Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

A wave carries energy: sunlight warms, a radio wave drives a current in a receiving antenna. This subsection introduces the vector that describes the flow of electromagnetic energy, derives the energy balance it obeys, and then applies it to the plane wave.

The Poynting vector

The power density associated with an electromagnetic wave, the Poynting vector, is defined in the time domain by

P (r; t) = E (r; t) \times H (r; t) [W/ m^{2}] . (6.1.32)

Its units are $[V/m] \times [A/m] = [W/ m^{2}]$ , a power per unit area. The surface integral of $P$ gives the total power passing through a surface $S$ in the direction of $d S$ (Figure 6.1.8):

P = \int_{S} P \cdot d S [W] . (6.1.33)

$P$ is perpendicular to both $E$ and $H$ . For a wave it points the way the energy travels, along $a_{k}$ in a plane wave, as Figure 6.1.6 already showed. That $E \times H$ really is a flow of energy, and not just a convenient product, is the content of Poynting's theorem below.

The Poynting vector \vec{\mathcal P}=\vec E\times\vec H is perpendicular to both fields. Its flux through a surface S, taken in the direction of d\vec S, is the power crossing S. — **Figure 6.1.8.** The Poynting vector $P = E \times H$ is perpendicular to both fields. Its flux through a surface $S$ , taken in the direction of $d S$ , is the power crossing $S$ .

For time-harmonic fields $P (r; t)$ is the product of two sinusoids of frequency $ω$ , so it oscillates at $2 ω$ . What an instrument measures, a bolometer, a thermometer or your skin, is its average over a period.

The time-average power density

In time-harmonic form the vector

P_{c} (r) = \frac{1}{2} E (r) \times H^{*} (r) (6.1.34)

is called the complex Poynting vector, the asterisk denoting the complex conjugate. Its real part is the average value of $P (r; t)$ over a period, the time-average power density:

P_{a v} (r) = \frac{1}{T} \int_{0}^{T} P (r; t) d t = Re {P_{c} (r)}, (6.1.35)

and its surface integral gives the time-average power through $S$ ,

P_{a v} = \int_{S} P_{a v} (r) \cdot d S . (6.1.36)

The real part of $P_{c}$ is active power: energy that really leaves through $S$ and does not come back. The imaginary part is reactive: energy flowing back and forth with no net transport, like the reactive power $Q$ of a circuit.

Why does the average carry the factor $\frac{1}{2}$ and the conjugate? Write the real part out with $Re {z} = \frac{1}{2} (z + z^{*})$ :

E (r; t) H (r; t) = \frac{1}{2} (E e^{j ω t} + E^{*} e^{- j ω t}), = \frac{1}{2} (H e^{j ω t} + H^{*} e^{- j ω t}) . (6.1.37)

Multiplying out, the four products pair into two complex numbers and their conjugates, $E \times H^{*} + E^{*} \times H = 2 Re {E \times H^{*}}$ and $E \times H e^{j 2 ω t} + E^{*} \times H^{*} e^{- j 2 ω t} = 2 Re {E \times H e^{j 2 ω t}}$ , so that

E (t) \times H (t) = constant \frac{1}{2} Re {E \times H^{*}} + oscillates at 2 ω \frac{1}{2} Re {E \times H e^{j 2 ω t}} . (6.1.38)

The second term averages to zero over a period, which leaves the constant one.

Time-average Poynting vector

P_{a v} P_{a v} = \frac{1}{2} Re {E \times H^{*}} [W/ m^{2}], = \int_{S} P_{a v} \cdot d S [W] . (6.1.39)

In plain words

The product of two sinusoids is a constant plus a sinusoid at twice the frequency. The constant is the average, and with phasors it is half the real part of $E \times H^{*}$ , exactly as the average power of a circuit is $\frac{1}{2} Re {V I^{*}}$ .

An example shows the role of the phase. Take $E = E_{0} a_{x}$ and $H = H_{0} e^{- j θ} a_{y}$ with real $E_{0}$ , $H_{0}$ . In the time domain

P_{z} (t) = E_{0} H_{0} cos ω t cos (ω t - θ) = \frac{1}{2} E_{0} H_{0} [cos θ + cos (2 ω t - θ)], (6.1.40)

so $P_{a v} = \frac{1}{2} E_{0} H_{0} cos θ$ , which is exactly $\frac{1}{2} Re {E_{0} H_{0} e^{j θ}} = \frac{1}{2} Re {E_{0} (H_{0} e^{- j θ})^{*}}$ . Figure 6.1.9 plots $P_{z} (t)$ for three phase differences.

