There are lots of posts about EM wave model of photons, but I haven't read one that covers the more specific question I am focusing on here.

Here How does energy transfer between B and E in an EM standing wave? david was concerned about the presence of a zero $E$ & $B$ field point in the wave, close, but that is not my concern.

An electromagnetic waves is propagated by the oscillations of the electric and magnetic fields. A changing electric field produces a changing magnetic field and a changing magnetic field produces a changing electric field. An electromagnetic wave is self propagating and does not need a medium to travel through.

But I can't overcome the idea that in order to achieve propagation an $\dot E$ or $\dot B$ in one place must be capable of inducing a $\dot B$ or $\dot E$ in a different place.

How do we understand a change in position to occur?

In Maxwells Vacuum Equations (such as $\nabla \times E = -\dot B$ ) does not curl E result in a vector located in the same place as E, suggesting that it is only at that point a B field may be induced?

We know that two different EM waves interfere, rather than interacting - yet propagation seems to require that a decaying EM wave at one point interacts with and generates more of itself, that is, an EM wave at another point. Something's missing from the picture.


The rest of the post is just a list of dead ends I considered.

2) If a $\dot E$ resulted in a distant $\dot B$ (or vice versa), energy would need to be carried between the locations. I suppose this might be by a propagating EM wave. However I am trying to understand a propagating EM wave in the first place, and it is difficult (although not impossible) to work with a recursive or circular explaination.

3) if constant speed is assumed one can easily turn a time dependent wave equation to a (space) spatially dependent one. However, what I'm looking for is a mechanism from which to derrive or at least justify propagation, and the assumption of any speed essentially skips over that step.

4) Matter waves, such as on a string exhibit a clear coupling in the form of tension along the string. Although as they are a fundamentally different kind of wave looking for something very similar to that may be flawed. Is there some conceptual aspect of field waves I've managed to miss or forget, I wonder?

5) Maybe I got it backwards, and photon propagation is the evidence of E to B induction over a distance/ I haven't got very far with that line of reasoning

6) Special relativity "explains" magnetic fields as the relativistic effect of charge motion. I always felt this was one of the greatest insights, so I started to wonder if there is any way to make use of it to develop an argument for motion of energy in an E field. Must be googling the wrong keywords though.

7) Another approach is to imagine that $\dot E$ is equivalent to the motion of a charge, and then try to think about the B-field that moving charge would induce. However E fields being present omnidirectionally around the charge seems to make this impossible.

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    $\begingroup$ I know on S.E. it's blasphemous to seriously consider photons as real particles that propagate the energy of light. Sure everyone says duality but they don't mean it. A so called EM wave can easily be derived mathematically and physically based on individual photons. EM waves and fields can't even begin to be explained without incorporating many coherent photons. Photons will explain what your looking for, especially the propagation part. $\endgroup$ Sep 7, 2018 at 6:10
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    $\begingroup$ @Bill no one serious ever objects to using the photon picture in places where it makes sense, nor to claims that phenomena can be described in the photon picture. What I object to is any kind of absolutist suggested that photons is either (a) the only way or (b) always the better way to address problems in E&M. $\endgroup$ Sep 8, 2018 at 17:17
  • $\begingroup$ @dmckee I appreciate you at least discussing it. What you are objecting to, is exactly what I am objecting to. 90% of The books or comments (probably You too) always claim that interference can only be caused by a wave and that particles cannot explain this. I object to that and it can be proven otherwise. The status quo is very biased on this subject when it comes to photons. They claim duality but they don’t really visualize it. $\endgroup$ Sep 8, 2018 at 18:34
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    $\begingroup$ 90% of books claim that classical particles can not explain interference which is exactly true. Photons are not classical particles. Photons are also not the complete quantum picture, as the represent unlocalized Fock states of the field with well defined momentum. That expansion is the correct one for asymptomatic states (i.e. the far field), but is not fully general. And you have repeatedly claimed that situation perfectly described by classical E&M have to be treated in terms of photons, which is simply not the case. $\endgroup$ Sep 8, 2018 at 19:16

4 Answers 4


Looking at matter waves on a string: As long as the oscillations aren't too big, we get the wave equation for $f(x, t)$:

$$ \frac{\partial^2 f}{\partial t^2} = c^2\frac{\partial^2 f}{\partial x^2} $$

Your question applies to this equation just as well as it does to the equations of electricity and magnetism. Don't both the terms in this equation apply just to a single point? How can disturbances propagate through space without violating locality?

The key is to go back to the definition of a partial derivative:

$$ \frac{\partial f(x, t)}{\partial x} = \lim_{h \to 0} \frac{f(x+h, t) - f(x,t)}{h} $$

Now this next bit is going to get a little hand-wavey, since that's the nature of this question. This partial derivative doesn't just care about $f$ at the point $(x,t)$. It also cares about the value of $f$ in a tiny, ever-shrinking neighborhood right around $x$. Similarly, the derivative $\frac{\partial^2 f}{\partial x^2}$ cares about an arbitrarily small, but not point-like, neighborhood around $x$.

