Is a black hole's surface area invariant for distant intertial observers? Let's imagine I'm very far from any massive objects, so my local space-time is Minkowskian. Off in the distance is a black hole, far enough away that it doesn't noticeably curve space-time near me, but close enough for me to see it. Its entropy is proportional to the area of its event horizon. Another observer is moving past me at a high relative velocity. Looking at the same distant black hole as me, does she see it as having the same surface area (and hence the same entropy) as I do?
Naïvely, the event horizon should Lorenz-transform into an oblate spheroid, contracted in the direction of motion but unchanged in perpendicular directions, so it should have a smaller surface area and a smaller entropy. Is this correct (which would suggest that entropy is not Lorenz-invariant after all), or does the event horizon transform in a different way that preserves its surface area?
 A: Yes it is, because null areas (and only null areas) are invariant to the time-slicing, although it isn't completely obvious to someone who isn't day-to-day familiar with Minkowki geometry. This is proved here in the first titled section of this answer: Second Law of Black Hole Thermodynamics 
Perhaps I should say how this proves the intended result: no matter what spacelike surface you slice a stationary black hole with, so long as matter didn't fall in (so that the geodesics on the black hole surface didn't spread apart), the area you cross with the slice is invariant. A boosted observer has a tilted simulteneity plane, so this observer crosses the horizon with a different natural coordinate (although which natural coordinate to use is not specified by GR--- you can use isotropic Schwarzschild coordinates, then boost these by a naive Lorentz transformation (this keeps the asymptotic metric flat), and then continue a little bit into the interior by any method, this requires bending the t-coordinate away from the symmetry direction so that it can cross the horizon). The area of each little triangle on the crossing surface can be slid up and down to match each little triangle on another crossing surface, so the area is independent of the frame.
