<p>Let $n$, $T$, and $x_i$ be the number density of hydrogen, the temperature of the gas, and $n_i/n$, where $n_i$ is the number density of the $i$th component of the interstellar medium. We can then write the criteria for thermal equilibrium as
$$n^2\Lambda(n,T,x_i)-n\Gamma(n,T,x_i)\equiv n^2\mathcal{L}=0$$
where $\Lambda$ and $\Gamma$ and the heating and cooling functions, respectively, and $\mathcal{L}$ is defined through those functions and $n$. If the equilibrium is unstable,
$$\left(\frac{\partial\mathcal{L}}{\partial S}\right)&lt;0$$
for entropy $S$. This leads to different instability conditions, termed the <em>isochoric</em> and <em>isobaric</em> instabilities (<a href="http://adsabs.harvard.edu/cgi-bin/bib_query?1965ApJ...142..531F" rel="noreferrer">Field (1965)</a>, $\text{Eq } 4a,4b$). These can be determined from the temperature, pressure, and density of the gas (also assuming that the gas can be approximated as an ideal gas).</p>

<p>In general, $\Lambda$ and $\Gamma$ are complicated to determine, although combinations of power-law and exponentially decaying factors may often be sufficient. An example curve that seems to pop up a lot as an example was calculated by <a href="http://adsabs.harvard.edu/abs/1972ARA%26A..10..375D" rel="noreferrer">Dalgarno &amp; McCray (1972)</a>, Figure 2:</p>

<p><a href="https://i.sstatic.net/tFGfT.png" rel="noreferrer"><img src="https://i.sstatic.net/tFGfT.png" alt="enter image description here"></a></p>

<p>More accurate measurements have been made since then, but the general shape is still applicable. The warm neutral medium occupies the area near the sharp change at around $\sim10,000\text{ K}$, and the cold neutral medium occupies the area in the left of the diagram. Another way to visualize this is on a $\log P/\log n$ diagram like this one (from <a href="http://w.astro.berkeley.edu/~ay216/08/NOTES/Lecture11-08.pdf" rel="noreferrer">these slides</a>, annotated from <a href="http://adsabs.harvard.edu/abs/1995ApJ...443..152W" rel="noreferrer">Wolfire et al. (1995)</a>):</p>

<p><a href="https://i.sstatic.net/7lPas.png" rel="noreferrer"><img src="https://i.sstatic.net/7lPas.png" alt="enter image description here"></a></p>

<p>In reality, the two-phase model is an oversimplification, and the ISM has more distinct components. However, the isochoric/isobaric instabilities still limit the range in which clouds can exist in stable equilibria, and explain the dearth of gas in the relevant temperature range.</p>

<hr>

<p>Let me elaborate on the terms <em>isobaric</em> and <em>isochoric</em>. In thermodynamics, it is sometimes convenient to assume that some thermodynamic variables remain constant in a certain situation. <a href="https://en.wikipedia.org/wiki/Isothermal_process" rel="noreferrer"><em>Isothermal</em> processes</a> occur at constant temperature; likewise, <a href="https://en.wikipedia.org/wiki/Isobaric_process" rel="noreferrer"><em>isobaric</em> processes</a> occur at constant pressure and <a href="https://en.wikipedia.org/wiki/Isochoric_process" rel="noreferrer"><em>isochoric</em> processes</a> occur at constant volume.</p>

<p>The equations for the two instabilities are
$$\left(\frac{\partial\mathcal{L}}{\partial T}\right)_{\rho}&lt;0\tag{Isochoric}$$
$$\left(\frac{\partial\mathcal{L}}{\partial T}\right)_{p}=\left(\frac{\partial\mathcal{L}}{\partial T}\right)_{\rho}-\frac{\rho_0}{T_0}\left(\frac{\partial\mathcal{L}}{\partial\rho}\right)_T&lt;0\tag{Isobaric}$$
In the first, we assume that the cloud is at constant volume, and since the total amount of matter in the system is constant, the (mean) density must also be constant. In the second case, we assume that the cloud is at constant pressure. Perturbations that lead to instabilities thus arise from perturbations of other thermodynamic variables.</p>

<p>As a final note on notation,
$$\left(\frac{\partial\mathcal{L}}{\partial A}\right)_B$$
means that we take the partial derivative of $\mathcal{L}$ with respect to $A$ while keeping $B$ constant. It's a common thermodynamical convention.</p>

CC BY-SA 3.0
2017-02-16T21:13:20.720Z
It is my understanding that the clouds are specifically not in equilibrium.  Is this calculation just based on the assumption that the timescale is sufficiently long that they can be considered in equilibrium?
Phiteros
CC BY-SA 3.0
2017-02-16T21:37:21.490Z
@Phiteros Yes, it's only an approximate equilibrium. Perturbations are always going to show up, and they may change the temperature of the cloud, but you won't see large-scale changes happen suddenly without outside influence.
HDE 226868
CC BY-SA 3.0
2017-02-17T05:20:40.903Z
Good question, good answer, now I've asked a [follow-up question](http://astronomy.stackexchange.com/q/20083/7982).
uhoh
CC BY-SA 3.0
2017-02-16T20:22:31.650Z
2017-02-17T14:52:58.293Z
2017-02-17T14:52:58.293Z
HDE 226868
<p>In the interstellar medium, there are several different diffuse phases of gas, distinguished by their density and temperature. Specifically, the cold neutral medium has temperatures from ~50-100 K and the warm neutral medium ranges from 6,000-10,000 K. According to my professor, there have been few gas clouds found with temperatures between these two.</p>

<p>Does anyone know why such a large gap exists?</p>

CC BY-SA 3.0
2017-02-16T21:58:09.053Z
Related: [How cold is interstellar space?](http://astronomy.stackexchange.com/a/15087/5264) and [density distributions of the interstellar medium](http://physics.stackexchange.com/a/310190/70207).
pela
<temperature><interstellar-medium>
Why are there no ISM clouds with temperatures between 100 and 6,000K?
Phiteros
# Why are there no ISM clouds with temperatures between 100 and 6,000K?

