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I am kind of new to GR but I have been familiar with the concepts for a long time, I am getting used to the mathematics just now. My question is, what would GR predict if we would have an empty universe? No energy whatsoever, only a spacetime continuum?

What I mean by energy is any type of energy, including mass, so a 0 curvature universe.

Conifold
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Darawan
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    That's Minkowski space! Either that or any other Ricci-flat spacetime, such as spacetimes with gravitational waves. – Slereah Aug 08 '15 at 20:17
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    So you want to know what $G_{\mu\nu}=0$ (or equivalently $T_{\mu\nu}=0$) represents? – Kyle Kanos Aug 08 '15 at 20:19
  • Yes, 0 curvature – Darawan Aug 08 '15 at 22:00
  • @Darawan 0 Riemann curvature? The unique solution to $\mathrm{Riem}[g]=0$ is the Minkowski metric. 0 Ricci curvature? See Wiki. (Note that because the metric is nondegenerate, the scalar curvature can only vanish iff the Ricci curvature vanishes.) – Ryan Unger Aug 09 '15 at 03:46
  • @Slereah OP is not asking for a vanishing energy momentum tensor, but rather for a spacetime without any kind of energy at all. It is possible for a vaccum solution to have nonzero energy, for instance the ADM energy of Schwarzschild spacetime is just the mass of the body. – Ryan Unger Aug 09 '15 at 03:52
  • Are you also considering the cosmological constant in the EFE? – Kyle Kanos Aug 09 '15 at 14:33

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Solutions to $G_{\mu\nu}=0$ are called vacuum solutions in GR, it follows mathematically that this happens if and only if the Ricci tensor vanishes, i.e. the solutions are exactly the Ricci flat Lorentzian manifolds. In most known explicit examples only some region is Ricci flat (e.g. around Schwarzschild or Kerr black holes), but some global vacuum solutions without singularities are known. Their existence contradicts the strong Mach principle, which implies that universe without matter (singularities are interpreted as degenerated matter) must be the flat Minkowski spacetime. One example is the Ozsváth–Schücking vacuum, describing a sinusoidal gravitational wave, others are given by a family of Kasner vacua, describing quaint expanding universes without matter. In Kasner universes the expansion can never be isotropic, in fact if the volume overall is expanding with time at least one spatial direction will be contracting.

Keep in mind however that "only a spacetime continuum" (zero stress-energy and hence Einstein tensor) does not imply "no energy whatsoever", because gravitational field itself can do work, and therefore carries energy. For instance, Ozsváth–Schücking waves transport energy just as electromagnetic waves would. "No energy whatsoever" means that not only $G_{\mu\nu}=0$ but even $\partial_{\alpha}g_{\mu\nu}=0$, i.e. the spacetime is locally flat Minkowski. Even those can be peculiar though, for instance the locally flat Deutsch–Politzer space contains closed timelike curves ("time machines").

Conifold
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  • Well doesn't this give a clue to what dark energy and dark matter is then? Thank you for you answer, I think I've got it. – Darawan Aug 09 '15 at 10:42
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    $g_{\mu\nu}=0$ is not Minkowski space. – Ryan Unger Aug 09 '15 at 11:56
  • @0celo7 Sorry, forgot the $\nabla$. – Conifold Aug 10 '15 at 19:19
  • Then what's $\nabla$? If it's the covariant derivative as usual, then $\nabla g_{\mu\nu} = 0$ in any spacetime whatsoever in GTR because the connection is Levi-Civita. Maybe it would work better in terms of vanishing Weyl curvature as the additional condition, $C_{mu\nu\sigma\rho} = 0$, because the Weyl curvature is part of the Ricci curvature not determined by the stress-energy of sources. – Stan Liou Aug 10 '15 at 19:42
  • @Stan Liou I meant the gradient in local coordinates, meaning the entries are constant, but $\nabla$ is confusing. Maybe $Dg_{\mu,\nu}=0$? – Conifold Aug 10 '15 at 22:41
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    @Conifold: the clear way of writing this would be $\partial_{a}g_{bc} =0$ – Zo the Relativist Aug 10 '15 at 23:23
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OP is looking for the vacuum solutions to the Einstein field equations. Including only the cosmological constant $\Lambda$, the EFE become the Lambdavacuum field equations, $$ R_{\mu\nu}=\left(\frac12R-\Lambda\right)g_{\mu\nu}\tag{1} $$ with $R_{\mu\nu}$ the Ricci curvature tensor and $R$ the Ricci scalar. Solutions to this depend on the sign/value of $\Lambda$.

For $\Lambda\neq0$, the spacetime must be treated as curved, as (1) does not admit flat spacetime solutions (cf. Padmanabhan (2003), pdf). Solutions to (1) result in either de Sitter space (for $\Lambda>0$) or anti-de Sitter space (for $\Lambda<0$). See also this NED article (also by Padmanabhan) and this Physics.SE post.

For $\Lambda=0$, the vacuum field equations can be solved with flat Minkowski space, Schwarzschild space or Kerr space (provided we are looking at the space outside some sphere of non-zero radius). Solutions of this case would be singularity free, Ricci flat but not necessarily Riemann flat. See also this Physics.SE post.

Community
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Kyle Kanos
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The Einstein Field equations are $G_{\mu\nu}=8\pi GT_{\mu\nu}$.

An empty universe would be one where $T_{\mu\nu}=0$ The Einstein field equations would than read

$G_{\mu\nu}=R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=0$.

The 00 term of this (for an FLRW metric) is

$\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G}{3}\rho-\frac{\kappa}{a^2}$.

You say that you want the density of the universe to be 0, so $\rho=0$. Setting the curvature to be flat, positive, or negative ($\kappa = 0/+1/-1$, respectively), gives you three differential equations to solve for the scale factor as a function of time ($a(t)=...$).

Bob
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