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If one creates an oscillating electric field and a magnetic field, transversal to each other, and oscillating at a given frequency belonging to the visible spectrum, and moving in a given direction of an observator, will an observer see the same as compared to the same experiment where he looks at the light ?

Said differently, is light nothing more than a pair of transverse electric and magnetic fields?
So would creating a pair of transverse electric and magnetic oscillating fields moving in a given direction be equivalent to create light?

Deschele Schilder
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Mathieu Krisztian
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An oscillating electric field and magnetic field propagates as light and Heinrich Hertz has already demonstrated that by creating radio waves in the laboratory.

So yes light is a pair of transverse oscillating electric and magnetic fields.

Lost
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  • check the question "nothing more than". – Mathieu Krisztian Aug 22 '20 at 18:15
  • What do you mean by "nothing more"? Maxwell's electromagnetic theory is sufficient to account for the wave nature of light. As far as the particle nature goes that does not discard the wave nature so in essence light is pair of transverse oscillating electric and magnetic fields. – Lost Aug 22 '20 at 18:18
  • ok : so if in a laboratory, I create a transverse electric and magnetic oscillating fields moving in a given direction : will we see light ? – Mathieu Krisztian Aug 22 '20 at 18:20
  • Yes. If the frequency is within the visible spectrum. You do see light out of a light bulb don't you? What do you think causes that? – Lost Aug 22 '20 at 18:23
  • ok thank you @Lost – Mathieu Krisztian Aug 22 '20 at 18:30
  • This doesn't answer the question: So would creating a pair of transverse electric and magnetic oscillating fields moving in a given direction be equivalent to create light?, to which the answer is "no". – Deschele Schilder Aug 22 '20 at 18:59
  • @descheleschilder The answer is "yes". – my2cts Aug 22 '20 at 19:06
  • @descheleschilder: I have given the answer by referring to the Hertz experiment which clearly implies that E-M waves can be and have been created in the Lab. Visible light is nothing unique and can be created similarly. So the answer is yes. – Lost Aug 23 '20 at 10:22
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To the question "is light nothing more than a pair of transverse electric and magnetic fields?" the answer is: "classically yes, indeed". However, these fields are a wave function, like Schrödinger's equation, that describes the probability of finding photons, the massless point particles that light consists of.

Deschele Schilder
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my2cts
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  • +1. I see the point of your comment to my answer now. – mmesser314 Aug 22 '20 at 19:29
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    This isn't an answer to the question. However these fields are a wave function, like Schrodinger's, that describes the probability of finding photons This simply not true. The fields are not wavefunctions. They have the form of a wave. There are mathematical expressions tough that give a description of these waves. These can derived from Maxwell's equations. – Deschele Schilder Aug 22 '20 at 19:40
  • So we meet again. I am sorry to have to tell you that you are again wrong, @descheleschilder. The famous two slit experiment with very low intensity proved that the classical intensity is to be interpreted as the probability of detecting a photon. – my2cts Aug 22 '20 at 21:56
  • You seriously must try (read!) to get some understanding of what classical and quantum physics are about. Read some books, or read online. It's what I did! Classical electromagnetics, of course, can explain the interference pattern. There is no need to interpreter it in a quantum mechanical way. It's how the pattern is formed where quantum mechanics kicks in: Dot, dot, dot... – Deschele Schilder Aug 22 '20 at 22:19
  • @descheleschilder what my2cts is saying is that classical electrodynamics is an average behaviour of the more accurate quantum description. If the double slit experiment is conducted with sufficiently low intensity, then classical electrodynamics fails to explain the “dot, dot...” – Superfast Jellyfish Aug 23 '20 at 06:17
  • @SuperfastJellyfish an average behavior of the more accurate quantum description. What exactly do you mean with that? – Deschele Schilder Aug 23 '20 at 06:27
  • Fields can entirely be described by charge and current distributions (Jefimenko’s equations). So under quantum mechanics, one replaces those with expectation of charge and current distributions. So the theory is accurate only on average. But if we have enough resolution to resolve between each event then we see quantum properties, like the low intensity double slit experiment. – Superfast Jellyfish Aug 23 '20 at 06:30
  • No. Under quantum mechanics (QED) one assumes the basic unit of an e.m. wave is a photon. You construct the Lagrangian corresponding to the wave-equation for charged particles (the Dirac equation), after which you make the assumption that the Lagrangian has to stay invariant under local gauge transformations. When doing this the four-vector $A_{\mu}$ joins the game which corresponds to the photon field. It's quite difficult to distill the classical wave from photons. – Deschele Schilder Aug 23 '20 at 06:58
  • The classical intensity doesn't represent an average. It's just an intensity, already worked out before the advent of QED. – Deschele Schilder Aug 23 '20 at 06:58
  • @descheleschilder "the basic unit of an e.m. wave is a photon" This statement is not mainstream physics, which holds that a photon is a massless point particle. – my2cts Aug 23 '20 at 09:55
  • @descheleschilder "The classical intensity doesn't represent an average." The classical intensity gives the probability density. This is validated by the famous low intensity experiments. – my2cts Aug 23 '20 at 10:01
  • @my2cts: Wave functions are solutions to the Schrodinger wave equation while E-M waves are solutions of the classical wave equation derived from Maxwell's equations. Hence they are not same. – Lost Aug 23 '20 at 10:17
  • @Lost I agree until "hence". Experiment is on my side. Also, Maxwell's equations are essentially massless wave equations, like the Klein-Gordon equation with m=0. – my2cts Aug 23 '20 at 10:30
  • my2cts: Please state which experimental indicate that E-M waves are matter waves. – Lost Aug 23 '20 at 12:30
  • Also Schrodinger's Differential Equation is different than Maxwell's where the there is an iota with the first derivative of time giving complex solns as opposed to Maxwell's wave equation which is identical to the classical wave equation where there are no complex contributions and is in second order with respect to time , the solution of which give real vector fields. – Lost Aug 23 '20 at 12:33
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That's not the way you can create classical light waves. Both the electric and magnetic fields can't be made to vary in time independently.
It's the easiest just to let an electric field vary periodically in time at visible light frequencies. The magnetic field will automatically be created, according to Maxwell's equations.

