# 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?

• The area is invariant--- this is the first thing one shows about it (although nobody in the literature does, it's the first thing you work out on your own), this is the reason people study it as a physical quantity. – Ron Maimon May 9 '12 at 20:43

• @Nathaniel: The metric for a boosted black hole can be found by using isotropic Schwarzschild (reparametrize r to make the spatial metric be $f(u)(du^2 + u^2 d\Omega^2)$, which gives $-g(u) dt^2 + f(u)(dx^2 + dy^2 + dz^2)$, where $u=\sqrt{x^2+y^2+z^2}$, and then boost the x,t coordinate at in special relativity--- this will give you a squashed horizon in these coordinates, although the whole thing is just described by the metric, and there's not much more to say (other than ray tracing, which is hopeless and mostly unilluminating). – Ron Maimon May 10 '12 at 18:29