I am looking for direct proofs that the total curvature $\int_0^{L_\gamma} \! \gamma''(s) \, \mathrm{d} s$ of any Jordan curve $\gamma$ resp. $\int_0^{L_\gamma} \! |\gamma''(s)| \, \mathrm{d} s$ of a convex Jordan curve equals $2\pi$. Direct proof means: a calculation and not a special case of the theorem of Gauss-Bonnet or of enter image description here

I have no idea even how to show that the integral

$$\int_0^1 \frac{\mathrm{d} x}{(1-(1-b^2)x^2)^{3/2}}$$

evaluates to $\frac{1}{b}$ which would prove the above at least for ellipses with major axis $a=1$ and minor axis $b$. Let alone for arbitrary (convex) curves/functions.


The following is a sketch of Hopf's own proof:

Assume that $\gamma$ is given in the form $$\gamma:\quad t\mapsto {\bf z}(t)=\bigl(x(t),y(t)\bigr)\qquad(0\leq t\leq T)$$ with $${\bf z}'(t)\ne {\bf 0}\quad\forall t,\quad {\bf z}(0)=(0,0),\quad {\bf z}'(0)=(1,0), \quad y(t)\geq 0\quad\forall t\ .$$

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Consider the triangular domain $$B:=\bigl\{(s,t)\ \bigm|\ 0\leq s\leq t\leq T\bigr\}$$ and on $B$ the function $${\bf g}(s,t):=\int_0^1{\bf z}'\bigl((1-\tau) s +\tau t\bigr)\ d\tau =\cases{{{\bf z}(t)-{\bf z}(s)\over t-s} \quad &$(s<t)$ \cr {\bf z}'(s)&$(t=s)$\cr}$$ From the assumptions on $\gamma$ it follows that ${\bf g}(s,t)\ne{\bf 0}$ on $B$. Since $B$ is simply connected the function ${\bf g}$ therefore possesses a continuous real-valued argument $\theta(\cdot,\cdot)$ on $B$, i.e., there exists a continuous function $\theta:\ B\to{\mathbb R}$ such that $$\bigl[\theta(s,t)\bigr]=\arg{\bf g}(s,t)\qquad\bigl((s,t\in B\bigr)\ .$$ We may assume $\theta(0,0)=0$. Since $\bigl[\theta(s,s)]=\arg{\bf z}'(s)$ we may consider $\hat\theta(s):=\theta(s,s)$ as a continuous real-valued argument of ${\bf z}'(s)$ along $\gamma$. It remains to prove that $\hat\theta(T)=\theta(T,T)=2\pi$.

To this end we look at the behavior of $$\theta(0,t)=\arg\bigl({\bf z}(t)-{\bf z}(0)\bigr)\qquad(0\leq t\leq T)$$ along the vertical edge of $B$. The limiting direction at $(0,T)$ is parallel to $(-1,0)$, and as $\arg\bigl({\bf z}(t)-{\bf z}(0)\bigr)\ne-{\pi\over2}$ for all $t\in[0,T]$ it follows that necessarily $\theta(0,T)=\pi$. Similarly we then look at the behavior of $$\theta(s,T)=\arg\bigl({\bf z}(T)-{\bf z}(s)\bigr)\qquad(0\leq s\leq T)$$ along the horizontal edge of $B$. The limiting direction at $(T,T)$ is parallel to $(1,0)$, and as $\arg\bigl({\bf z}(T)-{\bf z}(s)\bigr)\ne{\pi\over2}$ for all $s\in[0,T]$ it follows that necessarily $\theta(s,s)=2\pi$.

  • $\begingroup$ Finally, I found out that my question and your answer is about Hopf's so-called Umlaufsatz. Do you know its english name, eventually? (Hopf, by the way, ascribes the proposition and some earlier proofs to Rolle, Ostrowski, and Riemann.) $\endgroup$ – Hans-Peter Stricker Jan 11 '13 at 17:19
  • $\begingroup$ @Hans Stricker: The English name is Umlaufsatz; see, e.g. here: warwickmaths.org/files/Curvature.pdf $\endgroup$ – Christian Blatter Jan 12 '13 at 9:42

The integral above may be evaluated using a trig substitution: $x=\sin{\theta}/\sqrt{1-b^2}$. The result is

$$ \frac{1}{\sqrt{1-b^2}} \int_{0}^{\arccos{b}} d\theta \sec^2{\theta} $$

$$ = \frac{1}{\sqrt{1-b^2}} \tan{(\arccos{b})} $$

$$ = \frac{1}{\sqrt{1-b^2}} \frac{\sqrt{1-b^2}}{b} $$

$$ = \frac{1}{b} $$


The proof that for smooth Jordan curves $\gamma$ the total curvature is

$$\kappa = \int_\gamma r''(s)\text{d}s = 2\pi$$

goes along the same line as the proof that the total length is

$$l = \int_\gamma |r'(s)|\text{d}s $$

One just has to observe that the sum of the external angles of any polygon is $2\pi$. (The external angle pretty much corresponds to the curvature.)


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