Origin of the term “physical density parameter” for $\omega_i = \Omega_i h^2$ When measuring cosmological density parameters, the parameters that are actually determined are the combinations $\Omega_i h^2$, sometimes defined as $\omega_i = \Omega_i h^2$, rather than the “plain” density parameters $\Omega_i = \rho_{i, 0} / \rho_{\mathrm{crit}, 0}$, so I understand the relevance of $\omega_i$.
However, $\omega_i$ are often called the “physical densities”, and I don’t understand why that is (is there something “un-physical” about $\Omega_i$)?
Unfortunately, I haven’t been able to find an explanation of this term “physical densities” anywhere – it always just seems to be taken for granted.
What is “physical” supposed to mean in this context?
Examples of using the term “physical densities” for $\omega_i$:

*

*https://en.wikipedia.org/wiki/Lambda-CDM_model#cite_note-physical_density-88

*https://ned.ipac.caltech.edu/level5/Sept03/Peacock/Peacock8.html

*https://cmb.wintherscoming.no/equations.php

*https://lambda.gsfc.nasa.gov/education/graphic_history/parameters.html

*https://arxiv.org/abs/1212.5226 (WMAP)

*https://arxiv.org/abs/1401.1389

*https://arxiv.org/abs/1807.06209 (Planck)

*https://arxiv.org/abs/2003.08277
 A: There's nothing "unphysical" about $\Omega_i$. From a purely theoretical perspective, $\Omega_i$ is arguably a more natural quantity than $\omega_i$. However, the value of $\Omega_i$ depends on the Hubble constant $H_0$, which is not known very precisely. Therefore it is often useful to quote observational results in terms of $\omega_i=\Omega_i h^2$ instead of $\Omega_i$, where $H_0=100h\ {\rm km\ s^{-1}\ {Mpc}^{-1}}$. The reason is that $\omega_i$ does not depend on the value of the Hubble constant. The observational uncertainty for $\omega_i$ is therefore much less than the uncertainty for $\Omega_i$.
It is similar to the idea that the mass of the Earth $M_\oplus$ is only known to five or so decimal places, but $GM_\oplus$ is known to much better precision (ten or so), because $G$ is not known very precisely and $GM_\oplus$ is the quantity that is actually measured.

Under Eq 2, the paper https://arxiv.org/abs/2007.08991 says "The dimensionless quantity $\omega_x \equiv \Omega_x h^2$ is proportional to the physical density of component $x$ at the present day." This suggests the following as a possible interpretation of the term "physical density parameter": the physical density $\rho_x$ is given by
$$
\rho_x = \frac{3 H_{100}^2 c^2 \omega_x}{8 \pi G},
$$
where $H_{100} \equiv 100\ {\rm km s^{−1} Mpc^{−1}}$. Since all quantities on the right hand side are known to good precision, so is $\rho_x$. If you used $H_0$ instead of $H_{100}$ and $\Omega_x$ instead of $\omega_x$, you would introduce uncertainty into $\rho_x$.
A: I would like to contribute to @Andrew's answer.
As we know $$\Omega_{\rm i} \equiv \frac{\rho_{\rm i}}{\rho_{\rm c}}$$ where $\rho_{\rm c} = \frac{3H^2}{8\pi G}$. Thus the physical density parameter becomes
$$\omega_{\rm i} = \Omega_{\rm i}h^2 = \rho_{\rm i}\frac{8\pi G}{3H^2}\frac{H^2}{10^4}=c\rho_i$$ where $c$ is some constant. Thus, unlike $\Omega_{\rm i}$, $\omega_{\rm i}$ does not contain $H_0$ and thus it's model independent.
To give you an example, consider $\omega_{\rm r}$ parameter
$$\omega_{\rm r} = 2.469 \times 10^{-5} \times [1 + \frac{7}{8} (\frac{4}{11})^{4/3}N_{\rm eff}]$$ This parameter only depends on the $N_{\rm eff}$ and $T_{\rm CMB}$. So if your cosmological model does not change these parameters then $\omega_{\rm r}$ will be constant. However, depending  on the $H_0$, $\Omega_{\rm r}$ will be different.
For instance, most cosmological models do not change $\omega_{\rm r}$. So, if your new cosmological model predicts different $H_0$ (and does not change $N_{\rm eff}$), you’ll have different $\Omega_{\rm r,0}$ but same $\omega_{\rm r}$.
