Part 2 · Light

2.2 Electromagnetic waves in vacuum

Source: Postorino, week 2 slides, pp. 15–19, and week 3 slides, pp. 4–5

Objectives

Theorem. Maxwell’s equations in vacuum give a wave equation for $E$ and for $B$ , with speed $c = 1/ μ_{0} ε_{0}$ .
Recipe. On a plane wave, replace $\nabla$ with $i k$ and $\partial / \partial t$ with $- iω$ ; read off transversality and $B = \frac{1}{ω} k \land E$ .
Picture. $E$ , $B$ , $k$ form a right-handed set; the energy flows along $k$ , with intensity proportional to $∣ E ∣^{2}$ .
Vocabulary. Plane wave, wavefront, polarisation vector, Poynting vector, spherical wave, electromagnetic spectrum.

From Maxwell’s equations to the wave equation

Theorem 2.2.1Wave equations for the fieldsslides w2, p. 15

In vacuum, $\nabla^{2} E = μ_{0} ε_{0} \frac{\partial ^{2} E}{\partial t ^{2}}, \nabla^{2} B = μ_{0} ε_{0} \frac{\partial ^{2} B}{\partial t ^{2}} .$

Proof

Take the curl of Faraday’s law and use Maxwell’s law for the curl of $B$ (Theorem 2.1.6). The curl acts on space and $\partial / \partial t$ on time, so they can be swapped: $\nabla \land \nabla \land E = - \frac{\partial}{\partial t} [\nabla \land B] = - \frac{\partial}{\partial t} [μ_{0} ε_{0} \frac{\partial E}{\partial t}] = - μ_{0} ε_{0} \frac{\partial ^{2} E}{\partial t ^{2}} .$ By the curl-of-curl identity, $\nabla \land \nabla \land E = \nabla (\nabla \cdot E) - \nabla^{2} E = - \nabla^{2} E$ , since $\nabla \cdot E = 0$ in vacuum. Putting the two together gives the equation for $E$ .

For $B$ , start from the curl of Maxwell’s law and use Faraday’s law and $\nabla \cdot B = 0$ : $\nabla \land \nabla \land B = μ_{0} ε_{0} \frac{\partial}{\partial t} (\nabla \land E) = - μ_{0} ε_{0} \frac{\partial ^{2} B}{\partial t ^{2}} .$

Proposition 2.2.2Curl of a curlslides w2, p. 15

$\nabla \land \nabla \land E = \nabla (\nabla \cdot E) - \nabla^{2} E .$

Proofbeyond the lectures

The slide says the identity “can be shown”. Prove the $x$ component; $y$ and $z$ follow by relabelling. Let $W = \nabla \land E$ , so $W_{y} = \partial_{z} E_{x} - \partial_{x} E_{z}$ and $W_{z} = \partial_{x} E_{y} - \partial_{y} E_{x}$ . Then $(\nabla \land W)_{x} = \partial_{y} W_{z} - \partial_{z} W_{y} = \partial_{y} \partial_{x} E_{y} - \partial_{y}^{2} E_{x} - \partial_{z}^{2} E_{x} + \partial_{z} \partial_{x} E_{z} .$ Add and subtract $\partial_{x}^{2} E_{x}$ : $= \partial_{x} (\partial_{x} E_{x} + \partial_{y} E_{y} + \partial_{z} E_{z}) - (\partial_{x}^{2} + \partial_{y}^{2} + \partial_{z}^{2}) E_{x} = [\nabla (\nabla \cdot E)]_{x} - \nabla^{2} E_{x} .$

Key observation 2.2.3Light is an electromagnetic waveslides w2, p. 15

Compare with $\nabla^{2} Y = \frac{1}{v ^{2}} \partial_{t}^{2} Y$ (Theorem 1.5.1): $\frac{1}{v ^{2}} = μ_{0} ε_{0} \Rightarrow v = \frac{1}{μ _{0} ε _{0}} = c = 3 \times 1 0^{8} m/s .$ Two constants measured with charges, magnets and coils give the speed of light: light is an electromagnetic wave.

