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I'm taking E & M II this semester, and one question got me thinking. We know the formulation of the four Maxwell's equations, and that's okay so far. But in the absence of sources, they take the form:

$$\nabla \cdot \mathbf{E}= 0 \\ \nabla \cdot \mathbf{B}= 0 \\ \nabla \times \mathbf{E} = -\dfrac{\partial\mathbf{B}}{\partial t}\\ \nabla \times \mathbf{B} = \mu_{0}\epsilon_{0}\dfrac{\partial\mathbf{E}}{\partial t}$$

The electric field is caused by the presence of charged particles, which if it varies, causes the existence of a magnetic field. But if there aren't any sources, how can the Maxwell's equations have nontrivial solutions? Wouldn't the electric and magnetic field always be zero?

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    $\begingroup$ Take a parallel plate capacitor. Consider the volume between but not including the plates. There are no charges in that volume. Yet there clearly is an electric field. $\endgroup$
    – Jon Custer
    Nov 7, 2022 at 18:58

9 Answers 9

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Think about a simpler problem. You have the following differential equation: $$f'=0$$ Does this mean that $f$ has to be zero? No. It's a possible solution but not the only one. You need to provide the value of $f$ somewhere.

As for Maxwell's equations, zero fields are a possible solution, but not the only one. If the fields are non-zero somewhere (boundary conditions), then a solution with zero fields may no longer be acceptable.

For example, if you are in the dead of space (no charge and no current), you can still have, for example, wave-like solutions describing an electromagnetic wave coming from far away and crossing that space.

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    $\begingroup$ wave coming from far away ... you mean like from a source? $\endgroup$ Nov 6, 2022 at 17:43
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    $\begingroup$ @rafaelcastrocouto Yes. Your question is mathematical in nature, and those equations require boundary conditions to have a solution. Within classical electromagnetism, of course there won't be anything if the universe is empty. Keep in mind, however, that things are more complicated if you move to quantum electrodynamics. $\endgroup$
    – Miyase
    Nov 6, 2022 at 23:33
  • $\begingroup$ Field is either divergence or curl and which is zero in absence of field. So your photon can ionize through which it passes. $\endgroup$ Jan 21, 2023 at 3:05
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Please, remember that differential equations are not enough to describe a system, corresponding to the domain of the PDEs, but they need suitable boundary conditions on the boundary of the domain and initial conditions in the domain itself, if the problem is time-dependent.

If you have no charge in your domain, Maxwell's equations in absence of charges hold. If you measure a non-zero electromagnetic field this "enters" from the boundaries into the domain, and is generated somewhere else in the world, outside the domain of your interest.

You need to know the field on some regions of the boundary of the domain to prescribed the boundary conditions:

  • to physically represent your problem
  • to mathematically obtain a well defined mathematical problem.

Few considerations about boundary conditions:

  • Maxwell's equations in absence of charges and currents may reduce to homogeneous wave equations, a second-order linear PDE that shows an hyperbolic behavior; this qualitatively means that the information travels with characteristic directions and velocity; you may prescribe boundary conditions only in that regions of the boundary where the information is entering the domain, while the solution on the remaining regions of the boundaries depends on the solution in the domain;
  • in order to prescribe a meaningful boundary condition you should have a reliable estimate or a measurement of the physical quantities on the boundary; otherwise you could increase the dimension of your domain to include the sources of the fields acting on the boundary of the original fields.
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A natural interpretation that I got from Wald:

The fact that maxwells equations imply the existence of fields without the need for charges, imply that the EM field is an independant entity existing independant of charges and we should instead think of charges as interacting with the EM field rather than create it.

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    $\begingroup$ Is this a mathematical artifact of the equations? Perhaps, but the interpretation still holds, and there is nothing we can do that can determine what the true background solution is. $\endgroup$ Nov 4, 2022 at 13:45
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    $\begingroup$ Wonderfully explained! $\endgroup$ Nov 8, 2022 at 17:12
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Here is an analogy. If you throw a stone into a pond, you will create ripples in the water that travel outward from the stone. These ripples carry away some of the energy imparted to the water by the stone. Initially, energy was added to a bunch of water molecules near the stone by the stone pushing on them. The water molecules can then push and pull on other water molecules nearby, and so on and so forth, spreading the ripples outward from the entry point of the stone. So you will see ripples in the pond far away from where the stone entered.

