<p>The <a href="https://en.wikipedia.org/wiki/Schwarzschild_metric" rel="nofollow noreferrer">Schwarzschild metric</a> can be written as
<span class="math-container">$$ c^2 d\tau^2 = \left( 1 - \frac{r_s}{r}\right)\ dt^2 - \left(1 - \frac{r_s}{r}\right)^{-1}\ dr^2 - ...,$$</span>
where <span class="math-container">$r$</span> is the radial coordinate, <span class="math-container">$t$</span> is the coordinate time, <span class="math-container">$\tau$</span> is the proper time (that measured on an observer's own clock) and <span class="math-container">$r_s = 2GM/c^2$</span> is the Schwarzschild radius. I have left out the angular terms on the right hand side which contribute a further negative term independent of whether <span class="math-container">$r$</span> is greater or less than <span class="math-container">$r_s$</span>.</p>
<p>For an observer with mass, <span class="math-container">$d\tau&gt;0$</span>; for a massless particle <span class="math-container">$d\tau=0$</span> (e.g. a photon).</p>
<p>When <span class="math-container">$r&lt;r_s$</span> the first term on the RHS is negative, while the second term becomes positive. In order for the LHS to be <span class="math-container">$\geq 0$</span>, then
<span class="math-container">$$ \left(\frac{r_s}{r}-1 \right)^{-1}\ dr^2 \geq \left(\frac{r_s}{r}-1\right)\ dt^2 + ...$$</span>
<span class="math-container">$$ \left| \frac{dr}{dt}\right| \geq \left(\frac{r_s}{r} -1\right)$$</span></p>
<p>What this means is that <span class="math-container">$dr/dt$</span> can never be zero, which means the direction of radial travel can never reverse. ie. Anything that enters a black hole (i.e. for which <span class="math-container">$r&lt;r_s$</span>) and has <span class="math-container">$dr/dt&lt;0$</span>, can never have <span class="math-container">$dr/dt &gt;0$</span>.</p>
<p>There is a slightly more satisfactory &quot;proof&quot; using Eddington-Finkelstein coordinates, that shows that all future light cones point inwards and that <span class="math-container">$dr&lt;0$</span> when <span class="math-container">$r&lt;r_s$</span>.</p>

CC BY-SA 4.0
2020-07-06T03:20:59.033Z
If I understand this correctly, though (which is not at all certain), then the opposite is also true. That is, if the observer is inside the black hole then everything outside the black hole (r>rs) is 'invisible' since the time dilation inside the black hole means that light from outside the black hole takes longer than it takes light to travel that distance to travel that distance.
dmedine
CC BY-SA 4.0
2020-07-06T05:03:35.110Z
THanks @Rob. Big help!
Amartya
CC BY-SA 4.0
2020-07-06T06:41:56.127Z
@dmedine no, that isn't true. Light signals will continue to reach an observer falling into a black hole.
ProfRob
CC BY-SA 4.0
2020-07-05T19:48:25.623Z
2020-07-06T06:40:09.000Z
2020-07-06T06:40:09.000Z
ProfRob
<p>Mathematically, for black holes old enough that the stellar material has collapsed all the way into
the singularity, the region between the horizon and the singularity is occupied by a spacetime
where the time and space coordinates are reversed from those of the outside world. What this
means in terms of what you experience is unknown. Other more complex conditions can occur of
the black hole is rotating. In that case the singularity becomes a ring around the center of the
black hole. You can pass through the center, but the tidal gravitational field would be lethal in all
likelihood. In nearly all cases there would be gravitational radiation rattling about, and this would
cause distortions in spacetime that would probably lead to spectacular optical distortions.</p>
<p>My question is <em>Black hole absorbs everything.</em> <strong>Is there any mathematical proof regarding this?</strong></p>

