Proof of Euler's Theorem involving curvature.

Question

Theorem: Let $φ$ be the angle, in the tangent plane, measured counterclockwise from the direction of minimum curvature $\kappa_1$ . Then the normal curvature $\kappa_n(φ)$ in direction $φ$ is given by $$\kappa_n(\varphi)=\kappa_1 \cos^2\varphi+\kappa_2\sin^2\varphi.$$

How do I prove this?

Answer 1

The normal curvature can be defined in terms of the second fundamental form as follows: the normal curvature $k_n$ of a vector $v \in T_{p}S$ is defined as $k_n = \text{II}_{p}(v)=<-dN_{p}(v), v>$. Recall that the differential of the gauss map is a self-adjoint linear map and so there exists an orthonormal basis $\{e_1, e_2\}$ for $T_{p}S$ such that $-dN_{p}(e_1) = k_{1}e_{1}$ and $-dN_{p}(e_2) = k_{2}e_{2}$.

Thus for an arbitrary direction vector described by your $\varphi$, call it $v$ can we written as a linear combination of some orthonormal basis $\{e_1, e_2\}$ with the above properties as $v = e_{1}\cos\varphi + e_{2}\sin\varphi$.

Now following the definition we get: \begin{align} k_{n} & = \text{II}_{p}(v) = -<dN_{p}(v), v> \\ & =-<dN_{p}(e_{1}\cos\varphi + e_{2}\sin\varphi), e_{1}\cos\varphi + e_{2}\sin\varphi> \\ & =<k_{1}e_{1}\cos\varphi+k_{2}e_{2}\sin\varphi, e_{2}\sin\varphi+ e_{1}\cos\varphi> \\ & =k_{1}\cos^2\varphi + k_{2}\sin^2\varphi \\ \end{align}