advanced35 min studyLesson 279 in Physics

Energy density in dispersive media

Account for material energy and state the narrowband weak-loss limits of the dispersive energy formula.

Before you begin

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The question for today

What you’ll understand

Explain why nondispersive energy formulas omit material dynamics.
Use the derivative energy formula only within its stated validity range.
Separate reversible energy storage from cycle-averaged heating.

A dielectric can return energy to a field after the instantaneous electric amplitude has changed. Its moving charges and internal modes store energy that is absent from a vacuum-only expression. In a transparent dispersive band, a narrowband wave can be assigned a useful cycle-averaged energy density involving frequency derivatives of the material response. Applying that expression to a broad pulse or a strongly absorbing resonance, however, extends it beyond the approximation that justified it.

Start from the work done on matter

Poynting's theorem can be written with electric and magnetic terms weighted by the vacuum constants, while placing work on polarization and magnetization on the material side. That form keeps the energy transfer explicit without first deciding how to divide stored material energy from heat. In a bound oscillator, electric work changes the particle's kinetic and restoring energies; damping separately converts organized motion into other degrees of freedom. A frequency-dependent permittivity summarizes this dynamics but does not make it disappear.

For a nondispersive lossless linear dielectric, an instantaneous quadratic electric energy follows by integrating the work as the field is raised. With memory, polarization at one time depends on earlier fields, so the same instantaneous integration no longer supplies the complete energy. Simply replacing a real constant by a complex frequency-dependent permittivity in the old quadratic expression can even yield a complex energy, which is not a physical stored-energy density. A constitutive model or a controlled bandwidth approximation is required.

u_0=\frac{\epsilon_0|\mathbf E(t)|^2}{2}+\frac{\mu_0|\mathbf H(t)|^2}{2},\qquad \partial_tu_0+\nabla\cdot(\mathbf E\times\mathbf H)=-\mathbf J_f\cdot\mathbf E-\mathbf E\cdot\partial_t\mathbf P-\mu_0\mathbf H\cdot\partial_t\mathbf M

All fields are real and instantaneous. The reference term u0 is not the total stored energy in matter; it has J/m³, and the power terms have W/m³. Explicit free-current work must not duplicate response already absorbed into P or M.↔ Scroll wide equations horizontally.

Apply the derivative formula in a transparent band

Assume a local isotropic stationary material and a nearly monochromatic wave packet. Its envelope must vary slowly over a carrier cycle, and its occupied frequency band must be narrow enough that a first-order expansion of material response is accurate. In a transparent band, or a weak-loss band where attenuation is a small correction over the relevant evolution, the cycle-averaged reversible energy density includes derivatives of angular frequency times the real permittivity and permeability.

The factor of one quarter comes from using peak complex amplitudes and averaging quadratic real oscillations. Frequency derivatives act on dimensional permittivity and permeability, including their explicit factor of angular frequency. If response is approximately constant, the formula reduces to the familiar nondispersive time average. Near a strong absorption line, separating stored energy from irreversible material energy generally needs a more complete dynamical model; a derivative of the real part alone is not a universal definition of positive stored energy.

\overline u\simeq\frac14\frac{d[\omega\epsilon^{\prime}(\omega)]}{d\omega}|E|^2+\frac14\frac{d[\omega\mu^{\prime}(\omega)]}{d\omega}|H|^2

The dimensional ε′ has F/m and μ′ has H/m. The formula gives J/m³ for peak phasors E and H, in a narrowband transparent or suitably weak-loss regime; it is not an arbitrary lossy-medium energy formula.↔ Scroll wide equations horizontally.

Keep loss and energy transport distinct

For a local scalar monochromatic response, the imaginary constitutive components determine time-averaged heating. A high stored energy can coexist with small absorption when a transparent response is strongly dispersive. Conversely, a region with anomalous dispersion and strong absorption should not be assigned an energy-transport speed merely by differentiating its phase index. In an ideal homogeneous lossless propagation band, the ratio of energy flux to the derivative energy density agrees with group velocity for an ordinary single branch.

A useful consistency check comes from the lossless Drude limit away from its zero-frequency pole. Even when the real permittivity is negative, the derivative of angular frequency times permittivity is positive. The extra term represents carrier kinetic energy that the naive nondispersive substitution misses. The oscillator lab likewise makes energy in a material coordinate explicit, but its displacement and velocity are only a mechanical analogy. It contains no electromagnetic fields or measured energy-density readout, and damping changes its total mechanical energy over time.

