You know about open string harmonics, right? How you can lightly touch say the bottom E string in the region of the 12th fret (the one just right of the two dots in this Wikipedia diagram), and plucking it, obtain a curiously muted octave note? Somehow softer in timbre than actually playing the fret in the usual way, but unmistakably the same note. That's because the 12th fret is positioned at a point exactly half way up the length of the string, so playing it has the same result, in audio frequency terms, as exciting the string's second harmonic (the red curve marked 2), aka its first overtone, using the light touch trick.
Of course, I'm ignoring the slight stretching of the string caused by pushing it down and on to the fingerboard. So far, so good. And you probably know about the 5th and 7th fret tricks too, right? How the node at the 5th (yellow curve 4) gives you another E, yet another octave up. And how the 7th (orange 3) rings out with a perfect fifth B, this time one octave up from the result of actually playing that fret in the usual way?
Wrong!
Actually those nodes do work as advertised, and the harmonics they produce are just perfect octaves and fifths. However, they are not positioned above the 5th and 7th frets. Because western semitones, specifically the 12 frets in a guitar octave or the 12 keys in a keyboard one, are evenly tempered, i.e., distributed in frequency terms, everything else - everything that isn't an interval of one or more octaves - is just an approximation.
Let's see why. If the perfect fifth interval between E and B corresponded to exactly seven frets, and we strung together twelve such intervals, then we should end up a total of...
7 frets in an interval × 12 intervals = 84 fretsabove our starting point, corresponding to a leap of seven octaves:
12 frets in an octave × 7 octaves = 84 frets.Incidentally, if you find it difficult as I do to imagine a design for a guitar with 84 frets, you might find it helpful to switch to a piano analogy at this point...
Pythagorean Commas
Now consider the sum harmonically. Each perfect fifth interval, say from E up to B, represents a 50% increase in frequency. In other words, the B note, expressed in Hertz (cycles per second), is 1½ times the value of the E. String together twelve such intervals, and you have a frequency multiplication factor of
(3/2) ^ 12 = 3^12 ÷ 2^12 = 531441 ÷ 4096 = 129.75to two decimal places. That's quite different from the expected value for seven octaves,
2^7 = 128.00This difference is known as the Pythagorean Comma, and works out at about 1.36%. The implication is that western music, as rendered on guitars and pianos anyway, is not mathematically commutative! We use that seventh evenly tempered fret, or key, as a perfect fifth interval. But if twelve such intervals (12×7 semitones) amount to something different from seven octaves (7×12 semitones), then something has to give. Because it just doesn't add up. Or rather, divide out. To unity.
From Wikipedia: http://en.wikipedia.org/wiki/Pythagorean_comma |
perfect fifth = 3÷2Historical Note (heh)
seventh fret = 2 ^ (7÷12)
perfect fifth ÷ seventh fret = 1.0011298906275257736
Prior to being written about by me, the calculation of the Pythagorean Comma was also documented in the Chinese (Han dynasty, 122 BC) philosophical classic work The Huainanzi, subsequently extended by Jing Fang (50 BC), then agonized over by centuries' worth of frustrated piano tuners. It gets its contemporary name from its description in The Division of the Canon, an ancient Pythagorean treatise on the relationship between mathematics and acoustics.
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