Is sorting pairwise distances as hard as sorting arbitrary points? If we have $n$ points in $\mathbb{R^d}$, what is the complexity of sorting the $O(n^2)$ pairwise distances?  

Clearly the complexity is $\Omega(n^2)$ but is there a reduction to show it is as hard as sorting $n^2$ arbitrarily chosen numbers?

As a concrete sub question, is the complexity $\Theta(n^2\log{n})$ in the comparison model?
This is an open question even for one-dimensional point sets.  In this setting, the distance-sorting problem is equivalent to sorting X+Y, where $X$ is the input set and $Y=-X$.