A varying magnetic field would also create an electromagnetic wave, but that's much harder to realize because there is much less energy contained in a magnetic field.

In both cases (creating the varying fields) an electromagnetic wave will emerge.

So the answer is: No.

Deschele Schilder
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    Wait. I’m with you till the last line. Why do you say no and how is it implied from your previous statements? – Superfast Jellyfish Aug 22 '20 at 19:30
  • @SuperfastJellyfish So would creating a pair of transverse electric and magnetic oscillating fields moving in a given direction be equivalent to create light?. It's impossible to create a pair of oscillating fields. You either create an oscillating electric field or an oscillating magnetic field (which is more difficult). In both separate cases, there will appear an e.m. wave. When you create an oscillating magnetic field and an oscillating electric field you can't just put them together to create such a wave. – Deschele Schilder Aug 23 '20 at 00:36
  • Whatever fields we create obeys Maxwell’s equations. So if we have an oscillating charge source, it would create a pair of transverse electric and magnetic oscillating fields moving in a given direction. And yes. That is equivalent to light (em wave). – Superfast Jellyfish Aug 23 '20 at 06:12
  • That is not what is asked. The OP is under the impression one can create two fields, a periodically varying magnetic field and the same kind of an electric field, independently from each other. Then you put these two waves together and behold, we have created an e.m. wave. I think he doesn't realize that it takes only one of both to create an oscillating e.m. wave. – Deschele Schilder Aug 23 '20 at 06:21
  • Perhaps OP is under that impression and your answer clarifies the error. But I still don’t understand what the “no” in your answer is referring to. OP has 3 statements with a question mark. So what are you saying no to. Due to this confusion I voted -1 – Superfast Jellyfish Aug 23 '20 at 06:25
  • I said no to his last question: So would creating a pair of transverse electric and magnetic oscillating fields moving in a given direction be equivalent to create light? – Deschele Schilder Aug 23 '20 at 06:29
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    @descheleschilder:'In both cases (creating the varying fields) an electromagnetic wave will emerge.' And then writing 'no' to the question asked is contradictory. – Lost Aug 23 '20 at 10:17
  • @Lost I just stated that in the case of an oscillating electric field an e.m. wave will emerge (containing the two transverse waves) as well as in the case of an oscillating magnetic field. But OP thinks an electromagnetic wave can be made out of two independent oscillating fields: one magnetic and one electric, at two different places. So would creating a pair of transverse electric and magnetic oscillating fields moving in a given direction be equivalent to create light? I think OP wants to create an oscillating e.m. field by superimposing the two. No contradiction. – Deschele Schilder Aug 23 '20 at 10:46
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It can be hard to say what light really is. You are talking about the classical view of light. Descheleschilder's answer is correct. But if you look at a microscopic view of light, you need quantum mechanics.

This is like looking at what air pressure is. In a large scale view (classical), it is a smooth force that air exerts on the walls. But microscopically, it isn't smooth. It is individual air molecules bouncing off the wall. Each molecule gives the wall an individual kick. When you add up lots of these kicks, you get a smooth force. It is really the same explanation, but it looks totally different.

Light is the same. On a microscopic scale, light can be emitted by an individual electron in an atom, and absorbed by another electron in another atom. One atom gives another a kick. When you add up lots of atoms, you can see a smooth force that is described by an electromagnetic field.

An individual atom's worth of light has a name "photon", but that doesn't say what light is. A photon is sort of like a particle and sort of like a wave. For more on that, see my answer to How can a red light photon be different from a blue light photon?.

It can also get confusing if you take a careful look at the classical picture. What kind of thing is an electric field? See my answer to In what medium are non-mechanical waves a disturbance? The aether?

mmesser314
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