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I'm trying to get the mean curvature $K/2=\kappa_1 + \kappa_2$ of a cone with a half angle $\alpha$ at a distance $d$ from the apex.

The curvature along the generator is zero $\kappa_1=0$. As for the curvature of the normal section, I'm trying to calculate it with Meusnier's theorem but I'm stuck as I don't fully understand it. I'm not sure if there's a simpler way to get it.

Here's a schematic: enter image description here

Thanks to everyone that replied.

James Craft
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  • Here’s one way: In $xyz$-space, put the generator on the $x$-axis with the vertex at the origin and the cone lying above $xy$-plane. The bottom half of the cone is the graph of a function, $z=f(x,y)$. Find $f$. Each point on the generator is a minimum point and therefore a critical point. It follows that the Hessian of $f$ at these points is the second fundamental form. – Deane Oct 01 '24 at 03:46
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    Please include a picture/diagram of this. – Nate Oct 01 '24 at 04:20
  • The cone's metric is formally equal to the flat Euclidean metric in polar coordinates. Exercise: calculate Chrisoffel symbols and Riemann tensor. – Kurt G. Oct 01 '24 at 06:29
  • The normal section is a conic. For an ellipse the curvature at vertex is $a/b^2$, where $a$ and $b$ are the semiaxes which can be computed as explained here: https://math.stackexchange.com/questions/2177867/projecting-a-cone-on-a-surface/4077306?noredirect=1#comment8429285_4077306 – Intelligenti pauca Oct 01 '24 at 13:17
  • When you say curvature, do you mean the Gaussian curvature? Or are you wanting to find the principal curvatures individually? And, since there may be some confusion, I assume you mean a right circular cone? – Ted Shifrin Oct 01 '24 at 16:24
  • Thanks for the comments. I updated my question. – James Craft Oct 04 '24 at 22:21

2 Answers2

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According to Meusnier's theorem, if you consider all plane sections of a surface having a common tangent, then the radius of curvature of a section is $r\cos\gamma$, where $r$ is the radius of curvature of the normal section and $\gamma$ is the angle formed with the normal section.

In your case we can take as oblique section the one perpendicular to the axis of the cone, which is a circle with radius $r=d\sin\alpha$ forming an angle $\gamma=\alpha$ with the normal section. Hence the radius of curvature of the normal section is $$ r_N={r\over\cos\alpha}=d\tan\alpha. $$

Intelligenti pauca
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The cone is the submanifold of spherical coordinates with constant angle $\theta$: $$x= r \cos \phi\ \sin \theta, \ y = r \sin \phi \ \sin \theta,\ z=r \cos \theta.$$ It follows that its metrics written as the quadratic form

$$ds^2 = dr^2 + r^2 \sin^2 \theta\ d\phi^2$$

The Lagrangian formalism yields the Euler geodetic equations in terms of the arc parameter $t$

$$\frac{d}{dt} \partial_{\dot r}\ \left(\frac{1}{2} \dot r^2 \right) -\partial_{ r}\left(\frac{1}{2} \left( r^2 \sin^2 \theta \ \dot \phi^2\right)\right)=0 $$ and $$\frac{d}{dt} \partial_{\dot \phi}\ (\frac{1}{2} r^2 \sin^2 \theta \ \dot \phi ^2) =0$$

or $$\ddot r = r \ \sin^2 \theta \ \dot \phi^2$$ $$\ddot \phi = -r \ \sin ^2\theta \ \dot r \ \dot \phi$$ yielding the only two nonvanishing Christoffel symbols in

$${\ddot x}^k = {\Gamma^k}_{lm}\ {\dot x}^l \ {\dot x}^m$$

$${\Gamma^r}_{\phi\phi} = r \sin^2\theta, \quad {\Gamma^\phi}_{r \phi} = {\Gamma^\phi}_{\phi r}- \frac{1}{2} \ r\ \sin^2\theta$$

With these connection forms the Riemann tensor evaluates to zero, the cone is flat and evolvable on a euclidean plane as part of disk (the or lower upper half of the double cone).

Roland F
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  • Thank you for your reply. Unfortunately, I couldn't quite follow your derivation, but I am sure it will be useful for other people that may have the same question. – James Craft Oct 04 '24 at 22:20
  • Reading the question carefully, it appears that OP asks for the mean curvature (extrinsic), not the sectional curvature (intrinsic) – Didier Oct 05 '24 at 09:23