# 42  Poynting's Theorem in Linear Dispersive Media with Loss

Michael Fowler

Note: This Jackson section (6.8) was added in the 3rd edition, and is basically a summary of Section 80 of Landau and Lifshitz’ Electrodynamics of Continuous Media. I am here following Landau’s treatment a little more closely than Jackson does. Also, this section assumes some basic familiarity with plane electromagnetic waves in media, so might have been better placed and discussed after that topic, but I’m following Jackson’s ordering in presenting it here.

Notational trivia:  I will follow Landau’s normalization of Fourier transforms, that is, the integrals are over $d\omega /2\pi$ and $dt$ (standard physics usage). Jackson here just writes $d\omega .$ On the other hand, later, in equation (7.104) for example, Jackson uses $d\omega /\sqrt{2\pi }$ and $dt/\sqrt{2\pi },$ the usual mathematicians’ symmetrical usage.

## Introduction

In the previous lecture, we found that the Poynting vector

$\stackrel{\to }{S}={\mu }_{0}^{-1}\left(\stackrel{\to }{E}×\stackrel{\to }{B}\right)=\left(\stackrel{\to }{E}×\stackrel{\to }{H}\right)$

is the vacuum energy flow rate in an electromagnetic field.

If energy flows across a boundary from vacuum into a medium, this expression continues to be valid, because the tangential components of $\stackrel{\to }{E}$ and $\stackrel{\to }{H}$ are continuous at the boundary.

The divergence of this flow rate in a medium is easy to calculate from Maxwell’s equations, just write ${\partial }_{i}{S}_{i}={\partial }_{i}\left({\epsilon }_{ijk}{E}_{j}{H}_{k}\right),$ then differentiate term by term to find

$-\stackrel{\to }{\nabla }\cdot \stackrel{\to }{S}=\left(\stackrel{\to }{E}\cdot \frac{\partial \stackrel{\to }{D}}{\partial t}+\stackrel{\to }{H}\cdot \frac{\partial \stackrel{\to }{B}}{\partial t}\right),\text{ }\stackrel{\to }{D}=\epsilon \stackrel{\to }{E},\text{ }\stackrel{\to }{B}=\mu \stackrel{\to }{H}.$

If $\epsilon ,\mu$ are real constants, this is just the local rate of change of electromagnetic energy density $U=\frac{1}{2}\left(\epsilon {\stackrel{\to }{E}}^{2}+\mu {\stackrel{\to }{B}}^{2}\right),$ as expected. 

However, if there is dispersion, that is, frequency dependence, $\epsilon =\epsilon \left(\omega \right),\mu =\mu \left(\omega \right)$, there will also be absorption (we’ll prove this later), and electromagnetic energy will not be conserved: some can go into heat, and the divergence of the Poynting vector includes heat production.

To address this more general situation, we need to work in frequency space,

$\begin{array}{l}\stackrel{\to }{E}\left(\stackrel{\to }{r},t\right)={\int }_{-\infty }^{\infty }\stackrel{\to }{E}\left(\stackrel{\to }{r},\omega \right){e}^{-i\omega t}d\omega /2\pi ,\\ \stackrel{\to }{D}\left(\stackrel{\to }{r},t\right)={\int }_{-\infty }^{\infty }\stackrel{\to }{D}\left(\stackrel{\to }{r},\omega \right){e}^{-i\omega t}d\omega /2\pi ,\end{array}$

with

$\stackrel{\to }{D}\left(\stackrel{\to }{r},\omega \right)=\epsilon \left(\omega \right)\stackrel{\to }{E}\left(\stackrel{\to }{r},\omega \right),\text{ }\stackrel{\to }{B}\left(\stackrel{\to }{r},\omega \right)=\mu \left(\omega \right)\stackrel{\to }{H}\left(\stackrel{\to }{r},\omega \right).$

Since the fields $\stackrel{\to }{E}\left(\stackrel{\to }{r},t\right),\stackrel{\to }{D}\left(\stackrel{\to }{r},t\right)$ are real, the coefficients in the Fourier transform necessarily satisfy

$\stackrel{\to }{E}\left(\stackrel{\to }{r},-\omega \right)={\stackrel{\to }{E}}^{\ast }\left(\stackrel{\to }{r},\omega \right),\text{ }\stackrel{\to }{D}\left(\stackrel{\to }{r},-\omega \right)={\stackrel{\to }{D}}^{\ast }\left(\stackrel{\to }{r},\omega \right),\text{ }\epsilon \left(-\omega \right)={\epsilon }^{\ast }\left(\omega \right).$