\overline p_{\rm abs}=\frac\omega2\bigl(\epsilon^{\prime\prime}|E|^2+\mu^{\prime\prime}|H|^2\bigr),\qquad \epsilon=\epsilon_0\left(1-\frac{\omega_p^2}{\omega^2}\right)\Rightarrow\frac{d(\omega\epsilon)}{d\omega}=\epsilon_0\left(1+\frac{\omega_p^2}{\omega^2}\right)

The heating formula uses dimensional constitutive parameters and no separately double-counted conduction loss. The derivative on the right is for a lossless Drude model at nonzero ω; negative ε alone does not imply a propagating wave.↔ Scroll wide equations horizontally.

Work it through

Carrier storage at negative permittivity

At a nonzero frequency in a lossless Drude model, $\omega_p/\omega=2$ and the local peak electric amplitude is $E=10\,\mathrm{V/m}$ . Compute the electric contribution to the derivative energy density and compare it with the naive nondispersive expression.

The relative permittivity is $1-2^2=-3$ . Inserting this directly into a nondispersive electric time-average would give $-3\epsilon_0 E^2/4$ , a sign warning that material dynamics have been omitted.
Differentiate before evaluating: $d(\omega\epsilon)/d\omega=\epsilon_0(1+\omega_p^2/\omega^2)=5\epsilon_0$ . The derivative contains the carrier kinetic storage as well as the vacuum electric term.
The proper electric contribution is $5\epsilon_0(10)^2/4=125\epsilon_0\,\mathrm{V^2/m^2}\approx1.11\times10^{-9}\,\mathrm{J/m^3}$ . This is not the complete energy unless the magnetic contribution is supplied, and it does not assert a propagating wave in a nonmagnetic negative-permittivity bulk.

The result

The electric contribution is positive, approximately 1.11 nJ/m³, whereas naive nondispersive substitution would incorrectly make it negative.

The living notebook

Explore the model

Interactive lab

Interactive lab · oscillator

Expose an internal storage coordinate

Reset and set BOTH Damping ratio ζ = 0 and Natural frequency ω₀ = 1. Use displayed displacement and velocity to predict that (v² + x²)/2 remains 1/2. For a new run Reset and set ζ = 0.50 and ω₀ = 1; predict decreasing mechanical energy, then Step or Play. The unit-mass oscillator begins at x = 1 and v = 0. It is a mechanical storage analogy, not a calculation of dispersive electromagnetic energy or heating.

Displacement x(t)Current time

Displacement x(t): 1
Velocity v(t): 0
Regime: underdamped

Damping ratio ζ0.2Natural frequency ω₀1.4

t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA response coordinate can retain kinetic and restoring energy even when the driving field is no longer described instantaneously.

Pause & reason

Check your understanding

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Read the checkpoint explanation

The envelope expansion and separation of reversible storage from loss require controlled bandwidth and loss assumptions. A single real constitutive value or intensity does not provide that control.

Your turn

Try it yourself

In a transparent nonmagnetic band let $\epsilon_r^{\prime}=4$ and $\omega\,d\epsilon_r^{\prime}/d\omega=2$ at the carrier frequency. A plane wave has $|E|=2\,\mathrm{V/m}$ . Compute both energy contributions and their sum using $\mu=\mu_0$ .

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The electric derivative factor is εr′ plus ω times its derivative. The plane-wave impedance gives μ0|H|² = ε0 εr′|E|².

Compare with the solution

The electric contribution is $\epsilon_0(4+2)(2^2)/4=6\epsilon_0$ in J/m³ when the field units are included. The magnetic contribution is $\mu_0|H|^2/4=\epsilon_0(4)(2^2)/4=4\epsilon_0$ . Their sum is $10\epsilon_0\approx8.85\times10^{-11}\,\mathrm{J/m^3}$ . Omitting dispersion would instead give $8\epsilon_0$ , so the missing material storage is $2\epsilon_0$ . The equality used for H assumes one transverse propagating plane wave in the stated homogeneous transparent band.

Keep the idea

Take it with you

Material degrees of freedom store energy that a vacuum-only expression misses.
The derivative energy formula requires narrow bandwidth and transparent or controlled weak-loss response.
Heating depends on imaginary response and must be kept separate from reversible storage.

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