Inside a Schwarzschild black hole

Hello and welcome

 

 Ever wondered what lies beyond the horizon of a black hole? Current thinking asserts any one of numerous outcomes; time travel, wormholes, being crushed to a point, or instead, perhaps a fiery end in a wall of flame, making it very hard to know what to believe. We offer a somewhat more prosaic answer; not nearly so exciting, but needing no extensions to existing theory, and as such, so more believable.

This is a new vision of what lies beyond the event horizon of a black hole. New, and as we will show later, testable and demonstrated by the otherwise unexplained presence of supermassive black holes. This will never be entirely understood without a smidgen of mathematics, but, if you have a basic college level understanding, there should be nothing overly hard for you to follow. So let us just jump right in. 

Introduction

To begin with, here are a couple of basic facts about Einstein's theory of general relativity for any visitors who are new to this field. There is no dispute about these facts so I hope you will just accept them for now:

150px-Schwarzschild.jpg

Karl Schwarzschild (1873–1916)  

  1. The gravitational field around a non-rotating symmetrical body (such as a star, or a planet) is by the Schwarzschild solution, originally developed by Karl Schwarzschild in 1916, just a year after Einstein announced his general theory of relativity: 
    \[ c^2d\tau ^2=\left(1-\frac {r_s}{r}\right)c^2dt^2 -\left(1-\frac {r_s}{r}\right)^{-1}dr^2-r^2\left(d\theta ^2+\sin ^2\theta \,d\varphi ^2\right) \]The key fact to notice about this equation is the first term which seems to 'blow up' when \(r=r_s\). This is what gives rise to the event horizon.
  2. Birkhoff's theorem added that for any non-rotating spherically symmetric body, the exterior gravitational field in empty space must be static, with a metric given by a piece of the Schwarzschild metric. This sounds difficult but all this is really saying is that there is only one solution, the Schwarzschild solution and that it is static - or unchanging.

An immediate consequence of Birkhoff’s theorem is that the field inside a symmetric non-rotating spherically shell of matter must be flat, or Minkowski space (the only piece of the Schwarzschild metric possible in this circumstance as there is no enclosed mass).

Knowing just these two undisputed facts, we could, for instance, calculate the precise field at the bottom of a mine shaft -- just calculate the field due to the mass beneath our feet whilst ignoring all of the mass above our heads, and neglecting the effect of the relatively slow rotation of the earth. This much is standard stuff and fully confirmed by experiments, here on earth. 

Now, keeping these same two undisputed facts in mind, consider a large ball of matter, collapsing due to the force of gravity, where the forces involved have already exceeded those needed to halt the collapse at the size of a neutron star. (Such as during the final stage of collapse after a sufficiently large star goes supernova at the end of it's active life.) For simplicity, let the ball be spherically symmetric and nonrotating. The collapsing ball of matter, if of sufficient mass, will eventually form a black hole with an event horizon having a reduced radius, \(r_{e}\), given by this simple equation

\[r_{e}=\frac{2Gm}{c^2}\]

where the [r_e\] is the reduced radius of the event horizon, \(G\) is the gravitational constant, the same constant used in the gravity equation of Newton, \(m\) is the total mass enclosed by the event horizon, and \(c\) is the speed of light. In the following argument, all radii will be reduced radii.

Inside this event horizon the ball of particles will continue to collapse, heading relentlessly towards the origin. So far, we have not deviated in any way from established theories. To avoid repetition, all radii in these discussions will refer to reduced radii.

The new stuff: here it gets interesting

Now I want you to consider a single particle, P, inside this event horizon, at a distance \(r'\)  from the origin, on its way to the centre. In a thought experiment, compare the motion of this particle with that of another identical particle on the surface of an identical distribution of matter but with all matter at a greater distance than \(r'\)  somehow removed.

We know that in the second case - in our thought experiment - the exterior field at P must be given by the Schwarzschild metric because of Birkhoff's theorem, whilst in the first case the field at P must be completely unaffected by the symmetrical shell of matter further from the origin than P due to the second aspect of Birkhoff's theorem given above. Consequently, the motion of the two particles should be, must be, identical.

The particle at P will continue to head towards the origin in both cases, but a point will be reached in our thought experiment, say at a distance \(r,\) when a new event horizon would form, for a distant observer. At this point we have

\[r=\frac{2Gm_{r}}{c^2}\] where \(m_{r}\) is the mass enclosed by the spherical surface through the point P, but excludes any mass further out.

Although in the original case, an event horizon will never be observable, we have no reason to assume that it does not still have significance, and crucially, that for a distant observer the point P would never have crossed this newly formed event horizon in a finite time. That is if we were to entirely remove the outer layer, it would still be hovering above this new event horizon. If this were untrue in the interior of a black hole, then we would have to accept that what happened there depended entirely upon whether it was being observed or not. This would not be an acceptable modification to the existing theory of General Relativity.

As our test particle was at an arbitrary distance from the origin, this must be equally true for a particle at any point within the event horizon of the original black hole, and, as a consequence, the eventual distribution of mass must be such that for all \(r\) less than \(r_{e}\)

\[r=\frac{2G}{c^2}\int_0^r4\pi r^2\rho(r)dr\]

Where \(\int_0^r4\pi r^2\rho(r)dr\) is simply the mass enclosed by a sphere at radius \(r\) and with \(\rho(r)\) being the eventual mass distribution function. At this point, rearranging and solving the integral gives

\[\rho(r)=\frac{c^2}{8\pi Gr^2}\]

The conclusion

The black hole must have a density inside the outer event horizon that is inversely proportional to the square of the (reduced) distance from the origin.

 

Now, after all this heavy stuff, time to look at a more restful place, a place of absolute peace and tranquillity - the end of time itself!

Nirvana->

 

Agree or disagree, or have any questions or observations about this, and I would love to hear from you, so please This email address is being protected from spambots. You need JavaScript enabled to view it.. Your comments are always most welcome.

 

See this published paper for more information.

 

 

Add comment


Security code
Refresh