The drum that has no harmonics

A guitar string and a drumhead are both continuous objects held at a boundary, and both answer a strike with a discrete set of vibrations. The string’s set is 1, 2, 3, 4 times its lowest frequency, and that whole-number series is what a pitch is made of. The drumhead’s begins 1.000, 1.593, 2.136, 2.295 — and no arrangement of tension, size or material will ever make those numbers whole.

Everything below is where those six numbers come from, what they are made of that no drum contributes, and what has to be done to a real drum to get a note out of it anyway.

A string’s arithmetic, and why it does not survive a second dimension #

On a string the counting argument fits in a line. Both ends are clamped, so both must be nodes; consecutive nodes are half a wavelength apart; the length therefore holds a whole number of half-wavelengths, and the medium’s speed turns that into a frequency:

$$L = \frac{n\lambda}{2} \quad \Longrightarrow \quad f_n = \frac{nv}{2L}, \qquad n = 1, 2, 3, \dots$$

A string’s arithmetic works because its modes are pieces of one function. The first four patterns a clamped string holds have frequencies 1, 2, 3 and 4 times the lowest, and each integer is a count of half-wavelengths fitting between the ends. The zeros of a sine are evenly spaced and its humps are all the same height, so fitting a whole number of them into a fixed length gives a whole number of frequencies — and a harmonic series is that evenness, nothing more.

The integer arrives because the boundary condition on a string reads $\sin(kL) = 0$, and the zeros of a sine sit at exact multiples of $\pi$. That is the load-bearing fact, and it is a fact about the sine function rather than about strings.

Now stretch a membrane over a circular hoop. The transverse displacement obeys the wave equation in two dimensions,

$$\nabla^2 z = \frac{1}{c^2}\frac{\partial^2 z}{\partial t^2}, \qquad c = \sqrt{\frac{T}{\sigma}},$$

with $T$ the tension per unit length and $\sigma$ the mass per unit area. Separating it in polar coordinates gives an angular factor and a radial factor. The angular factor has to come back to itself after a full turn, which forces it to be $\cos(m\theta)$ with $m$ a whole number — so a whole number does appear, and it is not the one that decides the frequencies. The radial factor obeys Bessel’s equation, whose solution finite at the centre is $J_m(kr)$. The rim cannot move, so the condition is

$$J_m(kR) = 0 \quad \Longrightarrow \quad f_{mn} = \frac{j_{mn}}{2\pi R}\sqrt{\frac{T}{\sigma}},$$

where $j_{mn}$ is the $n$th positive zero of $J_m$. The structure is identical to the string’s. The difference is entirely that a Bessel function’s zeros are not evenly spaced and are not multiples of anything. The whole numbers $m$ and $n$ survive as labels; they have stopped being the frequencies.

That is precisely what a circle does not have. The functions that fit a disc are Bessel functions rather than sines, their zeros are not evenly spaced, and there is no reason for the ratios between them to be whole numbers — so the drum’s overtones are not harmonics, and the instrument has no pitch in the sense a string has one.

The modes, and the two kinds of line that do not move #

A mode is labelled by two integers because a disc has two directions to be still in. The label $(m, n)$ means $m$ nodal diameters and $n - 1$ nodal circles: $m$ counts the lines through the centre where $\cos(m\theta)$ vanishes, and $n$ counts the humps of the radial function, so $n - 1$ counts its interior zeros.

Those two pictures hold the failure in a form that can be seen rather than derived. $J_0$’s zeros at 2.4048, 5.5201 and 8.6537 are separated by 3.115 and then 3.134 — creeping towards the sine’s even spacing of $\pi$ and never landing on a multiple of the first, which is the one thing that would have produced a harmonic series.

Three modes, three patterns, and the two intervals between them have no common measure.

Six ratios, and not one of them a whole number #

The generator behind these figures computes both halves of every quotient from the power series rather than reading them off a table — because the claim of the essay is those numbers, and a figure handed them would be a drawing of a table.

Two features of that figure carry the argument. The first is that the interleaving is thorough — the partials fall between the harmonics at every height, and the near misses are not small. As intervals, (1,2)'s 2.76% shortfall is 48 cents, very nearly half a semitone; (2,1)'s 6.78% excess over the octave is 114 cents, more than a semitone. A quarter of a semitone is about 1%, so these are not roundable errors. They are intervals with no name.

