Wave energy: what actually travels

The puzzle: the wave arrives, and the water was always here¶

A storm in the Southern Ocean puts energy into the sea. Ten days later, swell arrives on a Californian pier and breaks a plank. Nothing crossed the Pacific: the water that hit the pier had been floating off California the whole time, and the water that the storm pushed on is still off Antarctica. Module 08 established that much on a string — a speck of dust rises, falls, and is exactly where it started once the pulse has gone.

Yet the plank broke. Whatever made the storm’s energy and whatever broke the plank are the same joules, and in between they were somewhere, in a medium that stayed put.

Predict before you calculate¶

Commit to an answer for each before running anything.

A sinusoidal wave travels along a string. Where is the energy per unit length largest — at the crests, at the zero crossings, or is it the same everywhere?
Can a piece of string move faster than the wave travelling along it?
You double the amplitude of a wave and change nothing else. The power it delivers rises by what factor?
A travelling wave and a standing wave have the same amplitude and the same frequency. Which one delivers more energy past a point in the middle of the string?

Explore the model¶

Three animations, all computed from the same wavelab.waves functions the laboratory calls.

A sinusoidal wave travelling right, over its own energy density. The orange marker rides a crest; the green marker rides the zero crossing a quarter wavelength behind it. The crest marker sits in a trough of the energy for the whole loop — never above $2.5\times10^{-4}$ of what the green marker stands on — and the energy humps come twice as often as the crests, because the density goes as the square of a sine. Both travel at the wave speed, so the picture below the string is as rigid as the picture above it.

A single Gaussian pulse, with its energy beneath. Two things are happening at once. The thick pale curve is $u_P$ and the thin dark one $u_K$ , and they lie on top of each other everywhere, at every frame, to $2\times10^{-4}$ of the peak. And one hump of string carries two humps of energy: both densities go as the square of the slope, and a pulse is flat on top. The dotted curve is the flux divided by $v$ , which lands on the grey total to $2\times10^{-8}$ — the energy moves at exactly the speed the shape does.

A travelling train and a standing mode of the same amplitude and frequency, each with a gate where the power crossing is read. The travelling wave’s flux (blue) never goes below zero: energy crosses the gate twice a cycle and keeps going. The standing wave’s flux (orange) swings symmetrically about zero and nets $1.4\times10^{-10}$ of what it moved. Its peak is a quarter of the travelling wave’s, not a thousandth — a standing wave pushes comparable energy about and simply puts it all back.

Model specification

System: module 08’s ideal string — transverse displacement $y(x,t)$ , tension $T$ , linear mass density $\mu$ — now carrying three derived fields: the kinetic and potential energy densities $u_K(x,t)$ , $u_P(x,t)$ , and the energy flux $P(x,t)$ .
Dynamics: $\mu\,y_{tt} = T\,y_{xx}$ as before. The energy quantities are diagnostics: they are computed from the solution and never fed back into it, so every agreement between them and the solver is evidence rather than construction.
Boundary: fixed ends. No energy crosses one, because $y_t = 0$ there and a force that does no work transports no power.
Ensemble: deterministic; detector noise, where added, is Gaussian, independent and seeded.
Ignored: dissipation, longitudinal motion, and the momentum a transverse wave may or may not carry — the last is a genuinely open question, flagged in the advanced section rather than waved away.
Valid when: slopes are small, $|y_x| \ll 1$ . The potential density needs this one order deeper than the equation of motion did: the equation dropped terms of order $y_x^2$ , and $u_P$ keeps them and drops $y_x^4$ .
Failure modes: the densities read as exact for a steep wave; a lossy string’s energy assumed conserved because the ideal one’s is; the “energy velocity” trusted once the medium becomes dispersive, where module 13 finds it is no longer $\omega/k$ .

Run the laboratory now, then come back. Four things are worth doing before you read on:

Read the three velocities off one pulse, and notice that two of them are the same number.
Plot the energy density under a sinusoidal wave and find the offset between the two curves.
Meter the power at a point, and check that what the meter counted is what ended up beyond it.
Measure the mean power twice, through the same noise, with two estimators that are both correct algebra. One of them will lie to you.

