2013-12-18 15:59:53 8 Comments

At the end of this nice video (https://youtu.be/XiHVe8U5PhU?t=10m27s), she says that electromagnetic wave is a chain reaction of electric and magnetic fields creating each other so the chain of wave moves forward.

I wonder where the photon is in this explanation. What is the relation between electromagnetic wave and photon?

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## 7 comments

## @my2cts 2018-11-04 20:25:48

You report that in the video it is stated that an electromagnetic wave is "a chain reaction of electric and magnetic fields creating each other so the chain of wave moves forward." I disagree with this view. There is just one wave, that of the vector potential or more generally of the four potential. The electric and magnetic fields are just derivatives of the vector potential and do not "create each other".

Rejecting this explanation we then arrive at your deeper question: "What is the relation between electromagnetic wave and photon?"

Until a few years ago I shared the opinion of Willis Lamb, that the photon is a fictive particle. I finally changed my mind because such an explanation cannot account for low intensity diffraction experiments. Indeed, how can a single atom or molecule absorb a wave that is much larger that it? Note that I don't intend to fork off a discussion on this here but want to give my interpretation. This is that the vector potential describes the

probabilityof a photon being absorbed, just like the Schrödinger and Dirac wave functions do for an electron. Indeed the Maxwell equations in vacuum can be written as a wave equation that closely resembles the Klein-Gordon equation. This interpretation implies that the photon indeed exists as a particle, much smaller than an atom and at least as small as a nucleon.## @hyportnex 2018-11-08 15:55:22

"how can a single atom or molecule absorb a wave that is much larger that it? ", the same question can be asked how can an electrically small antenna ($"dimension"<< \lambda$), say, a Hertz dipole absorb an essentially infinite plane wave. It can, I have seen it; all waves all the way down, no photons needed...

## @my2cts 2018-11-08 19:49:38

@hyportnex your argument can easily be used to support the photon concept.

## @hyportnex 2018-11-08 20:53:36

I have not seen any attempt neither do I believe that, say, a 5cm long ferrite loaded loop antenna's operation at around 550kHz can be usefully explained via photons and quantum physics but, please, go ahead.

## @my2cts 2018-11-08 23:30:22

@hyportnex your example pertains to the limit of many photons. That is why no QM is needed.

## @Moonraker 2018-11-04 19:47:25

In order to understand the wave particle dualism you have simply to understand what time is:

In 1905, the Newtonian unique time concept was replaced by a twofold time concept of observed coordinate time and proper time - the observed time is relative and observer-dependent, and it is derived from the intrinsic proper time of the observed particle ("The time measured by a clock following a given object"). Proper time is the more fundamental time concept.

You can understand the wave particle dualism if you consider the simplest case of a photon, that is a photon moving at light speed c. The spacetime interval of such photons (which corresponds to their proper time) is zero. That means that the event of emission and the event of absorption are adjacent in spacetime, the emitting mass particle is transmitting the momentum which is called photon directly to the absorbing mass particle, without any spacetime between them. That means that the particle characteristics are transmitted directly without need for any intermediate massless particle.

However, for observers the zero spacetime interval is not observable, e.g. between Sun and Earth are observed to be eight light minutes, even if the spacetime interval of the path of the photon is zero. In spite of the direct transmission of a momentum between two mass particles, observers observe an electromagnetic wave which is filling the gap of eight light minutes.

In summary, particle characteristics are transmitted directly according to the principles of spacetime intervals and proper time, whereas the wave is transmitted according the principles of the observed spacetime manifold.

Now you will ask: What about photons which are moving slower than c (through gravity fields and through transparent media)? The answer is that here quantum effects such as nonlocality are implied. But it is important to notice that the limit case of photons in vacuum moving at c may be explained and understood classically, without need for any quantum theory.

