Poynting theorem

An electromagnetic wave carries energy — that is why sunlight warms your skin and a laser can cut metal. But how much energy sits in the fields, and at what rate does it flow from place to place? The Poynting theorem answers both questions at once: it packages the energy stored in $\vec E$ and $\vec B$ into a density $u$ , defines a vector $\vec S$ that measures the power crossing each square metre, and links the two by a continuity equation — energy lost from a region must appear as flow through its boundary, nothing simply vanishes. This article builds $u$ and $\vec S$ from first principles, states the theorem, and turns it into the single working formula you will use again and again in this module: the intensity of a plane wave, $I = \tfrac12 c\varepsilon_0 E_0^2$ .

The core insight

The Poynting vector $\vec S = \frac{1}{\mu_0}\vec E \times \vec B$ carries the electromagnetic energy density $u$ through space exactly as a continuity equation demands, $-\partial u/\partial t = \nabla\cdot\vec S$ , and for a plane wave this collapses to the intensity $I = \tfrac12 c\varepsilon_0 E_0^2$ .

The core idea

Energy lives in the fields. An electric field stores energy with density $u_E = \tfrac12\varepsilon_0 E^2$ , and a magnetic field stores energy with density $u_B = \tfrac{1}{2\mu_0}B^2$ . Adding them gives the total electromagnetic energy density at a point in vacuum:

u = \frac12\varepsilon_0 E^2 + \frac{1}{2\mu_0}B^2.

where: $u$ is energy per unit volume (J/m $^3$ ), $E$ and $B$ are the instantaneous field magnitudes, $\varepsilon_0 = 8.85\times10^{-12}$ F/m and $\mu_0 = 4\pi\times10^{-7}$ T·m/A.

A plane wave splits this energy equally. For a plane EM wave in vacuum $E = cB$ with $c = 1/\sqrt{\mu_0\varepsilon_0}$ , and a short substitution shows $u_E = u_B$ always — the electric and magnetic contributions are identical, so the total simplifies to

u = \varepsilon_0 E^2.

Energy flows: the Poynting vector. Energy does not sit still; it streams through space at speed $c$ along the wave's direction of travel. That flow is captured by the Poynting vector

\vec S = \frac{1}{\mu_0}\,\vec E \times \vec B,

where: $\vec S$ is power per unit area (W/m $^2$ ), pointing along $\vec E \times \vec B$ — which, for a plane wave with $\vec E \perp \vec B \perp \vec k$ , points along the propagation direction $\vec k$ , exactly as it should.

Figure

A plane wave with electric field vertical and magnetic field horizontal showing their cross product pointing along the direction of travel

Poynting's theorem: energy is conserved locally. Built directly from Maxwell's equations, the theorem states that for a volume with no free currents doing external work, the energy density and the energy current obey a continuity equation:

-\frac{\partial u}{\partial t} = \nabla\cdot \vec S.

where: the left side is the rate at which stored energy at a point decreases, and the right side is the net outward divergence of the energy flow. Integrated over a volume $V$ bounded by surface $A$ , this reads: the rate of decrease of field energy inside $V$ equals the net rate at which energy leaves through $A$ — nothing is created or destroyed, only moved, exactly the same bookkeeping as charge conservation in the continuity equation for current.

Derivation AFrom field energy density to the intensity of a plane wave

The goal is to start from the two ingredients above — the plane-wave energy density $u = \varepsilon_0 E^2$ and the fact that $\vec S$ is an energy flow — and derive the single most-used number in this module: the time-averaged power per unit area, the intensity $I$ , expressed purely in terms of the wave's peak electric field $E_0$ ; along the way we confirm that $S$ and $u$ are linked by the wave speed $c$ , which is the physically transparent way to remember the whole result.

Step 1. Write the plane wave explicitly. Take $E = E_0\cos(kz-\omega t)$ propagating along $z$ , with $B = E/c$ at every instant (a defining property of a plane EM wave in vacuum). The instantaneous energy density is then $u = \varepsilon_0 E^2 = \varepsilon_0 E_0^2\cos^2(kz-\omega t)$ , oscillating between $0$ and $\varepsilon_0 E_0^2$ .

Step 2. Connect $S$ to $u$ through the speed of the wave. In a time $dt$ , the energy contained in a thin slab of cross-sectional area $A$ and thickness $c\,dt$ (everything the wave has swept past) crosses the area $A$ . So the power crossing unit area is energy density times the distance covered per unit time: $S = u\,c$ . This is a purely geometric argument — it does not need the cross-product formula for $\vec S$ at all, only that energy travels at speed $c$ .

Step 3. Substitute the instantaneous $u$ from Step 1: $S = c\,\varepsilon_0 E_0^2\cos^2(kz-\omega t)$ . This instantaneous flux oscillates between $0$ and $c\varepsilon_0 E_0^2$ at twice the wave's frequency (because $\cos^2$ has period $\pi/\omega$ , not $2\pi/\omega$ ) — it is never the number quoted as "the intensity of the wave."

Step 4. Average over one full cycle. Since $\langle\cos^2(kz-\omega t)\rangle = \tfrac12$ over a period, the time-averaged flux — the intensity — is

I = \langle S\rangle = \frac12\,c\,\varepsilon_0\,E_0^2.

