The total torsion of a unit speed curve γ : [a, b] → R3 is defined to be. ∫ τ(s) ds (bounds a to b), where τ(s) is the torsion of γ at the point γ(s). If γ is a closed curve on the unit sphere S2, show that γ has zero total torsion. I have a couple hints to solve this but I can't figure it out, the hints are use k(T)=K_γ cos θ where T=γ' for a unit speed curve γ, and differentiate cosθ=N · U

For the curve r=r(t), its tangent, normal, and binormal unit vectors, often called T, N, and B, satisfies the Frenet–Serret formulas. So we have:
$\displaystyle \frac{dN}{ds}=-\kappa T+\tau B$.
Let n be the unit normal vector of the sphere, and denote $\displaystyle \theta$ as the angle between n and N such that $\displaystyle \cos\theta=\langle N, n \rangle$
Differentiate the above equation against s the arc length parameter we get:
$\displaystyle -\sin\theta \frac{d\theta}{ds}=\langle \frac{dN}{ds}, n \rangle + \langle N, \frac{dn}{ds} \rangle$
For spheres we have n=r the position vector, so $\displaystyle \frac{dn}{ds}=\frac{dr}{ds}=T$
Plugin this one and the above Frenet formula we get
$\displaystyle -\sin\theta \frac{d\theta}{ds}=\langle -\kappa T+\tau B, n \rangle + \langle N, T \rangle = \tau \langle B, n \rangle$
Since B, N, n are all orthogonal to T, their three are in the same plane. And since the angle between N and n is $\displaystyle \theta$,
and B, N are orthogonal, the angle between B and n is then $\displaystyle \theta-\frac{\pi}{2}$.
So we have $\displaystyle \langle B, n \rangle=\sin\theta$. Plugin to the above equation we got:
$\displaystyle \tau=-\frac{d\theta}{ds}$
Integrate it we get $\displaystyle \int \tau ds = -\int d\theta = 0$

why does n=r?

You said it is a unit sphere. so n=r.

Originally Posted by xxp9

You said it is a unit sphere. so n=r.

Can you explain more about n=r? i still don't get this part. Thanks

for the unit sphere S, for any point p on it, the unit normal vector n at p has the same direction and length as p. that is, n=p.