Note that if we take a Fourier transform of $\stackrel{\to }{D}\left(\stackrel{\to }{r},\omega \right)=\epsilon \left(\omega \right)\stackrel{\to }{E}\left(\stackrel{\to }{r},\omega \right)$ from frequency back to time, we will get a convolution: $\stackrel{\to }{D}$ no longer comes from $\stackrel{\to }{E}$ instantaneously, but has the form $D\left(\stackrel{\to }{r},t\right)={\int }_{0}^{\infty }d{t}^{\prime }G\left(t-{t}^{\prime }\right)\stackrel{\to }{E}\left(\stackrel{\to }{r},{t}^{\prime }\right).$ (We take it the “nonlocality” is in time but not in space.) $G\left(t\right)$ presumably decays fairly rapidly as time increases.  This will be discussed in more detail later.

## Energy Loss for a Monochromatic Plane Wave

Following Landau, we first consider a plane wave of frequency $\omega$ traveling through a medium of permittivity $\epsilon \left(\omega \right)$ and magnetic susceptibility $\mu \left(\omega \right).$ We’ll follow the standard procedure of taking the fields complex, the physical fields being the real parts of the variables. The permittivity $\epsilon \left(\omega \right)$ is also in general complex (because $\stackrel{\to }{D},\stackrel{\to }{E}$ will not be in phase, and $\stackrel{\to }{D}\left(\omega \right)=\epsilon \left(\omega \right)\stackrel{\to }{E}\left(\omega \right).$ )

Using Landau’s notation, with ${\epsilon }^{\prime },{\epsilon }^{″}$ real, we write

$\epsilon \left(\omega \right)={\epsilon }^{\prime }\left(\omega \right)+i{\epsilon }^{″}\left(\omega \right).$

(Warning: some authors flip the sign of the imaginary part, and Jackson doesn’t use this notation.)

A plane wave traveling through a medium with complex permittivity (and/or susceptibility) will lose electromagnetic energy, producing heat at a local density rate  $Q=-\stackrel{\to }{\nabla }\cdot \stackrel{\to }{S},$ given by the equation for $\stackrel{\to }{\nabla }\cdot \stackrel{\to }{S}$ above with electric field $\frac{1}{2}\left(\stackrel{\to }{E}+{\stackrel{\to }{E}}^{*}\right)$ and $\partial \stackrel{\to }{D}/\partial t=\frac{1}{2}\left(-i\omega \epsilon \stackrel{\to }{E}+i\omega {\epsilon }^{*}{\stackrel{\to }{E}}^{*}\right).$

Here $\stackrel{\to }{E},\stackrel{\to }{D}$ have time dependence ${e}^{-i\omega t},$ so ${\stackrel{\to }{E}}^{*},{\stackrel{\to }{D}}^{*}$ go as ${e}^{+i\omega t}$ and, time-averaging, the nonzero terms are those where the time-dependence cancels in the product, so the heat production

$\begin{array}{c}Q=\frac{i\omega }{4}\left[\left({\epsilon }^{*}-\epsilon \right)\stackrel{\to }{E}\cdot {\stackrel{\to }{E}}^{*}+\left({\mu }^{\ast }-\mu \right)\stackrel{\to }{H}\cdot {\stackrel{\to }{H}}^{\ast }\right]\\ =\frac{\omega }{2}\left({\epsilon }^{″}{\left|\stackrel{\to }{E}\right|}^{2}+{\mu }^{″}{\left|\stackrel{\to }{H}\right|}^{2}\right).\end{array}$

Here $\stackrel{\to }{E},\stackrel{\to }{H}$ are the (complex) amplitudes of the fields.

This is more naturally written with mean square real field intensities,

$Q=\omega \left({\epsilon }^{″}\overline{{\stackrel{\to }{E}}^{2}}+{\mu }^{″}\overline{{\stackrel{\to }{H}}^{2}}\right).$

That is, the rate of transfer of electromagnetic energy from a monochromatic plane wave to heat at a given point in a lossy medium is proportional to the imaginary part of the corresponding permittivity/susceptibility.

