Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

1 Answer

up vote 3 down vote accepted

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.

share|improve this answer
    
Will the form of the BH be spherical for all observers? –  Anixx May 10 '12 at 4:43
    
Wow, thanks, that proof was very helpful. I'm unable to thoroughly understand the connection to general relativity, but for now I'm willing to accept that it works. I'll accept the answer if you can comment on what the black hole would look like through my telescope - am I right in guessing that it would appear to be an oblate spheroid, but that the area I would estimate from this formula en.wikipedia.org/wiki/Spheroid#Surface_area is simply different from the area as rigorously defined in GR? –  Nathaniel May 10 '12 at 8:36
1  
@Anixx: This question is really impossible to answer--- it is what it is. It depends on your coordinate system what you declare the "shape" of the black hole is, it doesn't appear to be a sphere visually even when standing still. The total cross-section of light from far behind that it absorbs is equal to its surface area (a surprising result of Susskind's), but it deflects light in crazy ways but in general, so you won't see something obvious even when stationary. When highly boosted, you will see absorption at a given energy further out, so it becomes t'Hoofts universal boosted solution. –  Ron Maimon May 10 '12 at 18:26
1  
@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
    
@RonMaimon Thanks! –  Nathaniel May 11 '12 at 8:45
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.