As noted in my original post, many electric universe folk have been claiming that black holes are incompatible with the theory of general relativity. The most straightforward way to refute that claim is to present mathematical proof that a certain spacetime manifold exists and has all of the properties that define a black hole.
I'll outline and then sketch that proof here. I'll leave some of the details as exercises, mainly because inclusion of all details would lengthen the proof and make it harder to read. If you're capable of understanding the details I omit, then you are probably capable of filling them in on your own, and many of you have already done homework of that sort. I'll give hints in blue.
Outline of proof
- Define a differentiable manifold M.
- Define the points of M.
- Define a topology for M.
- Prove the properties required of a topological manifold:
- locally Euclidean
- Define a set of charts (coordinate systems) that cover M.
- Prove those charts comprise an atlas for M.
- Define a spacetime manifold M.
- Define a Lorentzian pseudo-metric on M.
- Prove the metric satisfies Einstein's field equations for empty space.
- Prove M has the properties that define a black hole.
Definition of M as a differentiable manifold
It's convenient to have a distinct name for every point of the manifold we're defining. There are too many points to list each name individually, so we'll use a systematic naming process based on coordinates. The points of the manifold can be arbitrary mathematical objects, so we'll identify each point with its name (which will be an ordered list of 4 coordinates).
Let X0 be the set of all 4-tuples of the form <τ,r,θ,φ> where τ, r, θ, and φ are real numbers in the range
-∞ < τ < ∞
0 < r < ∞
0 < θ < π
-π < φ < π
and π is the Greek letter pi (representing the ratio between a Euclidean circle and its diameter). Let X1 be the set of all <τ,r,θ,π> where τ, r, and θ are real numbers in the range
-∞ < τ < ∞
0 < r < ∞
0 < θ < π
Let X2 be the set of all <τ,r,0,0> where τ and r are real numbers in the range
-∞ < τ < ∞
0 < r < ∞
and let X3 be the set of all <τ,r,π,0> where τ and r are as above. Define
X = X0 ∪ X1 ∪ X2 ∪ X3
The existence of those sets is an immediate consequence of the usual axioms for set theory and mathematics. Except where noted, the existence of everything we define will be an obvious consequence of basic axioms.
X0 will be the domain of our primary chart. We'll need a second chart whose domain is an open set that covers the meridian points in X1 and the polar points in X2 ∪ X3. Let the domain of that second chart be obtained by rotating the domain of the first chart as in exercise 7 of the exercises I formulated for Farsight. The domains of the two charts will then be connected by an obvious homeomorphism.
For every point p=<τ,r,θ,φ> in X0 and for every positive real number ε, define the open neighborhood N(p,ε) to be the set of all points q=<τ',r',θ',φ'> in X0 such that
τ - ε < τ' < τ + ε
r - ε < r' < r + ε
θ - ε < θ' < θ + ε
φ - ε < φ' < φ + ε
These are square neighborhoods. The round neighborhoods you might have been expecting can be obtained by taking an infinite union of these square neighborhoods. Similarly, square neighborhoods can be obtained by taking an infinite union of round neighborhoods.
We also need an analogous definition of open neighborhoods for points on the meridians and at the poles, but I'll leave that as an exercise for readers.
Points on the meridians and poles add a lot of detail without adding much insight, so I'll basically ignore X1 ∪ X2 ∪ X3 from now on.
Define the open sets of M as the set of all (finite and infinite) unions of open neighborhoods, together with the empty set (which can be regarded as a degenerate union of zero open neighborhoods).
Exercise: Prove the intersection of any two open sets is an open set.
The union of all open neighborhoods is X, so we have defined a topology on X. We must now prove that the topology is Hausdorff, paracompact, and locally Euclidean.
Exercise: Prove the topology is Hausdorff. (You have to show that for every pair of distinct points p and q, there's an open neighborhood U around p and an open neighborhood V around q such that U and V don't overlap.)
Exercise: Prove the topology on X is consistent with the usual Euclidean distance metric on X.
Exercise: Prove the topology on X is paracompact. (Hint: Every metrizable space is paracompact, and the previous exercise says the topological space we've defined is metrizable, so this is just modus ponens. You don't even have to know what paracompactness means!)
Let f: X0 → R4 be the function that maps <τ,r,θ,φ> to <τ,r,θ,φ>.
Exercise: Prove that if V is an open subset of R4, then the set of points that f maps into V is an open subset of X.
Exercise: Prove that if U is an open subset of X0, then f(U) is an open subset of R4
Exercise: Prove that f defines a one-to-one correspondence between X0 and f(X0).
Exercise: Prove that f is a homeomorphism between X0 and f(X0). (Hint: That's just a succinct summary of the three exercises above!)
