# Why is a black hole black?

In general relativity (ignoring Hawking radiation), why is a black hole black? Why nothing, not even light, can escape from inside a black hole? To make the question simpler, say, why is a Schwarzschild black hole black?

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It's surprisingly hard to explain in simple terms why nothing, not even light, can escape from a black hole once it has passed the event horizon. I'll try and explain with the minimum of maths, but it will be hard going.

The first point to make is that nothing can travel faster than light, so if light can't escape then nothing can. So far so good. Now, we normally describe the spacetime around a black hole using the Scharwzschild metric:

$$ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \left(1-\frac{2M}{r}\right)^{-1}dr^2 + r^2 d\Omega^2$$

but the trouble is that the Schwarzschild time, $t$, isn't a good co-ordinate to use at the event horizon because there is infinite time dilation. You might want to look at my recent post Why is matter drawn into a black hole not condensed into a single point within the singularity? for some background on this.

Now, we're free to express the metric in any co-ordinates we want, because it's co-ordinate independant, and it turns out the best (well, simplest anyway!) co-ordinates to use for this problem are the Gullstrand–Painlevé coordinates. In these co-ordinates $r$ is still the good old radial distance, but $t$ is now the time measured by an observer falling towards the black hole from infinity. This free falling co-ordinate system is known as the "rainfall" co-ordinates and we call the time $t_r$ to distinguish it from the Schwarzschild time.

Anyhow, I'm going to gloss over how we convert the Schwarzschild metric to Gullstrand–Painlevé coordinates and just quote the result:

$$ds^2 = \left(1-\frac{2M}{r}\right)dt_r^2 - 2\sqrt{\frac{2M}{r}}dt_rdr - dr^2 -r^2d\theta^2 - r^2sin^2\theta d\phi^2$$

This looks utterly hideous, but we can simplify it a lot. We're going to consider the motion of light rays, and we know that for light rays $ds^2$ is always zero. Also we're only going to consider light moving radially outwards so $d\theta$ and $d\phi$ are zero. So we're left with a much simpler equation:

$$0 = \left(1-\frac{2M}{r}\right)dt_r^2 - 2\sqrt{\frac{2M}{r}}dt_rdr - dr^2$$

You may think this is a funny definition of simple, but actually the equation is just a quadratic. I can make this clear by dividing through by $dt_r^2$ and rearranging slightly to give:

$$- \left(\frac{dr}{dt_r}\right)^2 - 2\sqrt{\frac{2M}{r}}\frac{dr}{dt_r} + \left(1-\frac{2M}{r}\right) = 0$$

and just using the equation for solving a quadratic gives:

$$\frac{dr}{dt_r} = -\sqrt{\frac{2M}{r}} \pm 1$$

And we're there! The quantity $dr/dt_r$ is the radial velocity (in these slightly odd co-ordinates). There's a $\pm$ in the equation, as there is for all quadratics, and the -1 gives us the velocity of the inbound light beam while the +1 gives us the outbound velocity. If we're at the event horizon $r = 2M$, so just substituting this into the equation above for the outbound light beam gives us:

$$\frac{dr}{dt_r} = 0$$

Tada! At the event horizon the velocity of the outbound light beam is zero so light can't escape from the black hole. In fact for $r < 2M$ the outbound velocity is negative, so not only can light not escape but the best it can do is move towards the singularity.

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Epiphany. Well deserved +1. –  CHM May 14 '12 at 19:12
Why is it that in the absence of light, our eyes see black and not, say, white? –  Pacerier Jul 8 '12 at 1:51
@Pacerier: That's biology, as it deals with the mechanismp of the eye . And it's an off-topic comment (both for this answer and this site). Please don't hijack the thread. –  Dimensio1n0 Aug 9 '13 at 12:27

There is a substantial amount of other physics going on here. Right now it seems that you're trying to apply Newtonian mechanics to a realm where it breaks down (high-speed, high-gravity), and it's just not going to give you as much insight.

I always hate giving answers like this one, but if you really want to understand black holes you need to dig a little into general relativity.

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Unfortunately I do not have a professional book on general relativity but I am familiar with the tensor calculus. $$G_{\mu \nu}=\frac {8\pi G}{c^4}T_{\mu \nu}$$ –  user8784 May 14 '12 at 22:22
If you're familiar with tensor calculus, John Rennie's answer should be fairly easy to follow. –  Colin Fredericks May 15 '12 at 14:16

We see "black" in the absence of light. Black holes absorb all radiation and do not emit anything, so the hole is effectively "black" to our eyes.

Note that the rest of your analysis isn't exactly correct-- you're using a Newtonian framework and analyzing a general relativistic object. The correct-ish formula for the radius of a black hole is $R_s=\frac{2GM}{c^2}$. I'm not going to delve deeper, snce I'm not an expert in this field. Hopefully someone more knowledgeable can answer.

Note that black holes aren't completely black. They radiate [Hawking Radiation](http://enwp.org/Hawking_radiation] which is a quantum-mechanical/quantum gravity effect. Hawking radiation isn't that noticeable and the black hole practically remaina "black"

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when we say nothings can move faster than the speed of light which means The effect of gravity can not be faster than light then how we can say that Black holes absorb all radiation and do not emit anything in the case –  user8784 May 14 '12 at 16:34
@BadBoy: because the creation of a black hole changes the enveloping geometry of the black hole. The speed of light limit applies only to things moving very near to you. It says nothing about the relative motions of two distant objects. In particular, a naive interpretation of the notion of relative velocity would tell you that an observer far from a black hole would report that an observer inside the black hole had a superluminal relative velocity if there was some way to detect the observer inside the black hole. –  Jerry Schirmer May 14 '12 at 16:55
Even a chemist knows about Hawking radiation ;) I wouldn't be so eager to affirm that black holes don't emit anything, but I'd be glad to be proven wrong. –  CHM May 14 '12 at 19:02
@CHM I was hoping to steer clear of that. Yes, it's no longer black, but I didn't want to go into details. Anyway, I'll edit it in. –  Manishearth May 14 '12 at 19:04
The pure pleasure of reading nice questions and answers. I've lurked here some time. –  CHM May 14 '12 at 19:13