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?