Poynting's Theorem

Statement

From the two dynamical Maxwell equations (Faraday and Ampère–Maxwell) and the vector identity for the divergence of a cross product, the electromagnetic field obeys a local energy-conservation law \(\ \partial_t u + \nabla\cdot\mathbf{S} = -\,\mathbf{E}\cdot\mathbf{J}\), where \(u=\tfrac{1}{2}(\mathbf{E}\cdot\mathbf{D}+\mathbf{H}\cdot\mathbf{B})\) is the field energy density and \(\mathbf{S}=\mathbf{E}\times\mathbf{H}\) (equal to \(\mathbf{E}\times\mathbf{B}/\mu_0\) in vacuum) is the Poynting vector giving the directed energy flux; the source term \(-\mathbf{E}\cdot\mathbf{J}\) is the rate at which the field does work on charges.

Why it matters

Poynting's theorem is the statement that electromagnetic energy is locally conserved: it cannot vanish here and reappear there, only flow continuously, carried by the field itself. It converts the abstract fields \(\mathbf{E}\) and \(\mathbf{B}\) into a bookkeeping of energy with a definite density and a definite current, exactly as the continuity equation does for charge.

Its most striking consequence is that energy flows through empty space and even along the outside of a wire, not down the conductor's core. It underlies the intensity of light, radiation pressure, antenna power, and the entire concept of the field as a physical system carrying energy and momentum.

Assumptions

Derivation

Result

Limiting cases

• Source-free region (\(\mathbf{J}=0\)): \(\partial_t u+\nabla\cdot\mathbf{S}=0\), an exact conservation law — field energy behaves like an incompressible-charge continuity equation.
• Static fields (\(\partial_t\to 0\)): \(\nabla\cdot\mathbf{S}=-\mathbf{E}\cdot\mathbf{J}\); steady Ohmic dissipation is supplied entirely by inward Poynting flux, e.g. the DC current-carrying wire.
• Plane wave in vacuum: \(E=cB\) so \(u=\varepsilon_0 E^2\) (equal electric and magnetic shares) and \(\mathbf{S}=c\,u\,\hat{\mathbf{k}}\); energy streams at the speed of light.
• Electrostatics only: with \(\mathbf{B}=0\) the theorem collapses to the electrostatic energy density \(u=\tfrac12\varepsilon_0 E^2\), consistent with the prior result.

Breaks when

Failure modes

• Sign slip in the identity. Writing \(\nabla\cdot(\mathbf{E}\times\mathbf{H})=\mathbf{E}\cdot(\nabla\times\mathbf{H})-\mathbf{H}\cdot(\nabla\times\mathbf{E})\) with the terms swapped flips the sign of \(\mathbf{E}\cdot\mathbf{J}\) and gives energy creation instead of dissipation.
• Using \(u=\tfrac12\varepsilon_0 E^2+\tfrac12\mu_0 H^2\). The magnetic term is \(\tfrac12\mathbf{H}\cdot\mathbf{B}=B^2/2\mu_0=\tfrac12\mu_0 H^2\); students often write \(B^2/2\mu_0\) as \(\mu_0 H^2\) with a wrong factor, or mix \(\mathbf{B}\) and \(\mathbf{H}\) so that units of \(u\) come out wrong.
• Confusing \(\mathbf{S}=\mathbf{E}\times\mathbf{H}\) with \(\mathbf{E}\times\mathbf{B}\). In vacuum \(\mathbf{S}=\mathbf{E}\times\mathbf{B}/\mu_0\); dropping the \(1/\mu_0\) gives units of \(\mathrm{V\,T\,m^{-1}}\), not \(\mathrm{W\,m^{-2}}\).
• Believing energy flows down the wire. Inside a resistor the Poynting vector points radially inward from the surrounding field; the axial energy transport is a myth that survives because it "feels" like current.
• Treating \(-\mathbf{E}\cdot\mathbf{J}\) as always negative (loss). For a battery/EMF source \(\mathbf{E}\cdot\mathbf{J}<0\) in the seat of EMF, so the field gains energy; the sign depends on whether charges do work on the field or vice versa.

Discussion

Poynting's theorem is the electromagnetic instance of a universal template: whenever a conserved quantity has a local density \(u\) and a current \(\mathbf{S}\), it obeys \(\partial_t u+\nabla\cdot\mathbf{S}=(\text{source})\). Here the source \(-\mathbf{E}\cdot\mathbf{J}\) is not a loss of total energy but a transfer channel between the field and matter: the mechanical work rate on charges is \(\mathbf{f}\cdot\mathbf{v}=\rho\mathbf{E}\cdot\mathbf{v}=\mathbf{E}\cdot\mathbf{J}\) per unit volume (the magnetic force does no work). Adding the mechanical energy density restores an exact global conservation law for field-plus-matter.

The deepest lesson is that the field is a physical system with its own energy stored in space at density \(u\), not merely a mathematical device for computing forces. The same fields carry momentum density \(\mathbf{g}=\mathbf{S}/c^2\) and stress described by the Maxwell stress tensor, so Poynting's theorem is the energy row of the field's full energy–momentum tensor. Radiation pressure, the recoil of an antenna, and the angular momentum of circularly polarised light all follow from taking the field's mechanical attributes seriously.

Because only \(\nabla\cdot\mathbf{S}\) enters the theorem, \(\mathbf{S}\) is defined only up to the curl of an arbitrary vector field, and \(u\) only up to a corresponding rearrangement. This gauge-like freedom means the local energy-flow picture is not experimentally forced by Poynting's theorem alone; it is fixed by demanding consistency with the symmetric energy–momentum tensor and with the requirement that a localized static configuration radiate no energy. The Abraham–Minkowski controversy over electromagnetic momentum in media is a modern echo of exactly this ambiguity.