CC BY-SA 4.0
2020-07-05T13:58:23.677Z
Black hole by definition is an object whose escape velocity exceeds the speed of light. Nothing is faster than light. So by definition nothing can escape. Add quantum gravity and all bets at off (see also hawking radiation)
planetmaker
CC BY-SA 4.0
2020-07-05T14:03:21.123Z
I don't think q black hole absorbs *everything*, Amartya. For example LIGO say they've detected black hole mergers which result in gravitational waves. Besides, I don't think a mathematical proof is enough to establish whether a claim is correct. You need hard scientific evidence for that.
John Duffield
CC BY-SA 4.0
2020-07-05T14:43:08.867Z
@planetmaker That is *not* the definition of a black hole.  A black hole is defined by an event horizon and escape velocity is not a useful concept in GR.
StephenG - Help Ukraine
CC BY-SA 4.0
2020-07-05T15:28:42.960Z
How do you think an event horizon is defined? It's commonly the surface from which light takes an eternity / is infinitely red-shifted to reach any external observer. It gets more complicated, if you add rotation... Either case, in GR nothing can escape from within its EH.
planetmaker
CC BY-SA 4.0
2020-07-05T16:33:54.890Z
@planetmaker The event horizon is absolutely not defined using escape velocity.  Using escape velocity as a concept here leads to confusion.
StephenG - Help Ukraine
CC BY-SA 4.0
2020-07-05T18:01:58.850Z
@StephenG maybe try to express your thought in a positive definition and answer the question of the OP instead of just saying what it is not?
planetmaker
CC BY-SA 4.0
2020-07-05T18:11:13.903Z
@planetmaker To be clear I am correcting *your* factually incorrect definition of a black hole.  I'd refer you to e.g. [Why can't you escape a black hole ?](https://physics.stackexchange.com/questions/25369/why-cant-you-escape-a-black-hole?noredirect=1&lq=1) on Physics SE and it's answers.  The geometry of spacetime is the issue, not velocity.
StephenG - Help Ukraine
CC BY-SA 4.0
2020-07-05T18:57:04.277Z
Is there  any theory which can give info  about the gravity of objects in Black Hole?
Amartya
CC BY-SA 4.0
2020-07-05T19:11:49.730Z
@StephenG a BH is defined as the region where nothing can escape from (e.g. see ref 6 in https://en.wikipedia.org/wiki/Black_hole ). Your link actually doesn't contradict that either. For the classical Schwarzschild BH, the event horizon is the sphere where the escape velocity equals the speed of light. If you add rotation etc, things get more complicated and the light paths might be more useful to describe properties and the EH - yet it doesn't change physics. Similar to that you can choose good or bad coordinate systems in order to solve a particular problem.
planetmaker
CC BY-SA 4.0
2020-07-05T19:46:53.260Z
It isn't defined by the escape speed The escape speed of the Earth is 11 km/s, but you can escape from Earth's gravity without getting anywhere near that speed. The same is not true of a black hole.
ProfRob
CC BY-SA 4.0
2020-07-06T06:36:06.520Z
Furthermore, if you launch an object at just less than the escape speed, it travels away, almost to infinity, before falling back. That also doesn't happen for a black hole.
ProfRob
<black-hole>
Mathematical Analysis of Blackhole
Amartya
# Mathematical Analysis of Blackhole

(Asked by: username_0 on 2020-07-05)

Mathematically, for black holes old enough that the stellar material has collapsed all the way into
the singularity, the region between the horizon and the singularity is occupied by a spacetime
where the time and space coordinates are reversed from those of the outside world. What this
means in terms of what you experience is unknown. Other more complex conditions can occur of
the black hole is rotating. In that case the singularity becomes a ring around the center of the
black hole. You can pass through the center, but the tidal gravitational field would be lethal in all
likelihood. In nearly all cases there would be gravitational radiation rattling about, and this would
cause distortions in spacetime that would probably lead to spectacular optical distortions.

