Poynting vector — energy flux in EM waves: Question bank

Quick vocabulary refresher, so no symbol below is a surprise:

$S$ — Poynting vector, energy flux: watts per square metre, a vector pointing where energy flows.
$u$ — energy density: joules per cubic metre, energy stored in the fields at a point.
$I = ⟨ ∣ S ∣ ⟩$ — intensity: the time-average of the magnitude of $S$ . For a plane wave $S$ never reverses direction, so this equals $∣ ⟨ S ⟩ ∣$ ; we treat $I$ as a scalar (power per area) throughout.
$E, B$ — electric and magnetic field vectors; in a plane wave they are perpendicular and locked by $E = c B$ .
$J$ — current density: charge crossing unit area per second (A/m²), the flow of moving charges. It appears in Poynting's theorem through the term $- J \cdot E$ , the rate at which fields do work on those charges.

To keep the geometry in front of us, three pictures anchor this page: the right-hand rule for $E \times B$ , the standing-wave nodes where $S = 0$ , and the surprising radial energy inflow into a resistor.

Figure — Poynting vector — energy flux in EM waves

Look at the figure: fingers point along blue $E$ , curl toward yellow $B$ , and the green thumb $S$ sticks out perpendicular to both — this is why $S$ is a cross product and why it lands along the propagation direction.

True or false — justify

True or false:

S

points in the direction the wave travels.

True for a plane wave in vacuum —

E \times B

lands along the propagation direction. But

S

generally points wherever energy flows, which need not be the wave direction (e.g. near a resistor it points radially inward).

True or false: The Poynting vector can be non-zero even when no wave is propagating.

True. Static crossed fields (a charged capacitor sitting inside a magnet) give

E \times B \neq = 0

, so

S \neq = 0

— energy can circulate in steady fields without any wave.

True or false: If

E

and

B

are parallel,

S = 0

True. The cross product of parallel vectors is zero, so

∣ S ∣ = \frac{1}{μ _{0}} E B sin 0 = 0

— no net energy transport in that direction.

True or false:

u = S c

, i.e. energy density equals flux times speed.

False — it's the other way round:

S = u c

. Flux is bigger by a factor

c

because in time

d t

the energy in a slab of length

c d t

crosses the face, so flux

= u \times c

True or false: In a plane wave the electric and magnetic contributions to

u

are equal.

True. Using

B = E / c

and

c^{2} = 1/ (μ_{0} ε_{0})

\frac{B ^{2}}{2 μ _{0}} = \frac{E ^{2} / c ^{2}}{2 μ _{0}} = \frac{1}{2} ε_{0} E^{2}

— the two halves of

u

match exactly at every instant.

True or false: Intensity of a wave is

ε_{0} c E_{0}^{2}

False — that's twice too big. The missing

\frac{1}{2}

comes from time-averaging

cos^{2}

I = \frac{1}{2} ε_{0} c E_{0}^{2}

. See Intensity and amplitude of waves.

True or false: Doubling the amplitude

E_{0}

doubles the intensity.

False. Intensity

\propto E_{0}^{2}

, so doubling

E_{0}

quadruples

I

True or false: Poynting's theorem is a statement of energy conservation.

True.

\frac{\partial u}{\partial t} + \nabla \cdot S = - J \cdot E

says: energy lost from the fields either flows out (

\nabla \cdot S

) or is spent doing work on the moving charges (

J \cdot E

, current density dotted with field). Nothing vanishes. It follows straight from Maxwell's equations.

Spot the error

Spot the error: "

S = \frac{1}{μ _{0}} E \cdot B

A dot product gives a scalar with no direction — but energy flows somewhere, so

S

must be a vector. The correct operation is the cross product

E \times B

Spot the error: "In a wave

B_{0} = E_{0}

, they're equal partners."

They oscillate in phase, but the amplitudes obey

E_{0} = c B_{0}

, so

B_{0}

is smaller by

c \approx 3 \times 1 0^{8}

. "Equal partners" refers to equal energy density, not equal numerical value.

Spot the error: "

S

has units of J/m³, since it measures field energy."

That's the unit of energy density

u

S

is energy per area per time, W/m² = J/(m²·s) — a flux, not a density.

Spot the error: "For radiation pressure on a perfect absorber,

P_{r a d} = 2 I / c

Absorption gives

P_{r a d} = I / c

; the factor

2

applies to a perfect reflector, where the momentum change is doubled. See Radiation pressure.

