Derive $\tau = \frac{\langle a' \times a'', a''' \rangle}{||a' \times a''||^2}$ (torsion) directly instead of proving it.

Question

Let $a(t)$ be a parametrised curve. Then the torsion of the curve can be calculated as

$$\tau = \frac{\langle a' \times a'', a''' \rangle}{||a' \times a''||^2}$$ I have seen a proof of this formula. But nevertheless I wonder whether it is possible to derive this formula directly. I think I need to clarify my intention: Suppose you know the definition of torsion but you don't know the formula for torsion as in the above equation. Then, someone asks you to give a formula for the torsion in terms of $a(t)$ and its derivatives. How would you proceed, and how would you find this above equation? I think, it has to come from somewhere...

I know how the proof works, but what if you don't know the formula to be proven beforehand..

related question: http://math.stackexchange.com/q/346603/9464 — , Mar 27 '16 at 13:43

score 3 · Accepted Answer · 2016-03-27T13:43:33.097

3

This is clearly done in Elementary Differential Geometry by Pressley clearly (see Proposition 2.3.1).

Denote the identity as (2.14), then we have the following calculations.

edited Mar 27 '16 at 13:43

answered Mar 27 '16 at 13:37

Wow, that is impressive. Just got a question: Why don't more authors derive this equation this way. There is so much more intuition behind it. Why do many authors just prove the equation, without stating where it comes from? – user50224 Mar 27 '16 at 13:47
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This would be good to ask at http://matheducators.stackexchange.com/ – Mar 27 '16 at 13:59

Derive $\tau = \frac{\langle a' \times a'', a''' \rangle}{||a' \times a''||^2}$ (torsion) directly instead of proving it.

1 Answers1