity Momentum#Linear momentum of a particle), but for photons that doesn't work. I think that Richard Feynman said that the formula works, but he doesn't know what the reality behind the formula is, or if there even is anything behind the formula. Probably the math you are thinking of is the Planck relation. --Hroðulf (or Hrothulf) (Talk) 16:26, 28 September 2011 (UTC)

If photons have mass, is the earth getting heavier as the sun relentlessly bombards the planet with photons?--137.186.211.70 (talk) 17:28, 8 October 2009 (UTC)

Photons interact strongly with matter and will thus be in thermal equilibrium with the Earth. The capture of photons by the Earth will simply shift the equilibrium value of the mass in the form of captured photons. Count Iblis (talk) 18:09, 8 October 2009 (UTC)

A simple-minded way to say it is that, since the Earth's average temperature is nearly constant (not easy to measure precisely, of course), it must be losing energy as both reflected sunlight and thermal IR (heat) at almost exactly the same rate that it is gaining it from the Sun. Then the change in mass/energy must be very small, ≈ zero up to within the changes due to global warming (which should take a very long time to affect more than the surface layers in any case). Actually the Earth gains many tons of meteoroidal cosmic dust per year, and loses some hydrogen, He, and other gases from the top of the atmosphere, so the mass of all those should also be included in the accounting, and may well dominate the total. Wwheaton (talk) 22:05, 8 October 2009 (UTC)

Well, that doesn't really answer the question, now does it? What about the moon? The sun bombards it too with photons --- does it increase in mass because of the photons? Also, I suggest Earth's average temperature is the wrong thing to measure and thus it is the wrong proxy to deduce that photons don't have an effect. Why? Because Earth makes its own heat in the core - perhaps Earth is cooling at the same rate the sun is warming it, ya? In any event, if Earth is really getting heavier (granted as you claim, the increase is not from photons but from cosmic dust) what are the rammifications? If the earth gets heavier, its gravity will increase, and the consequences are that the moon will get closer to Earth and Earth will get closer to the sun. Or am I drunk?--137.186.235.152 (talk) 18:15, 9 October 2009 (UTC)

Same answer as for the Earth, E = mc², so if the total energy of the Moon increases, so does its mass, but you have to bookkeep the energy lost too, as the Moon radiates thermal energy into space. The essential thing is that all the photon's mass is due to its energy, and you have to reckon all the energy processes alike. The average temperature (making its contribution to the total energy, and thus mass) of the Moon is obviously not changing very fast. And again, you would have to include accretion of cosmic dust, etc (not to mention spacecraft, lately...) for an accurate answer. The important thing is that the effects are tiny. The mean distance from the Earth to the Moon is accurately measured, using light bounce travel times from the Apollo laser reflectors; it is increasing on the order of a few cm per year I believe, reasonably consistent with the tidal torque on the Moon due to the Earth. That energy comes from the Earth's rotation. Wwheaton (talk) 22:15, 10 October 2009 (UTC)

I came to this page via one of these: photon field, in the introduction on the Magnetic field page. But a search of the term "photon field" on the photon page comes up with nothing. Should one of the two pages be changed to deal with this inconsistency? Kansaikiwi (talk) 00:08, 16 November 2009 (UTC)

Good catch; this is a particularly bad usage of the "pipe" feature. I've changed it to the much more intuitive "photon field".

If the author really wanted to link a concept of photon field, but there just isn't such an article yet, it might have been defensible to use a redlink, photon field, or even put a redirect to photon at photon field as a placeholder, with the idea that an article would be written later but in the meantime readers would get where they would at least get some information. But just to turn both words blue, instead of one? What's the point of that? --Trovatore (talk) 04:56, 16 November 2009 (UTC)

Fixed this bad pipelink by introducing the idea of the virtual photon in the lede, which is the quantum of near fields, induction fields, electrostatic fields, static magnetic fields, and so on. "Photon field" is not a common term, and it's misleading anyway, as the photons being referred to, are not the "real photons" that we have experience with, in radiating EM waves. SBHarris 18:51, 20 April 2010 (UTC)

The electromagnetic photon field is discussed in the "Photon structure" subsection below.--HCPotter (talk) 09:55, 20 November 2011 (UTC)

Isn't it true that, from the point of view of a photon, the entire universe looks like a singularity? A photon travels EVERYWHERE in the universe in ZERO time from its own point of view. So doesn't that mean that from a photon's point of view, the universe is a singularity? 68.200.98.166 (talk) 03:26, 25 December 2009 (UTC)

I think that is true, but why did no one reply to this? Can someone more versed in physics answer this one way or the other? 173.168.177.217 (talk) 20:54, 20 March 2010 (UTC)

Talk pages are not really intended for answering generalized questions about the subject (see WP:TALK) although in practice small amounts of such questions are often tolerated. A better place to ask would be WP:RD/Science.

I think the usual answer from physicists would be "there is no such thing as a photon's point of view". That answer might be a little reductive (culturally, physicists have limited patience with questions they don't see a physical way to answer, which is not always a virtue) but it's probably the answer whose reasons you should be aware of before you try to figure out if there's a way around it. --Trovatore (talk) 19:26, 21 March 2010 (UTC)

Yes, it's true, but good look talking to a photon to get its impression of the experience. And you'll need an infinite amount of energy to ever get "up to speed" with one of them to even ask the question, and even if you could, you'd find that once in its rest frame, it's got no mass, and you will probably find that talking to massless objects is pretty boring. But you'd certainly have an interesting perspective. Unfortunately, you'd also find that the universe has come to an end when you try to return to Earth to tell us about the experience. Birge (talk) 00:35, 6 April 2010 (UTC)

It doesn't matter how much energy you have at your disposal. If you have mass you cannot get to c. Only massless particles follow null trajectories in spacetime, so the only way to travel at c is to become massless. 67.171.217.43 (talk) 17:33, 8 January 2011 (UTC)

You send an EMR to the moon and 2.5 seconds later you get a reflection of it back and you think no time has passed? Are you kidding? The only question that can be asked is as to the properties of the returned photons. And as far as I can see the ones that get back haven't changed a bit.WFPM (talk) 01:26, 17 April 2010 (UTC)

This was not about how a stationary observer sees it. 2.5 seconds? A photon does it in 0 seconds. The OP's question was about the point of view of a photon. I think you misunderstood the question... 173.168.177.217 (talk)

Well let's just compare the two noted events in the space-time continuum. 1: We're on earth and emit a light emission at T = 0 seconds. Then at T = + 2.5 seconds a second event occurs, namely that some part of the light emission returns and is noted. Now as concerns the measurement of the time continuum entity, what do you want to say that it is, other than that it's T = 2.5 seconds in your frame of measurement? And do you actually believe that it took zero seconds in your time frame of reference for the light to go to the moon and then return? Boy!!!WFPM (talk) 17:37, 22 May 2010 (UTC) And comes the question as to: could there exist a positionally located frame of reference, such that the time interval between the two noted events would be zero? and the answer is No!!. So were dealing with errors in the measurement of transient phenomena.WFPM (talk) 18:12, 22 May 2010 (UTC)

"A photon has two possible polarization states and is described by exactly three continuous parameters: the components of its wave vector, which determine its wavelength λ and its direction of propagation."

The polarization state of a photon is described as having three parameters, described by the Jones Vector, being two orthogonal electric field components and a relative phase between the two. With the wavelength, this would appear to be four parameters. However, I think this photon article is correct, but the fact that the polarization state can be described with only two parameters should be noted in the Photon polarization article. Is the two-parameter polarization state described anywhere on Wikipedia? cojoco (talk) 03:05, 26 March 2010 (UTC)

This is really confusing, I think. It makes a photon sound like a billiard ball. For example: what is the state of a photon created by the spontaneous relaxation of a single atom in free space? I would argue it's a superposition of an continuous set of modes spanning all of k space on a shell whose radius is determined by the energy of the relaxation, weighted to have a dipole spatial distribution, and constrained to have a photon number of one. Sure, after you measure it, the photon will be found to have a sharp k vector, but are we to believe the photon didn't exist before it was measured? Quantum mechanically, I would think you'd have to allow for a photon to be a superposition of k vectors, and thus it is an oversimplification to say that a photon is described by a single three dimensional k vector plus the polarization state. Am I missing something? Birge (talk) 00:43, 6 April 2010 (UTC)

I'm not as familiar with photon state as I'd like, hence the questions. It sounds like the "polarization state" of the photon can be described using two parameters, and three if you include the phase, and four if you include the frequency. The "wavefunction" of the photon, prior to detection, is a dipole spatial distribution. Currently, the article doesn't make any of this clear, as it does not distinguish between the wavefunction and the detected photon, and does not explain how it is possible to express the polarization state (without phase or frequency) as only two parameters. cojoco (talk) 00:46, 7 April 2010 (UTC)

You have ${\displaystyle \mathbf {E} (t)=\left(E_{x}(t),E_{y}(t)\right)=\left(E_{0x}e^{i\phi _{x}},E_{0y}e^{i\phi _{y}}\right)}$. One phase is arbitary, leaving you with three non-trivial parameters. If you don't care about absolute amplitude, you can get away with two parameters, but the description is incomplete. Frequency has nothing to do with descriptions of polarization. Pick up Eugene Hecht's Optics (ISBN 978-0805385663) if you need a solid reference on polarization. Headbomb {talk / contribs / physics / books} 00:43, 17 April 2010 (UTC)

I think there are still one or two problems with the article. You say that "Frequency has nothing to do with descriptions of polarization", yet this is contradicted by the article, which includes λ in the description. You also state "One phase is arbitary, leaving you with three ... If you don't care about absolute amplitude, you can get away with two parameters" While this article states the same thing, in neither this article, nor the article on photon polarization, is shown how photon polarization can be described using two parameters. cojoco (talk) 21:19, 17 April 2010 (UTC)

This is all very vague, and unless somebody with better knowledge steps in, we owe it to readers to just delete this whole vague business about the number of parameters (whatever that even means) that is needed. For one, saying there are two polarization states is referring to the fact that a photon is spin-one and can have the spin vector in one of two directions (i.e. right handed circular or left handed). What about linear polarization? Well, that would be a superposition of the two states with the same quantum phase. Is that a photon, or a superposition of two photons? I have no clue, and to a certain extent it's just arbitrary convention to say "what is a photon." As far as can tell, the concept of a photon is one of the most abused in all of physics. —Preceding unsigned comment added by Birge (talk • contribs) 17:24, 23 April 2010 (UTC)

I Want to complain about this same quoted passage. It has two polarization states but is completely described by the 3 parameters of its wave function? —Długosz (talk) 19:24, 5 May 2010 (UTC)

I read the "number of parameters" as "degrees of freedom". I found these concepts very important I was learning what photon polarization meant, so I'd like to see this article fixed. The Jones vector seems to be really clunky, as it seems to have one more parameter than it needs, and it changes markedly and confusingly with coordinate system. All polarization states can be be describable as the projection of a great circle onto a plane, and this requires only two parameters. Surely something like this must exist in the literature. cojoco (talk) 23:58, 5 May 2010 (UTC)

In the absence of any consensus, I suggest we delete this discussion. Nobody with a better understanding came along to help, and it's clear that the people writing the article (including me) are not really sure enough about this to write authoritatively. For the record, my misgivings is about the definition of a single photon, not what constitutes an allowable solution of Maxwell's equations. I understand what people are getting at when they want to say a photon can be expressed by a 3D k-vector and a polarization state, but nobody has offered a good explanation of why it's helpful to readers to consider the number of free parameters in a spatially infinite plane wave to be the degrees of freedom of a single photon. My current understanding of quantum optics (which is poor, admittedly) is that this is a very academic definition of a photon as it corresponds to EM in free space. It's certainly not in line with the more common usage of "photon" which would consider the quanta emitted by a single electronic transition in an atom, for example, to be referred to as a single photon (according to the academic definition I posited above, this photon would be a superposition of an infinite sum of photons constrained such that the total photon number is one). See what I mean about this being a mess? One thing I'm certain of is that the murky definition of "photon" is one of the more messed up things about modern physics, and people much smarter than me have written as much. Can we just delete this whole issue from the article? —Preceding unsigned comment added by Birge (talk • contribs) 20:52, 11 October 2010 (UTC)

I don't think this discussion should be deleted! Maybe "summarised"? Someone should certainly fix that sentence. I don't see how the article can have such a high rating when it contains such a sentence. Mal (talk) 12:21, 29 January 2012 (UTC)

I would like to add a layman's description of particle wave duality. Please let me know what you think?

