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Parabola: the set of points in the plane that are equidistant from a line, called the directrix, and a point, called the focal point, not on the line.

Suppose we try to replicate this on a sphere: let the directrix be given as a great circle $D$. Let $C$ be a great circle perpendicular to $D$. Let $P$ be our focal point, lying on $C$ (but not lying on the directrix $D$). Let e be an arbitrary point on $D$, and let $E$ be the great circle passing through e, perpendicular to $D$. Let a great circle through $P$ meet $E$ at point f such that f is equidistance from e and $P$. Finally, let our "parabola" consists of all such points equidistant from focal point $P$ and directrix $D$ (measured along the "straight lines" of great circles).

From what I can tell, there are only two possibilities for the shape of this "parabola": either (1) $P$ will lie on one of the apexes of the circle (that is, one of the two points where great circles perpendicular to $D$ intersect), in which case the "parabola" will be a small circle; otherwise, (2) $P$ will lie at some other point along $C$, in which case I'm uncertain about the shape, other than that it will be symmetrical with respect to $C$ and occupy one hemisphere (since it will lie entirely above $D$).

What will be the shape of the curve under condition (2)?

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  • $\begingroup$ This strikes me as a pretty interesting question which might be handled well by some curve/surface differential geometry in $\Bbb R^3$! $\endgroup$ – Robert Lewis Apr 6 '14 at 3:51
  • $\begingroup$ @Ryan Cool, great question! $\endgroup$ – user98602 Apr 6 '14 at 3:54
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We can set this up with only a little algebra and trigonometry. Let $P$ be the north pole of the sphere $x^2+y^2+z^2=1$, also parameterized with polar angle $\theta\in[0,\pi]$ (where $\theta=0$ is the north pole), and radial angle $\phi\in[0,2\pi)$. The planes $x=0$ and $y\cos\alpha+z\sin\alpha=0$ are perpendicular, so the great circles $C$ and $D$ formed by their intersections with the sphere are also perpendicular. We want the angle from the north pole to equal the angle to $D$. The angle to $D$ is $\frac\pi2$ minus the angle to the perpendicular to $D$, and the cosine of this angle can be calculated by a dot product. The cosine of $\frac\pi2$ minus the angle to the north pole is easy to calculate; this is just $\sin\theta$. Thus we want:

$$\sin\theta=(0,\cos\alpha,\sin\alpha)\cdot(\cos\phi\sin\theta,\sin\phi\sin\theta,\cos\theta)=\cos\alpha\sin\phi\sin\theta+\cos\theta$$

Solving this yields $\theta=\frac\pi2-\tan^{-1}(\csc\alpha-\sin\phi\cot\alpha),$ so a full parametric description of the "parabola" is this equation substituted into $(x,y,z)=(\cos\phi\sin\theta,\sin\phi\sin\theta,\cos\theta)$.

The resulting figure is roughly circular, except near $\alpha=0$, when it narrows to an ellipse with foci at the pole and at the perpendicular to $D$. Here's an animation for varying values of $\alpha\in(0,\pi)$:

                                     

EDIT: If we project this figure onto the plane $z=1$, we transform $(\cos\phi\sin\theta,\sin\phi\sin\theta,\cos\theta)\mapsto(\cos\phi\tan\theta,\sin\phi\tan\theta,1)$, which is the equation for a figure in polar coordinates given by $r(\phi)=\tan\theta(\phi)$. Given our known value for $\theta$, this amounts to

$$r(\phi)=\frac{\sin\alpha}{1-\cos\alpha\sin\phi},$$

which is the equation of a (plane) ellipse. Thus this really is the intersection of an elliptical cone with the sphere.

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  • $\begingroup$ Looks very nice, +1! $\endgroup$ – Robert Lewis Apr 6 '14 at 7:20
  • $\begingroup$ @1950RobertLewis Updated due to incorrect derivation. (I calculated the point halfway between the focus and the line for various points on the line; doing this in euclidean space would just get a line halfway between the directrix and the focus.) $\endgroup$ – Mario Carneiro Apr 6 '14 at 9:31
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    $\begingroup$ +100 if I could. I'm unfamiliar with spherical coordinates, so having the animation there to reify that math really makes this answer come alive for me. Thanks so much. $\endgroup$ – Ryan Apr 6 '14 at 14:56
  • $\begingroup$ Do you have a simple description of the shape? Is it actually an ellipse in 3-space? I suspect it's the intersection of a cone (maybe an elliptical cone) with the sphere, but I haven't ground out the details yet. $\endgroup$ – Hurkyl Apr 6 '14 at 15:06
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    $\begingroup$ @Hurkyl I just checked, and it really is the intersection of an elliptical cone with the sphere. See my edit. (It is not an ellipse in 3-space, because it is clearly non-planar - the only planar figures on a sphere are circles.) $\endgroup$ – Mario Carneiro Apr 7 '14 at 5:49
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We can set up the problem as an algebra problem.

Lines on the circle are great circles; we can choose our coordinates so that the line under question is the equator, and the point has coordinates $(0,a,b)$ with $a,b > 0$.

The closest point from $(x,y,z)$ to the equator lies in the same radial direction from the $z$ axis: that is, it is $\left( \frac{x}{\sqrt{x^2 + y^2}}, \frac{y}{\sqrt{x^2 + y^2}}, 0 \right) $.

Thus, you seek the curve that simultaneously solves

$$ x^2 + (y-a)^2 + (z-b)^2 = \left(x - \frac{x}{\sqrt{x^2 + y^2}} \right)^2 + \left( y - \frac{y}{\sqrt{x^2 + y^2}} \right)^2 + z^2$$ $$ x^2 + y^2 + z^2 = 1$$

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