Part of The Beauty of Physics, Mathematics, and Music (and Maybe Cats).

C, C♯, E and F♯. Then C, C♯, E♭ and G. These are different collections of four notes, yet each contains exactly one instance of every nonzero interval class available in twelve-tone equal temperament: semitone, tone, minor third, major third, fourth and tritone.

The inventory is complete. The inventory also fails to identify the collection.

Music theorists call this a Z-relation. The corresponding mathematical idea, homometry, also appears in crystallography: different arrangements can give the same diffraction intensities. The connection is exact for a specified model. It does not mean that these two chords sound alike, or that an X-ray experiment on a musical instrument would somehow discover its harmony. [[1]]

The calculation needs nothing more than counting pairs and arithmetic that wraps around at twelve. A Fourier transform then shows why the same ambiguity reaches beyond music.

The Octave Turns Distances Into Classes

Number the pitch classes from C = 0 to B = 11. Octave-equivalent notes receive the same number; enharmonic spellings such as C♯ and D♭ are identified. Our collections become

$$ A=\{0,1,4,6\},\qquad B=\{0,1,3,7\}. $$

This representation retains which pitch classes occur. It discards register, doublings, order, rhythm, dynamics and timbre. A low C and a high C are one entry. These are collections that can be voiced as chords, not complete descriptions of a performance.

For two distinct entries \(a\) and \(b\), define their interval class as

$$ \operatorname{ic}(a,b) =\min\bigl((b-a)\bmod12,(a-b)\bmod12\bigr). $$

Both remainders lie between zero and eleven. Taking the smaller identifies an interval with its inversion: seven semitones becomes class five. A perfect fifth and a perfect fourth therefore contribute to the same category.

Four distinct notes have \(\binom42=6\) unordered pairs. In \(A\), the ordinary positive differences between the listed numbers are

$$ 1,4,6,3,5,2. $$

In \(B\), they are

$$ 1,3,7,2,6,4. $$

Replacing seven with its complementary distance five gives the same inventory. In the conventional order of interval classes one through six, both interval vectors are

$$ (1,1,1,1,1,1). $$

The equality depends on octave identification. As finite sets of points on an ordinary line, the collections have different distance inventories: one contains a distance five, the other a distance seven.

Moving the Whole Collection Does Not Explain It

Transposition adds the same number to every pitch class, modulo twelve. Inversion replaces every \(x\) with \(c-x\), also modulo twelve, for some fixed \(c\). Both operations preserve interval classes. The group-theory treatment of Messiaen’s modes uses these same operations.

Our pair is more interesting because neither operation turns one collection into the other. The circular gaps provide a short proof.

Read each set in ascending order, and include the gap returning from its last entry to the first:

$$ A:\ (1,3,2,6),\qquad B:\ (1,2,4,5). $$

A transposition rotates this gap list. An inversion reverses its circular order, possibly followed by a rotation. Neither can change which gap sizes occur. One collection has gaps three and six; the other has four and five. Their gap multisets differ, so they cannot be related by transposition or inversion.

This is the nontrivial ambiguity meant by the Z-relation: equal interval content between collections outside the same transposition-and-inversion family. Jedrzejewski and Johnson discuss precisely this pair in their treatment of Z-related sets. [[1]]

Counting Directed Pairs Makes the Physics Visible

The six-entry interval vector combines directions. Fourier analysis works more naturally with all twelve directed differences.

Let \(x_A(n)\) be one when pitch class \(n\) belongs to \(A\), and zero otherwise. Define the cyclic autocorrelation

$$ C_A(d)=\sum_{n=0}^{11}x_A(n)x_A(n+d), $$

where indices wrap modulo twelve. It counts ordered pairs whose second entry is \(d\) steps beyond the first. The product contributes one only when both positions are occupied.

Our calculation gives, in the order \(d=0,1,\ldots,11\),

$$ C_A=C_B=(4,1,1,1,1,1,2,1,1,1,1,1). $$

The zero entry counts four self-pairs. Each unordered pair at interval class one through five contributes once in each direction, at \(d\) and \(12-d\). The tritone is special: both directions have difference six, so its single unordered pair contributes twice to that entry.

Thus the familiar interval vector and this autocorrelation carry the same pair information for fixed-size sets, but they count it differently. Dividing the tritone entry by two is essential. [[1]]

Notice what the count has removed: which pairs share a note. An inventory can record every distance without recording how the distances fit together.

