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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 Schwarzschild 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 coordinate 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 coordinates we want, because it's coordinate independent, and it turns out the best (well, simplest anyway!) coordinates to use for this problem are the Gullstrand–Painlevé coordinates. In these coordinates $r$ 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 coordinate system is known as the "rainfall" coordinates 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 coordinates). 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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  • $\begingroup$ I have difficulty in understanding how an actual experimenter would calculate $t_r$. Is it that we have arranged some clocks to be placed everywhere and these clocks are influenced by non-gravitational fields so that they do not change either of $r,\theta,\phi$ coordinates? (This will match the meaning given to $t_r$ via the metric and the observer will notice reading from these nearby available clocks.) But this requires a synchronization among clocks and clearly, such a sync would be impossible inside the EH as signaling can't be done in the out-radial direction. $\endgroup$ – Feynmans Out for Grumpy Cat Jan 1 '17 at 20:20
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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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  • $\begingroup$ Unfortunately I do not have a professional book on general relativity but I am familiar with the tensor calculus. $$G_{\mu \nu}=\frac {8\pi G}{c^4}T_{\mu \nu}$$ $\endgroup$ – user8784 May 14 '12 at 22:22
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    $\begingroup$ If you're familiar with tensor calculus, John Rennie's answer should be fairly easy to follow. $\endgroup$ – Colin Fredericks May 15 '12 at 14:16
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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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  • $\begingroup$ 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 $\endgroup$ – user8784 May 14 '12 at 16:34
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    $\begingroup$ @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. $\endgroup$ – Jerry Schirmer May 14 '12 at 16:55
  • $\begingroup$ 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. $\endgroup$ – CHM May 14 '12 at 19:02
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    $\begingroup$ @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. $\endgroup$ – Manishearth May 14 '12 at 19:04
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    $\begingroup$ The pure pleasure of reading nice questions and answers. I've lurked here some time. $\endgroup$ – CHM May 14 '12 at 19:13
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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.

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  • $\begingroup$ I think answer isn't quite right. In the Gullstrand–Painlevé coordinates, a stellar black hole is way too small for it to take a year for you to observe the object as half the distance from the event horizon. In one dimension, you would receive half the rate of photons and the photons would be lower energy so you would receive 1/4 of the wattage when it's half as close. In the third dimension, once the object is close to the event horizon, only the photons reflected in a precise direction escape so the object appears even fainter. $\endgroup$ – Timothy Aug 20 '18 at 3:22
  • $\begingroup$ Whait, where you say 'say' you actually mean 'saw', is that correct? Perhaps you are typing on a German keyboard or something? Or do I completely misunderstand and it really is about saying (rather that seeing) things? $\endgroup$ – Vincent May 2 at 14:54
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The reason a black hole is black is purely because light con not escape. The reason there is an event horizon is because you are just seeing the point where light can no longer escape. The only thing that can escape a black hole are protons on the event horizon due to Hawking Radiation. A black hole is black because the color black absorbs all light and reflects none, so since no light can escape, it is all absorbed and we see it as the color black.

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  • $\begingroup$ That's not actually the reason they appear black. When an object falls into a black you, you never see it reach the event horizon. After more time, photons reflacted off of it escape at a lower rate and are of lower energy so the rate of escape of energy of an electromagnetic wave approaches zero so it gets very close to black after enough time. $\endgroup$ – Timothy Aug 20 '18 at 3:32

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