The Poynting Vector: Energy and Momentum
Step into sunlight and you are standing in a river. Every second, roughly
1360\ \text{joules} of energy pour through each square metre facing the Sun
— the solar constant — after an eight-minute crossing of empty space. Nothing solid
made that trip: it was carried by the electric and magnetic fields of a
light wave, streaming
outward from the Sun at c. The fields do not merely store energy;
they move it, and they carry momentum too — enough that a large,
thin, mirrored sheet in space feels a steady push and can sail on sunlight with no fuel at all.
This whole page rests on one vector, named for John Henry Poynting (and the pun is deserved): the
Poynting vector
\mathbf{S} = \frac{1}{\mu_0}\,\mathbf{E}\times\mathbf{B}.
It points in the direction electromagnetic energy is flowing, and its magnitude is the
power crossing unit area perpendicular to that flow, in
\text{W/m}^2. Get comfortable with \mathbf{S} and
you can answer three questions at once: where is field energy going, how much is
going, and how hard does it push when it gets there.
First: fields hold energy
Before energy can flow it must be somewhere. Both fields store energy, spread through space with an
energy density (joules per cubic metre):
u = \tfrac{1}{2}\varepsilon_0 E^2 \;+\; \frac{B^2}{2\mu_0} \;=\; \tfrac{1}{2}\!\left(\varepsilon_0 E^2 + \frac{B^2}{\mu_0}\right).
The first term is the energy in the electric field (you can dig it out of a charged capacitor), the
second is the energy in the magnetic field (stored in an inductor's coil). These are not bookkeeping
tricks — pull the charges apart or run down the current and that energy comes back out. In a light
wave the two terms turn out to be exactly equal, because B = E/c
makes B^2/\mu_0 = \varepsilon_0 E^2: the electric and magnetic halves of a
wave carry the same energy, always.
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Energy is stored in the fields themselves, with density
u = \tfrac{1}{2}(\varepsilon_0 E^2 + B^2/\mu_0), measured in
\text{J/m}^3.
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In vacuum a travelling wave splits this energy evenly: the electric and magnetic contributions are
equal at every instant.
The Poynting vector: energy on the move
Now the flow. The direction is fixed by a cross product: curl the fingers of your
right hand from \mathbf{E} towards \mathbf{B} and
your thumb points along \mathbf{S}=\tfrac{1}{\mu_0}\mathbf{E}\times\mathbf{B}
— the way the energy travels. For a wave heading towards you with its electric field pointing up and
its magnetic field pointing sideways, \mathbf{S} points straight at you.
Reveal the triad below to watch \mathbf{E},
\mathbf{B} and their product \mathbf{S} click
into a right-handed set of three mutually perpendicular arrows.
Two features are worth pinning down. First, the units:
[\,E\,]\cdot[\,B\,]/\mu_0 works out to
\text{W/m}^2, a power per area — an intensity. Second,
\mathbf{S} is perpendicular to both fields, so in a light wave the
energy flows in the one direction the fields are not oscillating: sideways-shaking fields,
forward-moving energy.
Poynting's theorem: energy conservation, written locally
Why should \tfrac{1}{\mu_0}\mathbf{E}\times\mathbf{B} be the energy flow?
Because it makes the energy books balance at every point in space. Start from Maxwell's equations, dot
\mathbf{E} into Ampère's law and \mathbf{B} into
Faraday's law, subtract, and a vector identity
(\nabla\cdot(\mathbf{E}\times\mathbf{B}) = \mathbf{B}\cdot(\nabla\times\mathbf{E}) - \mathbf{E}\cdot(\nabla\times\mathbf{B}))
collapses the mess into a single clean statement:
-\frac{\partial u}{\partial t} = \nabla\cdot\mathbf{S} + \mathbf{J}\cdot\mathbf{E}.
Read it as an accountant would. The left side is the rate at which field energy is disappearing
from a tiny box. It goes to exactly two places: \nabla\cdot\mathbf{S}, energy
flowing out through the walls of the box, and \mathbf{J}\cdot\mathbf{E},
the rate at which the field does work on charges inside (the power that heats a
resistor). Nothing vanishes and nothing appears from nowhere — this is the
continuity equation
idea applied to energy, with \mathbf{S} playing the role of the energy
current.
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Local energy conservation:
-\dfrac{\partial u}{\partial t} = \nabla\cdot\mathbf{S} + \mathbf{J}\cdot\mathbf{E}.
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Field energy lost from a region equals what streams out as \mathbf{S}
plus what is delivered to charges as \mathbf{J}\cdot\mathbf{E}.
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In empty space (\mathbf{J}=0) the theorem says field energy only ever
moves — it is transported by \mathbf{S}, never destroyed.
