Bounding Geodesic Curvature

Question

Let $\Sigma$ be a smooth surface, let $p,q \in \Sigma$ and let $n_{p}$ be the normal at $p$. Suppose that $d(p,q) < c$ for some constant $c$. That is $p$ and $q$ are pretty close to each other on $\Sigma$. Consider the plane $h = \operatorname{span}\{n_{p}, \vec{pq}\}$. The intersection $h \cap \Sigma$ defines a curve $\gamma$ from $p$ to $q$. I want to bound the geodesic curvature of $\gamma$. I have a bound on the principal curvatures in a neighborhood of $p$. This allows me to bound the normal curvature of $\gamma$. I want to bound the geodesic curvature of $\gamma$, so that I can bound the total curvature of $\gamma$. Is there a way to leverage the principal curvatures at $p$, the fact that $\gamma$ lies in a plane defined above, and the fact that $p$ and $q$ are close on $\Sigma$ to bound the geodesic curvature $k_{g}$ of $\gamma$?

Answer 1

If $A$ is the unit normal to the plane (pointing in the direction of $n_p\times \overrightarrow{pq}$), then $$k_g = \kappa(A\cdot n).$$ (One way to see this is to use the formula $(A\times T)\cdot(n\times T) = \left|\begin{matrix} A\cdot n & A\cdot T \\ T\cdot n & 1 \end{matrix}\right|$. Here $A\times T$ will be the principal normal of the plane curve $\gamma$.) This checks since the geodesic curvature of $\gamma$ at $p$ is in fact $0$.

Now, if we define $\theta$ (up to sign) by $A\cdot n = \cos\theta$, then we will have $$k_g = \kappa \cos\theta \quad\text{and}\quad k_n = \kappa\sin\theta,$$ so $$\kappa = \frac{k_n}{\sin\theta} = \frac{|k_n|}{\sqrt{1-(A\cdot n)^2}}.$$

Is this something useful for you?

Answer 2

It is better to remember curvature relations.

The component curvature vectors are given by k , where k = kg + kn, the latter two vectores are orthogonal

They are like forces in vector combination / resolution on the surface points of curve in $\mathbb R^3$.

$$k^2 = k_g^2 + k_n^2 $$

which are the curvature, geodesic and normal curvature components. Non-geodesy when $(\theta =0) $ the curve becomes a geodesic. Components can be found as:

$$ \tan \theta = \frac{ k_g}{k_n} \quad \text{or} \quad \sin \theta = \frac{ k_g} {k}. $$