Spatial Curvature and Critical Density

Critical density is the density that makes an FLRW model spatially flat for a given nonzero expansion rate. It is an instantaneous reference, $\rho_c=3H^2/(8\pi G)$, not a new material component. Comparing the total physical density with $\rho_c$ converts the Friedmann constraint into a curvature sign test. The resulting density parameters are useful only after their component convention is declared and become singular when $H=0$, even though the underlying geometry and Friedmann equation can remain regular.

Required background. The Einstein Equation supplies the Einstein Equations and curvature-source relation; Flat-Background Hilbert Stress-Energy from Metric Variation supplies the Stress–Energy and Conservation meaning of physical density.

Helpful background. Relativistic Matter and Compact Objects supplies the matter closures whose densities enter the component sum.

Critical density from the Friedmann constraint

Use a dimensionless scale factor $a_0=1$ and a dimensionful constant $K$ such that the spatial Ricci scalar is $^{(3)}R=6K/a^2$. Treat the cosmological constant, if present, as the physical effective density $\rho_\Lambda=\Lambda/(8\pi G)$. Then

$$H^2 =\frac{8\pi G}{3}\rho-\frac{K}{a^2},$$

where $\rho$ is the sum of all physical component densities, including $\rho_\Lambda$. For $H\ne0$, define

$$\rho_c\equiv\frac{3H^2}{8\pi G}, \qquad \Omega_i\equiv\frac{\rho_i}{\rho_c}, \qquad \Omega\equiv\sum_i\Omega_i.$$

Rearranging the constraint gives

$$\boxed{ \Omega-1=\frac{K}{a^2H^2} }.$$

Therefore

$$\begin{array}{c|c|c} K\gt0 & \Omega\gt1 & \text{positive spatial curvature},\\ K=0 & \Omega=1 & \text{spatially flat},\\ K\lt0 & \Omega\lt1 & \text{negative spatial curvature}. \end{array}$$

This sign relation assumes that $\Omega$ includes every physical contribution used on the source side. If $\Lambda$ remains geometric instead, define $\Omega_\Lambda=\Lambda/(3H^2)$ and write $\Omega_{\rm matter}+\Omega_\Lambda-1=K/(a^2H^2)$. Switching conventions mid-calculation changes what the symbol $\Omega$ means.

Curvature density parameter and closure

It is often convenient to leave curvature on its own side and define

$$\Omega_K\equiv-\frac{K}{a^2H^2}.$$

The closure relation is then

$$\boxed{\Omega+\Omega_K=1}.$$

The minus sign is conventional and important: a positively curved model has $\Omega_K\lt0$. Curvature can be assigned the effective algebraic density $\rho_K=-3K/(8\pi G a^2)$, but it is not stress-energy. It has no material rest frame, perturbation sound speed, or local equation of state. Tong derives the critical-density and closure notation in Tong 2019, §1.4, pp. 47–48, eqs. (1.65)–(1.67) (PDF), where the symbols denote present-day values; here the same definitions are evaluated at an arbitrary event with $H\ne0$.

For $K\ne0$, the physical curvature radius is

$$R_{\rm curv}=\frac{a}{\sqrt{\lvert K\rvert}},$$

so

$$\lvert\Omega_K\rvert =\frac{1}{H^2R_{\rm curv}^2}.$$

Small $\lvert\Omega_K\rvert$ says that the curvature radius is large compared with the instantaneous Hubble length $\lvert H\rvert^{-1}$. It does not determine the global topology: distinct quotients can share the same local curvature.

The definitions and closure relation are collected in Particle Data Group 2026, §22.1.4, p. 3, eqs. (22.11)–(22.15) (PDF).