Intensity of a plane wave
Put a real light wave into the formula. With B = E/c and
c = 1/\sqrt{\mu_0\varepsilon_0}, the instantaneous Poynting magnitude for a
plane wave simplifies beautifully:
S = \frac{EB}{\mu_0} = \frac{E^2}{\mu_0 c} = \varepsilon_0 c\,E^2 = c\,u.
The energy flow is just the speed of light times the energy density — energy sitting at density
u and sweeping past at speed c, exactly as you
would guess. But E oscillates, so the useful quantity is the
time-averaged value, the intensity I.
Averaging \cos^2 over a cycle gives a factor of one half:
I = \langle S\rangle = \tfrac{1}{2}\varepsilon_0 c\,E_0^2 = c\,u_{\text{avg}},
where E_0 is the peak field. The graph shows how the intensity
climbs as the square of the field amplitude — double the field and you quadruple the
power delivered.
We can run the intensity formula backwards. Sunlight above the atmosphere delivers
I \approx 1360\ \text{W/m}^2. Solving
I = \tfrac{1}{2}\varepsilon_0 c\,E_0^2 for the peak field:
E_0 = \sqrt{\frac{2I}{\varepsilon_0 c}} = \sqrt{\frac{2(1360)}{(8.85\times10^{-12})(3\times10^{8})}} \approx 1.0\times10^{3}\ \text{V/m}.
About a thousand volts per metre — and the magnetic partner is
B_0 = E_0/c \approx 3.4\times10^{-6}\ \text{T}, a few millionths of a
tesla, roughly a tenth of Earth's own field. Modest numbers, yet they light the entire daytime
world.
Momentum and radiation pressure
Here is the surprise that makes solar sails possible: an electromagnetic wave carries not just energy
but momentum. The momentum stored per unit volume — the momentum density
— is the energy flow divided by c^2:
\mathbf{g} = \frac{\mathbf{S}}{c^2}.
When that momentum lands on a surface, it exerts a radiation pressure. If the surface
absorbs the light (a black sheet), it soaks up all the incoming momentum, and the pressure is
P_{\text{absorbed}} = \frac{I}{c}.
If the surface reflects the light straight back (a mirror), the light reverses its momentum,
so the surface must supply twice the impulse — the pressure doubles:
P_{\text{reflected}} = \frac{2I}{c}.
This is precisely why solar sails are shiny: a mirror gets double the push of a black cloth. Let us put
a number on it. Sunlight at I = 1360\ \text{W/m}^2 hitting a perfect mirror
gives
P_{\text{reflected}} = \frac{2(1360)}{3\times10^{8}} \approx 9.1\times10^{-6}\ \text{Pa}.
Nine millionths of a pascal — feeble by everyday standards (a footstep is a hundred thousand times
more). But it never stops, needs no propellant, and scales with area. Spread a sail over a square
kilometre (10^{6}\ \text{m}^2) and the total force is
F = P\!\cdot\!A \approx 9\ \text{N} — a gentle but ceaseless shove that, over
months, can carry a lightweight craft across the Solar System.
S points where the energy is going, not where the field is strongest. It is
tempting to imagine \mathbf{S} living "inside" the wave wherever
\mathbf{E} is biggest, and pointing along the field. Neither is right.
\mathbf{S}=\tfrac{1}{\mu_0}\mathbf{E}\times\mathbf{B} is a cross
product, so it is perpendicular to both fields — it can never point along
\mathbf{E}. And its job is to show the direction of energy
transport. Two crossed fields with \mathbf{E}\parallel\mathbf{B}
store plenty of energy but carry \mathbf{S}=0 — the energy just sits
there. Strong fields are not the same as flowing energy; only the cross product tells you the flow.
Track the Poynting vector around a current-carrying resistor and you meet one of the loveliest
shocks in electromagnetism. Inside the wire the electric field
\mathbf{E} points along the wire (that is what drives the
current), while the magnetic field \mathbf{B} from the current
wraps around it. Their cross product
\mathbf{S}=\tfrac{1}{\mu_0}\mathbf{E}\times\mathbf{B} therefore points
radially inward, straight into the wire through its cylindrical surface. Integrate
\mathbf{S} over that surface and you get exactly
I^2R — the heating power.
So the energy that warms a resistor does not flow along the copper like water down a pipe;
it flows through the empty space beside the wire, in the surrounding fields, and dives in
from the sides. The same story holds for a charging capacitor: as the field between the plates
builds, \mathbf{S} points inward through the gap at the rim, delivering
the stored energy from the outside. The wires guide the fields; the fields carry
the energy.