The origin of energy density formulas in Maxwell's electromagnetism

Question

I am looking at the derivation of Poynting theory in Electrical Engineering texts and they usually start by the two statement:

$$W_e = \int_V\frac{1}{2}(\vec{E}.\vec{D})\,\,dv$$ $$W_m = \int_V\frac{1}{2}(\vec{H}.\vec{B})\,\,dv$$ as being the electric and magnetic stored energies in a volume $V$. The quantities inside the integrals are the energy densities. Most of time, the fields are sinusoidal and these values are averaged over one period and the phasor notation is used...etc. Notice the equations above are in instantaneous form.

What I couldn't find is a general proof of the energy density formulas. At best, I just find a kind of proof for electrostatic stored energy, and for some reason, similar formulas are used for dynamic fields (electric and magnetic).

Is there a proof that starts from the definitions of the field quantities, or uses convincing physics argument to arrive at these formulas?

Reason for asking: I am an Electrical Engineer, and I am looking for a deeper understanding of Poynting theory and its derivation.

Answer 1

It's important to realize Poynting's theorem, stated in terms of the fields, is valid regardless of what you call $u = \frac{1}{2}\vec E\cdot\vec D + \frac{1}{2}\vec H\cdot\vec B$. In other words, you don't have to assume that this is the electromagnetic energy density (or even use or define the concept of energy density) to derive Poynting's theorem in the form $$-\frac{\partial}{\partial t}\left({\frac{1}{2}\vec E\cdot\vec D + \frac{1}{2}\vec H\cdot\vec B}\right)=\vec\nabla\cdot(\vec E \times \vec H)+\vec J\cdot\vec E.$$ To put it yet another way, Poynting's theorem doesn't say that energy is distributed according to this density. Recall for instance that in electrostatics we sometimes view energy as being stored in the charges as opposed to the field.

That said, if you do accept $\frac{1}{2}\vec E\cdot\vec D + \frac{1}{2}\vec H\cdot\vec B$ as energy density, then Poynting's theorem becomes a simple and elegant statement of local conservation of electromagnetic energy, with the Poynting vector as the energy flux. If nothing else, this strongly motivates viewing $\frac{1}{2}\vec E\cdot\vec D + \frac{1}{2}\vec H\cdot\vec B$ as the energy density. It should be mentioned however that there are other choices of energy density and Poynting vector that turn Poynting's theorem into a continuity equation for energy density, unless you impose certain constraints on these choices.

To borrow from an earlier answer, Griffiths writes in Introduction to Electrodynamics:

Where is the energy stored? Equations 2.43 and 2.45 offer two different ways of calculating the same thing. The first is an integral over the charge distribution; the second is an integral over the field. These can involve completely different regions. For instance, in the case of the spherical shell (Ex. 2.9) the charge is confined to the surface, whereas the electric field is everywhere outside this surface. Where is the energy, then? Is it stored in the field, as Eq. 2.45 seems to suggest, or is it stored in the charge, as Eq. 2.43 implies? At the present stage this is simply an unanswerable question: I can tell you what the total energy is, and I can provide you with several different ways to compute it, but it is impertinent to worry about where the energy is located. In the context of radiation theory (Chapter 11) it is useful (and in general relativity it is essential) to regard the energy as stored in the field, with a density $$\frac{\epsilon_0}{2}E^2 = \text{energy per unit volume.}$$ But in electrostatics one could just as well say it is stored in the charge, with a density $\frac{1}{2}\rho V$. The difference is purely a matter of bookkeeping.

In short, it's possible that there is no proof of the kind you are looking for, nor am I aware of experimental evidence for $\frac{1}{2}\vec E\cdot\vec D + \frac{1}{2}\vec H\cdot\vec B$ as the energy density other than the observation that general relativity seems to work well.

Answer 2

Let us consider the case fields in vacuum, i.e. ${\bf D}\equiv {\bf E}$ and ${\bf B}\equiv {\bf H}$. The case of fields in medium seems similar (it is my expectation). Let us write down Maxwell equations in $$\nabla\times{\bf H}=\frac{1}{c}\frac{\partial{\bf E}}{\partial t}+\frac{4\pi}{c}{\bf j},\tag{1}$$ $$\nabla\times{\bf E}=-\frac{1}{c}\frac{\partial{\bf H}}{\partial t}.\tag{2}$$ Then multiply both sides of $(1)$ by ${\bf E}$ and multiply both sides of $(2)$ by ${\bf H}$ and add equations side by side, $$\frac{1}{c}{\bf E}\cdot\frac{\partial{\bf E}}{\partial t}+\frac{1}{c}{\bf H}\cdot\frac{\partial{\bf H}}{\partial t}=-\frac{4\pi}{c}{\bf j}\cdot{\bf E}-\left({\bf H}\cdot\nabla\times{\bf E}-{\bf E}\cdot\nabla\times{\bf H}\right).\tag{3}$$ Next, notice that $$\nabla\cdot\left({\bf a}\times{\bf b}\right)={\bf b}\cdot\nabla\times{\bf a}-{\bf a}\cdot\nabla\times{\bf b}.\tag{4}$$ With $(4)$ the eq. $(3)$ can be rewritten as $$\frac{1}{2c}\frac{\partial}{\partial t}\left(E^2+H^2\right)=-\frac{4\pi}{c}{\bf j}\cdot{\bf E}-\nabla\cdot\left({\bf E}\times{\bf H}\right).\tag{5}$$ Let ${\bf S}=c({\bf E}\times{\bf H})/(4\pi)$, then the eq. $(5)$ becomes $$\frac{\partial}{\partial t}\frac{E^2+H^2}{8\pi}=-{\bf j}\cdot{\bf E}-\nabla\cdot{\bf S}.$$ Now integrate both sides over finite volume and apply Gauss theorem for the second term in right hand side, $$\frac{\partial}{\partial t}\int dV\,\frac{E^2+H^2}{8\pi}=-\int dV\,{\bf j}\cdot{\bf E}-\oint\,d{\bf f}\cdot{\bf S},$$ where $d{\bf f}$ is element of surface, which enclosed the considered volume. Notice that discrete analog of the first integral in the right hand side is $$\int dV\,{\bf j}\cdot{\bf E}\rightarrow \sum_{i}e_i{\bf v}_i\cdot{\bf E},$$ where we replace integration by sum over charges $e_i$ with velocities ${\bf v}_i$. Now take into account that $$e{\bf v}\cdot{\bf E}=\frac{d}{dt}\mathcal{E}_{\text{kin}}.$$ If we integrate over all space, the surface integral vanishes because we assume that at infinity all fields are zero. We conclude that $$\frac{d}{dt}\left\{\int dV\,\frac{E^2+H^2}{8\pi}+\sum\mathcal{E}_{\text{kin}}\right\}=0.\tag{6}$$ Eq. $(6)$ says that for the closed system consisting of charges and EM field the quantity in brackets is conserved. The second term in brackets is simply kinetic energy of particles. So, we can say that the first terms is the energy of electromagnetic field. We call the quantity $$W=\frac{E^2+H^2}{8\pi}.$$ the energy density of electromagnetic field.

References:

1. Landau & Lifshitz Course in Theoretical Physics, vol. 2, paragraphs 27-31
2. Landau & Lifshitz Course in Theoretical Physics, vol. 8 (for the discussion of Maxwell equations in medium)