My question is *Black hole absorbs everything.* **Is there any mathematical proof regarding this?**

**Tags:** black-hole

> **Comments:**
> **username_1**: Black hole by definition is an object whose escape velocity exceeds the speed of light. Nothing is faster than light. So by definition nothing can escape. Add quantum gravity and all bets at off (see also hawking radiation)  
> **username_2**: I don't think q black hole absorbs *everything*, username_0. For example LIGO say they've detected black hole mergers which result in gravitational waves. Besides, I don't think a mathematical proof is enough to establish whether a claim is correct. You need hard scientific evidence for that.  
> **username_3**: @username_1 That is *not* the definition of a black hole.  A black hole is defined by an event horizon and escape velocity is not a useful concept in GR.  
> **username_1**: How do you think an event horizon is defined? It's commonly the surface from which light takes an eternity / is infinitely red-shifted to reach any external observer. It gets more complicated, if you add rotation... Either case, in GR nothing can escape from within its EH.  
> **username_3**: @username_1 The event horizon is absolutely not defined using escape velocity.  Using escape velocity as a concept here leads to confusion.  
> **username_1**: @StephenG maybe try to express your thought in a positive definition and answer the question of the OP instead of just saying what it is not?  
> **username_3**: @username_1 To be clear I am correcting *your* factually incorrect definition of a black hole.  I'd refer you to e.g. [Why can't you escape a black hole ?](https://physics.stackexchange.com/questions/25369/why-cant-you-escape-a-black-hole?noredirect=1&lq=1) on Physics SE and it's answers.  The geometry of spacetime is the issue, not velocity.  
> **username_0**: Is there  any theory which can give info  about the gravity of objects in Black Hole?  
> **username_1**: @StephenG a BH is defined as the region where nothing can escape from (e.g. see ref 6 in https://en.wikipedia.org/wiki/Black_hole ). Your link actually doesn't contradict that either. For the classical Schwarzschild BH, the event horizon is the sphere where the escape velocity equals the speed of light. If you add rotation etc, things get more complicated and the light paths might be more useful to describe properties and the EH - yet it doesn't change physics. Similar to that you can choose good or bad coordinate systems in order to solve a particular problem.  
> **username_4**: It isn't defined by the escape speed The escape speed of the Earth is 11 km/s, but you can escape from Earth's gravity without getting anywhere near that speed. The same is not true of a black hole.  
> **username_4**: Furthermore, if you launch an object at just less than the escape speed, it travels away, almost to infinity, before falling back. That also doesn't happen for a black hole.  

---

## 1 answers

### Answer by username_4 on 2020-07-05. Score: 10 | Accepted answer

The [Schwarzschild metric](https://en.wikipedia.org/wiki/Schwarzschild_metric) can be written as
$$ c^2 d\tau^2 = \left( 1 - \frac{r\_s}{r}\right)\ dt^2 - \left(1 - \frac{r\_s}{r}\right)^{-1}\ dr^2 - ...,$$
where $r$ is the radial coordinate, $t$ is the coordinate time, $\tau$ is the proper time (that measured on an observer's own clock) and $r\_s = 2GM/c^2$ is the Schwarzschild radius. I have left out the angular terms on the right hand side which contribute a further negative term independent of whether $r$ is greater or less than $r\_s$.

For an observer with mass, $d\tau>0$; for a massless particle $d\tau=0$ (e.g. a photon).

When $r the first term on the RHS is negative, while the second term becomes positive. In order for the LHS to be $\geq 0$, then
$$ \left(\frac{r\_s}{r}-1 \right)^{-1}\ dr^2 \geq \left(\frac{r\_s}{r}-1\right)\ dt^2 + ...$$
$$ \left| \frac{dr}{dt}\right| \geq \left(\frac{r\_s}{r} -1\right)$$

What this means is that $dr/dt$ can never be zero, which means the direction of radial travel can never reverse. ie. Anything that enters a black hole (i.e. for which $r) and has $dr/dt<0$, can never have $dr/dt >0$.

There is a slightly more satisfactory "proof" using Eddington-Finkelstein coordinates, that shows that all future light cones point inwards and that $dr<0$ when $r.

> **Comments:**
> **username_5**: If I understand this correctly, though (which is not at all certain), then the opposite is also true. That is, if the observer is inside the black hole then everything outside the black hole (r>rs) is 'invisible' since the time dilation inside the black hole means that light from outside the black hole takes longer than it takes light to travel that distance to travel that distance.  
> **username_0**: THanks @Rob. Big help!  
> **username_4**: @username_5 no, that isn't true. Light signals will continue to reach an observer falling into a black hole.  

---