Instantaneous power density \mathcal P_z(t) for \vec E=E_0\vec a_x and \vec H=H_0\mathrm e^{-\mathrm j\theta}\vec a_y, with the averages dashed. In phase (\theta=0) the power flows one way only and averages \tfrac12E_0H_0, the familiar \tfrac12VI of circuits. At \theta=60^\circ it flows back for part of each cycle and the average halves. In quadrature (\theta=90^\circ) it only sloshes back and forth: zero average, purely reactive. — **Figure 6.1.9.** Instantaneous power density $P_{z} (t)$ for $E = E_{0} a_{x}$ and $H = H_{0} e^{- j θ} a_{y}$ , with the averages dashed. In phase ( $θ = 0$ ) the power flows one way only and averages $\frac{1}{2} E_{0} H_{0}$ , the familiar $\frac{1}{2} V I$ of circuits. At $θ = 6 0^{\circ}$ it flows back for part of each cycle and the average halves. In quadrature ( $θ = 9 0^{\circ}$ ) it only sloshes back and forth: zero average, purely reactive.

Common mistake · Forgetting the conjugate or the half

$\frac{1}{2} Re {E \times H}$ , without the conjugate, depends on the arbitrary phase reference and is not the average of anything. $Re {E \times H^{*}}$ , without the half, is twice the average. The half belongs to amplitude phasors, the convention of this book. With rms phasors, $E_{rms} = E_{0} / 2$ , the average is $Re {E_{rms} \times H_{rms}^{*}}$ and the half disappears.

Application 6.1.1 · Sunlight as a Poynting vector

Satellites measure the power of sunlight crossing $1 m^{2}$ held normal to the rays above the atmosphere, at the mean Sun–Earth distance of one astronomical unit ( $1 AU = 1.496 \times 1 0^{11} m$ ). This total solar irradiance, the solar constant, is $S \approx 1361 W/ m^{2}$ , and it is the magnitude of $P_{a v}$ . Sunlight is broadband and incoherent: its countless frequency components have random phases, and their powers simply add. We can still describe it by one equivalent plane wave that carries the same power density. The part of this subsection on the power carried by a plane wave, below, shows that a plane wave in free space has $∣ P_{a v} ∣ = E_{0}^{2} /2 η_{0}$ , so

S E_{0} H_{0} = ∣ P_{a v} ∣ = \frac{E _{0}^{2}}{2 η _{0}} ⟹ = 2 η_{0} S = 2 \times 377 \times 1361 = 1.01 kV/m, = \frac{E _{0}}{η _{0}} = 2.69 A/m . (6.1.41)

In empty space nothing absorbs the light, so $P_{a v}$ is radial and its flux is the same through every sphere round the sun. Hence $S \propto 1/ r^{2}$ , and the power that leaves the sun is $4 π r^{2} S = 4 π (1.496 \times 1 0^{11})^{2} \times 1361 = 3.8 \times 1 0^{26} W$ . Figure 6.1.10 shows where this power sits in the spectrum. Modelled as a black body at $5772 K$ , the sun has its peak at $502 nm$ , $37 %$ of $S$ lies in the visible band from $400$ to $700 nm$ and $51 %$ beyond $700 nm$ , in the infrared. At the ground about $1000 W/ m^{2}$ remain on a clear day, and a solar panel of $20 %$ efficiency turns $200 W/ m^{2}$ of it into electricity. Worked example 6.1.4 carries the numbers to Mars.

**Figure 6.1.10.** Left: the spectrum of sunlight above the atmosphere, modelled as a black body at $5772 K$ and scaled to the solar constant. The peak is at $502 nm$ , and $37 %$ of the power lies in the visible band. Right: $P_{a v}$ is radial, so its flux through every sphere centred on the sun is the same and $S$ falls as $1/ r^{2}$ .