## Energy Loss for a Non-Monochromatic Field of Finite Time Duration

Next Landau considers an electromagnetic disturbance of finite time duration in a lossy medium. Perhaps a transmitter embedded in the medium is switched on for a finite time, so the signal must have some spread in frequency space, perhaps a wave packet. Here we derive the total energy dissipation over all time, there being no energy initially and none left finally. Inserting the Fourier transformed expressions from above,

$\underset{-\infty }{\overset{\infty }{\int }}\stackrel{\to }{E}\cdot \frac{\partial \stackrel{\to }{D}}{\partial t}=-i\underset{-\infty }{\overset{\infty }{\int }}\omega \epsilon \left(\omega \right){\stackrel{\to }{E}}_{\omega }\cdot {\stackrel{\to }{E}}_{{\omega }^{\prime }}{e}^{-i\left(\omega +{\omega }^{\prime }\right)t}\frac{d\omega d{\omega }^{\prime }}{{\left(2\pi \right)}^{2}}dt.$ and remembering $\stackrel{\to }{E}\left(\stackrel{\to }{r},-\omega \right)={\stackrel{\to }{E}}^{\ast }\left(\stackrel{\to }{r},\omega \right),$ and adding in the corresponding magnetic term,

$\underset{-\infty }{\overset{\infty }{\int }}Qdt=\underset{-\infty }{\overset{\infty }{\int }}\omega \left[{\epsilon }^{″}\left(\omega \right){\left|{\stackrel{\to }{E}}_{\omega }\right|}^{2}+{\mu }^{″}\left(\omega \right){\left|{\stackrel{\to }{H}}_{\omega }\right|}^{2}\right]\frac{d\omega }{2\pi }.$

## Transparency Ranges: Motion of a Wave Packet

Although ${\epsilon }^{″},{\mu }^{″}$ are strictly nonzero except at $\omega =0,$ for many materials (like glass) there are ranges of $\omega$ where they are very small compared with ${\epsilon }^{\prime },{\mu }^{\prime }.$ These are called transparency ranges. There is still dispersion, of course, but a small enough loss rate that it is meaningful to talk about internal energy of the material in the electromagnetic field.  Think of Lorentz’ model of static dielectric polarization: the molecules are electrons attached to simple harmonic oscillator “springs”. The “internal energy” in this electrostatic case is the extra potential energy in the spring from being stretched by the applied field. The idea can clearly be extended to the dynamic case, where the oscillating applied field excites the electrons into more energetic states. This picture is reasonable provided the damping is sufficiently small that the internal energy is not quickly dissipated as heat. (It will be, eventually, if the driving field is turned off, but that is on a longer timescale$—$by definition of a transparency range.)

We found the polarization energy in the static case by gradually adding charge to build the field up from zero. The analogous procedure in the dynamic case is to send in a long wave packet, made of waves close in frequency to ${\omega }_{0},$ and find how the energy density varies with the (slowly varying) wave amplitude. So we take

$\stackrel{\to }{E}={\stackrel{\to }{E}}_{0}\left(t\right){e}^{-i{\omega }_{0}t},\text{ }\stackrel{\to }{H}={\stackrel{\to }{H}}_{0}\left(t\right){e}^{-i{\omega }_{0}t},$

with  ${\stackrel{\to }{E}}_{0}\left(t\right),\text{ }{\stackrel{\to }{H}}_{0}\left(t\right)$ varying very slowly in time compared with ${e}^{-i{\omega }_{0}t}.$ Averaging over a cycle (as before) cuts out the ${e}^{±2i{\omega }_{0}}$ terms, leaving

The tricky part is $\partial \stackrel{\to }{D}/\partial t=\left(\partial /\partial t\right)\epsilon \stackrel{\to }{E}.$ We need to handle this in frequency space, so we choose a wave packet close to monochromatic, that is, we take

$\stackrel{\to }{E}\left(t\right)={\stackrel{\to }{E}}_{0}\left(t\right){e}^{-i{\omega }_{0}t},\text{ }{\stackrel{\to }{E}}_{0}\left(t\right)=\int {\stackrel{\to }{E}}_{0\alpha }{e}^{-i\alpha t}d\alpha ,$

where the ${\stackrel{\to }{E}}_{0\alpha }$ are constants, and $\alpha \ll {\omega }_{0},$ so the amplitude variation is slow compared to the oscillation time.