Derive the result¶

The two densities¶

A short piece of string between $x$ and $x + \mathrm{d}x$ has mass $\mu\,\mathrm{d}x$ and moves sideways at $y_t$ . Its kinetic energy is $\tfrac12(\mu\,\mathrm{d}x)\,y_t^2$ , and per unit length that is

u_K = \tfrac12\,\mu\left(\frac{\partial y}{\partial t}\right)^2 .

(1)

The potential energy takes one more step. A string stores energy by being longer than it was: tilted to a slope $y_x$ , a piece of rest length $\mathrm{d}x$ becomes $\sqrt{1 + y_x^2}\;\mathrm{d}x$ , and the work done against the constant tension $T$ is $T$ times that extra length. Expanding the square root,

T\left(\sqrt{1 + y_x^2} - 1\right)\mathrm{d}x = \tfrac12\,T\,y_x^2\,\mathrm{d}x + O(y_x^4)\,\mathrm{d}x ,

(2)

u_P = \tfrac12\,T\left(\frac{\partial y}{\partial x}\right)^2 .

(3)

Read what $u_P$ says about where a wave’s energy is. It is not displacement that stores energy — pick up a straight stretch of string, hold it a metre higher, and you have stretched nothing. It is slope. And $u_K$ needs motion. On a travelling sine $y = A\sin(kx - \omega t)$ the crests have neither: there the string is momentarily flat and momentarily at rest, both densities vanishing together. The crest is the emptiest place on the string, and the energy piles up at the zero crossings, where the string is steepest and moving fastest. That is prediction 1, and the laboratory measures the crest holding $4\times10^{-5}$ of what the zero crossing holds.

The flux: force times velocity, as always¶

Cut the string at $x$ and ask what the left piece does to the right one. It pulls along the string’s own direction with tension $T$ , whose transverse component is $-T\,y_x$ for small slopes. That force acts on a point moving at $y_t$ , and power is force times velocity:

\boxed{\;P = -T\,\frac{\partial y}{\partial x}\,\frac{\partial y}{\partial t}\;}

(5)

in watts, counted positive when energy flows toward $+x$ . There is no new physics in it — it is the same $F v$ that module 01 used on a single mass, applied at a cut rather than to a body.

The sign is the direction, and both signs occur. For a right-mover, module 08’s $y_t = -v\,y_x$ gives $P = T v\,y_x^2 \ge 0$ everywhere: energy flows the way the wave goes, whether the string at that point happens to be rising or falling. A left-mover has $y_t = +v\,y_x$ and $P \le 0$ everywhere. A negative $P$ is not a paradox; it is a wave going the other way.

The first local conservation law¶

Now watch a short stretch of string and account for its energy. Differentiate the density in time:

\frac{\partial u}{\partial t} = \mu\,y_t\,y_{tt} + T\,y_x\,y_{xt} .

(6)

Use the wave equation, $\mu\,y_{tt} = T\,y_{xx}$ , on the first term:

\frac{\partial u}{\partial t} = T\,y_t\,y_{xx} + T\,y_x\,y_{xt} = T\,\frac{\partial}{\partial x}\!\left(y_x\,y_t\right) = -\,\frac{\partial P}{\partial x} .

(7)

Continuity, and what a conservation law looks like locally

\boxed{\;\frac{\partial u}{\partial t} + \frac{\partial P}{\partial x} = 0\;}

(8)

The energy in any stretch $[a,b]$ changes only by what crosses its two ends:

\frac{\mathrm{d}}{\mathrm{d}t}\int_a^b u\,\mathrm{d}x = P(a) - P(b) .

(9)

Energy does not vanish here and reappear there. It flows, continuously, through every point in between — which is the statement the ocean needed.

This is the course’s first local conservation law, and the template for every one that follows. Wherever a quantity is conserved, the same two objects appear: a density that says how much is here, and a flux that says how much is going past. Module 15 builds the Poynting vector this way for electromagnetic energy; probability current in quantum mechanics is the same sentence again, with $|\psi|^2$ for $u$ .

Take the two fixed ends of a string as $a$ and $b$ . A fixed end cannot move, so $y_t = 0$ there, so $P = 0$ : nothing crosses, and the string’s total energy is constant. That is the conservation module 08’s solver was already demonstrating, now derived rather than observed.

