# Mathematical Analysis of Blackhole

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?

• 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 5 at 13:58
• @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 Jul 5 at 14:43
• @planetmaker The event horizon is absolutely not defined using escape velocity. Using escape velocity as a concept here leads to confusion. – StephenG Jul 5 at 16:33
• @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 Jul 5 at 18:11
• 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. – Rob Jeffries Jul 5 at 19:46

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 $$r the first term on the RHS is negative, while the second term becomes positive. In order for the LHS to be $$\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.

• 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 6 at 3:20
• THanks @Rob. Big help! – Amartya Roy Jul 6 at 5:03
• @dmedine no, that isn't true. Light signals will continue to reach an observer falling into a black hole. – Rob Jeffries Jul 6 at 6:41