Perhaps a simple and close to accurate description of a photon's or any particle's wave-particle duality in the standard model of quantum physics is as follows below. Note that this description is designed to ease one into the understanding of this dual behaviour not fully explain it.

At one point Bohmian mechanics was a promising field. It eventually got derailed due to a scientist named Bell proving that it could not be true in it's advertised form. That is not important so much as to understand what it was Bohm thought.

Bohm and others tried to think of a photon as a particle that had a wave "in front" of it, guiding it. He called it the pilot wave. This pilot wave acted just like a wave in water acts, capable of interfering with itself and refracting off of other matter. The places where the pilot wave constructively interfered the photon had a good chance of traveling through. The places where it destructively interfered it did not have a good chance of traveling through.

If this pilot wave was put through a beam splitter, it would split into two even though there was only one photon. At some point down the line from that beam splitter there would be a detector, one on each path. When the pilot wave reaches those detectors the laws of physics tell us that if the photon happens to be there it will get absorbed by the material in the detector and be detected.

But how does one reconcile that now there is only one photon and two pilot waves?

Since the pilot wave only says where the photon may be one can treat it as just a probability phenomenon. It is then a wave that directs the photon's path by giving it more and less preferreable paths to take; a probability distribution wave. Physicists then say that the probability wave has split equally into two and now there is a 50/50 chance the photon will "appear" at one detector or the other. So although the pilot or probability wave does indeed take both paths, the photon takes...?

This statement can't be completed with absolute certainty and may be regarded as one of the limits of our current understanding of quantum physics. Does the photon exist at both places at once? Did the photon just take one path and not the other? Is the photon the particle or the pilot wave guiding it, or both?

Current research suggests the pilot wave is the actual photon with no actual seperate "particle" existing at that moment. The pilot wave upon encountering an atom in the detector collapses accross space and time instantaneously to a single point in that detector/atom with a 50% probability. Though this is somewhat easy to think of for a massless and chargeless particle such as a photon, it is not so easy to think of a mass and charge containing particle such as an electron existing as a continuous probability wave, but this is what current experimental results show to be true.

--Ldussan (talk) 04:00, 9 April 2010 (UTC)

I don't know the mathematics of quantum probability, so for what it is worth, you offer a clear description that jibes with those I have read in other sources.

A couple of thoughts come to mind:

1. Would it be possible to adapt this to omit the convoluted reference to Bohm?

2. Since (according to Feynman, for example) all the wave is a probability wave, is there a clearer way to introduce the concept of a probability function before the phrase "although the pilot or probability wave"? This sentence seems to introduce the hard concept of a probability wave a little abruptly.

--Hroðulf (or Hrothulf) (Talk) 11:22, 6 April 2010 (UTC)

Thanks I'll work on it, but what do you mean by probability function?--Ldussan (talk) 02:12, 9 April 2010 (UTC)

Wave function. In other words, the probability that an event occurs as a function of space and time. Be patient with me, I still don't know the mathematics :( --Hroðulf (or Hrothulf) (Talk) 05:48, 10 April 2010 (UTC)

I suggest you post it, and we can collaborate on it in the article.

I think the two sentences about wavefunction collapse prejudge the outcome of research on the quantum measurement problem (see also quantum measurement and Interpretation of quantum mechanics). From what I have read, I propose this opening to the final paragraph:

The pilot wave is one of the less popular of several alternative interpretations. One popular view regards the probability wave, not as Bohm's pilot wave, but as the actual photon, with no distinct "billiard ball-like" particle existing until the photon interacts with a detector. The probability wave, upon encountering an atom in the detector, collapses across space and time instantaneously to a single point in that detector/atom with a 50% probability.

There is also a possibility of confusing the reader with quantities in the final paragraph: are we discussing one pilot wave or two? (I know the answer, but I don't know how to copyedit the paragraph to make it more clear.)

--Hroðulf (or Hrothulf) (Talk) 07:21, 10 April 2010 (UTC)

Well as you know it is just one pilot wave but two distinct directions or wave vectors. I'm just not happy with it though. Gotta keep thinking. The double slit experiment is the classic explanation but it's just so long. I want something short, that grabs peoples attention, and gets them to understand the double slit experiment better.Ldussan (talk) 04:52, 21 April 2010 (UTC)

I suggest that you examine the Wiki image of the Whirlpool Galaxy and see if the details of that image can be turned into a modulated wave front after 25,000,000 light years of propogation.WFPM (talk) 20:20, 16 April 2010 (UTC) I think what you are getting at is that the galaxy in actuality doesn't look like that because after many years of traveling some sort of diffraction pattern alters its image from what it really is to what we see. Is that what you mean? --Ldussan (talk) 04:52, 21 April 2010 (UTC)

We talk of wave and particle natures to help general readers understand the concepts. Ldussan's proposal is not intended to convey the full modern physical description of the photon or of light. --Hroðulf (or Hrothulf) (Talk) 20:39, 17 April 2010 (UTC)

It sounds to me that he's talking about the details of a mathematical theory about light that is way over my head. And I'm looking at the blown up pictures of the Whirlpool Galaxy and wondering if he thinks his detailed description can explain how I'm seeing what I'm seeing. Consider the implication of a hypothetical pilot wave for each photon that is 25,000,000 years in radius. Do you expect to get much detail out of that?WFPM (talk) 21:01, 17 April 2010 (UTC) Maybe if you posted an image of the whirlpool galaxy in the photon article and labeled it "photons from the Whirlpool Galaxy" you'd get some alternate theories other than direct photon propagation, but they would be harder to justify while you're looking at the image.WFPM (talk) 21:18, 17 April 2010 (UTC)

Well fooey if WFPM doesn't understand it then it's probably not going to be of any help. WFPM I just want to help people understand how light can be a particle and a wave at the same time. Can you read it again and see if it makes sense to you. Hrooulf can you read it again since I've changed it. WFPM I'm not sure what you mean by the Whirlpool Galaxy photons. Try asking me in a different way on my talk page and I'll give you my best answer if I understand your question.--Ldussan (talk) 04:52, 21 April 2010 (UTC)

Ldussan, please post it to the article. To my mind, it reads much better than the wave-particle duality article.

I hate to dismiss a thoughtful enquiry into physics, but as far as I can tell, WFPM wants to engage in dialectic about the nature of energy: a good thing but there are other wikis for that. If I interpret the comment correctly, WFPM understands your description but disagrees with the standard model of quantum mechanics. There are some attempts to restate WFPM's question at User talk:WFPM#Whirlpool Galaxy.

--Hroðulf (or Hrothulf) (Talk) 09:21, 21 April 2010 (UTC)

I added it Hrothulf. Let's see what happens. WFPM i think if you studied quantum mechanics and the wavefunction you probably would understand your question and have your answer.Ldussan (talk) 05:35, 5 August 2010 (UTC)

I'm afraid that as far as the "Photon" concept is concerned, I can only see it as the description of the relative magnitude of EMR energy related to the method of EMR emission and propagation, and the only thing I can see is a stream of small polarized particles. And I don't do mathematics sufficiently to be able to use their inference to describe the properties of my limited concept. But it's too bad that we have to obfuscate the issue of both the EMR energy and/or the electromagnetic force vector issues by trying to name and simultaneously discuss 1 real physical entity that can accomplish either or both purposes. In politics they call that coopting an issue and it doesn't seem to be a very clarifying method of scientific discussion.WFPM (talk) 10:32, 5 August 2010 (UTC)

=Refraction

I want to throw in here a comment about refraction and note that I can't see any problem with the bending of a light beam within a refractive media, except for how it manages to speed back up and keep going in the right direction after it exits the media. Take a diamond media for instance, with a 40%c internal velocity of propogation.WFPM (talk) 18:13, 8 May 2010 (UTC)

WFPM I'm not sure what you mean here but a quick lesson in snell's law would probably clear this up. Regardless, take a look at your statement. "How does it manage to speed back up and keep going in the right direction after it exists the media" That is actually saying a lot. First of all, it doesn't have to go back in the same direction after exiting the media i.e. a lens or prism and it doesn't have to speed back up unless it's going from the media directly back to the original media i.e. air. What you in fact are really describing is a photon traveling through a uniform thickness piece of media such as a microscope slide or window. Air-window-Air. You are not describing Water-Air-Water or Air-Lens-Air. Hope that helps. Ldussan (talk) 16:37, 14 August 2010 (UTC)

What I was thinking about was the passage of EMR through a medium confined between 2 parallel planes where it enters and departs at the same angle. If it slows down upon entry, it would then have to speed back up upon exit. And the angle of refraction would be the same, only in the reverse direction. And I used the refraction index of a diamond as a presumably worst case scenario. I thought such an incident only resulted in a sideways displacement of the path of the EMR, and not a change in direction. I call this the "last atom paradox", because I don't understand why the velocity of the photon should speed back up. Am I missing something somewhere?WFPM (talk) 01:00, 16 August 2010 (UTC)

You seem to be missing that light travels at c always. However, inside of an interacting medium it isn't just light that's traveling. Commonly in a polarizable material the energy does not stay confined to the electromagnetic field. It will drive a polarization wave in the medium, which in turn drives an electromagnetic wave and so on. The energy sloshes back and forth between the two and you get the polaritons mentioned in the article. This quasi particle travels at less than c due to its effective mass. However, upon exit from the material, the electromagnetic wave resumes propagation without any attending polarization wave and the observed velocity will be c. 67.171.217.43 (talk) 18:14, 8 January 2011 (UTC)

The section Photons in matter has the phrase: "this polariton has a nonzero effective mass, which means that it cannot travel at c." Currently this links to a disambiguation page, effective mass.

Should it link to Effective mass (solid-state physics)?

--Hroðulf (or Hrothulf) (Talk) 09:28, 31 March 2010 (UTC)

Yes, it should. 67.171.217.43 (talk) 18:15, 8 January 2011 (UTC)

I attempted to address the WildBot dab issue, but ran into some difficulties:

• Degeneracy – "...are the degeneracy of the state i and that of j, respectively..." – Is this the same as "Degenerate energy levels"?
• Effective mass – "...this polariton has a nonzero effective mass, which means that it cannot travel at c." – I wasn't certain whether any of the topics on the dab article included this definition of "effective mass".

RJH (talk) 17:12, 18 April 2010 (UTC)

Good article on Nature: http://www.nature.com/milestones/milephotons/pdf/milephotons_all.pdf --Nevit (talk) 21:23, 2 May 2010 (UTC)

The mass of a photon is usually given as zero, but gravity bends light and also prevents it from escaping from black holes, therefore moving photons have mass. Anything with a finite mass when stationary has an infinite mass when traveling the speed of light. Clearly photons do not have infinite mass when traveling at the speed of light. They must therefore have a finite mass when traveling at the speed of light and an infinitely small mass (the zero figure that is usually quoted) when stationary. Can a figure be provided for the mass (or energy) of a photon in motion? —Preceding unsigned comment added by 82.71.43.37 (talk) 14:04, 1 June 2010 (UTC)

If the carrier of the electromagnetic energy were a stream of particles whose individual translational energy carrying capacity were the planck's constant value of 6.62x10e-27 erg second, then the particle would only have a rest mass of approximately 10e-47 grams, and which value would not be significantly increased by the relativity factor until it nearly reached the velocity of light.WFPM (talk) 20:08, 4 June 2010 (UTC)

To the original poster: yes. The volunteers at the reference desk sometimes answer this kind of question. --Hroðulf (or Hrothulf) (Talk) 14:03, 7 June 2010 (UTC)

In dealing with this problem we have to deal with the definition of the subject matter which, in this case is the photon. Since the formula for the photon (as an energy bundle) is 1: a Planck's constant value of action times 2: a multiple of that value that occurs per second. We are therefor left with the Planck's constant energy value as the only entity that could possibly have a mass value. And, unless we want to modify, (multiply? or divide?) the Planck's value by a mathematical number for some reason, we are thus left with the value of 6.62x10e-27 erg as the only value worthy of consideration. and as the amount of mass that will have approximately a planck's constant. amount of kinetic energy (of translation velocity), that is the amount of rest mass proposed for a particle that will have a planck's amount of translational kinetic when moving at the velocity of light.WFPM. (talk) 17:34, 7 June 2010 (UTC).

