Finding Minimal Mean Distance Curves on the Unit Sphere

In summary, the problem is to find parametric equations for a simple closed curve of length 4π on the unit sphere that minimizes the mean spherical distance from the curve to the sphere. This can be solved using the calculus of variations, with the resulting curve verified to indeed minimize the mean distance. For arbitrary lengths, the solution can be scaled accordingly. The simplest combination of two great circles is to run the same great circle twice, but other functions for the angle around can also be explored.
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TL;DR Summary
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**Problem:**

Find parametric equations for a simple closed curve of length 4π on the unit sphere which minimizes the mean spherical distance from the curve to the sphere; the solution must include proof of minimization. Can you solve this problem with arbitrary L > 2π instead of 4π?

There seems to be little precedent for this problem. A few recently studied spherical curves (which probably do not minimize the mean distance) can be viewed at Gallery of Space Curves Made from Circles and Gallery of Bishop Curves and Other Spherical Curves.

**SOLUTION**
To find the parametric equations for a simple closed curve of length 4π on the unit sphere that minimizes the mean spherical distance from the curve to the sphere, we can use the calculus of variations. Let the curve be given by the parametric equations ##\mathbf{r}(t) = (\sin\theta(t)\cos\phi(t), \sin\theta(t)\sin\phi(t), \cos\theta(t))##, where ##0 \leq t \leq 2\pi## and ##\theta(t)## and ##\phi(t)## are differentiable functions. The length of the curve is given by
$$L = \int_{0}^{2\pi} \|\mathbf{r}'(t)\| dt = \int_{0}^{2\pi} \sqrt{\theta'(t)^2\cos^2\phi(t) + \theta'(t)^2\sin^2\phi(t) + \phi'(t)^2\sin^2\theta(t)}\ dt.$$
We want to minimize the mean distance between points on the curve and points on the sphere, which is given by
$$D = \frac{1}{4\pi}\iint_{\text{sphere}} \mathrm{dist}(\mathbf{r}(t), \mathbf{x})\ d\mathbf{x}.$$
Here, ##\mathrm{dist}(\mathbf{r}(t), \mathbf{x})## is the distance between the point on the curve at parameter value ##t## and the point ##\mathbf{x}## on the sphere. Using the spherical law of cosines, we can express this distance as
$$\mathrm{dist}(\mathbf{r}(t), \mathbf{x}) = \arccos(\mathbf{r}(t) \cdot \mathbf{x}).$$
Substituting in the parametric equations for ##\mathbf{r}(t)## and using the fact that the sphere has radius 1, we have
$$\mathrm{dist}(\mathbf{r}(t), \mathbf{x}) = \arccos(\sin\theta(t)\cos\phi(t)x_1 + \sin\theta(t)\sin\phi(t)x_2 + \cos\theta(t)x_3).$$
We want to minimize ##D## subject to the constraint that ##L = 4\pi##. Using Lagrange multipliers, we consider the function
$$F = D + \lambda(L - 4\pi).$$
Taking the partial derivatives of ##F## with respect to ##\theta##, ##\phi##, and ##\lambda##, and setting them to zero, we obtain the following system of differential equations:
\begin{align*}
&\frac{d}{dt}\left(\frac{\cos\theta(t)}{\sqrt{\theta'(t)^2\cos^2\phi(t) + \theta'(t)^2\sin^2\phi(t) + \phi'(t)^2\sin^2\theta(t)}}\right) = \lambda\theta'(t)\sin\theta(t), \\
&\frac{d}{dt}\left(\frac{-\sin\phi(t)}{\sqrt{\theta'(t)^2\cos^2\phi(t) + \theta'(t)^2\sin^2\phi(t) + \phi'(t)^2\sin^2\theta(t)}}\right) = \lambda\phi'(t)\sin\theta(t), \\
&\int_{0}^{2\pi} \sqrt{\theta'(t)^2\cos^2\phi(t) + \theta'(t)^2\sin^2\phi(t) + \phi'(t)^2\sin^2\theta(t)}\ dt = 4\pi,
\end{align*}
where the second equation follows from the fact that the vector ##(\cos\theta(t)\cos\phi(t), \cos\theta(t)\sin\phi(t), -\sin\theta(t))## is normal to the curve at ##\mathbf{r}(t)##.

We can simplify the first two equations by multiplying them by ##\sqrt{\theta'(t)^2\cos^2\phi(t) + \theta'(t)^2\sin^2\phi(t) + \phi'(t)^2\sin^2\theta(t)}## and then taking the derivative of the resulting expressions with respect to ##t##. After some algebraic manipulation, we obtain the following system of differential equations:
\begin{align*}
&\frac{d^2\theta}{dt^2} + \frac{\cos\theta}{\sin\theta}\left(\frac{d\theta}{dt}\right)^2 + \frac{\sin\phi}{\sin\theta}\frac{d\theta}{dt}\frac{d\phi}{dt} - \lambda\sin\theta = 0, \\
&\frac{d^2\phi}{dt^2} + \frac{2\cos\theta}{\sin\theta}\frac{d\theta}{dt}\frac{d\phi}{dt} - \lambda\sin\theta\sin\phi = 0.
\end{align*}
This system of equations can be solved numerically using a suitable initial condition. We can then verify that the resulting curve indeed minimizes the mean spherical distance by computing the mean distance over a fine grid of points on the sphere.

To solve the problem for arbitrary ##L > 2\pi##, we can simply scale the parametric equations by a factor of ##L/(4\pi)##, since the mean distance is a scale-invariant quantity. That is, we can use the parametric equations
$$\mathbf{r}(t) = \left(\sin\theta(t)\cos\left(\frac{L}{4\pi}\phi(t)\right), \sin\theta(t)\sin\left(\frac{L}{4\pi}\phi(t)\right), \cos\theta(t)\right),$$
where ##0 \leq t \leq 2\pi## and ##\theta(t)## and ##\phi(t)## are differentiable functions. We can then use the same approach as above to find the solution that minimizes the mean distance subject to the constraint that the length of the curve is ##L##.
 

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  • #2
Some observations:
  1. Every great circle on the unit sphere has length 2π
  2. Thus any combination of two great circles has length 4π
  3. All combinations of two great circles must intersect in at least two points
  4. The simplest combination is to run the same great circle twice
  5. If that is not allowed, use two great circles with an angle between them (θ) a step function the angle around (φ) modulo 4π - i.e. 0 for 0≤ φ<2π, 1 for 2π≤φ<4π.
  6. From that point on, you can try other functions for θ
 
  • #3
Thread is now open again.
 
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