Poynting Vector and Power Flow
A wave is not an abstraction — it carries energy, and that energy can light a room, cook a meal, or warm the ocean. So far we have tracked where the fields point and how fast they fade; now we ask the engineer's question: how much power crosses a square metre, and where does it go when the wave dies? The answer is a single elegant vector, \(\vec{E}\times\vec{H}\), discovered by John Henry Poynting, that points the way power flows and measures exactly how much. Behind it stands a conservation law as strict as any in physics.
- The energy density stored in the electric and magnetic fields.
- Poynting's theorem as a statement of energy conservation for fields.
- The instantaneous Poynting vector \(\vec{\mathcal{P}}=\vec{E}\times\vec{H}\).
- Time-average power and the complex Poynting vector \(\tfrac12\mathrm{Re}\{\vec{E}_s\times\vec{H}_s^{*}\}\).
- Power carried by a plane wave and its \(e^{-2\alpha z}\) decay.
- Ohmic dissipation \(\tfrac12\sigma|\vec{E}_s|^2\) and the power balance.
Energy Stored in the Field
Before energy can flow, it must be somewhere. The fields themselves hold it: an electric field stores energy in proportion to \(E^2\), and a magnetic field in proportion to \(H^2\). These densities, first met in electrostatics (Chapter 8) and magnetostatics (Chapter 14), now stand side by side:
For a plane wave the two are not just comparable — they are equal at every instant, because \(E/H=\eta=\sqrt{\mu/\varepsilon}\) forces \(\tfrac12\varepsilon E^2=\tfrac12\mu H^2\). Energy sloshes equally between the electric and magnetic forms as the wave advances, the field-theory echo of an LC circuit trading energy between capacitor and inductor.
Poynting's Theorem
Take the dot product of Maxwell's curl equations with \(\vec{H}\) and \(\vec{E}\), subtract, and apply a vector identity. Out drops a bookkeeping statement of breathtaking generality — Poynting's theorem, the energy-conservation law for electromagnetic fields:
Read it left to right: the net power flowing into the closed surface \(S\) (the minus sign makes inflow positive) equals the rate at which energy is stored in the fields, plus the rate at which it is dissipated as ohmic heat \(\sigma E^2\). No energy appears or vanishes — it only flows, accumulates, or turns to heat. The same statement in differential form is \(-\nabla\cdot(\vec{E}\times\vec{H})=\partial w/\partial t+\sigma E^2\).
The Poynting Vector
The surface integrand in the theorem is the quantity we have been chasing. The Poynting vector is the power crossing unit area, and — crucially — it points in the direction that power flows:
Its direction is that of wave travel \(\hat{a}_k\) — exactly the right-handed \(\vec{E}\times\vec{H}\) triad from Chapter 18 — and its magnitude is the instantaneous power density. The total power through any surface is the flux \(P=\oint\vec{\mathcal{P}}\cdot d\vec{S}\).
Time-Average Power
The instantaneous \(\vec{\mathcal{P}}\) pulses at twice the wave frequency; what a meter — or an engineer — cares about is the average over a cycle. Using the half-real-part rule for products of phasors from Chapter 17, the average power density is the real part of the complex Poynting vector:
The split mirrors AC circuit theory exactly. The real part is the average power that actually travels and can be delivered to a load; the imaginary part is reactive power that merely sloshes back and forth, stored and returned each cycle without net transport. The conjugate \({}^{*}\) on \(\vec{H}_s\) is what encodes the phase angle between the fields — and in a lossy medium that angle is precisely the \(\theta_\eta\) of Chapter 19.
Power in a Plane Wave
Apply the rule to the plane wave \(\vec{E}_s=E_0e^{-\alpha z}e^{-j\beta z}\hat{a}_x\), whose companion \(\vec{H}_s\) lags by \(\theta_\eta\) and has amplitude \(E_0/|\eta|\). The cross product lands along \(\hat{a}_z\), and the result captures everything at once:
Three features deserve a pause. The power falls as \(e^{-2\alpha z}\) — twice as fast as the field, because power goes as amplitude squared. The factor \(\cos\theta_\eta\) is a power factor identical to \(\cos\phi\) in AC circuits: a lossy medium's field-phase lag throttles the deliverable power. And in the lossless limit \(\alpha=0\), \(\theta_\eta=0\), it reduces to the clean free-space result:
Power Loss and Decay
Where does the missing power go? Straight into heat. The average power dissipated per unit volume in a conducting medium is set by the same half-real-part rule applied to \(\vec{E}\) and the conduction current \(\sigma\vec{E}\):
The two pictures must agree, and they do. Differentiating the average power density gives \(-d\mathcal{P}_{\text{avg}}/dz=2\alpha\,\mathcal{P}_{\text{avg}}\): the power lost per metre is exactly the heat deposited per cubic metre, \(\tfrac12\sigma E_0^2 e^{-2\alpha z}\). Attenuation is not the wave "running out of steam" — it is energy being handed, joule by joule, to the medium. That closed loop between \(\alpha\), \(\eta\), and dissipation is Poynting's theorem made local.