You could ask, "how can there be ripples in the pond way out here where there was no stone to create them?" And the answer is just that the water pushing and pulling on itself can drive ripples through the pond, even if no stones collided anywhere nearby.

In this analogy, the water in the pond is like the electromagnetic field, the ripples in the pond are electromagnetic waves, and the collision of the stone with the water is like the acceleration of a charged particle. An accelerating charge creates electromagnetic radiation in a small region near the charge, but away from the charge the propagation of the electromagnetic waves is driven by the internal dynamics of the electromagnetic field, not by any direct influence from the accelerating charge.

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The sources of divergent electric field are charges. Divergenceless electric field can be generated by (or more accurately, accompanied by) a changing magnetic field - as seen in Faraday's law. Certain solutions to the vacuum Maxwell's equations - electromagnetic waves - do exist where both the E- and B-fields are non-zero.

But you are right, if the vacuum version of Maxwell's equations applied everywhere in the universe, then unless one applies a boundary condition where the fields already exist at $t=0$ then there would be no E- or B-fields. Equivalently we can imagine a situation where the vacuum equations are true in some domain, but beyond that there are charges and currents that generate fields. The differential equations you have written apply at a point in space and time, they do not necessarily have to be true everywhere (well, the second one probably does).

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When Maxwell was in the process of trying to undrerstand electromagnetism he came to hypothesize a capacity of empty space that he named 'displacement current'.

At the time the existence of atoms was expected, on the basis of circumstantial evidence; the technology of the time did not yet provide the means for direct confirmation.

It was also expected that matter can exist in the form of ions. (But it was not known whether of the two forms of charge, positive and negative, both were mobile, or only one of them.)

The capacity of a capacitor is dependent on the composition of the dielectric medium. Maxwell offered the hypothesis that some forms of dielectric medium have ions among their constituent parts, and that these ions can be displaced.

When a capacitor is charged up there is a buildup of electric field between the plates. This electric field displaces ions. This displacement offsets the electric field between the plates. This allows the capacitor to store more charge per unit of voltage.


In Maxwell's time the experimentors could already create very good vacuum, so they could also measure the dielectric properties of vacuum.

Most intrigingly, the dielectric property of empty space is such that you have to conclude that something physical is happening. The effect is small, but measurable.

Maxwell considered that if the Luminiferous Aether consists to some extent of electric charge, then it too can support displacement current.

This would allow propagation of electromagnetic waves.

Further reading:

Maxwell's displacement (From Kevin Brown's website mathpages . com)

James Clerk Maxwell:
Wikisource:
On physical lines of force 1861

(This essay 'On physical lines of force' preceded the larger work 'A treatise on electricity and magnetism' by quite a few years. The 'treatise' was published in 1873.




To avoid misunderstanding:
Maxwell wasn't necessarilty committed to his 1861 model of the Luminiferous Aether.
The mechanical model was not featured in the 'Treatise'.

Whatever the case, the existence of propagating electromagnetic ways, that keep propagating when the original source has ceased emitting, strongly suggests that the electromagnetic field has an existence of its own.

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This is an example of the math not fully capturing the physics. What we observe in reality is that every EM field in the universe has a source somewhere. The source may be a long way from where we detect the field, but there's always a source. The math allows for source-free fields, but none are seen.

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  • $\begingroup$ Realistic view. Fine. $\endgroup$ Nov 7, 2022 at 14:41
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You have written the derivative form of Maxwell's equations, also known as the "local" form, because it gives a relationship between $\vec E$ and $\vec B$ near a single point in space. As you know from the one dimensional case, the definition of the derivative asks what is happening only between $x$ and $x+ \mathrm{d}x $. Between there, there may not be a source, the source may be somewhere else, but there may be a field there which was caused by a source elsewhere.

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Nov 5, 2022 at 7:34
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Maxwell's equations specify relations between the space and time derivatives of fields, not the fields themselves.

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