(Asked by: username_0 on 2017-02-16)

In the interstellar medium, there are several different diffuse phases of gas, distinguished by their density and temperature. Specifically, the cold neutral medium has temperatures from ~50-100 K and the warm neutral medium ranges from 6,000-10,000 K. According to my professor, there have been few gas clouds found with temperatures between these two.

Does anyone know why such a large gap exists?

**Tags:** temperature, interstellar-medium

> **Comments:**
> **username_1**: Related: [How cold is interstellar space?](http://astronomy.stackexchange.com/a/15087/5264) and [density distributions of the interstellar medium](http://physics.stackexchange.com/a/310190/70207).  

---

## 1 answers

### Answer by username_2 on 2017-02-16. Score: 10 | Accepted answer

Let $n$, $T$, and $x\_i$ be the number density of hydrogen, the temperature of the gas, and $n\_i/n$, where $n\_i$ is the number density of the $i$th component of the interstellar medium. We can then write the criteria for thermal equilibrium as
$$n^2\Lambda(n,T,x\_i)-n\Gamma(n,T,x\_i)\equiv n^2\mathcal{L}=0$$
where $\Lambda$ and $\Gamma$ and the heating and cooling functions, respectively, and $\mathcal{L}$ is defined through those functions and $n$. If the equilibrium is unstable,
$$\left(\frac{\partial\mathcal{L}}{\partial S}\right)<0$$
for entropy $S$. This leads to different instability conditions, termed the *isochoric* and *isobaric* instabilities ([Field (1965)](http://adsabs.harvard.edu/cgi-bin/bib_query?1965ApJ...142..531F), $\text{Eq } 4a,4b$). These can be determined from the temperature, pressure, and density of the gas (also assuming that the gas can be approximated as an ideal gas).

In general, $\Lambda$ and $\Gamma$ are complicated to determine, although combinations of power-law and exponentially decaying factors may often be sufficient. An example curve that seems to pop up a lot as an example was calculated by [Dalgarno & McCray (1972)](http://adsabs.harvard.edu/abs/1972ARA%26A..10..375D), Figure 2:

[![enter image description here](https://i.sstatic.net/tFGfT.png)](https://i.sstatic.net/tFGfT.png)

More accurate measurements have been made since then, but the general shape is still applicable. The warm neutral medium occupies the area near the sharp change at around $\sim10,000\text{ K}$, and the cold neutral medium occupies the area in the left of the diagram. Another way to visualize this is on a $\log P/\log n$ diagram like this one (from [these slides](http://w.astro.berkeley.edu/~ay216/08/NOTES/Lecture11-08.pdf), annotated from [Wolfire et al. (1995)](http://adsabs.harvard.edu/abs/1995ApJ...443..152W)):

[![enter image description here](https://i.sstatic.net/7lPas.png)](https://i.sstatic.net/7lPas.png)

In reality, the two-phase model is an oversimplification, and the ISM has more distinct components. However, the isochoric/isobaric instabilities still limit the range in which clouds can exist in stable equilibria, and explain the dearth of gas in the relevant temperature range.

---

Let me elaborate on the terms *isobaric* and *isochoric*. In thermodynamics, it is sometimes convenient to assume that some thermodynamic variables remain constant in a certain situation. [*Isothermal* processes](https://en.wikipedia.org/wiki/Isothermal_process) occur at constant temperature; likewise, [*isobaric* processes](https://en.wikipedia.org/wiki/Isobaric_process) occur at constant pressure and [*isochoric* processes](https://en.wikipedia.org/wiki/Isochoric_process) occur at constant volume.

The equations for the two instabilities are
$$\left(\frac{\partial\mathcal{L}}{\partial T}\right)\_{\rho}<0\tag{Isochoric}$$
$$\left(\frac{\partial\mathcal{L}}{\partial T}\right)\_{p}=\left(\frac{\partial\mathcal{L}}{\partial T}\right)\_{\rho}-\frac{\rho\_0}{T\_0}\left(\frac{\partial\mathcal{L}}{\partial\rho}\right)\_T<0\tag{Isobaric}$$
In the first, we assume that the cloud is at constant volume, and since the total amount of matter in the system is constant, the (mean) density must also be constant. In the second case, we assume that the cloud is at constant pressure. Perturbations that lead to instabilities thus arise from perturbations of other thermodynamic variables.

As a final note on notation,
$$\left(\frac{\partial\mathcal{L}}{\partial A}\right)\_B$$
means that we take the partial derivative of $\mathcal{L}$ with respect to $A$ while keeping $B$ constant. It's a common thermodynamical convention.

> **Comments:**
> **username_0**: It is my understanding that the clouds are specifically not in equilibrium.  Is this calculation just based on the assumption that the timescale is sufficiently long that they can be considered in equilibrium?  
> **username_2**: @username_0 Yes, it's only an approximate equilibrium. Perturbations are always going to show up, and they may change the temperature of the cloud, but you won't see large-scale changes happen suddenly without outside influence.  
> **username_3**: Good question, good answer, now I've asked a [follow-up question](http://astronomy.stackexchange.com/q/20083/7982).  

---