## @Maxwell 2018-11-04 18:42:04

The photon dilemma

It is postulated by Planck that energy is quantized. Owing to classical electromagnetic theory light is an electromagnetic field. This field satisfies a wave equation traveling at the speed of light. Hence, light is an electromagnetic wave. Light consists of photons; and thus each photon carries a unit of energy. This behavior is demonstrated by the photoelectric and Compton effects. Since light is an electromagnetic energy, photons must also carry electromagnetic field and a unit of it. While photons are quantum objects, light still governed by Maxwell's classical theory. The photon model is not critically consistent with Maxwell equations, since it has a dual nature. In fact light as a wave is well described by Maxwell. Recall that Maxwell's equations don't involve the Planck's constant, and thus can not describe the particle nature of the photon. A complete Maxwell's equations should involve this missing element. In quantum electrodynamic paradigm, the photon is brought to interact with the electrons by invoking the idea of minimal coupling where electrons and photons exchange momentum. The photon appears as a mediator between charged particles.

At the same time while a moving charged particle has its self electric field, and magnetic field that depend on the particle velocity, the photon, the carrier of the electromagnetic energy is void of these self-fields because it has no charge and mass. Thus, a charge-less photon can't have electric and magnetic fields accompanying its motion.

The appropriate Maxwell's equations should then incorporate the photon linear momentum as well as its angular momentum. In such a case the new Maxwell's equations can then describe the dual nature of the photon. Like electric charge, the angular momentum is generally a conserved quantity. The question is how one can correct for these photon proprieties? One way to achieve that is to employ quaternions that generically allow many physical properties to be joined in a single equation. This is so because the quaternion algebra is so rich, unlike the ordinary real numbers. To this end we employ the position-momentum commutator bracket, and invoked a photon wavefunction. This wavefunction is constructed from the linear complex combination of the electric and magnetic fields.

The outcome of the bracket yields three equations defining the photon electric and magnetic fields in terms of its angular momentum. These equations turn out to be very similar to those fields created by a moving charge. Thus, the electric and magnetic fields of the photon doesn't' require a charge for the photon. It is intriguing that the photon has no charge and mass but has electric and magnetic fields as well as energy. These fields should also satisfy Maxwell's equations. Doing so, yields additional electric and magnetic charge and current densities for the photon. The emergent Maxwell's equations are now appropriate to describe the photon as a quantum particle. These additional terms in Maxwell's equations are the source in describing the photon quantum electrodynamics behavior. Some emergent phenomena associated with topological insulator, Faraday's rotation effect, Hall effect and Kerr's effect could be examples of this contribution terms to Maxwell's equations.

Here are the quantized Maxwell's equations incorporating the photon linear and angular momentum. These are the electric and magnetic fields due to the photon as a particle: \begin{equation} \vec{L}\cdot\vec{E}=-\frac{3\hbar c}{2}\,\Lambda\,, \qquad\qquad \vec{L}\cdot\vec{B}=0\,, \end{equation} and \begin{equation} \vec{B}=-\frac{2}{3\hbar c}\,(\vec{L}\times\vec{E})\,,\qquad\qquad\vec{E}=\frac{2 c}{3\hbar}(-\Lambda\,\vec{L}+\vec{L}\times\vec{B})\,. \end{equation} And these are the new Maxwell's equations: \begin{equation} \vec{\nabla}\cdot\vec{E}=-\frac{4c}{3\hbar}\,\,(\vec{B}-\frac{1}{2}\,\mu_0\vec{r}\times\vec{J})\cdot\vec{p}+\frac{2}{3\hbar c}\,\vec{E}\cdot\vec{\tau}+\frac{\partial \Lambda}{\partial t}\,,\qquad \vec{\nabla}\cdot\vec{B}=\frac{4}{3\hbar c}\,\,\vec{E}\cdot\vec{p}+\frac{2}{3\hbar c}\,\vec{B}\cdot\vec{\tau}\,, \end{equation} and \begin{equation} \vec{\nabla}\times\vec{B}=\frac{1}{c^2}\,\frac{\partial\vec{E}}{\partial t}+\frac{2}{3\hbar c}\left(\Lambda\vec{\tau}+\vec{B}\times\vec{\tau }-\frac{\vec{P}}{\varepsilon_0}\times\vec{p}\right)-\vec{\nabla}\Lambda\,, \end{equation}

\begin{equation} \vec{\nabla}\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}-\frac{2c}{3\hbar}\left(\mu_0\vec{J}\times\vec{L}+\frac{\vec{\tau}}{c^2}\times\vec{E}+2\Lambda\,\vec{p}\right)\,, \end{equation} where \begin{equation} -\Lambda=\frac{1}{c^2}\,\frac{\partial\varphi}{\partial t}+\vec{\nabla}\cdot\vec{A}=\partial_\mu A^\mu\,. \end{equation} In the standard electrodynamics $\Lambda=0$ represents the Lorenz gauge condition.