An equivalent wave

$E_{0} = 1.01 kV/m$ is not a field that anybody measures in sunlight: no single component of the light has this amplitude. It is the amplitude of the one monochromatic plane wave that would carry the same $1361 W/ m^{2}$ . A kilovolt per metre sounds large, yet free space stores only a few microjoules per cubic metre at this field. It is a poor store of energy but a fast carrier.

Poynting's theorem

That $E \times H$ is a flow of energy follows from Maxwell's equations themselves. Consider a volume $V$ bounded by the closed surface $S$ with outward $d S$ , filled with a linear medium $(ε, μ, σ)$ that does not change in time, and containing sources described by an impressed current density $J_{i}$ (Figure 6.1.11). The total current density is then $J = σ E + J_{i}$ .

Step 1. A vector identity. From Section 2.2,

\nabla \cdot (E \times H) = H \cdot (\nabla \times E) - E \cdot (\nabla \times H) . (6.1.42)

Step 2. Insert Maxwell's curl equations, $\nabla \times E = - \partial B / \partial t$ and $\nabla \times H = σ E + J_{i} + \partial D / \partial t$ :

\nabla \cdot (E \times H) = - H \cdot \frac{\partial B}{\partial t} - E \cdot \frac{\partial D}{\partial t} - σ E^{2} - E \cdot J_{i} . (6.1.43)

Step 3. Recognise the stored energies. In a linear medium that does not change in time, $B = μ H$ and $D = ε E$ with constant $μ$ , $ε$ , so

H \cdot \frac{\partial B}{\partial t} E \cdot \frac{\partial D}{\partial t} = μ H \cdot \frac{\partial H}{\partial t} = \frac{\partial}{\partial t} (\frac{1}{2} H \cdot B), = \frac{\partial}{\partial t} (\frac{1}{2} E \cdot D) . (6.1.44)

These are the time derivatives of the magnetic and electric energy densities $w_{m} = \frac{1}{2} H \cdot B$ and $w_{e} = \frac{1}{2} E \cdot D$ of Section 4.2 and Section 3.4.

Step 4. Integrate over $V$ and turn the left-hand side into a surface integral with the divergence theorem (Section 2.4), $\int_{V} \nabla \cdot P d V = \oint_{S} P \cdot d S$ . Rearranging,

Poynting's theorem in the time domain

- \oint_{S} P \cdot d S - \int_{V} J_{i} \cdot E d V = \frac{d}{d t} \int_{V} [\frac{1}{2} E \cdot D + \frac{1}{2} H \cdot B] d V + σ \int_{V} ∣ E ∣^{2} d V, P = E \times H . (6.1.45)

Every term is a power in watts, an energy per unit time. Read term by term:

the first term on the left is the power pumped into $V$ through $S$ by external sources,
the second term on the left is the power provided by the internal sources, the impressed current density $J_{i}$ ,
the first term on the right is the rate of increase of the electromagnetic energy stored in $V$ , with the densities $w_{e} = \frac{1}{2} E \cdot D$ and $w_{m} = \frac{1}{2} H \cdot B$ ,
the second term on the right is the power dissipated in $V$ as heat, Joule's law $p = σ E^{2}$ of Section 4.2.

In plain words

The field carries energy like a fluid. What the sources give is either stored in the field, turned into heat, or carried out through the boundary at the rate $\oint_{S} P \cdot d S$ . Nothing is created or lost: Poynting's theorem is the conservation of energy, written locally.

**Figure 6.1.11.** Poynting's theorem. Left: a volume $V$ with sources $J_{i}$ , a lossy region, stored energy, and the flux of $P$ out through the boundary $S$ . Right: the energy budget. The power supplied by the sources (here with no power entering through $S$ ) is shared between the outflow, the growth of the stored energy and the heat.

The sign of the source term

A source whose impressed current $J_{i}$ points along $E$ absorbs power, like a resistor or a charging battery. One with $J_{i}$ against $E$ delivers power, which is why $- \int_{V} J_{i} \cdot E d V$ is the power supplied. A battery driving a current through a load does exactly this inside itself: the chemical forces push the charges against the field.