Suppose that we stop the shrinking of these neighborhoods at some point, so they have size $\epsilon$. Then instead of behaving like a string, our model behaves like a bunch of point masses connected by springs of length $\epsilon$. However, as $\epsilon$ goes to 0, this behavior approximates that of a truly continuous string.

So the short version of this answer is that having spatial derivatives in your PDE is what allows things happening in one point in space to affect other points in space. This has something to do with derivatives existing in a strange twilight zone where, on one hand they are local, but on the other they care about change over spatial distance.

And spatial derivatives appear in Maxwell's equations in the terms $\nabla\cdot E$, $\nabla\cdot B$, $\nabla\times E$, and $\nabla\times B$, so it shouldn't be too much of a mystery that light travels through space.

  • $\begingroup$ I know I quoted Maxwell in the title. It seems like a good way in for those more mathematically inclined. Anyway with physical waves there are forces relating different physical points on the wave; whether pressure or dispalcement, and that is why I have no (well, I'll say "much less") to question about physical waves. (Your answer so far has not addressed this, but just explained calculus really.) Look at it this way; somebody else might say "I believe", they would go, "I believe the quantity E is discontinuous over space, why shouldn't I? Therefore differentiate smitherentiate." $\endgroup$
    – JMLCarter
    Sep 18, 2018 at 23:26

This answer is general, but too long for a comment.

When modeling physical behavior with mathematical functions one has to be clear:

Are we talking : a) mathematics creates reality or b)mathematics models reality.

a) is the platonic view and b) the realist view.

Under a) the predictive power of mathematics leads to questions as the ones above and they are answered by the other answers

Under b) one is not surprised by the underlying quantum mechanical level, that is modeled with a quantized maxwell equation, which eventually builds up the classical electrodynamics equations, as both depend on the same mathematics with different application. How the classical light description emerges from a confluence of probabilistic photons in quantum field theory is outlined here

Imo it is the realist view that physicists should have, using mathematics as a tool of modeling new data and asking questions of the theories that fit them. That is how science has progressed since the time of Newton.

A rough analogy for the classical electromagnetic wave:

If one maps a dry river bed mathematically, the shape of the water flow when it rains can be predicted immediately, long before the water reaches the bends. In a similar sense, Maxwell's equations map space time, and given the initial conditions (light beam) the "flow" is beautifully predictable.

edit after comments:

To carry my analogy to a map further, a map is static. That is because time is a parameter and does not enter in the functions describing a map, which can be very accurate and used to predict motion through it, it is the context for a flow.

Maxwell's equations are a four dimensional map. Static has no meaning in four dimensions , because time is one of the dimensions. Given the initial conditions , there is a complete solution. The predictive in time possibility is because one deals with time and space separately. but the solution for given boundary conditions,is unique in four dimensional spacetime.

The underlying quantum mechanical level of photons also explains why E and B can be zero at the same spacetime point, what happens to the energy ?


Electromagnetic waves can be imagined as a self-propagating transverse oscillating wave of electric and magnetic fields. This 3D animation shows a plane linearly polarized wave propagating from left to right. Note that the electric and magnetic fields in such a wave are in-phase with each other, reaching minima and maxima together.

The superposition of the zillions of photons making up the classical wave, also give a wavefunction for the whole bundle. The zeros mean that there is zero probability for a photon to exist at those points where both E and B are zero, thus energy conservation can be carried out at all space time points.

  • $\begingroup$ regarding your last paragraph, would it be correct to say, we can replace "flow" with "the propagation of photons through space"? $\endgroup$
    – undefined
    Sep 7, 2018 at 7:22
  • $\begingroup$ @undefined it s a qualitative analogy, so it depends how far you take it. $\endgroup$
    – anna v
    Sep 7, 2018 at 7:29
  • $\begingroup$ I know, I meant it for my understanding of it $\endgroup$
    – undefined
    Sep 7, 2018 at 8:22
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    $\begingroup$ Well I am usually found firmly in camp b), but that said; understanding theories (especially mathematically expressed) does seem capable of providing some insight; albeit, that is second hand news. On that basis I interpret your answer to be that there is no insight into light propogation to be gained from Maxwells Equations. I'm not clear if you are saying there is some insight in quantum field theory... probably not? Nor anywhere else I suppose. $\endgroup$
    – JMLCarter
    Sep 18, 2018 at 22:59
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    $\begingroup$ "Imo it is the realist view that physicists should have, using mathematics as a tool of modeling new data and asking questions of the theories that fit them." This is what the OP is asking about. How does the mathematical model describe the propagation of waves when the model seems to only describe what is happening at each isolated point in space. $\endgroup$ Sep 19, 2018 at 0:06

You are (roughly speaking) correct that “in order to achieve propagation an E˙ or B˙ in one place must be capable of inducing a B˙ or E˙ in a different place”. This can be seen directly in Maxwell’s equations.