## @Zorawar 2019-11-20 17:29:42

This answer is very confused. The photon does not "carry" the electric or magnetic fields, it is the Standard Model mediator of the EM interaction. It looks like you've confused classical and quantum concepts. Maxwell's equations do not

needto incorporate anything quantum mechanical -- they are purely classical equations. I also don't know where those equations have come from.## @HolgerFiedler 2016-05-04 16:58:17

What are photons?Photons get emitted every time when a body has a temperature higher 0 Kelvin (the absolute zero temperature). All bodies, surrounding us (except black holes) at any time radiate. They emit radiation into the surrounding as well as they receive radiation from the surrounding. Max Planck was the physicist who found out that this radiation has to be emitted in small portions, later called quanta and even later called photons. Making some changes in the imagination of how electrons are distributed around the nucleus, it was concluded that electrons get disturbed by incoming photons, by this way gain energy and give back this energy by the emission of photons. And photons not only get emitted from electrons. The nucleus, if well disturbed, emits photons too. Such radiations are called X-rays and gamma rays.

What is electromagnetic radiation?EM radiation is the sum of all emitted photons from the involved electrons, protons and neutrons of a body. All bodies emit infrared radiation; beginning with approx. 500°C they emit visible light, first glowing in red and then shining brighter and brighter. There are some methods to stimulate the emission of EM radiation. It was found out that beside the re-emission of photons there is a second possibility to generate EM radiation. Every time, an electron is accelerated, it emits photons. This explanation helps to understand what happens in the glow filament of an electric bulb. The electrons at the filament are not moving straight forwards, they bump together and running zig-zag. By this accelerations they lose energy and this energy is emitted as photons. Most of this photons are infrared photons, and some of this photons are in the range of the visible light. In a fluorescent tube the electrons get accelerated with higher energy and they emit ultraviolet photons (which get converted into visible light by the fluorescent coating of the glass). Higher energy (with higher velocity) electrons reach the nucleus and the nucleus emits X-rays. As long as the introduced energy is a continuous flow, not one is able to measure an oscillation of EM radiation.

What are EM waves?Using a wave generator it is possible to create oscillating EM radiation. Such radiations are called radio waves. It was found out that a modified LC circuit in unit with a wave generator is able to radiate and that it’s possible to filter out such a modulated radiation (of a certain frequency) from the surrounding noisy EM radiation.

So the wave generator has a double function. The generator has to accelerate forward and backward the electrons inside the antenna rod and by this the photons of the radio wave get emitted, and the generator makes it possible to modulate this EM radiation with a carrier frequency. It has to be underlined that the frequency of the emitted photons are in the IR range and sometime in the X-ray range. There is an optimal ratio between the length of the antenna rod and the frequency of the wave generator. But of course one can change the length of the rod or one can change the frequency of generator. This changes the efficiency of the radiation to the needed energy input only. To conclude from the length of the antenna rod to the wavelength of the emitted photons is nonsense.

What is the wave characteristic of the photon?Since the electrons in an antenna rod are accelerated more or less at the same time, they emit photons simultaneous. The EM radiation of an antenna is measurable and it was found out that the nearfield of an antenna has two components, an electric field component and a magnetic field component. This two components get converted in each other, the induce each other. At some moment the transmitting energy is in the electric field component and otherwise the energy is in the magnetic field component. So why not conclude from the overall picture to the nature of the involved photons? They are the constituents which make the radio wave.

## @Luaan 2016-07-21 09:20:31

The two components do not

induceeach other, though it's a common misconception (that's what I've been taught in school as well :-). Because of how wide that misconception is, animations now usually show both the electric and magnetic field in phase, to prevent confusion.## @rob 2016-12-15 19:50:19

The final figure here shows the $E$ and $B$ fields oscillating a quarter-turn out of phase. For waves in vacuum that's incorrect; $E$ and $B$ should be in phase.

