# Gaussian curvature of an ellipsoid proportional to fourth power of the distance of the tangent plane from the center?

Is it true that the Gaussian curvature of an ellipsoid is proportional to fourth power of the distance of the tangent plane from the center? I can verify that it holds at the places where the major axes intersect the surface. (Mathworld has an equation for the Gaussian curvature, which simplifies at those points.) But verifying that it holds elsewhere seems like it would get ugly.

The motivation for this question is that Lord Kelvin proved that the charge density on a conducting ellipsoid is proportional to the distance of the tangent plane from the center, while McAllister (I W McAllister 1990 J. Phys. D: Appl. Phys. 23 359 doi:10.1088/0022-3727/23/3/016) finds that under certain assumptions, the charge density on a conducting surface is proportional to the fourth root of the absolute value of the Gaussian curvature. However, I think the assumptions of McAllister's result fail for the ellipsoid (actually I only have access to the abstract, so I'm not sure), so it would be nontrivial to learn that this held for the ellipsoid. (The proportionality is definitely not universal. For a pair of conducting spheres that are far apart and connected by a wire, the exponent is not 1/4. For a deep concavity, all of this definitely fails -- you get a a Faraday cage, which excludes the electric field almost completely.)

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I obtained a copy of the McAllister paper. His proof holds only in the following five cases: ellipsoid, hyperboloid of two sheets, elliptic paraboloid, hyperboloid of one sheet, hyperbolic paraboloid. These are the cases in which the Laplace equation is separable in orthogonal coordinates and the Gaussian curvature is nonzero. I don't get any deeper insight from his paper into why it's the Gaussian curvature that matters. He actually begins the paper by explaining why it would normally be the mean curvature you'd care about for this type of problem. – Ben Crowell Aug 24 '12 at 23:14

I think you have a winner. I did it from the recipe at your link, with restrictions $b=a=1$ and $u=0,$ the part about $u$ changing nothing because I made it a surface of revolution. So $$x = \sin v, \; \; y = 0, \; \; z = c \cos v,$$ The normal vector at some point with a chosen value of $v$ is $$(c \sin v, 0, \cos v).$$ The tangent plane there is, allowing $x,y,z$ to vary but $v$ fixed, $$c x \sin v + z \cos v = c.$$ The point of closest approach to the origin within the plane occurs when $x = c z \tan v,$ or $$\left( \frac{c^2 \sin v}{\cos^2 v + c^2 \sin^2 v}, \; \; 0, \; \; \frac{c \cos v}{\cos^2 v + c^2 \sin^2 v}. \right)$$ The distance from the origin is $$d \; \; = \; \; \sqrt{ \frac{c^2}{\cos^2 v + c^2 \sin^2 v} }.$$ The fourth power is $$d^{ \, 4} = \frac{c^4}{(\cos^2 v + c^2 \sin^2 v)^2}$$ Meanwhile, $$K \; \; = \; \; \frac{c^2}{(\cos^2 v + c^2 \sin^2 v)^2}.$$ That is, for an ellipsoid of revolution as described, $$K \; \; = \; \; \frac{d^4}{c^2}$$

Should be a bit harder while restricting only $b=1,$ partly because it is no longer legitimate to restrict $u.$ But I bet it works.

EDIT: now that I have seen this, I suspect there may be an argument purely by multiplying the length of one axis at a time, a linear transformation. Worth a try as well as brute calculation.

EDIT TOOOOO: I finished it, not so bad. You just need to know how to get the normal vector, the plane, and the point of the plane closest to th origin. Given the implicit equation we know the normal is a multiple of $$\left( \frac{x}{a^2}, \; \; \frac{y}{b^2}, \; \;\frac{z}{c^2} \right)$$ or $$\left( bc \cos u \sin v, \; \; ca \sin u \sin v, \; \; a b \cos v \right)$$ and tangent plane $$bc x \cos u \sin v \; + \; ca y \sin u \sin v \; + \; a b z \cos v = a b c.$$ The point in the plane closest to the origin is $$\frac{ abc \left( bc \cos u \sin v, \; \; ca \sin u \sin v, \; \; a b \cos v \right)}{b^2 c^2 \cos^2 u \; \sin^2 v \; + \; c^2 a^2 \sin^2 u \; \sin^2 v \; + \; a^2 b^2 \cos^2 v }$$ with distance $$d \; = \; \frac{ abc}{ \sqrt{b^2 c^2 \cos^2 u \; \sin^2 v \; + \; c^2 a^2 \sin^2 u \; \sin^2 v \; + \; a^2 b^2 \cos^2 v} }$$ and $$d^4 \; = \; \frac{ a^4b^4c^4}{ \left({b^2 c^2 \cos^2 u \; \sin^2 v \; + \; c^2 a^2 \sin^2 u\; \sin^2 v \; + \; a^2 b^2 \cos^2 v}\right)^2 }$$ Since we are told that the Gauss curvature is $$K \; = \; \frac{ a^2b^2c^2}{ \left({b^2 c^2 \cos^2 u\; \sin^2 v \; + \; c^2 a^2 \sin^2 u \; \sin^2 v \; + \; a^2 b^2 \cos^2 v}\right)^2 }$$ we see that $$K = \frac{d^4}{a^2 b^2 c^2}.$$
So there.