Equipartition, and the speed of the energy¶

For a travelling wave everything collapses. Put $y_t = -v\,y_x$ into the two densities:

u_K = \tfrac12\,\mu\,v^2\,y_x^2 = \tfrac12\,T\,y_x^2 = u_P ,

(10)

since $\mu v^2 = T$ by definition of $v$ .

Do not read this as the familiar equipartition of a mass on a spring. Module 01’s oscillator is all potential at the turning points and all kinetic at the middle, and its halves are equal only after averaging over a cycle. Here nothing is averaged. The two densities are the same function of position, at the same instant, because every element of the string is a little behind its neighbour: what one mass does in sequence, the string does all at once, laid out along its length.

And a standing wave does not do it either. For $y = A\sin kx\,\cos\omega t$ the string is all potential at the moment of release and all kinetic a quarter period later, so $u_K$ and $u_P$ are as unequal pointwise as two quantities can be. Only the cycle averages match. Pointwise, instantaneous equipartition is a property of a travelling wave, not of waves in general.

The power a sinusoidal wave carries¶

For the course’s travelling wave,

y(x,t) = \Real\!\left[A\,e^{\ii(kx - \omega t)}\right] = |A|\cos(kx - \omega t + \arg A) ,

(13)

the transverse velocity is $y_t = |A|\,\omega\sin(kx - \omega t + \arg A)$ , so $u_K = \tfrac12\mu|A|^2\omega^2\sin^2(\cdot)$ and, doubling for the equal potential part, $u = \mu|A|^2\omega^2\sin^2(\cdot)$ . The flux is $v$ times that, and the average of $\sin^2$ over a cycle is $\tfrac12$ :

\boxed{\;\langle P\rangle = \tfrac12\,\mu\,v\,\omega^2 |A|^2\;}

(14)

Every factor earns its place. Doubling the amplitude quadruples the power, and so does doubling the frequency — that is prediction 3, and both squares have one cause: the flux is quadratic in the transverse velocity $|A|\omega$ , and the two symbols enter only through that product. A wave of half the amplitude and twice the frequency delivers exactly what it did before, on a string that looks nothing like it did.

Written with the string’s impedance $Z = \sqrt{T\mu} = \mu v$ , the same result is

\langle P\rangle = \tfrac12\,Z\,\left(|A|\omega\right)^2 ,

(15)

a medium constant times the square of a velocity amplitude. Module 10 makes $Z$ the whole subject; from there it runs to the Fresnel equations and to the irradiance of a light beam, which is this formula with fields in place of displacements.

Three velocities, told apart¶

What moves	How fast	Chosen by
a piece of the medium, sideways	$\partial y/\partial t = -v\,y_x$	whoever made the wave
the pattern	$v = \sqrt{T/\mu}$	the medium
the energy	$P/u = v$	the medium

Two of the three are the same number and the third is unrelated. The medium’s speed is $|y_t| = v|y_x|$ : the wave speed times the slope, and since the slope is small, the string always crawls compared with its wave. In the laboratory’s pulse the string’s fastest is 1.145 m/s while the pattern runs at 20 m/s.

But nothing forbids $|A|\omega$ from exceeding $v$ — they are independent knobs, and that is prediction 2. What stops it is the model, not the physics: $|A|\omega > v$ is the same statement as $|A|k > 1$ , a slope of order one, which is exactly where the small-slope derivation of everything above gives out. A string whose pieces keep up with its own wave is a string this module cannot describe.

Verify computationally¶

Each check below is also a test in the project’s suite, so these claims cannot silently rot.

1. The densities are equal, pointwise. A right-moving Gaussian pulse on a 1600-cell grid: the largest gap between $u_K$ and $u_P$ anywhere is $1.1\times10^{-4}$ of the peak density, and the integrated totals agree to $1.6\times10^{-4}$ . Both fall fourfold for every halving of the spacing, so the disagreement is the grid’s.

2. The energy rides at $v$ . On the same runs, $P/u$ is within $1.1\times10^{-6}$ of $v$ across the pulse.

The energy velocity is better than second order

Refining the grid closes $u_K = u_P$ fourfold per halving — ordinary second-order behaviour — but closes $P/u = v$ sixteenfold: $1.7\times10^{-5}$ , $1.1\times10^{-6}$ , $6.9\times10^{-8}$ , $4.4\times10^{-9}$ at 400, 800, 1600 and 3200 cells.

