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Hope this doesn't come off as too pedantic or overinterpreted. I've been working on revisiting electrostatic and electrodynamic energy from first principles, and I have the following stumbling block. I'm working within the framework of classical electrodynamics only (mostly ignoring the interaction of radiation and matter), and this is on purpose. It's about the classical theory and how it is developed.

We can say that a charge density $\rho$ and a current density $\mathbf J$ are the fundamental sources of the electromagnetic field. We will then consider a point charge to be a sufficiently small and localized charge distribution which has a monopole moment and negligible multipole moments, for all observation points of interest.

Suppose we impose electrostatic conditions, such that all motions of charges must be infinitely slow (adiabatic, in the late 19th-century sense). This is a valid approximation e.g. when charges are confined to a conductor and are in equilibrium. We define the electrostatic potential due to the charge density $\rho$ by $$ \Phi(\mathbf r) = \frac{1}{4\pi\epsilon_0}\int \frac{\rho(\mathbf r')}{|\mathbf r - \mathbf r'|} d^3 r'. $$ And note that the electric field $\mathbf E = -\nabla \Phi$, and that $$ \Phi(\mathbf r) = -\int_{\mathbf r_0}^\mathbf r \mathbf E(\mathbf r') \cdot d\mathbf r' $$ Where $\mathbf r_0$ is a point of zero potential.

Question: How can the electrostatic energy of a continuous charge distribution be derived without heuristic arguments or generalizations from configurations of point charges, in the framework of classical electrostatics (as laid out above)?

Remarks. Energy is not a strictly defined quantity in classical physics. It does not have an exact definition, and in fact the concept is only as useful as its features allow (conservation, modes of energy transfer). The success of the common definition of electrostatic energy is not evidence for an accurate, rigorous, or unique derivation. If one were to formulate electrostatic energy from first principles in the classical model, the formulation is open to some interpretation (e.g. any constant term can be added to the total energy of a system without changing the energy balance, including terms calculated from point charge self-energy, and there is no way to transfer this part of the energy so it makes no difference to the theory and should be eliminated from it).

To make this more exact, then, I say that any definition of energy should have the following features:

  1. It must have units of joules, and any work done in the system between charges must be convertible from energy in the system.
  2. The relationship between electric potential $\Phi$ and the energy would ideally follow from the product of $\Phi$ and an infinitesimal charge $\rho dV$
  3. In the case of "point charges", the energy of the configuration should be obtained by substituting Dirac delta functions for each of the charges, so long as the monopole approximation holds, although the concept of a point charge can be discarded if singularities come up.
  4. The energy should include as few "untransferrable" components as possible, meaning all the energy in the system would ideally be assignable to a physical, reversible process, and there should be a mechanism of transferring whatever energy is stored in the system. Since work is the only mechanism I'm aware of for energy transfer in electrostatics, all energy should arise from forces required to (adiabatically) bring the system into its equilibrium state.

My point is to remove some of the hand-waving steps in the usual derivations. A self-consistent theory should be purely classical, and based only on physically realizable quantities and processes. Thus a pair of charged conducting spheres being brought slowly together is a good example, while point charges floating in space which are bounded by an unspecified constraining force are discouraged. Most derivations determine the energy in a configuration of point charges, then generalize to charge densities, but this introduces problems like infinite point charge self-energy, which are not really necessary to a classical theory (which in the above framework treats point charges as approximations of localized charge densities). The resulting energy from these derivations is $$ U_E = \frac{1}{2}\int_V \rho(\mathbf r)\Phi(\mathbf r)\ d^3 r $$ But this derived from $q\Phi \rightarrow \rho dV \Phi$ and the concept of work performed on point charges being equal to $q\Phi$, coming from the expression $$ U_E(\mathbf r) = -\int_{\mathbf r_0}^{\mathbf r} q\mathbf E(\mathbf r')\cdot d \mathbf r' $$ And substituting the expression for $\Phi$ in terms of $\mathbf E$ given above. The meaning of "bringing an infinitesimal volume charge $\rho dV$ from zero potential to its location in the configuration" is not immediately physically meaningful, because there is no mechanism I'm aware of by which this can occur adiabatically, but I'm not completely against the idea.

A resolution to this is to simply define the electrostatic energy by the expression above, and to demonstrate that it meets our criteria. But this would be admitting that the electrostatic energy is not unique (unless a proof exists that it is). Perhaps a better way forward is to determine the work which can be performed by the charge distribution upon itself until no more work can possibly be done. But I'm not sure of the "right" approach, subjective as that might be.

