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Perhaps a bit of a strange question. I know the traditional way of finding a black hole's surface area is through the Schwarzschild radius,

$$ r_{\rm{Schwarzschild}}=\frac{2GM}{c^2} $$

however, that equation is formed from the assumption that there is no preference on direction; so it seems odd to do something like M'=gamma M or a length contraction, even when assuming that the observer is far enough away that space-time could be considered flat.

Any help on this would be greatly appreciated!

Qmechanic
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Nathan Stone
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1 Answers1

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Let's say we have a big box. The volume and the surface area of the box is the largest in its rest frame.

Let's say there is a small black hole inside said box.

The black hole shines Hawking radiation equally to all faces of the box. This is an invariant.

Now we can see that the black hole must Lorentz-transform the same way as the box. Right?

stuffu
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  • I suspect you are right, and the black hole should transform the same way as the box. The problem is that this is leading me to a discrepancy in Hawking radiation that I can't seem to fix unless the surface area of the black hole transforms by a factor of $\gamma^{2/3}$, but I can't seem to find where that would fit in – Nathan Stone Jan 22 '22 at 19:23
  • From a naive perspective discounting where the radius equation comes from, I would suspect that the black hole would grow perpendicular to the direction of propagation, and have no change parallel to it since length contraction balances relativistic energy. – Nathan Stone Jan 22 '22 at 19:29