The second is that the ordering of the modes is a result rather than a list. Mode (0,2) arrives at 2.295, between (2,1) at 2.136 and (3,1) at 2.653, and no ordering by $m$ or by $n$ would have put it there. Which modes are the lowest eight has to be found by computing zeros and sorting them.

What follows is not that a drum sounds bad but that it has no pitch, in the specific sense that a pitch is what the ear extracts when a set of partials shares a common divisor. Presented with 1.000, 1.593, 2.136, 2.295 there is no fundamental to be found, because there is none: the lowest partial divides the others in no approximate sense either. Sustained, that is a wash. Brief, it is a thud.

The ratios belong to the boundary, not to the drum #

Here is the part that makes the six numbers more than a fact about drums. Compare what is in $f_{mn} = j_{mn}\sqrt{T/\sigma}\,/2\pi R$ with what is in the ratio $f_{mn}/f_{01} = j_{mn}/j_{01}$.

The ratio contains no tension. No density, no thickness, no material, no radius. Every physical property of the membrane sits in the factor $\sqrt{T/\sigma}/2\pi R$, which is common to all the modes and therefore absent from every quotient of two of them. The ratios are pure numbers fixed by one thing: that the boundary is a circle.

The factor that cancels is easiest to see in one dimension. The same 50 Hz on two strings at the same tension travels at different speeds and arrives at different wavelengths — but the ratios between a string’s own modes are unchanged, because the speed multiplies every frequency equally. That is why the ratios belong to the boundary and not to the material, and why a drum retuned to a different pitch is still inharmonic in the same way.

So a concert timpano, a tabla, a tambourine and a dinner plate covered in cling film all have the same partial ratios, to whatever accuracy their rims are circular and their heads uniform. A tuning gauge changes $T$ and slides the whole spectrum: going from 1,900 to 2,100 newtons per metre raises every frequency by 5.13% and alters no ratio in the eleventh decimal place. No adjustment on any drum reaches the shape of its spectrum, because that shape is the shape of the hoop.

That is strong enough to invite its converse, which is one of the better questions in twentieth-century mathematics. If the boundary decides the frequencies, do the frequencies decide the boundary? Mark Kac put it in 1966 as can one hear the shape of a drum? — and the answer, found in 1992, is no. Two differently shaped membranes can have every one of their infinitely many frequencies in common.

The boundary decides the ratios, and the ratios do not decide the boundary. Both halves are true and the second was not obvious to anybody for twenty-six years.

What a missing harmonic series costs, and how a timpano is tuned anyway #

An instrument that has to get a definite note out of a circular membrane is working against the arithmetic above, and what it does about it is specific and countable.

The fundamental has to be got rid of. A timpanist strikes about a quarter of the way in from the rim, not at the centre. At $r = 0.75R$ the (0,1) mode’s radial function has fallen to 0.339 of its central value while the (1,1) mode’s is at about two-thirds of its own maximum, so the strike weights the head’s motion away from the mode that stands in no divisor relationship to anything. A centre strike, which excites (0,1) hardest, is the way to make a timpano sound like a drum rather than like a note.

And what survives of it does not survive long. The (0,1) mode moves the whole head the same way at once, making it an efficient monopole source; it therefore loses its energy to the air far faster than the diametral modes, whose opposite-moving regions largely cancel each other’s radiation at a distance. Its decay is a fraction of a second where the (1,1) mode rings for several, so the sustained sound of a kettledrum is (1,1), (2,1), (3,1) and (4,1), and the fundamental is gone.

The air in the kettle moves the rest into line. The head drags air, which adds effective mass and lowers frequencies. The loading is largest for the modes that shift the most net volume of air, which are the lowest, so it does not scale the spectrum, it reshapes it. Relative to (1,1), the ideal membrane gives 1.000, 1.340, 1.665 and 1.980 for the first four diametral modes. On a real kettledrum they come out near 1 : 1.5 : 2 : 2.5.

Which is a harmonic series with its fundamental removed. Those four numbers are 2 : 3 : 4 : 5 of a frequency an octave below the (1,1) mode — a frequency the drum does not produce at all. The ear is handed the second, third, fourth and fifth harmonics of a note that is not there, finds the divisor those four do share, and hears a pitch below the lowest thing on the head. A timpano’s note is manufactured by suppressing a mode and loading the survivors until they nearly agree.