Common misconceptions. (i) That \(\mathbf{S}\) represents "where the energy really is" — it represents flux, and its non-uniqueness means the streamline picture is a convention, not an observable. (ii) That a static crossed \(\mathbf{E}\) and \(\mathbf{B}\) (e.g. a charge next to a magnet) has "no energy flow because nothing changes" — in fact \(\mathbf{S}=\mathbf{E}\times\mathbf{H}\neq 0\) circulates, though \(\nabla\cdot\mathbf{S}=0\) so no net energy accumulates. (iii) That the source term is always dissipative — in an EMF it is the field being charged up.

Worked examples

Example 1 — Field amplitudes of sunlight. The solar irradiance at Earth is \(S_{\text{avg}}=1360\ \mathrm{W\,m^{-2}}\). Find the peak electric and magnetic field amplitudes of the (assumed) plane wave.

Example 2 — Energy flow into a resistive wire. A cylindrical wire of radius \(a=1.0\ \mathrm{mm}\) and length \(L=1.0\ \mathrm{m}\) carries a steady current \(I=2.0\ \mathrm{A}\) and has resistance \(R=0.50\ \Omega\). Compute the Poynting flux into its surface and compare with \(I^2R\).

Problems

1. (A) Radiation pressure. A perfectly absorbing black surface receives sunlight of intensity \(S=1360\ \mathrm{W\,m^{-2}}\) at normal incidence. Find the radiation pressure.
   Solution

   For an absorber, momentum flux equals \(S/c\): \(P=S/c=1360/(3.00\times10^{8})=4.5\times10^{-6}\ \mathrm{Pa}\). (A perfect reflector would double this to \(9.1\times10^{-6}\ \mathrm{Pa}\).)
2. (A) Energy density of a laser. A \(5.0\ \mathrm{mW}\) laser is focused to a spot of area \(1.0\ \mathrm{mm^2}\). Find the intensity, the peak \(E\)-field, and the energy density.
   Solution

   \(S=P/A=5.0\times10^{-3}/1.0\times10^{-6}=5.0\times10^{3}\ \mathrm{W\,m^{-2}}\). \(E_0=\sqrt{2S/\varepsilon_0 c}=\sqrt{2(5000)/(8.85\times10^{-12})(3\times10^{8})}=\sqrt{3.77\times10^{6}}\approx1.9\times10^{3}\ \mathrm{V\,m^{-1}}\). Mean energy density \(u=S/c=5000/3\times10^{8}=1.7\times10^{-5}\ \mathrm{J\,m^{-3}}\).
3. (B) Charging capacitor. A parallel-plate capacitor of radius \(a\) and gap \(d\) is charged so that the uniform field between the plates is \(E(t)\), increasing at rate \(\dot E\). Show that the Poynting flux through the cylindrical edge accounts for the rate of change of stored energy.
   Solution

   Displacement current gives \(H\) at the rim: \(\oint\mathbf{H}\cdot d\boldsymbol\ell=\varepsilon_0\dot E\,\pi a^2\Rightarrow H(a)=\tfrac12\varepsilon_0 a\dot E\). With axial \(E\), \(S=E H=\tfrac12\varepsilon_0 a E\dot E\), directed inward. Flux in \(=S(2\pi a d)=\tfrac12\varepsilon_0 a E\dot E\,(2\pi a d)=\varepsilon_0 E\dot E\,(\pi a^2 d)\). The stored energy is \(U=\tfrac12\varepsilon_0 E^2(\pi a^2 d)\), so \(dU/dt=\varepsilon_0 E\dot E(\pi a^2 d)\) — identical. Energy enters through the sides.
4. (B) Coaxial cable power. A coaxial line has inner radius \(a\), outer radius \(b\), carries current \(I\) on the inner conductor, with voltage \(V\) between the conductors. Show that the total Poynting flux through the annular cross-section equals \(VI\).
   Solution

   Between conductors \(E_r=\lambda/2\pi\varepsilon_0 r\) with \(V=\frac{\lambda}{2\pi\varepsilon_0}\ln(b/a)\Rightarrow E_r=\frac{V}{r\ln(b/a)}\); and \(H_\phi=\frac{I}{2\pi r}\). Then \(S_z=E_r H_\phi=\frac{VI}{2\pi r^2\ln(b/a)}\). Integrate: \(P=\int_a^b S_z\,2\pi r\,dr=\frac{VI}{\ln(b/a)}\int_a^b\frac{dr}{r}=\frac{VI}{\ln(b/a)}\ln(b/a)=VI\). All the power flows through the dielectric, not the metal.
5. (C) Non-uniqueness of S. Consider a point charge at rest next to a bar magnet, both static. Show \(\mathbf{S}\neq 0\) yet no energy accumulates anywhere, and comment on the "circulating energy" paradox.
   Solution

   With static \(\mathbf{E}\) (from the charge) and static \(\mathbf{H}\) (from the magnet), \(\mathbf{S}=\mathbf{E}\times\mathbf{H}\) is generally nonzero and forms closed loops. Since nothing changes, \(\partial_t u=0\) and \(\mathbf{J}=0\), so Poynting's theorem gives \(\nabla\cdot\mathbf{S}=0\): the flux is divergence-free, energy circulates without accumulating. The resolution is that \(\mathbf{S}\) is defined only up to \(\nabla\times\mathbf{G}\); the circulating "flow" carries no observable energy transport but does carry real field angular momentum \(\int \mathbf{r}\times(\mathbf{S}/c^2)\,dV\), which is recovered when the fields are switched off (Feynman's disk paradox).