How departure from flatness evolves

Consider one separately conserved constant-$w$ fluid with $\Lambda=0$. Since $\rho\propto a^{-3(1+w)}$,

$$\frac{\Omega-1}{\Omega} =\frac{3K}{8\pi G\rho a^2} \propto a^{1+3w}.$$

This relation is exact under the one-fluid assumptions. Only near flatness, where $\Omega\simeq1$, may it be shortened to

$$\lvert\Omega-1\rvert\propto a^{1+3w}.$$

Thus departure from flatness grows with expansion during radiation domination as $a^2$ and during matter domination as $a$, while an accelerating component with $w\lt-1/3$ drives it down. For mixtures or time-dependent $w$, use the full Friedmann history rather than one global exponent. The Particle Data Group discusses this evolution in Particle Data Group 2026, §22.2.4, pp. 9–10, eq. (22.34) (PDF).

The statement is about relative curvature, not about curvature being dynamically generated. $K$ is constant in the FLRW ansatz; the changing quantity is the ratio of $K/a^2$ to $H^2$.

Where density parameters fail

At a turnaround, bounce, or static event, $H=0$. Then $\rho_c=0$ and $\Omega_i$ and $\Omega_K$ either diverge or become undefined ratios such as $0/0$. This is a poor coordinate choice on the space of background variables, not necessarily a spacetime singularity. Use the undivided constraint

$$\frac{K}{a^2} =\frac{8\pi G}{3}\rho$$

in the effective-vacuum convention at that event. A closed model can reach $H=0$ at finite positive $a$ while every dimensional term remains finite.

Likewise, “critical” does not mean dynamically stable or observationally preferred. It means only the density corresponding to $K=0$ for the measured $H$ within the assumed general-relativistic FLRW model and chosen component budget.

Common pitfalls

Nearly flat is not exactly flat. A small measured $\lvert\Omega_K\rvert$ places the curvature radius far beyond the Hubble length, but no finite-precision bound establishes $K=0$ as a mathematical identity.

Curvature sign does not determine global topology. A complete positive-constant-curvature space form is compact but may be a non-simply-connected spherical quotient. Flat and hyperbolic geometries also admit compact quotients, so sign alone fixes neither topology nor simple connectedness.

Critical density is model dependent. The numerical value uses the inferred $H$, the gravitational coupling $G$, and the FLRW/general-relativistic interpretation. At fixed $H$ and $G$, the algebraic value $3H^2/(8\pi G)$ is independent of how the source budget is partitioned; that budget instead changes the inferred parameters and the interpretation of each $\Omega_i$.

Exercises

Derive the exact one-fluid relation $(\Omega-1)/\Omega\propto a^{1+3w}$ without first assuming $\Omega\simeq1$.

Solution

From the definitions,

$$\Omega-1=\frac{K}{a^2H^2}, \qquad \Omega=\frac{8\pi G\rho}{3H^2}.$$

Taking their ratio cancels $H^2$:

$$\frac{\Omega-1}{\Omega} =\frac{3K}{8\pi G\rho a^2}.$$

Separate conservation gives $\rho=\rho_\star(a/a_\star)^{-3(1+w)}$. Since $K$ is constant, the ratio is proportional to $a^{3(1+w)-2}=a^{1+3w}$. Replacing the ratio by $\Omega-1$ is an additional near-flat approximation, not part of the exact derivation.

A model has $\Omega_K=-0.04$ at an event with $H\ne0$. Determine the local curvature sign and $\lvert H R_{\rm curv}\rvert$. What cannot be inferred about topology?

Solution

Since $\Omega_K=-K/(a^2H^2)$ is negative, $K$ is positive. The spatial geometry is locally positively curved. Its magnitude gives

$$\lvert H R_{\rm curv}\rvert =\frac{1}{\sqrt{\lvert\Omega_K\rvert}} =5.$$

This determines only a local curvature scale relative to the Hubble length. It does not determine whether the spatial slice is simply connected or which discrete quotient and global topology, if any, has been chosen.

References

• Particle Data Group. “Big-Bang Cosmology.” In Review of Particle Physics 2026, revised August 2025. Official PDF.
• Tong, David. Cosmology. University of Cambridge Part II Mathematical Tripos lecture notes, Michaelmas 2019. Official PDF.