# Is anything lost by simply saying that a charge is nothing but the divergence of a field?

If we propose that 1) there are fields and 2) that some fields can have places where they converge (that is to say, places where del dot the field is not zero) then we have the definition of such a region, and we might as well call it a charge. In other words, instead of proposing that a a divergence is non-zero due to a charge, why should we not simply define a charge as a divergence / convergence of a field? Is there any aspect or property of a charge (of any kind, not just electric) that is not adequately described by this definition?

• conservation thereof? – JEB Jul 2 '19 at 0:09
• @JEB Charge conservation is a consequence of the Maxwell equations. It'd be interesting to think about charge quantization in this picture, though. – rob Jul 2 '19 at 0:12
• One assumes you are talking about electric charge only? The space integral of the field divergence? – Cosmas Zachos Jul 2 '19 at 0:27
• I suppose that you could define charge as being nothing but the divergence of an E-field, but what would be the advantage of looking at things in that way? For example, suppose we have an electric current in a wire. Instead of viewing this as being the movement of electrical charges in a wire, according to your picture we should view it as some sort of movement or evolution of an electric field which is the sum of individual electric fields, each of which has a non-zero divergence at a point which moves with time. Seems like an unnecessarily complicated picture. – user93237 Jul 2 '19 at 2:29
• @rob- yeah, quantization is what I’m getting at. – vwcanter Jul 3 '19 at 15:54

This approach does not look complete - you need something extra, for example, to describe the mass of the charge.

EDIT(7/3/2019): As I said, the approach in the question does not look complete: it does not take into account the mass of the charges and thus does not describe the motion of charges. However, this issue can be overcome in some cases, though this is not easy. Let us look at the problem from a more general point of view. If we define charges as divergence of field, we actually eliminate charges from our equations. So how far can we proceed in this direction? Can we do just with fields, for example, just with electromagnetic fields, but still describe the physics behind motion of charges? It turns out it can be done, for example, for the case of scalar electrodynamics, where minimally interacting electromagnetic and matter fields are described by the Maxwell and Klein-Gordon equations, respectively. The following is taken from my article (Eur. Phys. J. C (2013) 73:2371, open access).

The equations of scalar electrodynamics are as follows: $$(\partial^\mu+ieA^\mu)(\partial_\mu+ieA_\mu)\psi+m^2\psi=0,$$ $$\Box A_\mu-A^\nu_{,\nu\mu}=j_\mu,$$ $$j_\mu=ie(\psi^*\psi_{,\mu}-\psi^*_{,\mu}\psi)-2e^2 A_\mu\psi^*\psi.$$ The complex charged matter field $$\psi$$ in scalar electrodynamics can be made real by a gauge transform (at least locally), and the equations of motion in the relevant gauge (unitary gauge) for the transformed four-potential of electromagnetic field $$B^{\mu}$$ and real matter field $$\varphi$$ are as follows ( Schrödinger, 1952): $$\Box\varphi-(e^2 B^\mu B_\mu-m^2)\varphi=0,$$ $$\Box B_\mu-B^\nu_{,\nu\mu}=j_\mu,$$ $$j_\mu=-2e^2 B_\mu\varphi^2.$$ It turns out that the equations obtained from the previous system after natural elimination of the matter field form a closed system of partial differential equations and thus describe independent dynamics of electromagnetic field. The detailed wording is as follows: if components of the four-potential of the electromagnetic field and their first derivatives with respect to time are known in the entire space at some time point, the values of their second derivatives with respect to time can be calculated for the same time point, so the Cauchy problem can be posed, and integration yields the four-potential in the entire space-time.

