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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 $\omega_i$:

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2 Answers 2

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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 $\rho_x$.

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  • $\begingroup$ +1 Good analogy re: $GM$ accuracy/measure compared to $G$ and $M$ seperate. Also, $M_\odot$ represents the sun's mass and usually $M_\oplus$ that for earth? Just nitpicking. $\endgroup$
    – joseph h
    Commented Jul 25, 2022 at 1:15
  • $\begingroup$ @josephh Of course you are right, thank you! I'll fix that. $\endgroup$
    – Andrew
    Commented Jul 25, 2022 at 1:22
  • $\begingroup$ No problems Andrew. Cheers. $\endgroup$
    – joseph h
    Commented Jul 25, 2022 at 1:24
  • $\begingroup$ Sure, this is what I understood. I’m mostly confused about the term “physical”, since it doesn’t really seem like an appropriate term to describe this difference (to go with your analogy, we don’t call $GM_\oplus$ the “physical mass of the Earth”). But if you’re saying that this is all there is to it, I can live with that. $\endgroup$
    – Socob
    Commented Jul 25, 2022 at 10:28
  • $\begingroup$ @Socob To be completely up front, I've never heard anyone use the term "physical" for this quantity, so it's possible they have something specific in mind that I'm not aware of. But, I don't see how introducing $h$ could be part of a "physically natural" quantity, given that its value depends on an arbitrary reference scale of $100\ {\rm km\ s^{-1}\ Mpc^{-1}}.$ $\endgroup$
    – Andrew
    Commented Jul 25, 2022 at 11:23
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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 $\omega_{\rm r}$.

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  • $\begingroup$ Well, this is all very reasonable, but as I was trying to say in the comments, it’s still not clear to me why one would call this “physical density”. With your explanation, I could see it being called “model-independent density” or something similar, but not “physical”. $\endgroup$
    – Socob
    Commented Jul 26, 2022 at 12:00

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