Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 41 down vote accepted

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.

share|cite|improve this answer
2  
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
6  
@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. – centralcharge Aug 9 '13 at 12:27
    
Sorry guys for continuing the hijack, but... @Pacerier: you might be interested: Your black is highly improbable to be my black !!! See this and this. – AneesAhmed777 Feb 19 at 13:42
    
+1, especially for the "Tada" at the end: I get the impression you thoroughly enjoyed this problem; one of the main things that I love about MTW btw is that the writing style exudes this kind of enthusiasm constantly. – WetSavannaAnimal aka Rod Vance Jun 22 at 23:44

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.

share|cite|improve this answer
    
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
7  
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"

share|cite|improve this answer
    
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
2  
@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
1  
@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
2  
The pure pleasure of reading nice questions and answers. I've lurked here some time. – CHM May 14 '12 at 19:13

A black hole is black because you can't see it. You can't see it because it's not in your past. To you, it simply hasn't happened yet. The things that become the black hole are slowed relative to you, and slowed so badly that you just don't ever see them become a black hole.

When you look at your hand you see your hand from a nanosecond ago. When you look at the moon you see it from a second ago. When you look at Mars you see it from ten minutes ago. You always see things from your past.

Those examples you see them from farther in the last becasue they were farther away, it took more time for the light to get to you.

Near a compact body something different happens. Time itself slows down relative to you, it takes longer for things to happen. So there was a star and thibgs started going slower and so you say something that would take a day on earth and it you watched it happen in slow motion, it took a year. And during that year you saw the star contract and get a bit smaller and that made it more compact and the problem got worse. Over the next year you say something happen on the star that normally takes 12 hours to happen. But the star got even smaller in that time and that made it even worse. Over the next year you say something happen on the star that normally takes 6 hours to happen. And it got worse. The next year you watched something happen there that should have taken 3 hours. And the year after you saw something that should have taken 90 minutes. The next hear you saw something that should have taken 45 minutes. And it keeps getting smaller and the problem gets worse. The next year you saw something that should have taken 22.5 minutes. Then the year after you saw something that should have taken 11.25 minutes. And it's not just getting boring. It's faint. The same amount of light is leaving the star as normally left in 11.25 minutes but now you have wait a year to receive it. That means it is 1/128th as faint as it should have been as well as being a slow motion moving at 1/128th speed.

And thirteen more years later at this rate and its a million times slowler than normal and a million times fainter. That fainter is why it looks similar to black. But you saw that first day of stuff go by that first year. How long does it take you to see the next day?

Then the next year you saw 12 hours more (so 1/2 of that day in 1 year) then 6 hours more (so 3/4 of that day in 2 years) then 3 hours more (so 7/8 of that day in 3 year) then 90 minutes hours more (so 15/16 of that day in 4 years) then 45 minutes more (so 31/32 of that day in 5 years) then 22.5 minutes hours more (so 63/64 of that day in 6 years) then 11.25 minutes more (so 127/128 of that day in 7 years). And you never actually see that second day. That second day is never in your past.

And that's becasue in relativity your past is relative. It's literally the things you can see. And to you, the black hole is never in your last. Someone on the star might see it, but you never do. And that's life. The time slow down makes it do those things that happen on the third day are actually just never in your past. In fact we don't know if there is a third day.

Your sense of now just swoops down to before that fateful event. And it does that every day and every year. That black hole is more like the future to you than the past. Sometimes you could be in their past. For instance on the third day for them, they might see you doing something, so one second after that thing you did it would then be too late to try to join them for the eventful moment the second day becomes the third day. But you'd still be getting slower down faint images from thibgs that hadn't gotten too close to the compact parts.

You can't see something if its not in your past. And when things are moving slower and slower and slower than you then a finite amount of their time might fill all the past you are going to have. It's just harder to know when things are relative becasue it will depend on how you move and where you go.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.