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I'm not sure if there is some wrong manipulation in this. This is not homework, just some simple manipulations I've made here that led me to an identity, and I want to know if this is valid. If not, where is wrong?

Consider a curve $\beta:(a,b)\rightarrow \mathbb{R}^3$ parametrized by arc length with curvature $\kappa >0$ and torsion $\tau > 0$. We know that $|\beta'| = 1$ and $\beta'' = \kappa\cdot n$, where $n$ is the normal vector. From the Frenét equations we have $b' = \tau\cdot n$, where $b$ is the binormal vector. Note that we can write $n = \frac{1}{\tau}b'$, with this we have that $\beta'' = \frac{\kappa}{\tau}b'$ and $\beta' = n\times b = \frac{1}{\tau}b'\times b $.

Now we can do the following calculations

$$\kappa = \frac{|\beta'\times\beta''|}{|\beta'|^3} = |\beta'\times\beta''| = \left| \left(\frac{1}{\tau}b'\times b\right)\times\left(\frac{\kappa}{\tau}b'\right)\right| = $$ $$= \frac{\kappa}{\tau^2}\left| \left(b'\times b\right)\times\left(b'\right)\right| = \frac{\kappa}{\tau^2}\left| \left(b'\cdot b'\right)\cdot b - \left(b\cdot b'\right)\cdot b'\right| = $$ $$= \frac{\kappa}{\tau^2}\left||b'|\cdot b\right|= \frac{\kappa}{\tau^2}|b'|\cdot |b| = \frac{\kappa}{\tau^2}|b'|$$

Therefore,

$$\kappa = \frac{\kappa}{\tau^2}|b'| \implies \tau = \sqrt{|b'|} $$

Thanks.

Integral
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    Why are you assuming $\tau>0$? The sign of the torsion is a meaningful concept and tells you whether the curve is twisting in a left- or right-handed manner. But your mistake is that $b'\cdot b' = |b'|^2$, not $|b'|$. – Ted Shifrin Jun 09 '14 at 17:20

1 Answers1

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The equation

$\vert \tau(s) \vert = \sqrt{\Vert B'(s) \Vert} \tag{0}$

is in fact incorrect, viz:

In the usual notation, the Frenet-Serret equation for the binormal vector is

$B'(s) = -\tau(s) N(s), \tag{1}$

where $s$ is the arc-length along $\beta$. Since

$\Vert N(s) \Vert = 1, \tag{2}$

(1) leads directly to

$\Vert B'(s) \Vert = \vert -\tau(s) \vert \Vert N(s) \Vert = \vert \tau(s) \vert, \tag{3}$

not

$\vert \tau(s) \vert = \sqrt{\Vert B'(s) \Vert}; \tag{4}$

as Ted Shifrin pointed out in his comment,

$B'(s) \cdot B'(s) = \Vert B'(s) \Vert^2 \ne \Vert B'(s) \Vert \tag{5}$

unless, of course, $\Vert B'(s) \Vert = 1$; not, though, in general.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Robert Lewis
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