This is the working formula: given a peak field $E_0$ , you get the intensity directly, and given a measured intensity you can solve backward for $E_0 = \sqrt{2I/(c\varepsilon_0)}$ . Verified: with $S = c\,u$ and $\langle u\rangle = \tfrac12\varepsilon_0 E_0^2$ , $I = \langle S\rangle = c\langle u\rangle = \tfrac12 c\varepsilon_0 E_0^2$ , matching the direct cross-product calculation of $\vec S$ for a plane wave.

Worked example

A plane EM wave in vacuum has peak electric field $E_0 = 600$ V/m. Find (a) the peak magnetic field, (b) the cycle-averaged energy density, and (c) the intensity of the wave.

Step 1. Peak magnetic field from $E_0 = cB_0$ , with $c = 3\times10^8$ m/s:

B_0 = \frac{E_0}{c} = \frac{600}{3\times10^8} = 2.0\times10^{-6}\ \text{T}.

Step 2. Averaged energy density. The instantaneous $u = \varepsilon_0 E^2$ averages to $\langle u\rangle = \tfrac12\varepsilon_0 E_0^2$ because $\langle E^2\rangle = E_0^2/2$ :

\langle u\rangle = \frac12(8.85\times10^{-12})(600)^2 = \frac12(8.85\times10^{-12})(3.6\times10^5) = 1.59\times10^{-6}\ \text{J/m}^3.

Step 3. Intensity, using either $I = \tfrac12 c\varepsilon_0 E_0^2$ directly or the shortcut $I = c\langle u\rangle$ from Step 2 of the derivation:

I = c\langle u\rangle = (3\times10^8)(1.59\times10^{-6}) \approx 4.8\times10^{2}\ \text{W/m}^2.

Peak field: $B_0 = 2.0\times10^{-6}$ T. Averaged energy density: $\langle u\rangle = 1.59\times10^{-6}$ J/m $^3$ . Intensity: $I \approx 4.8\times10^2$ W/m $^2$ — comparable to bright sunlight ( $\sim 1000$ W/m $^2$ at Earth's surface), a good sanity-check scale to keep in mind.

Competitive stretch

Sunlight arrives at the top of the atmosphere with $I = 1360$ W/m $^2$ (the solar constant). Invert the intensity formula: $E_0 = \sqrt{2I/(c\varepsilon_0)} = \sqrt{2(1360)/[(3\times10^8)(8.85\times10^{-12})]} = \sqrt{1.024\times10^6} \approx 1.01\times10^3$ V/m, so $E_{\text{rms}} = E_0/\sqrt2 \approx 720$ V/m. Sunlight's instantaneous peak field is around a kilovolt per metre — imperceptible to us only because its frequency ( $\sim10^{15}$ Hz) is far too fast to sense directly; what you feel as warmth is the time-averaged $S$ , i.e. $I$ .

Common traps

Writing $u = \varepsilon_0 E^2 + \tfrac{1}{2\mu_0}B^2$ (dropping the $\tfrac12$ on the electric term or double-counting) instead of correctly summing $\tfrac12\varepsilon_0E^2 + \tfrac{1}{2\mu_0}B^2$ , which are individually equal for a plane wave and together give $\varepsilon_0E^2$ .
Confusing instantaneous $S$ with intensity $I = \langle S\rangle$ : $S$ oscillates at twice the wave frequency because it involves $\cos^2$ , and only the time-average is the reported "intensity."
Mixing up units: $S$ is power per unit area (W/m $^2$ ), while $u$ is energy per unit volume (J/m $^3$ ); the bridge between them is $S = uc$ , not $S = u$ .
Assuming $\vec E\times\vec B$ is automatically nonzero for any two fields; it is nonzero here specifically because a plane wave in vacuum has $\vec E$ and $\vec B$ in phase and mutually perpendicular — a feature of this particular solution, not a universal identity.

Ask the tutor: Ask the tutor for extra practice converting between intensity, peak field, and peak magnetic field, or a walkthrough of how Poynting's theorem follows term-by-term from Maxwell's equations.

Why does intensity use E naught squared and not just E naught?

Because intensity is built from energy density, which itself comes from $E^2$ (energy in a capacitor-like field scales as the square of the field, not linearly). Averaging $\cos^2$ over a cycle gives the factor of one half, but the fundamental $E^2$ dependence survives — so doubling the field quadruples the intensity, not doubles it. This square-law scaling is why a modest increase in field amplitude (say, near a focused laser) can produce a dramatic jump in delivered power.

Is the Poynting vector only defined for plane waves?

No — $\vec S = \frac{1}{\mu_0}\vec E\times\vec B$ is a completely general definition that applies to any electromagnetic field configuration, and Poynting's theorem $-\partial u/\partial t = \nabla\cdot\vec S$ holds everywhere Maxwell's equations hold. The plane wave is simply the simplest case, where the geometry makes $\vec S$ constant in direction and lets us reduce everything to the single scalar formula $I = \tfrac12c\varepsilon_0E_0^2$ . For more complicated fields (near an antenna, inside a capacitor being charged) $\vec S$ still tracks energy flow correctly, just with a more complicated shape.

Why do we average over a cycle instead of just using the peak S?

Because nothing you can measure directly — a light meter, your skin, a solar panel — responds fast enough to track oscillations at $10^{14}$ - $10^{15}$ Hz; every practical detector reports the time-average. Reporting the peak $S = c\varepsilon_0E_0^2$ would overstate the delivered power by a factor of two, since $S$ spends equal time near zero and near its peak. The averaged quantity $I = \langle S\rangle$ is therefore the physically meaningful, measurable number, and it is what "intensity" always means in this context.

Poynting theorem

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads