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?

| cite | improve this question | | | | |

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}

| cite | improve this answer | | | | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.