To eliminate the matter field $$\varphi$$ from the previous system, let us use a substitution $$\Phi=\varphi^2$$ first. For example, as $$\Phi_{,\mu}=2\varphi\varphi_{,\mu},$$ we obtain $$\Phi_{,\mu}^{,\mu}=2\varphi^{,\mu}\varphi_{,\mu}+2\varphi\varphi^{,\mu}_{,\mu}= \frac{1}{2}\frac{\Phi^{,\mu}\Phi_{,\mu}}{\Phi}+2\varphi\varphi^{,\mu}_{,\mu}.$$ Multiplying the equation for the matter field (the Klein-Gordon equation in the unitary gauge) by $$2\varphi$$, we obtain the following equations in terms of $$\Phi$$: $$\Box\Phi-\frac{1}{2}\frac{\Phi^{,\mu}\Phi_{,\mu}}{\Phi}-2(e^2 B^\mu B_\mu-m^2)\Phi=0,$$ $$\Box B_\mu-B^\nu_{,\nu\mu}=-2e^2 B_\mu\Phi.$$ To prove that these equations describe independent evolution of the electromagnetic field $$B^\mu$$, it is sufficient to prove that if components $$B^\mu$$ of the potential and their first derivatives with respect to $$x^0$$ ($$\dot{B}^\mu$$) are known in the entire space at some time point $$x^0=\rm{const}$$ (that means that all spatial derivatives of these values are also known in the entire space at that time point), the previous system yields the values of their second derivatives, $$\ddot{B}^\mu$$, for the same value of $$x^0$$. Indeed, $$\Phi$$ can be eliminated using the Maxwell equation for $$\mu=0$$, as this equation does not contain $$\ddot{B}^\mu$$ for this value of $$\mu$$: $$\Phi=(-2e^2 B_0)^{-1}(\Box B_0-B^\nu_{,\nu 0})= (-2e^2 B_0)^{-1}(B^{,i}_{0,i}-B^i_{,i 0})$$ (Greek indices in the Einstein sum convention run from $$0$$ to $$3$$, and Latin indices run from $$1$$ to $$3$$). Then $$\ddot{B}^i$$ ($$i=1,2,3$$) can be determined by substitution of this equation into the Maxwell equations for $$\mu=1,2,3$$: $$\ddot{B}_i=-B^{,j}_{i,j}+B^\nu_{,\nu i}+(B_0)^{-1} B_i(B^{,j}_{0,j}-B^j_{,j 0}).$$ Thus, to complete the proof, we only need to find $$\ddot{B}^0$$. Conservation of current implies $$0=(B^\mu \Phi)_{,\mu}=B^\mu_{,\mu}\Phi+B^\mu\Phi_{,\mu}.$$ This equation determines $$\dot{\Phi}$$, as spatial derivatives of $$\Phi$$ can be found from the equation used for elimination of $$\Phi$$. Differentiation of the previous equation with respect to $$x^0$$ yields $$0=(\ddot{B}^0+\dot{B}^i_{,i})\Phi+(\dot{B}^0+B^i_{,i})\dot{\Phi}+ \dot{B}^0\dot{\Phi}+\dot{B}^i\Phi_{,i}+B^0\ddot{\Phi}+B^i\dot{\Phi}_{,i}.$$ After substitution of the expressions for $$\Phi$$, $$\dot{\Phi}$$, and $$\ddot{\Phi}$$ (from the previous equation) into the Klein-Gordon equation we obtain an expression for $$\ddot{B^0}$$ as a function of $$B^\mu$$, $$\dot{B}^\mu$$ and their spatial derivatives (again, spatial derivatives of $$\Phi$$ and $$\dot{\Phi}$$ can be found from the expressions for $$\Phi$$ and $$\dot{\Phi}$$ as functions of $$B^\mu$$ and $$\dot{B}^\mu$$). Thus, if $$B^\mu$$ and $$\dot{B}^\mu$$ are known in the entire space at a certain value of $$x^0$$, then $$\ddot{B}^\mu$$ can be calculated for the same $$x^0$$, so integration yields $$B^\mu$$ in the entire space-time. Therefore, we do have independent dynamics of electromagnetic field.

Thus, modified Maxwell equations for electromagnetic field only describe the evolution of both the electromagnetic field and the matter field.

A similar conclusion is derived in the same article (with some caveats) for spinor electrodynamics (the Dirac-Maxwell electrodynamics).

• "But why?" This answer as it is (v1) does not show any reasoning. – Quantumness Jul 2 '19 at 1:31
• @Quantumness : And what reasoning do I need? If charge is just "divergence of a field", we have no information about its mass. If you disagree, you need to indicate the source of such information. – akhmeteli Jul 2 '19 at 2:08
• I agree that one must know the mass to analyze many situations, but I do not see how that ties in to OP's question: Gauss's law provides no information about mass. What do you mean by "complete" in this context, and why would you need something to describe the mass? – Quantumness Jul 2 '19 at 2:50
• @Quantumness Gauss’s law don’t tell you anything about electrons as the source of an electric field. The approach that charges are disturbances of a overall field, maybe, is good for the description of atomic processes but it was applied to macroscopic processes and this without any need. – HolgerFiedler Jul 2 '19 at 3:04
• @HolgerFiedler If I understood your comment correctly, that is my point. Gauss's law does not give much information other than the charge (density). I am analyzing my thoughts but AFAIK right now there is nothing inconsistent with OP's definition. – Quantumness Jul 2 '19 at 3:17

One purpose of electric charge is to explain the movement of particles in a field. If you look at charges just as "the divergence of a field", or (more generally) as a property solely of the fields, then you can't explain the movement of particles that carry this charge.