In the article Quantum teleportation it is reported as being possible for a photon of light energy to spontaneously change into 2 photons, with each photons having a lower amount. Thus, it must necessarily be possible for for the components of an electromagnetic photon to not only incrementally vary in energy content, but also be itself subdivided into 2 lesser energy value electromagnetic energy carrying packages.WFPM (talk) 02:29, 8 June 2010 (UTC)

To OP: Ignore all WFPM answers. They are all wrong so far. Dauto (talk) 03:30, 8 June 2010 (UTC)

I was not talking about a precise mass value, but only about the logic related to an approximate mass value. So if you have a better estimate of the precise mass value, I would very much like to know what it is.WFPM (talk) 04:44, 8 June 2010 (UTC)

The mass of the photon is zero. Dauto (talk) 23:16, 8 June 2010 (UTC)

And what about E = M times c Squared? That's Einstein's conversion factor.WFPM (talk) 01:58, 9 June 2010 (UTC)

That formula gives you the relativistic mass. The standard convention is to use the 'unqualified' word "mass" to refer to the rest mass and the rest mass of the photon is zero. —Preceding unsigned comment added by Dauto (talk • contribs) 04:15, 9 June 2010

In some sense the relativistic mass of a photon of energy E is E/c². However the unmodified term mass is generally understood to mean rest mass. In either case it has nothing to do with the photon's angular momentum. --Trovatore (talk) 02:28, 9 June 2010 (UTC)

The original question was about The mass (or energy), of a photon in motion. and didn't have to do with angular momentum. My answer had to do with the energy of a photon in motion, and with a supplementary estimate of the amount of rest mass that would have a planck's constant amount of kinetic energy of motion when translating at the velocity of light, with it being understood re the photon, that it would consist of some multiple of this mass value in proportion to the number of planck's constant units that were contained within any given photon.WFPM (talk) 03:22, 9 June 2010 (UTC)

Ah, I assumed you were talking about angular momentum because the only attribute of a photon that is anything remotely related to "one Planck's constant" is its angular momentum. "One Planck's constant of kinetic energy" has nothing to do with anything. In fact it doesn't even make sense dimensionally — the units of Planck's constant are energy times time. --Trovatore (talk) 03:39, 9 June 2010 (UTC)

Yes the units of the photon entity are energy (in planck's contstant units) times time. But when you're receiving them during a time interval, you are at the end of the time interval in the position of having accumulated the time interval's value multiple of the Planck's energy unit value.WFPM (talk) 23:39, 9 June 2010 (UTC)

And I understand that the questioner was trying to question and understand the paradox about how anything, (like the sun for instance) can convert matter into fusion energy and then emit the resultant EMR, without that EMR being the carrier of the lost mass value away from the sun? And to solve that paradox, he merely has to avoid it by allotting the lost mass value to the EMR radiation, on a per planck's constant or some other equitable basis, and let you purists worry about the (mathematical?) theory that nothing with a real value of mass can travel with the velocity of light.WFPM (talk) 12:19, 9 June 2010 (UTC)

It is not so complicated, and can be answered with some high school physics. BUT it really should have been discussed at the reference desk, not here, where there are advanced physicists who could explain how to calculate the actual gravitational bending of light. The relativistic mass-energy of the photon is, as you say, calculated by ${\displaystyle E=h\nu .\,}$ (The Planck relation, nu is the frequency, h is the Planck constant.) To express this as a mass that you can use to calculate gravitational bending, use E=mc². --Hroðulf (or Hrothulf) (Talk) 11:38, 10 June 2010 (UTC)

But if you're going to have a separate article about as important a subject matter as the Photon, which is the name of nature's method of emission and propagation of electromagnetic energy and/or angular momentum, it is pretty much a cop-out make a bunch of categorical statements about these matters and then refer interested questioners to some reference desk without further discussion. The challenge is, of course, to better organize the summary description of the photon in such a manner as to better segregate the different conceptual characteristics and functional properties of the photon entity. And with regard to this matter, it is to be noted that when the photon "quanta" was first proposed by Max Planck, as the means of explaining the photoelectric effect, there wasn't the confusion that presently exists about its properties. The confusion then arose when the photon name was also adopted to be the force carrying particle related to the the electromagnetic wave emission theory per the QM analysis method procedure, and when mixed in explanation with the planck's quantum theory has resulted in practically everybody becoming confused, including me.WFPM (talk) 17:58, 10 June 2010 (UTC)

The article is confusing. Lets discuss improvements here. --Hroðulf (or Hrothulf) (Talk) 20:08, 10 June 2010 (UTC)

Well as a means of improving the article, that partitioning of concept should be better explained, and then separately dealt with, say by separating the definition of the name "photon" into 2 different subconcepts, like one for photonA as a quantity of EMR per both of the emission and propagation theories, and the other one about photonB as a force carrying particle component of the standard model theory. Maybe if we can get the individual concepts of those two subdivisions straight, we might then learn what they have in common that would allow us to merge the concepts.WFPM (talk) 01:55, 21 June 2010 (UTC)

We might then consider what happens to mass value calculations when using the Lorentz-Fitzgerald adjustment formula which involves us in a mass value calculation formula where the denominator of the formula goes to zero as the square of the velocity value approaches the limit c. We then interpret this to mean that there is no possibility that any emitted minute mass particles can carry any of the lost fusion energy mass away from the sun, and results in our present conceptual paradox.WFPM (talk) 02:57, 21 June 2010 (UTC)

We might then note in this matter, that the paradox that we are dealing with in this case is a mathematical paradox, rather than a physical paradox, since by examination of the details concerning the energy required to result in an increase in velocity of a physical object it is apparent that that a small physical force could still increase the velocity of a physical object regardless of its velocity, but that the mathematical formula relative to the process implies that in infinite amount of energy would be required. However this is because our physical system does not have the physical capability to create the required velocity of motion of the required impulsive force in order to allow it to add to the momentum of the impelled particle. This can be determined by analyzing the incremental "square" area of a rectangle relative to the differential rate of increase of the two side measurements.WFPM (talk) 12:34, 28 June 2010 (UTC)

Yes, Gravity does bends light, could that be a secondary, indirect effect. Since, regardless of the nature of light massless or not; in the case of massless, something has to "carry" the massless electromagnetic wave. This way the medium, the vacuum does "Vacuum energy is involved". Therefore the Gravity effects the medium which the light travel through. Gravity bends that medium and bending light waves with it. On the other hand, if light has mass, then gravity effect light directly. Examples: gravitational lensing, event horizon, and interaction photon and other particles. An interesting point comes up if photons have mass, which if it does that should gives the universe a center of gravity. The universe should have a center of gravity since there is mass in the universe. Another argument arise is that photons or electromagnetic waves will not be able to escape gravity, Newton law of gravity and E = M C^2, This way the energy in the universe is within the universe borders. Since, energy so far is in the form of matter and electromagnetic waves. On the other hand the suggested existence of antimatter particles will be to make energy escape the universe at a certain rate. --e:Y,?:G 05:40, 6 September 2010 (UTC)

I feel the phrase "their normally straight trajectories may be bent by warped spacetime" is slightly misleading. Spacetime is bent, but the light still follows a straight path, even though it doesn't look straight from a distance. The key is that spacetime is non-Euclidian (and positively curved by mass). Saying the light trajectories themselves are bent might lead someone to believe that light is bent the same way as the path of a stone thrown in the air is curved by gravity. I hope someone can correct me if I'm wrong. TvN —Preceding unsigned comment added by 188.38.130.159 (talk) 19:27, 24 July 2010 (UTC)

does anyone know a suitable comparison for 'photons' and the phenomena of visible light in general when placed relative to, for example, lightning? i think it might be useful when thinking of 'photons' and other measurements that are very, very small to place them within a context. might also help to disambiguate a conceptual unit of energy (?) from the measurements of same energy? -6627. 131.230.224.28 (talk) 23:10, 1 August 2010 (UTC)

Lightning is one of the numerous categories of events which result in the emission of "photons" of EMR as the result of a physical activity of the atom or of it's electrons. A study of the visible light and/or other EMR emessions of this phenomena has still not exactly determined the character of the individual basic unit of EMR emission and/or propagation. A "photon" quantity of EMR energy emission is a variable amount of energy that must be able to stimulate a physical or chemical detection process with a very short time period, such that a numerical measurement of the EMR's energy intensity per unit time may be determined. Since the initial EMR energy intensity values can be modified subsequent to emission, it must be that all presently determined "photon" quantities of EMR still are a multiple quantity of the basic light energy units, whose basic physical identity has yet to be determined. In the case of lightning, the emitted light energy has to do with events of emissions related to the disassociation of water molecules in the path of the lightning strike.WFPM (talk) 21:02, 3 August 2010 (UTC) Thus the relationship of an event like lightning to a photon is like the relationship of a waterfall to a molecule of water.

My Wiley Atomic Physics book says that a barely visible level of sodium flame light, the photon intensity impacting a square centimeter of surface area is approximately 2 million quanta per second.WFPM (talk) 17:14, 8 March 2011 (UTC)

The first paragraph of the article states that this is zero. Since some of the references of the article put an upper bound on this rest mass, the first paragraph seems to be claiming more than is known. —Preceding unsigned comment added by 80.229.247.11 (talk) 16:30, 2 October 2010 (UTC)

You're dealing with the result of mathematical logic, which says that it always moves at the speed of light, it could not have a rest mass, because then while in motion, it would have infinite mass.WFPM (talk) 19:20, 2 April 2011 (UTC) So now we have the logical concept that nothing with rest can move with the velocity of light and yet it's possible for some portions of the universe to be moving away from us at the speed of light.WFPM (talk) 23:00, 2 April 2011 (UTC)

Elementary particle of light

WP:NOTFORUM

Since there is no such thing as an elementary particle of light, this discussion of the subject matter is never going to get off the ground!! What there is, is a concept of an elementary quantity of energy that can be emitted and absorbed by the electron as well as a medium of propagation that allows these quantities to travel over long distances. For a succinct and understandable discussion of this subject, I suggest you read the discussion in the Micropedia volume 7 section of the 15th edition of the Encyclopaedia Britannica.WFPM (talk) 15:48, 24 March 2011 (UTC)

your understanding of this topic is outdated by roughly a century. Dauto (talk) 04:10, 25 March 2011 (UTC)

Did you read the Micropedia article? And you don't believe the Photon is a quantity? Then how can it get to be more or less. I appreciate your quandary, because I'm trying to understand Feynman's QED too. But lets be fair and open minded. And thanks for the comment.WFPM (talk) 12:39, 25 March 2011 (UTC)

There is no need to read the Micropedia article. If what you related from it is an accurate reflection of Micropedia, that shows just how unreliable Micropedia is. An no, the photon is not a "quantity", and yes it's the "elementary particle of light" (although that's a very awkward way of saying things). Headbomb {talk / contribs / physics / books} 14:48, 25 March 2011 (UTC)

Well after you read in Feynman's QED about how a Photon can turn into a Positron + an Electron it's conforting to go back to my outdated Wiley's "Atomic Physics" and read the chapter on "The corpuscular nature of Radiant Energy" and see how this subject matter was originally approached and figured out. So if you want to keep the concept of a unit particle in the face of Feynman's discussion I can only wish you luck. In the mean time I think that the Microopedia type historical discussion would very much help in in getting people to get past the Fog Factor in your first paragraph.WFPM (talk) 15:19, 25 March 2011 (UTC)

Yes, QED describes how three elementary particles - the photon, the electron, and the positron - interact with each other. Dauto (talk) 16:02, 25 March 2011 (UTC)

Technically, QED describes the electromagnetic interaction of every particle, not just electrons and antielectrons, with the photon as a mediator. Feynman just focuses on electrons and antielectrons because everything behaves as electrons and antielectrons when it comes to the electromagnetic interaction. Headbomb {talk / contribs / physics / books} 17:02, 25 March 2011 (UTC)

