This is a puzzle based on a purported "refutation of GR" on this webpage, which is also the subject of a subthread in this Hacker News discussion. (Btw, the Hacker News discussion contains my answer to the puzzle if you scroll down enough, so you're on your honor not to look at it until you've tried to answer the puzzle yourself. ) I emphasize that I am not saying this puzzle actually does "refute" GR; of course it doesn't. But it does pose an interesting question about how inertial frames centered on the black hole's horizon work, and I would like to see how people respond to it. Here's the puzzle: an astronaut free-falls radially into a black hole. Just before he reaches the horizon, he launches a probe radially outward at just over the escape velocity at the launch altitude (which will be a bit less than the speed of light because the launch point is above the horizon). Just after he crosses the horizon, he launches a second probe radially outward at a slightly larger speed, relative to him, than the first probe. For concreteness, suppose the first probe is launched at a speed [itex]( 1 - \epsilon ) c[/itex] relative to the astronaut, and the second probe is launched at a speed [itex]( 1 - 1/2 \epsilon ) c[/itex] relative to the astronaut, where [itex]\epsilon << 1[/itex]. Now, it seems obvious that these two probes should separate, because the one launched above the horizon will increase its radius with time (since it's moving at greater than escape velocity), while the one below the horizon will decrease its radius with time (since every object below the horizon does). However, if we look at this scenario in a local inertial frame whose origin is the event of the astronaut crossing the horizon, and in which the astronaut is at rest, we should see the two probes moving closer together! This is because the second probe is moving at closer to the speed of light, relative to the astronaut, than the first probe, so in a local inertial frame in which the astronaut is at rest, the second probe should catch up to the first; with the given speeds above, it should catch up at a rate [itex]1/2 \epsilon c[/itex]. What gives? Btw, I should also emphasize that "tidal gravity" is *not* the answer. Assume the hole's mass is large enough that tidal gravity is negligible; that is, assume that we can set up a local inertial frame of sufficient size to realize the given scenario. (It's fairly easy to show mathematically that this is possible for a hole with large enough mass.) Comments welcome.