Of what I understand, current observations indicate the universe curvature is closely a flat one. This is made possible since adding up the energy densities of regular matter, dark matter and dark energy sums to just the critical density required for a flat geometry.

However, the model also describes how the densities change with scaling: matter and radiation decay to 3rd and 4th powers while dark energy density remains constant in scale.

My question - doesn't that mean that the overall density decays with expansion, specifically to below the critical density- which means the curvature will drift from flat to negative with time?

Is the flat geometry temporary and incidental with our time? Or is it constant, but if so, how does that agree with the changing energy density?


1 Answer 1


The overall density decreases as the universe expands, but so does the critical density. Specifically, one of the Friedmann equations can be written as $$ \dot{a}^2 -\frac{8\pi G}{3}\rho a^2= -kc^2, $$ where the scale factor $a(t)$ describes the expansion of the universe, $\rho$ is the total mass density (radiation, baryonic matter, dark matter, and dark energy) and the constant integer $k$ can have the values $k = 1$, $0$ or $-1$ for positive, zero, or negative curvature respectively. We can rewrite this as $$ H^2\left(1 - \frac{\rho}{\rho_\text{c}}\right) = -\frac{kc^2}{a^2},\tag{1} $$ with $H(t) = \dot{a}/a$ the Hubble parameter and $$ \rho_\text{c}(t) = \frac{3H^2(t)}{8\pi G} $$ the critical density. So $\rho_\text{c} \sim H^2$. Since $k$ is a constant, the sign of the curvature of the universe cannot change. For a flat universe, $k=0$ and $\rho \equiv \rho_\text{c}$ at all times.

What happens if the universe is not flat? For this, we need to know how $H(t)$ changes over time. The present day corresponds with $a=1$, so that $$ H_0^2\left(1 - \frac{\rho_0}{\rho_\text{c,0}}\right) = -kc^2, $$ and therefore $$ H^2\left(1 - \frac{\rho}{\rho_\text{c}}\right) = H_0^2\left(1 - \frac{\rho_0}{\rho_\text{c,0}}\right)a^{-2}, $$ which can be written in terms of the radiation, matter, and dark energy densities as $$ H^2\left(1 - \Omega_{R} - \Omega_{M} - \Omega_{\Lambda}\right) = H_0^2\left(1 - \Omega_{R,0} - \Omega_{M,0} - \Omega_{\Lambda,0}\right)a^{-2}. $$ I will assume a cosmological constant dark energy. Using the shorthand notation $\Omega_{K}= 1 - \Omega_{R} - \Omega_{M} - \Omega_{\Lambda}$, the equation takes the form $$ H^2\,\Omega_{K} = H_0^2\,\Omega_{K,0}\,a^{-2}.\tag{2} $$ A flat universe corresponds with $\Omega_{K} = 0$. Since $$\begin{align} \Omega_{R} &= \frac{\rho_{R}}{\rho_{c}} = \frac{\rho_{c,0}}{\rho_{c}}\frac{\rho_{R,0}\,a^{-4}}{\rho_{c,0}} = \frac{H_0^2}{H^2}\Omega_{R,0}\,a^{-4}\\ \Omega_{M} &= \frac{\rho_{M}}{\rho_{c}} = \frac{\rho_{c,0}}{\rho_{c}}\frac{\rho_{M,0}\,a^{-3}}{\rho_{c,0}} = \frac{H_0^2}{H^2}\Omega_{M,0}\,a^{-3}\\ \Omega_{\Lambda} &= \frac{\rho_{\Lambda}}{\rho_{c}} = \frac{\rho_{c,0}}{\rho_{c}}\frac{\rho_{\Lambda}}{\rho_{c,0}} = \frac{H_0^2}{H^2}\Omega_{\Lambda,0}, \end{align} $$ we can solve equation (2) for $H^2$: $$ H^2 = H_0^2\left(\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}\right), $$ so that $$ \Omega_{K} = \frac{\Omega_{K,0}\,a^{-2}}{\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0}\,a^{-2} + \Omega_{\Lambda,0}}. $$ This has two remarkable consequences:

  1. At early times, $a\rightarrow 0$, and $$ \Omega_{K} \approx \frac{\Omega_{K,0}}{\Omega_{R,0}}a^2 \rightarrow 0. $$ In other words, the early universe must have been very nearly flat. This is known as the flatness problem, and is one of the motivations for inflation theory.

  2. At late times, $a\rightarrow \infty$, and $$ \Omega_{K} \approx \frac{\Omega_{K,0}}{\Omega_{\Lambda,0}}a^{-2} \rightarrow 0, $$ so dark energy is actually pushing the universe towards a flat geometry (again).

Conclusion: the geometry of the universe is either exactly flat and will remain so, or is becoming more and more flat because of dark energy.


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