Okay. So now we've got the atomic nucleus with it's protons and neutrons. And we've got the protons interacting with the electrons by interchanging photons. And then the electron starts swallowing photons and reemitting them, so it must be larger than the photon. And then the photon starts emitting electrons and positrons. And as a quantity of energy I can go along with that, except for the identity of the electron, which I have defended mightily in my talk section. But Einstein said that a photon was an amount of radiant energy minus an amount of work, and that could still do additional work. So I cant see a particle identity in any of this, just varying quantities of radiant energy.WFPM (talk) 17:34, 25 March 2011 (UTC)

The electron is neither larger nor smaller than the photon. They are both elementary particle which means they are not made of smaller parts. The proton for instance isn't an elementary particle because it is made of three quarks. When an electron emits or absorbs photons, the photons are being created and destroyed. I think you may be quoting Einstein out of context and that quote is likely from his work on photoelectric effect from 1905, more than a century ago and before our modern understanding of elementary particles based on quantum field theory had been developed. Dauto (talk) 17:47, 25 March 2011 (UTC)

Well I base my concept on the work function of a photoelectron, whose stopping potential Ev = plancks constant times frequency minus the work lost in getting out of the emitting material, whose validity I think is still current. And I'm pretty sure that the electron is much larger than most photons that I know of. But the electron that I know of has an identity and a rest mass, and a limit quantity related to each atomic nucleus. And I find no such limitation on the identity or quantity of photons related to the atom other than that they transfer energy as needed. So I think that your attempt to establish the (1926)word "photon" as an individual identity of anything other than a "package" of energy is obscuring the rationale of the functional physical processes of the atom.WFPM (talk) 18:30, 25 March 2011 (UTC)

You may be pretty sure that the electron is much larger than most photons that you know of, but that only makes you pretty wrong. The only meaningful way to measure the size of a elementary particle such as the photon or electron is their wavelength and elementary particle may have any wavelength depending on its momentum. See De Broglie relation. You are also wrong about the electron having an identity. See identical particles, and note how both the electron and the photon are included in the list of particles. Also note the important distinction that electrons are listed as a fermion while photons are listed as bosons. That accounts for some of the difference in behaviour between them that seems to be confusing you into believing that photons are not particles. The photon also has a rest mass (which happens to be equal to zero). Dauto (talk) 19:05, 25 March 2011 (UTC)

Well there's nothing like introducing humor into a situation, like that a particle can have a rest mass of zero. And I'm like Socrates, in that I know that I don't know. And as I read on through the article, I learn more about what I don't know. But I know about radio vacuum tubes, and thermionic emission and radio theory, and simple things like that and I'm convinced about electrons. But an individual photon? With a wavelength? Then it has to both manage the light energy and the magnetic energy at the same time, I guess because the 2 waves run perpendicular to each other, except that they don't exist because the thing is a particle? I'm just sorry the subject matter cant be split up and analyzed like it was originally before we loaded all the additional properties onto the photon.WFPM (talk) 19:49, 25 March 2011 (UTC)

Well, physics can be quite unintuitive at times, specially modern physics. But that is the modern view. Photons are elementary particles with zero rest mass and with a wavelength that can be much larger then the size of the atom that produced them. The typical size of an atom is given by the Bohr radius, or a few Bohr radii (10^-10m)while the typical wavelength of the radiation being produced by atomic transitions is given by the inverse of the Rydberg constant or a few of them (10^-7m). So the photon is about 1000 times larger than the atom that absorbed it. Talk about a minnow swallowing a whale. Dauto (talk) 14:08, 26 March 2011 (UTC)

Yes and in the Wiley Atomic Physics the De Briglie "matter waves" are discussed and it is noted that the wavelength of an electron can vary from 1.22x 10^-8cm for a 100volt electron to 7.27cm for a 1 volt electron. They also show the wave of a 45.9 gm golf ball as being 5.71x10^-32cm. But the concept is not that the entity has that dimension, but that the "associated" or "director", or "ghost" wave that can be detected have that wavelength dimension. So particulate entities can be associated with electromagnetic waves. But that doesn't mean that the electromagnetic wave is a particle.WFPM (talk) 14:42, 26 March 2011 (UTC)

Now you're just splitting hairs. Ultimately it comes to what is the definition of a particle and does the photon meet that definition? For us physicists it makes sense to go with the most useful definition. You would be hard pressed to find a definition that both includes the electron and excludes the photon while still being self consistent because photons and electrons behave in such a similar ways. As you said, particulate entities can be associated with electromagnetic waves and those particulate entities are useful enough a concept to deserve a name. Physicists call the photons. Dauto (talk) 18:12, 26 March 2011 (UTC)

Yes but!! In understanding the atom we need the neutrons and the protons (and the deuterons) plus the electrons as entities to formalize a physical concept. Which we haven't yet adequately done!. So now we have the energy management problem which involves radiation. And you want to say it's particles! But like Feynman said, it's a mad mad world when it comes to radiant energy concepts. And I can't see particles of energy like particles like I can see the protons, neutrons and electrons. So I want to know about radiant energy (and magnetism) but I don't see them as particles. They're more like mathematical unit entities. And I don't see the atom like some kind of sphere inside a cloud of electrons. I know it has a structure, that I'd like to know better. So if you're willing to have everything be some kind of particle, I guess you can get away with that. But I don't see how you're ever going to come to a rational concept of what the operational functions of the atom are if it were big enough for you to see them. That's one of the amusing problems that Feynman was having with his diagrams.WFPM (talk) 21:15, 26 March 2011 (UTC)

Well, you may have some trouble imagining radiation as made of particles. That's a common difficulty. That is just a conceptual inertia - that is a natural tendency to resist changing the way we visualize concepts. Dauto (talk) 05:02, 27 March 2011 (UTC)

I can live with that! What I think we're discussing is the article explanation where you start out saying it's a basic particle (which it isn't) and then go to mix up the two concepts within the discussion. And all in the first paragraph! That's why I like the Micropedia article much better, and I think you should read it. Please excuse my candor, because I'm interested in helping accumulate information in this subject matter.WFPM (talk) 12:58, 27 March 2011 (UTC) It is interesting to note that if a person can get by the first paragraph, and down into the historical section, the article then becomes an interesting discussion in advancement in concepts and experimental data. So why cant you intrigue people in the beginning?WFPM (talk) 13:27, 27 March 2011 (UTC)

If you post a link to the article you are talking about, I read it and tell you what I think of it. Let me just reiterate a very important point: The photon is indeed a particle. That's what physicists call it and that's what any good encyclopedia should call it. Dauto (talk) 14:21, 27 March 2011 (UTC)

I don't know how to post a link to the EB (15th edition, 1979). But the article starts out with "Photon, a quantum, or minute package of electromagnetic radiation. The concept originated (1905) in Einstein's explanation of the photoelectric effect....", which is a reasonable way of leading you into a historical development of the subject to whatever level of complications that you're interested in. Then it has reference to about 40 sub articles for further interests. The trouble with categorical statements (mine included) is that they start out by turning a lot of people off.WFPM (talk) 15:34, 27 March 2011 (UTC) This is consistent with my theory that all information falls into either and maybe both of 2 categories: namely : 1 FACT or 2 OPINION. And the thing that I have to do first is properly classify any information into its proper category/s. And, your initial assertion that a wave is a particle defeats my ability to do what I want to do, which is to sort out the facts.WFPM (talk) 16:07, 27 March 2011 (UTC)

Note that if you consider the image of the light/magnet wave in the article, you understand Maxwell's concept that that the waves intensity (height) is the important factor. But you can also see the concept where the energy of the photon is contained in the integral of a certain number of the waves, and which may be the more important energy transport factor. And the 1 second of accumulation value seems to be the important time period for the photon concept.WFPM (talk) 16:31, 27 March 2011 (UTC)

Is the picture I added the one you are talking about? If so, let me start by clarifying that it is not called a "light/magnet wave". It is called an electromagnetic wave AKA simply as light. Unfortunately that picture is inaccurate because it doesn't capture the particle-like nature of the photon. It is still a very useful picture but the reader must keep in mind that it gives only a partial view of the true nature of light. I know that the wave particle duality may be quite hard to understand and that it can be frustrating trying to sort fact from imagination. You are not alone there. That is a normal part of the process of learning physics. Dauto (talk) 19:44, 27 March 2011 (UTC)

Yes. And I studied electrical and electromagnetic waves in college as part of my EE degree. But this is the first one where I've seen the electrical and magnetic wave forms orthogonal to each other, which sounds like a good idea, if true. But the electromagnetic (radiant) energy formula involves a certain number of waves per second, so you would have to cut up the energy of the transmission into links, like in sausage, with the 1 seconds quantity being the photon value quantity. And, of course, the higher the frequency, the more the energy. And how you do that is a good question. And according to Feynman, the probability of the location of the presumed photon particle is somewhere near the middle of the length of the accumulation time period. I made my best grades in Physics.WFPM (talk) 21:22, 27 March 2011 (UTC) I'm in favor of imaginative ideas. But the important things are facts. And remember that we can take pictures in milliseconds, which complicates the concept. So in order for your photon particle to make available the photon quantum quantity of energy, it's going to have accumulated 1 second's work into one point, and then make it available at any instant in time during that second within a say 1 millisecond time interval!

There is no getting around the fact that light must exhibit the properties of individual particles during low-intensity Compton scattering. That fact that these "particles of light" are not conserved like electrons, only means light is different kind of beast (one where the particles are not conserved, and can melt and split at will).

However, there's no getting around light's particle-like aspect. It isn't just radiated and absorbed in lumps-- sometimes it interacts with electrons in lumps, and there's no other way to explain it. However, the lump idea gives the correct quantitative answers. We call those lumps of light "photons." If you call them lumps of light that would be fine also, but they already have a shorter name!

After Compton, physicists quit trying to say that light is one thing or the other (wave or lump), and simply said that it's like nothing we know of at our scale. It's not like water-waves, and it's not like baseballs. It exhibits interference and a sinusoidal phase behavior like water and sound waves. But it interacts like electrons sometimes as lumps. You just have to accept that both things are true at the same time, and that it's not visualizable, since there's nothing like it at our normal level of experience. All particles of low or zero mass behave in observably wave-like ways, sometimes-- that's the message of quantum mechanics. Particles with rest-mass behave more lumpily than particles with no mass. And fermions behave more "lumpily" than do bosons. Light is a boson and has no mass, so it's hard to get it to display lumpiness except at high energies. Gravity is even worse, and has yet to display any lumpy properties at all (physics has failed to find a good lump theory of gravity). For all the rest of the forces and particles of nature, however, the dual-aspect is there, and can't be avoided. SBHarris 01:04, 28 March 2011 (UTC)

Well thank you for the comment. But I'm not interested in light, in itself, what I'm interested is what light is as related to its function in connection with the atom. See I have a one track mind. And I want to know about the atom. So I'm limiting myself to factual considerations related to the atom. And I think that radiant energy is the energy carrying entity related to the atom's functions. But I don't think they are particles like energy quanta that we determine from other applications. They are entities that make up gamma rays between nucleons and x rays between electrons and nucleons, and of course light, and infra red. And I think that they are all made up of the same raw material and must have originated in the same manner (that's a guess) so if I can get the facts adequately organized I should able to formulate a single rational concept. And it wont be a dual matter/wave concept. So I'm not interested in a special name for this or that light energy quantity because naming doesn't accomplishing anything, except for communication purposes. That limitation fact keeps me in the conceptual realm of a small basic matter particle because I'm like Dr Le Bon and can't conceive of energy without matter. That's why I am considering the possibility of a Planck particle as a basic energy unit and could use some help. But there has to be a rational system of organization of the knowledge about the atom that's better than what we have and I see you looking for that too, and lots of luck!.WFPM (talk) 03:42, 28 March 2011 (UTC) Have you determined the amount of energy in the basic lump? Note that if there were a "planck particle" with an energy content (Mv^2/2) of (6.6 x 10^-27 erg), and that was moving at (10e^10.5 cm/sec), It's v squared value would then be 10e^21 cm^2/sec^2 and its mass would be (2 x 6.6x10^-48 grams), or say 10e-47 grams! So what we would be doing is managing the activities of various quantities of 10e-47 gram photon (planck) particles. How about using the string theory to accomplish that?WFPM (talk) 18:22, 1 April 2011 (UTC) I Guess that our main problem with this would be the management difficulty related how the appropriate number of planck particles could be fitted to the determined photon quantities.

