Normal curvature visualization

I am currently studying Meusnier's theorem that tells that the normal curvature of a curve $\gamma$ drawn on a surface $S$ depends only on the direction of $\dot{\gamma}$. I completely agree with the proof but I have no visualization/intuition of what it represents geometrically.

And I have this problem not only for the normal curvature, geodesic curvature and for curvature in general, I know it represents the "tendency to curve" but I still have a weak visualization of what it means

Solution 1:

Signed Curvature. Think of a particle traveling along a curve parameterized at unit-speed. Its velocity at time $t$ is normal to its acceleration. Let $L$ be the line passing through the particle and parallel to its acceleration. Let the time elapse by $\delta t$ and repeat the same construction as before getting a new line $L'$. The intersection $L\cap L'$ depends on $t$ and $\delta t$ and converges to a point $\mathbf c$ on $L$ as $\delta t\to0$. The curvature $\kappa$ is the reciprocal of the distance between $p$ and $\mathbf c$. It is positive if $\mathbf c$ is in the direction of the normal and negative otherwise. The circle with center $\mathbf c$ and radius $1/|\kappa|$ has the same tangent at the particle than the curve.

Normal Curvature. Consider a plane $\Pi$ in $\mathbb R^3$ including the normal vector $\mathbf n(p)$ at a given point $p\in S$. The intersection $\Pi+p\cap S$ can be visualized as a curve on $S$. The intersection $\Pi\cap T_p(S)$ is a one-dimensional subspace, i.e., a line. Let $\mathbf v$ be a unit vector on $\Pi\cap T_p(S)$. Note that $\{\mathbf v,\mathbf n(p)\}$ is an orthonormal basis of $\Pi$. In this regard, having $\Pi$ can be considered equivalent to having $\mathbf v$. In addition, $\mathbf v$ is an element of the tangent to the curve $\Pi+p\cap S$. If $\mathbf v_1$ and $\mathbf v_2$ are the principal directions of $S$ at $p$ and we are assuming that $\|\mathbf v\|=1$, we have $$ \mathbf v = (\cos\theta)\mathbf v_1 + (\sin\theta)\mathbf v_2 $$ for some convenient value of $\theta$. Since we have some freedom to fix the origin of $\theta$ we may consider it to be the angle measured from $\mathbf v_1$ to $\mathbf v_2$. In other words, $\theta=0$ corresponds to $\mathbf v_1$ and $\theta=\pi/2$ to $\mathbf v_2$. By definition the normal curvature at $\mathbf v$ is \begin{equation}\label{eq119-1} \kappa_{\mathbf n} = \mathbb{II}_p(\mathbf v) = k_1\cos^2\theta + k_2\sin^2\theta. \end{equation} Using this information it can be shown that $\kappa_{\mathbf n}$ is the signed curvature of the (plane) curve $\Pi+p\cap S$ at $p$. As $\Pi$ rotates from $0$ to $\pi$, the intersecting curves run over all the possible values of $k_{\mathbf n}$ (which get repeated between $\pi$ and $2\pi$).

In sum, the normal curvature of $\gamma$ at $t_0$ is the signed curvature of the plane curve defined by the intersection of $S$ with the plane spawn by $\mathbf n$ and $\gamma'(t_0)$.

Osculating Plane. A particle traveling along a spatial curve would experience acceleration on any turn of said curve. This acceleration would be caused by the force impelled by the curve on the particle, aimed at annihilating the particle's natural inertia and deviate the direction of its velocity vector so as to constrain it to the trajectory dictated by the curve. Since (in the unit-speed case) the curve can only push in directions that are normal to it, the acceleration of the particle must be in a normal direction too. In this regard, we can see that the motion of the particle is instantaneously confined to the plane defined by the velocity and the acceleration vectors. The osculating plane of the (spatial) curve at a given point is the plane spawn by its tangent and normal vectors.

Geodesic Curvature. In the case where the spatial curve lies on a regular surface, we have two planes: the osculating one and the tangent to the surface. Even though both planes include the vector tangent to the curve, they are in general different because the normal to the curve is not the normal to the surface: the former points to the center of the curvature circle of the curve, the latter is influenced by all the other curves laying on the surface and passing through the point. In particular, we can project the derivative of the tangent vector (i.e., the acceleration of the particle traveling along the curve) onto the tangent plane. The geodesic curvature is the component (a.k.a. shadow) of the acceleration vector $T'$ onto the tangent plane.

For lots of wonderful geometrical insights, explanations and proofs refer to Visual Differential Geometry and Forms by Tristan Needham.