# Maintaining local thermodynamic equilibrium (LTE) in radiating gas with a broad atomic transition line

## Definitions / Background

In LTE, Kirchoff's law for radiation holds:

$$\frac{j_{\nu}}{\alpha_{\nu}} = B_{\nu} (T)$$

where $j_{\nu}$ is the specific radiative emissivity, $\alpha_{\nu}$ is the monochromatic radiative absorption, and $B_{\nu} (T)$ is the Planck function evaluated at the temperature $T$.

Consider a gas of two-level atoms with energies $E_u$ and $E_l$, with $E_u > E_l$, statistical weights $g_u$ and $g_l$, and number densities $n_u$ and $n_l$. The transition between these states has Einstein coefficients $A_{ul}$, $B_{ul}$ and $B_{lu}$ that we can use to write the emissivity and absoprtion of the transition:

$$j_{\nu} = \frac{h \nu}{4 \pi} n_u A_{ul} \psi({\nu})$$ $$\alpha_{\nu} = \frac{h \nu}{4 \pi} [ n_l B_{lu} \phi({\nu}) - n_u B_{ul} \chi({\nu})]$$

where $\psi$, $\phi$, and $\chi$ are line profile functions accounting for line-broadening mechanisms such as thermal motion.

Then, making use of the standard relations between the Einstein coefficients, we have $$\frac{j_{\nu}}{\alpha_{\nu}} = \frac{2 h \nu^3}{c^2} \frac{\frac{\psi}{\phi}}{\frac{g_u n_l}{g_l n_u} - \frac{\chi}{\phi}}$$

## Question

I want to understand how the right-hand side of the last equation simplifies to the Planck function under conditions of LTE, over the entire width of the line, without assuming that the line is narrow.

A standard discussion of this topic, as found in e.g. the astrophysics textbook by Rybicki and Lightman, does not seem to achieve this. In my reading, they proceed by making the following observations and/or assumptions:

1) There is complete redistribution of frequency between absorption and all types of emission, so that $\psi = \chi = \phi$.

2) In LTE at temperature T, the fraction $g_u n_l / (g_l n_u)$ is equal to $\exp[h \nu_0/(kT)]$, where $\nu_0 = (E_u - E_l)/h$.

If those are true, then we have

$$\frac{j_{\nu}}{\alpha_{\nu}} = \frac{2 h \nu^3}{c^2} \frac{1}{\exp[h \nu_0/(kT)] - 1}$$

But this does not have the correct exponential term in the denominator to match the Planck function, because $\nu_0$ is constant (frequency-independent). Or from another perspective, the problem is that $g_u n_l / (g_l n_u)$ is frequency-independent.

So, what is going on? Do we have to break either of the assumptions/observations 1 or 2 above? If so, how? If not, then does Kirchoff's law simply not apply on a frequency-by-frequency basis over the width of a broad line, although it still might apply in a line-averaged sense? Is there some other possibility or detail I have overlooked?

• The Einstein way to arrive at the Planck function is to assume there is a molecule with appropriate pair of energy levels for any radiation frequency $\nu$ desired. So first you choose some $\nu$ and only then you assume there is a molecule with levels that radiates spontaneously at this frequency. Gas of two-level atoms with one emission frequency does not work with this derivation, I think. To obtain radiation with the Planck spectrum, a gas radiating at any possible frequency is needed in this derivation. – Ján Lalinský Jul 31 '14 at 19:21
• @JánLalinský Even if there are only 2 levels in the atom, the line profile is not a delta function - there are a still a range of frequencies over which the atom can radiate for the single transition. (I think in Einstein's derivation he implicitly assumed that the line profile is a delta function, but I don't want to make that approximation). I'd expect based on thermodynamic arguments that you'd still get the Planck source function over all of those frequencies in the width of the line, not just the center of the line. Is that the case? – kleingordon Jul 31 '14 at 20:10
• The Einstein derivation is very simplistic. Line profile does not enter into it at all. Assuming molecules with wide emission lines is alright, but I do not think that is essential in the Einstein derivation. Presence of different pairs of levels with different associated frequencies is. Just one pair of levels with broad emission line will not suffice in this derivation. You will have to find another reasoning to arrive at the Planck function with such assumption. – Ján Lalinský Aug 1 '14 at 0:05
• @JánLalinský Thanks. In that case, my question is "what is the other line of reasoning needed to arrive at the Planck function for a single level transition?" And where exactly does the simplistic derivation in my question break down? – kleingordon Aug 1 '14 at 0:49
• LTE is only ever an approximation. I'd go with your latter explanation that the source function equals the Planck function averaged over a line profile. Would this matter only if the radiation field changes appreciably over the width of a line? In which case could you ever expect LTE? Great question. – Rob Jeffries Aug 3 '14 at 19:53