In a vacuum Photons are moving constantly with lightspeed. Thus they will not get older, correct ? HH 93.216.65.141 (talk) 11:32, 21 May 2011 (UTC)

Layman here. What exactly is the nature of the photon's frequency. Wherever I go, I see frequency defined with a relation to momentum or wavelength or energy, and when I followed these concepts, they are defined with relation to frequency. So what is it? Is it a property of a bunch of photons as detected by a photon receiver? Does a single photon possess a certain frequency? If so, what's "ticking" in it according to the frequency? If two photons move towards the same direction at the speed of light, what exactly makes them different? What makes one "red" and the other "blue"? 46.116.141.119 (talk) 09:36, 2 June 2011 (UTC)

Such questions are better the reference desk, unless you are proposing improvements to the article. --Hroðulf (or Hrothulf) (Talk) 15:36, 26 August 2011 (UTC)

Actually, it might be a worthy entry to the article, as each photon has a specific energy level/frequency/"size". Hence, perhaps one better able to explain that item and also delineate when a frequency of EM energy is no longer a photon, but merely a RF wave.Wzrd1 (talk) 05:03, 28 September 2011 (UTC)

If I was trying to answer this at the reference desk, I might try to reproduce Feynman's description of a spinning arrow representing the probability of an event. For different frequencies of light, the arrow spins at a different speed. Meanwhile the color red or blue is perceived when the energy of the photon matches the difference between energy levels in the red or blue receptor in your eye. I, personally, don't think I could make such an explanation concise enough, or clear enough to a layman, to be a section of this article.

Perhaps readers will find it worthwhile if somebody attempts such an explanation.

As for differentiating between a photon and 'merely a RF wave', I can't see how an encyclopedia can do that, as the probability equations of QED give the same results as wave equations, except for specially designed experiments. I understand that such specially designed experiments have found evidence of photons at frequencies as low as those of microwaves / radar, yet theoretical physics doesn't, as far as I know, offer a concept of 'merely a RF wave', to cover lower frequencies such as AM/MW and LW radio. In theory, I think that all those waves are also photons.

--Hroðulf (or Hrothulf) (Talk) 17:06, 28 September 2011‎

I'm getting very confused reading you, it's easy to define a photon. It's intrinsically timeless, massless, expends no energy in its 'propagation', always following geodesics. Has no charge, meaning that it does not bend to any electric field. Will leave a recoil as it propagates from a source, due to the demand of symmetry of 'conservation of momentum'. Does not exist until measured, that just means that if compared to a ball coming at you in the air, a photon is no ball and does - NOT - leave itself to being observed, except in its annihilation. It has no acceleration, can't be defined as having a 'rest frame' as it never can be 'at rest', always moving at the invariant speed of 'c'. Unchanging in, and from, any frame measured. It is one part of the wave/particle duality that signifies light/radiation. Although some, using strategically placed out clocks in a accelerating frame, may find 'c' to be questionable comparing A-B against B-A that is a expression of 'gravity' acting on the clocks, as per Einsteins GR (equivalence principle). Even if ignoring that you will find light to always show 'c' locally, when measured. And if you want to discuss the whole spectrum use 'radiation' please. Leave hypothesizes outside the definitions, you might want to make a section called 'alternative hypothesizes' if you can't control those introducing 'pilotwaves', quantum teleportations, tunnelings, entanglements and whatever..

As Birge said "I understand what people are getting at when they want to say a photon can be expressed by a 3D k-vector and a polarization state, but nobody has offered a good explanation of why it's helpful to readers to consider the number of free parameters in a spatially infinite plane wave to be the degrees of freedom of a single photon." It would be highly theoretical, not defined experimentally, and embarrassingly stupid to put such into a factual description of what we think us to actually 'know' about a photon. A photon does not exist, except in the recoil, aka symmetry with SpaceTime, and in its subsequent annihilation. The ball can also be defined this way actually, if you consider what communicates its motion, namely 'light'. But that would just mess with peoples heads. Concentrate on what we know, not what we guess.

"What exactly is the nature of the photon's frequency" The photon does not have a frequency. It has a energy, but that energy can through Einstein explanation of the photoelectric effect, via Planck's constant 'h', (E=hf) be presented as that 'e'nergy of a photon is proportional to a waves 'f'requency. So there is a equivalence through that. And equivalences and symmetries are important phenomena in SpaceTime, but intensity and amplitude doesn't apply to a single photon at all. If you look at Maxwell's equations light becomes a electromagnetic radiation consisting of oscillations (waves) in the electric and magnetic fields, 'perpendicular' (at a right angle) to each other. Waves describe polarization, refraction, interference (quenching and reinforcing itself, via two waves interfering) etc, but they do not tell you about photons. And that's where 'equivalences' becomes important. And so this, to me that is, is all about trying to find a common ground for the concept of photon fitting the concept of waves. This is a good description of that.

"The frequency of the oscillations in a beam of light is proportional to the energy in each photon, as demonstrated by the photoelectric effect, and in the case of light is related to the color of the light. The intensity of the beam is proportional to the number of photons. The polarization of light (that is explained by Maxwell) is related to the quantum-mechanical concept of spin. You can see the photon as a little top spinning around an axis that coincides with the direction of propagation. But while in classical mechanics an object can spin only in one direction at a time, in quantum mechanics you have the paradoxical and counter-intuitive fact that an object can spin lets say clockwise and counterclockwise at the same time.

It is like having two "realities" existing at the same time. It takes a while to get used to this new idea and to accept it. A photon spinning in one direction corresponds to a rotating electric field, and to what is called circular polarization. A photon that spins in both directions at the same time gives you, under the right circumstances, plane polarization, which means the electric field is oriented always in the same direction."

Two realties huh :) Or a particle/wave duality. — Preceding unsigned comment added by 178.30.31.138 (talk) 15:56, 13 June 2011 (UTC)

Yoron. 178.30.5.142 (talk) 01:19, 13 June 2011 (UTC)

The article currently says "The photon is currently understood to be strictly massless, but this is an experimental question." I don't understand what this means. In the context of a theory that predicts/explains a certain mass for a photon, it is a theoretical question too. There is even a footnote in the article that says "The mass of the photon is believed to be exactly zero, based on experiment and theoretical considerations described in the article." 81.159.105.50 (talk) 11:37, 15 June 2011 (UTC)

How about this?

The photon is understood to have a rest mass of zero, within current limits of experimental accuracy.

--Hroðulf (or Hrothulf) (Talk) 13:53, 15 June 2011 (UTC)

Thanks, that's much clearer. I had no idea that's what it meant. 86.181.168.5 (talk) 17:25, 15 June 2011 (UTC).

There are some subtle issues involved here. You need to have some theory to be able to correctly interpret experimental results that set the limit on the mass, and the experimental limit then depends on the assumed theory, see e.g. here. Now, if you consider the charge of the photon and take the lesson from the case of the mass serious, then this means that one cannot make any statements on limits on the charge. Count Iblis (talk) 17:44, 15 June 2011 (UTC)

Thanks. Perhaps we should be more circumspect. Is this even half way right?

Informally speaking, the Standard Model considers the photon to have a rest mass of zero, which is confirmed by experiment within current accuracy limits (as of 200x?) .

--Hroðulf (or Hrothulf) (Talk) 20:28, 15 June 2011 (UTC)

The only comment I would have about this is that "informally speaking" seems an unexpected caveat for a statement that seems so straighforward. The expression is usually used in cases when the full explanation is long, technical and/or hard to understand. In this case, my question would be "does this mean that, formally (and by implication more correctly), the model considers the proton to have a non-zero rest mass, and, if so, why doesn't the article just say so?" 86.181.168.5 (talk) 21:01, 15 June 2011 (UTC).

I suggested the caveat, because more formally speaking, quantum electrodynamics (QED) can be modified to determine an upper bound on the photon rest mass based on experimental data. By the way, my latest proposal is

Quantum electrodynamics, the generally accepted theoretical foundation of photon physics, considers the photon to have a rest mass of zero. FOOTNOTE: As of 2010, the upper bound of the photon rest mass is an unsolved problem, and depends on modifications of QED to interpret experimental results and astronomical observations. In 2010, the Particle Data Group indicated that the upper bound was in the range 3 x 10^-27 eV to 1.4 x 10^-7 eV depending on theoretical and experimental methods. http://pdg.lbl.gov/2010/listings/rpp2010-list-photon.pdf

Any use? --Hroðulf (or Hrothulf) (Talk) 18:34, 16 June 2011 (UTC)

By the way, the 2010 PDG listing suggests that the article section Photon#Experimental checks on photon mass is somewhat out of date in detail, though not in spirit. ---Hroðulf (or Hrothulf) (Talk) 19:08, 16 June 2011 (UTC)

There is no charge that I've seen proved for a photon? It does not get influenced by a electric field. Before stating something not proven you would do well to admit to it being a theory. Or is this a place for people wanting to prove their pet theories? Making them real by stating them in a wiki? QCD gives no charge to a photon, neither did Einstein, neither do QED as far as I've seen? As I said, it's simple to prove a 'charge' if it would exist. Just measure a 'photon propagation' through a electromagnetic field and show me that it will be 'bent'. And as you wrote, it has no known mass. What it has is theoretical limitations for what a 'possible mass' might be. A little like defining mine theoretical limitation of being the one and only Santa. It's very small, if I may add so :) I find it simpler if you use what we know first, and then go into theory. I will not take anything less than a experiment proving a charge seriously. And that one, does not exist.

Yoron. 178.30.123.207 (talk) 22:36, 16 June 2011 (UTC)

I agree, Yoron. One goal of physicists is to put upper limits on things that probably don't exist at all. PDG 2010[1] that I mentioned earlier says that it is very tiny indeed ( < 8.5 × 10^{− 17} e or much smaller), and appears (to me) to question the whole enterprise of measuring an upper bound:

"OKUN 06 has argued that schemes in which all photons are charged are inconsistent. He says that if a neutral photon is also admitted to avoid this problem, then other problems emerge, such as those connected with the emission and absorption of charged photons by charged particles. He concludes that in the absence of a self-consistent phenomenological basis, interpretation of experimental data is at best difficult."

I don't see why Wikipedia can't attempt to summarize honest peer-reviewed attempts to understand how accurate humanity's knowledge of photons is, but yes, the article should make clear to a lay reader that neither rest mass nor charge have been observed. (My guess is that it's an open question what they would even mean if they did exist) .

I also agree that the talk page has become a home for pet theories, though I don't think this conversation has gone down that slippery slope.

--Hroðulf (or Hrothulf) (Talk) 14:54, 18 June 2011 (UTC)

That's right, we can say that neither charge nor mass has been observed, that some limits have been placed on these quantities based on their non-observation, but also that drawing conclusions from the non-observations is complicated in case of the mass and especially in case of the charge because of model dependence. In case of the charge, there are no models that are not already ruled according to Okun, so there is no meaningful way to give upper bounds for the charge. Count Iblis (talk) 15:21, 18 June 2011 (UTC)

We can also say that the ability of photons to transfer momentum has been observed, see Solar Sail. Can we have a transfer of momentum without having a mass?WFPM (talk) 17:59, 12 March 2012 (UTC)

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.68.171.143.254 (talk) 16:55, 16 August 2011 (UTC)

This book includes papers from 1923 on and uses the word photoelectrons for the one particle and electrons for the other. All he did was invent a new word by combining two words then later he shortened the new word so as not to confuse it with the electron by eliminating electro. The new particles thus became the photons from a shortening of photoelectrons. Read the book. — Preceding unsigned comment added by 68.171.143.254 (talk) 14:17, 26 August 2011 (UTC)

Which book? --Hroðulf (or Hrothulf) (Talk) 15:52, 26 August 2011 (UTC)

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. All he did was invent a new word by combining two words then later he shortened the new word so as not to confuse it with the electron by eliminating electro. The new particles thus became the photons from a shortening of photoelectrons. Read the book. — Preceding unsigned comment added by 68.171.143.254 (talk) 12:48, 29 August 2011 (UTC)

This reference [2] says that in 1923 Compton called the photons "energy quanta." I'm quite sure he never called them "photoelectrons" because photoelectrons are electrons (that are ejected by light) and not photons. No danger of anybody confusing the two. By 1926 Gilbert Lewis had already proposed the word "photon." If Compton does use it in a 1926 book for a light quantum, you have to show it was before Gilbert's paper came out. SBHarris 01:40, 22 October 2011 (UTC)

This is the year that Arther H. Compton won the Nobel Prize for proving that the Photon exists from his 1923 work. He shortens photoelectron to photon by eliminating electro. — Preceding unsigned comment added by 12.97.244.195 (talk) 17:59, 21 October 2011 (UTC)

Thanks, I missed that when reverted your first try. Adding a word in the "edit summary" box helps avoid such reverts. Cheers. Materialscientist (talk) 00:36, 22 October 2011 (UTC)

1925

From the Wikipedia article Wave-Particle Duality the year 1925 was when light quanta were first called photons. Dsmith7707 (talk) 19:55, 9 November 2011 (UTC)

Unfortunately, the Wave-particle duality article only mentions this date in passing, and doesn't offer a reference for it. --Hroðulf (or Hrothulf) (Talk) 19:28, 10 November 2011 (UTC)

This page is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page.

The following sentence is missing a reference: """Photons inside superconductors do develop a nonzero effective rest mass; as a result, electromagnetic forces become short-range inside superconductors.""" — Preceding unsigned comment added by 213.136.58.100 (talk) 10:40, 29 July 2012 (UTC)

 tagged - DVdm (talk) 10:54, 29 July 2012 (UTC)

I removed a longish paragraph in edit http://en.wikipedia.org/w/index.php?title=Photon&oldid=509261917 just now. Please see http://physics.stackexchange.com/a/34966/8841 for my description of why. In addition to what I wrote there, you can read discussions of how the EM waves ARE very close to "Schrodinger" wave functions in ^[1], and ^[2]. Briefly, a "complex" electromagnetic field is used as a matter of course in quantum electronics and engineering, in the dynamic field case, the imaginary part of the dynamic electric field is within a constant of the dynamic magnetic field (and vice verse). The localization of the photon is completely possible, but the localized states cannot be stationary in free space. But is a car moving 100 kph any less localized at any instant in time than a car parked?

Mwengler (talk) 15:50, 26 August 2012 (UTC)

Regardless of validity of the paragraph removed (and reinserted by an unreg), the electromagnetic field is not a wave function. This does not preclude the photon to have its wave function, but one just has to realize that it is not the same as a field strength, in the same way as a quantum mechanics is not a quantum field theory. There exists a simple thought experiment to check the difference. Wave function is a quantity those linear combinations correspond to quantum superposition. If a particle's wave function is, say, doubled (in some piece of a space), this does not mean that we now have more particles. And what does a doubled field strength mean? IMHO it means 4 times more photons. Incnis Mrsi (talk) 10:18, 29 August 2012 (UTC)

If a (bosonic) particle's wave function is doubled in all of space, you have four particles instead of one particle, just as with the photon. I am not claiming that the E-field is the Schrodinger wave. I am claiming that the exact same thing that allows a wave packet description of the electron to demonstrate so clearly the Heisenberg uncertainty principle allows it to be demonstrated for the photon. The photon cannot be localized and stationary, but every time a photodetector is triggered, it is with a photon quite localized in x, y, z, and t. If one of the fans of non-localisability of photons could explain the very obvious and trivially achieved "real" localization of em energy in that context, that would be very helpful.

On a related subject, just how much should I respect an uncommented untalked about undo by an unreg? Mwengler (talk) 19:23, 29 August 2012 (UTC)

In a wave the particles do not displace. But according to this article EM waves have photons and they displace at speed of 3.0 X 10⁸ m/s. I assume that photons do not travel but only show their energy through vibrations like ordinary matter. So only energy is travelling at the speed of 3.0 X 10⁸ m/s. Therefore the whole universe would be filled with photons and they show different energy intensities by their vibrations just like ordinary matter. I think all should consider this. — Preceding unsigned comment added by G.Kiruthikan (talk • contribs) 11:07, 2 February 2013 (UTC)

[3] This (second) IP addition

1. is redundant and off-topical (I already said: the infobox shows the invariant mass);
2. written (and added) in poor English;
3. uses a translated text from Spanish Wikipedia without attribution.

If one is willing to investigate what “Algunas fuentes” should mean in Spanish and to fix the IP’s “wavelenght” and so, then feel free to reintroduce it, but in some less prominent spot than the infobox. BTW, in es:Fotón this footnote was referred only from the body text for many years, and only recently was included to infobox. Incnis Mrsi (talk) 05:48, 14 April 2013 (UTC)

The photon mass is already discussed in the Physical properties section, with a nearly identical footnote, which I suspect is where the Spanish version came from. There's no need for an attempt to translate the footnote back (even if that was successful), and there's no need to link it in the infobox. The word "Mass" in the infobox already links to Invariant mass, and readers can be expected to look in the main article for further details and explanations regarding the material summarized in the infobox. — HHHIPPO 10:16, 14 April 2013 (UTC)

Thanks Hhhippo, now I learned that they accustomed to infringe Wikipedians’ copyright in both directions, and probably inside their wiki as well. Sorry, I can’t deter myself from this attack. Incnis Mrsi (talk) 10:38, 14 April 2013 (UTC)

Would it be acceptable for me to add [4] to the other references page section? The work is a significant advance with relevance to many other pages. Reference to it in those pages would augment this project's value. HCPotter (talk) 08:25, 23 October 2011 (UTC)]

• Talk:Planck's law(-->Standard photon)[[[User:HCPotter|HCPotter]] (talk) 10:20, 13 November 2011 (UTC)]
• Talk:Photon(-->Photon Field)(HCPotter (talk) 10:25, 25 December 2011 (UTC))
• Talk:Doppler effect(-->Photon volume) (HCPotter (talk) 10:18, 18 December 2011 (UTC))
• Talk:Hubble's law(-->Photon expansion)(HCPotter (talk) 08:55, 22 January 2012 (UTC))
• Hubble's law(-->Acceleration)(HCPotter (talk) 09:32, 19 February 2012 (UTC))
• Talk:Physical cosmology(-->Photon effects)(HCPotter (talk) 09:06, 26 February 2012 (UTC))
• Physical cosmology(HCPotter (talk) 08:47, 4 March 2012 (UTC))
• Talk:EPR paradox(-->Hidden variables)(HCPotter (talk) 08:03, 11 March 2012 (UTC))
• EPR paradox(-->Hidden variables) (HCPotter (talk) 08:11, 18 March 2012 (UTC))
• Talk:Dirac equation(-->Pair production)(HCPotter (talk) 08:43, 1 April 2012 (UTC))
• Dirac equation(HCPotter (talk) 09:19, 8 April 2012 (UTC))

This article proposes a new theory, but is published in Apeiron, which is not indexed by Web of Science (thus has no impact factor) and is considered as a "dissident journal". We need reliable secondary sources establishing the notability of this theory. Materialscientist (talk) 08:34, 23 October 2011 (UTC)

Withouth reading the paper in detail, (I pretty much stopped when I saw this was published in Apeiron, which isn't considered a reliable source by our standards), the jist of it is that this is original material, thus shouldn't be included per our policy on original research. First get those claims published in a reputable journal. Then it needs to be reviewed/endorsed as mainstream by the physics community. Then it could be included. But not before. This is not a judgment on whether your paper has merits or not, simply that its inclusion in Wikipedia would be premature at this point in time. Headbomb {talk / contribs / physics / books} 10:43, 6 November 2011 (UTC)

On 17:11, 25 May 2013, I gave a contribution which was reverted by Materialscientist on 18:18, 25 May 2013‎ by the reason Not a reliable source yet. Maybe in a few years. I wish to inform that an English translation version of the source reference was posted on http://arxiv.org/abs/1305.3602 (same link of the cited reference). Now anyone can read and verify that the physics therein is consistent and supports that the photon may be composed by pairs of flux quanta. The source reference at arXiv.org reproduces the version accepted for publication on the next issue of Rev. Bras. Ens. Fis., ISSN 1806-1117 (print), 1086-9126 (online). It is a peer reviewed publication, the source reference for this add to Photon at Wikipedia was submited to analysis under neutral point of view and fully accepted. User:Celso ad 23:27, 5 June 2013 (UTC)

So now it has become a wp:primary source, but we need wp:secondary sources. Is the paper significantly cited elsewhere yet? - DVdm (talk) 07:02, 6 June 2013 (UTC)

Sure, I realize that author(s) of Photon #Physical properties thought about the momentum representation but forgot to say it explicitly. BTW, they actually forgot polarisation which is not derivable from a 3-vector, is it? But for a person not advanced in quantum mechanics all this should look very confusing, in any case. Imagine an analogous text for an arbitrary massive particle ¤ (such as electron, proton, neutron, pi-meson…):

Would it be useful for a reader who does not know what is a quantum state, wave function, and Fourier transform? Incnis Mrsi (talk) 08:56, 6 June 2013 (UTC)

This was a recently created article that, based on it current content and sourcing, might be better merged here. Do folks have any particular objections to including its material here? I, JethroBT ^{drop me a line} 16:38, 27 September 2013 (UTC)

In my opinion, Photon molecule should definitely not be merged here. It is not particularly relevant to this article, and if it were to be merged, there would be hundreds or thousands of other esoterica that would qualify to be merged on the same criteria: photons are fundamentally involved in almost every physical phenomenon that we observe. — Quondum 17:30, 27 September 2013 (UTC)

Agree. PM is a new form of matter, not a detail to a particle. I'll be adding more detail over the coming days. Lfstevens (talk) 17:52, 27 September 2013 (UTC)

What about Photonic matter? — Reatlas (talk) 02:58, 28 September 2013 (UTC)

Every photon is a form of matter. Lfstevens (talk) 07:36, 28 September 2013 (UTC)

I have proposed that Photon molecule be merged into Photonic matter, since this was mentioned by Reatlas. I feel that the original merge proposal into Photon should be removed. — Quondum 02:11, 29 September 2013 (UTC)

Interestingly, the two pieces were created within one hour of each other. (GMTA) I chose molecule rather than matter for the reason stated above: the discovery wasn't that photons are a type of matter. It's that they can act as a molecule. Since the two discuss the same phenomenon some merger is clearly called for. Hopefully, more will chime in. Cheers! Lfstevens (talk) 04:50, 29 September 2013 (UTC)

One must forgive a certain amount of hype and grantsmanship on the part of the inventors. It's all well to say you've discovered a new state of matter but it's something else to get you colleagues to agree. Matter is not well defined and when it is , photons are most often given as examples of non- matter . It's not even clear here if the reference is to photon " molecules " in free space, or to their theoretical existence as part of an ensemble (like a Cooper pair of electrons in a superconductor ). In the latter case a whole lot of atomic matter is also part of the package . The idea of "crystals" of pre photons holding together is hyperbole and approaches what you might hear in an ad campaign for a product or politician. Come, now. A little less Star Wars light saber, before I gag. Some scientist needs to step away from his rubidium and perhaps try a few clarifying weeks of lithium! SBHarris 06:28, 29 September 2013 (UTC)

I think this is better discussed at state of matter. Paradoctor (talk) 15:47, 30 September 2013 (UTC)

I just found out that "photonic molecule" seems to have an established second meaning, I'll make a few edits. Paradoctor (talk) 15:47, 30 September 2013 (UTC)

Looks like merge is out of the question now. If nobody beats me to it, I'll move to photonic matter what belongs there. Paradoctor (talk) 16:23, 30 September 2013 (UTC)

This whole discussion of "matter" versus "molecules" (media speak) completely misses the point. The original title of the nature paper was "Attractive photons in a quantum nonlinear medium". So this is about photon interactions, similar to stimulated emission. It probably belongs in a subsection of Two-photon physics or Rydberg atom instead. — Preceding unsigned comment added by 91.176.150.45 (talk) 17:50, 30 September 2013 (UTC)

I'm beginning a sort of WP:Expert review process for articles independent of the featured article system which I've realized has problems. As such, I've rated this article a level 'A' which means it is of the quality that would be expected from a professional reference work on the subject. I say this as a person with graduate degrees in astrophysics, but I encourage others who have similar qualifications to make comments if they believe my judgement to be incorrect.

