Part 2 · Light

2.4 Adding waves: interference, standing waves, beats

Source: Postorino, week 3 slides, pp. 6–11 and 16

Objectives

Recipe. Add two waves with the sum-to-product formula (or by factoring out the mean phase): a fast carrier times a slow amplitude.
Picture. Three cases of one formula: interference (amplitude set by the phase difference), standing waves (nodes and antinodes, no energy transport), beats (an envelope moving at the group velocity).
Key. Fields add; intensities don’t. The velocity that matters is the group velocity, at which energy moves.
Vocabulary. Constructive and destructive interference, node, antinode, carrier, envelope, phase velocity, group velocity, beat frequency.

Adding two waves

Take two plane waves, the first with an extra phase $ϕ$ : $E_{1} (r, t) = E_{01} e^{i [k_{1} r - ω_{1} t + ϕ]}, E_{2} (r, t) = E_{02} e^{i [k_{2} r - ω_{2} t]}, α_{1} = k_{1} r - ω_{1} t, α_{2} = k_{2} r - ω_{2} t .$ For simplicity: equal amplitudes, $E_{01} = E_{02} = E_{0}$ ; and $k_{1} ∥ k_{2}$ , $E_{01} ∥ E_{02}$ : same direction of travel, same polarisation, so the fields add as numbers along one direction.

Key observation 2.4.1How complex fields are usedslides w3, p. 6

Calculations: with the complex field.
Solution: the real part of the complex field.
Energy: the square of the real part.

The field at $r$ is the sum of the fields (superposition). The intensity there goes as the squared modulus of the total field, $∣ E_{1} + E_{2} ∣^{2}$ . Fields add; intensities don’t.

Theorem 2.4.2Sum of two wavesslides w3, p. 6

$Re {E} = 2 E_{0} cos [\frac{( k _{1} + k _{2} ) r}{2} - \frac{( ω _{1} + ω _{2} ) t}{2} + \frac{ϕ}{2}] cos [\frac{( k _{1} - k _{2} ) r}{2} - \frac{( ω _{1} - ω _{2} ) t}{2} + \frac{ϕ}{2}] .$ The sum is a product of two waves: one at the average wave vector and frequency, one at half the difference.

Proof

$Re {E_{1} + E_{2}} = E_{0} Re {e^{i [α_{1} + ϕ]} + e^{i α_{2}}} = E_{0} [cos (α_{1} + ϕ) + cos α_{2}]$ . Use the sum-to-product (prosthaphaeresis) formula $cos p + cos q = 2 cos (\frac{p + q}{2}) cos (\frac{p - q}{2})$ with $p = α_{1} + ϕ$ , $q = α_{2}$ . The formula itself beyond the lectures: write $p = a + b$ , $q = a - b$ with $a = \frac{p + q}{2}$ , $b = \frac{p - q}{2}$ ; then $cos (a + b) + cos (a - b) = 2 cos a cos b$ , since the $sin a sin b$ terms cancel.

Alternative: complex fields only

Factor out the mean phase: $E = E_{0} [e^{i [α_{1} + ϕ]} + e^{i α_{2}}] = E_{0} e^{i [α_{1} + α_{2} + ϕ] /2} [e^{i [α_{1} - α_{2} + ϕ] /2} + e^{- i [α_{1} - α_{2} + ϕ] /2}] = E_{0} e^{i [α_{1} + α_{2} + ϕ] /2} 2 cos [\frac{α _{1} - α _{2} + ϕ}{2}],$ since $e^{i θ} + e^{- i θ} = 2 cos θ$ . The real part gives the same result. Here it is clear which factor is the fast carrier, $e^{i [α_{1} + α_{2} + ϕ] /2}$ , and which the real amplitude, $2 cos [\dots]$ .

The intensity, from the complex form beyond the lectures: the carrier has modulus 1, so $∣ E ∣^{2} = 4 E_{0}^{2} cos^{2} [\frac{α _{1} - α _{2} + ϕ}{2}] .$ The three cases below just read Theorem 2.4.2.

Case 1: interference

Same direction, same frequency: waves along $x$ in the same sense, $ω_{1} = ω_{2} = ω$ , $k_{1} = k_{2} = (k, 0, 0)$ . Then $α_{1} = α_{2}$ and the difference factor is constant.

Proposition 2.4.3Interferenceslides w3, p. 7

$Re {E} = 2 E_{0} cos [\frac{ϕ}{2}] cos [k x - ω t + \frac{ϕ}{2}] .$ The same wave, with an amplitude $2 E_{0} cos (ϕ /2)$ set by the phase difference:

$ϕ = 0$ (in phase): amplitude $2 E_{0}$ , intensity $4∣ E_{0} ∣^{2}$ : constructive interference;
$ϕ = π$ (in antiphase): amplitude 0, intensity 0: destructive interference.

