<p>Your misunderstanding is in the second paragraph. If $n&lt;n_c$ then there are insufficient collisions to excite the molecule into the upper level. Thus in equilibrium there are very few excited molecules and since the luminosity is proportional to the number of molecules in the excited level then little radiation is seen.</p>

<p>The critical density is defined as that (temperature-dependent) density at which the rate of collisional excitation to the upper level of a molecular (or atomic, I suppose) transition, equals the rate of radiative de-excitation.</p>

<p>The rate of de-excitation is given by $n_2\ A_{21}$, where $n_2$ is the number density of the molecular species <em>in the excited state</em> and $A_{21}$ is the relevant spontaneous emission coefficient.</p>

<p>The rate of excitation is given by $\gamma_{12} n_1 n$,
where $\gamma_{12}$ is the collision probability (which depends on the speed of molecules in the gas and hence temperature), $n_1$ is the number density of the molecule of interest in the lower energy state and $n$ is the number density of <em>all molecules</em> in the gas.</p>

<p>There is also a rate of collisional de-excitation given by $\gamma_{21}\ n_2 n$. However there must be a relationship between $\gamma_{12}$ and $\gamma_{21}$ because for a gas in thermodynamic equilibrium, we know that 
$$n_2/n_1 = \exp(-E_{21}/k_BT),$$ and so if collisions are in detailed balance then
$$\gamma_{21}\ n_1 \exp(-E_{21}/k_BT) n = \gamma_{12} n_1 n$$
$$ \gamma_{21} = \gamma_{12} \exp(E_{21}/k_BT)$$</p>

<p>OK, so now consider a gas where there are radiative losses from state 2 and allow it to come into equilibrium such that
$$\frac{dn_2}{dt} = 0$$
$$ -A_{21}n_2 - \gamma_{21}n_2 n + \gamma_{12} n_1 n = 0$$
$$ \frac{n_2}{n_1} = \exp(-E_{21}/k_BT)\ \frac{1}{1 + A_{21}/\gamma_{21}n}$$</p>

<p>If now we say that $A_{21} \ll \gamma_{21}n$, then radiative de-excitation is (relatively) negligible compared with collisional de-excitation then the $n_2/n_1$ ratio approaches the level expected by the Boltzmann ratio.</p>

<p>If on the other hand $A_{21} \gg \gamma_{21}n$, then the ratio of $n_2/n_1$ becomes much <em>lower</em> and radiation removes excited molecules faster than collisions create them.</p>

<p>Now we ask how much radiation we are going to see from this gas.</p>

<p>(1) If $A_{21} \ll \gamma_{21}n$, which we can restate as $n \gg n_c$, the "critical density".</p>

<p>The luminosity from the molecules will be 
$$L \propto n_2 A_{21} \simeq n_1 A_{21} \exp(-E_{21}/k_BT)$$
If $k_BT &lt; E_{21}$ this means that $n_1 \gg n_2$ and the luminosity per molecule in the gas has a fixed value that is independent of density $n$. There is just a fixed amount of radiative losses per molecule of the species in question.</p>

<p>(2) If $A_{21} \gg \gamma_{21}n$, which we can restate as $n \ll n_c$, The luminosity from the molecules will be 
$$L \propto n_2 A_{21} \simeq n_1 A_{21} \exp(-E_{21}/k_BT) \frac{\gamma_{21} n}{A_{21}}$$
and so this is <em>smaller</em> than for the first case by a factor of $n/n_c$. So now there is less radiation per molecule of the interesting species because fewer of those molecules are in an excited state.</p>

<p>NB This is a different thing to the "quenching" density for <em>forbidden</em> lines that I refer to in my comment.</p>

CC BY-SA 3.0
2017-09-06T16:29:04.133Z
2017-09-06T16:29:04.133Z
ProfRob
<p>I do not understand completely the meaning of critical density when talking about atomic or molecular line emission in astrophysical gas clouds: I know that it is way more complex than this, but just taking the two levels approximation (i.e. an atom/molecule with two quantum levels, 1 and 2, where 2 is the upper one) the critical density would be
$n_c = A_{12}/\gamma_{12}$, where $A_{12}$ is the Einstein coefficient for spontaneous radiative emission (i.e. de-excitation) and $\gamma_{12}$ the collisional de-excitation probability.</p>

<p>It seems to me that if the gas density is $n &lt; n_c$ the radiative de-excitations dominate, so I would expect to see the photons; conversely if $n&gt;n_c$ collisions dominate, so the particles will lose energy through collisions before having enough time to spontaneously decay and emit photons.</p>

<p>This argument of mine, however, seems to be wrong, because everybody use $n_c$ exactly in the opposite way: I see the photons only if $n&gt;n_c$. I would expect that for $n \ll n_c$ the particles are experiencing almost no collisions, so neither collisional excitation, and they do not decay because they are not excited in the first place; but I feel that something wrong has been written in the previous lines, so I would like to read someone's opinion about it before to make even more pie in the sky conjectures.</p>

CC BY-SA 3.0
2017-09-05T14:52:57.877Z
Give context or examples. If you are talking about forbidden lines, then yes, the radiative de-excitation is quenched above some critical density.
ProfRob
CC BY-SA 3.0
2017-09-06T14:10:31.027Z
I am more interested in molecular rotational lines, but I thought the meaning of the critical density was the same of, for instance, forbidden atomic lines. Am I wrong?
Federico Esposito
CC BY-SA 4.0
2024-09-28T11:24:27.977Z
Following the answer by ProfRob, I want to ask what the physical explanation of detailed balance is. Sometimes we just use it, but can we have more physical support for that?
Fengwei Xu
<radiative-transfer>
Meaning of critical density in radiative transfer
Federico Esposito
# Meaning of critical density in radiative transfer

