This question was motivated by Definition of an ellipsoid based on its focal points .

I'd like to avoid terms like ellipsoid, so I'll use terms like one-dimensional ellipse (a normal ellipse in the plane) and two-dimensional ellipse (an ellipsoid in space).

One description for a one-dimensional ellipse in the $x_1x_2$-plane captures the geometry that for any point on the ellipse, the sum of the distances to the foci is constant: $$E_1=\left\{\bar{x}\,\left|\,\sum_{i=1,2}\|\bar{x}-\bar{f}_i\|=c\right.\right\}$$ where $\bar{f_1}$ and $\bar{f}_2$ are the foci and $c$ is some constant. It occurred to me that the set $\{\bar{f}_1,\bar{f}_2\}$ is a zero-dimensional ellipse. So I thought to rephrase the description above as $$E_1=\left\{\bar{x}\,\left|\,\int_{\textrm E_0}\|\bar{x}-\bar{e}\|\,de=c\right.\right\}$$ where $E_0$ is the zero-dimensional ellipse $\{\bar{f}_1,\bar{f}_2\}$ and each point in $E_0$ has measure 1.

What do you get when you upgrade a dimension? In words, what surface is the collection of all points in space where for each of those points, the integrated distance between that point and a one-dimensional ellipse is constant? I.e. what surface is $S$ in $\mathbb{R}^3$ where $$S=\left\{\bar{x}\,\left|\,\int_{\textrm E_1}\|\bar{x}-\bar{e}\|\,de=c\right.\right\}$$ where $E_1$ is a one-dimensional ellipse in $\mathbb{R}^3$? Answers could be implicit equations in $x_1$, $x_2$, and $x_3$, or something else.

At first I thought it would be neat if this gave a generic two-dimensional ellipse, providing an answer to the question linked above, but I have convinced myself that this is not so.

  • $\begingroup$ If $C$ is small enough, in the special case that $E_1$ is a circle, I think you should get a torus. Does that sound right? $\endgroup$ – Simon May 21 '11 at 23:09
  • $\begingroup$ @Simon: That keeps a constant minimum distance from points on the surface to the circle. I'd like to keep the integrated distance (ID) constant. A torus won't be the surface, since points on the outer rim have a greater ID from the circle than points on the inner rim. Some examples: the center of a circle of radius 1 has an ID of $2\pi$ ($\int_0^{2\pi} 1\,dt$) from the circle, and I'm pretty sure $2\pi$ is the minimal possible ID. A point on the circle has ID $4\pi$ ($\int_0^{\pi} 2\sin(t)\,dt$), so with $c=4\pi$ the surface would contain the circle, but also other points in space. $\endgroup$ – alex.jordan May 22 '11 at 0:08
  • $\begingroup$ Also, you could divide integrated distance by the length of $E_1$ and have the more common average distance. Keeping the average distance from $E_1$ constant would give the same shapes as keeping the integrated distance constant, just using different values of $c$. I prefer to use integrated distance, since that is the usual form of the equation for the one-dimensional ellipse. $\endgroup$ – alex.jordan May 22 '11 at 0:15
  • $\begingroup$ What do you mean by an integral of the form $\int_{E_1} \ldots de\ $? Are you integrating over the interior or along the curve, or what? $\endgroup$ – Christian Blatter May 22 '11 at 10:00
  • $\begingroup$ @Christian: Along the curve, with respect to arc length. $\endgroup$ – alex.jordan May 22 '11 at 23:37

(I tried to post this as a comment but it mangled the equation.)

Even for a degenerate $E_1 = \{(x,0): -1\le x\le1\}$, the implicit form of $E_2$ is already pretty complicated. Mathematica gives me, after some massaging, $$\begin{align}&\int_{-1}^1 \sqrt{(x-t)^2 + y^2}\ \mathrm dt \\ &\qquad = (1-x)\sqrt{(1-x)^2+y^2}+(1+x)\sqrt{(1+x)^2+y^2} \\ &\qquad +\ y^2\log\left(\frac{\left(1-x+\sqrt{(1-x)^2+y^2}\right)\left(1+x+\sqrt{(1+x)^2+y^2}\right)}{y^2}\right)\end{align}$$ which you want the level sets of. This is for the restriction to the $xy$-plane, but since there is rotational symmetry in this case, one can just replace $y^2$ with $y^2+z^2$ for the full 3D surface.

When $E_1$ is nondegenerate and the arc length of the ellipse comes into play, the equation will probably get much worse.


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