The Fourier Magnitudes Agree Because the Pairs Agree

Define the discrete Fourier transform of the indicator by

$$ X_A(k)=\sum_{n=0}^{11}x_A(n)e^{-2\pi i kn/12}, \qquad k=0,\ldots,11, $$

where \(i^2=-1\). Each occupied position contributes a complex phase; the sum records both magnitude and phase.

Transforming the autocorrelation and substituting \(m=n+d\) gives

$$ \begin{aligned} \sum_{d=0}^{11}C_A(d)e^{-2\pi i kd/12} &=\sum_{n,m=0}^{11}x_A(n)x_A(m) e^{-2\pi i k(m-n)/12}\\ &=X_A(k)\overline{X_A(k)}\\ &=|X_A(k)|^2. \end{aligned} $$

The bar denotes complex conjugation. Since our autocorrelations agree, the squared Fourier magnitudes agree at every \(k\). Conversely, the discrete Fourier transform is invertible, so equal squared magnitudes imply equal cyclic autocorrelations.

For this pair the numbers are especially simple. The autocorrelation is a constant row of ones, with three extra units at zero and one at six. The constant row transforms to twelve at \(k=0\) and zero elsewhere. Therefore

$$ |X_A(k)|^2=|X_B(k)|^2= \begin{cases} 16,&k=0,\\ 3+(-1)^k,&k\ne0. \end{cases} $$

The nonzero even indices give four; the odd indices give two. At zero, four occupied positions add coherently, giving \(4^2=16\).

The phases need not agree. If the full complex transforms agreed everywhere, their inverse transforms would give identical sets. Equal magnitudes leave precisely the missing information that allows our two different collections to survive the comparison.

A Diffraction Experiment Needs a Different Physical Object

To make the crystallographic connection literal, use a one-dimensional periodic arrangement of identical scatterers. Put scatterers at positions \(a/12\) within a unit cell of length one, for each \(a\) in the chosen set, and repeat the cell along a line.

At an integer reciprocal-lattice index \(k\), its structure factor is

$$ F_A(k)=f(k)\sum_{a\in A}e^{-2\pi i ka/12}, $$

where \(f(k)\) is the common scattering amplitude of each constituent. In the ideal kinematic model, the corresponding diffraction intensity is proportional to \(|F_A(k)|^2\). The two periodic arrangements therefore have identical intensities at all those indices, despite their different contents within the cell.

Identical scattering weights, periodic repetition and the loss of phase are the assumptions doing the work. Unequal weights or additional structural information can change the inverse problem. Nor does this establish equality at every wavevector for two isolated finite clusters.

Patterson’s 1939 paper already used the term homometric structures for this kind of structural ambiguity. The historical connection predates the often-cited 1944 work on ambiguities in X-ray analysis. [[2]]

A musical sound has a different spectrum. Equal-tempered frequencies follow \(f_n=f_0 2^{n/12}\); they are not equally spaced physical scatterer positions. The Fourier transform above acts on a twelve-entry pitch-class indicator, rather than on acoustic pressure measured over time. Equal interval fingerprints give no equality of actual sound spectra.

Three Notes Can Recover What Pairs Forget

The triple \(\{0,1,4\}\) inside \(A\) has interval classes one, three and four. None of the four triples inside \(B\) has that combination: their inventories are \((1,2,3)\), \((1,5,6)\), \((3,4,5)\) and \((2,4,6)\).

For these particular collections, information about three-note groupings distinguishes what the complete two-note inventory cannot. This calculation does not prove that triple information uniquely reconstructs every possible set.

An interval vector is useful because it compresses a collection into a small, transposition-invariant description. The compression necessarily asks a narrower question than identifying every musical relationship, still less explaining consonance or aesthetic value. These two collections expose the boundary without requiring a large example.

The fingerprint has all six distances. It has forgotten who was holding hands.

Sources and the worked calculations checked through 2026-10-12.

References

[1] Jedrzejewski, F., and Johnson, T. (2013). The Structure of Z-related Sets. Preprint, arXiv:1304.6608, version 1, §§1 and 4. Full text. DOI: 10.48550/arXiv.1304.6608.

[2] Patterson, A. L. (1939). Homometric Structures. Nature, 143, 939–940. DOI: 10.1038/143939b0.

Changelog

  • 2026-10-12: Research and calculation review. Derived and checked the pair counts, inequivalence, Fourier magnitudes and distinguishing triples; separated cyclic diffraction from acoustic spectra.