Isochrone on a Spherical surface

The Cycloid in all its grandeur scooped all the glory for itself by way of being both the brachistochrone curve and the tautochrone curve.

But, seemingly out of some dark magic, the semi-cubical parabola makes its way as the (vertical) isochrone curve, a curve on which a bead sliding without friction covers equal vertical distances in equal intervals of time.

In my last post, we found the differential equation of a tautochrone curve on a spherical surface and used it to find the curve. In this post, we kind of continue the same discussion.

Our aim in this post is to find a curve on the surface of a sphere such that a bead sliding (without friction) under the influence of gravity covers equal polar angles at equal intervals of time. In other words, we are trying to find the (polar) Isochrone curve on the spherical surface.

For reasons that'll be apparent later, we modify our problem construct a little. Let's assume that the bead starts at the north pole and slides (without friction) slowly along the $\phi=0$ plane from $\theta=0$ to $\theta=\theta_0$ where the bead seamlessly enters the curve.

Let $\omega=d\theta/dt$ be the constant polar velocity as the bead enters the isochrone curve. Also, note that, because of the way the bead travels before entering the curve, the azimuthal velocity will be zero at the entry point.

Using the differential arclength of a curve on a  (unit) spherical surface, we have,

$\displaystyle \left(\frac{ds}{dt}\right)^2 =\left(\frac{d\theta}{dt}\right)^2+\sin^2\theta\left(\frac{d\phi}{dt}\right)^2=\left(\frac{d\theta}{dt}\right)^2\left(1+\sin^2\theta\left(\frac{d\phi}{d\theta}\right)^2\right)$

Let the plane tangent to the sphere at the south pole be the 'base' from which we measure the gravitational potential energy. Then, using the law of conservation of energy at $\theta=\theta_0$ (just after the bead entered the isochrone) and any point beyond,

$\displaystyle \omega^2\left(1+\sin^2\theta\left(\frac{d\phi}{d\theta}\right)^2\right)+2g(1+\cos\theta)=\omega^2+2g(1+\cos\theta_0)$

Simplifying this,

$\displaystyle \frac{d\phi}{d\theta}=\frac{\sqrt{2g}}{\omega}\frac{\sqrt{\cos\theta_0-\cos\theta}}{\sin\theta}$

It is easy to express $\omega$ in terms of $\theta_0$. Using the law of conservation of energy at $\theta=0$ and $\theta=\theta_0$ (just before the bead enters the isochrone), it is easy to see,

$\omega=\sqrt{2g(1-\cos\theta_0)}$

Using this in the expression above, we finally have,

$\displaystyle \frac{d\phi}{d\theta}=\frac{1}{\sin\theta}\sqrt{\frac{\cos\theta_0-\cos\theta}{1-\cos\theta_0}}$

(WARNING: Now is the time where I give you a fair warning to keep your mind in a secured place because it is about to be blown.)

Looking at the above differential equation, it is clear that it is the same equation we found in our previous post for the tautochrone problem. The same curve solves both the problems just like the cycloid does in plane geometry!!!

This says, if you place the bead at any point on our curve, the time it takes to reach the south pole is the same and hence becomes tautochrone. But if you place the bead at the north pole and let it slide itself into the curve, then it covers equal polar angles at equal intervals of times and hence becomes (polar) isochrone.

Truly, some dark magic stuff going on. I would have never expected in any way that such a weird coincidence would happen in the spherical case. I truly enjoyed this. Hope you did too. See you in the next post.

Until Then

Yours Aye

Me