From this,

$\stackrel{\to }{D}\left(t\right)=\int \epsilon \left({\omega }_{0}+\alpha \right){\stackrel{\to }{E}}_{0\alpha }{e}^{-i\left({\omega }_{0}+\alpha \right)t}d\alpha ,$

and, keeping only the leading order term in $\alpha ,$

$\begin{array}{c}\frac{\partial \stackrel{\to }{D}\left(t\right)}{\partial t}=-i\int \left({\omega }_{0}+\alpha \right)\epsilon \left({\omega }_{0}+\alpha \right){\stackrel{\to }{E}}_{0\alpha }{e}^{-i\left({\omega }_{0}+\alpha \right)t}d\alpha \\ =-i\int \left[{\omega }_{0}\epsilon \left({\omega }_{0}\right)+\alpha \frac{d}{d\omega }{\left(\omega \epsilon \left(\omega \right)\right)}_{\omega ={\omega }_{0}}\right]{\stackrel{\to }{E}}_{0\alpha }{e}^{-i\left({\omega }_{0}+\alpha \right)t}d\alpha .\end{array}$

Then reversing the Fourier transforms,

$\frac{\partial \stackrel{\to }{D}\left(t\right)}{\partial t}=-i{\omega }_{0}\epsilon \left({\omega }_{0}\right)\stackrel{\to }{E}\left(t\right)+\frac{d}{d\omega }{\left(\omega \epsilon \left(\omega \right)\right)}_{\omega ={\omega }_{0}}\frac{\partial {\stackrel{\to }{E}}_{0}\left(t\right)}{\partial t}{e}^{-i{\omega }_{0}t}.$

At this point, Landau drops the subscript, so ${\omega }_{0}$ becomes $\omega ,$ since we are taking a very slow variation of amplitude, like a long wave packet, so almost monochromatic,

$\frac{\partial \stackrel{\to }{D}\left(t\right)}{\partial t}=-i\omega \epsilon \left(\omega \right)\stackrel{\to }{E}\left(t\right)+\frac{d\left(\omega \epsilon \right)}{d\omega }\frac{\partial {\stackrel{\to }{E}}_{0}\left(t\right)}{\partial t}{e}^{-i\omega t}.$

Making the approximation that $\epsilon \left(\omega \right)$ is real,

$\frac{1}{4}\frac{d\left(\omega \epsilon \right)}{d\omega }\left({\stackrel{\to }{E}}_{0}{}^{*}\cdot \frac{\partial {\stackrel{\to }{E}}_{0}\left(t\right)}{\partial t}+{\stackrel{\to }{E}}_{0}\cdot \frac{\partial {\stackrel{\to }{E}}_{0}{}^{*}\left(t\right)}{\partial t}\right)=\frac{1}{4}\frac{d\left(\omega \epsilon \right)}{d\omega }\frac{\partial }{\partial t}\left(\stackrel{\to }{E}\cdot {\stackrel{\to }{E}}^{*}\right).$

Adding the corresponding magnetic term, the rate of change of energy per unit volume is $d\overline{U}/dt,$ where

$\overline{U}=\frac{1}{4}\left[\frac{d\left(\omega \epsilon \right)}{d\omega }\stackrel{\to }{E}\cdot {\stackrel{\to }{E}}^{*}+\frac{d\left(\omega \mu \right)}{d\omega }\stackrel{\to }{H}\cdot {\stackrel{\to }{H}}^{*}\right],$

or in terms of the real fields,

$\overline{U}=\frac{1}{2}\left[\frac{d\left(\omega \epsilon \right)}{d\omega }\overline{{E}^{2}}+\frac{d\left(\omega \mu \right)}{d\omega }\overline{{H}^{2}}\right].$

## Quasi-Monochromatic Wave Packet: Energy Flows at Group Velocity

The above formula (due to Brillouin) is not further explained by Jackson (or Landau), but Garg provides some insight (p 473). He assumes a wave packet made of a very narrow band of frequencies, so the Poynting flow vector is (now using Garg’s notation to emphasize that it’s almost monochromatic)

$\overline{S}=\frac{1}{4}\left({\stackrel{\to }{E}}_{\omega }×{\stackrel{\to }{H}}_{\omega }^{*}+{\stackrel{\to }{E}}_{\omega }^{*}×{\stackrel{\to }{H}}_{\omega }\right)$

(setting ${\epsilon }^{″},{\mu }^{″}=0$ ).  For this almost plane wave, ${\stackrel{\to }{E}}_{\omega },{\stackrel{\to }{H}}_{\omega }\sim {e}^{i\stackrel{\to }{k}\cdot \stackrel{\to }{r}}$ and Maxwell’s macroscopic equations are

$\begin{array}{l}\stackrel{\to }{k}×{\stackrel{\to }{E}}_{\omega }=\omega \mu \left(\omega \right){\stackrel{\to }{H}}_{\omega },\\ \stackrel{\to }{k}×{\stackrel{\to }{H}}_{\omega }=-\omega \epsilon \left(\omega \right){\stackrel{\to }{E}}_{\omega }.\end{array}$

