Relating Einstein’s A and B coefficients

We can write the rate equation for $N_2$ in steady state, in which case the time derivative goes to zero. Rearranging to solve for the ratio $N_2 / N_1$ we obtain:

$\displaystyle \frac{N_2}{N_1} = \frac{B_{12} \rho(\nu)}{A_{21} + B_{21} \rho(\nu)}$

If we consider the particular case of equilibrium, we also know that

$\displaystyle \frac{N_2}{N_1} = \frac{g_2}{g_1} \exp{\left( - \frac{E_2 - E_1}{k_B T} \right)}$

$\displaystyle = \frac{g_2}{g_1} \exp{\left( - \frac{h \nu}{k_B T} \right)}$

We can set these two equations equal to one another and solve for the electromagnetic energy spectral density $\rho(\nu)$ . Doing so yields

$\displaystyle \rho(\nu) = \frac{A_{21}}{B_{21}} \frac{1}{(\frac{B_{12}g_1}{B_{21}g_2} \exp{(h \nu / k_B T)} - 1)}$

Let’s consider a blackbody, which is an idealized object that is capable of absorbing or emitting any wavelength of light. One real-world approximation to a blackbody is a small hole in the side wall of a box made of light absorbing material. It can be shown that a blackbody always emits a characteristic spectrum that depends only on its temperature; fundamentally this is because the electronic energy states in the blackbody are filled according to thermodynamic principles. The electromagnetic spectral energy density produced by an ideal blackbody is:

$\displaystyle \rho(\nu) = \frac{8 \pi h \nu^3}{c^3} \frac{1}{\exp{(h \nu / k_B T)} - 1}$

Comparing this to the equation for $\rho$ we derived above, we see that the following relationships must hold amongst the $A$ and $B$ coefficients:

$\displaystyle \frac{B_{12}}{B_{21}} \frac{g_1}{g_2} = 1$

$\displaystyle g_2 B_{21} = g_1 B_{12}$

also

$\displaystyle \frac{A_{21}}{B_{21}} = \frac{8 \pi h \nu^3}{c^3}$

$\displaystyle A_{21} = \frac{8 \pi h \nu^3}{c^3} B_{21}$

these are the Einstein relations, which hold for the A and B coefficients between any pair of states. One thing to note is that making an x-ray laser would be extremely difficult, as $B / A$ decreases with photon frequency cubed $\nu^3$ .