Prove that the integral $\int\frac{\tau(s)}{\kappa(s)}ds=0$

Let $$\vec r=\vec r(s)$$ be a closed regular curve on $$S^2\subset\mathbb R^3$$, parametrized by arclength. Prove that $$\int\frac{\tau(s)}{\kappa(s)}ds=0$$ where $$\tau$$ and $$\kappa$$ are respectively the torsion and the curvature of the curve.

My progress: I think I am supposed to write $$\tau/\kappa$$ as a derivative something else. But the only information I can obtain from $$\vec r$$ lying on the unit sphere is that $$\vec r(s)=0\vec t-\frac{1}{\kappa}\vec n-\frac{1}{\tau}\frac{d}{ds}\Big(\frac{1}{\kappa}\Big)\vec b$$ where $$\vec t,\vec n,\vec b$$ is the Frenet frame of $$\vec r$$. Since $$\vec r$$ has length $$1$$ I can write $$\frac{\tau}{\kappa}=\pm\sqrt{1-\bigg(\frac{d}{ds}\Big(\frac{1}{\kappa}\Big)\bigg)^{\!2}}$$ which does not help me solve the integral.

Any suggestions?

• If your curve is parametrized by arc length, it has constant speed, and therefore, doesn't it hold that the curvature is constantly 1 since we are on $S^2$? – Richard Jensen Oct 8 at 12:09
• @RichardJensen No, $S^2$ is the unit sphere in $\mathbb R^3$. Are you confusing it with the unit circle? Check this out: math.stackexchange.com/questions/533028/… If what you say is true then this question is trivial. – trisct Oct 8 at 13:16
• I meant $S^2 \subset \mathbb{R}^3$ aswell. But thinking about it a second time, I'm not sure if it's even true. I'll take a further look when I'm home from work. – Richard Jensen Oct 8 at 13:27

Let $$\omega$$ denote a period for the curve $$\vec{r}$$. It is useful to know that for a curve $$\vec{r}$$ lying on $$S^{2}\subset\mathbb{E}^{3}$$, the curvature $$\kappa$$ and torsion $$\tau$$ can be expressed in terms of the function $$j(s):=\ddot{\vec{r}}(s)\cdot\left(\vec{r}(s)\times\dot{\vec{r}}(s)\right).\tag{*}$$ Namely, I claim that \begin{align} &\kappa(s)=\sqrt{1+j(s)^{2}},\tag{1}\\ &\tau(s)=\frac{j'(s)}{1+j(s)^{2}}\tag{2}. \end{align} Assuming for now that the equalities $$(1)$$ and $$(2)$$ are true, we get $$\int_{0}^{\omega}\frac{\tau(s)}{\kappa(s)}ds=\int_{0}^{\omega}\frac{j'(s)}{(1+j(s)^{2})^{3/2}}ds=\int_{j(0)}^{j(\omega)}\frac{1}{(1+u^{2})^{3/2}}du,\tag{3}$$ substituting $$u:=j(s)$$ in the last equality. Now, since $$\vec{r}(s)=\vec{r}(s+\omega)$$, we know in particular that $$\vec{r}(0)=\vec{r}(\omega)$$, $$\dot{\vec{r}}(0)=\dot{\vec{r}}(\omega)$$ and $$\ddot{\vec{r}}(0)=\ddot{\vec{r}}(\omega)$$. So from $$(*)$$ it is then clear that $$j(0)=j(\omega)$$, which implies that the integral $$(3)$$ is indeed zero.

Now it remains to prove the equalities $$(1)$$ and $$(2)$$.