## @HolgerFiedler 2016-12-15 20:13:14

@rob Is there any experimental evidence that the shift is existing only in the near field of antenna radiation?

## @rob 2016-12-16 06:51:12

@HolgerFiedler If the fields are a quarter-turn out of phase, the average value for the Poynting vector is zero and the wave is not transmitting any energy.

## @HolgerFiedler 2016-12-16 08:07:33

@rob Than how the energy transfer in the near field of an antenna work? And how a standing EM wave inside a box work?

## @rob 2016-12-18 05:42:22

Those would make good follow-up questions; I don't know if I can answer completely in a comment.

## @HolgerFiedler 2016-12-18 18:23:24

@rob I ask this question now physics.stackexchange.com/questions/299408/…. (The question was edited and I'm not sure that it matches what I wanted to ask.)

## @Cham 2017-08-10 14:20:50

The monochromatic plane wave described by this picture is actually a

standing wave, i.e a superposition of two identical plane waves propagating in opposite directions. This is why the average energy flux is 0. For a simplemonochromatic plane wavepropagating in vacuum, both components should be in phase.## @thermomagnetic condensed boson 2018-05-18 20:42:20

-1 because of "All bodies, surrounding us (except black holes) at any time radiate." Do you really believe black holes do not radiate? Terrible.

## @HolgerFiedler 2018-05-19 06:27:59

@lobotomized_sheep_99 I’m referring to this statement:”A black hole is a region of spacetime exhibiting such strong gravitational effects that nothing—not even particles and electromagnetic radiation such as light—can escape from inside it.” You are referring to this prediction: “Moreover, quantum field theory in curved spacetime predicts that event horizons emit Hawking radiation, with the same spectrum as a black body of a temperature inversely proportional to its mass.” WP Do you think in relation to the original question this has to be mentioned?

## @thermomagnetic condensed boson 2018-05-19 07:44:44

I am absolutely not interested in answering a question whose answer is entirely subjective. I prefer to focus on the absolute truth and the sentence I quoted from your answer is false.

## @hyportnex 2013-12-18 19:26:32

In 1995 Willis Lamb published a provocative article with the title "Anti-photon", Appl. Phys. B 60, 77-84 (1995). As Lamb was one of the great pioneers of 20th century physics it is not easy to dismiss him as an old crank.

He writes in the introductory paragraph:

He finishes with these comments:

## @electronpusher 2017-03-23 07:54:43

Wow, Lamb is actually making my rethink me admittedly amateur perspective on the matter. This quote blew my mind: " It is very likely that Bohr never, by himself, made a significant quantum-mechanical calculation after the formulation of quantum mechanics in 1925-1926."

## @anna v 2018-08-03 03:54:39

This is not within the mainstream physics models at present, but a peculiar proposal not validated or supported by model calculations and predictions.

## @hyportnex 2018-08-03 19:54:28

@anna_v to the limited extent I understand it, I believe that if you read the whole paper and not just the snippet I quoted here you would agree that Lamb's is mainstream physics with mainstream interpretation.

## @Helen - down with PCorrectness 2018-09-30 08:01:05

@annav, then again, the chosen answer interpreting everything as fields is not necessarily mainstream physics for many physicists (or, more importantly, not necessarily correct). I think this reference deserves a reading.

## @anna v 2018-09-30 08:49:47

@Helen Inmy opinion quantum field theory has very many calculational successes in describing particle physics, where it is mainstream . One could argue about its region of validity, as with many mathematical models. For example QCD has more success with lattice QCD as the expansions of perturbative field theory do not work. I do not think that there is a problem with photons in the standard model, and photons are their own antiparticle. So I will not go to the trouble of reading the paper ( no link provided so it means a library or a paywall) where a prominent physicist discusses new theory

## @Helen - down with PCorrectness 2018-09-30 10:49:46

@annav About the field theory part and the chosen answer: Calculational successes do not justify an interpretation of "there are no waves, there are no particles, but there are fields". As much as we have to be careful with new proposals, we also have to be accurate about interpretations passing as mainstream while they are not necessarily correct. Thus my pointing it out.