Poynting's theorem in time-harmonic form

The same steps work with phasors. Start from the identity with the conjugate field, $\nabla \cdot (E \times H^{*}) = H^{*} \cdot (\nabla \times E) - E \cdot (\nabla \times H^{*})$ , and insert $\nabla \times E = - j ω B$ and the conjugate of the second Maxwell equation, $\nabla \times H^{*} = J_{i}^{*} + σ E^{*} - j ω D^{*}$ :

\nabla \cdot (E \times H^{*}) = - j ω B \cdot H^{*} - E \cdot J_{i}^{*} - σ E \cdot E^{*} + j ω E \cdot D^{*} . (6.1.46)

Multiply by $- \frac{1}{2}$ , so that the left-hand side becomes $- \nabla \cdot P_{c}$ , integrate over $V$ , use the divergence theorem and take the real part:

Poynting's theorem in time-harmonic form

- \oint_{S} Re {P_{c}} \cdot d S - \int_{V} Re {\frac{1}{2} E \cdot J_{i}^{*}} d V = - \int_{V} Re {\frac{1}{2} E \cdot j ω D^{*} - \frac{1}{2} j ω B \cdot H^{*}} d V + \int_{V} Re {\frac{σ}{2} E \cdot E^{*}} d V . (6.1.47)

In plain words

Averaged over a period, the power that enters through the boundary or is supplied by the sources can go to one place only: heat. The stored energy rises and falls twice per period but, on average, takes nothing.

The left-hand side is the time-average power entering $V$ through $S$ plus the time-average power supplied by the sources inside $V$ . The second term on the right, $\frac{1}{2} σ ∣ E ∣^{2}$ per unit volume, is the time-average Joule heat, half the peak value of $σ E^{2}$ . The first term on the right deserves a closer look. Over a full period the stored energy returns to its starting value, so on average it neither grows nor falls. Indeed, with real $ε$ and $μ$ the products $E \cdot D^{*} = ε ∣ E ∣^{2}$ and $B \cdot H^{*} = μ ∣ H ∣^{2}$ are real, the bracket is imaginary, and its real part is zero. Its imaginary part,

\int_{V} (\frac{1}{2} ω μ ∣ H ∣^{2} - \frac{1}{2} ω ε ∣ E ∣^{2}) d V = 2 ω (⟨ W_{m} ⟩ - ⟨ W_{e} ⟩), ⟨ W_{m} ⟩ = \int_{V} \frac{1}{4} μ ∣ H ∣^{2} d V, ⟨ W_{e} ⟩ = \int_{V} \frac{1}{4} ε ∣ E ∣^{2} d V, (6.1.48)

is the reactive power: it appears in the imaginary part of the same balance, which is not written out here. If the permittivity is complex, $ε = ε^{'} - j ε^{''}$ , then $E \cdot D^{*} = (ε^{'} + j ε^{''}) ∣ E ∣^{2}$ and the real part of the first term becomes $\int_{V} \frac{1}{2} ω ε^{''} ∣ E ∣^{2} d V$ : the dielectric loss, heat just like that of $σ$ . This is how a microwave oven heats water (Section 5.6).

Quick check

A region is filled with a lossless medium and contains no sources. A standing wave lives in it. What does the time-harmonic theorem say about the net power through its boundary?

With $σ = 0$ , no $J_{i}$ and real $ε$ , $μ$ , every term but the first vanishes on average: $\oint_{S} Re {P_{c}} \cdot d S = 0$ . Energy may slosh in and out, but no net power crosses $S$ (Home exercise 6.1.5).

The power carried by a plane wave

For $E = E_{0} cos (ω t - β z) a_{x}$ in a lossless medium, $H = (E_{0} / η) cos (ω t - β z) a_{y}$ and the instantaneous Poynting vector is

P = E \times H = \frac{E _{0}^{2}}{η} cos^{2} (ω t - β z) a_{z} . (6.1.49)

It pulsates at $2 ω$ between zero and $E_{0}^{2} / η$ but never points backwards. Its time average follows from the phasors for any uniform plane wave. With $H^{*} = a_{k} \times E^{*} / η^{*}$ and the identity $A \times (B \times C) = B (A \cdot C) - C (A \cdot B)$ ,

P_{a v} = \frac{1}{2} Re {E \times \frac{a _{k} \times E ^{*}}{η ^{*}}} = \frac{1}{2} Re {\frac{a _{k} ( E \cdot E ^{*} ) - E ^{*} ( E \cdot a _{k} )}{η ^{*}}} = \frac{1}{2} Re {\frac{1}{η ^{*}}} ∣ E ∣^{2} a_{k}, (6.1.50)

because $E \cdot a_{k} = 0$ .