(Asked by: username_0 on 2017-09-05)

I do not understand completely the meaning of critical density when talking about atomic or molecular line emission in astrophysical gas clouds: I know that it is way more complex than this, but just taking the two levels approximation (i.e. an atom/molecule with two quantum levels, 1 and 2, where 2 is the upper one) the critical density would be
$n\_c = A\_{12}/\gamma\_{12}$, where $A\_{12}$ is the Einstein coefficient for spontaneous radiative emission (i.e. de-excitation) and $\gamma\_{12}$ the collisional de-excitation probability.

It seems to me that if the gas density is $n < n\_c$ the radiative de-excitations dominate, so I would expect to see the photons; conversely if $n>n\_c$ collisions dominate, so the particles will lose energy through collisions before having enough time to spontaneously decay and emit photons.

This argument of mine, however, seems to be wrong, because everybody use $n\_c$ exactly in the opposite way: I see the photons only if $n>n\_c$. I would expect that for $n \ll n\_c$ the particles are experiencing almost no collisions, so neither collisional excitation, and they do not decay because they are not excited in the first place; but I feel that something wrong has been written in the previous lines, so I would like to read someone's opinion about it before to make even more pie in the sky conjectures.

**Tags:** radiative-transfer

> **Comments:**
> **username_1**: Give context or examples. If you are talking about forbidden lines, then yes, the radiative de-excitation is quenched above some critical density.  
> **username_0**: I am more interested in molecular rotational lines, but I thought the meaning of the critical density was the same of, for instance, forbidden atomic lines. Am I wrong?  
> **username_2**: Following the answer by username_1, I want to ask what the physical explanation of detailed balance is. Sometimes we just use it, but can we have more physical support for that?  

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## 1 answers

### Answer by username_1 on 2017-09-06. Score: 6 | Accepted answer

Your misunderstanding is in the second paragraph. If $n

The critical density is defined as that (temperature-dependent) density at which the rate of collisional excitation to the upper level of a molecular (or atomic, I suppose) transition, equals the rate of radiative de-excitation.

The rate of de-excitation is given by $n\_2\ A\_{21}$, where $n\_2$ is the number density of the molecular species *in the excited state* and $A\_{21}$ is the relevant spontaneous emission coefficient.

The rate of excitation is given by $\gamma\_{12} n\_1 n$,
where $\gamma\_{12}$ is the collision probability (which depends on the speed of molecules in the gas and hence temperature), $n\_1$ is the number density of the molecule of interest in the lower energy state and $n$ is the number density of *all molecules* in the gas.

There is also a rate of collisional de-excitation given by $\gamma\_{21}\ n\_2 n$. However there must be a relationship between $\gamma\_{12}$ and $\gamma\_{21}$ because for a gas in thermodynamic equilibrium, we know that
$$n\_2/n\_1 = \exp(-E\_{21}/k\_BT),$$ and so if collisions are in detailed balance then
$$\gamma\_{21}\ n\_1 \exp(-E\_{21}/k\_BT) n = \gamma\_{12} n\_1 n$$
$$ \gamma\_{21} = \gamma\_{12} \exp(E\_{21}/k\_BT)$$

OK, so now consider a gas where there are radiative losses from state 2 and allow it to come into equilibrium such that
$$\frac{dn\_2}{dt} = 0$$
$$ -A\_{21}n\_2 - \gamma\_{21}n\_2 n + \gamma\_{12} n\_1 n = 0$$
$$ \frac{n\_2}{n\_1} = \exp(-E\_{21}/k\_BT)\ \frac{1}{1 + A\_{21}/\gamma\_{21}n}$$

If now we say that $A\_{21} \ll \gamma\_{21}n$, then radiative de-excitation is (relatively) negligible compared with collisional de-excitation then the $n\_2/n\_1$ ratio approaches the level expected by the Boltzmann ratio.

If on the other hand $A\_{21} \gg \gamma\_{21}n$, then the ratio of $n\_2/n\_1$ becomes much *lower* and radiation removes excited molecules faster than collisions create them.

Now we ask how much radiation we are going to see from this gas.

(1) If $A\_{21} \ll \gamma\_{21}n$, which we can restate as $n \gg n\_c$, the "critical density".

The luminosity from the molecules will be
$$L \propto n\_2 A\_{21} \simeq n\_1 A\_{21} \exp(-E\_{21}/k\_BT)$$
If $k\_BT < E\_{21}$ this means that $n\_1 \gg n\_2$ and the luminosity per molecule in the gas has a fixed value that is independent of density $n$. There is just a fixed amount of radiative losses per molecule of the species in question.

(2) If $A\_{21} \gg \gamma\_{21}n$, which we can restate as $n \ll n\_c$, The luminosity from the molecules will be
$$L \propto n\_2 A\_{21} \simeq n\_1 A\_{21} \exp(-E\_{21}/k\_BT) \frac{\gamma\_{21} n}{A\_{21}}$$
and so this is *smaller* than for the first case by a factor of $n/n\_c$. So now there is less radiation per molecule of the interesting species because fewer of those molecules are in an excited state.

NB This is a different thing to the "quenching" density for *forbidden* lines that I refer to in my comment.


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