Worked Examples
Problem. A free-space wave has \(E_0=100\ \text{V/m}\). Find the peak electric and magnetic energy densities.
Solution. Use \(w_E=\tfrac12\varepsilon_0E_0^2\); for a plane wave \(w_H=w_E\):
Problem. Find the average power density of the same wave (\(E_0=100\ \text{V/m}\), free space).
Solution. Use \(\mathcal{P}_{\text{avg}}=E_0^2/(2\eta_0)\):
Problem. How much power does that wave carry through a \(2\ \text{m}^2\) window normal to it?
Solution. Power is density times area, \(P=\mathcal{P}_{\text{avg}}A\):
Problem. A wave with \(E_0=100\ \text{V/m}\) enters a medium with \(|\eta|=87.8\ \Omega\), \(\theta_\eta=38.7^\circ\), \(\alpha=5.6\ \text{Np/m}\). Find \(\mathcal{P}_{\text{avg}}\) at \(z=0.1\ \text{m}\).
Solution. Use the full plane-wave result with the \(e^{-2\alpha z}\) and \(\cos\theta_\eta\) factors:
Problem. A plane wave in free space carries the same average density as sunlight, \(\mathcal{P}_{\text{avg}}\approx 1360\ \text{W/m}^2\). Estimate \(E_0\) and \(H_0\).
Solution. Invert \(\mathcal{P}=E_0^2/2\eta_0\): \(E_0=\sqrt{2\eta_0\mathcal{P}}\), \(H_0=E_0/\eta_0\):
Problem. For the lossy medium of Example 4 (\(\sigma=0.1\ \text{S/m}\)), verify that the power lost per metre at \(z=0\) equals the heat dissipated per cubic metre.
Solution. Compare \(2\alpha\,\mathcal{P}_{\text{avg}}(0)\) with \(\tfrac12\sigma E_0^2\):
Chapter Summary
\(w=\tfrac12\varepsilon E^2+\tfrac12\mu H^2\); for a plane wave \(w_E=w_H\).
\(P_{\text{in}}=\tfrac{d}{dt}(\text{stored})+\text{dissipated}\); energy conservation for fields.
\(\vec{\mathcal{P}}=\vec{E}\times\vec{H}\) (W/m²); points along \(\hat{a}_k\).
\(\vec{\mathcal{P}}_{\text{avg}}=\tfrac12\mathrm{Re}\{\vec{E}_s\times\vec{H}_s^{*}\}\); real flows, imaginary stores.
\(\mathcal{P}_{\text{avg}}=\tfrac{E_0^2}{2|\eta|}e^{-2\alpha z}\cos\theta_\eta\); free space \(=E_0^2/2\eta\).
\(p_d=\tfrac12\sigma|\vec{E}_s|^2\); equals \(2\alpha\mathcal{P}_{\text{avg}}\), the \(e^{-2\alpha z}\) loss.
Problems
For each item, decide whether you need an energy density, a power density, a total power, or a balance check — then reach for the matching relation. Difficulty rises down the list.
- A free-space wave has \(E_0=200\ \text{V/m}\). Find \(H_0\) and the average power density.
- Find the total power radiated through a \(0.5\ \text{m}^2\) area normal to the wave of Problem 1.
- Show that for a plane wave \(\tfrac12\varepsilon E^2=\tfrac12\mu H^2\) using \(\eta=\sqrt{\mu/\varepsilon}\).
- A wave in free space carries \(50\ \text{W/m}^2\). Find \(E_0\) and \(H_0\).
- Write the differential form of Poynting's theorem and identify each term physically.
- A lossy medium has \(|\eta|=120\ \Omega\), \(\theta_\eta=20^\circ\), \(\alpha=0.5\ \text{Np/m}\), \(E_0=60\ \text{V/m}\). Find \(\mathcal{P}_{\text{avg}}\) at \(z=2\ \text{m}\).
- By what factor does the average power of Problem 6 drop between \(z=0\) and \(z=2\ \text{m}\)?
- Find the average dissipated power density at \(z=0\) for a medium with \(\sigma=2\times10^{-3}\ \text{S/m}\) and \(E_0=40\ \text{V/m}\).
- Explain why the imaginary part of the complex Poynting vector carries no net power.
- Show that \(\cos\theta_\eta\to 1\) recovers the free-space power, and \(\cos\theta_\eta\to\cos45^\circ\) for a good conductor.
- For the good-conductor limit (\(\theta_\eta=45^\circ\)), what fraction of the "apparent" power \(\tfrac{E_0^2}{2|\eta|}\) is actually delivered?
- Verify the local power balance \(-d\mathcal{P}_{\text{avg}}/dz=\tfrac12\sigma E_0^2 e^{-2\alpha z}\) starting from \(\mathcal{P}_{\text{avg}}=\tfrac{E_0^2}{2|\eta|}e^{-2\alpha z}\cos\theta_\eta\) and \(\alpha=\tfrac12\sigma|\eta|\cos\theta_\eta\) (low-loss).