Exercise: Define an analogous function g for the subset of X obtained by rotating X0 as in exercise 7. (Hint: Define g as the composition of f with that rotation's inverse.)
The exercises above prove that the topological space we have defined is locally Euclidean. They also provide us with two homeomorphisms f and g whose combined domains cover all of X.
Exercise: Let U be the intersection of X0 with the domain of g, and define h: f(U) → g(U) by h(x) = g(f-1(x)). Prove that h is a one-to-one correspondence.
Exercise: Prove that h is differentiable.
Repeat both of the above exercises for h: g(U) → f(U) defined by h(x) = f(g-1(x)).
The exercises above prove that f and g have the composition property required of charts. Taken together, the domains of f and g cover all of X.
Define M to be the differentiable manifold whose set of points is X, whose topology is as defined above, and whose atlas consists of the two charts f and g.
Definition of a Lorentzian manifold
We have now defined a Riemannian (locally Euclidean) manifold M. What we really wanted, however, was a pseudo-Riemannian (locally Lorentzian) manifold M. Instead of repeating the above definition with the Minkowski pseudo-metric on R4 substituted for the Euclidean metric throughout, I'm going to leave that as an (uninteresting) exercise for the reader. Almost all of that definition is exactly the same as for a locally Euclidean manifold, and the few steps that aren't exactly the same are almost exactly the same.
The more important fact is that we have not yet defined a metric (or pseudo-metric) on the manifold.
Definition of M as a spacetime manifold
To define a Lorentzian pseudo-metric tensor field on M, it suffices to define the tensor field for both the domain of the chart f (which is X0) and for the domain of the chart g. With f and g the same up to a spatial rotation, it's enough to define the tensor field on X0.
Define the Lorentzian pseudo-metric tensor field on X0 by
where m is a non-negative parameter to be discussed later and
The definition above assigns a pseudo-metric tensor to every point of M. The nonzero components of that tensor are
To prove that g is a pseudo-metric tensor, we must prove it's bilinear, symmetric, and nondegenerate.
Bilinearity is obvious because we've written it as a 4x4 matrix with only 6 nonzero elements. Symmetry is obvious because the matrix is symmetric.
Exercise: Prove g is nondegenerate. (Hint: Calculate the determinant of the matrix of coefficients, and show that it's nonzero everywhere in X0.)
The pseudo-metric defined above is not static, because its coefficient for dτ dr is nonzero.
The next step is to prove g satisfies Einstein's field equations for empty space:
That's just a calculation, but it's a chore to do the straightforward way (by using g to calculate 40 Christoffel symbols, using those to calculate 20 components of the Riemann tensor, and using those to calculate the Ricci tensor Rμν and the Ricci scalar R).
I'm going to cheat. In his paper cited earlier, Abrams proves that all metrics of a certain form satisfy the vacuum field equations. Similar proofs are found in standard references such as Misner/Thorne/Wheeler (section 23.2 and Box 23.3) or Wald (section 6.1).
I'm not going to cheat very much, though. Exercise 25, in the "Black holes" thread, proves that the Painlevé-Gullstrand metric defined above is equal to the Schwarzschild metric outside the event horizon, and everyone seems to agree that the Schwarzschild metric satisfies Einstein's equations.
That proof also reveals that the r, θ, and φ coordinates of our spacetime M are identical to the corresponding Schwarzschild coordinates.
Exercise 26, in the "Black holes" thread, proves that the Schwarzschild manifold (when restricted to the part outside the Schwarzschild radius) is a proper submanifold of the spacetime manifold we've defined above.
That completes our proof that the Schwarzschild manifold is extensible, contrary to the central claim made by Abrams and by Crothers.
Proof that M has the properties of a black hole
The free parameter m represents the total mass that's present at the center of each spatial slice of M. That can be confirmed by examining the asymptotically Newtonian gravitational field implied by the manifold M, or we can cheat again by using the meaning of m in the Schwarzschild metric, which is equivalent to the one we defined wherever both are defined.
Let's calculate the coordinate speed of radial light at the event horizon of our manifold. (That coordinate speed of light is coordinate-dependent, but we already know that the r coordinate of our coordinate system is identical to the r of Schwarzschild coordinates, and we also know that a positive dτ represents a positive (future-directed) increment of time.) Light travels along null geodesics, so ds=0 for light. For radial light, dθ=dφ=0 as well. At the event horizon, β = 1. Hence
At the event horizon, therefore, dr=0 (for outgoing light) or dr=-2 dτ (for ingoing light). The coordinate speed of outgoing light is dr/dτ=0, and the coordinate speed of ingoing light is dr/dτ=-2.
Outgoing light isn't making any outward progress at all. Looks like a black hole to me.