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From reading about general relativity, the event horizon and the cosmological radius are the radius when $f(r)=0$, in the metric $$ ds^{2}=-f(r)dt^{2}+\frac{dr^{2}}{f(r)}+r^{2}d\Omega^{2} $$ However, how to distinguish between these two quantities and what happens if $f(R)=0$ has only one solution $r=2M$ or when it has two solution $r=2M$ and $r=R$ where $R>2M$.

MrDi
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  • Related questions by OP: http://physics.stackexchange.com/q/191013/2451 , http://physics.stackexchange.com/q/203541/2451 and http://physics.stackexchange.com/q/203977/2451 – Qmechanic Sep 09 '15 at 14:45

2 Answers2

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For a spherically symmetric system there are only two possibilities for the function $f(r)$:

$$ f(r) = 1 - \frac{2a}{r} $$

$$ f(r) = 1 - br^2 $$

Though you can combine the two to get:

$$ f(r) = 1 - \frac{2a}{r} - br^2 $$

Any of these forms is a perfectly valid solution of the Einstein equations, and the exact form of $f(r)$ is determined by specifying the initial conditions. For example if we have a non-zero mass but no cosmlogical constant then $a = M$ and $b = 0$ and we get the Schwarzschild metric. If we have no mass but a non-zero cosmological constant then $a = 0$ and $b = \Lambda$ and we get a de Sitter metric. Non-zero $a$ and $b$ give the de Sitter-Schwarzschild metric.

So the existance of one or two zeros for $f(r)$ just depends on what initial conditions you specify. Specifically it requires non-zero mass and non-zero cosmological constant.

John Rennie
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  • In modified $f(R)$ gravity, a solution may have one or more solutions to $f(R)$, independent of the initial conditions. – MrDi Sep 09 '15 at 14:56
  • @MrDi: But your question was about GR was it not? – John Rennie Sep 09 '15 at 15:03
  • My question is about how to distinguish between cosmological radius and event horizon if $f(r)=0$ has more than one solution where $f(r)$ is component of the metric. I don't think it matters whether its general relativity metric or modified gravity metric, does it? – MrDi Sep 09 '15 at 15:17
  • @MrDi: I'm not sure I have grasped exactly what you're unsure of. If you're asking why solution is which then the event horizon is at the smaller value of $r$ and the cosmological horizon at the larger value of $r$. Obviously (at least I think it's obvious) the cosmological horizon cannot be inside the event horizon. – John Rennie Sep 09 '15 at 15:41
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To calculate event horizon...

  1. It is the radius sphere around a black hole through which even light fails to escape

  2. Therefore, light energy$(E)=mc²$
    Energy needed to escape
    i.e potential energy=$\frac{GMm}{R}$

    Therefore, equating $$mc²=\frac{GMm}{R}$$ $$C²=\frac{GM}{R}$$ $$R=\frac{GM}{c^2}$$ Where, $M$ = mass of the black hole notsure if my logic is correct

SchrodingersCat
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elnino rosario
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