Check beyond the lectures: $μ_{0} = 4 π \times 1 0^{- 7}$ H/m and $ε_{0} = 8.85 \times 1 0^{- 12}$ F/m give $1/ μ_{0} ε_{0} = 3.00 \times 1 0^{8}$ m/s.

Plane waves

$E$ and $B$ are both waves, but not independent. Consider $E$ . Its equation is linear, so superposition and Fourier apply (Key observation 1.6.8): the general field is built out of elementary plane waves, its Fourier components, $E (r, t) = E_{0} e^{i (kr - ω t)} .$

Proposition 2.2.4Plane-wave argumentslides w2, p. 16–17

As in the 1D solution, the argument must be a function of $r - v t$ , with $r = (x, y, z)$ , $v = (v_{x}, v_{y}, v_{z})$ , $k = (k_{x}, k_{y}, k_{z})$ and $k = ∣ k ∣ = 2 π / λ$ . Then $k \cdot (r - v t) = kr - ω t, ω = k \cdot v = \frac{2 π}{T} (dispersion relation) .$ For $v$ along $k$ this is $ω = k v$ , so $\frac{2 π}{T} = \frac{2 π}{λ} v$ : $λ = v T$ .

(In this part the slides also put an arrow on the bold vectors, $k$ , “only in this section, for clarity”.)

Definition 2.2.5Monochromatic plane wave and wavefrontslides w2, p. 17

A monochromatic plane wave has a fixed frequency $ω = ω^{*}$ . Its wavefronts are infinite parallel planes, perpendicular to the direction of propagation and spaced by $λ$ .
A wavefront is the set of points at constant phase: $kr - ω^{*} t = const$ .

RemarkWhy planesbeyond the lectures

At a fixed time, $k \cdot r = const$ is the equation of a plane perpendicular to $k$ . As $t$ grows the constant grows, so each plane moves along $k$ at speed $ω / k$ . Planes whose phases differ by $2 π$ are $2 π / k = λ$ apart.

Transversality, and B from E

Take a plane wave for both fields: $E = E_{0} e^{i (kr - ω t)}$ , $B = B_{0} e^{i (kr - ω t)}$ .

Key observation 2.2.6Derivatives of a plane wavebeyond the lectures

On a plane wave, $\nabla$ becomes $i k$ and $\partial / \partial t$ becomes $- iω$ : each derivative of $e^{i (kr - ω t)}$ just multiplies it, as for the oscillator (Key observation 1.2.6). So $\nabla \cdot E \to i k \cdot E$ and $\nabla \land E \to i k \land E$ .

Theorem 2.2.7The electromagnetic wave is transverseslides w2, p. 18

$E ⊥ k, B ⊥ k, B = \frac{1}{ω} k \land E, E = c B (magnitudes) .$ $E$ , $B$ , $k$ form a right-handed set.

Proof

Gauss’s law gives $E ⊥ k$ : $\nabla \cdot E = (\partial_{x}, \partial_{y}, \partial_{z}) \cdot (E_{0 x}, E_{0 y}, E_{0 z}) e^{i (k_{x} x + k_{y} y + k_{z} z - ω t)} = i k_{x} E_{0 x} e^{i (\dots)} + i k_{y} E_{0 y} e^{i (\dots)} + i k_{z} E_{0 z} e^{i (\dots)} = i k \cdot E = 0.$ The same for $\nabla \cdot B = 0$ gives $B ⊥ k$ .

Faraday’s law gives $B$ : $\nabla \land E = - \frac{\partial B}{\partial t} \Rightarrow i k \land E = iω B \Rightarrow B = \frac{1}{ω} k \land E,$ perpendicular to both $E$ and $k$ .

Magnitudes. $k ⊥ E$ , so $∣ k \land E ∣ = k E$ , and $k E = ω B$ : $E = \frac{ω}{k} B = c B$ .

The wave is transverse: the fields oscillate across the direction of travel, never along it.

