Stability of Spacetime Structures

<< Great.m

Don't forget…

We'll work in spherical coordinates:

x = {t, r, θ, ϕ}

The Metric and its Curvature

Let's look at the well-known Schwarzschild (black hole) metric is:

(met = 
          DiagonalMatrix[{-(1 - 2  M/r), 1/(1 - 2  M/r), 
              r^2, (r Sin[θ])^2}]) // MatrixForm

The Riemann curvature of this metric is non-trivial:

Riemann[met, x]

Riemann[met, x] // MatrixForm

The fact that the Riemann tensor has nonzero (and nontrivial) entries does not, by iteself, imply that the Schwarzschild metric does not—in a very convoluted way—simply describe flat spacetime.

To check that it doesn't, we have to come up with an invariant measure of the curvature, and this is what SCurvature[met,x], SqRicci[met,x] and SqRiemann[met,x], Since the Ricci tensor already vanished, so will its invariant square (= norm square) and the trace (= scalar curvature). However, the square of the Riemann tensor need not:

SqRiemann[met, x]

Indeed! Not only is it not zero, but it diverges at r = 0. This means that this point ( r = 0 ) is a real, physical singularity—akin to the location of an electrostatic charge being a real, physical singularity of its Coulomb field.

Technically, locations where the metric, the Christoffel symbol, or any of the curvature tensors diverge are called "putative singularities".
To prove that these are in fact physical singularities, you must demonstrate that at least one invariant measure of the curvature diverges there. The above computation of \(TraditionalForm\`R\_μνρ\%σ·
          R\_σ\%μνρ = \ \(\(||\)\(R\_μνρ\%σ\)\( || \^2\)\)\) proves that r = 0 (the location of the black hole's center) is a physical singularity. Also, with some normalization, the parameter M acquires the meaning of the mass of this black hole.

For fun, we can also compute:

Christoffel[met, x]

IMetric[met] // MatrixForm

Confirm that the Einstein tensor of the Schwarzschild solution vanishes, i.e. that the Schwarzschild solution satisfies Einstein's vacuum (that is, there's no matter anywhere around!) field equations:

EinsteinTensor[DiagonalMatrix[{-(1-2M/r),1/(1-2M/r),
                                r^2, (r Sin[θ])^2}],
               {t,r,θ,ϕ}] // Simplify

Check that the Einstein tensor of flat spacetime vanishes:

EinsteinTensor[DiagonalMatrix[{-1,1,1,1}],{x1,x2,x3,x4}]

Notice also that not only is the Einstein tensor, (Subscript[G, μ])ν := 
        (Subscript[R, μ])ν - (1/2) (Subscript[g, μ])ν R, zero, but in fact the Ricci tensor, (Subscript[R, μ])ν itself vanishes too:

Ricci[met, x]

The scalar curvature, \(TraditionalForm\`R := \(\(R\_μν\) 
            g\^μν := ∑\_\(μ = 0\)\%3\(∑\_\(ν = 0\)\%3\( R\_μν\) g\^μν\)\)\), then trivially vanishes…

…which is kind of funny, since the Einstein-Hilbert action for general relativity is \(TraditionalForm\`𝒮\_EH = \(∫d\^4 
                x\ R = ∫dt\ L\)\), where \(TraditionalForm\`L\_EH := ∫\_0\%∞\( 
                r\^2\) dr \(∫sin\ θ dθ \(∫dϕ\ \(R(t, 
                      r, θ, ϕ)\)\)\)\) is the Einstein-Hilbert Lagrangian. (Rememer that a Lagrangian itself is not an observable quantity, such as the total energy. So, the fact that it vanishes doesn't mean anything in particular. One should calculate the Hamiltonian, \(TraditionalForm\`H := ∫\_0\%∞\( 
                r\^2\) dr \(∫sin\ θ dθ \(∫dϕ \(∑\_\(μ = 0\)\%3\([\((\(x\& . \^μ\) ∂\ L\/∂\ x\& . \^μ)\) - R]\)\)\)\)\), and see what that amounts to, for a physical observable.)

