Re: [math-fun] rolling a ball along a fractal
Dan Asimov wrote:
This MR abstract of a paper by Henry McKean from 1960 looks relevant:
<< MR0133869 (24 #A3693) 60.62 McKean, H. P., Jr. Brownian motions on the 3-dimensional rotation group. Mem. Coll. Sci. Univ. Kyoto Ser. A Math. 33 1960/1961 25–38 . . . An application describes the motion of a sphere rolling without slipping on a plane while its center performs Brownian motion [see also C. D. Gorman, Trans. Amer. Math. Soc. 94 (1960), 103–117; MR0115218 (22 #6020)].
Thanks, Dan! I wasn't able to find that in JSTOR, but I did find Gorman's article, which constructs Brownian motion on the 3-dimensional rotation group precisely by the tactic of rolling a sphere on the plane so that the point of tangency follows a polygonal approximation to a 2-dimensional Brownian motion on that plane; the main result is that you get a well-defined limit as the mesh of the approximation goes to 0. Note, though, that this doesn't answer the question about the graph of a 1-dimensional Brownian motion. Incidentally, I think there is still more to be said about Dan's nice sphere-rolling puzzle:
PUZZLE: Suppose a unit sphere rolls on the plane, once around the unit circle about the origin.
What net spatial rotation (as a 3x3 matrix) does this impart to the sphere?
Gene Salamin submitted a detailed algebraic solution, culminating in the conclusion that the rotation angle is 2 pi sqrt(1 + r^2). But isn't there a purely geometrical solution, where you somehow unroll the motion of the sphere to see a right triangle with legs of length 2 pi and 2 pi r, sort of like the way we measure the length of a geodesic on a cylinder by unrolling it, but different? Here's a start towards such a picture: Stick a spit through the sphere that pierces it at the antipodal points (r+1,0,1) and (r-1,0,1) so that one end of the spit is at (0,0,1). Now spin the spit around (0,0,1), giving a little bit of a twirl as you do. (Think of the way you would have an ox walk in a circle so as to grind wheat at the center of the circle.) When the rolling sphere has completed one circuit, the spit is still going through it, so we can see that computing the axis of net rotation is easy. But how do we see what the rotation angle is? Jim
I haven't been following most of this discussion, but there's an easy picture for this: On Aug 26, 2008, at 9:22 PM, James Propp wrote:
Gene Salamin submitted a detailed algebraic solution, culminating in the conclusion that the rotation angle is 2 pi sqrt(1 + r^2). But isn't there a purely geometrical solution, where you somehow unroll the motion of the sphere to see a right triangle with legs of length 2 pi and 2 pi r, sort of like the way we measure the length of a geodesic on a cylinder by unrolling it, but different?
Here's a start towards such a picture: Stick a spit through the sphere that pierces it at the antipodal points (r+1,0,1) and (r-1,0,1) so that one end of the spit is at (0,0,1). Now spin the spit around (0,0,1), giving a little bit of a twirl as you do. (Think of the way you would have an ox walk in a circle so as to grind wheat at the center of the circle.) When the rolling sphere has completed one circuit, the spit is still going through it, so we can see that computing the axis of net rotation is easy. But how do we see what the rotation angle is?
Jim
For any circle on the sphere, just look at the cone tangent to the sphere along that circle. If you roll the sphere along the circle, it's the same as rolling the cone on this circle. The cone develops into the plane, so you see that the circle unrolls to a circle in the plane whose radius is its distance to the apex of the cone. This geometric picture is better than the algebraic formula, but of course the formula is immediate. The net angle is given by ratio of arc lengths, i.e. 2 pi * (radius of circle in space)/(distance to apex of cone). ---- Also, once I did (and recommend) the very easy computer experiment of drawing computer-generated Brownian curves equivalent to Brownian rolling of a sphere. One way to do it is take a Brownian path in the plane, then transform by using the same X-coordinates but use the Y- coordinate as a slope. When you see the picture it becomes intuitively clear. This is equivalent to Brownian rolling of a surface because all contact structures are locally isomorphic --- the only variable is the metric on the contact plane fields, which doesn't change the small- scale picture. Bill
participants (2)
-
Bill Thurston -
James Propp