Key observation 2.2.8Only E needs studyingslides w2, p. 18

Once $E (r, t)$ is given, $B (r, t)$ is completely determined: magnitude, direction and sense. From here on, only the electric field is studied.

Polarisation and energy

A plane wave travelling along $k ∥ \hat{x}$ : $E$ oscillates in one plane (say $x$ – $y$ ), $B$ in the perpendicular plane ( $x$ – $z$ ), in phase, with wavelength $λ$ .

Definition 2.2.9Polarisation vectorslides w2, p. 19

One degree of freedom remains: the direction of the electric field, set by $E_{0}$ . This polarisation vector fixes the direction along which $E$ oscillates, and so the one for $B$ too.

Loading the demo…

Figure 2.2.1 A plane wave travelling along

k

E

and

B

at right angles, in phase, both across

k

. Drag the background to turn the view. Change the polarisation angle, or switch to circular: two perpendicular linear waves

90°

apart, whose

E

tip draws a helix. Press Play to send the wave along.

Theorem 2.2.10Poynting vectorslides w2, p. 19

The Poynting vector is the instantaneous flux of electromagnetic energy carried by the wave: $S = \frac{1}{μ _{0}} E \land B .$ $∣ S ∣$ is the energy carried per unit time and per unit area (W/m²); the direction of transport is $\hat{S}$ . For a plane wave, $∣ S ∣ = \frac{1}{μ _{0}} ∣ E ∣ ∣ B ∣ = \frac{1}{μ _{0} c} ∣ E ∣^{2} = \frac{ε _{0}}{μ _{0}} ∣ E ∣^{2} .$

Proof

$E ⊥ B$ , so $∣ E \land B ∣ = ∣ E ∣∣ B ∣$ , and $B = E / c$ . Then $\frac{1}{μ _{0} c} = \frac{μ _{0} ε _{0}}{μ _{0}} = \frac{ε _{0}}{μ _{0}}$ . For a plane wave $E \land B$ points along $k$ : energy flows the way the wave travels.

It is proportional to the square of the electric field’s amplitude, like the energy of an oscillator (Key observation 1.3.8).

WarningSlide slip

The slide’s last step is $μ_{0} / ε_{0} ∣ E ∣^{2}$ . With $c = 1/ μ_{0} ε_{0}$ , $\frac{1}{μ _{0} c} = ε_{0} / μ_{0}$ . ( $μ_{0} / ε_{0} \approx 377 Ω$ is the impedance of free space, which divides $∣ E ∣^{2}$ .)

Proposition 2.2.11Intensitybeyond the lectures

The physical field is real, $E_{0} cos (kr - ω t)$ , and $⟨ cos^{2} ⟩ = \frac{1}{2}$ . The intensity, the mean energy flux, is $I = ⟨ ∣ S ∣ ⟩ = \frac{1}{2} \frac{ε _{0}}{μ _{0}} E_{0}^{2} = \frac{1}{2} c ε_{0} E_{0}^{2} .$ With complex fields, $∣ E ∣^{2}$ stands for $E \cdot E^{*}$ , as $∣ \overset{x}{^} ∣^{2}$ did for the oscillator.

Plane and spherical waves

Plane wave, $E (r, t) = E_{0} e^{i (kr - ω t)}$ : the wavefronts are planes spaced $λ = c t_{0}$ ( $t_{0}$ the period), moving along the direction of propagation; the amplitude is the same everywhere.

Spherical wave, from a point source O: $E (r, t) = \frac{E _{0}}{r} e^{i (k r - ω t)},$ with $r$ the distance from O. The wavefronts are spheres centred on O, spaced $λ$ , moving outwards at $c$ . $E$ and $B$ are still transverse to the outgoing ray.

RemarkWhy 1/rbeyond the lectures

The intensity goes as $∣ E ∣^{2} \propto 1/ r^{2}$ . The total power through a sphere, intensity × $4 π r^{2}$ , is then the same at every radius: energy is conserved. Far from the source a small patch of a spherical wave looks like a plane wave.