Fluctuating Parametrization

But, how about if we make the one parameter in the Scwarzschild metric, the mass of the black hole, M, into a fluctuating parameter: M ->  M + a(t), and see what the Lagrangian would turn out to be for this fluctuation, a(t):

(mFluct = 
          DiagonalMatrix[{-(1 - 2  M/r), 1/(1 - 2  M/r), 
                r^2, (r Sin[θ])^2}] /. {M -> (M + 
                    a[t])}) // MatrixForm

(Ricci[mFluct, x]) // MatrixForm

dLag = SCurvature[mFluct, x]

First of all, note that the result is independent of angles. That means that the angular integration just produces a simple, over-all factor of 4π. Radial dependence is, however, nontrivial, and the fact that this must be integrable is a condition on the fluctuation, a[t]. Since we can't use that condition before we solve for a[t], we must proceed working "inside the radial integral", and deal with that condition later.

This is not quite in the format of a standard Lagrangian, like \(TraditionalForm\`\(m(a(t))\)\/2\ \((\(a\& . \)(t))\)\^2\), where m(a(t)) is the a[t]-dependent effective mass of the fluctuation. We need to do some t-integration by parts.

We may rewrite dLag as a sum of two terms, one already proportional to \(TraditionalForm\`\((\(a\& . \)(t))\)\^2\), and the other, proportional to (Overscript[a, ¨])(t) . The coefficient of this last part is:

b[t_] = Coefficient[dLag, a''[t]]

When integrating [b(t)][(Overscript[a, ¨])(t)] by part, we'll get two terms:
• an integral of the total derivative of \(TraditionalForm\`\([b(t)]\)[\(a\& . \)(t)]\), which is a "boundary term", to be evaluated at t = ±∞ (infinite future – infinite past);
• the negative integral of the "other way around", \(TraditionalForm\`\([\(b\& . \)(t)]\)[\(a\& . \)(t)]\).

And in \(TraditionalForm\`\(\([\)\(\(b\& . \)(t)\)\(]\)\)\), we need the coefficient of \(TraditionalForm\`\(a\& . \)(t)\):

bCoeff[t_] = Coefficient[D[b[t], t], a'[t]]

So, we can now assemble:

mLagrangian = (Coefficient[dLag, (a'[t])^2] - 
            bCoeff[t]) ((a'[t])^2)

So, this is what the Lagrangian for the fluctuation of the mass of the black hole turns out to be. Notice that:
• There is no potential.
• There is only a kinetic term.
• The kinetic term looks non-relativistic, and the coefficient multiplying the "velocity-squared" must be identified as half of the effective mass of this fluctuation.

So, how does this effective mass density (recall, we are yet to integrate over r) depend on time? Glad you asked:

Simplify[mLagrangian]

While this seems like a trivial result, it speaks volumes: it means that there is nothing in the classical dynamics of black holes that would (dynamically or otherwise) settle the mass of the black hole to any particular value. In other words, isolated (there's nothing else around, remember?) black holes of any one mass are as stable as black holes of any other mass.

Another Track

We could have also computed

(mFluct = 
          DiagonalMatrix[{-(1 - 2  M/r), 1/(1 - 2  M/r), 
                r^2, (r Sin[θ])^2}] /. {M -> (M + 
                    a[t])}) // MatrixForm

(EinsteinTensor[mFluct, x]) // MatrixForm

The following is the Einstein Tensor with one index raised, \(TraditionalForm\`G\_\(\(\ \ \ \)\(ν\)\)\%μ\  := \(g\^μλ\) G\_λν\), which is sometimes preferred when discussing the energy-momentum distribution of matter:

(IMetric[mFluct] . EinsteinTensor[mFluct, x]) // 
      MatrixForm

In any case, the vanishing of the terms in the top-left corner requires a'[t]=0, i.e., that the mass is a constant, whereupon the elements in the bottom-right corner vanish too.

This then ammends the previous conclusion: that the mass is an arbitrary parameter, but for the Einstein equations to be satisfied, it must be a constant, albeit arbitrary.

Conclusions

The general principle we've followed here is however… well, …general: It applies to analyzing the stability of the solution of any differential equation:
	1. Obtain the general solution of a system of differential equations;
		here, the Schwarzschild metric.
	2. This solution depends on a number of constants;
		here, M.
	3. Shift these constants (and thereby the colution) by a fluctuation;
		shift M ⟶ M + a[t].
At this point you have two options:
	4.a. Compute a function(al) of the so shifted function (here the Lagrangian)
		that is to be discussed (minimized, optimized, whatever), or
	4.b. Compute the shift in the Einstein tensor, and set that equal to zero.
		Here, this says that a[t] must be a constant.