There is field without source. In the case at hand, the Maxwell equations allow for solutions in the absence of source.

For the general case, consider the linear problem K f = g subject to some boundary conditions. The solution of this problem is unique only up to a superposition of the solutions of the homogeneous problem K f = 0 that satisfies the relevant boundary conditions. This arbitrariness is absent only when det(K) ≠ 0, whereby K f = 0 is only satisfied for f ≡ 0. The problem of electromagnetic fields in free space being a linear problem, it is not exempt from these considerations.

I recall that in the theory of linear differential equations, the general solution of the problem, in the presence of source, is expressed as a linear superposition of a particular solution of the differential equation and the solution(s) of the source-free differential equation. The sought-after solution (the one that satisfies both the relevant differential equation and the boundary conditions) is determined by imposing the relevant boundary conditions on this linear superposition. (I attach a relevant link below.)

http://tutorial.math.lamar.edu/Classes/DE/NonhomogeneousDE.aspx

Dear dr. Prokhorenko,

the question you pose is somewhat tricky. In the example of light, you usually can perceive both, light sources as well as lit objects. Sometimes though, even if you believe that a source must be there, it may go unnoticed. That happens, for instance, if a laser pointer isn't directed toward your eyes. Again, if you only concentrate on the behavior of lit objects, then you may or may not take sources into account.

From a theoretical viewpoint, if electromagnetic radiation is conceived as a property of matter, then whatever frequency range is detected, one deals only with signals received from luminous and lit bodies. And one calls electromagnetic field those received signals, once they are displayed as (spatial) images.

By far the most common way to conceive of an electromagnetic field is however to attribute to itself space-time properties. As Faraday's lines of force were substituted by electromagnetic properties such as vacuum energy, the latter is also ascribed to void space.

Really I don't understand Spanish and Google don't translate good :(

Sara, by sources I understand the following: L*A=j - Maxwell equations. Their general solution is: A=A0+G*j, where A0 - general solution of L*A0=0, G - Green function. No field with sources means A0=0 and Green function is fixed. But for photons we only have A0 without j!

Other approach is to allow A0 to depend on j, may be it is equivalent of choosing definite Green function, if dependency is linear.

Dear dr. Prokhorenko,

OK. My answer was too generic. There are many approaches to electromagnetic fields, and among them there is also the formalism of second quantization. Perhaps the formalism of Green's function is incompatible with vacuum fluctuations.

1. Here second quantization is just the only quantization of classical field - photon description. If we dequantize classic field we will get just classical massless particle.

2. Vacuum fluctuations just derives from Green function.

Eugene,

don't mix up an algorithm with the physics that the algorithm is intended to describe. Field quantization belongs to such an algorithm. In setting up the perturbation algorithm of QED, Feynman started from a "free" quantized electromagnetic field and introduced an interacting term with charged particles. However, when he finally arrived at the formulation of the Feynman rules, he had eliminated all field operators of the electromagnetic field. In his formulation of QED, the charged particles interact "at a distance, albeit retarded in time", while the quanta of the electromagnetic field act only as a kind of bookkeeping variables. (Have a look at his papers of 1949/50.) Therefore, in Feynman's formulation of QED (which is in use up to the present day) "photons" are described only by their action on charged particles. Therefore, they always have a source and an absorber. (This does not prevent us from calculating transition amplitudes with "incoming" or "outgoing" photons without knowing the sources or absorbers.)

I am not sure whether I understand the question correctly, but even classically, the Maxwell equations do allow for electromagnetic fields without sources. This means, if one sets the charges and currents in the Maxwell equations to zero, there are still nonvanishing solutions for E and B, which describe electromagnetic waves. See, for example, Chapter 16 in [A. Zangwill, Modern Electrodynamics, Cambridge University Press, 2012].

Exactly. Things are also complicated because solutions depends on space-time manifold  topology. Walls on the border can be considered as sources.

Let us write down Maxwell's equations in differential form ( ħ=c=1 units)

One definitely needs sources to produce fields. However, fields can also exist in source free regions. To obtain electric and magnetic fields in source free regions, one has to set the RHS of inhomogeneous Maxwell's equations to zero, and then solve the set of coupled differential equations.

Dear Dr. Brahmachari, one can set even ρ and j to zero in the whole three-dimensional space (not just in some region), and the resulting coupled differential equations for E and B will still have nonvanishing solutions (see e.g. the reference mentioned in my previous answer). These are electromagnetic fields without sources.

No, because that will violate cause effect relationship. Source is the cause and field is the effect. Technically, a theory cannot violate causality principle.

Maxwell's equations are just a synthesis of four known laws 1. Coulombs Law 2. Gauss's law of magnetism 3. Faraday's law of induction 4. Ampere's law. All of them perfectly respect cause effect relationship.

Even in the quantum version of the theory (QED) causality is strictly guaranteed.

"Source is the cause and field is the effect" - this is precisely the question under debate. I would not generally agree with this statement, because the Maxwell equations without sources still have nonvanishing wave solutions for the electric and magnetic fields.

