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In other news... recently a friend named William Torres (ordinarily more musical than mathematical) asked me an intriguing question:
Can you rotate a rigid, hollow sphere in space, holding its center fixed (but possibly varying the axis of rotation), in such a way that every point moves the same nonzero total amount?
Note that we're not referring to a point's net displacement; in fact, the net motion of the sphere will always correspond to a rotation about some axis (you shouldn't necessarily find this obvious... see wiki), hence two points will always have zero net displacement. What we're interested in is equalizing the path lengths of points' trajectories--how much 'exercise' they get. Of course, the amount should be finite.
I don't know the answer to William's question, but this doesn't mean it's hard. As a warm-up, see if you can prove the following (which I think I can show): It is possible for each point on Earth (understood as a sphere) for each point to have the same nonzero average wind speed over a 24-hour period (where wind is understood as a continuous vector field on the sphere, with vectors tangent to the sphere's surface, and varying continuously over time).
This contrasts with Brouwer's classical 'Hairy Ball Theorem', which states that at any given time, the wind speed will be zero at two or more points on Earth.
In other news... recently a friend named William Torres (ordinarily more musical than mathematical) asked me an intriguing question:
Can you rotate a rigid, hollow sphere in space, holding its center fixed (but possibly varying the axis of rotation), in such a way that every point moves the same nonzero total amount?
Note that we're not referring to a point's net displacement; in fact, the net motion of the sphere will always correspond to a rotation about some axis (you shouldn't necessarily find this obvious... see wiki), hence two points will always have zero net displacement. What we're interested in is equalizing the path lengths of points' trajectories--how much 'exercise' they get. Of course, the amount should be finite.
I don't know the answer to William's question, but this doesn't mean it's hard. As a warm-up, see if you can prove the following (which I think I can show): It is possible for each point on Earth (understood as a sphere) for each point to have the same nonzero average wind speed over a 24-hour period (where wind is understood as a continuous vector field on the sphere, with vectors tangent to the sphere's surface, and varying continuously over time).
This contrasts with Brouwer's classical 'Hairy Ball Theorem', which states that at any given time, the wind speed will be zero at two or more points on Earth.
Labels: puzzles
posted by Andy D at
3:35 PM
3 Comments:
I'm clearly confused about something, because the points on the axis of rotation (two of them, the eigenvectors of the transformation), will not move at all, and other points move by some nonzero amount.
By
Suresh Venkatasubramanian, at 5:35 PM
This is the issue I tried to address in the paragraph after the statement. If just one rotation is used, then you're right.
But the motions we're considering are more general: for instance, we might first rotate 30 degrees around the x-axis, then 30 degrees around the right axis.
Sure, some two points have zero net displacement from initial to final state (since the composite transformation is also a rotation); but we're not concerned about a point's net displacement thru the experiment, but about the length of the arc in space it traces out as the two rotations are executed as continuous motions.
Since every point is moved by at least one of the two rotations, every point traces out an arc of nonzero length.
We also consider more general motions in which the axis of rotation varies continuously, and the goal is to equalize all the arclengths.
By
Anonymous, at 6:46 PM
I guess I should've said 'move' the sphere, not 'rotate' it, but the fixity of the center is key since otherwise a translation would be a trivial solution.
Maybe 'twiddle' is the best word for this type of motion.
By
Anonymous, at 7:45 PM
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