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When we write $$\Omega_m = \frac{\rho_M}{\rho_{\mathrm{crit,o}}}$$ what density does $\rho_M$ refer to? As in, suppose our universe was made with just hydrogen. Then would $\rho_M$ mean just the average (usual) density of hydrogen gas (so $0.08988 \; g/m^3$)?

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We split the energy content of the universe into three parts that scale differently as the universe expands. Radiation and other light highly relativistic components have an energy density that scales as $a^{-4}$ as the universe expands. Dark energy has an energy density that is constant and doesn't scale at all, and what's left scales as $a^{-3}$ i.e. its energy density is inversely proportional to volume. This is the component referred to by $\rho_M$.

This matter includes visible matter and also dark matter - whatever that is. For the visible matter the dominant contribution comes from hydrogen and helium, and this density is just the number of hydrogen and helium atoms per cubic metre averaged over some suitably large scale. So it is just density in the usual meaning of the term. In principle the same is true of dark matter, though since we know next to nothing about dark matter it's hard to say anything definitive.

The density of hydrogen you quote, $0.08988$ g/m$^3$, is the density at STP. The actual average density of hydrogen in the universe at the present time is around 1 atom per cubic metre. The density of dark matter is equivalent to around 4 hydrogen atoms per cubic metre.

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