Do black holes ever die? I mean does their mass decrease as they bleed off X-rays and gamma rays? Or does such EM radiation only occur as a result of things being pulled in?
Why I ask is because I'm puzzled at how they factor into entropy and the evenutal Heat Death of the universe. How can the gravitational energy of a singularity ever dissipate?
Stupid Black Hole Question
Moderator: Alyrium Denryle
Eventually, like Omega said. Hawking radiation means they radiate energy at a rate inversely proportional to their mass.
Large (stellar mass and above) black holes radiate energy so slowly that their effective temperature is lower than the cosmic background temperature. So even with no matter around them they should gain mass slowly.
So the supermassive black holes in the centres of galaxies will be with us for a very long time.
Large (stellar mass and above) black holes radiate energy so slowly that their effective temperature is lower than the cosmic background temperature. So even with no matter around them they should gain mass slowly.
So the supermassive black holes in the centres of galaxies will be with us for a very long time.
-
Wyrm
- Jedi Council Member
- Posts: 2206
- Joined: 2005-09-02 01:10pm
- Location: In the sand, pooping hallucinogenic goodness.
The temperature of a black hole is
T = \hbar c³ / 8πkGM
(and, as anything with temperature has entropy associated with it, the entropy of a the black hole is S = Akc³ / 4\hbar G, where for a Schwartzchild black hole, A = π R² where R is the Schwartzchild radius of 2GM/c²)
The luminosity of the black hole is AσT^4, and since luminosity is a power, we can convert this to a mass loss per second by the equation L = c²dM/dt. Area is A = π(2G/c²)² M² = a M² (where we collapse the constant into a). T^4 = (\hbar c²/8πkG)^4 M^{-4} = b M^{-4} (again collapsing the constant into b). Then c²dM/dt = a M² σ b M^{-4} = abσ M{-2}.
Rearrange the differential equation to get:
(c²/abσ) M² dM = dt
Integrate both sides (the first from 0 to M_0 and the second from 0 to τ) and you get
τ = (c²/abσ) M³
where a = π(2G/c²)² and b = (\hbar c²/8πkG)^4. You can then use this equation to crank out a lifetime for a black hole.
T = \hbar c³ / 8πkGM
(and, as anything with temperature has entropy associated with it, the entropy of a the black hole is S = Akc³ / 4\hbar G, where for a Schwartzchild black hole, A = π R² where R is the Schwartzchild radius of 2GM/c²)
The luminosity of the black hole is AσT^4, and since luminosity is a power, we can convert this to a mass loss per second by the equation L = c²dM/dt. Area is A = π(2G/c²)² M² = a M² (where we collapse the constant into a). T^4 = (\hbar c²/8πkG)^4 M^{-4} = b M^{-4} (again collapsing the constant into b). Then c²dM/dt = a M² σ b M^{-4} = abσ M{-2}.
Rearrange the differential equation to get:
(c²/abσ) M² dM = dt
Integrate both sides (the first from 0 to M_0 and the second from 0 to τ) and you get
τ = (c²/abσ) M³
where a = π(2G/c²)² and b = (\hbar c²/8πkG)^4. You can then use this equation to crank out a lifetime for a black hole.
Darth Wong on Strollers vs. Assholes: "There were days when I wished that my stroller had weapons on it."
wilfulton on Bible genetics: "If two screaming lunatics copulate in front of another screaming lunatic, the result will be yet another screaming lunatic.
"
SirNitram: "The nation of France is a theory, not a fact. It should therefore be approached with an open mind, and critically debated and considered."
Cornivore! | BAN-WATCH CANE: XVII | WWJDFAKB? - What Would Jesus Do... For a Klondike Bar? | Evil Bayesian Conspiracy
wilfulton on Bible genetics: "If two screaming lunatics copulate in front of another screaming lunatic, the result will be yet another screaming lunatic.
"SirNitram: "The nation of France is a theory, not a fact. It should therefore be approached with an open mind, and critically debated and considered."
Cornivore! | BAN-WATCH CANE: XVII | WWJDFAKB? - What Would Jesus Do... For a Klondike Bar? | Evil Bayesian Conspiracy