Karl Schwarzschild (1873–1916)

# Inside a Schwarzschild black hole

Hello and welcome.

This is a new and exciting vision about what lies beyond the event horizon of a black hole. This cannot be achieved entirely without a smidgen of mathematics but there should be nothing overly hard to follow. So let us jump in.

To begin, here are a couple of basic facts about Einstein's general relativity for any visitors who are new to the this field.

1. The gravitational field around a non-rotating symmetrical body (such as a star, or a planet) is given by the Schwarzschild solution, originally developed by Karl Schwarzschild in 1916, just a year after Einstein published his general theory of relativity.
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 non-rotating spherically symmetric shell of matter must be flat, or Minkowski space (the only piece of the Schwarzschild metric possible in this circumstance as there in no enclosed mass).

Knowing just these two undisputed facts, we can calculate the 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 ignoring the effect of the relatively slow rotation of the earth . This much is standard stuff and fully confirmed by experiments, here on earth.

The maths bit

Now, with these same two facts in mind, consider a spherically symmetric ball of matter collapsing due to 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.) The collapsing ball of matter, if of sufficient mass, will eventually form a black hole with an event horizon given by this simple equation

$r_{e}=\frac{2Gm}{c^2}$

where the reduced radius $$r_{e}$$, is the radius of the event horizon, $$G$$ is the gravitational constant, $$m$$ 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 theory.

The new stuff

Now I want you to consider a single particle, P, inside the event horizon, at a reduced 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 $$r'$$ 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.

Although in the original case, an event horizon will never be observable, we have no reason to assume that it does not still have meaning, 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 of the existing theory of General Relativity.

As the test mass was at an arbitrary distance from the origin, this must be equally true for every 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 has a density inside the outer event horizon that is inversely proportional to 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 !

## Nirvana->

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