The reason is worth having. Write the departure from a pure right-mover as $\varepsilon = (y_t + v\,y_x)/(v\,y_x)$ ; the grid is slightly dispersive, so a pulse made of many wavelengths does not stay perfectly one-directional, and $\varepsilon$ is small but not zero. The two densities differ at first order in $\varepsilon$ . In the ratio $P/(vu)$ the first-order terms cancel between numerator and denominator, leaving $\varepsilon^2$ . The claim that a wave’s energy travels at exactly $v$ is the most numerically robust statement in the module.

3. The local law holds where nothing imposed it. simulate_string propagates displacements; it has never heard of $u$ or $P$ . Reading $\partial u/\partial t + \partial P/\partial x$ off its output with centred differences gives a residual of $1.1\times10^{-3}$ of $|\partial P/\partial x|$ on the laboratory’s grid, falling as the spacing refines.

The scheme obeys a discrete conservation law exactly

There is a stronger statement available, and it is about the scheme rather than the string. Assign the kinetic energy to the grid points, where the masses are, and the potential energy and the flux to the cells between them, where the springs are — the slope of a cell times the mean velocity of its two ends. Then differentiating in time and substituting the solver’s own update makes every term cancel against the difference of two neighbouring cell fluxes, identically, for any $\Delta x$ .

So the residual cannot be improved by refining the grid and must be improved by refining the step. Measured on one fixed grid: $3.3\times10^{-3}$ , $8.3\times10^{-4}$ , $2.1\times10^{-4}$ at $S = 0.5$ , 0.25, 0.125 — fourfold per halving, which is the velocity Verlet time error and nothing else. The same law read pointwise instead sits at $1.0\times10^{-2}$ over those three runs and does not move, because its error belongs to the spacing.

This is module 08’s closing loop again, one level up. There the stencil was the mass chain; here the chain’s exact energy bookkeeping is what the grid inherits.

4. The wattmeter balances the books. A pulse is launched at 1 m and metered at 2 m. The integral of $P\,\mathrm{d}t$ at the gate comes to 4.177317 mJ; weighing the string beyond the gate at the end of the run gives 4.177487 mJ, a ratio of 0.999959, and that is also exactly the energy the pulse set out with. Three formulas that share nothing agree to four parts in 10⁵ — and the laboratory repeats the measurement on a different pulse, to the same figure.

5. Mean power, and its two squares. A windowed sinusoidal train passes a gate and the flux there is averaged over whole cycles. Against $\tfrac12\mu v\omega^2A^2$ the measurement lands within 0.2% across a factor of four in amplitude and a factor of four in frequency; the fitted exponent in amplitude is 2.000000 and in frequency 1.995.

The frequency exponent’s shortfall is instructive and is not the string’s. The slope at the gate is recovered by a centred difference, which reads the slope of a sine low by $(k\Delta x)^2/6$ — 0.6% for the shortest wave in the sweep, where a wavelength spans 32 grid points. The amplitude sweep has no such term and comes out exact. As in module 08’s photogates, what you measure depends on how you measure it.

6. The standing wave delivers nothing. Four periods of the fourth mode of a fixed string: the largest time-averaged flux anywhere is $2.6\times10^{-7}$ of the largest instantaneous flux, while the instantaneous flux itself peaks at 12.6 mW. The cycle-averaged kinetic and potential totals agree to $1.6\times10^{-4}$ , and the pointwise difference between the densities reaches the entire density — the contrast with the travelling wave, in one run.

7. And what a real string can be asked. Two of this module’s three velocities are within reach of a phone camera. Film a pulse on a stretched slinky, time its arrival at two tape marks as module 08’s problem 5 does, and then track a single taped coil frame by frame: two velocities out of one video, differing by more than a factor of ten. The flux is another matter. Measuring $P$ at a point means knowing the slope and the transverse velocity there at the same instant and multiplying them — two small, separately noisy numbers, with the bias the laboratory’s last part measures waiting underneath. No affordable apparatus reads it directly, so a home experiment measures the flux’s consequences instead: energy arriving where it was not before. The problem set sets that up.

