An Interesting Result in the Theory of Well-Temperaments

Theorem

Let consider the set of all circulating temperaments having only two sizes of fifth. We may describe such a temperament using a binary vector of length 12, 1s being associated with one size of fifth, and 0s with the other. For example, Vallotti's temperament could be represented by the vector 111111000000. Then, the theorem may be stated thus:

The temperament 111110100000 is alone (barring reflections and rotations) among temperaments having two sizes of fifth in that merely knowing the sizes of two successive (along the circle of fifths) major thirds uniquely determines one's position within the circle.

Put another way: in this and no other temperament having only two sizes of fifth, the sound of a V - I cadence, as determined by the amount the major thirds of the triads are tempered, is different in each key.

Proof

From the primary temperament/vector F, we may derive three related vectors (all subscripts are understood to be modulo 12):

We define T such that T(n) = F(n) + F(n+1) + F(n+2) + F(n+3).

We define TP such that TP(n) = the ordered pair [T(n), T(n+1)].

We define TV such that TV(n) = T(n) + T(n+1).

We further define a function S such that for any vector v, S(v) = summ(v(n), n = 1 ... 12).

To satisfy the given conditions, it is both necessary and sufficient that TP consist of twelve different ordered pairs. (This is perhaps the most concise way to state the theorem.)

Let us consider all the possible duplets that could be elements of TP. Since the two elements of the duplet cannot differ by more than one, the following list of thirteen is exhaustive:

  1. [0,0]
  2. [1,1]
  3. [2,2]
  4. [3,3]
  5. [4,4]
  6. [0,1]
  7. [1,2]
  8. [2,3]
  9. [3,4]
  10. [1,0]
  11. [2,1]
  12. [3,2]
  13. [4,3]
To construct an F such that its associated TP contains twelve distinct elements, one and only one of these thirteen must be eliminated. But which? We can find out with a little number theory. It is obvious that for any temperament F,

S(TV) = 2S(T) = 8S(F)

The sum of all the integers in the list above is 52. We know, however, that for any actual temperament F, S(TV) must be a multiple of 8. 52 mod 8 is 4, therefore we must eliminate [2,2], the only pair which adds up to 4.

We have now demonstrated that a necessary condition of any vector F which satisfies the desired conditions must have a TP which contains both [4,4] and [0,0]. A [4,4] can only occur in a vector which contains the subvector 11111. However, a vector containing the subvector 111111 would result in to two successive [4,4]s in the TP, which violates the conditions. The 11111 must therefore be flanked by 0s, and by symmetrical argument there must also be a 00000 flanked by 1s. Attempting to fit together 1000001 and 0111110 within a circle of 12 can only be accomplished with the vector 111110100000 and its associated rotations and reflections.

This demonstrates that no vector other than the given one could satisfy the desired conditions; verifying that it actually does so is a simple exercise which is left to the reader. Q.E.D.

Discussion

Note that this vector, strictly speaking, does not represent a single temperament but an entire class of them. Not only is there a group of temperaments which can be generated by rotations and reflections, but we can also vary the two sizes of fifths. As long as the average of the 12 fifths equals an equal-tempered fifth, everything else works. However, the simplest solution would be to

  1. Let 0s represent just fifths and the 1s fifths tempered by 1/6 Pythagorean comma. This allows for the maximum variety of chords and keys possible without harmonic waste. (Harmonic waste occurs in a temperament if there are fifths wider than just, or major thirds narrower than just.)
  2. Map the given vector to the circle of fifths beginning at F, i.e. temper the fifths FC, CG, GD, DA, AE, and BF#. This keeps the thirds which are closer to just in the keys with fewer sharps and flats.
Using these assumptions, the resulting temperament is that proposed as Bach's preferred keyboard temperament by John Barnes in his article in Early Music, v.7, no.2 (1979). Barnes derives this temperament based on statistical analysis of Das Wohltemperierte Klavier, but he does not remark on the above-described property, nor has anyone else in the literature with which I am familiar.

Further Thoughts

[Under construction.]

There are two possible temperaments in which each major third is of a distinct size (given some assumptions I don't have time to lay out right now). They require fifths of four different sizes; the vectors are 233331210000 and 133332120000.

Please email me, Paul Hahn, at manynote@library.wustl.edu, if you have any comments regarding the above.