Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Formulating my question seems to have given me the answer: that the point will continue getting closer to the pole but never reach it. Am I correct?

Edit in response to Martin Argerami:
I see your point. So let's make the point always move at an angle x to the latitude line that includes the point. Then will the point reach a pole?

share|cite|improve this question
What's your definition of "angle to the equator"? The analog of a straight line on a sphere is a maximum circle, which of course will never touch the poles if it is not perpendicular to the equator. – Martin Argerami Jul 13 '13 at 5:39
If we travel on a rhumb line, we sure will get there, if we don't slow down too much as things get cold. – André Nicolas Jul 13 '13 at 5:46
@AndréNicolas I was about to correct you when I remembered Zeno's paradox. You will get to the pole, but you will walk an infinite number of times around it first, within finite time. – Arthur Jul 13 '13 at 8:54
up vote 1 down vote accepted

Spherical coordinates $(a,b)$ of a point on the unit sphere are related to its Cartesian coordinates $(x,y,z)$ through the relations $$ (x,y,z)=(\cos a\cos b,\sin a\cos b,\sin b). $$ In the Northern hemisphere, the longitude $b$ is an angle in $[0,\frac\pi2]$ and the latitude $a$ is an angle in $\mathbb R/2\pi\mathbb Z$. Assume that the point is at $(a(0),b(0))=(0,0)$ at time $0$, is at $(a(t),b(t))$ at time $t\geqslant0$, and moves at constant speed, crossing parallels at a constant angle.

Then, writing down the direction of the parallel at point $(x,y,z)$ as $(-y,x,0)$, that is, the direction of the parallel at point $(a,b)$ as $(-\sin a,\cos a,0)$, and the speed vector at time $t$ as $$ (-\sin a,\cos a,0)\cos b\cdot a'+(-\cos a\sin b,-\sin a\sin b,\cos b)b', $$ one gets the conditions that $$ (\cos b)^2(a')^2+(b')^2,\qquad\cos b\cdot a', $$ should both be constant. That is, $b(t)=\beta t$ for some $\beta\gt0$ and $a'(t)=\alpha/\cos(\beta t)$ for some positive $\alpha$, the ratio $\alpha/\beta$ characterizing the angle the particle crosses parallels at.

Thus, the point is at the North pole at time $t_N=\pi/(2\beta)$, which is finite. Since the point moves at constant speed, the total distance it moved when it reaches the North pole is finite. But the number of turns around the North-South axis is described by $$ a(t_N)=\int_0^{t_N}\frac\alpha{\cos(\beta t)}\mathrm dt=\frac\alpha\beta\int_0^{\pi/2}\frac{\mathrm dt}{\cos t}, $$ which diverges, hence it is infinite.

share|cite|improve this answer
The point moving at a constant speed was not built into my question. The answer to my question remains "yes" even if the speed isn't constant. It could even stop repeatedly for finite lengths of time, no? Just wanting to touch all the bases.... – NotSuper Jul 13 '13 at 11:46
The point moving at a constant speed was not built into my question... I know, but this case allows to deduce every other case where the speed $s(t)$ at time $t$ is prescribed, by a change of time. – Did Jul 13 '13 at 11:53
Yes, I see. Thanks. – NotSuper Jul 13 '13 at 14:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.