Transfer the idea¶

One flux, many fields

The pattern established here — a density, a flux, and a continuity equation tying them — is how every conserved quantity in physics is accounted for locally. The wave equation just happens to be the simplest place to meet it.

Electromagnetic energy. Module 15 derives $\partial u/\partial t + \nabla\cdot\mathbf{S} = 0$ with $u = \tfrac12(\varepsilon_0 E^2 + B^2/\mu_0)$ and the Poynting vector $\mathbf{S} = \mathbf{E}\times\mathbf{H}$ . The equal split of $u_K$ and $u_P$ becomes the equal split of the electric and magnetic energy densities in a light wave — the same theorem, the same proof.
Intensity. From module 23 onward, “intensity” means the time-averaged flux of a wave, and every interference and diffraction pattern in this course is a map of $\langle P\rangle$ . That it goes as the square of an amplitude is decided here.
Sound. Acoustic intensity is $\langle p\,u\rangle$ — pressure times particle velocity, the same product of a force-like and a velocity-like quantity.
Quantum mechanics. $\partial|\psi|^2/\partial t + \partial j/\partial x = 0$ defines the probability current. Nothing is conserved differently; only the density changes its name.

Standing waves. 11-standing-waves puts walls at both ends and asks where the energy goes when the flux averages to zero. The answer — that it is stored, sloshing between kinetic and potential a quarter wavelength at a time, rather than transported — starts from this module’s last verification.
Impedance. Module 10 sends a wave at a junction between two media and divides its energy in two. The reflection and transmission coefficients are fixed by requiring exactly what this module built: that the flux arriving equals the flux leaving.
Dispersion. The energy velocity $P/u$ came out equal to $v = \omega/k$ here because the ideal string is non-dispersive. 13-dispersion finds that when $\omega/k$ varies with wavelength, the energy travels at the group velocity $\mathrm{d}\omega/\mathrm{d}k$ instead, and the two can differ by a lot — the reason a phase velocity above $c$ breaks nothing.
Cables and antennas. Power sent along a transmission line is $\tfrac12 Z |I|^2$ , the same algebra with current for transverse velocity. An unterminated cable reflects it, for the reason module 10 gives.

