Energy, Intensity and the Poynting Vector

The Poynting vector is simply an arrow that tells you which way energy is flowing in an electromagnetic field, and how much of it, in watts per square meter. Despite how it sounds, the name has nothing to do with pointing: it's named after John Henry Poynting, the English physicist who worked it out in 1884 (the pun is a happy accident). It's what you use to size a solar panel, work out a radio's range or check a phone against safety limits. The formula $\mathbf{S} = \frac{1}{\mu_0}\mathbf{E}\times\mathbf{B}$ is just a few pieces: $\mathbf{E}$ and $\mathbf{B}$ (the two fields), the cross product (which points the arrow along the direction the wave travels) and $\frac{1}{\mu_0}$ (a constant that makes the units come out in watts). It's one cross product and a constant; bear with it.

The scene. Sunlight warms your face after crossing 150 million km of empty space. On the way there's no material to hold the energy, so where is it? It's in the fields.

Energy density, from circuits you know. A capacitor stores $U = \tfrac12CV^2$. With $C = \varepsilon_0A/d$ and $V = Ed$, that's $U = \tfrac12\varepsilon_0E^2\,(Ad)$: energy $\tfrac12\varepsilon_0E^2$ in each cubic meter between the plates (capacitors). A long solenoid stores $U = \tfrac12LI^2$, and with $L = \mu_0n^2A\ell$ and $B = \mu_0nI$ that's $\dfrac{B^2}{2\mu_0}(A\ell)$ (inductance). Notice that neither answer mentions the plates or the coil anymore, only the field. That's the hint: these densities belong to the fields themselves, wherever they are, circuit or not.

In a wave the two shares are equal. With $B = E/c$ and $c^2 = 1/(\mu_0\varepsilon_0)$,

The magnetic field looks tiny next to $E$ (it's $E/c$), yet it carries exactly half the energy. The total is $u = \varepsilon_0E^2$.

Where the energy goes. Picture a box with end area $A$ and length $c\,dt$ along the direction of travel. In time $dt$ everything inside it, energy $u\,A\,c\,dt$, crosses the far end. So the power per unit area is $uc = c\varepsilon_0E^2$, and since $E = cB$, that's $c^2\varepsilon_0EB = EB/\mu_0$. It flows along the direction of travel, which is $\mathbf{E}\times\mathbf{B}$, so

Why it holds for any fields: Poynting's theorem. Differentiate $u$ and replace $\partial\mathbf{E}/\partial t$ and $\partial\mathbf{B}/\partial t$ using Maxwell's curl equations:

The product rule $\nabla\cdot(\mathbf{E}\times\mathbf{B}) = \mathbf{B}\cdot(\nabla\times\mathbf{E}) - \mathbf{E}\cdot(\nabla\times\mathbf{B})$ turns the bracket into $-\nabla\cdot(\mathbf{E}\times\mathbf{B})$, so

Field energy at a point drops only by flowing away ($\nabla\cdot\mathbf{S}$, the outflow per unit volume, a divergence) or by doing work on charges ($\mathbf{J}\cdot\mathbf{E}$, the heating in a resistor). It's conservation of energy, written for fields.

Energy enters a resistor from the side. Take a wire of radius $a$ and length $L$ carrying current $I$ with voltage $V$ across it. Inside, $\mathbf{E}$ points along the wire with $E = V/L$, and at its surface $B = \mu_0I/(2\pi a)$ circles it (Ampère's law). So $\mathbf{S} = \mathbf{E}\times\mathbf{B}/\mu_0$ points radially into the wire, with $S = \dfrac VL\cdot\dfrac{I}{2\pi a}$, and over the surface area $2\pi aL$ the inflow is exactly $VI$, the power the resistor turns into heat. The energy travels through the space around a circuit; the wires guide it.

Intensity. For $E = E_0\cos(kx - \omega t)$, $S = c\varepsilon_0E_0^2\cos^2(kx - \omega t)$, and the average value of $\cos^2$ over a cycle is $\tfrac12$. So the intensity is $I = \tfrac12c\varepsilon_0E_0^2 = c\varepsilon_0E_{\text{rms}}^2$, the same ½ that turns a peak voltage into an RMS one (AC power).

Spreading out. A source radiating power $P$ equally in all directions sends all of it through every sphere around itself, so $I = P/(4\pi r^2)$. Double the distance and the intensity drops to a quarter, while the field, which goes as $\sqrt I$, drops by half.

Pushing. In an absorber, $\mathbf{E}$ drives charges sideways, and the magnetic force $q\mathbf{v}\times\mathbf{B}$ on those moving charges points along the beam. The force per watt absorbed is $B/E = 1/c$, so an absorbed beam pushes with pressure $I/c$. A mirror sends the momentum back and feels twice that, $2I/c$. Sunlight at Earth ($I \approx 1360$ W/m²) presses on an absorber with only $4.5\ \mu$Pa, but a large, light mirror in space, a solar sail, needs no fuel.

A common mistake. It's tempting to picture energy flowing down the inside of a wire, like water in a pipe. It doesn't. As the resistor example showed, it travels through the space around the circuit and enters the wire from the side. A few smaller slips: forgetting the ½ (the peak of $S$ is twice the intensity), using $I/c$ for a mirror when it's $2I/c$, and mixing up $E_0$ and $E_{\text{rms}}$.

Solar engineers size panels from intensity (about 1000 W/m² of sunlight reaches the ground on a clear day) and radio engineers budget links by how intensity thins out as $1/r^2$. Safety limits for phones, routers and radar are written as maximum power densities in W/m². Laser cutters and welders concentrate kilowatts onto spots a fraction of a millimeter wide, and solar sails ride radiation pressure. On the power grid, the Poynting vector shows energy traveling through the fields around transmission lines, guided by the wires rather than carried inside them.