jps (talk) 02:19, 12 September 2013 (UTC)

I don't agree. It is a mash-up which in my opinion lacks clarity. Level 'A' means it has the 'quality' of a 'professional reference work'? Please provide references to this claim. ThanksAbitslow (talk) 23:41, 12 December 2013 (UTC)

I tried to correct the table of properties to correctly link to the Charge parity article. My change resulted in the entire line no longer appearing. Specifically, the table links c parity to the article on c symmetry, not to the article on Charge_parity (c parity). I also tried to change the Parity line to Intrinsic Parity which should appear there (either in addition to or instead of "Parity"), but was unable to do that also. Several other problems with the article: 1) reference for these two lines is to a cryptic web page that does not provide any more information than that in the table, and in addition the page (in my web browser,(FireFox)) contains garbled characters. 2) Also, the link if you click on Parity (intrinsic) links you NOT to the intrinsic parity article, but to the article on parity {in (quantum) physics}. This also needs to be fixed. Other problems include 3) the statement that photons are stable. Is a virtual photon "stable"?? 4) Virtual photons are common enough to require the article to distinguish between the various 'flavors' of photons. Two other issues I found which need some attention (imho) are: 5) the implied claim that Maxwell's equations are only appropriate for semiclassical treatments. As is probably obvious, I'm not an expert on this but: its my understanding that appropriately modified, Maxwell's equations are perfectly compatible with quantum field theory. 6) I don't think this article appropriately or sufficiently handles the fact that in modern applications electromagnetic theory must include Maxwell's equations. 7) Another error is the claim that the pictured light cone (2d representation of a 3d cone) IS the light cone. The cone is 4d, of course, best depicted as a series of (3d) spheres. 8) Finally, this article has NO mention of classical (optical) rays, and the relationship between rays and ((9) classical waves) and photons ((10)not to mention the 'classical' quantum mechanics and quantum field theory differences).Abitslow (talk) 00:10, 13 December 2013 (UTC)

In the "Physical Properties" section it says "A photon is massless", and in Note 2 that "The mass of the photon is believed to be exactly zero, based on experiment and theoretical considerations described in the article"

The article quoted does not say any of this. http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html

The article actually says that "Photons are traditionally said to be massless. This is a figure of speech". It also states "It is almost certainly impossible to do any experiment that would establish the photon rest mass to be exactly zero."

You can never measure a zero mass, that idea is absurd because there's nothing to measure!

All they can do is say the mass is less than a very small amount. It could turn out that the mass is much smaller than the currently measurable limit. Effectively, you can never say it is zero until you have infinite measuring capability, which we will never have.

I wrote all this into the section but it has been reverted.

The article is being used as a false reference for an author's personal views.88.203.90.14 (talk) 18:14, 6 January 2014 (UTC)

I expected to find something about lazy photons on wikipedia, but could not find anything. There was a reference to lazy photons at LHC in New Scientist. Could some professional/amateur particle physicist please explain for me... 80.4.63.73 (talk) —Preceding undated comment added 01:19, 19 February 2014 (UTC)

Hi 80.4.63.73. Please ask this question on Wikipedia:Reference desk/Science, and I'm sure someone will be happy to help you. This page is for discussing what should appear in the photon article; it's not for talking about photons generally. --Trovatore (talk) 04:06, 19 February 2014 (UTC)

The light cone diagram seems very strange, or at least the caption does. The caption says the green and indigo areas represent polarization of light, but it sure looks to me like they represent past and future. Can someone clear this up? — Preceding unsigned comment added by 172.251.246.55 (talk) 14:20, 26 April 2014 (UTC)

The diagram is a little confusing, but if you read the section carefully, you'll see that it makes some sense. The axes in the diagram are mislabelled, though: it is not time and space, but rather frequency in the time and spatial directions respectively. It is a matter of convention which direction of the frequency-in-time axis would correspond with which circular rotation. Ideally the axes should be relabelled, and the caption made clearer. —Quondum 15:38, 26 April 2014 (UTC)

I've tried to make this clearer in the caption, but the axes in the diagram should really be relabelled. Well spotted on the part of the OP. —Quondum 16:00, 26 April 2014 (UTC)

Acoording to this arxiv paper the history of term is a bit more involved than currently described in the article.--~~ — Preceding unsigned comment added by Kmhkmh (talk • contribs) 10:17, 5 May 2014 (UTC)

Is it possible to encrypt a photon?The main problem in long range communication is it takes lot of time.Really a lot of time.If we somehow managed to encrypt the photon and decrypt it with some kind of transmitter and receiver system by making slight or no change in its properties that will become easiest way to communicate cause we all know,FOR NOW, nothing travels faster than light.So at long range distance communication will be much easier.sidsandyy (talk) 06:37, 8 June 2015 (UTC)

You could ask at the wp:reference desk/science. Here we discuss the article, not the subject—see wp:TPG. Good luck. - DVdm (talk) 08:55, 8 June 2015 (UTC)

The image is not clear, and it is not connected with the text. The photon does is not a packet of electomagnetic waves, so the image indicate a wave packet of what? What means "x" cohordinate) what "y"?— Preceding unsigned comment added by CocisIt (talk • contribs) 21:25, 17 June 2015 (UTC)

Please see http://www.gizmag.com/light-quantum-spin-trinity-college/43382/ PAR (talk) 16:50, 21 May 2016 (UTC)

Yes, I looked at that. I think either it's wrong, or more likely dependent on an esoteric treatment having to do with "reduced dimensionality" which I don't understand. Unless you can find an editor who understands that and is willing to make the contested content understandable at least to a physicist, then this doesn't belong in the article. Sorry, Interferometrist (talk) 15:37, 27 May 2016 (UTC)

This article no longer satisfies the Featured Article criteria. The main problem is the lack of citations; there are way too many unsourced statements throughout the article. It is a candidate for WP:FAR unless improvements are made. Graham Beards (talk) 14:54, 18 January 2017 (UTC)

@Graham Beards: I am willing to add references and citations into this. Are there any specific areas that you feel need the most attention? Also, any other elements (ex. more pictures) that should also be included, to keep this in featured article status? Popcrate (talk) 05:15, 20 February 2017 (UTC)

If you could add more citations, that would suffice. Thanks. Graham Beards (talk) 08:40, 24 February 2017 (UTC)

I have removed the graphic at right from the article; I don't think that, as it stands, it's a useful explanation. Please see my detailed criticisms at commons:file talk:Wave Particle Duality Dark.png. No offense intended to User:ThreePhaseAC; the effort is appreciated. --Trovatore (talk) 11:15, 2 December 2017 (UTC)

I agree that image is way too misleading to be useful. The particle is not associated with wave peaks. Dicklyon (talk) 22:19, 16 April 2018 (UTC)

The properties of a photon from the m=0, s=1 irreducible representation of the Poincare Group are: 1. rest mass= 0; 2. no position observables; 3. momentum; 4. angular momentum; 5. energy; 6. spin=1; 7. no speed observables. David edwards (talk) 09:39, 25 March 2018 (UTC)David edwards; http://alpha.math.uga.edu/~davide/

@David edwards: Can you explain what your comment is in reference to? Did you mean to start a new section? I don't really see what your comment has to do with the graphic or my criticisms of it, and even if it's not related to that, I still don't understand why you posted it. --Trovatore (talk) 21:44, 16 April 2018 (UTC)

Obviously unrelated, so I added a section header. Dicklyon (talk) 22:19, 16 April 2018 (UTC)

In the lede of the article it is said: "A single photon may be refracted by a lens and exhibit wave interference with itself, and it can behave as a particle with definite and finite measurable position or momentum, though not both at the same time."

In the section on Photon#Wave–particle_duality_and_uncertainty_principles it is said: "A key element of quantum mechanics is Heisenberg's uncertainty principle, which forbids the simultaneous measurement of the position and momentum of a particle along the same direction."

But as I understand the Heisenberg uncertainty principle, there is a trade-off in the precision with which measurements of momentum and position can be made. It's not the black-and-white, either-or distinction being made in the two sentences I've quoted above from the article. Perhaps I just need to be straightened out on this, or perhaps the article needs repair. Thanks. Attic Salt (talk) 01:46, 26 May 2018 (UTC)

The statements don't disallow the intermediate tradeoff. They reminds readers that, as a quantum object, a photon has to be considered as both wave and particle. Both also lead into Interpretations_of_quantum_mechanics. I don't know either way, if the more general explanation would help here. Gah4 (talk) 15:19, 26 May 2018 (UTC)

I think the only thing that makes sense is the intermediate trade off. So, for example, what exactly is the meaning of "definite and finite measurable" (as said in the lede)? I would say that the whole notion of definite location or definite momentum have no practical meaning, and neither are definitely measurable. At least that is my understanding. Attic Salt (talk) 15:58, 26 May 2018 (UTC)

I suspect people tend to believe, when they see the word particle, of something that can have a position. The statement is a gentle introduction to the need for quantum mechanics in the description. The actual details of the QM description come later. Most people have heard about wave-particle duality, but haven't actually thought about it. This reminds them to start thinking about it, as they read the rest of the article. (Which hopefully explains it with enough detail.) Gah4 (talk) 16:13, 26 May 2018 (UTC)

In the lede it is said that a photon is "the quantum of the electromagnetic field". I find this confusing. My understanding is that a photon can, in principle, have any amount of energy, since light can have any frequency. In this sense, light is not quantized in a general sense. On the other hand, an atom (say) has discrete quantum states and, therefore, emits photons with discrete energies or "quanta". This is due to the nature of the atom, but it is not an intrinsic property of light in general, since, like I say, light can have any frequency. Anyway, this is why I find it confusing to say that the photon is "the" quantum of "the electromagnetic field". I suppose I would prefer to say that "a photon" is "a quantum of an electromagnetic field". Thanks, Attic Salt (talk) 02:00, 26 May 2018 (UTC)

I am not at all an expert in Quantum_field_theory but I believe the statement is correct. As with momentum-position duality, there is also an energy-time duality. Yes you can have any energy, and so frequency, but then you are restricted by uncertainty, or by Fourier, on the time. It is the usual way to express in field theory. Gah4 (talk) 15:27, 26 May 2018 (UTC)

I think what is written is not correct. It should say "a photon is a quantum of an electromagnetic field" so as to distinguish this from the generalisation. Attic Salt (talk) 15:33, 26 May 2018 (UTC)

If you say a quantum, that would imply that there are others. Some quantum fields are described with more than one particle, but EM has only one. The Z boson is a quantum of the weak force, but not the only quantum of it. Yes it is easier to think of quantization in terms of things we naturally think of a particles, like electrons and neutrons, but the same rules apply. As Feynman says, nobody understands quantum mechanics. But we try, anyway. Gah4 (talk) 16:21, 26 May 2018 (UTC)

Of course there are others. Light of one frequency has a quantum, light of another frequency has another quantum. Attic Salt (talk) 16:25, 26 May 2018 (UTC)

The usual notion of a photon is as stated, "the quantum of the electomagnetic field". There are alternative ways to describe it, where field is not really a thing on its own and the quantum properties are associated with the states of the atoms and electrons, as in Collective electrodynamics. But that's not the usual approach. Dicklyon (talk) 16:27, 26 May 2018 (UTC)

I went ahead and made my changes to the article, but if a consensus is developed to stay with "the", then we can easily revert my minor edits. Thanks. Attic Salt (talk) 16:29, 26 May 2018 (UTC)

I've reverted my changed due to lack of consensus on this. Note, I remain concerned about the description of Heisenberg uncertainty in this article.

The expression

${\displaystyle {\boldsymbol {p}}=\hbar {\boldsymbol {k}},}$

can not be true in general, since it also holds that the momentum is proportional to the Poynting vector

${\displaystyle {\boldsymbol {S}}={\boldsymbol {E}}\times {\boldsymbol {H}},}$

which is, in general for anisotropic media, not parallel to the wave vector. Steak (talk) 11:52, 6 July 2018 (UTC)

In media is different. It is not one photon, but a series of photons absorbed and emitted along the way. Macroscopically, the result looks like you say. Gah4 (talk) 18:35, 6 July 2018 (UTC)

A photon is made of one electron and one anti-electron so how can photons be described as elementry? EveryThingIsRelative (talk) 10:07, 20 April 2012 (UTC)

Questions like that are best put at the wp:reference desk/science. Article talk pages are for discussions about the content and format of the article, not for discussions or questions about the subject. See wp:talk page guidelines. Cheers - DVdm (talk) 11:15, 20 April 2012 (UTC)

The premise however is transparently false. A particle and its anti-partner annihilate each other, they don't then coexist as a unity. It's the transformation that produces the photon. 98.4.124.117 (talk) 06:53, 29 July 2018 (UTC)

The article states "...the modern concept of the photon was developed gradually by Albert Einstein..."