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Figure 2.4.1 Waves

1

and

2

, same frequency and direction, and their

sum

. Drag the phase difference

ϕ

. At

ϕ = 0

the sum has twice the amplitude and four times the intensity; at

ϕ = π

nothing is left.

RemarkEnergy is not createdbeyond the lectures

Two waves of intensity $∣ E_{0} ∣^{2}$ each give $4∣ E_{0} ∣^{2}$ in phase, not $2∣ E_{0} ∣^{2}$ . Averaged over all phases, $4 cos^{2} (ϕ /2)$ is 2: interference only redistributes energy, more in some places and less in others.

Case 2: standing waves

Same frequency, opposite directions. The slide takes $ω_{1} = ω$ , $ω_{2} = - ω$ and $k_{1} = k_{2} = (k, 0, 0)$ . Flipping the sign of $ω$ is the same as flipping the sign of $v$ : by $ω = k v$ , $[k x - ω t]$ is the progressive wave and $[k x + ω t]$ the regressive one. Then $α_{1} + α_{2} = 2 k x$ and $α_{1} - α_{2} = - 2 ω t$ .

Proposition 2.4.4Standing waveslides w3, p. 8

$Re {E} = 2 E_{0} cos [k x + \frac{ϕ}{2}] cos [ω t - \frac{ϕ}{2}] .$ The dependence on $t$ separates from the dependence on $x$ : the wave no longer travels.

Each point oscillates with an amplitude $2 E_{0} ∣ cos (k x + ϕ /2) ∣$ that depends only on its position. That amplitude, and so the energy at that point, does not change.
There is no transport of energy with the wave disturbance. The energy is localised: maximal at the antinodes (ventri), zero at the nodes (nodi).

WarningSlide slip

The slide writes the second factor as $cos [ω t + ϕ /2]$ . From Theorem 2.4.2 it is $cos [- ω t + ϕ /2] = cos [ω t - ϕ /2]$ . The sign of the phase changes nothing below.

Nodes are where $cos (k x + ϕ /2) = 0$ . They are $λ /2$ apart, and so are the antinodes beyond the lectures.

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Figure 2.4.2 Two equal waves running in opposite directions, and their

sum

. Press Play. The travelling waves move; the sum only swells and shrinks in place, with

nodes

that never move,

λ /2

apart.

A rope fixed at one end and shaken at the other. The wave and its reflection form standing waves with 1, 2, 3, … loops. Both ends must be nodes, so only lengths $L = mλ /2$ fit beyond the lectures. That is why a string, or a cavity (§5.1 Line spectra and blackbody radiation), resonates only at a discrete set of frequencies: the “vibrating string” of Key observation 1.6.11.

Case 3: beats and group velocity

Waves along $\hat{x}$ with nearly the same frequency and nearly the same wave vector, $ϕ = 0$ . Define the means and half-differences: $ω = \frac{ω _{1} + ω _{2}}{2}, Δ ω = \frac{ω _{1} - ω _{2}}{2} ≪ ω; k = \frac{k _{1} + k _{2}}{2}, Δ k = \frac{k _{1} - k _{2}}{2} ≪ k .$

Proposition 2.4.5Beatsslides w3, p. 9

$Re {E} = 2 E_{0} cos [k x - ω t] cos [Δ k x - Δ ω t] .$ A fast wave $cos [k x - ω t]$ is multiplied by a slow envelope $cos [Δ k x - Δ ω t]$ that modulates its amplitude. Where the two original waves are in phase the sum is large; where they are in antiphase it almost vanishes. These are beats (battimenti).

Each factor is a travelling wave with its own speed: $ω / k$ for one, $Δ ω /Δ k$ for the other.

Definition 2.4.6Phase and group velocityslides w3, p. 9

The crests move at the phase velocity $v_{φ} = ω / k$ .
The envelope, which modulates the amplitude and so carries the energy, moves at the group velocity $v_{g} = \frac{Δ ω}{Δ k} \to v_{g} = \frac{d ω}{d k} in the limit.$

Key observation 2.4.7The relevant velocityslides w3, p. 9

The relevant velocity is the group velocity: the one at which energy is transferred.