None of it is available to a flat-mounted head. A frame drum has no air to reshape anything and keeps the ideal ratios, which is why it has a thud where a timpano has a note — and why a kettle is a heavy stiff shell rather than a resonator: any resonance of its own would add frequencies nobody asked for.

Where the model stops #

The ideal membrane is an idealisation in five specific ways, and each is large enough to be audible.

No bending stiffness. A membrane resists being stretched and not being bent. A real head is a thin plate, its flexural rigidity adds a restoring term that grows with curvature, and the more curved higher modes are pulled up in frequency. On a 0.19 mm mylar head at playing tension the shift is a fraction of a per cent on the low modes and of order a per cent by the tenth — which puts the high partials sharp of the ideal ratios rather than flat.

No air. The kettle’s air load is the largest correction in the list: a few per cent for the higher modes and considerably more for the lowest. It is not a perturbation to be mentioned and dropped; it is the reason the previous section works. The six ideal ratios in the hero figure are therefore not the ratios a timpano sounds, and stating them as though they were is the commonest error in accounts of this subject.

Small amplitude. The derivation assumes the head’s slope is small everywhere, so that the tension is unchanged by the displacement. A hard strike on a 0.65 m head moves it several millimetres, the extra stretch raises the tension, and the pitch is momentarily sharp — the same linear approximation running out that gives a pendulum an amplitude-dependent period. The fractional tension change goes as the square of the slope, so a slope of 3% buys about 0.1%.

No damping. Every mode above was computed for a membrane vibrating for ever, and a drum is built to radiate. A mode with decay time $\tau$ has a frequency width of order $1/\pi\tau$, so the low modes of a kettledrum — tenths of a second at around 100 Hz, quality factors of order tens — have fractional widths of a few per cent. That is comparable to the 2.76% by which the nearest partial misses a harmonic. The partials are not merely inharmonic; they are broad enough that asking whether one coincides with a harmonic barely has an answer, which is what any short-lived oscillation pays in frequency definition.

A uniform circle. A real head is pulled by six or eight tuning rods and its tension varies round the hoop by a per cent or two, splitting mode pairs that should be degenerate; and the rim is not rigid, since one that reflected perfectly would let no sound out.

The same number in a telescope #

The eigenvalue problem above is the Laplacian on a disc with a vanishing boundary condition, and it does not care what is oscillating. Two other places it turns up are worth the space, and the second is the surprise.

Replace the wave equation with the Schrödinger equation and the problem becomes a particle in a circular box: $-(\hbar^2/2m)\nabla^2\psi = E\psi$, with $\psi = 0$ on the rim. The allowed wavenumbers are identical, the same $j_{mn}/R$, but energy goes as the square of the wavenumber rather than the first power, so the level ratios are the squares of the drum’s: 1, 2.539, 4.561, 5.269, 7.039. That is the same step the one-dimensional box takes when the string’s 1, 2, 3, 4 becomes 1, 4, 9, 16.

Then the number. The first zero of $J_1$ is 3.831706, which is what puts the drum’s second mode at 1.593 times the fundamental — and it also sets the size of a telescope’s smallest possible star image. Light through a circular aperture of diameter $D$ gives an amplitude proportional to $2J_1(v)/v$ with $v = \pi D\sin\theta/\lambda$, so the first dark ring sits where $J_1$ first vanishes:

$$\sin\theta_1 = \frac{j_{11}}{\pi}\frac{\lambda}{D} = 1.2197\,\frac{\lambda}{D}.$$

The 1.22 in every account of the diffraction limit is 3.8317 divided by $\pi$. It is the drum’s number, and it is the same number because it is the same geometry: a condition on a circle, separated in polar coordinates, reduced to Bessel’s equation, and answered by asking where $J_1$ is zero. The membrane uses the circle as the place the displacement must vanish; the aperture uses it as the place the light stops. Both hand the arithmetic to $J_1$, and $J_1$ has one first zero.

The identical boundary problem, on the identical circle, appears as an optical instrument. The pattern a circular aperture makes has its first zero at the same place the drum’s fundamental has its first nodal circle, because both are the first zero of the same Bessel function — the 1.22 in every telescope’s resolution formula and the 2.405 in every drum’s mode table are the same mathematics answering two questions.

