Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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? –  Simon May 21 '11 at 23:09
    
@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. –  alex.jordan May 22 '11 at 0:08
    
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. –  alex.jordan May 22 '11 at 0:15
    
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? –  Christian Blatter May 22 '11 at 10:00
    
@Christian: Along the curve, with respect to arc length. –  alex.jordan May 22 '11 at 23:37
show 2 more comments

1 Answer

(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.

share|improve this answer
    
+1 Neat! I'll think about this some more. –  alex.jordan Jun 12 '12 at 17:17
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.