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Let $\sigma$ denote surface measure on a sphere on $S^2$ and $\nu$ volume measure on the unit ball. The Poisson mapping theorem can be proved directly here: counts of image points in disjoint sets are counts of the original process in their disjoint preimages. They are therefore independent Poisson variables, and their means are the volumes of those preimages.

For the radius map, a measurable set $A\subset [0,1]$ has meanThis follows from spherical coordinates or by subtracting concentric-ball volumes. The disjoint-preimage argument proves that the radius image is a Poisson process with this mean measure of a point process. There is no mass at zero or at one.

The original process has no point at the origin almost surely, since its mean count there is zero. Thus division by $r$ in the direction map is well-defined almost surely. For a measurable surface set $C\subset S^2$, spherical coordinates giveAgain disjoint surface sets have disjoint cone preimages, giving the required independent Poisson counts. Both pushforward measures are diffuse, so there are almost surely no image collisions; they describe the indicated point sets as well as counting measures.

The two image processes are not independent. They have exactly the same total number of points,Indeed,This alone disproves process independence.

More generally, split the original points into those whose radius lies in $A$ and direction lies in $C$, those satisfying only the first condition, and those satisfying only the second. Counts in these three disjoint regions are independent. The count shared by both images is Poisson with mean $\sigma (C){\int }_A{r}^{2}dr$, soThe fact that Poisson polar projections share their total count explains the dependence despite the product polar intensity $r^{2}dr\sigma (d\omega )$. Conditional on $N$, the points are independent uniform samples from the ball. Their radius density is $3r^2$ and their direction law is $\sigma /(4\pi )$, independently for each point. Thus the two image configurations are independent conditional on $N$; unconditionally the shared nonconstant count couples them.