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Reading Portis's Electromagnetic Fields: Sources and Media I came across this expression for the stored electric energy in a volume in a general medium: $$ U = \int dV \int \mathbf{E}\cdot d\mathbf{D} $$ Where $V$ is the volume under consideration and $ \mathbf{D} = \epsilon \mathbf{E}$ where $\epsilon$ may be a tensor permittivity. This same expression can be found e.g. here.

The idea is that the material may be nonlinear and anisotropic. In the case of linear media, the above integral apparently reduces to $$ U = \frac{1}{2} \int \mathbf{E} \cdot \mathbf{D}\ dV $$

My question is, how does the above integral expression work?


Here is my best attempt.

First note that we can expand the differential $d\mathbf{D}$ by its definition, $$ d\mathbf{D} = (d\mathbf{r} \cdot \nabla)\mathbf{D} $$ And this is a tensor product which can be rewritten $$ d\mathbf{D} = d\mathbf{r} \cdot \nabla\mathbf{D} $$

Now from the integral $ \int \mathbf{E} \cdot d\mathbf{D}$ we can write: $$ \int (\mathbf{E} \cdot \nabla \mathbf{D}) \cdot d\mathbf{r} $$ Or $$ \int (\mathbf{E} \cdot \nabla) \mathbf{D} \cdot d\mathbf{r} $$ Some points of note here. Most references on vector calculus will present vector integrals as either line, surface, or volume integrals. In the above integral, the position vector $\mathbf{r}$ takes on all values in the volume, and so this is neither a line, surface, nor volume integral. To evaluate it, we must expand the expression.

As well, note that we were able to use the commutativity of the dot product only after rearranging the tensor product, as the del operator $\nabla$ is not generally commutative.

For the identity, we have: $$ \int \mathbf{E} \cdot d\mathbf{D} = \frac{1}{2} \mathbf{E} \cdot \mathbf{D} $$ In the particular case that $$ \mathbf{D} = \epsilon \mathbf{E} $$ And $\epsilon$ is a linear transformation.

Because $\epsilon$ is a linear transformation, we can pull it out of the integral: $$ \epsilon \int (\mathbf{E} \cdot \nabla \mathbf{E}) \cdot d\mathbf{r} $$ Now we can use the identity $$ (\mathbf{A}\cdot \nabla)\mathbf{A} = \frac{1}{2} \nabla (\mathbf{A} \cdot \mathbf{A}) - \mathbf{A} \times (\nabla \times \mathbf{A}) $$ And we take advantage of the fact that the electric potential energy expression given assumes a static arrangement of external charges, so that $\nabla \times \mathbf{E} = 0$, and we have: $$ \epsilon \int (\mathbf{E} \cdot \nabla \mathbf{E}) \cdot d\mathbf{r} = \frac{\epsilon}{2} \int \nabla (\mathbf{E} \cdot \mathbf{E}) \cdot d\mathbf{r} $$ Now it might help to note that $\mathbf{E} \cdot \mathbf{E} = ||\mathbf{E}||^2$ = $E_x^2 + E_y^2 + E_z^2$, but other than this I fail to see how this is going to resolve itself. So I'm stuck!

Side question: In general, what reference is there for this sort of vector calculus?

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  • $\begingroup$ Does your first equation exactly match what appears in the book? I would expect something like $$\delta U = \int dV \int \mathbf{E}\cdot \delta\mathbf{D} $$ as in equation (851) here. $\endgroup$ – G. Smith Sep 9 '19 at 17:15
  • $\begingroup$ Yes it does, but the derivation you link to is essentially the same so we might chalk it up to difference in notation. What exactly does the $\delta$ represent? I remember something like that notation from thermo $\endgroup$ – Sam Gallagher Sep 9 '19 at 19:05
  • $\begingroup$ It represents a small variation in the vector field across the volume. So the integral is just an integral over volume, not “over $\mathbf{D}$”, which is good because I have no idea what the latter would mean. In my comment above I accidentally added a second integral sign which isn’t in (851) but is in your book. It is this second integral that doesn’t make sense to me. $\endgroup$ – G. Smith Sep 9 '19 at 19:16

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