Check your understanding¶

A sinusoidal wave travels along a string. A photograph freezes it. Where along the string, in that photograph, is the energy per unit length largest?
A. At the zero crossings, where the string is steepest and moving fastest. B. At the crests and troughs, where the displacement is largest. C. Uniformly along the string — a travelling wave carries its energy evenly. D. A quarter wavelength ahead of each crest, in the direction of travel.
Check your answer
Correct answer: A — At the zero crossings, where the string is steepest and moving fastest.
- A. Right. A string stores energy two ways and neither of them is displacement: by being stretched, which needs slope, u_P = (1/2) T (dy/dx)^2, and by moving, which needs transverse velocity, u_K = (1/2) mu (dy/dt)^2. At a zero crossing of a travelling sine both are at their largest, and at a crest both vanish at once — the string there is momentarily flat and momentarily at rest. Measured on the simulated string, the crest holds 4e-5 of what the zero crossing holds. Notice the consequence: the energy humps come twice as often as the crests, because a square has half the period of the sine.
- B. This is the module’s misconception, and it comes from treating displacement as if it were stored energy. It is not. Lift a straight stretch of string bodily to one side and hold it there: you have displaced every point of it and stored nothing, because you stretched nothing and nothing is moving. Energy needs slope or speed, and a crest has neither at the instant the photograph was taken.
- C. The energy density of a travelling sine goes as sin^2(kx - omega t), which is far from uniform: it falls to zero twice per wavelength. What is uniform is the average over a whole wavelength, and that is the quantity (1/2) mu v omega^2 A^2 measures when it is multiplied by v. Uniform on average is not uniform in the photograph.
- D. The energy maxima do sit a quarter wavelength from the crests, but there are two of them per wavelength, not one, and neither has a preferred side: u goes as sin^2(kx - omega t), whose maxima lie a quarter wavelength either side of every crest and every trough alike. Nothing in the density distinguishes forwards from backwards — the direction of travel shows up in the sign of the flux, not in where the energy is.
On a string with v = 20 m/s, a sinusoidal wave has amplitude 2.0 mm and frequency 50 Hz. Which statement about its three velocities is correct?
A. The pattern and the energy both move at 20 m/s; the string’s pieces move sideways at up to 0.63 m/s. B. All three are 20 m/s — it is one wave, so it has one speed. C. The energy moves at half the pattern speed, since only half of it is kinetic. D. The transverse speed is 20 m/s and the energy is what moves at 0.63 m/s.
Check your answer
Correct answer: A — The pattern and the energy both move at 20 m/s; the string’s pieces move sideways at up to 0.63 m/s.
- A. Right, and the two coincidences are not accidents. The pattern moves at v = sqrt(T/mu) because the medium says so. The energy moves at P/u, and for a travelling wave P = v(u_K + u_P) exactly, so that ratio is v too. The transverse speed is A omega = 0.002 x 2 pi x 50 = 0.63 m/s, set entirely by whoever made the wave, and it is sideways rather than along. Two numbers are forced and one is a free knob.
- B. Two of them are, and the third is 0.63 m/s. The transverse velocity dy/dt is what a single piece of string does, and it is related to the wave speed by y_t = -v y_x: the wave speed times the slope. Since the slope is small — 0.031 here — the string crawls compared with its own wave. Confusing the two is the same error as thinking a stadium wave requires anyone to run.
- C. The two densities are each half the total, which is true and is this module’s surprise — but the flux carries the whole density, not half of it: P = T v y_x^2 = v(u_K + u_P). Both halves travel, and they travel together at v. The factor of one half that does appear in this module is in the time average of a sinusoidal wave’s power, and it comes from averaging sin^2, not from splitting the energy in two.
- D. This has the two exactly backwards, which is worth noticing because it is what the misconception “the wave carries the medium along” would predict. Module 08’s dust speck settles it: the string’s own motion is small, sideways, and returns to zero. The energy’s motion is the full wave speed and goes somewhere.
A travelling wave and a standing wave have the same amplitude and frequency on identical strings. Over one full period, which delivers more energy past a point halfway along?
A. The travelling wave. The standing wave delivers exactly zero, though its instantaneous flux is not small. B. The standing wave, because its energy is concentrated at the antinodes rather than spread out. C. They deliver the same, since they have the same energy on the string. D. Neither: a wave transports no energy, only a disturbance.
Check your answer
Correct answer: A — The travelling wave. The standing wave delivers exactly zero, though its instantaneous flux is not small.
- A. Right, and the second clause is the interesting half. For y = A sin(kx) cos(omega t) the flux is proportional to sin(2kx) sin(2 omega t), which is as large as a quarter of the travelling wave’s peak at the same amplitude — but it changes sign twice a cycle and averages to zero at every x. Simulated over four periods, the largest time-averaged flux anywhere is 2.6e-7 of the largest instantaneous one. A standing wave moves comparable energy about and puts all of it back.
