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A black hole is a volume from which matter cannot escape. More formally, the coordinate speed of light at the event horizon - the boundary of a black hole - is zero, as measured by a sufficiently separated observer.

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I recently learned that the BMS supertranslations actually add hairs to black holes and physically change the metric. In such a scenario, I am very curious to know if the black hole can be overcharged …
asked Dec 22 '17 by Feynmans Out for Grumpy Cat
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2answers
Consider a black hole of some radius $R$. Now, suppose a cosmohiker (with her extremely powerful rockets and a glass bowl) enters the black hole. The mass of the cosmohiker along with her belongings i …
asked Jun 21 '17 by Feynmans Out for Grumpy Cat
4
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2answers
The entropy of a Schwarzschild black hole is proportional to $m^2$ where $m$ is the mass of the black hole. The volume of the black hole would be proportional to $m^3$ and the area would be proportion …
asked Jun 28 '17 by Feynmans Out for Grumpy Cat
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3answers
The famous "No-Hair Conjecture" states that a blackhole can have only 3 hairs: Mass, Angular Momentum, and Electric Charge. It occurred to me that the basic underlying reason behind this might be that …
asked Jun 8 '17 by Feynmans Out for Grumpy Cat
0
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1answer
We know that the temperature $T$ of an uncharged non-rotating black hole is inversely proportional to the radius $r$ of the event horizon of the black hole, i.e., $$T \propto r^{-1}$$ Therefore, $$\d …
asked Nov 12 '17 by Feynmans Out for Grumpy Cat
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1answer
It is always said that Schwarzschild black holes are solutions of the vacuum Einstein equation, but I don't quite understand why. I understand that $R_{{\mu}{\nu}}=0$ at each and every spacetime point …
asked Mar 20 '17 by Feynmans Out for Grumpy Cat
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0answers
In "Entropy in Black Hole Pair Production" (arXiv:gr-qc/9306023), Strominger et al. notes The issue of whether (1.2) can be taken literally has bearing on the vexing question of what happens to …
asked Jun 17 '17 by Feynmans Out for Grumpy Cat
12
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1answer
The Vaidya Metric is the metric that can be used to describe the spacetime geometry of a varying mass black hole. This metric reads $$d\tau^2=\bigg(1-\dfrac{2M(\nu)}{r}\bigg)d\nu^2+2d\nu dr - r^2 d\O …
asked Nov 26 '17 by Feynmans Out for Grumpy Cat
1
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0answers
Consider a pair of identical extremal black holes at rest. It is a fact that they exert no net force on each other and remain in static equilibrium. Now, clearly, each of the black holes is in equilib …
asked Sep 13 '17 by Feynmans Out for Grumpy Cat
6
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0answers
Since the universe has a positive cosmological constant, there is an upper limit on the mass of the black holes as evident from the so-called Schwarzschild-de Sitter metric: $ds^2 = -f(r)dt^2 + \dfr …
asked Jun 6 '17 by Feynmans Out for Grumpy Cat
7
votes
1answer
This might be a "metaphysical" question with no actual physics content and if such is the case then I apologize. Actually, to figure out whether or not it is an actual physics question is one of the m …
asked Sep 13 '17 by Feynmans Out for Grumpy Cat
14
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1answer
The famous Hawking calculation is done with an assumption that the background is static, i.e. the evaporation doesn't change the mass parameter in the metric. Thus, we simply describe the geometry usi …
asked Nov 27 '17 by Feynmans Out for Grumpy Cat
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vote
0answers
When we put the BPS condition and the extremality condition together on the most general black hole solution in $AdS_5$ (with minimally gauged supergravity), we get that the relation between the horiz …
asked Jul 8 '17 by Feynmans Out for Grumpy Cat
5
votes
2answers
The Reissner Nordstrom metric $$ds^2 = - \bigg(1-\dfrac{2M}{r}+\dfrac{4\pi Q^2}{r^2}\bigg)dt^2 + \dfrac{1}{\bigg(1-\dfrac{2M}{r}+\dfrac{4\pi Q^2}{r^2}\bigg)} dr^2 + r^2 d\Omega_2^2 $$ is a solution …
asked Jun 10 '17 by Feynmans Out for Grumpy Cat