The electromagnetic spectrum

The same equations describe waves over more than fifteen orders of magnitude of wavelength.

name	wavelength	typical source
radio	≳ 1 m (up to km)	AM, FM radio
microwaves	~1 mm – 1 m	microwave oven, radar
infrared	~1 µm – 1 mm	people, warm bodies
visible	~400–700 nm	light bulb, the Sun
ultraviolet	~10–400 nm
soft, hard X-rays	~10 nm – 10 pm	synchrotrons, X-ray machines
gamma rays	≲ 10 pm	radioactive elements

$λ ν = c, c = 300000 km/s = 3 \times 1 0^{8} m/s .$ Frequencies run from about $1 0^{6}$ Hz (radio) to $1 0^{20}$ Hz (gamma); the energy of a photon, $E = h ν$ , from about $1 0^{- 9}$ eV to $1 0^{6}$ eV, with the visible around 1 eV. The slide compares each wavelength with an object: a football field, a house, a cell, a virus, a protein, a water molecule.

$E = h ν$ is the photon energy of §5.3 The photoelectric effect, with $h = 6.63 \times 1 0^{- 34}$ J·s. Green light at 500 nm has $ν = 6 \times 1 0^{14}$ Hz and $E = 2.5$ eV: the scale of atomic energies (Energies) beyond the lectures.

Loading the demo…

Figure 2.2.2 The electromagnetic spectrum on a logarithmic scale. Drag the

wavelength

: the readouts give its frequency, its photon energy in eV, the region it belongs to, and an object of that size. Each step to the left is ten times shorter.

Exercises

Exercise 2.2.1beyond the lectures

A plane wave in vacuum has $E = E_{0} \hat{y} cos (k x - ω t)$ , $E_{0} = 100$ V/m. Find $B$ and the mean intensity.

Solution

$B = \frac{1}{ω} k \land E = \frac{E _{0}}{c} (\hat{x} \land \hat{y}) cos (k x - ω t) = \frac{E _{0}}{c} \hat{z} cos (k x - ω t)$ , so $B_{0} = 3.3 \times 1 0^{- 7}$ T. $I = \frac{1}{2} c ε_{0} E_{0}^{2} = \frac{1}{2} \cdot 3 \times 1 0^{8} \cdot 8.85 \times 1 0^{- 12} \cdot 1 0^{4} = 13$ W/m².

Exercise 2.2.2beyond the lectures

Sunlight at the Earth carries about 1360 W/m². Estimate the amplitude of its electric field.

Solution

$E_{0} = 2 I / (c ε_{0}) = 2 \cdot 1360/ (3 \times 1 0^{8} \cdot 8.85 \times 1 0^{- 12}) = 1.0 \times 1 0^{3}$ V/m.

Exercise 2.2.3beyond the lectures

A wave has $k = k (1, 1, 0) / 2$ . Which of these are allowed polarisation vectors: $(0, 0, 1)$ , $(1, - 1, 0)$ , $(1, 0, 0)$ ?

Solution

Need $k \cdot E_{0} = 0$ . $(0, 0, 1)$ : yes. $(1, - 1, 0)$ : yes. $(1, 0, 0)$ : $k \cdot E_{0} = k / 2 \neq = 0$ , no.

Exercise 2.2.4beyond the lectures

A 1 W isotropic point source: what is the intensity 2 m away, and the field amplitude?

Solution

$I = P / (4 π r^{2}) = 1/ (4 π \cdot 4) = 0.020$ W/m². $E_{0} = 2 I / (c ε_{0}) = 3.9$ V/m.

Exercise 2.2.5beyond the lectures

Find the frequency and photon energy of a 10 µm infrared wave and of a 0.1 nm X-ray.

Solution

10 µm: $ν = c / λ = 3 \times 1 0^{13}$ Hz, $E = 1240/1 0^{4} = 0.124$ eV. 0.1 nm: $ν = 3 \times 1 0^{18}$ Hz, $E = 1240/0.1 = 12.4$ keV.