## @anna v 2018-09-30 11:13:27

@Helen It is a model. The main answerer believes in it, it is a platonic view. I just see it as a successful model. I started doing field theoretical models back in 1962, for nuclear physics!! My doctorate thesis uses Regge poles ( which btw are having a revival with string theory) Field theory is just a tool . afaik it is just the data that exists. If a model fits it, fine . If not, go to the another one

## @hyportnex 2018-09-30 15:28:05

@anna_v here is Lamb's article page by page for your enjoyment but I will remove it in a few days

## @hyportnex 2018-09-30 15:29:50

@anna_v imgur.com/teIkC10 ; imgur.com/SQ091JQ ; imgur.com/hFrEV6G ; imgur.com/amaZrSl ; imgur.com/buqCx0E ; imgur.com/ns7DuwR ; imgur.com/yqoLgb4 ; imgur.com/wR4tReQ ;

## @Zorawar 2019-11-20 16:25:49

The fundamental point of Lamb's paper is not entirely clear to me. The point that he is making, I think, is that using "photon" to mean a particle was an attempt at understanding what was happening in the early days of QM in terms of a "real" particle, whereas now we have a proper field-theoretic basis to QM, and where we really have "excitations" or normal modes of the field. Thus, using "photon" and thinking of a particle is not really valid in either the quantum or classical worlds. This is a point of semantics and teaching, and is a good point, but is not a groundbreaking argument on QM.

## @anna v 2013-12-18 16:16:31

In this link there exists a mathematical explanation of how an ensemble of photons of frequency $\nu$ and energy $E=h\nu$ end up building coherently the classical electromagnetic wave of frequency $\nu$.

It is not simple to follow if one does not have the mathematical background. Conceptually watching the build up of interference fringes from single photons in a two slit experiment might give you an intuition of how even though light is composed of individual elementary particles, photons, the classical wave pattern emerges when the ensemble becomes large.

## @John Rennie 2013-12-18 17:00:28

Both the wave theory of light and the particle theory of light are approximations to a deeper theory called Quantum Electrodynamics (QED for short). Light is not a wave nor a particle but instead it is an excitation in a quantum field.

QED is a complicated theory, so while it is possible to do calculations directly in QED we often find it simpler to use an approximation. The wave theory of light is often a good approximation when we are looking at how light propagates, and the particle theory of light is often a good approximation when we are looking at how light interacts i.e. exchanges energy with something else.

So it isn't really possible to answer the question

where the photon is in this explanation. In general if you're looking at a system, like the one in the video, where the wave theory is a good description of light you'll find the photon theory to be a poor description of light, andvice versa. The two ways of looking at light are complementary.For example if you look at the experiment described in Anna's answer (which is one of the seminal experiments in understanding diffraction!) the wave theory gives us a good description of how the light travels through the Young's slits and creates the interference pattern, but it cannot describe how the light interacts with the photomultiplier used to record the image. By contrast the photon theory gives us a good explanation of how the light interacts with the photomultiplier but cannot describe how it travelled through the slits and formed the diffraction pattern.

## @Val 2013-12-20 18:09:37

This is news because all QM teachers told me that photons abstractions, proposed by QED, which is more exact than wave discription. However, this should not stop us from figuring out how two are related. Actually quanta = particles.

## @Kevin Driscoll 2013-12-20 19:26:57

@Val The way we actually calculate things in QED is with a perturbative expansion that involves photons. The underlying exact theory is one of several completely quantum fields.

## @Prahar 2016-05-04 18:01:52

There is a sense in which the classical description of light is retrieved as the classical limit of a coherent state of photons. I would say that this would be an appropriate answer to "where is the photon in the classical wave theory of light?"

## @Luaan 2016-05-05 11:32:09

@Prahar Yes, but you just said it yourself - that's not the reality. That's just "how it fits in the models"- it doesn't help you outside of the constraints of the models, and that's exactly what the OP is asking here. In the classical wave theory of light... there's no photons. Not one per wave, not "infinite amounts" per wave, just no photons, period.

## @Helen - down with PCorrectness 2018-09-30 07:56:54

I think that "excitation of a field instead of waves and particles" is one interpretation, and probably not the most popular one. Many people view fields only as a handy mathematical tool.