# Are normal curvature and geodesic curvature independence of choice of curves?

By intuition, if the direction of tangent of a point $P$ is given, I think the curve passing through $P$ on the surface have only one choice (locally). So, does $T'(s)$ only depend on the given tangent direction but not the choice of curves? And if so, does it mean both normal curvature and geodesic curvature are independence of choice of curves? (it seems not but why?)

Let $T(s)$ be an arc length parametrized curve passing through point $P$ on the surface $r=r(u,v)$. The curve can be parametrized as $r(u(s),v(s))$.

Then, $$T(s)=\frac{d}{ds}r=r_u\frac{du}{ds}+r_v\frac{dv}{ds}$$

$$T'(s)=\frac{d}{ds}\left(r_u\frac{du}{ds}+r_v\frac{dv}{ds}\right)=r_{uu}\left(\frac{du}{ds}\right)^2+2r_{uv}\frac{du}{ds}\frac{dv}{ds}+r_{vv}\left(\frac{dv}{ds}\right)^2$$

From the definition of normal curvature (let $n$ be the unit normal vector of the tangent plane):$$\kappa_n=T'(s)\cdot n\left(u(s),v(s)\right)$$

$$=r_{uu}\cdot n\left(\frac{du}{ds}\right)^2+2r_{uv}\cdot n\frac{du}{ds}\frac{dv}{ds}+r_{vv}\cdot n\left(\frac{dv}{ds}\right)^2$$ $$=\frac{r_{uu}\cdot n\left(du\right)^2+2r_{uv}\cdot n dudv+r_{vv}\cdot n (dv)^2}{ds^2}$$

$$=r_{uu}\cdot n\left(\frac{du}{ds}\right)^2+2r_{uv}\cdot n\frac{du}{ds}\frac{dv}{ds}+r_{vv}\cdot n\left(\frac{dv}{ds}\right)^2$$ $$=\frac{r_{uu}\cdot n\left(du\right)^2+2r_{uv}\cdot n dudv+r_{vv}\cdot n (dv)^2}{r_{u}\cdot r_{u}\left(du\right)^2+2r_{u}\cdot r_{v} dudv+r_{v}\cdot r_{v} (dv)^2}$$

As seen from above, since $r_u$, $r_v$, $r_{uu}$, $r_{vv}$, $r_{uv}$, $n$ are fix at a point, normal curvature depends only on $du$ and $dv$, which is the direction of the tangent vector, but not the choice of the curve. THAT IS THE CASE.

Form the definition of geodesic curvature: $$\kappa_g=T'(s)\cdot (n\times T)$$ the only difference from above is changing $n$ into $n\times T$. Since $T(s)$ is the tangent vector which is given, in the same sense, geodesic curvature also depends only on $du$ and $dv$, which is the direction of the tangent vector, but not the choice of the curve. However, IT IS NOT THE CASE.

Why?

"By intuition, if the direction of tangent at a point $P$ is given, I think the curve passing through $P$ on the surface have only one choice (locally)." This is wrong. Already in the $(x,y)$-plane there are many different curves passing through $(0,0)$ and having a horizontal tangent there. The curvature $\kappa$ of such curves at $(0,0)$ may take any values between $0$ (inclusive) and $\infty$ (exclusive). Think of the circles $x^2+(y-r)^2=r^2$, resp. the limiting line $y=0$.

Both depend on choice of the functional dependence in 3-space. Curvature $\kappa$ has two components $\kappa_n$ and $\kappa_g$ ; $\kappa^2 = \kappa_n^2 + \kappa_g^2$.

Geodesic curvature $\kappa_g$ is an isometric invariant, depends only upon first fundamental form coefficients $E$, $F$ and $G$ their partial derivatives, Christoffel coefficients. It (along with tangential rotation and Integral curvature in Gauss-Bonnet Theorem) do not vary in isometric transformation. They are all natural or intrinsic.

Normal curvature is given by the second fundamental form, changes much by isometry.

Gauss, by his Egregium Theorem showed that product of principal normal curvatures however is an isometric invariant, in a proof that needs some patience to follow.

• At a point P normal curvature is same for a given tangent direction, but depending on $k_g$ the curves are many.They have a common tangent at P. Departure from geodesy is measured by geodesic curvature. – Narasimham Aug 29 '14 at 23:31

The geodesic curvature measures intuitively how much a curve on a surface deviates from the geodesic with the same initial direction, so it is an invariant tied in with the second derivative of the curve. In fact, in the plane the geodesic curvature is the same as the ordinary curvature of a curve, so you should understand what is going on in the flat (plane) case first.

Note that $T'(s)$ is the second derivative of the curve itself (with respect to the arclength parameter), so one does not expect to be able to account for it in terms of first derivatives.

• Actually I understand the meaning and calculation of geodesic curvature. However I am confused by the above derivation. As [Wolfram] shown: $\kappa_n=\frac{I}{II}$ which its derivation is just like what I write above. How come that the same derivation for geodesic curvature is not true? :mathworld.wolfram.com/NormalCurvature.html – Y.H. Chan May 9 '14 at 1:21
• Because it is not "the same derivation", as shown by the formula you cited: $\kappa_g=T'(s)\cdot (n\times T)$. – Mikhail Katz May 9 '14 at 7:07