You might say now that you're not interested in how particles are affected by the fields anyways. Still then, the movement of these same particles or (if you distribute them evenly) the "movement" of a "charge density" affects the fields, and this is something you want to describe with your theory. If you don't attribute charge to be a property of something else than a field, something that can move, then you'll have a hard time making sense of certain phenomena you'll observe in the fields.

TLDR: Properties of charge that are not adequately described:

• Its ability to move from one position to another, and to therefore affect the field in a sensible pattern.
• The movement of charge to be (conversely) affected by the fields
• But it does explain the motion of charges. Does it not? Because two convergences should repel, and two divergences should repel, and a convergence should attract a divergence. Am I mistaken? – vwcanter Jul 2 '19 at 14:00
• @vwcanter: Of course you can propose rules to the time-evolution of the field (what you word as "two divergences should repel"). Question is, how easy will these rules be - I can't answer that question. The other question is: What do you do with the situation that a charged particle is moved by an exterior force (which is independent from the field). How would you model this? – Quantumwhisp Jul 2 '19 at 15:33
• I think you can show that they should follow the same law as the basic Coulomb force. If a charge is moved by a mechanical force, usually that would mean an electromagnetic force, so I that case, it would just be the E field interacting with charges, and nothing else. If it’s due to another force, not the EM force, then in that case, there be another field and another charge that corresponded to the field. – vwcanter Jul 2 '19 at 17:57
• It’s not just a proposed rule that would happen to account for observes behavior- the same rule applies already to fluids, for example, two divergences of fluid will repel, and so on. – vwcanter Jul 3 '19 at 16:06
• @vwcanter how would you describe behavior of two particles with the same charge and different masses? You'll have to introduce mass in addition to simple divergence of the field. – Ruslan Jul 4 '19 at 9:46

Basically, electromagnetism requires (a) electric and magnetic fields and (b) sources for those fields, i.e. charge density and current. The point of Maxwell's equations then is to relate the fields to the charges so that you can determine $$\vec E$$ and $$\vec B$$ from $$\rho$$ and $$\vec j$$ (up to vacuum solutions). In other words, the equation $$\vec\nabla \cdot \vec E=\rho$$ gives a connection between two quantities, and you know one of those, you can deduce something about the other one. Furthermore, you can deduce charge conservation as a nontrivial condition on $$\rho$$ and $$\vec j$$, and that conservation really motivates the interpretation of current as flow of charge.

On the other hand, if you simply define $$\tilde\rho:=\vec\nabla \cdot \vec E$$ (and presumably analogously for the current $$\vec{\tilde{j}}$$), Maxwell's equations are empty: You have fields $$\vec E$$ and $$\vec B$$, and the divergence of $$\vec E$$ is an arbitrary quantity you have just decided to call $$\tilde \rho$$. (Also, charge conservation becomes an empty identity: $$\partial_t \tilde\rho =\vec\nabla \cdot \vec{\tilde j}$$, but that is just shorthand for $$\partial_t \vec{\nabla} \cdot \vec{E} =\vec\nabla\cdot \left(\vec\nabla\times\vec B +\partial_t \vec E\right)$$, i.e. an identity and not a relation between physical quantities.

The situation is maybe somwhat similar to the equation of motion inclassical mechanics: in $$m\vec a=\vec F$$, you need an independent expression for $$\vec F$$ to solve for the trajectory of the particle.

• Wouldn’t that just give you a conservation of divergences? – vwcanter Jul 2 '19 at 15:23
• @vwcanter I have added a comment on the conservation law. Does that answer your question? – Toffomat Jul 2 '19 at 16:21
• Toffomat- yes, that’s exactly my main concern. I believe there is missing piece, and that’s what I’m looking for. You put it exactly right. – vwcanter Jul 3 '19 at 16:03

A mathematical theory starts with axioms. It is trivial and you should know that in a complete mathematical theory one is allowed to use theorems proven by axioms , as axioms, and then one can prove the axioms as theorems. Usually one picks up as axioms the simplest statements, and the rest are theorems proven using mathematics and logic.

A physics theory takes a complete mathematical theory with its axioms, and imposes extra axioms by laws , postulates and principles, and also by specific definitions of the variables used to associate to specific measurement variables.

Take Eucledian geometry, which is the first physics model in the western tradition, through the parallel postulate it fitted observation on the fields and land distribution.But one could take one of the theorems and make it an axiom instead of the parallel lines. In general any theorem in a rigorous mathematical model can turn into an axiom, and then the axiom can be proven as a theorem. One chooses the simplest for the axioms.

The same is true for more complicated physics models, the theorems provable using the mathematical axioms and the postulates can become axioms, and then the former axioms have to be proven as theorems. One chooses the simplest for axiomatic position.