Stephen Lavelle
Messiaen was a fair-sound composer with some fair-sound ideas.
I’m basing the body of this analysis around one of the things he was big into, though I’m approaching it from a
mathematical point of view.
I won’t speak of him again in this article.
ok, here’s the notation:
| C |
C# |
D |
D# |
E |
F |
F# |
G |
G# |
A |
A# |
B |
That’s how I’m going to represent the scale, I’m not going to put the sharps above the other normal notes or anything.
Ok, most scales we use don’t have any transformational symmetries. Why? very simple – we know where we stand. Here’s the major scale (x mark the notes):
It has no symmetries.
So you always know exactly where you are and where the main note should be. You feel secure.
Look at the whole tone scale however:
Because the scale looks the same whether you start it on C or D or F# or whatever, you
can’t tell where you are – it leaves you feeling very confused and helpless.
Or, in other words, it has transformational symmetries.
I’m going to try and see if i can catalogue these symmetris now if i can. (I’m just seeing what happens as i go along).
The 12 note scale has 12 different symmetries, so when you’re listening to the scale, you can’t tell absolutely where you are, you could be in any one of twelve positions:
The 11 note scale, though it’s hard to differentiate it from a 12 note scale unless you have some really explicit runs, has no symmetries. It is non-symmetrical.
The following 10 note scale has two symmetries, so you can’t tell whether you’re in the top group of five or the bottom group.
The 9-note scale has symmetries too, three of them, (you see now the niceness inherant in having a 12 note scale? so many divisors!).
The 8-note scale has a fair few symmetries, but there are lots of different possible 8-note scales, but they all have the same two symmetries as the 10-note one, except for one version:
This one has two symmetries
the following one has four symmetries! (innit it great?)
Seven-note scales arn’t symmetrical, they have no symmetries, and are the scales we use mostly (eg. the major/harmonic-minor scales)
Six-note scales have a fair few symmetries, the main one being the whole-tone scale from the beginning. It has 6 symmetries:
These ones have two symmetries:
And this one has three:
I hypothesise that given a scale with
n notes (in this case 12), and a subscale of
k notes, the number of
symmetries depends on the number of common factors of the two numbers. I won’t prove it till I’ve slogged through the rest of the scales.
Five note scales don’t have any symmetries.
Four note scales have lots of symmetries and possible scales. This one has four symmetries:
These ones have only 2:
Three note scales arn’t that common (or versatile!)but there’s one scale that fits the bill with three symmetries:
There’s only one two-note scale with a symmetry(it has two transformational symmetries):
And the one note scale has none.
Right!, that’s the scales done out. Now to look for a pattern :)
| does it have scale with these symmetries? |
2 note |
3 note |
4 note |
6 note |
8 note |
9 note |
10 note |
12 note |
| 2 |
y |
|
y |
y |
y |
|
y |
y |
| 3 |
|
y |
|
y |
|
y |
|
y |
| 4 |
|
|
y |
|
y |
|
|
y |
| 6 |
|
|
|
y |
|
|
|
y |
| 12 |
|
|
|
|
|
|
|
y |
see that there’s a sort of pattern emerging? :)
Notice how scales with a length that has no common factors with 12 (eg. 5 or 7) have no symmetries, and how no scale has a number of transformational symmetries equal to a number that doesn’t divide into 12.
The reason that, say, a scale with 7 notes doesn’t have any symmetries in the 12 note system is that it has to be able to divide the 12 notes into pieces of equal sizes. But it can’t. So it has no symmetries.
Here’s a rewritten version of the same table:
| # of scales with these symmetries |
2 note |
3 note |
4 note |
6 note |
8 note |
9 note |
10 note |
| 2 |
1 |
|
2 |
2 |
2 |
|
1 |
| 3 |
|
1 |
|
1 |
|
1 |
|
| 4 |
|
|
1 |
|
1 |
|
|
| 6 |
|
|
|
1 |
|
|
|
(I left out the 12 note scale because it’s not part of the pattern. Also, i dount count scales with 6 symmetries as also having 2 and 3 (though they do))
Given a segment of 12 notes, it can be divided into lengths of its factors greater than 1, i.e.2,3,4,6 (we’ll ignore 12).
Each of these segments can be filled in many different ways, though we won’t count some of them (i.e. xx– is same as –xx, and x–x is same as -xx- when they’re
stuck beside eachother). So given a length
n, and
k “x”s to put in it, there are (n-1)!/(k!(n-k)!) ways to fill it up. (! means
factorial.
e.g. 5!=5*4*3*2*1)
Now, lets say you’re counting the scales with 6 notes. You know that you can get at least one scale with 3 symmetries because 3 divides 12 and 6. The question is, “how many can you get?”. The answer is pretty simple. You have 6 notes, and three identicle areas, each of length 4, each with 6/3 notes in it.
3!/(2!*1!)=3
So this tells me that there are 3 different scales with 6 notes and 3 symmetries. But my table only hints at two…why? because a scale with 6 symmetries also has two so it must also be included.
Here’s a modified version of the original table taking this into account:
| # of scales with these symmetries |
2 note |
3 note |
4 note |
6 note |
8 note |
9 note |
10 note |
| 2 |
1 |
|
3 |
3 |
3 |
|
1 |
| 3 |
|
1 |
|
2 |
|
1 |
|
| 4 |
|
|
1 |
|
1 |
|
|
| 6 |
|
|
|
1 |
|
|
|
So, given an
L note superscale (in this article
L=12), and a subscale of length
l, for each common factor
f of
L and
l,
the number of different scales
S, of symmetry
f is:
S=((L/f)-1)!/((l/f)!*(L/f–l/f)!)
I didn’t explain every step in the proof but it generates the grid above … basically if you don’t understand how i deduced it by this stage you wouldn’t
understand the proof. (all to do with cyclic permutations, nothing too interesting).
