3

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?

Amartya
  • 153
  • 6
  • 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 Jul 05 '20 at 13:58
  • 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 Jul 05 '20 at 14:03
  • 1
    @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 Jul 05 '20 at 14:43
  • 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 Jul 05 '20 at 15:28
  • 1
    @planetmaker The event horizon is absolutely not defined using escape velocity. Using escape velocity as a concept here leads to confusion. – StephenG - Help Ukraine Jul 05 '20 at 16:33
  • @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 Jul 05 '20 at 18:01
  • 1
    @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 ? on Physics SE and it's answers. The geometry of spacetime is the issue, not velocity. – StephenG - Help Ukraine Jul 05 '20 at 18:11
  • Is there any theory which can give info about the gravity of objects in Black Hole? – Amartya Jul 05 '20 at 18:57
  • @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 Jul 05 '20 at 19:11
  • 3
    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 Jul 05 '20 at 19:46
  • 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 Jul 06 '20 at 06:36

1 Answers1

10

The 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 $r<r_s$ 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 $r<r_s$) 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 $r<r_s$.

ProfRob
  • 151,483
  • 9
  • 359
  • 566
  • 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 Jul 06 '20 at 03:20
  • THanks @Rob. Big help! – Amartya Jul 06 '20 at 05:03
  • 3
    @dmedine no, that isn't true. Light signals will continue to reach an observer falling into a black hole. – ProfRob Jul 06 '20 at 06:41