- B. Its energy is indeed unevenly distributed, but distribution is not transport. Ask what crosses the halfway point: on a standing wave the flux there swings positive and negative in equal measure, so the net over a cycle is zero no matter how the energy is arranged. Concentration and transport are separate questions, and this module’s flux answers only the second.
- C. They do have comparable energy on the string — that is what makes the contrast sharp. Energy stored and energy transported are different quantities, and a standing wave is a device for having the first without the second. Module 11 builds its entire budget on that distinction.
- D. A travelling wave transports energy at exactly v, and that is this module’s central result: P = v(u_K + u_P). The thing a wave does not transport is the medium — module 08’s dust speck goes nowhere. Energy and medium are different things, and separating them is the point of both modules.
A rope with mu = 0.080 kg/m is held at a tension of 120 N. A shaker drives one end at 8.0 Hz with an amplitude of 25 mm. What average power, in watts, does the shaker supply?
Check your answer
Correct answer: 6.1 (± 0.2)
The wave speed is v = sqrt(120 / 0.080) = sqrt(1500) = 38.7 m/s, and the angular frequency is omega = 2 pi x 8.0 = 50.3 rad/s. Then = (1/2) mu v omega^2 A^2 = 0.5 x 0.080 x 38.7 x 50.3^2 x 0.025^2 = 6.1 W. Two checks worth doing as you go. The dimensions: (kg/m)(m/s)(1/s^2)(m^2) = kg m^2/s^3 = W. And the scalings: halve the amplitude and the answer falls to 1.5 W, quarter of what it was, because the flux is quadratic in the transverse velocity A omega and both symbols enter only through that product.
At one point on a string with tension T = 20 N, a snapshot shows the slope dy/dx = +0.015 and the transverse velocity dy/dt = -0.40 m/s. What is the energy flux P at that point, in watts?
Check your answer
Correct answer: 0.12 (± 0.005)
P = -T (dy/dx)(dy/dt) = -20 x 0.015 x (-0.40) = +0.12 W. The sign is the direction: the energy is flowing toward +x. Read the geometry rather than trusting the arithmetic alone — the string here slopes up to the right and is moving down, so the piece on the left is pulling the piece on the right downward while the right piece moves down, and force through displacement is positive work done on the right. Note also what the snapshot alone cannot tell you: whether this is a right-moving wave. For a right-mover y_t = -v y_x would give v = 0.40/0.015 = 27 m/s, which is only consistent if the string’s sqrt(T/mu) is 27 m/s. Otherwise both directions are present here.
A right-moving pulse on a string is measured at one instant. Its kinetic energy density and its potential energy density are found to be equal at every point, to the accuracy of the measurement. What does this tell you?
A. Nothing beyond what a travelling wave must do — it follows from y_t = -v y_x and mu v^2 = T. B. The pulse happens to be at the moment in its cycle when the two densities cross. C. The string must be non-dispersive, or the equality would fail. D. The measurement must be wrong, since the total energy would then be double-counted.
Check your answer
Correct answer: A — Nothing beyond what a travelling wave must do — it follows from y_t = -v y_x and mu v^2 = T.
- A. Right. For a right-mover the chain rule gives y_t = -v y_x, so u_K = (1/2) mu v^2 y_x^2 = (1/2) T y_x^2 = u_P, the last step because mu v^2 = T is what v means. It holds at every point and every instant, with nothing averaged. The statement is not empty, though: it fails for a standing wave, where the string is all potential at one moment and all kinetic a quarter period later, and it fails for any superposition of two directions. Equality at every point is a signature of one-way travel.
- B. That is the right picture for an oscillator, and the wrong one for a travelling wave. A mass on a spring trades kinetic for potential twice a cycle and they are equal only at instants. A travelling wave has every element a little behind its neighbour, so what the mass does in sequence the string does all at once, laid out along its length: the equality holds at every instant, not at special ones.
- C. Dispersion is not what is being tested here. The equality follows from a single travelling shape obeying y_t = -v y_x, which is true of f(x - vt) whatever the medium does to other wavelengths. What dispersion breaks is the shape — a pulse spreads, so it stops being a rigid f(x - vt) — and only then does the equality go with it. Module 13 takes that up.
- D. Nothing is double-counted: u_K and u_P are different energies in the same place, one of motion and one of stretch, and the total is their sum. If each is (1/2) T y_x^2 then the total is T y_x^2, and the flux P = T v y_x^2 is exactly v times that — which is the module’s statement that the energy travels at the wave speed. The bookkeeping closes.
Explain the local conservation law du/dt + dP/dx = 0 in words: what each term means, what the equation forbids, and why a law of this shape says more than “energy is conserved”. Then say what the corresponding statement is for a stretch of string between two fixed ends.
Check your answer
The first term is the rate at which the energy per unit length at a point is changing. The second is the rate at which the flux — the power crossing a point, P = -T y_x y_t — changes as you move along the string. The equation says these two always cancel: energy at a point can only change because more is arriving from one side than is leaving on the other.