This conflicts with another Wikipedia article on Electromagnetic radiation that states, "An anomaly arose in the late 19th century involving a contradiction between the wave theory of light on the one hand, and on the other, observers' actual measurements of the electromagnetic spectrum that was being emitted by thermal radiators known as black bodies. Physicists struggled with this problem, which later became known as the ultraviolet catastrophe, unsuccessfully for many years. In 1900, Max Planck developed a new theory of black-body radiation that explained the observed spectrum. Planck's theory was based on the idea that black bodies emit light (and other electromagnetic radiation) only as discrete bundles or packets of energy. These packets were called quanta." — Preceding unsigned comment added by 70.189.252.150 (talk) 00:02, 1 February 2013 (UTC)

Please, read Photoelectric effect and Bose–Einstein statistics articles. It may help to realize the difference between Planck’s raw thoughts about “discrete bundles” and the actual theory of photons. Incnis Mrsi (talk) 08:46, 1 February 2013 (UTC)

Actually, the article on the Photoelectric effect backs up the previous posters assertion that photon theory starts with Plank. While I'm sure Einstein's contributions are important I feel as though this article gives him too much credit. — Preceding unsigned comment added by 2600:1700:69C1:2A00:2C28:C63E:62FD:BFD1 (talk) 08:29, 18 June 2018 (UTC)

As I understand it, though from when I learned it many years ago, Plank didn't really believe in photons and light quantization, but only that it was a way to get the solution. It might be, in modern terms, he believed in virtual photons but not real ones. It took longer to realize that light is always quantized. Gah4 (talk) 04:21, 15 October 2018 (UTC)

I don't mean to be disrespectful to everyone, but the photoelectric effect may be misinterpreted. The observed effect really arises out of the quantum mechanical based band structure of materials. Photons cannot be absorbed until their energy matches the difference between two quantum states in the material, that allow for a transition. You may think that this is naïve, but this explains why glass and fused silica are transparent in the visible light region, but they can emit electrons under x-ray radiation. The one interesting property of a photon is that it can possess energy in any amount, and is an inherent continuous property of the photon. That is a very important point of its characteristics.

Another interesting point is the character of the photon's wavelength. When an electron bound in an atom drops into a lower energy orbit, then a photon is emitted. Take mercury for an example of an element that emits visible light, the "size" of mercury atoms is about .3 nanometers (3.0E-10 meters). But the characteristic length of the emitted photon is about 0.7 E-6 meters. Thus the wave length is 2,000 times the size of the atom. I have not seen anyone really address this huge factor in size.

I will follow up with another very serious question about the photon, but less discuss these two statements first.

Respectfully, James T Clemens — Preceding unsigned comment added by Jtclemens (talk • contribs) 19:34, 12 October 2014 (UTC)

I suspect this is what surprised Plank and others at the time. The photoelectric effect comes as, for example, two photons with half the energy can't do what one photon with the whole energy can. But yes, band structure is important, and it took some time before that was understood. It has surprised me, that since silver bromide photography was invented in the mid-1800's, and was known to be blue sensitive, that the mechanism wasn't considered until Einstein. Silver bromide is fine under bright red light, but only needs a tiny bit of blue light. As above, it took a while to understand that light is always quantized, not just in the interactions in question. Gah4 (talk) 04:36, 15 October 2018 (UTC)

If photons are particles, then do they have a diameter, or at least a theoretical diameter like other particles? But since they travel at the speed of light, then I suppose they can have a diameter of zero. 86.149.134.65 (talk) 02:07, 18 August 2015 (UTC)

Photons have the dimensions required by Heisenberg uncertainty. As for other particles, electrons are considered to have no size, other than, again, Heisenberg uncertainty. Protons and neutrons have internal structure, and so a minimum diameter. Gah4 (talk) 04:39, 15 October 2018 (UTC)

The article says: always moves at the speed of light within a vacuum. As far as I know, photons always move at the speed of light. In refractive materials, absorption and re-emission of photons is responsible for the measured, lower, speed. It is a convenience of the not so easy to describe (Feynman does it in his lecture series) physics that the result of such absorption/emission, or the same thing in the Maxwell equations form, results in what looks like slower moving light. Or, in another sense, nature is close enough to linear that the simple explanation works. Gah4 (talk) 04:46, 15 October 2018 (UTC)

This diagram needs correcting, fails to show 90 degree phase shift between E-M waves. I'll replace sometime, unless someone else does first. Here's a correct diagram: [5]. Tom Ruen (talk) 04:09, 25 January 2012 (UTC)

Correct me if I'm wrong - but I think both images are incorrect in that they give the incorrect wave direction. I'm assuming that the arrow and word 'distance' in the present image is referring to the direction of wave propagation. The proposed image is much nicer looking, except for the direction which should be the cross product of E and B. The animation on the electromagnetic radiation page is correct.PhySusie (talk) —Preceding undated comment added 20:32, 25 January 2012 (UTC).

There is no phase shift (see e.g. Sinusoidal plane-wave solutions of the electromagnetic wave equation or Electromagnetic wave equation), but the propagation direction is indeed wrong in most images on Commons (as they are derivatives of this one). Correct images are File:Tem wave.gif File:Onda e y ma 20.JPG File:Onda electromagnetica1.jpg [6] [7], etc. I might have known why there are two types of images, but completely forgot. Materialscientist (talk) 07:45, 27 January 2012 (UTC)

If the diagram is correct, and E & B are 'in phase', then the energy formula for a photon (e = hv) is plainly wrong and must be re-written as a time varying formula. Further, 'why' do the E & B fields ever 'recover' from the point at which both fields are zero (explanations that invoke the 'storage' of energy 'in the aether' are not acceptable :-) ) and if the energy varies how do we explain the photo-electric effect ? — Preceding unsigned comment added by 62.3.239.168 (talk) 14:49, 23 March 2013 (UTC)

Your question has a hidden assumption, which is that one field (E or B) or a change in it, causes the other field (B or E) to "recover." Not your fault, as generations of students are told that a varying E "causes" a B, and vice versa, and that is how EM waves are made. Wrong. But Maxwell's equations for B and E are not causal. They demand a certain association between E and B, but associations between two variables can be caused by a third Main Cause, and in this case, they are. In Maxwell's equations, E and B are both caused by charges and they literally have nothing to do with each other. They are both caused by an accelerated charge at the source, and by the time you get very far away from the charge, E and B are in fixed ratios to each other because they are both a product of this same acceleration. And not the charge's density or velocity (current), which can be arbitrary, and produce arbitrary E/B ratios close to the source charges and currents. But that doesn't mean E causes B in an EM wave (EMR), which is made of photons.

There's also an enormous amount of confusion caused by the fact that displacement currents in a capacitor (the term in Maxwell's equations that suggests an EM wave) give E and B fields that are NOT in phase. In fact, E and B in a capacitor gap are exactly out of phase, even though described by the same equations. But the changing EM field inside a capacitor (mostly what we call the near-field) is not the same changing EM radiation that you see far away from the capacitor (the far-field), and the phase is not the same. Inside the capacitor the ratio of E to B can be anything you like (depends on the design or the capacitance of the capacitor, even if you specify vacuum dielectric). Far away from a capacitor, in the direction of x propagation in vacuum it must be the case for linearly polarized plane waves that [dx/dt]^2 dE/dx = dE/dt, and the same equation for B, so that E = (dx/dt) B where dx/dt is some constant wave speed determined by the permittivity and permiability of vacuum. This velocity turns out to be dx/dt = c. Please note the form of the equation. It is not true that E = dB/dt or that B = dE/dt, which are the Faraday's law looking things, even though you will be told that this is the "mechanism" by which EM waves are made. For sine or cosine waves this would give a phase difference, as sine of B would be cosine of E, and vice versa. But the correct single differentiation on both sides (one for space and one for time) gives the same phase of E and B in the far-field.

As for the precise "where" in space that the energy of EM radiation is stored, that is related to the uncertainty principle. You can't point to a specific place on a wave antinode and say "what is there no energy THERE?" The energy of the wave is spread out along the EM wave (many wave peaks), so you can ignore places where the E and B fields go to zero. That's true even for a single photon, which isn't composed of just one peak-- it's a whole train of them. If you want to squeeze a photon in space (along the direction of travel) the frequency of the waves making it up get uncertain, and so does its energy. SBHarris 23:40, 14 April 2013 (UTC)

I suppose that works, but I tend to think of it as a relativistic effect. You might also look at the case of standing waves, with photons going opposite directions. Gah4 (talk) 18:43, 5 March 2019 (UTC)

The article says: that light was an electromagnetic wave—which was confirmed experimentally in 1888 by Heinrich Hertz's detection of radio waves. I believe this isn't quite right. Hertz did show that electromagnetic waves satisfying Maxwell did exist, but didn't show that those were the same as light. Also, Hertz experiments showed the photoelectric effect, as the spark in his receiver would change with light. Gah4 (talk) 18:47, 5 March 2019 (UTC)

Light beams are distributions of photons. Usually, photons can be considered to be point-like particles. They are known to obey the Einstein-Planck relation E = h v. The model of their structure must explain the fact that photons that are emitted by a nearby star can be detected by the naked eye. The path that this photon must travel must be supported by the carrier field and the field excitation that represents the photon must keep the integrity of the emitted photon. The field must exist everywhere and always along the traveling path. The field must be sufficiently flat. The field that is generated by electrons requires the nearby existence of electrons. Between a star and the earth exist no waveguides that support a steady carrier. Wave packages would disperse during the long travel through empty space. Spherical waves quickly diminish their amplitude and will not deliver sufficient energy, such that the naked eye can capture the photon. Still, the field excitation can be a solution of the same second order partial differential equation that supports waves. Indeed since several centuries solutions of the wave equation are known that support pulse responses in odd dimensions that act as shock fronts. During travel, the shock fronts keep their shape. One-dimensional shock fronts keep their shape as well as their amplitude. If a string of equidistant one-dimensional shock fronts obeys the Einstein-Planck relation, then the string can implement the functionality of a photon. It means that the one-dimensional shock fronts each carry a standard amount of energy and the emission duration must be the same for all photons independent of the frequency of the photon. The wave equation has a quaternionic equivalent and that equation goes together with a quite similar quaternionic second order partial differential equation that splits into two first-order partial differential equations. The homogeneous version of that equation also supports shock fronts, but no waves. This equation supports one-dimensional shock front solutions that feature polarization, which is determined by the emitter. This solution can be used to generate the full behavior of photons. The field that represents the universe and that acts as our living space can act as the carrier field. Spherical shock fronts can temporarily deform and permanently expand this field. Please read https://www.opastonline.com/wp-content/uploads/2018/11/tracing-the-structure-of-physical-reality-by-starting-from-its-fundamentals-atcp-18.pdf and https://www.opastonline.com/wp-content/uploads/2018/12/the-behavior-of-basic-fields-atcp-18.pdf The light beams can be distributions of angular direction, energy, emission location, and phase of photons. These distributions may show wave nature. The notion of the Optical Transfer Function applies this effect by supposing that the distribution has a Fourier transform. This means that the photon distribution is a detection probability density distribution. — Preceding unsigned comment added by HansVanLeunen (talk • contribs) 13:51, 17 December 2018 (UTC) --HansVanLeunen (talk) 14:57, 17 December 2018 (UTC)

You are not allowed to ask that question. Photons are quantum particles, obeying the rules of quantum mechanics, and QM restricts which questions you can ask. You might look at interpretations of quantum mechanics to start understanding which questions you are allowed to ask. If you try to only consider photons as point-like, you will get confused. Just remember, quoting Richard Feynman: Nobody understands quantum mechanics. Gah4 (talk) 15:02, 13 March 2019 (UTC)