Qmechanic
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Sam Gallagher
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    If the question stays closed I'll need some input about why so I can clean it up and make it more answerable. – Sam Gallagher Nov 07 '22 at 20:28
  • See: Poyntings Theorem – jensen paull Nov 07 '22 at 21:56
  • @jensenpaull Poynting's theorem only describes power flow, for electrostatic energy it doesn't (and cannot) say anything – Sam Gallagher Nov 08 '22 at 01:52
  • Electrostatics is a subset of maxwells equations when db/dt = 0 and de/dt = 0. Set these derivatives and you have your desired formulas. – jensen paull Nov 08 '22 at 09:16
  • Or further, ( set J and B = 0) and you arrive at d/dt ( U_{E}) = 0 the energy density is constant which is what you'd expect for electrostatics when nothing is moving – jensen paull Nov 08 '22 at 09:24
  • The energy density of the EM field in the absence of moving charges is the same as the energy density derived in electrostatics. See :https://physics.stackexchange.com/questions/721794/why-is-the-total-electrostatic-energy-bigger-than-the-sum-of-separate-energies/721877#721877 this derivation is slightly different from what you want to show but it can easily be translated to your problem – jensen paull Nov 08 '22 at 09:27
  • @jensenpaull I was going to link to this question: https://physics.stackexchange.com/questions/625816/validity-of-e-field-energy-density-equation but then I realized it's yours! The answer there is good, because it derives a conserved quantity with units of energy, but note that any time-independent term can be added to the quantity (E^2+B^2) without changing the result. We can add the self-energy of point charges, potential energy of a static system, etc, so long as they have units of joules. – Sam Gallagher Nov 08 '22 at 13:12
  • @jensenpaull The argument there would probably be that the energy quantities that could be added to (E^2+B^2) are not localized, and so including them isn't physically useful. What I had in mind when I said it "doesn't say anything" about the electrostatic case is that the equation as-written becomes 0=0 for electrostatics, but if we follow the argument there, then it's not the energy simply because we can evaluate the equation, but because it has the same form as a continuity equation, and hence we can infer (as we would with rho in the charge continuity) that it is the energy in all cases – Sam Gallagher Nov 08 '22 at 13:20
  • My last concern then is maybe philosophical, so not really a show-stopper, but: we derive energy density and not total energy, and the former tacitly assumes energy is stored in the field, while any good electrodynamics book will tell you there is no classical reasoning that can distinguish between energy stored in fields vs energy being local to the charge density. We're left to wonder, how can we identify this "energy density" with a field-free concept of energy? Does it matter if it's just for problem solving? And can we obtain an energy balance that includes $\rho$ and $\mathbf{J}$? – Sam Gallagher Nov 08 '22 at 13:32
  • $$\frac{d}{dt}(U_{E}) = 0$$ Does imply 0=0, integrating however implies $$U_{E} = c$$ There are freedoms in poyntings theorem: Let $$U_{E+B} \rightarrow U_{E+B} -\nabla \cdot \vec{f}$$ $$ \vec{S} \rightarrow \vec{S} + \frac{\partial \vec{f}}{\partial t}$$ These new descriptions of energy and the poynting vector satisfy poyntings theorem. Even though the energy densitys value is different, the same amount of work is done on charges by the field , and the same amount of energy required to change from state 1 to state 2 is the same. – jensen paull Nov 08 '22 at 17:36
  • The total "energy" of a system is changed, however energy is not measured, changes in energy is the physical thing, which is a constant for a specific situation , regardless of how you define the energy density and poynting vector, aslong as they satisfy poyntings theorem. This is similar to redefining the zero of potential to another location, the value of potential is not significant , but the change in potential is, and is a constant .p.s the freedom in poynting theorem is fully defined and is locatable to a point in space – jensen paull Nov 08 '22 at 17:40
  • Poyntings theorem in integral form derives the total energy, which as a subset is described in terms of it'd energy density, not sure what you mean by that last thing – jensen paull Nov 08 '22 at 17:41
  • Ammendment : first comment should be $$\frac{\partial U_{E}}{\partial t} = 0$$ $$U_{E} = c(x,y,z)$$ – jensen paull Nov 08 '22 at 17:43
  • And the resolution I got from my old question is that A. Use poyntings theorem instead of worrying about discrete charges initially , and then apply the relevant restrictions to obtain the static case . B. Poyntings theorem isnt correct for point charges since it assumes the contribution of each $\rho dv$ element is negligible, a point charge following Q\delta^3(r) clearly isn't negligible. Secondly for the point charge case from electrostatics, using the full potential instead of the potential-potential of the charge is just plainly wrong. Hence the incorrect result of infinite energy – jensen paull Nov 08 '22 at 17:51
  • The energy of a system of point charges is just the potential energy between the charges, this does not have the standard energy density, since that is derived under the assumption the charges can be broken down into infinitesimal parts. We can think of charges as balls of a finite density, with a specific radius such that its energy matches the experimental energy – jensen paull Nov 08 '22 at 17:55
  • Let us continue this discussion in chat. – Sam Gallagher Nov 08 '22 at 19:00

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