1
$\begingroup$

The standard parameterization of the solid sphere of radius $r$ centered at the origin in $3$-space is $$(\rho\cos\theta\sin\phi,\rho\sin\theta\sin\phi,\rho\cos\phi)\quad\rho\in[0,r],\theta\in[0,2\pi],\phi\in[0,\pi].$$ But this has Jacobian $\rho^2\sin\phi$ - among other things, this means that we cannot randomally pick a point on the sphere in an "unbiased" manner (by choosing random values for the parameters $\rho,\theta,\phi$ in their intervals) because some areas of the sphere (the points near the center) are more likely to be picked than others.

My question is: how can we construct an "orthonormal" parameterization from this one, which is still based on spherical coordinates? In other words, can we change this to an "arc-length parameterization" as is commonly done with one-dimensional curves, so that the Jacobian is everywhere $1$?

$\endgroup$
0
$\begingroup$

You can't! If you could, the sphere would be flat (have $0$ Gaussian curvature).

$\endgroup$
  • $\begingroup$ +1 Thanks - would I learn that in differential geometry? $\endgroup$ – user142299 Jun 18 '14 at 20:58
  • $\begingroup$ Absolutely! One of the most basic results called Gauss's Theorema Egregium. $\endgroup$ – Ted Shifrin Jun 18 '14 at 21:10
  • $\begingroup$ We both initially read OP's question as about a sphere, the 2D object. But OP is actually asking about the 3D ball, which is flat, no? $\endgroup$ – alex.jordan Jun 19 '14 at 4:53
  • $\begingroup$ My interpretation — perhaps flawed — was that the OP sought to keep the same parameter curves but change coordinates to reparametrize them all by arclength. That's what I was addressing. ("still based on spherical coordinates") $\endgroup$ – Ted Shifrin Jun 19 '14 at 10:46
0
$\begingroup$

To choose a point uniformly randomly on the surface of a sphere, see this.

The paramatrization here is not nice at the caps, so differential geometry rules don't apply there. And it may bother you that the caps are not "as likely" to be chosen randomly as other individual points on the sphere. (Their $0\%$ likelihood is different from other points $0\%$ likelihood.) But integrating this distribution over any area of size $A$ will give you the same result.

$\endgroup$
  • $\begingroup$ Quick question: they re-write the surface area element $d\Omega=\sin\phi\,d\theta\,d\phi=-d\theta\,d(\cos\phi)$. Why can we not write $dV=\rho^2\sin\phi\,d\rho\,d\phi\,d\theta=-(1/3)d(\rho^3)\,d(\cos\phi)\,d\theta$, suggesting that we take $\rho=\sqrt[3]{R}$, $\phi=\cos^{-1}(2v-1)$, and $\theta=2\pi u$? $\endgroup$ – user142299 Jun 18 '14 at 21:08
  • $\begingroup$ Are you asking about a way to distribute the volume inside a ball of a certain radius? That sounds like a valid thing to do as long as you bound an normalize the radial variable. Now points along the $z$-axis are omitted from consideration for lack of smoothness. And the center of the sphere would be doubly so due to the cube root function's behavior at $0$. $\endgroup$ – alex.jordan Jun 18 '14 at 21:15
  • $\begingroup$ Why isn't that inconsistent with @TedShifrin 's answer? $\endgroup$ – user142299 Jun 18 '14 at 21:21
  • $\begingroup$ Because of the point I added to my answer about the lack of smoothness at the caps. Ted's answer is for your direct question about a constant Jacobian. Mine is to get to your underlying purpose of selecting a point uniformly randomly. $\endgroup$ – alex.jordan Jun 18 '14 at 21:22
  • $\begingroup$ The method at the link has constant Jacobian everywhere except at the poles where it is undefined. What Ted cites applies to parametrizations where the Jacobian is everywhere defined. $\endgroup$ – alex.jordan Jun 18 '14 at 21:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy