I was going through these pages one more time recently and I thought "All of this is fine. But is there a tautochrone curve on a sphere?". Turns out the answer is yes and this post is all about finding such a curve.
The simplest way to attack this problem is to use Able's solution idea. It says,
$\displaystyle \frac{ds}{dy}=\pm\sqrt{2g}\frac{T_0}{\pi}\frac{1}{\sqrt{y}}$
where $T_0$ is the constant time of travel, $g$ is the gravitational constant, $s$ is the arc length of the curve and $y$ is the measure of the gravitational potential energy.
Let's consider a unit sphere centered at the origin. Also, Let the plane tangent to the south pole of the sphere be the 'base' for measuring the gravitational potential.
As we are working on a spherical surface, we'll naturally use spherical geometry for this problem with $\theta$ as the polar angle and $\phi$ as the azimuthal angle. We have,
$y=1+\cos\theta$ which implies $dy=-\sin\theta d\theta$.
Therefore, Abel's equation becomes,
$\displaystyle ds=\sqrt{2g}\frac{T_0}{\pi}\frac{\sin\theta}{\sqrt{1+\cos\theta}}d\theta=\sqrt{2g}\frac{T_0}{\pi}\sqrt{1-\cos\theta}\text{ }d\theta$
The differential arc length in a spherical coordinate system is given by,
$\displaystyle \frac{ds}{d\theta}=\sqrt{1+\sin^2\theta\left(\frac{d\phi}{d\theta}\right)^2}$
Using this and simplifying, we get
$\displaystyle \sin^2\theta\left(\frac{d\phi}{d\theta}\right)^2=2g\left(\frac{T_0}{\pi}\right)^2(1-\cos\theta)-1$
Now, let $2g(T_0/\pi)^2=1/(1-\cos\lambda)$ where $\lambda$ is a constant. Then the above equation with further simplication becomes,
$\displaystyle \frac{d\phi}{d\theta}=\frac{1}{\sin\theta}\sqrt{\frac{\cos\lambda-\cos\theta}{1-\cos\lambda}}$
This is the differential equation that must be satisfied by the tautochrone curve.
We can solve this with the same substitution used in the Mathworld page. We then have,
$\displaystyle \phi=\cot(\lambda/2)\tanh^{-1}\left(\sqrt{\frac{\cos\lambda-\cos\theta}{1+\cos\lambda}}\right)-\tan^{-1}\left(\sqrt{\frac{\cos\lambda-\cos\theta}{1-\cos\lambda}}\right)$
The above expression should also have an arbitrary constant of integration, but let's say we only care about the shape of the curve and hence leave it out of this analysis.
Also, from the expression above, we see that the curve stats at $\theta=\lambda$ and spirals its way down to the south pole. We thus know, once we specify $T_0$, we indirectly specify where the curve should start from and vice versa.
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| Tautochrone curve with $\lambda=\pi/8$ |
We know the time of descent of the tautochrone curve is $T_0$. The arc length $L$ of this tautochrone curve is given by,
$\displaystyle L(\theta_0)=\int ds=\int \frac{ds}{d\theta}d\theta=\sqrt{2g}\frac{T_0}{\pi}\int\limits_{\theta_0}^\pi \sqrt{1-\cos\theta}d\theta=\sqrt{2g}\frac{T_0}{\pi}\cdot2\sqrt{2}\cos(\theta_0/2)$
or in terms of $\lambda$, we have,
$\displaystyle L(\theta_0)=2\frac{\cos\theta_0/2}{\sin\lambda/2}$
All of this absolutely amazing. But the story isn't complete yet. There is something more to this tautochrone curve that truly blows our mind which I'll share in the next post. Hope you enjoyed the post.
Mathematica code for the above curve:
Clear["Global`*"];
\[Alpha] = 45 Degree/2;
Phif[\[Theta]_] :=
Cot[\[Alpha]/2] ArcTanh[Sqrt[(Cos[\[Alpha]] - Cos[\[Theta]])/(
1 + Cos[\[Alpha]])]] -
ArcTan[Sqrt[(Cos[\[Alpha]] - Cos[\[Theta]])/(1 - Cos[\[Alpha]])]];
Cartf[\[Theta]_] := {Cos[Phif[\[Theta]]] Sin[\[Theta]],
Sin[\[Theta]] Sin[Phif[\[Theta]]], Cos[\[Theta]]};
Show[Graphics3D[{Opacity[1/5], White, Sphere[]}],
ParametricPlot3D[Cartf[\[Theta]], {\[Theta], 0 Degree, 180 Degree},
PlotStyle -> Thick, ColorFunction -> (Hue[0.8 #4] &)],
ParametricPlot3D[{Cos[t], Sin[t], 0}, {t, 0, 2 Pi}],
ParametricPlot3D[{Cos[t], 0, Sin[t]}, {t, 0, 2 Pi}],
ParametricPlot3D[{0, Cos[t], Sin[t]}, {t, 0, 2 Pi}],
ImageSize -> Large, Boxed -> False]
See here to paste this code into Mathematica with proper formatting.
Until then
Yours Aye
Me
