Do things move towards you or away from you in a black hole? If one is falling towards the singularity inside a black hole along with a collection of disconnected objects surrounding you, would they appear to move towards you (because everything is contracting and falling towards the singularity) or would they appear to be moving away from you (because of the differential acceleration)?
I'm inclined to think they'd appear to be moving away from you because that's how spaghettification happens, with one end of you being pulled in faster than the other end. But what about for objects that are at the same radial distance from the singularity as you are. Would they move away, get closer, or appear not to be moving?
Furthermore, if the answer is that things appear to move away from you, is that movement uniform in all directions? If you were surrounded by a cloud of dust falling in with you, if you looked in any direction would it all appear to be moving away from you at the same rate (proportional to the distance from you) for a sufficiently large black hole?
 A: To answer this question I will use the geodesic deviation equation. This describes the relative motion of two test masses separated by a vector $u^\alpha$. This equation is
$$
\frac{dU^\alpha}{ds}~+~R^\alpha_{\mu\beta\nu}V^\mu U^\beta V^\nu~=~0.
$$
For the Schwarzschild metric the Riemann curvature tensor components are
$$
R_{trtr}~=~-\frac{2mc^2}{r^3},~R_{t\theta t\theta }~=~\frac{(r~-~2m)mc^2}{r^2},~R_{t\phi t\phi }~=\frac{(r~-~2m)mc^2sin^2\theta}{r^2}
$$
$$
R_{r\theta r\theta}~=~-\frac{m}{r~-~2m},~R_{r\phi r\phi}~=~-\frac{m~sin^2\theta}{r~-~2m}, R_{\theta\phi\theta\phi}~=~2mr~sin^2\theta.
$$
We first consider the deviation between two masses separated by a radial distance so the geodesic deviation equation is
$$
\frac{dr}{ds}~+~g^{rr}(R_{rtrt}V^trV^t~+~R_{r\theta r\theta}V^\theta r V^\theta~+~R_{r\phi r\phi}V^\phi r V^\phi)~=~0.
$$
I also recognize that the first term in $O(c^2)$ larger, so I really only need to consider this term. For a frame falling at say the center of mass of the two test charges we can set $V^t~=~(1,~0,~0,~0)$ and reduce this differential equation to
$$
\frac{dr}{ds}~+~g^{rr}R_{rtrt}r~=~0,
$$
or
$$
\frac{dr}{ds}~-~\left(1~-~\frac{2m}{r}\right)\frac{2mc^2}{r^3}U^r~=~0.
$$
This is easily solved by recognizing
$$
\frac{\frac{dr}{ds}}{r}~=~\frac{d~lnr}{ds}~=~\left(1~-~\frac{2m}{r}\right)\frac{2mc^2}{r^3},
$$
It is then not hard to see that the radial vector is exponential in the proper interval $s$. Particularly near or in the black hole the separation is huge. This is the basis for so called spaghettification.
Now consider the separation of a vector by $\theta$ coordinate. The geodesic equation again has the relevant term $O(c^2)$ larger with
$$
\frac{d\theta}{ds}~=~-g^{\theta\theta}R_{\theta t\theta t}\theta^2~=~-r^2\frac{(r~-~2m)mc^2\theta^2}{r^2}.
$$
The actual distance between the test masses is of course is given by arclength $r\theta$. It is then not hard to see the distance decreases. This means a sphere of test masses falling radially into a black hole will elongate along the radial direction into an ellipsoid that quickly turns into a line of particles.