From these, ${k}^{2}=\mu \left(\omega \right)\epsilon \left(\omega \right){\omega }^{2},$ and $\mu {\stackrel{\to }{H}}_{\omega }^{2}=\epsilon {\stackrel{\to }{E}}_{\omega }^{2}.$

The Poynting energy density flow vector is then

$\begin{array}{c}\overline{S}=\frac{1}{4}\left(\sqrt{\frac{\epsilon }{\mu }}{\stackrel{\to }{E}}_{\omega }\cdot {\stackrel{\to }{E}}_{\omega }^{*}+\sqrt{\frac{\mu }{\epsilon }}{\stackrel{\to }{H}}_{\omega }\cdot {\stackrel{\to }{H}}_{\omega }^{*}\right)\\ =\frac{1}{4\sqrt{\epsilon \mu }}\left(\epsilon {\stackrel{\to }{E}}_{\omega }\cdot {\stackrel{\to }{E}}_{\omega }^{*}+\mu {\stackrel{\to }{H}}_{\omega }\cdot {\stackrel{\to }{H}}_{\omega }^{*}\right)\\ =\frac{1}{2}\sqrt{\frac{\epsilon }{\mu }}{\stackrel{\to }{E}}_{\omega }\cdot {\stackrel{\to }{E}}_{\omega }^{*}.\end{array}$

Now we’ll write the energy density using Brillouin’s result:

$\begin{array}{l}\overline{U}=\frac{1}{4}\left[\frac{d\left(\omega \epsilon \right)}{d\omega }{\stackrel{\to }{E}}_{\omega }\cdot {\stackrel{\to }{E}}_{\omega }^{*}+\frac{d\left(\omega \mu \right)}{d\omega }{\stackrel{\to }{H}}_{\omega }\cdot {\stackrel{\to }{H}}_{\omega }^{*}\right]\\ =\frac{1}{4}\left[\left(\epsilon +\omega \frac{d\epsilon }{d\omega }\right){\stackrel{\to }{E}}_{\omega }\cdot {\stackrel{\to }{E}}_{\omega }^{*}+\left(\mu +\omega \frac{d\mu }{d\omega }\right){\stackrel{\to }{H}}_{\omega }\cdot {\stackrel{\to }{H}}_{\omega }^{*}\right]\\ =\frac{1}{4}\left[\left(\epsilon +\omega \frac{d\epsilon }{d\omega }\right)+\frac{\epsilon }{\mu }\left(\mu +\omega \frac{d\mu }{d\omega }\right)\right]{\stackrel{\to }{E}}_{\omega }\cdot {\stackrel{\to }{E}}_{\omega }^{*}\\ =\frac{1}{2}\left(\epsilon +\frac{\omega }{2}\frac{d\epsilon }{d\omega }+\frac{\epsilon }{\mu }\frac{\omega }{2}\frac{d\mu }{d\omega }\right){\stackrel{\to }{E}}_{\omega }\cdot {\stackrel{\to }{E}}_{\omega }^{*}\\ =\frac{1}{2}\sqrt{\frac{\epsilon }{\mu }}\left(\sqrt{\epsilon \mu }+\frac{\omega }{2}\sqrt{\frac{\mu }{\epsilon }}\frac{d\epsilon }{d\omega }+\frac{\omega }{2}\sqrt{\frac{\epsilon }{\mu }}\frac{d\mu }{d\omega }\right){\stackrel{\to }{E}}_{\omega }\cdot {\stackrel{\to }{E}}_{\omega }^{*}.\end{array}$

Now, the group velocity for this wave packet is $d\omega /dk$ so its inverse, recalling ${k}^{2}=\mu \left(\omega \right)\epsilon \left(\omega \right){\omega }^{2},$

${v}_{G}^{-1}=\frac{dk}{d\omega }=\sqrt{\epsilon \mu }+\frac{\omega }{2}\sqrt{\frac{\mu }{\epsilon }}\frac{d\epsilon }{d\omega }+\frac{\omega }{2}\sqrt{\frac{\epsilon }{\mu }}\frac{d\mu }{d\omega }.$

Hence,

$\overline{S}={v}_{G}\overline{U}.$

The Poynting energy flow rate equals the energy density multiplied by the group velocity. However, this simple picture is only good for weak absorption and low dispersion. As we’ll see later, there are sometimes dispersion relations such that a wave packet will fall apart before it is useful for analyzing energy flow.