Solutions in general relativity

In summary, the conversation discusses the relationship between solutions to Einstein's field equations and Lorentzian manifolds. It is mentioned that a Lorentzian manifold is a solution to the field equations, but the metric tensor g_{\mu\nu} is not unique and can correspond to multiple differential manifolds. It is also noted that in general relativity, the metric together with the manifold provides the complete solution to the field equations, and the metric-compatible connection gives the specific curvature expressed in G_{\mu\nu}. The conversation ends with a discussion about the uniqueness of solutions in Newtonian gravity compared to general relativity.
  • #1
Eredir
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Hi everyone, this is my first post in this nice forum. :smile:

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 [tex]T_{\mu\nu}[/tex] the equations [tex]G_{\mu\nu} = 8\pi T_{\mu\nu}[/tex] determine the metric tensor [tex]g_{\mu\nu}[/tex]. But a Lorentzian manifold [tex](M,g)[/tex] is a differential manifold [tex]M[/tex] together with a pseudo-Riemannian metric [tex]g[/tex], so it's not clear to me how that manifold is related to the metric tensor [tex]g_{\mu\nu}[/tex].
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?
 
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  • #2
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, I am not an expert on differential geometry!)
 
  • #3
element4 said:
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, I am not an expert on differential geometry!)

But being a Lorentzian manifold is just a condition on the metric, that is having signature [tex](1,3)[/tex] or [tex](3,1)[/tex]. 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
Eredir said:
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?
In general a solution need not be unique in order to be a solution.
 
  • #5
DaleSpam said:
In general a solution need not be unique in order to be a solution.

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
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
Are you asking about something like this?

"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
Eredir,
You are absolutely correct that the solution to Einstein's field equations are a differentiable manifold [tex]M[/tex] together with a pseudo-Riemannian metric (usually a [tex](3,1)[/tex] 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 [tex]\mathbb{R}\times S^2[/tex]. 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 [tex]\mathbb{R}^4[/tex] (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 [tex]S^1\times S^1[/tex] a flat metric and also give [tex]\mathbb{R}^2[/tex] 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 [tex]G_{\mu\nu}[/tex] uses the metric-compatible connection - this gives you a specific [tex]G_{\mu\nu}[/tex] given your metric.
 
  • #9
DaleSpam said:
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?

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
Thanks for your helpful reply Jaunty.

Jaunty said:
You could do something like giving the torus [tex]S^1\times S^1[/tex] a flat metric and also give [tex]\mathbb{R}^2[/tex] a flat metric but I wouldn't really call those the same metric (although they are certainly similar).

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.

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

I haven't fully undestood this, I'll give it some more thought tomorrow!
 
  • #11
Eredir said:
Thanks atyy, something like "metrics related to spacetime diffeomorphisms are physically equivalent" might be what I'm searching for.

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.
 
  • #13
Eredir said:
I haven't fully undestood this, I'll give it some more thought tomorrow!

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, [tex]\nabla_\rho g_{\mu \nu}=0[/tex]. 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
[tex]\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})[/tex]
and don't usually mention that there are other connections that are possible.
 
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  • #14
Thanks guys, you have been very helpful!
 

FAQ: Solutions in general relativity

1. What is general relativity?

General relativity is a theory of gravity proposed by Albert Einstein in 1915. It describes how massive objects affect the fabric of space-time, and how this curvature of space-time causes objects to move on a curved path.

2. How do solutions in general relativity differ from solutions in classical physics?

The main difference between solutions in general relativity and solutions in classical physics is that general relativity takes into account the curvature of space-time, whereas classical physics assumes a flat space-time. This leads to different predictions for the behavior of massive objects and the effects of gravity.

3. What are the most well-known solutions in general relativity?

The most famous solution in general relativity is the Schwarzschild solution, which describes the space-time around a non-rotating, spherically symmetric mass. Other well-known solutions include the Kerr solution for a rotating mass and the Friedmann-Lemaître-Robertson-Walker solution for the expanding universe.

4. How are solutions in general relativity tested and confirmed?

Solutions in general relativity are tested and confirmed through various observations and experiments. For example, the predictions of general relativity for the orbit of Mercury were confirmed by observations, and the bending of light near massive objects has been observed. Additionally, the existence of gravitational waves, predicted by general relativity, was confirmed by the LIGO experiment in 2015.

5. Are there any unsolved problems or limitations of solutions in general relativity?

While general relativity has been incredibly successful in explaining the behavior of gravity on a large scale, there are still some unsolved problems and limitations. For example, general relativity does not fully account for the behavior of gravity on a quantum level. Additionally, it cannot fully explain the acceleration of the expansion of the universe, leading to the need for the concept of dark energy. These are areas of ongoing research and development in the field of general relativity.

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