Preservation of metric signature in Cauchy problem for the Einstein equations

Question

In Choquet-Bruhat's solution to the Cauchy problem for Einstein's equation, one reduces the Einstein equations to a quasidiagonal quasilinear hyperbolic system on $ M := [0, T] \times \bar M$ where $T > 0$ and $\bar M$ is some initial spacelike 3-manifold for unknowns $g_{\alpha\beta}$ on $M$. Hyperbolic PDE theory then shows that one can, given first-order initial data, solve these equations forward in time to find smooth solutions $g_{\alpha\beta}$. Thus we get a metric $g$ on $M$.

1. How do we know that the resulting metric $g = (g_{\alpha\beta})$ is Lorentzian? Is there a reason why the signature cannot change, or that it cannot become degenerate?

At first I thought continuity of the metric would solve this problem - if it is Lorentzian at $t = 0$ then it should be Lorentzian shortly afterwards. However if $\bar M$ is noncompact (i.e. in the natural case of $\bar M = \mathbb{R}^3$) then we may not get a uniform $T > 0$ on which $g$ is Lorentzian, as given $x \in \bar M$ the time interval $I_x$ on which $g(\cdot, x)$ is Lorentzian could get small as $x$ varies, making $|I_x| \to 0$ as $x \to \infty$. But from what I can tell, the results on this problem do give us a uniform $T > 0$, so I am guessing that this is not the reason.

I have not studied carefully the proof of well-posedness to hyperbolic systems of this form; perhaps the answer is part of this theorem. Any help is much appreciated.

Answer 1

1. The dynamic metric is the metric for the quasidiagonal system; and so for self consistency the solution can only be proven to exist (and be unique) when the metric (i.e. the solution itself) is hyperbolic/Lorentzian.

2. Noncompactness has zero impact whatsoever. Hyperbolic equations have finite speed of propagation and hence "local local uniqueness" (one local for in time, and one local for in space).

3. Finally, about $T$: time functions in Relativity theory is not god given. With diffeomorphism invariance you can always re-coordinate an open spacetime neighborhood of the spacelike hypersurface $M$ as $[0,T]\times M$. On the other hand, there are times where you can prove that on your coordinate system $|g(\partial_t, \partial_t)|$ has a lower bound; in this case usually what happened is that you assumed enough decay on the initial data (in some weighted Sobolev space, say) that you can in fact prove using energy estimates that you have good global foliations. In situations like this you automatically also get uniform boundeness on, say, the connection coefficients, and then your continuity argument will have no problem.

Answer 2

I will add a pessimistic answer. You are right that Choque-Bruhat's (and any related local-in-time) existence result only guarantees that the solution metric exists and is sufficiently regular (including remaining of Lorentzian signature) only in some open neighborhood of the Cauchy surface $\bar{M}$, without any guarantee that this neighborhood will be of uniform thickness over $\bar{M}$ with respect to a pre-determined time coordinate, if $\bar{M}$ is non compact.

To change signature, say from Lorentzian to Riemannian, the metric must first become degenerate (non-invertible). Such a degeneration of the metric is considered a singularity, so in the PDE approach is is treated as a "blow up" of the solution. As we know, blow ups are just a fact of life that we must live with for some equations. For example, a cosmological metric $\mathrm{d}s^2 = -\mathrm{d}t^2 + f(t) \mathrm{d}\mathbf{x}^2$ changes signature when $f(t)$ changes sign, or more conservatively you can only say that it degenerates when $f(t)=0$. There are certainly solutions with a Big Bang singularity that have this feature. If you choose your Cauchy surface $\bar{M}$ to asymptote to this Big Bang singularity in some directions, then you produce exactly the situation you've described: the solution metric exists and remains Lorentzian for a a positive time that depends on where you are on $\bar{M}$, but the metric degenerates arbitrarily quickly in those asymptotic directions.

Answer 3

It's easy to come up with examples of spacetimes, expressed in a certain coordinate system, such that the signature changes. An example is the spacetime defined by the metric $ds^2=-tdx^2-dx^2-dy^2-dz^2$. A calculation shows that this spacetime is flat, and therefore it's a perfectly fine model of the spacetime we live in (since the gravitational fields we live in are weak). It's a solution of the vacuum Einstein field equations.

However, since this example is flat, it follows that there is no experimental observable that would tell us when the catastrophe at $t=0$ is actually going to occur. By a change of coordinates, you can make this spacetime into Minkowski spacetime, with the usual expression for the metric. This change of coordinates, however, will have to be singular. When we study differential geometry, we get told over and over that coordinates are an arbitrary choice, but that's not quite true. It's only true that coordinate systems related by a diffeomorphism are equivalent.

General relativity can actually be expressed in more than one formalism. In the standard formalism, using tensors and the Einstein field equations, there is an implicit assumption that the metric never becomes degenerate, and therefore can never change signature. The machinery breaks down when the metric becomes degenerate. For example, we can't find the norm of a vector when the metric is degenerate.

So it's not actually a well-defined question whether a Cauchy problem can lead to a change of signature, because a Cauchy problem is something we think of as being posed in some absolute way, but the question of whether the signature can change depends on both the coordinates and the choice of formalism.