(1) Since $$\vec{r}(s)$$ lies on the unit sphere and has speed $$1$$, we have that $$\{\vec{r}(s),\dot{\vec{r}}(s),\vec{r}(s)\times\dot{\vec{r}}(s)\}$$ is an orthonormal frame. Expressing $$\ddot{\vec{r}}(s)$$ in this frame gives \begin{align*} \ddot{\vec{r}}(s)&=(\ddot{\vec{r}}(s)\cdot\vec{r}(s))\vec{r}(s)+(\ddot{\vec{r}}(s)\cdot\dot{\vec{r}}(s))\dot{\vec{r}}(s)+(\ddot{\vec{r}}(s)\cdot \vec{r}(s)\times\dot{\vec{r}}(s))\vec{r}(s)\times\dot{\vec{r}}(s)\\ &=(\ddot{\vec{r}}(s)\cdot\vec{r}(s))\vec{r}(s)+j(s)\vec{r}(s)\times\dot{\vec{r}}(s)\\ &=-\vec{r}(s)+j(s)\vec{r}(s)\times\dot{\vec{r}}(s).\tag{4} \end{align*} Here we use that $$\vec{r}(s)\cdot\vec{r}(s)=1$$, so that $$\dot{\vec{r}}(s)\cdot\vec{r}(s)=0$$ and therefore $$\ddot{\vec{r}}(s)\cdot\vec{r}(s)=-\dot{\vec{r}}(s)\cdot\dot{\vec{r}(s)}=-1$$. We also use that $$\ddot{\vec{r}}(s)\cdot\dot{\vec{r}}(s)=0$$ since $$\dot{\vec{r}}(s)\cdot\dot{\vec{r}}(s)=1$$. Now finally, since $$\kappa(s)=\|\ddot{\vec{r}}(s)\|$$, the expression $$(4)$$ gives $$\kappa(s)=\|\ddot{\vec{r}}(s)\|=\sqrt{1+j(s)^{2}}.$$

(2) We can rewrite the expression (*) as follows: \begin{align*} j(s)&=\ddot{\vec{r}}(s)\cdot\left(\vec{r}(s)\times\dot{\vec{r}}(s)\right)=\vec{r}(s)\cdot\left(\dot{\vec{r}}(s)\times\ddot{\vec{r}}(s)\right)=\vec{r}(s)\cdot\left(\vec{t}(s)\times\kappa(s)\vec{n}(s)\right)\\&=\kappa(s)\vec{r}(s)\cdot\vec{b}(s). \end{align*} Differentiating the equality $$\vec{r}(s)\cdot\vec{r}(s)=1$$ repeatedly and using the Frenet-Serret formulas, we get \begin{align*} \vec{r}(s)\cdot\vec{t}(s)=0 &\Rightarrow \vec{t}(s)\cdot\vec{t}(s)+\kappa(s)\vec{r}(s)\cdot\vec{n}(s)=0\\ &\Rightarrow \vec{r}(s)\cdot\vec{n}(s)=-1/\kappa(s). \end{align*} So, expressing $$\vec{r}(s)$$ in the orthonormal frame $$\{\vec{t}(s),\vec{n}(s),\vec{b}(s)\}$$ gives \begin{align*} \vec{r}(s)&=(\vec{r}(s)\cdot\vec{t}(s))\vec{t}(s)+(\vec{r}(s)\cdot\vec{n}(s))\vec{n}(s)+(\vec{r}(s)\cdot\vec{b}(s))\vec{b}(s)\\ &=\frac{-1}{\kappa(s)}\vec{n}(s)+\frac{j(s)}{\kappa(s)}\vec{b}(s). \end{align*} Differentiating this equality and using the Frenet-Serret formulas gives $$\vec{t}(s)=-\left(\frac{1}{\kappa(s)}\right)'\vec{n}(s)-\frac{1}{\kappa(s)}(-\kappa(s)\vec{t}(s)+\tau(s)\vec{b}(s))+\left(\frac{j(s)}{\kappa(s)}\right)'\vec{b}(s)-\frac{j(s)}{\kappa(s)}\tau(s)\vec{n}(s)\tag{5}$$ So the coefficient of $$\vec{b}(s)$$ in $$(5)$$ has to be zero, which gives $$\tau(s)=\kappa(s)\left(\frac{j(s)}{\kappa(s)}\right)'.\tag{6}$$ But we know from $$(1)$$ that $$\kappa(s)=\sqrt{1+j(s)^{2}}$$. So $$\left(\frac{j(s)}{\kappa(s)}\right)'=\frac{j'(s)\sqrt{1+j(s)^{2}}-\frac{j(s)^{2}j'(s)}{\sqrt{1+j(s)^{2}}}}{1+j(s)^{2}}=\frac{j'(s)(1+j(s)^{2})-j(s)^{2}j'(s)}{(1+j(s)^{2})^{3/2}}=\frac{j'(s)}{(1+j(s)^{2})^{3/2}}.\tag{7}$$ So we conclude from $$(6)$$, $$(7)$$ and $$(1)$$ that $$\tau(s)=\frac{j'(s)}{1+j(s)^{2}}.$$

This is a theorem of B. Segre from 1947. See reference 6. in the page visible here: