Geodesic and Normal Curvatures

1. Geodesic and Normal Curvatures: Curvature is a key concept in understanding differential geometry. In particular, for curves on a curved surface, geodesic and normal curvatures are crucial to understand. It is also essential to understand how they are connected to the intrinsic and the extrinsic geometries of a curved surface. Let’s dive into understanding these two concepts.

Suppose $S$ is a smooth surface embedded in $\mathbb R^3$ . Suppose $p$ is a point on the surface. (e.g., consider $\mathbb R^3$ to be the 3-dimensional space of our universe and $S$ to be the earth’s surface or if you wish, some other planet’s surface with a different shape like a squash or a melon).

Your fancy home planet might look like this

Suppose you are an inhabitant standing at the position $p$ on $S$ , of the above planet. In this context, here’s a way to get a feel of what we exactly mean by geodesic curvature and normal curvature of a curve $\gamma$ passing through $p$ and lying on $S$ .

Now, imagine yourself running on the surface $S$ , traversing along the curve $\gamma$ . Assume that the curve $\gamma$ is regular, that is, $\gamma'(t)\neq 0$ for every $t$ , so that we can assume without loss of generality that $\gamma$ can be made unit speed (by reparametrization, if needed).

Imagine yourself running along a curve $\gamma$ on the surface of the home planet $S$ , passing through a point $p$ on the surface.

Assuming that $\gamma(t)$ is a unit speed parametrization i.e. you are travelling at a constant unit speed,

Let’s now analyze your acceleration vector

as you travel along the curve $\gamma$ .

Consider the following frame,

i.e. $\mathbf T$ the unit tangent vector / velocity vector of the unit speed curve, pointing in the direction you are headed towards. The vector,

is the unit normal to the surface $S$ and,

Clearly, the vector $\mathbf N$ lies on the tangent plane of the surface $S$ and it is perpendicular to the velocity vector $\mathbf T$ of the curve $\gamma$ .

Now, If you note that,

then you see that this frame is an orthonormal frame (shown in the picture below).

Focusing back on your acceleration vector, i.e.

decompose it with respect to the orthonormal frame to get,

However, since the curve is constant speed, it means that there is no tangential acceleration. This is clear intuitively, and is also easy to prove (as done below),

Thus, it turns out that,

Consequently, we have,

i.e. the acceleration vector decomposes completely along two orthogonal directions $\mathbf n$ and $\mathbf N$

For you, an inhabitant of the surface $S$ , you have the capability to perceive the intrinsic geometry of $S$ only, (unless you’re not a flatlander being and you smart enough to make spaceships and go to the outer space to see the geometry of the planet from outside).

Assuming so, you can sense only the intrinsic geometry. As you move along the surface, the curvature you can sense during the motion is in terms of your deviation from a straightest path (geodesic) on the surface, and through the point. This is precisely what the

captures, this is the component of your acceleration lying on the tangent plane of the surface, the projection on the tangent plane at that point.

Thus, the signed scalar quantity,

is the intrinsic curvature of the curve of my motion on $S$ and the corresponding vector is called the geodesic curvature vector

In summary, the geodesic curvature of the unit speed curve $\gamma$ is defined by,

This is in fact also the covariant acceleration,

along the curve $\gamma$ .

Thus, we end up getting,

The other component

of the acceleration vector measures how your path of motion bends in space with respect to the tangent plane $T_pS$ and it is perceivable only by intelligent outer-space creatures or astronauts, watching you from the outer space. This component is the so called,

the normal curvature of the curve $\gamma$ .
This is clearly a part of the extrinsic geometry of the surface $S$ .

In conclusion, for a unit speed curve $\gamma$ , we have the total curvature

Decomposition of the ***total curvature*** of the ***unit speed*** curve $\gamma$ into **geodesic curvature** and ***normal curvature***, along the vectors $\mathbf n$ and $\mathbf N$ respectively.

In the case when $\gamma$ is a geodesic, the geodesic curvature of $\gamma$ vanishes, i.e.

Thus, the total curvature is contributed by the normal curvature

in the case of a geodesic curve. This corresponds to the fact that for an inhabitant on the surface, a geodesic is always the straightest of all possible curves on the surface, through a given point.

The geodesic curvature of a curve on a surface is independent of how the surface has been embedded in the ambient Euclidean space. It remains unchanged if we transform the surface by an isometric isomorphism, i.e. bending a surface rigidly without stretching, compressing or shearing etc. (i.e. keeping the angles and lengths unchanged.)