Space Warps + Black-holes

I realized I had a misunderstanding about how space is warped in gravity. I am trying to gain intuition from the Schwarzschild solution for a blackhole. This can be used to see how space reacts far away from a kind of ideal massive object. I am using wikipedia as references:

https://en.wikipedia.org/wiki/Schwarzschild_metric
https://en.wikipedia.org/wiki/Deriving_the_Schwarzschild_solution

The initial solution presented however is not isotropic in the coordinates, which makes it hard to understand how space is really behaving from a measurable point of view. If these equations don't make any sense to you I will try to explain so don't freak out. I will copy here for reference, but I will explain why this one is deceiving:

\[ds^2 = \left (1-\frac{2Gm}{c^2 r} \right)^{-1} dr^2 + r^2 \left ( d\theta^2 + sin^2(\theta)d\phi^2 \right) - \left (1-\frac{2Gm}{c^2 r} \right )c^2 dt^2 \]

The solution (or metric) in the second page for isotropic coordinates I found more revealing and I will copy here.

\[ds^2 = \left (1+\frac{Gm}{2c^2 r_1} \right)^4 \left (dr_1^2 + r_1^2 \left ( d\theta^2 + sin^2(\theta)d\phi^2 \right)  \right) - \frac{\left (1-\frac{Gm}{2c^2 r_1} \right )^2}{\left (1+\frac{Gm}{2c^2 r_1} \right )^2}c^2 dt^2 \]


The \(dt\) represents a tick of a reference clock, and \(dr\), \(rd\theta\), and \(r sin(\theta) d\phi\) represent reference lengths. The factors multiplying them represent a scaling factor of what the actual clock ticked, or length is, at that radius relative to the reference.

The factor on the time coordinate gives the time dilation.

Time dilation as a function of radius: \(\frac{Gm}{2c^2}=1\)

The factors are a function of radius, \(r\), but the radius does not mean the same thing that it means in flat space. The solution is defined from imagining space being built up from a series of concentric spherical shells. The coordinate \(r\) tells which shell you are sitting on by equating the surface area of the shell to \(4 \pi r^2 \). It does not, however, necessarily tell you how far from the center you are. And actually it doesn't even tell you what the actual surface area is, since you also have to look at the scaling factor for those dimensions.

In flat space, each shell has to be bigger than the one inside of it, and smaller than the one outside, by a fixed amount. This is the limit where \(m = 0\), and so the scaling factor is a constant. As m increases, the scaling factor becomes a function of radius.

Length expansion as function of radius: \(\frac{Gm}{2c^2}=1\) 

There is a limit to this solution at \(r = \frac{Gm}{2c^2}\). The metric at that radius is

\[ds^2 = 4^2 \left (dr_1^2 + r_1^2 \left ( d\theta^2 + sin^2(\theta)d\phi^2 \right) \right)\]

This is called the event horizon because the time component vanishes, which means nothing can ever cross this boundary from the point of view of someone outside. Now, from what I have read, other coordinate systems allow the solution to progress past the event horizon. It's not important right now whether this is physically real since I only care about events far away from this limit. I think what may be more important is to see that nothing too crazy is happening to the spatial coordinates here.

However, in the first solution it looks like the length factor in the radial coordinate blows up as the event horizon is approached. This is because of the choice of radial coordinate. The problem is that the surface area of each shell as measured by a distant observer starts to approach a constant value as the event horizon is approached. That is, each concentric shell has about the same surface area as the one just outside, and the one just inside. This means space is not flat.

The problem with this solution is that in order to get to a spherical shell of a smaller surface area, one must drop a much further distance toward the event horizon. And at the event horizon itself the areas become constant, which makes it look like the radial coordinate blows up. However, the distance to the event horizon is actually a finite distance through space.

Radial factor accounting for length expansion near horizon

The second solution scales the surface areas as one gets close to the event horizon which is what one would actually experience. We can see that the radial length factor is actually only 4x that of a reference length far away. But also every dimension is 4x bigger there, not just the radial lengths.

For things that are not blackholes, what this means is that essentially there is slightly more space inside and around a planet than one would expect from far away, in addition to time running slightly slower.