Power density of a uniform plane wave

For a uniform plane wave $E = E_{0} e^{- j k \cdot r}$ of any polarisation in a lossless medium,

P_{a v} ∣ E_{0} ∣^{2} = \frac{∣ E _{0} ∣ ^{2}}{2 η} a_{k} = \frac{η ∣ H _{0} ∣ ^{2}}{2} a_{k} [W/ m^{2}], = E_{0} \cdot E_{0}^{*} = ∣ E_{0 x} ∣^{2} + ∣ E_{0 y} ∣^{2} + ∣ E_{0 z} ∣^{2} . (6.1.51)

With loss, $η = ∣ η ∣ e^{j θ_{η}}$ and $Re {1/ η^{*}} = cos θ_{η} /∣ η ∣$ , so $P_{a v} = \frac{∣ E _{0} ∣ ^{2}}{2∣ η ∣} e^{- 2 α z} cos θ_{η} a_{k}$ for a wave travelling along $z$ .

In plain words

A plane wave carries the power density $E^{2} /2 η$ along its direction of travel, the field analogue of $V^{2} /2 R$ . Every component of $E$ contributes its own share, whatever their relative phases.

Two more facts complete the picture. The electric and magnetic energy densities are equal at every point and every instant, since $H = E / η$ gives $w_{m} = \frac{1}{2} μ H^{2} = \frac{1}{2} μ E^{2} ε / μ = \frac{1}{2} ε E^{2} = w_{e}$ . And the energy moves at the speed

v_{E} = \frac{P _{a v}}{⟨ w ⟩} = \frac{E _{0}^{2} /2 η}{\frac{1}{2} ε E _{0}^{2}} = \frac{1}{η ε} = \frac{1}{μ ε} = v_{p}, (6.1.52)

where $⟨ w ⟩ = ⟨ w_{e} ⟩ + ⟨ w_{m} ⟩ = \frac{1}{4} ε E_{0}^{2} + \frac{1}{4} μ H_{0}^{2} = \frac{1}{2} ε E_{0}^{2}$ . In a lossless, non-dispersive medium energy and phase travel together.

Figure 6.1.12 compares a linearly and a circularly polarised wave with the same peak field $E_{0}$ per component. The circular wave of Problem 6.1.4 has two components of amplitude $E_{0}$ in time quadrature, its $∣ E ∣$ never changes, and its power density is constant, twice the average of the linear one. Polarisation is the subject of the next subsection.

The instantaneous power density at t=0 along the direction of travel. The linearly polarised wave E_0\cos(\omega t-\beta z)\vec a_x carries its energy in pulses, two per wavelength, with the average E_0^2/2\eta. The circular wave E_0[\cos(\omega t-\beta z)\vec a_x+\sin(\omega t-\beta z)\vec a_y] carries the constant E_0^2/\eta. — **Figure 6.1.12.** The instantaneous power density at $t = 0$ along the direction of travel. The linearly polarised wave $E_{0} cos (ω t - β z) a_{x}$ carries its energy in pulses, two per wavelength, with the average $E_{0}^{2} /2 η$ . The circular wave $E_{0} [cos (ω t - β z) a_{x} + sin (ω t - β z) a_{y}]$ carries the constant $E_{0}^{2} / η$ .

Quick check

A plane wave in free space has $E_{0} = 1 V/m$ . Find $P_{a v}$ and $H_{0}$ .

$P_{a v} = 1/ (2 \times 377) = 1.33 mW/ m^{2}$ and $H_{0} = 1/377 = 2.65 mA/m$ . One volt per metre is a strong signal, found only within a few kilometres of a broadcast transmitter, and it carries about a milliwatt per square metre.

6.1.5 Power flow and the Poynting vector

The Poynting vector

The time-average power density

Poynting's theorem

Poynting's theorem in time-harmonic form

The power carried by a plane wave