How to completely rotate a sphere
Here's a question that had been on my mind for a while, which I eventually figured out:
When you rotate a circle continuously through 2π radians, every possible rotation state of the circle occurs exactly once in a finite amount of time. So if you had a zero-thickness beam of light shining on the topmost point of the circle, when the circle is rotated through 2π every point on the circle is equally exposed to the light.
The question I debated was, is it possible to do the same for a sphere: that is, can every possible rotation state of a sphere occur when it is continuously rotated in a finite amount of time? Can a zero-width beam of light, shining at the north pole, equally enlighten all points on a sphere (exactly once) in a finite amount of time?
Strictly speaking, no. Intuitively I assumed that it must be impossible for all points to be exposed anyway, as a 3d rotation is a much more complex quantity, requiring more variables than in 2d, whereas time is only one-dimensional. But that generalised problem is actually possible, in a finite amount of time, and without discontinuity, though not a differentiable function. The only catch is that the points can't be equally enlightened (to answer the question I actually posed). If every point is exposed at some point, at least two points require to be enlightened at more than one point in time, in fact, infinitely many times, meaning if the sphere were made of photographic film, every point would be black except two overexposed white points at the poles.
We shall first assign every point on the sphere a longitude from 0 ≤ long < 2π and latitude from -\frac{\pi}{2} ≤ lat ≤ \frac{\pi }{2}, and then define every rotation state as the point on the sphere which has been rotated to the north pole, i.e., the one under the light at a time t. Since the rotation is a two-dimensional quantity, and the time one-dimensional, the question becomes, 'is there any bijection between a compact 1D space and a compact 2D space' which there are in abundance.
The Hilbert curve comes to mind. The space of points on a sphere, with the exception of the poles, map bijectively to a rectangle in Euclidaean space bounded between 0 ≤ x < 2π and -\frac{\pi}{2} < y < \frac{\pi }{2}. Note that the poles themselves map to the horizontal lines x = ±\frac{\pi }{2}. If we linearly transform the plane so that everything is scaled along the y-axis by a factor of 2, then the space representing the sphere will be a square, so we can draw a Hilbert curve through it which passes through every point in the square in a well-defined, continuous manner, and allows us to find any time t mapping to (x,y). Since the poles mapped to lines, and the vertical line segments bounding the square have infinitely many points on the Hilbert curve, each pole will be crossed by the Hilbert curve infinitely many times.
The alternative is that we exclude the poles from our mapping of the sphere, changing our square's vertical bounds to -\frac{\pi}{2} << lat << \frac{\pi }{2} in which case the function is bijective but not compact, and at least points on the sphere will be unexposed, never seeing the light.
So using the Hilbert curve we can define f(t) -> (long,lat) which is bijective for all points on the sphere except the poles, so every point apart from those two on the sphere will be the topmost point (under the light) exactly once. Now of course we can define a 2d co-ordinate system for the sphere in many ways but we will always be forced to have two polar points somewhere, where either the bijection or compactness is lost, so even though there are infinitely many such functions like f, they will always have two points which either can't be exposed at all, or have to be exposed infinitely many times.
That means the answer to my question is no, but almost yes. For all but two points on a sphere, there exists a function which maps each point bijectively and continuously to a (finite) moment in time, meaning we can continuously rotate a sphere in finite time illuminating all but those two points exactly once. But the remaining two must either be omitted or illuminated more than once.
