Arc length parameterization lying on a sphere

Show that if $\alpha$ is an arc length parameterization of a curve $C$
which lies on a sphere of radius $R$ about the origin then $$R^2 =

I know I can write the unit tangent $T$, normal $N$, and the binormal $B$ as a linear combination $\alpha(s) = a(s)T(s)+b(s)N(s)+c(s)B(s)$ and try to determine what $a, b, c $ are, but how can I continue solving this?

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First, we need that $\kappa$ and $\tau$ vanish nowhere. Clearly we have $\langle \alpha, T\rangle=0$, thus $\langle \alpha, T’\rangle+\langle T’,T’\rangle=0$, so we derive
Now show that
$$\left(\frac{1}{\kappa}\right)’=\langle\alpha,\tau B\rangle,

now evaluate $\|\alpha\|^2$.


The key is to construct a vector $\alpha$ as @Michael Hoppe did.:

If $R$ is on a sphere. Then think of the vector $\alpha$ as exactly the vector that points from the centre of sphere to the curve $R$. So $\alpha$ is different from the original curve $R$ by just by a translation, hence they have the same tangent, curvature and torsion.

We can always express $\alpha$ as a linear combination (as suggested by Shifrin) of:

$$\alpha(s) = \lambda(s)T(s) + \mu(s)N(s)+\nu(s)B(s)$$

for some functions $\lambda$, $\mu$, $\nu$

(Note that $\alpha$ itself is not necessarily arc length parametrised.)

Then since the $R$ is on the sphere, and that $\alpha$ is from the centre to the sphere, we require that $\alpha$ is tangent to its tangent vector (notice that if $\alpha$ does not point from the centre to the sphere, it will not be tangent to its tangent vector $T$), that is,

$$<\alpha, T>=0$$

By differentiating above, we will find the expression for $\alpha$ as shown in @Michael Hoppe’s post:


$\alpha$ must be of constant length. This implies hat the relationship in the problem holds.