## Energy density and the Poynting vector

The units of energy density are energy per volume and the units of energy flux density are energy per time area.  For a propagating wave, these two quantities are related to each other by a velocity.  What is that velocity?  Well, that depends on the medium through which the electromagnetic wave propagates.

### Free space

In free space, the electromagnetic field energy density is

$u_0=(\epsilon_0 E^2+\mu_0 H^2)/2,$

and the electromagnetic energy flux density is

$\mathbf S_0=\mathbf E \times \mathbf H$,

where $\mathbf E$ is the electric field, $\mathbf H$ is the magnetic field, $\epsilon_0$ is the permittivity of free space, and $\mu_0$ is the permeability of free space.

For a plane wave in free space, these quantities are simple to relate.  Since the magnetic field and electric field vectors are orthogonal and their amplitudes are related according to $H = E \sqrt{\epsilon_0/\mu_0}$, the contributions of the electric and magnetic portions of the energy are equal, and we can write $u_0=\epsilon_0 E^2=\mu_0 H^2$ and $S_0 = E^2 \sqrt{\epsilon_0/\mu_0}$.

We now find that the Poynting vector and the energy density of a plane wave in free space are indeed related by a velocity, the speed of light.

$\frac{S_0}{u_0}=\frac{1}{\sqrt{\epsilon_0 \mu_0}}=c$.

### Nondispersive medium

Here we look at a transparent medium far from resonance. In this case, the electromagnetic field energy density is

$u=(\epsilon E^2+\mu H^2)/2,$

and the electromagnetic energy flux density is still

$\mathbf S=\mathbf E \times \mathbf H$,

where $\epsilon$ is the electric permittivity, and $\mu$ is the magnetic permeability.

For a plane wave in a transparent, isotropic medium, the magnetic and electric field amplitudes are related according to $H = E \sqrt{\epsilon/\mu}$. For this reason $u=\epsilon E^2=\mu H^2$ and $S = E^2 \sqrt{\epsilon/\mu}$.

We now find that the Poynting vector and the energy density of a plane wave in free space are indeed related by a velocity, the phase velocity.

$S/u=1/\sqrt{\epsilon \mu}=c/n$.

### Dispersive but ‘lossless’ medium

Of course, every medium is dispersive. In the same way, every medium has gain or loss. (If you find an exception, let the fiber optics community know about it. Someone might be able to figure out a way to use it.) But it may be possible for a medium to simultaneously have non-negligible dispersion and negligible gain and absorption. I believe that this is only possible over a narrow spectrum so all the thoughts that follow apply only to the case where the waves of interest fall in the necessary spectral window.

Also, while in any medium it makes sense to take a cycle average (an average over a full period, in a dispersive medium this becomes more necessary than ever because energy storage for a dispersive medium is largely associated with the pulse envelope and therefore decoupled from the cyclical field changes associated with the frequency. So now we are talking about time averaged quantities.

With those two caveats, we can write the energy density as

$u_d=\left(\frac{d\omega\epsilon}{d\omega}E^2+\frac{d\omega \mu}{d\omega}H^2 \right)/2$.

The interesting thing is that without loss, this can be rewritten. How might it be rewritten? Well, consider a pulse that leaves a lossless nondispersive medium and enters a lossless, dispersive medium with the same refractive index and the same impedance.

Because the group index changes, the pulse speed changes and the pulse is compressed (or lengthened if the new medium has a faster group velocity than the old one) longitudinally. Because the impedance remains the same, there is no reflection of energy to worry about. Because the refractive index is also matched, there is no discontinuity in E, B, H, or D. Because there is no loss and no reflection, the total energy associated with the pulse must be conserved.

So let’s underscore the point: the pulse is scaled longitudinally, but the total pulse energy is not changed. That means that the energy density of the pulse must be changed by the same factor as the longitudinal scaling. Because the fields are all continuous as the pulse enters the medium, this scaling cannot reside in the fields. One way to interpret it is in terms of temporary absorption (or emission) of field energy by the medium. Anyway, conservation of pulse energy tells us that the energy density must scale inversely with the length of the pulse. In the nondispersive medium, the group velocity is equal to the phase velocity. In the dispersive medium, it is not. Taking the ratio between the group velocity in and out of the medium tells us that the energy density must be scaled by the factor $n_g/n$, where $n_g$ is the group index of the dispersive medium and $n$ is the group index and phase index of the nondispersive medium. Thus, we expect

$u_d=u \frac{n_g}{n},$

where $u$ is the energy density associated with the same squared fields for the nondispersive medium.

Let’s show that this intuition is correct. First, note that
$d(\omega \epsilon)/d\omega=\epsilon+\omega d\epsilon/d\omega=\epsilon(1+d\ln\epsilon/d\ln\omega),$

and similarly for the magnetic terms. This allows us to rewrite $u_d$ as

$u_d=(1+d\ln\epsilon/d\ln\omega)\epsilon E^2/2+(1+d\ln\mu/d\ln\omega)\mu H^2/2.$

But for a plane wave in a lossless medium, $\epsilon E^2=\mu H^2$, and each term is also equal to the total nondispersive energy density $u$. This lets us factor out the nondispersive energy density, giving

$u_d=u(1+d\ln\epsilon/d\ln\omega)/2+d\ln\mu/d\ln\omega/2)$

Combining log terms gives us

$u_d=u(1+d\ln\sqrt{\epsilon\mu}/d\ln\omega).$

For a lossless medium, $d\ln \sqrt{\epsilon\mu}=d\ln n$, so that

$u_d=u(1+d\ln n/d\ln\omega)$.

Well, $n_g/n=(n+\omega dn/d\omega)/n=(1+d\ln n/d\ln \omega)$. Thus,

$u_d=u \frac{n_g}{n}$.