<p><strong>tl;dr</strong>:
For you, <span class="math-container">$\mu\simeq0.6$</span>.</p>
<hr />
<h3>Mean molecular mass</h3>
<p>The term you're looking for is the mean molecular mass <span class="math-container">$\mu$</span>, more often called the mean molecular <em>weight</em>, despite not being a weight, despite usually not involving molecules, and despite being a unitless number, not a mass.</p>
<p>It is the average mass per <em>particle</em>, measured in terms of the proton mass (or sometimes the hydrogen mass which is almost the same, and sometimes the atomic mass unit <span class="math-container">$m_\mathrm{u} \equiv m_{^{12}\mathrm{C}}/12$</span>), and it includes any free electrons.</p>
<p>That is, it is defined as
<span class="math-container">$$
\mu = \frac{\langle m \rangle}{m_p},
$$</span></p>
<h3>Neutral gas</h3>
<p>For a neutral gas (i.e. no free electrons) of mass fractions <span class="math-container">$X$</span>, <span class="math-container">$Y$</span>, and <span class="math-container">$Z$</span> of hydrogen, helium, and heavier elements (called &quot;metals&quot; in astronomy), you have
<span class="math-container">$$
\frac{1}{\mu_\mathrm{neut.}} = X + \frac{Y}{4} + \langle 1/A_n \rangle Z,
$$</span>
because helium weighs four times as much as hydrogen, and where <span class="math-container">$\langle 1/A_n \rangle$</span> is the weighted average of all metals. For Solar abundances <span class="math-container">$\{X,Y,Z\}\simeq\{0.70,0.28,0.02\}$</span>, and <span class="math-container">$\langle 1/A_n \rangle \simeq 1/15.5$</span> (e.g. <a href="https://ui.adsabs.harvard.edu/abs/1996ima..book.....C/abstract" rel="nofollow noreferrer">Carroll &amp; Ostlie 1996</a>), so
<span class="math-container">$$
\mu_{\mathrm{neut.,\odot}} = \frac{1}{0.70+0.28/4+0.02/15.5} \simeq 1.30.
$$</span></p>
<p>For a primordial (i.e. metal-free) gas, <span class="math-container">$\{X,Y,Z\}\simeq\{0.75,0.25,0\}$</span> so
<span class="math-container">$$
\mu_{\mathrm{neut.,prim.}} = \frac{1}{0.75+0.25/4} \simeq 1.23.
$$</span></p>
<h3>Ionized gas</h3>
<p>For an ionized gas you have roughly twice the amount of particles, but the mass of half of them can be neglected, so <span class="math-container">$\mu$</span> is roughly half of the above:
<span class="math-container">$$
\frac{1}{\mu_{\mathrm{ion.}}} \simeq 2X + \frac{3Y}{4} + \frac{Z}{2}.
$$</span></p>
<p>Thus, a fully ionized primordial gas has
<span class="math-container">$$
\mu_{\mathrm{ion.,prim.}} = \frac{1}{2\times0.75 + 3\times0.25/4} \simeq 0.59.
$$</span>
while a fully ionized metal-rich gas has
<span class="math-container">$$
\mu_{\mathrm{ion.,\odot}} = \frac{1}{2\times0.70 + 3\times0.28/4 + 0.02/2} \simeq 0.62.
$$</span></p>
<h3>The answer for you</h3>
<p>Since you're dealing with galaxy clusters, the bulk of the gas is ionized. As you can see above, the exact value of <span class="math-container">$\mu$</span> depends on the metallicity, but as an astronomer you will most likely not offend anyone by just using <span class="math-container">$\mu=0.6$</span>.</p>
<p>Or even <span class="math-container">$\mu\sim1$</span>. I mean if we can't even make a proper definition, why bother about anything more precise than an order-of-magnitude estimate…</p>

CC BY-SA 4.0
2023-12-28T18:13:17.343Z
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?
Youngsub Yoon
CC BY-SA 4.0
2023-12-28T20:29:12.170Z
@YoungsubYoon Ah yes, I'm sorry, good catch. Too much copy-pasting :) I'll edit!
pela
CC BY-SA 4.0
2023-12-28T17:15:16.417Z
2023-12-28T20:30:16.413Z
2023-12-28T20:30:16.413Z
pela
<p>The quantity <span class="math-container">$\rho/(\mu m_p)$</span> is approximately (or can be defined as) the particle number density (including ions and electrons). The interpretation of <span class="math-container">$\mu$</span> would therefore be the number of proton masses per particle in the gas.</p>
<p>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).</p>
<p><span class="math-container">$$\mu \simeq 0.92 \times \frac{1}{2} + 0.08 \times \frac{4}{3} =  0.57$$</span></p>
<p>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 <span class="math-container">$\mu \simeq 2$</span> 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
<span class="math-container">$$\mu \simeq 0.91 \times \frac{1}{2} + 0.08 \times \frac{4}{3}  + 0.01 \times 2 = 0.58$$</span></p>

CC BY-SA 4.0
2023-12-28T17:14:18.660Z
2023-12-28T17:23:23.517Z
2023-12-28T17:23:23.517Z
ProfRob
<p>I am studying hydrostatic equilibrium in galaxy clusters and encountered the following expression:</p>
<p><span class="math-container">$P=(kT/\mu m_p)\rho$</span></p>
<p>The interpretation of this formula is obvious. It is just the ideal gas law expressed slightly differently. It is apparent that <span class="math-container">$\mu m_p$</span> here is the average mass of molecules in galaxy clusters, where <span class="math-container">$m_p$</span> is the mass of proton. However, the problem is that I do not know what <span class="math-container">$\mu$</span> is called or where to look to find the value of <span class="math-container">$\mu$</span>.</p>
<p>Could somebody help me?</p>

<galaxy><cosmology><galaxy-cluster><nucleosynthesis><hydrostatic-equilibrium>
Where can I find the $\mu$ value in galaxy clusters for ideal gas law?
Youngsub Yoon
# Where can I find the $\mu$ value in galaxy clusters for ideal gas law?

(Asked by: username_0 on 2023-12-28)

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?

**Tags:** galaxy, cosmology, galaxy-cluster, nucleosynthesis, hydrostatic-equilibrium


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## 2 answers

### Answer by username_1 on 2023-12-28. Score: 1 | Accepted answer

**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](https://ui.adsabs.harvard.edu/abs/1996ima..book.....C/abstract)), 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…

> **Comments:**
> **username_0**: 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?  
> **username_1**: @username_0 Ah yes, I'm sorry, good catch. Too much copy-pasting :) I'll edit!  

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### Answer by username_2 on 2023-12-28. Score: 1

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
$$\mu \simeq 0.91 \times \frac{1}{2} + 0.08 \times \frac{4}{3} + 0.01 \times 2 = 0.58$$


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