question about geodesic curvature of geometry

Question

I have questions. Can anyone help me to get the idea or figure out this problem? What is the formula of geodesic curvature and what is the easy formula to compute geodesic curvature for any surface? Thanks!

Answer 1

If we have a Darboux basis $(T(t), N(t), U(t))$, where

• $T(t) = \frac{\gamma ' (t)}{|\gamma'(t)|}$ is the unit tangent of a space curve on a surface patch: $\gamma(t) = X(u(t), v(t))$.
• $N(t) = N(u(t), v(t))$ is the unit normal of $X$, i.e $N(u,v) = \frac{(X_u \times X_v)(u,v)}{(|X_u \times X_v|)(u,v)}$ and,
• $U(t) = N(t) \times T(t)$ is the normal in the tangent plane,

then we can define the geodesic curvature in terms of the dot product as

$$\kappa_g = \frac{T' \cdot U}{|\gamma'|} = \frac{-T \cdot U'}{|\gamma'|}.$$

A curve on $X$ will be called a geodesic curve is $\kappa_g(t) = 0$ for all $t$.

One problem with the above dot products is that in order to calculate $\kappa_g$,we are using unit length vectors, and this dividing by the length can become very messy. So we ask, is it possible to take $\tilde{T}, \tilde{N}$ and $\tilde{U}$ as multiples of $T, N$ and $U$ and define $\kappa_g$ (and also the normal curvature and geodesic torsion)? To cut a long proof short, the answer is YES. We can. We therefore get another formula to work out the geodesic curvature to be

$$\kappa_g = \frac{\tilde{T}' \cdot \tilde{U}}{|\gamma'| |\tilde{T}||\tilde{U}|} = \frac{-\tilde{T}\cdot \tilde{U}'}{|\gamma'||\tilde{T}||\tilde{U}|}.$$

Notice how both formulas take very similar forms. In the second set, we have divided through by the length of the multiples of $T$ and $U$, instead of using the unit length vectors to start off with and then calculate the curvature.

Answer 2

It is $\mathbb{R}^3$ generalisation of tangential rotation rate in tangent plane. In $\mathbb{R}^2$ the well known $k_g$ formula : $y^{''} /(1+ y'^2 )^ {3/2}$. Liouville's formula is one among several forms.