What is true, however, is the following: If we consider electromagnetic fields produced in the laboratory, such that all fields and all sources vanish before some initial "switching on", then the general solution of the initial value problem for the Maxwell equations is indeed given in terms of a retarded Green function. Hence, in this case, there is indeed a causal relationship between the electromagnetic fields and the sources. Please allow me to provide here a link to our recent article, where we have discussed this problem in detail in Section 3.2:

https://www.researchgate.net/publication/275208133_Functional_Approach_to_Electrodynamics_of_Media

Article Functional Approach to Electrodynamics of Media

BTW, simplest way to solve homogeneous equations is to make Fourier transformation for x,y,z. This will definitely restrict class of solutions and basic solutions will have infinite energy. Solutions will be of oscillator type.

There is field without source. In the case at hand, the Maxwell equations allow for solutions in the absence of source.

For the general case, consider the linear problem K f = g subject to some boundary conditions. The solution of this problem is unique only up to a superposition of the solutions of the homogeneous problem K f = 0 that satisfies the relevant boundary conditions. This arbitrariness is absent only when det(K) ≠ 0, whereby K f = 0 is only satisfied for f ≡ 0. The problem of electromagnetic fields in free space being a linear problem, it is not exempt from these considerations.

I recall that in the theory of linear differential equations, the general solution of the problem, in the presence of source, is expressed as a linear superposition of a particular solution of the differential equation and the solution(s) of the source-free differential equation. The sought-after solution (the one that satisfies both the relevant differential equation and the boundary conditions) is determined by imposing the relevant boundary conditions on this linear superposition. (I attach a relevant link below.)

http://tutorial.math.lamar.edu/Classes/DE/NonhomogeneousDE.aspx

One can understand the existence of electromagnetic waves in a system where there are no charges and currents at all in the plane English. This is because a change in an electric field can produce a magnetic field and vice-versa. So, electromagnetic waves can be viewed as a self-sustaining system that does not need any charge. We can imagine a universe where there is no charge but where there are electromagnetic waves. This is what we mean by the mathematical statement that the Maxwell equations admit a solution in absence of charges.

No, pure electromagnetic wave system will pair produce electrons and positrons in MeV energy scales or higher. This is an U(1)em gauge invariant interaction. Therefore a system of pure electromagnetic waves is not "self sustaining" as such. An universe where there are no charges and which contains purely radiation is not feasible at all. This is precisely why universe entered matter dominated era at the end of radiation dominated era.

Sanjeev Kumar Verma: in other words, you can think of EM waves as of elastic string without masses on the ends. This string can vibrate.

'Biswajoy Brahmachari: You are right if in your theory electron charge "e" is not zero. In my world I just have photons, not electrons. Theory is simple and linear :)

Dear Behnam,

indeed, the solutions of the homogeneous (source-free) differential equations are - in the case of the Maxwell equations - precisely the wave solutions which I had mentioned in my previous answer. I would also like to add that one has to distinguish between (spatial) boundary conditions and (temporal) initial conditions on the electromagnetic fields. By imposing appropriate initial conditions, the solution of the Maxwell equations becomes unique as discussed in our article mentioned before.

Giulio, I find out many interesting things in your article! Unfortunately from mathematical point of view, we are loosing some nontrivial solutions (singular distributions) when do Fourier transformation. The same is true for longitudinal/transverse decomposition, where space-time topology is involved.

Dear Giulio, my earlier response was to the main question on this page, and did not amount to a commentary on the various responses by others, including yours. As for the initial conditions, that was taken for granted in my comment. In fact, when using a two-sided time-Fourier transform, we take account of it by adding an appropriate infinitesimal imaginary part to the frequency. In this way, we transform the problem into an eigenvalue one, very much in the spirit of dealing with the time-dependent Schrödinger equation in terms of the solutions of the time-independent one. Once one has all the eigenmodes, one can separately deal with any desired initial condition by appropriately superimposing these eigenmodes.

Dear Eugene,

thank you very much. I agree with you, and this is why we have provided the general solution of the wave equation with appropriate initial conditions in real space, see Eq. (3.11) in our article which coincides with the so-called Duhamel formula. In Fourier space, one has to keep track e.g. of the retardation by including the infinitesimal η in the denominator of the retarded Green function (compare Eqs. (3.8) and (3.9) of our article).

Dear Biswajoy Brahmachari,

I am referring to your reply below:

"No, pure electromagnetic wave system will pair produce electrons and positrons in MeV energy scales or higher. This is an U(1)em gauge invariant interaction. Therefore a system of pure electromagnetic waves is not "self-sustaining" as such. A universe where there are no charges and which contains purely radiation is not feasible at all. This is precisely why universe entered matter dominated era at the end of radiation dominated era."

Here you are talking about our physical universe where EM field is described by QED. However, my viewpoint and that of this question was classical. In classical ED, you can have a solution of Maxwell Eqs. with no charge (j_\mu = 0). However, due to quantum mechanics and gauge invariance, we have the minimal EM coupling (between charges and photon in QED that you were referring to). This arises automatically in QED.

But, in classical ED, we can have EM waves without any j_mu anywhere!

As I understand the question, it does not necessarily imply that we should restrict ourselves to classical electrodynamics. But in any case, it implies that we should disregard the sources (which in reality are always present). In the case of classical electrodynamics, this means that we should set ρ = 0 and j = 0 on the right hand side of the Maxwell equations; in the case of quantum electrodynamics, this means that we should disregard the matter fields as well as the coupling term between the electromagnetic and matter fields in the QED Lagrangean. In both cases, there will still be electromagnetic fields present in the theory: in the form of classical wave solutions, or in the form of stable excited states ("photons") of the quantized electromagnetic field.