(Indian Classical) Violin note-spacing vs. Guitar note spaces

Question

I play the Indian classical violin. In that, when you have a song in A, for instance, you would tune the first (lowest) and third strings to A, an octave apart, and the second and fourth to the perfect fifth, again an octave apart (that's E, if I'm not mistaken). Similarly, if your song is in F, you tune the first and third to F, and the second and fourth to the perfect fifth, C. After that, you use the Hindi/Urdu equivalents of Do, Re, Mi, Fa, So, La, Ti, and Do.

However, since I believe that this platform has primarily western musicians, I'll try to translate stuff to your lingo. Tell me if something doesn't make sense; I'll review the little I know of western notation.

Let's consider a song in C major. The notes should be C, D, E, F, G, A, B, C; with my tuning, I will play the first C on the third string with no fingers on the neck of the violin, and G on the fourth string with no fingers, and I'll press the fourth sting somewhere around the middle of the neck for the final C.

No, I switch to some other scale within the same string tuning, so I play the notes C, C#, E, F, G, G#, B, C (I don't know the western name for this, but it certainly exists and is very common in Indian music). To play C# and G# on the C and G strings respectively, I place my pointer finger right next to the end of the neck, so the length of the vibrating string is changed very slightly. This creates an increase by a semitone. However, if I were to play a different common Indian scale, with the violin tuned to CGCG again, such that I had C, D, E, F#, G, A, B, C, I would shift my ring finger (which I use to press both F and F#) up but almost a whole inch compared to the F position to get F#. And if you change the tuning of the violin, as long as you keep the alternation of root-perfect fifth-root-perfect fifth, the finger positioning doesn't change at all for corresponding Do, Re, etc.

On the other hand, with a guitar (which I don't know how to play or tune), I think each fret represents a semitone, and for the most part, they seem equidistant. What's going on here? This somehow seems confusing and it doesn't add up. Is it because of some physics, or is it because of the way notes' frequencies differ somehow?

Answer 1

If you start at an open string on your violin, and progressively finger each of the eight notes to arrive at the note one octave above that start note, you will have a finger exactly half way along the string. (From nut to bridge).

Notice that each next step you take is smaller than the last. By the same percentage, actually.

This time, measure how far re is from do. Then keep that same measured step for eight times, going up from open. Notice how 1. the notes are out of tune. 2. you run out of fingerboard.

You are not familiar with guitars. But be certain that each fret going up from the open string is smaller than the previous one! It works exactly as the violin does, except with a guitar in your hand, you can see and measure that fact quite easily. The octave position is 12th fret,(half way), the second octave comes along at 24th fret,(three quarters way), so it's clear that the frets diminish as the notes get higher. That second octave is half as long (along the fingerboard), so they must get smaller!

Answer 2

The guitar frets are certainly anything but equidistant. We can use some simple physics to see how they need to be spaced.

The frequency of the (fundamental = lowest mode) of the string is given by f = c/2L, where c is the speed of wave propagation in the string (some constant) and L is the length of the string. In the equal temperament, each step up a semitone equals multiplying the frequency by a factor of 2 to the power of 1/12. Stopping the strings with our fingers is the same as making them shorter. So in order to increase the tone by one semitone, we need to multiply f by 2^1/12, and that amounts to multiplying L by 2^-1/12.

For guitar, that means that the first fret has to be in 5.6% of the length of the string, the second in 10.9%, the third in 15.9% etc. These seem to be equally spaced, but the differences are actually becoming smaller and smaller. For instance, the 7th fret has to be in 33.3%, the 8th in 37%, the 9th in 40.5%, etc., and the 12th exactly in 50%.

Moreover, this physics is just the same for the violin. So if you want to play exactly one semitone (in equal temperament) above the open string, you need to stop it in 5.6% of its length, etc., the numbers are exactly the same.

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So as far as the physics are concerned, guitar and violin are just the same. Of course without hearing you playing I cannot say why you think that it is different. However, as I said, the guitar frets are laid out so as to make the guitar play in the equal temperament. And I highly doubt that the European/Western 12-tone equal temperament is used in the Indian classical music, so what you call a semitone step is probably quite different from increasing of the frequency by the factor of 2^1/12, which is what you get in the equal temperament.