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December 4

Is an infinite sphere circumnavigatable at infinite speed?

In abstract geometry that seems reasonable at first but at infinite radius it becomes a plane which isn't connected to itself anymore. Is there some weird thing in advanced mathematics that allows this? Or is the answer undefined like division by zero? Or it's impossible or both at the same time? Is it possible to travel infinite distance at infinite speed in finite non-zero time? (or infinite time?) Sagittarian Milky Way (talk) 06:15, 4 December 2015 (UTC)[reply]

Infinity divided by infinity is considered an undefined quantity, with the exception where one is dealing with the ratio of two well-defined expressions that approach infinity in reasonably clean ways, in which case the limit can be determined using l'Hopital's Rule. Robert McClenon (talk) 06:29, 4 December 2015 (UTC)[reply]

So if the infinite sphere/infinite speed case was the limit of travelling around a finite sphere in a specified time (which is constant for all spheres), then it would be possible (taking the same specified time), but if it was the limit of travelling around a finite sphere at a specified speed (so it takes longer to circumnavigate larger spheres), it would not be (or at least would be undefined), because it would take infinite time? Is that roughly correct? MChesterMC (talk) 14:53, 4 December 2015 (UTC)[reply]

Wouldn't even an infinite cube not be connected to itself either ergo not circumnavigatable? So at infinite size cubes are not geometrically similar anymore but are either a plane, one with a reachable edge, or one with a reachable corner (which would be 0.125∞ liters and ∞L at the same time). Also, there's absolutely no difference between a sphere, polygon, and one kind of cube at infinite size. Sagittarian Milky Way (talk) 15:33, 4 December 2015 (UTC)[reply]

A plane plus a point at infinity is sometimes treated as a sphere; the Riemann sphere is an important example. Mathematics doesn't really have "speed", but you could parametrize a great circle of the Riemann sphere containing 0 and ∞ by angle; then you'd have a path that left the origin, passed through infinity, and returned to the origin from the opposite direction in a total "time" of 2π (but always a finite speed, except at ∞). -- BenRG (talk) 18:29, 4 December 2015 (UTC)[reply]

@Sagittarian Milky Way: You need to be more precise about what you mean by an "infinite sphere" and "infinite speed" in this context. Also, when you say "connected", I am not sure what you mean. The problem as-stated is ill-posed.--Jasper Deng (talk) 08:09, 6 December 2015 (UTC)[reply]

Infinite miles radius, and infinite miles per hour. By connected to itself I meant something which is "attached to itself" allowing circumnavigation like the surfaces of finite globes, as opposed to a plane which has no way to get back to the starting point without a fairly huge change from the initial compass direction at some point (minimum: infinitesimally over 90° I believe). Sorry for the misunderstanding. Something like closed surface and open surface but not quite as I wouldn't consider holes that go through all the great circle paths but are small enough to not bend the paths that avoid the holes much to make a sphere uncircumnavigatable for instance. Sagittarian Milky Way (talk) 17:40, 6 December 2015 (UTC)[reply]

The problem that we do not have a "path towards infinity" still stands:

Short version: what Robert McClenon wrote.

Long version: Say for example that the sphere does not exert any force, and that we want to limit our acceleration to 1 m/s². That would imply a travel time of 2 pi = 6.283... seconds on a sphere, for we'd have to limit our speed to 1m/s, lest the acceleration needed to stick to the sphere would exceed 1m/s². For a 100-meter sphere, the possible speed would increase to 10m/s, and the time needed would be 20 pi. Increasing the radius without limit would lead to a sequence of ever increasing speeds, yet ever increasing trip times. Neither time nor speed would increase as fast as the radius; both would be proportional to the square root of r.

But what if we didn't have that limitation, but could for some reason pick a speed proportional to r no matter how large r is? You can see that in that case, the angular velocity would no longer depend on r, and the time would be constant, and it would be reasonable to call that the circumnavigation time.

But wait, there's less! If the speed was proportional to r², we could see that the bigger the sphere, the shorter the trip time would become; for an infinite sphere, the trip times would approach zero.

Moral of the story: there is no one-and-only answer to the problem "What happens when speed and radius both approach infinity?" , one has to specify how they do to get a mathematically sound problem. - ¡Ouch! (^{hurt me} / _{more pain}) 10:15, 7 December 2015 (UTC)[reply]

Essentially, this explanation is an excellent elaboration on what it means for the limit I mentioned below to not exist.--Jasper Deng (talk) 18:33, 7 December 2015 (UTC)[reply]

Or alternatively, the curvature of the surface becomes less and less curved and the horizon dip gets smaller and smaller at any given eye height (purely geometrical, no air refraction, bent spacetime, gravitational deflection of light or anything like that) until it becomes completely flat. Sagittarian Milky Way (talk) 17:49, 6 December 2015 (UTC)[reply]

Then this problem essentially reduces to an indeterminate form. To traverse a great circle of radius r at an average speed s requires a time of

{\frac {2\pi r}{s}}

. The two-variable limit

\lim _{(r,s)\to (\infty ,\infty )}{\frac {2\pi r}{s}}

does not exist.--Jasper Deng (talk) 23:36, 6 December 2015 (UTC)[reply]

Ah, okay. I was wondering if there was some math I haven't heard of and might not be able to understand that'd make this possible despite the obvious handicap of the sphere becoming a plane. Sagittarian Milky Way (talk) 00:24, 7 December 2015 (UTC)[reply]

If you choose the speed to be numerically equal to the circumference, $s=2\pi r$ , then the time for circumnavigation ${\frac {2\pi r}{s}}={\frac {2\pi r}{2\pi r}}=1$ is independent of the size of the sphere, and in the limit you have $\lim _{r\to \infty }{\frac {2\pi r}{s}}=\lim _{r\to \infty }{\frac {2\pi r}{2\pi r}}=\lim _{r\to \infty }1=1$ , so you can circumnavigate an infinite sphere in a finite time. If you choose the speed to be numerically equal to the square of the radius, then the infinite sphere is circumnavigated in no time at all. The undefined two-variable limit $\lim _{(r,s)\to (\infty ,\infty )}{\frac {2\pi r}{s}}$ does not influence these results. The fact that neither infinite spheres nor infinite speeds exist does not change the answer either, because logically any conclusion may be drawn from a false premise. In any case the answer to the OP's question is YES. Bo Jacoby (talk) 07:28, 10 December 2015 (UTC).[reply]

@Bo Jacoby: His question does not specify what "kind" of infinity, thus implying the general case, including if I chose the speed to grow slower than linearly with the radius. See Robert McLenon's answer above. This becomes especially clear if we perform this calculations using the hyperreal numbers.--Jasper Deng (talk) 16:40, 10 December 2015 (UTC)[reply]

I'm not sure what "infinite speed" could mean, but it is possible to travel in a finite amount of time an infinite distance by going a finite but unbounded speed. Sławomir
Biały 13:44, 10 December 2015 (UTC)[reply]
- The whole problem is ill-posed, simply put.--Jasper Deng (talk) 18:12, 10 December 2015 (UTC)[reply]

The word "circumnavigatable" means "can it be circumnavigated". It does not mean "will it be circumnavigated in all cases". So the question is well posed and the answer is yes. Bo Jacoby (talk) 00:30, 11 December 2015 (UTC).[reply]