# Where can I find the $\mu$ value in galaxy clusters for ideal gas law?

I am studying hydrostatic equilibrium in galaxy clusters and encountered the following expression:

$$P=(kT/\mu m_p)\rho$$

The interpretation of this formula is obvious. It is just the ideal gas law expressed slightly differently. It is apparent that $$\mu m_p$$ here is the average mass of molecules in galaxy clusters, where $$m_p$$ is the mass of proton. However, the problem is that I do not know what $$\mu$$ is called or where to look to find the value of $$\mu$$.

Could somebody help me?

## 2 Answers

tl;dr: For you, $$\mu\simeq0.6$$.

### Mean molecular mass

The term you're looking for is the mean molecular mass $$\mu$$, more often called the mean molecular weight, despite not being a weight, despite usually not involving molecules, and despite being a unitless number, not a mass.

It is the average mass per particle, measured in terms of the proton mass (or sometimes the hydrogen mass which is almost the same, and sometimes the atomic mass unit $$m_\mathrm{u} \equiv m_{^{12}\mathrm{C}}/12$$), and it includes any free electrons.

That is, it is defined as $$\mu = \frac{\langle m \rangle}{m_p},$$

### Neutral gas

For a neutral gas (i.e. no free electrons) of mass fractions $$X$$, $$Y$$, and $$Z$$ of hydrogen, helium, and heavier elements (called "metals" in astronomy), you have $$\frac{1}{\mu_\mathrm{neut.}} = X + \frac{Y}{4} + \langle 1/A_n \rangle Z,$$ because helium weighs four times as much as hydrogen, and where $$\langle 1/A_n \rangle$$ is the weighted average of all metals. For Solar abundances $$\{X,Y,Z\}\simeq\{0.70,0.28,0.02\}$$, and $$\langle 1/A_n \rangle \simeq 1/15.5$$ (e.g. Carroll & Ostlie 1996), so $$\mu_{\mathrm{neut.,\odot}} = \frac{1}{0.70+0.28/4+0.02/15.5} \simeq 1.30.$$

For a primordial (i.e. metal-free) gas, $$\{X,Y,Z\}\simeq\{0.75,0.25,0\}$$ so $$\mu_{\mathrm{neut.,prim.}} = \frac{1}{0.75+0.25/4} \simeq 1.23.$$

### Ionized gas

For an ionized gas you have roughly twice the amount of particles, but the mass of half of them can be neglected, so $$\mu$$ is roughly half of the above: $$\frac{1}{\mu_{\mathrm{ion.}}} \simeq 2X + \frac{3Y}{4} + \frac{Z}{2}.$$

Thus, a fully ionized primordial gas has $$\mu_{\mathrm{ion.,prim.}} = \frac{1}{2\times0.75 + 3\times0.25/4} \simeq 0.59.$$ while a fully ionized metal-rich gas has $$\mu_{\mathrm{ion.,\odot}} = \frac{1}{2\times0.70 + 3\times0.28/4 + 0.02/2} \simeq 0.62.$$

### The answer for you

Since you're dealing with galaxy clusters, the bulk of the gas is ionized. As you can see above, the exact value of $$\mu$$ depends on the metallicity, but as an astronomer you will most likely not offend anyone by just using $$\mu=0.6$$.

Or even $$\mu\sim1$$. I mean if we can't even make a proper definition, why bother about anything more precise than an order-of-magnitude estimate…

• You mean for fully ionized primordial gas and fully ionized metal-rich gas, the left-handsides must be mu_ion, not the inverse of mu_ion, right? Dec 28, 2023 at 18:13
• @YoungsubYoon Ah yes, I'm sorry, good catch. Too much copy-pasting :) I'll edit!
– pela
Dec 28, 2023 at 20:29

The quantity $$\rho/(\mu m_p)$$ is approximately (or can be defined as) the particle number density (including ions and electrons). The interpretation of $$\mu$$ would therefore be the number of proton masses per particle in the gas.

The vast majority of the gas in the intracluster medium will be primordial gas from the big bang and it is hot and fully ionised. It is therefore (by number) 92 per cent hydrogen (1 proton [1 mass unit] and 1 electron) and 8 per cent helium (1 alpha particle [4 mass units] and 2 electrons).

$$\mu \simeq 0.92 \times \frac{1}{2} + 0.08 \times \frac{4}{3} = 0.57$$

If the gas has been slightly enriched with heavier elements (1 per cent is about an upper limit), then there would be a small upward correction, since $$\mu \simeq 2$$ for most ionised elements heavier than helium. e.g. fully ionised oxygen has 16 mass units and 9 particles, ionised iron has 56 mass units and 27 particles. Thus $$\mu \simeq 0.91 \times \frac{1}{2} + 0.08 \times \frac{4}{3} + 0.01 \times 2 = 0.58$$