What it forbids is energy disappearing here and reappearing there. A global statement, “the total energy of the string is constant”, would permit that: it constrains only a sum. The local statement constrains every point, and so requires the energy to have been somewhere in between, continuously, at every moment of the crossing. That is exactly the question the module opens with — a storm’s energy crossing an ocean whose water never moves — and only a local law answers it. The integral form makes the difference concrete. Integrating over a stretch [a, b] gives d/dt of the energy in the stretch = P(a) - P(b): what comes in at the left minus what goes out at the right. Take the stretch to be a whole string with two fixed ends. A fixed end cannot move, so y_t = 0 there, so P = 0 at both: nothing crosses either end, and the total energy is constant. Conservation for the whole string is a consequence of the local law plus a fact about the boundaries, rather than a separate assumption — which is why the same machinery will handle a string with an open end, a junction, or a driver, just by changing what P is at the edges. Worth adding: this shape — a density, a flux, and a continuity equation — is how every conserved quantity is accounted for locally. The Poynting vector in module 15 and the probability current in quantum mechanics are the same sentence with a different density.

Exam-style problems for this module — the ones worth doing with a pen — are collected in the problem set.

Explain it in your own words¶

Answer in a few sentences each. No number will save you.

Explain why the crest of a wave is the least energetic part of the string, to someone who is certain the energy is where the displacement is. Use only the two things a string can do to store energy.
Energy flows through a medium that goes nowhere. Describe how, using the word “neighbour” and without using the word “carry”.
A wattmeter on a string reads -0.4 W. Say what is happening, and what would have to change for it to read +0.4 W with the string looking exactly the same in a photograph.
A travelling wave and a standing wave of the same amplitude have the same energy on the string and the same instantaneous flux, to within a factor of four. One delivers power and the other delivers none. Explain the difference without writing down a formula.

Advanced: the momentum question, and what “energy velocity” will stop meaning¶

Energy velocity, and its expiry date. Defining the energy transport velocity as $P/u$ is more than a convenience: it is the definition that survives when $\omega/k$ stops being the answer. On the ideal string all three of $\omega/k$ , $\mathrm{d}\omega/\mathrm{d}k$ and $P/u$ coincide, so the module can afford to be casual. In a dispersive medium they separate, $P/u$ follows the group velocity $\mathrm{d}\omega/\mathrm{d}k$ , and the phase velocity $\omega/k$ — which can exceed $c$ without embarrassment — stops describing the motion of anything at all. 13-dispersion does this properly; the reason it can is that the quantity to track was defined here as a ratio of two things a measurement can reach.

Does a transverse wave carry momentum along the string?

A question the ideal string cannot answer

The energy bookkeeping above is settled physics. The momentum bookkeeping is not, and it is worth knowing which is which.

Transverse momentum is easy: the string’s pieces move sideways, $\int \mu\,y_t\,\mathrm{d}x$ is their total transverse momentum, and for a pulse that starts and ends at rest it is zero. The hard question is longitudinal momentum — does a pulse push the string along its own length? The answer is second order in the slope, $O(y_x^2)$ , which is precisely the order at which the ideal string model was built by throwing things away. Whether an element is carried forward or back, and by how much, depends on constitutive details the model discarded: whether the string is inextensible or Hookean, whether the tension is held constant or the length is, what the ends are attached to. Papers in the American Journal of Physics have gone back and forth on it for decades — a representative entry is D. R. Rowland, “The potential energy density in transverse string waves depends critically on longitudinal motion,” Eur. J. Phys. 32, 1475 (2011), which shows that even the potential-energy density is not unambiguous once longitudinal motion is taken seriously.

What survives that argument is everything this module actually used. The total energy, the flux, the continuity equation and the mean power are all first-order-in- $y_x^2$ quantities that every version of the model agrees on. The course stays at the flux level on purpose, and says so rather than quietly choosing a side.

The corresponding question for light — whether a photon in a medium carries momentum $n\hbar\omega/c$ or $\hbar\omega/(nc)$ — is the Abraham–Minkowski controversy, a century old and resolved only in the sense that the two answers turn out to account for different halves of the same total. 16-light-in-matter mentions it in passing, for the same reason it appears here: knowing where a model’s authority ends is part of knowing the model.

Why the potential density has that form, exactly. The derivation above expanded $\sqrt{1 + y_x^2}$ and kept one term. A fair objection is that the discarded $-\tfrac18 y_x^4$ is of the same order as corrections to the equation of motion that were also discarded, so what licenses keeping $\tfrac12 T y_x^2$ ? The answer is that both are the leading term of their own quantity: the equation of motion’s leading term is $O(y_x)$ and the energy’s is $O(y_x^2)$ , so each is the first non-vanishing contribution, and the ratio of neglected to kept is $y_x^2$ in both. The two approximations are consistent, which is why the continuity equation closes exactly rather than to some order — as the discrete version above shows, it closes exactly even on a grid.