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?

  • 1
    $\begingroup$ 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 $S^2$? $\endgroup$ – Richard Jensen Oct 8 at 12:09
  • $\begingroup$ @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. $\endgroup$ – trisct Oct 8 at 13:16
  • $\begingroup$ 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. $\endgroup$ – 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:


  • $\begingroup$ I did find this article before asking. But the reference is not available anywhere (that I could find). The digital versions go back to only 2006, and the paper copies in my school library do not include the 1947 issue. And the google scholar link in the page you cited links to a blank page. $\endgroup$ – trisct Oct 9 at 0:10
  • $\begingroup$ I saw the formula but no proof. And there's no available copy of Segre's paper. When I said "the reference" I meant Segre's paper. $\endgroup$ – trisct Oct 9 at 9:33

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