2-D scales and matrices

I've stated before that, in my matrix formalism, a meantone
temperament can be defined:

H' = H + (k2)p
(k3)

The key to this is that H is a column matrix of logs of primes:

(log2(2))
H = (log2(3)) oct
(log2(5))
( ... )

I did say as much on the list a while back, but I don't think
anyone was paying attention. Any logarithm of a rational number
can be specified using a large enough H. H' is a matrix that
behaves like H, but isn't, p is a real number corresponding
to the pitch of a comma, and k2 and k3 are real numbers.

Using that equation at the top, once you've defined an interval in
terms of H', you can find it's corresponding pitch difference.
However, the inverse problem also deserves attention: how to
approximate an interval (w x y)H to (w' x')H'. This requires a
conversion matrix, C, such that (w x y)C = (w' x'). In the case
of syntonic meantone, where p=(-4 4 -1)H, C can be defined by the
following equation:

( 1 0 0) (1 0)
( 0 1 0)C = (0 1)
(-4 4 -1) (0 0)

Which will work for any comma (i j k)H where |k|=1. For a 2-D
scale approximating 4-D (7 prime limit) harmony, you need to
define two commas (i j k l)H and (i' j' k' l')H where |k*l'- k'*l|
= 1. As the matrix here is unitary, we can solve for C really
easily:

( 1 0 0)(1 0) ( 1 0)
C = ( 0 1 0)(0 1) = ( 0 1)
(-4 4 -1)(0 0) (-4 4)

With this matrix, any just interval can be converted into any
syntonic meantone expressed using the 2-D matrix H'.


Another application of matrix formalism to 2-D scales is the
method of defining the scale from a comma being zero and two
intervals being perfect. As an example, set the octave and the
major 7th (-3 1 1) to be perfect, and zero the syntonic comma.
These conditions can be summarised:

( 1 0 0) ( 1 0 0)
(-3 1 1)H' = (-3 1 1)H
(-4 4 -1) ( 0 0 0)

To solve for H':

( 1 0 0)-1( 1 0 0) 1( 5 0 0)
H' = (-3 1 1) (-3 1 1)H = -( 4 1 1)H
(-4 4 -1) ( 0 0 0) 5(-4 4 4)

My computer did the inverting and multiplication, which would take
a while by hand. If you're expecting a fractional comma meantone,
you can go straight to the denominator by calculating the
determinant of the matrix on the left. Rearranging for the form
above:

1( 5 0 0) (1 0 0) 1( 0 0 0) ( 0 )
H' = -( 4 1 1)H = (0 1 0)H + -( 4 -4 1)H = H + (-1/5)(-4 4 -1)H
5(-4 4 4) (0 0 1) 5(-4 4 -1) ( 1/5)

It's quite alright to throw away the third dimension: it means
taking the first two prime to dimensions to be a basis. Then,
this is 1/5 comma meantone from the general form: k2=0, k3=-1/5,
p(k) = (-4 4 -1)H.

As in this case octaves are perfect, the calculation can be
simplified by using the octave invariant matrix L:

L' = (1 1)-1(1 1)L = (-1 -1)(1 1)L/(-1-4) = (1 1)L/5
(4 -1) (0 0) (-4 1)(0 0) (4 4)

No need for a computer there. The only dimension we're now
interested in is then (1 1)L/5 = (1 0)L + k*(4 -1)L, or (1 1) =
(5 0) + k*(20 -5). The solution of this is k=-1/5, consistent
with k3 above. Again, this is the reciprocal of the determinant
of the defining matrix.


Just some of the wondrous things you can do with interval space
matrices!



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