Faradays law says $\nabla \times E = -\dot B$. Note that the curl is a spatial derivative, and the dot is a time derivative. So a B field changing in time gives an E field changing in space. This introduces coupled changes over time and space.

Similarly with Ampere’s law.

  • $\begingroup$ $\nabla \times E$ only gives an infinitesimally "rotating E field projection" at the same point as E. Is there an equation or phyiscal theory that tells me that the same quantity in a "neighbouring" point cannot be a discontinuous value, no stronger, must be a related value? $\endgroup$
    – JMLCarter
    Sep 18, 2018 at 23:13
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    $\begingroup$ I don’t know what you mean by “infinitesimally rotating projection”. If you expand the curl operator then you will get spatial derivatives. Spatial derivatives describe how something varies from place to place. Neighboring values are related to each other by the derivatives. That is the whole purpose of a spatial derivative, to describe how something changes with respect to space. $\endgroup$
    – Dale
    Sep 19, 2018 at 18:08
  • $\begingroup$ Grad gives spatial derrivatives, Curl gives rotational spatial derrivatives. Curl is a vector operator that describes the infinitesimal rotation of a vector field in three-dimensional Euclidean space. At every point in the field, the curl of that point is represented by a vector, neighbouring curls can be discontinuous. $\endgroup$
    – JMLCarter
    Oct 15, 2018 at 1:24
  • $\begingroup$ Anyway, I am looking to move beyond the existance of a relation and understand what can be the reason for it, such as is readily available in the analysis of matter waves in a medium. $\endgroup$
    – JMLCarter
    Oct 15, 2018 at 1:50
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    $\begingroup$ I don’t know what more can be done for you here. Maxwell’s equations already do that explicitly in the spatial derivatives, as I have pointed out already. It seems like your question is based on a problem understanding differentiation rather than specifically Maxwell’s equations. Again, spatial derivatives describe how the function in one location is related to the function in neighboring locations $\endgroup$
    – Dale
    Oct 17, 2018 at 2:28


Answering my own question? Well yes; My view has developed.

Thought experiment; a wave of matter that does not propagate through a medium.

Consider a gun in vacuum oscilating sinusoidaly perpendicular to the axis along which it is repeatedly firing. The bullets form a sine wave in space. The wave moves through space. There is a relation between each element of the wave and the next. This is wave type "A".

enter image description here

Whilst this is a matter wave it is a mediumless wave and different from a wave propagating through a medium, such as a sound wave, wave on the sea or wave on a string. These are examples of wave type "B".

I understand that light is like wave "A" (above), it is a static entity moving through space, whose form was defined at the point of generation. Any relation between wave properties at two points is a consequence of the mechanism that was used to generate it, it's not a restriction upon free space.

In matter waves through a medium ?(wave "B") the waveform is affected by the generation mechanics, but ultimately as it propagates the medium properties will dominate and degrade the waveform towards sinusoidal.

Perhaps this is what I have missed all along, that in the sense described above "field vectors" themselves, like matter, move through space.

  • $\begingroup$ well, you have just "accepted" the existence of photons. $\endgroup$
    – anna v
    Oct 15, 2018 at 4:00
  • $\begingroup$ This is wrong. In your terminology, light is medium-less but still of type B. $\endgroup$ Oct 15, 2018 at 6:37
  • $\begingroup$ @anna v I didn;t think I had. All the above is saying is that something that is not matter and is not the influence of fields, but more the fields themselves is propagating. Anyway apparently it's the wrong track, maybe. I'm fine to park it for now. $\endgroup$
    – JMLCarter
    Oct 16, 2018 at 18:34
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    $\begingroup$ @JMLCarter Yes. It is also known as "physics as we know it". I'm not sure what gave you the impression that you can describe the propagation of EM fields without Maxwell's equations, but if you try to do that then you're not doing physics. $\endgroup$ Oct 16, 2018 at 18:40
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    $\begingroup$ That's already been explained in the existing answers. (Basically: Maxwell's equations don't "relate fields at the same point", because they contain spatial derivatives.) The solution you present in this answer is to ditch established physics and to make up nonsense as a substitution; since that seems to make you happy, I'm not here to debate with you. I commented exclusively because it's important that incorrect content be noted as such. If you want to take that incorrectness seriously and learn from the existing correct answers, great, if not, then it's your choice. $\endgroup$ Oct 16, 2018 at 21:54

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