Deriving Frenet-Serret Formulas: Why -τᴿ?

[Arya Prasetya]

Hi, I'm trying to derive the Frenet Serret Formulas, but I am having trouble to understand why, after some checking, that the derivative of binormal vector is:

\frac{d\hat{b}}{ds}=-\tau\hat{n}

I understand that, \hat{t}\wedge\frac{d\hat{n}}{ds}\parallel\hat{n} and |\frac{d\hat{b}}{ds}|=\tau, but why the negative sign? Isn't it equally possible that it has a positve sign?

 

[jedishrfu]

Wikipedia has a nice write up on the Frenet Serret formulas:

https://en.wikipedia.org/wiki/Frenet–Serret_formulas

B is the cross-product of T and N specifically:

B = T x N so this specifies a particular direction via the right-hand rule.

http://galileo.math.siu.edu/mikesullivan/Courses/251/S11/torsion.pdf

In part (c) of the proof, It looks like the $-\tau$ is chosen so that the formula for N' comes out without two negative terms in it.

 

[Charles Link]

Since $\tau$ is simply a constant of proportionality, I believe the choice is arbitrary, but whatever sign that is used for $\tau$ needs to be consistent with the other Frenet equation involving $\tau$. The equations with their choice of the sign of $\tau$ is apparently somewhat standard. It's a somewhat specialized topic, but I think you might find most of the textbooks use the same sign convention.

 

[lavinia]

Hi, I'm trying to derive the Frenet Serret Formulas, but I am having trouble to understand why, after some checking, that the derivative of binormal vector is:

$\frac{d\hat{b}}{ds}=-\tau\hat{n}$

I understand that, ##\hat{t}\wedge\frac{d\hat{n}}{ds}\parallel\hat{n}[/itex] and |\frac{d\hat{b}}{ds}|=\tau##, but why the negative sign? Isn't it equally possible that it has a positive sign?

<br /> <br /> $\frac{d\hat{b}}{ds}= \frac{d\hat{t×n}}{ds}= \frac{d\hat{t}}{ds}×\hat{n} + d\hat{t}×\frac{d\hat{n}}{ds} = 0 + \hat{t}×κ\hat{b} = κ\hat{t}×\hat{t}×\hat{n}$<br /> <br /> Now use the cross product identity $T×T×N = T(T⋅N) - N(T⋅T)$