Spot the error: "The magnitude

∣ E \times B ∣ = E_{x} B_{x} + E_{y} B_{y} + E_{z} B_{z}

That expression is the dot product. The cross-product magnitude is

E B sin θ

; for perpendicular wave fields it is simply

E B

Spot the error: "Since energy flows down the wire,

S

points along the wire inside a resistor."

Inside/around a resistor

E

is along the wire and

B

circles it, so

E \times B

points radially inward — energy enters through the sides from the surrounding fields, delivering exactly

I^{2} R

. See figure below.

Spot the error: "We assumed the form of

S

and plugged it into Maxwell's equations."

No —

S

emerged from the derivation. Combining Maxwell's curl laws with a vector identity forces the flux term to be

\frac{1}{μ _{0}} E \times B

; we never posited it.

Why questions

Why must

S

be perpendicular to both

E

and

B

Because it is defined as a cross product, and any cross product is perpendicular to both its inputs — which matches the physical fact that a plane wave carries energy in the one direction perpendicular to both fields.

Why does the

\frac{1}{2}

appear in the intensity but not in the instantaneous

S

Instantaneously

S = ε_{0} c E^{2}

with

E = E_{0} cos (k x - ω t)

. Intensity is the time average of the magnitude, and

⟨ cos^{2} ⟩ = \frac{1}{2}

, so the average is half the peak.

Why do we time-average at all instead of quoting instantaneous

S

Optical frequencies (

\sim 1 0^{15}

Hz) are far faster than any detector; what we actually measure is the energy delivered over many cycles, which is the average.

Why is the vector identity

\nabla \cdot (E \times B) = B \cdot (\nabla \times E) - E \cdot (\nabla \times B)

the key step?

It repackages the two separate curl terms (from the two Maxwell curl laws) into the divergence of a single vector — and that single vector is exactly

S

. The parent derivation uses it to fold

\frac{1}{μ _{0}} [E \cdot (\nabla \times B) - B \cdot (\nabla \times E)]

into

- \nabla \cdot S

; see Vector calculus identities for the general form.

Why is the magnetic field of light usually ignored despite being an "equal partner"?

Equal in energy, not in force-per-charge: the magnetic force

q v B

is smaller than the electric

q E

by roughly

v / c

, and

B_{0} = E_{0} / c

makes it tiny (e.g.

\sim 3 \times 1 0^{- 6}

T for sunlight).

Why does

S = u c

hold specifically for a plane wave?

In a wave the energy simply rides along at speed

c

: energy density

u

swept through area

A

over length

c d t

gives flux

u c

. It won't hold for static fields where energy isn't translating at

c

. See Energy density of electric and magnetic fields and EM wave equation.

Edge cases

Edge case: What is

S

for a purely static, uniform electric field (no magnetic field)?

Zero. With

B = 0

E \times B = 0

— stored electric energy just sits there, no flux.

Edge case: What is

S

at an instant when

E = 0

(a node of the standing wave, or a zero-crossing)?

Zero at that instant, since

S \propto E B

and

E = 0

. For a standing wave the time-average

⟨ S ⟩

is also zero everywhere — no net energy is transported, only sloshed back and forth. The figure below marks these nodes.

Edge case: In a standing wave, is the intensity

\frac{1}{2} ε_{0} c E_{0}^{2}

No. That formula is for a travelling wave. A pure standing wave carries no net energy flow, so its net intensity is zero even though local

u

is large.

Edge case: Two beams of equal intensity cross at a point — is the total

S

just the sum of magnitudes?

No; you add the field vectors first, then form

S

from the totals. Because

S

is quadratic in the fields, interference (constructive or destructive) changes the result — magnitudes don't simply add.

Edge case: What happens to

S

if you reverse the propagation direction of a wave?

Reversing direction flips the sign of

B

relative to

E

(or of

E

relative to

B

), so

E \times B

reverses —

S

now points the new way, as it should.

Edge case: In an absorbing medium (complex permittivity), what does the time-averaged

S

do as the wave penetrates?

Its magnitude decays exponentially with depth — the "lost" flux is exactly the energy the term

- J \cdot E

dumps into the material as heat. So

\nabla \cdot ⟨ S ⟩ \neq = 0

: the divergence measures local absorption.

Edge case: In a dispersive but lossless medium, does energy still travel at the wave's phase speed?

No — energy (and

⟨ S ⟩

) moves at the group velocity, which can differ sharply from the phase velocity. The relation

S = u c

must use the actual energy-transport speed, not

c

Recall One-line summary of the traps

Flux not density ( $S$ is W/m²), cross not dot, $E = c B$ not $E = B$ , and the intensity's $\frac{1}{2}$ is the hero you keep forgetting.