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Short version of my question is as follows: What is curvature of spacetime (rather than space) in Friedman cosmological models?

Now the long version, including evolution of my thoughts and background of the question:

Quite recently I read in a tweet by Quanta Magazine: "If the universe's density is equivalent to 5.7 atoms per cubic meter, the universe will lie flat like a sheet of paper. If that number rises to 6, it will curve around like a sphere".

I have seen diagrams illustrating three types of geometry (hyperbolic, euclidean and elliptic or spherical) depending on the density of universe many times before. But only now I completely realized that quite empty universe would be negatively curved, not flat. In other words, Euclid geometry is not so natural choice as it would seem at first thought. If you want school geometry, you need matter and gravity.

Than I found that quite empty universe is described by Milne model, and realized that curvature of space is curvature of spacetime are two different things. And that it is spacetime which is flat (with Minkowski geometry as the natural choice).

My hunch is that Friedman models (without dark energy) have curvature of spacetime zero (no matter, Milne model) or positive (matter with gravity). Am I right?

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FLRW models have nonzero curvature of spacetime, because that's how general relativity describes gravity, and they have gravity in them. (The empty Milne universe is the exception.)

FLRW models can have either zero or nonzero spatial curvature.

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  • $\begingroup$ "FLRW models have nonzero curvature of spacetime". By nonzero you mean positive, right? $\endgroup$ – Leos Ondra Dec 27 '19 at 23:00
  • $\begingroup$ It can be either positive or negative, depending on if 1-Ωt=Ωk is smaller or larger than zero. If it is exactly 0 then the curvature is 0. If the overall spatial curvature is flat once it stays flat forever, but locally you can have it close in on itself by creating black holes, whose interior can also be desribed as a closed FLRW universe in the Oppenheimer Snyder Collapse. $\endgroup$ – Gendergaga Dec 28 '19 at 0:43
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    $\begingroup$ @LeosOndra: By nonzero you mean positive, right? The Riemann tensor is a rank-4 tensor with 20 independent components. Tensors aren't positive or negative. This is similar to vectors (which are rank-1 tensors): vectors aren't positive or negative. $\endgroup$ – user4552 Dec 28 '19 at 1:11
  • $\begingroup$ @Yukterez I think that you write 1/ about curvature of space, not spacetime, 2/ local inhomogenities in the homogeneous universe (which I am interested in) $\endgroup$ – Leos Ondra Dec 28 '19 at 12:49
  • $\begingroup$ @BenCrowell I thought that in homogeneous and isotropic universe (with smooth distribution of energy, without galaxies or stars) it is possible to describe curvature of spacetime in simpler terrms. But I will have a look at some textbook and will be back soon. $\endgroup$ – Leos Ondra Dec 28 '19 at 12:57
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You can't describe the curvature of spacetime with a single number; you need the full 20 components of the Riemann tensor for a complete description. The same is true for 3D space (though you need less numbers), it's just that the assumption of homogeneity and isotropy lets us get away with using a single number.

Riemann tensor

Thanks to the simplicity of FRLW spacetime, though, we can describe the Riemann tensor with just two equations. If I did everything right, in an orthonormal basis they are

$$R_{0i0j} = -(\dot{H} + H^2)\delta_{ij} \quad\text{and}\quad R_{ijkl} = \left(H^2 + \frac{k}{a^2}\right) (\delta_{ik}\delta_{jl} - \delta_{il}\delta{jk}),$$

with Latin indices taking values in $\{1,2,3\}$ and all other components (with an odd number of zero indices) vanishing. From this, one thing you can do is calculate the Ricci and Kretschmann scalars,

$$R = R^{\mu\nu}{}_{\mu\nu} \quad\text{and}\quad K = R^{\mu\nu\alpha\beta}R_{\mu\nu\alpha\beta},$$

which are sort of the trace and the square of the Riemann tensor, to get an idea of what's going on. Again, if I haven't made any mistakes, they are

$$R = 6\left( \dot{H} + 2H^2 + \frac{k}{a^2} \right) \quad\text{and}\quad K = 12 \left[\left(\dot{H} + H^2\right) + 8 \left(H^2 + \frac{k}{a^2}\right)\right].$$

Friedmann equations

To relate these to the matter content of the universe, we use the Friedmann equations

$$H^2 + \frac{k}{a^2} = \frac{8\pi}{3} \rho$$

$$\dot{H} + H^2 = -\frac{4\pi}{3} (\rho + 3p).$$

Using these, you can see that the components of the Riemann tensor are just the sides of the equations, and the scalars turn out to be

$$R = 8\pi(\rho-3p) = 8\pi (1-3w)\rho$$

$$K = \frac{64\pi^2}{3} \left[(\rho+3p)^2 + 32\rho^2\right] = \frac{64\pi^2}{3} \left[(1+3w)^2 + 32\right] \rho^2,$$

where I've also included the standard cosmological equation of state $p = w \rho$ for a single fluid.

Analysis

So what can we get from this? You can see that if there is any energy content at all in the universe ($\rho \neq 0$), the Riemann tensor is nonzero. This means that spacetime is flat (zero Riemann tensor) if and only if it is empty: the Milne model is just a section of Minkowski spacetime, and it is (spacetime-)flat even if it doesn't look like it.

For the three most common fluids considered in cosmology, we have $w = 0$ (dark matter), $w = 1/3$ (radiation) or $w = -1$ (dark energy). You can see that in all cases the scalars are positive, except for $w = 1/3$ when the Ricci scalar is zero. But you also shouldn't pay too much attention to the latter's sign, because it depends on the signature for the metric; if I had used $(+\ -\ -\ -)$ signs, $R$ would have come out with the opposite sign.

The Quanta Magazine quote

So if we found that $\text{empty} \iff \text{flat}$, why does the quote claim that for low densities the universe is negatively curved? That's because we observe that the universe is expanding, and so we're forced to use a nonzero $H$ in the equations, and this "moves the zero" of curvature, so to speak. An empty universe would actually not expand and it would be flat, but this contradicts observations.

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