Let's suppose the broadening mechanism is van der Waals or Stark broadening - something where the energy levels of individual atoms are perturbed.

In this case you could use the following argument.

Divide the line profile up into groups of atoms which share the same perturbation and treat each of these as a subpopulation with a different energy gap and hence a perturbed $\nu_0^{\prime}$. You can go through the usual argument involving the relationship between the Einstein coefficients and populate the two levels according to the Boltzmann factor, but with $h\nu_0^{\prime}$ in the exponential argument. At the end of this you find the source function for each subpopulation follows the Planck function at the same temperature $T$. Hence LTE means that the source function equals the Planck function at each frequency.

• Hmm. Perhaps. I appreciate the contribution, although I am also not entirely sure this is the correct explanation. If this could be backed up with a more detailed derivation, or a reference, that would be most helpful. – kleingordon Aug 4 '14 at 21:26
• I am agreeing with one of your suggestions: Kirchoff's law will not (exactly) apply if the line is very broad. I don't think you can use the Boltzmann factor as you have done in the case where the energy levels are significantly broadened (which is what a broad line profile implies). – Rob Jeffries Aug 5 '14 at 0:26
• Another issue - aren't the "standard relations between the Einstein coefficients" also derived by assuming that the radiation energy density is represented by some mean value over a narrow line profile? – Rob Jeffries Aug 5 '14 at 0:29
• Yes, the relations between the Einstein coefficients are often derived with the assumption of negligible line broadening. If these relations need to be refined for broad lines, that would be good to know. So now, which do you think is the most promising explanation here: the relations between the Einstein coefficients need to be refined, the application of the Boltzmann factor for the level population densities needs to be reconsidered, or both? I'd like to pinpoint in detail how this all works out. – kleingordon Aug 5 '14 at 7:39
• Doh! I was hoping someone else would weigh in. Do you have an example system where the broadening is a significant fraction of the central frequency? – Rob Jeffries Aug 7 '14 at 23:10

I'm not sure if this explanation is right, so please feel free to point out inconsistencies.

A gas with two energy levels absorbs a range of frequencies centered around $\nu_{0}$ because the atoms of the gas are in motion, making the photons appear appropriately doppler-shifted in their frame of reference.

In the laboratory frame, where the atoms of the gas are whizzing around, one could interpret this as a manifestation of the spread in energy levels of the gas. Therefore, any gas has fundamentally more than two energy levels. The relationship between Einstein's coefficients of any pair of energy levels (that are separated by a frequency $\nu$) is still:
\begin{equation} \frac{A_{21}}{B_{21}} = \frac{8\pi h\nu^{3}}{c^{3}} \\ \frac{B_{21}}{B_{12}} = \frac{g_{1}}{g_{2}} \end{equation}

Note that thermal spreading wouldn't disturb the degeneracies. This spread in energy levels has consequences, as you pointed out in the comments, for the Boltzmann factor. For every frequency difference $\nu$, the Boltzmann factor would now be:

\begin{equation} \frac{n_{u}}{n_{l}} = e^{-h\nu/KT} \end{equation}

Therefore, in principle, you could derive the source function for each frequency independently. It would then collectively hold that Kirchoff's law is applicable to any gas in LTE, regardless of what physical mechanisms contributed how much towards line spreading.

Of course, the following assumption is still utilized for all transition frequencies: \begin{equation} \phi(\nu) = \chi(\nu) = \psi(\nu) \end{equation}

These line profile functions represent three very specific processes: Spontaneous Emission, Absorption, Stimulated Emission. These processes are ideally centered around a specific frequency $\nu_{0}$ so that I would expect thermal broadening to affect them equally.