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Sweet and lovely! – Ben Crowell Feb 24 '12 at 4:29
Impressive derivation, and a beautiful theorem! – Joseph O'Rourke Feb 24 '12 at 15:53
@Joseph, it also works for hyperboloids, second answer posted. – Will Jagy Feb 24 '12 at 22:02
@Ben, thanks. Good intuition. It also works for hyperboloids. I do not know what status infinite shapes have in electrostatics, but for this problem they are fine. – Will Jagy Feb 24 '12 at 22:04

[This was written before Will generalized his answer to the asymmetric case.]

The following is a partial answer to my own question. It's not a complete answer, though.

The Yacas code below calculates the ratio $q=K/d^4$ of the Gaussian curvature to the fourth power of the distance of the tangent plane from the center. The notation is the same as in the MathWorld article and Will Jagy's answer. I evaluated $q$ at two randomly chosen points on the surface of the same randomly chosen ellipsoid, and it was the same to 40 decimal places. This suggests that the result holds when the ellipsoid is not axially symmetric.

/* Pick an ellipsoid with no special properties, and a point on it with no special properties: */
a := 1.0;
b := 1.23456;
c := 1.31416;

u := 0.789;
v := 0.345;

prec := 40; /* digits of precision */

x := N(a*Cos(u)*Sin(v),prec);
y := N(b*Sin(u)*Sin(v),prec);
z := N(c*Cos(v),prec);

/* Gaussian curvature: */
k := N(a^2*b^2*c^2/(a^2*b^2*Cos(v)^2+c^2*(b^2*Cos(u)^2+a^2*Sin(u)^2)*Sin(v)^2)^2,prec);

/* Find a normal vector: */

nx := N(x/a^2,prec);
ny := N(y/b^2,prec);
nz := N(z/c^2,prec);

/* Make it a unit vector: */
n := N((nx^2+ny^2+nz^2)^(1/2),prec);
nx := N(nx/n,prec);
ny := N(ny/n,prec);
nz := N(nz/n,prec);

/* Distance of tangent plane from center: */
d := N(nx*x+ny*y+nz*z,prec);

/* Is this ratio invariant? */
q := N(k/d^4,prec);

Write(q); NewLine();

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I did the same thing for the "elliptic hyperboloids," of one sheet or two sheets, see http://mathworld.wolfram.com/EllipticHyperboloid.html where they have some pictures and formulas. There is enough detail, it is not necessary to use the parametrizations.

I am going to use a fixed number $$\delta = \pm 1$$ and surface $$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \delta = \frac{z^2}{c^2},$$ or $$\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = - \delta.$$

It is a bit of a mess, but after you put back in all three variables you get

$$K = \frac{\delta a^6 b^6 c^6}{ \left( b^4 c^4 x^2 + c^4 a^4 y^2 + a^4 b^4 z^2 \right)^2}$$

If we eliminate $z$ as on the website, we get

$$K = \frac{\delta a^6 b^6 c^6}{ \left(\delta a^4 b^4 c^2 + b^4 c^2 (c^2 + a^2) x^2 + c^2 a^4 (b^2 + c^2) y^2 \right)^2}$$ which uses, and I kept missing factors in this one, $$a^4 b^4 z^2 = \delta a^4 b^4 c^2 + a^2 b^4 c^2 x^2 + a^4 b^2 c^2 y^2.$$

For the distance, fix some point $(x_0, y_0, z_0)$ on the hyperboloid, find the normal vector $\left(\frac{x_0 }{a^2}, \; \frac{y_0 }{b^2}, \; - \frac{z_0 }{c^2} \right)$ using the gradient of the defining equation, write the tangent plane $$\frac{x_0 x}{a^2} + \frac{y_0 y}{b^2} - \frac{z_0 z}{c^2} = - \delta.$$ Find the multiple of the normal vector (leaving the origin) lying in the tangent plane. Oh, the multiplier is $$t = \frac{-\delta a^4 b^4 c^4}{ b^4 c^4 x_0^2 + c^4 a^4 y_0^2 + a^4 b^4 z_0^2 }$$ So the closest point to the origin is that $t$ times the given normal vector, and the distance of that point from the origin is $$d = \frac{ a^2 b^2 c^2}{ \sqrt{ b^4 c^4 x_0^2 + c^4 a^4 y_0^2 + a^4 b^4 z_0^2} }$$ where at one point we use $|\delta| = 1.$

Alright, dropping the $0$ subscripts, we get $$d^4 = \frac{ a^8 b^8 c^8}{ \left( b^4 c^4 x^2 + c^4 a^4 y^2 + a^4 b^4 z^2 \right)^2}$$ and $$K = \frac{ \delta \; d^4}{a^2 b^2 c^2}$$ A few features to notice: the hyperboloid of one sheet has negative Gauss curvature. The hyperboloid of two sheets has positive Gauss curvature again. The tangent plane, in either case, never passes through the origin itself.

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