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This question might be naive for the practitioers of GR and Cosmology. But as an outsider in the field, I could not resist the temptation of asking this.

  1. The theory of General Relativity dictates that whenever there is matter or energy density, it curves spacetime. But even though the matter and energy densities in the Universe are nonzero, the Universe is spatially flat. Why is that?

  2. What determines the curvature parameter $k$ in the Friedman-Robertson-Walker metric? In particular, I want to know whether $k$ depends on (and therefore, is determined by) the various energy densities in the Universe or is it a parameter independent of various energy densities?

EDIT: From Friedman equations, $\frac{k}{a^2}$ appears as another component of energy density in addition to $\rho_{mat},\rho_{rad}$ and $\rho_{\Lambda}$. Does it therefore mean $\frac{k}{a^2}$ is an independent component of energy density? And like other energy components ($\rho_{mat},\rho_{rad}$ and $\rho_{\Lambda}$), is $k$ also an independent experimental input?

However, according to GR energy densities should fix the curvature, and it's not independent. Then what is that happens here?

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  • $\begingroup$ Spacetime is curved, but the spatial slices are not. $\endgroup$
    – Javier
    Commented Apr 12, 2017 at 19:02
  • $\begingroup$ curvature of universe is not the same as curvature due to mass/gravity. An open/closed/flat universe will always be open/closed/flat regardless what the energy densities become. The amount of curvature can change, but the type doesn't $\endgroup$
    – Jim
    Commented Apr 12, 2017 at 19:38

3 Answers 3

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I have been old that a non zero energy density will always curve spacetime but that curved spacetime can still be spatially flat.

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I think I have figured out the answer. Since $$\Omega_{mat}+\Omega_{rad}+\Omega_{\Lambda}-\frac{k}{H^2a^2}=1,$$ the parameter $k$ is not independent. And if $\Omega_{mat}+\Omega_{rad}+\Omega_{\Lambda}=1$, then $k=0$.

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  • $\begingroup$ $\Omega \equiv \rho/\rho_c$, where $\rho_c$ is the critical density that results in a spatially flat universe, ie $\sum_i \Omega_i = 1 \iff k = 0$ by definition $\endgroup$
    – Christoph
    Commented Apr 12, 2017 at 18:46
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  1. If matter is homogeneously distributed (aka perfect fluid), there can be no spatial curvature on large scales (more precisely, homogeneity and isotropy leads to 3 cases with $k=\pm 1, 0$ with maximal symmetry). Curvature appears when you consider smaller structures, like stars and galaxies: in these cases matter is concentrated in a region of space.

  2. Yes, k is determined from observation of density parameters, and it's very close to 0.

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