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Figure 2.4.3 Two waves of nearly equal

k

and

ω

add to a

carrier

inside an

envelope

. Press Play and follow the two markers: a crest moves at

v_{φ}

, the envelope at

v_{g}

. Set

v_{g} / v_{φ}

to 1 and the packet moves rigidly; otherwise crests slide through the envelope, as in the slide’s figure, which compares

v_{g} = 2, 1, 0.5

times

v_{φ}

RemarkWhen they differbeyond the lectures

For an elementary wave the dispersion relation is $ω = k v$ . If $v$ is the same for every $k$ , $ω$ is proportional to $k$ , so $d ω / d k = ω / k$ and $v_{g} = v_{φ}$ : the whole packet moves rigidly. In vacuum this holds for light. In matter $v = c / n$ with $n$ depending on frequency (dispersion): $ω (k)$ is not a straight line and $v_{g} \neq = v_{φ}$ (§3.4 Dispersion and group velocity).

RemarkBeats you can hearslides w3, p. 9, 16

A tuning fork (diapason) gives 440 Hz; the ear hears 20 Hz – 20 kHz. Two forks a few hertz apart, sounding together, swell and fade. The loudness peaks wherever the envelope’s $∣ cos ∣$ peaks, twice per envelope period, so the beat frequency is $2Δ ω /2 π = (ω_{1} - ω_{2}) /2 π$ , the difference of the two frequencies. Musicians tune by making it vanish. Try it: mathlets: beats and beats with sound.

The three cases side by side

	waves	sum	name
1	same frequency, along $x$ , same sense	$2 E_{0} cos [\frac{ϕ}{2}] cos [k x - ω t + \frac{ϕ}{2}]$	interference
2	same frequency, along $x$ , opposite senses	$2 E_{0} cos [k x + \frac{ϕ}{2}] cos [ω t + \frac{ϕ}{2}]$	standing wave
3	nearly equal $k$ and $ω$ , along $x$	$2 E_{0} cos [Δ k x - Δ ω t] cos [k x - ω t]$	beats

In each row the first factor is the amplitude of the resulting wave (in red on the slide). Row 2 keeps the slide’s sign; see the slip under case 2.

Key observation 2.4.8No wave is monochromaticslides w3, p. 16

Beats are not a peculiar case. No wave is truly monochromatic. A real wave has a beginning and an end, so it is a combination of infinitely many monochromatic waves (Key observation 1.7.5).

Exercises

Exercise 2.4.1beyond the lectures

Two waves of equal amplitude and phase difference $ϕ = π /3$ add. What fraction of the maximum intensity results?

Solution

$I / I_{max} = cos^{2} (ϕ /2) = cos^{2} (π /6) = 3/4$ .

Exercise 2.4.2beyond the lectures

Two tuning forks give 440 Hz and 444 Hz. What do you hear?

Solution

A tone at the mean, 442 Hz, whose loudness swells and fades 4 times per second, the difference of the two frequencies.

Exercise 2.4.3beyond the lectures

Show that in a standing wave the time-averaged Poynting flux is zero.

Solution

Each travelling wave carries intensity $I$ , in opposite directions, so the net average is $I - I = 0$ . Directly: $E \propto cos k x cos ω t$ , and by Faraday’s law $B \propto sin k x sin ω t$ , so $S \propto sin 2 k x sin 2 ω t$ , which averages to zero.

Exercise 2.4.4beyond the lectures

A guitar string 65 cm long, fixed at both ends, carries waves at 400 m/s. Find its lowest three frequencies.

Solution

$L = mλ /2$ gives $λ_{m} = 2 L / m$ and $ν_{m} = v / λ_{m} = m v /2 L = m \cdot 308$ Hz: 308, 615, 923 Hz.

Exercise 2.4.5beyond the lectures

Deep-water waves have $ω = g k$ . Find $v_{φ}$ and $v_{g}$ , and say which is larger.

Solution

$v_{φ} = ω / k = g / k$ ; $v_{g} = d ω / d k = \frac{1}{2} g / k = v_{φ} /2$ . The crests run through a group twice as fast as the group, appearing at its back and vanishing at its front.

Exercise 2.4.6beyond the lectures

An anti-reflective coating of index $n_{c}$ on glass must make the two reflected waves (from its top and bottom) cancel at normal incidence. If both reflections have the same phase change, what thickness $t$ is needed for wavelength $λ$ in air?

Solution

The wave reflected at the bottom travels an extra $2 t$ in the coating, a phase $2 π \cdot 2 t / (λ / n_{c})$ . Cancellation needs this to be $π$ : $t = λ / (4 n_{c})$ , a quarter wavelength in the coating. For 550 nm and $n_{c} = 1.38$ , $t = 100$ nm.