Chladni’s sand, and the question that took twenty-six years #

The nodal lines were seen long before they were computed. Ernst Chladni, from 1787, drew a bow along the edge of a metal plate scattered with fine sand, and the sand collected into curves — a grain on a moving part of the surface is thrown about until it arrives somewhere that does not move. Chladni’s plates are not membranes, since a plate’s restoring force is bending rather than tension and its equation is fourth order, but the idea that a two-dimensional resonator has still curves rather than still points is his.

Sand finds a nodal line because a nodal line is the one place that is not moving. The waves keep moving everywhere else and the grains are thrown about until they arrive somewhere still, so the pattern that emerges is a map of the zero set — which is why Chladni’s figures are a direct photograph of a mode rather than an inference about one.

Napoleon saw Chladni demonstrate in 1809 and put up a 3,000-franc prize for a theory of vibrating surfaces; Sophie Germain won it in 1816, at the third attempt, with the plate equation. The membrane had been done earlier and elsewhere: Euler wrote down the circular case in 1764 and found the functions that solve it, some sixty years before Bessel tabulated them for a problem in planetary motion and lent them his name. Rayleigh’s Theory of Sound of 1877 collected the whole of it, including the observation that a drum’s frequencies are not harmonic.

So by 1877 the six ratios were known and so was their independence of the drum. Kac’s 1966 question — whether the boundary can be recovered from them — sat open until Carolyn Gordon, David Webb and Scott Wolpert produced a pair of eight-sided polygons with identical spectra. Their example is no near miss: the two shapes are provably isospectral, which forces equal area and equal perimeter, and that is as much as the frequencies are able to say.

What the picture cannot show #

The nodal diagrams are a catalogue of shapes, and four things they leave out are the four things a real drumhead is mostly doing.

They cannot show amplitude. Every panel is drawn at the same size. In a struck head the mixture is decided by where the strike landed, through the value of each mode’s radial function there, and the mixture is the whole difference between a note and a thud.

They cannot show decay. All six patterns are drawn as though permanent, and their lifetimes differ by more than an order of magnitude — (0,1) radiating away in a fraction of a second while the diametral modes ring on. A drawing that gives the fastest-dying mode top billing is backwards for a real drum.

They cannot show degeneracy or orientation. Every mode with $m \geq 1$ is two modes at the same frequency, $\cos(m\theta)$ and $\sin(m\theta)$, differing only by a rotation; the figure picks an orientation and there is nothing to pick it. A slight asymmetry splits the pair by a fraction of a per cent, and the two split modes then exchange energy back and forth in a slow beat — what two nearly identical coupled oscillators always do, audible as a wobble on a badly seated head.

And they cannot show that the real motion is a sum. No single panel is ever what a head is doing: a strike excites many modes at once, each with its own amplitude, frequency and decay, and the shape at any instant is everything present added together, which has no fixed nodal lines at all.

There is one more absence and it is the largest. Every figure here is drawn on a perfect circle, and every number quoted is the circle’s. The ratios of a square membrane, or a triangle, or a stadium, are different sets from different transcendental conditions. The circle’s are the ones with a name.

The ladder from here #

Later rungs on this anchor: the rectangular membrane, whose spectrum $\sqrt{(p/a)^2 + (q/b)^2}$ has two independent integers and produces accidental degeneracies whenever the sides are commensurate; Weyl’s law, which counts how many modes lie below a frequency and is how the area of a shape is heard; the clamped plate, whose fourth-order equation makes a cymbal a different instrument from a drum; the topology of nodal lines, and Courant’s bound on how many regions the $n$th mode can divide a shape into; and the annulus, where a second boundary renegotiates the whole spectrum.

The neighbouring ladders are close by. Resonance is what happens when a drum is driven at one of these frequencies rather than struck, and the point where rays stop being sufficient is where the same Bessel zero comes back wearing an optical hat.

The claim worth carrying forward is narrow and does a great deal of work. Discreteness comes from a boundary, and the shape of the resulting spectrum comes from the shape of that boundary — so a whole-number series is not what confinement generally produces. It is what confinement between two points produces, and two points are a special case of nothing much.