# Rate Equations and Detailed Balance

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2 at frequency ν of J ν = (1/4π) I ν dω. Note that the difference in energies between level 2 and level 1 can t be defined with infinite precision, because of the uncertainty principle. Let us therefore be careful and define a line profile φ(ν) that is sharply peaked at the line center ν 0 and is normalized so that φ(ν)dν = 1. (1) 0 We can define an average intensity over this profile, J = 0 J ν φ(ν)dν. Note that if J ν varies slowly over the line profile (which it almost always does), this practically reduces to J ν0, but it s nice to be careful. Ask class: what effects can the external radiation field produce on the atom? Absorption or stimulated emission. Now, we d like to assume that the absorption rate and stimulated emission rate are both proportional to J, i.e., they are linearly proportional to intensity. As always, though, we want to be sure we know when this simplifying assumption can become invalid. This is a tough one, though: Ask class: when might this be incorrect? The linear relation could break down if the radiation is self-interacting, which might happen at extremely high intensities or with high photon energies (when processes such as pair production enter in). In almost all applications this is insignificant, but if you re analyzing a very high energy environment (e.g., gamma-ray bursts) you should be cautious. Anyway, let s apply that assumption. Then we have a transition rate of B 12 J for absorption and B 21 J for stimulated emission, where B12 and B 21 are the Einstein B coefficients. How are the three coefficients related to each other? We ll start by analyzing the situation in thermodynamic equilibrium. For this analysis we assume that the upper level has a statistical weight of g 2 and the lower level has a statistical weight of g 1. We also assume that in equilibrium there are n 2 atoms in level 2 and n 1 in level 1. Ask class: what relation does equilibrium imply? It implies that the rate from level 1 to 2 equals the rate from level 2 to 1, or n 1 B 12 J = n2 A 21 + n 2 B 21 J. (2) Therefore, A 21 /B 21 J = (n 1 /n 2 )(B 12 /B 21 ) 1. (3) In equilibrium, the level populations n 1 and n 2 are given by the Boltzmann distribution, which says that the number is proportional to the statistical weight times exp( E/kT), where E is the energy of the level. We re taking a ratio, so the zero point of the energy doesn t matter and therefore J = A 21 /B 21 (g 1 B 12 /g 2 B 21 ) exp(hν 0 /kt) 1. (4)

4 boson), so that two or more neutrinos cannot occupy the same state. As a result, instead of stimulated emission you d have suppressed emission: In addition, in thermal equilibrium the mean intensity is n 1 B 12 J = n2 A 21 n 2 B 21 J. (8) J ν = 2hν 3 /c 2 exp(hν/kt) + 1. (9) Ask class: if we carry through the same analysis as before, what would they guess would be the neutrino Einstein relations? As you can verify directly, you get exactly the same expressions as we had before! This is a dramatic indication that the Einstein relations are properties of the atomic physics and have nothing whatsoever to do with the properties of an external radiation field! Essentially, bringing in the field is just a convenient way to get the answer, but it isn t necessary. We can express the emission and absorption coefficients using the Einstein coefficients. Let s assume that the line profile for emission is the same as it is for absorption (a very good assumption most of the time). Then, as derived in the book, the emission coefficient is and the absorption coefficient is j ν = (hν 0 /4π)n 2 A 21 φ(ν) (10) α ν = (hν/4π)φ(ν)(n 1 B 12 n 2 B 21 ). (11) Note that again stimulated emission is acting like a negative absorption. For the record (and again from the book, equation 1.6), the transfer equation then becomes and the source function is di ν /ds = (hν/4π)(n 1 B 12 n 2 B 21 )φ(ν)i ν + (hν/4π)n 2 A 21 φ(ν) (12) S ν = n 2 A 21 /(n 1 B 12 n 2 B 21 ) = (2hν 3 /c 2 )/ ( g 2 n 1 g 1 n 2 1 ) 1 (13). Note that the A coefficient has different units from the B coefficients (the latter, multiplied by I ν, give the units of A 21 ). Also, note that it is pointless to memorize equations such as this. You can get an idea of what the answer should be by asking yourself how the intensity should change. Clearly, the intensity will increase if there is spontaneous emission or stimulated emission, and decrease if there is absorption. Also, remember that absorption and stimulated emission are proportional to I ν whereas spontaneous emission isn t. That should allow you to piece together most of the important parts. If the matter is in thermal equilibrium with itself, but not necessarily with the radiation, we have local thermodynamic equilibrium and n 1 /n 2 = (g 1 /g 2 ) exp(hν/kt). You can verify

5 that this makes S ν = B ν. In all other cases (nonthermal emission), this equality does not hold. A particularly cool way in which matter can be out of thermal equilibrium is in a laser or maser (the same phenomena but different wavelengths; however, masers are found in space whereas lasers tend to be laboratory creations). Let s divert with these a bit. Suppose for simplicity that g 1 = g 2 (equal multiplicity of the two states). Then B 12 = B 21. Ask class: if n 1 > n 2, which dominates, absorption or stimulated emission? Absorption. This is the normal case, because E 2 > E 1 so in thermal equilibrium (or even most forms of non-thermal distributions) the lower-energy state is more populated. Ask class: but what if n 2 > n 1? Then the intensity increases exponentially along the path of the beam. Moreover, although it isn t clear from our discussion here, bosons (e.g., photons) like to occupy the same quantum state, so the direction and phase of the emitted photon is the same as that of the original photon. This means tremendous intensity! Just for fun, let s think about what conditions might allow a laser/maser. Ask class: what are the requirements? We need an inverted population, which has higher energy than the normal population, meaning we need some supply of energy. Another, trickier, requirement is that the upper state must persist for a while (otherwise spontaneous emission will reduce the number of atoms in the upper state). In practice this often means that the level structure has to be complicated: excitation can occur to another state, say state 3, which decays to state 2. State 2 is then metastable, meaning it has a long time before it will decay. In the lab it takes lots of special preparation to get all this to work, mainly because the number densities (of gas, liquid, or solid) are high enough that thermal equilibrium (specifically a normal population) can be established rapidly. In space, the number densities are much less and therefore it is a lot easier for inverted populations to exist. The result is that masers are common in astronomy: around very young stars, around old stars, in the centers of galaxies, and so on. The tiny size of masing regions, plus their high intensity and extreme frequency coherence, makes them remarkably good probes of astrophysical conditions.

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