# Solutions in general relativity

1. Jan 18, 2009

### Eredir

Hi everyone, this is my first post in this nice forum.

I have some confusion regarding solutions of Einstein's field equations. I have read in several places that an exact solution to the field equations is a Lorentzian manifold. Now given a stress-energy tensor $$T_{\mu\nu}$$ the equations $$G_{\mu\nu} = 8\pi T_{\mu\nu}$$ determine the metric tensor $$g_{\mu\nu}$$. But a Lorentzian manifold $$(M,g)$$ is a differential manifold $$M$$ together with a pseudo-Riemannian metric $$g$$, so it's not clear to me how that manifold is related to the metric tensor $$g_{\mu\nu}$$.
There might exist several differential manifolds which could be given the same metric, so in which sense do we say that a Lorentzian manifold is a solution to the field equations?

2. Jan 18, 2009

### element4

I may have misunderstood your question. But in general relativity, loosely speaking, one has locally minkowski/pseudo-euclidean geometry (what einstein would call "freely falling elevator") and therefore by assumption spacetime is described by a Lorentzian manifold (and the solutions of Einstein's field equation should obey this). This is just how spacetime is modeled.
(It must said, im not an expert on differential geometry!)

3. Jan 18, 2009

### Eredir

But being a Lorentzian manifold is just a condition on the metric, that is having signature $$(1,3)$$ or $$(3,1)$$. The metric is an additional structure you put on a differentiable manifold to get a (pseudo-)Riemannian manifold, it does not identify a particular differentiable manifold except for general properties like its dimension. If this reasoning is correct then solving the field equations does not identify a particular Lorentzian manifold, but maybe a class of them up to diffeomorphism (but still there could be exceptions like exotic spheres and such).

4. Jan 18, 2009

### Staff: Mentor

In general a solution need not be unique in order to be a solution.

5. Jan 18, 2009

### Eredir

Of course this is true, but still not having a unique solution is quite peculiar, because in the weak field limit general relativity reduces to newtonian theory where the solution is unique.
Anyway, from the original question stems this new one: given that we can find several differential manifolds that are compatible with a metric tensor, how can we classify those solutions?

6. Jan 18, 2009

### Staff: Mentor

I am not sure I understand your point above. If you change the coordinate system in Newtonian gravity you get a different solution. Similarly in GR. In what sense is the GR solution non-unique while the Newtonian solution is unique?

7. Jan 18, 2009

### atyy

"Solutions to the EFE correspond to geometries rather than metrics given in a specific coordinate system, since the metrics related to spacetime diffeomorphisms are physically equivalent. As a consequence of this, equation (1) understood as a partial differential equations (PDE) system for the metric components gμν constitutes an underdetermined (incomplete) system that cannot produce a unique solution. An extra ingredient, namely a gauge fixing procedure, must be provided to establish a standard PDE problem—note this remark applies to any geometrically
introduced PDE." Jaramillo et al, http://arxiv.org/abs/0712.2332

8. Jan 18, 2009

### Jaunty

Eredir,
You are absolutely correct that the solution to Einstein's field equations are a differentiable manifold $$M$$ together with a pseudo-Riemannian metric (usually a $$(3,1)$$ one). A priori, there is not much of a relation between the metric and the underlying manifold - mainly the manifold provides the topological structure of you spacetime. Different metrics on the same manifold will yield different solutions to the field equations. For example, constant radius slices of both the Kerr metric (describing a rotating, uncharged black hole) and the Reissner-Nordstrom metric (describing a charged, non-rotating black hole) are two different metrics on the same manifold - namely $$\mathbb{R}\times S^2$$. You can also have more complicated examples such as black holes in an asymptotically anti-deSitter background (anti-deSitter is just a maximally symmetric space with constant negative curvature) or things like taub-NUT spacetime that have the topology of $$\mathbb{R}^4$$ (and also serves as a counterexample to many intuitive sounding statements in the theory of global structure). I'm not sure how you'd give different manifolds the "same metric". You could do something like giving the torus $$S^1\times S^1$$ a flat metric and also give $$\mathbb{R}^2$$ a flat metric but I wouldn't really call those the same metric (although they are certainly similar). So the point is it's the metric together with the manifold that provides the complete solution to the field equations.

The only other thing to remember is that the curvature that's expressed in $$G_{\mu\nu}$$ uses the metric-compatible connection - this gives you a specific $$G_{\mu\nu}$$ given your metric.

9. Jan 18, 2009

### Eredir

In newtonian gravity the spacetime is fixed, so changing the coordinate system merely changes the parametrizations for the wordlines. In this case events take place in a unique and defined manifold which is indipendent of the dynamic, while in general relativity determining the manifold constitutes part of the problem. If we can find a metric tensor by solving the field equations, but can't assign a unique manifold to it, then I would say that there isn't a unique solution to the problem. Maybe this idea is poorly worded or just plain wrong, as I said before I have some confusion regarding this argument.

Thanks atyy, something like "metrics related to spacetime diffeomorphisms are physically equivalent" might be what I'm searching for.

10. Jan 18, 2009

### Eredir

Yes, I wouldn't say that they are the same metric, given that those two spaces are not homeomorphic but only locally homeorphic. But given two spaces that are homeomorphic, or better diffeomorphic, I would be inclined to say that the solutions should be similar, and this seems hinted by the part atyy quoted of that article.

I haven't fully undestood this, I'll give it some more thought tomorrow!

11. Jan 18, 2009

### atyy

Oh, ok, d'Inverno mentions something about this in his book, apparently Cartan solved the problem, and it involves computing very high order derivatives (10th or something). I've read elsewhere on these forums that it's also in Peter Olver's book "Equivalence, Invariants, and Symmetry", but I haven't studied it myself.

12. Jan 18, 2009

### atyy

13. Jan 19, 2009

### Jaunty

Oh, all it means is that of the many connections one could choose - many ways of transporting vectors between tangent spaces - you pick the one that makes the covariant derivative of the metric zero, $$\nabla_\rho g_{\mu \nu}=0$$. This is only a point worth making if you've already studied differential geometry before learning GR. In normal GR classes, they introduce the (Christoffel) connection coefficients as
$$\Gamma^\sigma_{\mu \nu}=\frac{1}{2}g^{\sigma \rho}(\partial_\mu g_{\nu \rho} + \partial_\nu g_{\rho \mu} - \partial_\rho g_{\mu \nu})$$
and don't usually mention that there are other connections that are possible.

Last edited: Jan 20, 2009
14. Jan 20, 2009

### Eredir

Thanks guys, you have been very helpful!

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