The following is a summary of my reading of https://www.feynmanlectures.caltech.edu/I_42.html#Ch42-S5


  • $N_{i}$ Population of molecules in state $i$
  • $R_{i\to j}$Transition rate from state $i$ to state $j$
  • $A_{mn}$ Coefficient of spontaneous emission
  • $B_{mn}$ Coefficient of induced emission
  • $B_{nm}$ Coefficient of absorption
  • $E_{m}-E_{n}=\Delta E=\hbar\omega>0$ Transition energy
  • $\mathcal{I}(\omega)$ Radiation intensity profile
  • $N_{m}=N_{n}e^{-\frac{\Delta E}{\mathit{k}T}}$Boltzmann relation

Feynman's equation 42-12

$$ \mathcal{I}(\omega)d\omega=\frac{\hbar\omega^{3}d\omega}{\pi^{2}c^{2}\left(e^{\frac{\hbar\omega}{\mathit{k}T}}-1\right)}. $$


Write the expressions for transition rates and set them equal using argument by footnote $$\begin{aligned} R_{n\to m}&=N_{n}\mathcal{I}(\omega)B_{nm}\\ R_{m\to n}&=N_{m}\left(A_{mn}+\mathcal{I}(\omega)B_{mn}\right)\\ R_{n\to m}&=R_{m\to n}. \end{aligned}$$

Combining expressions and applying basic algebra we get $$ \mathcal{I}(\omega)=\frac{A_{mn}}{B_{nm}e^{\frac{\hbar\omega}{\mathit{k}T}}-B_{mn}}=\frac{\hbar\omega^{3}}{\pi^{2}c^{2}\left(e^{\frac{\hbar\omega}{\mathit{k}T}}-1\right)}. $$

Therefore we can deduce something: First, that $B_{nm}$ must equal $B_{mn}$, since otherwise we cannot get the $(e^{\hbar\omega/kT} - 1)$. So Einstein discovered some things that he did not know how to calculate, namely that the induced emission probability and the absorption probability must be equal.

Clearly, setting $B_{nm}=B_{mn}$ gives a compelling result, but I don't believe that follows from the algebra. Does the "necessity" of the result follow from a variation of $\omega$ or some other method of differential calculus?

If $\omega$ were a continuous real number parameter with all other terms constant, the result would be obvious. But in this case $\omega$ is a discrete value determined by the transition energy.

I also observe that in this recent and more detailed application of these ideas, the equation $B_{nm}=B_{mn}$ does not, in general, hold. See equation 14



The expression


is a function of $\omega$. If it is supposed to hold for more than one specific value of $\omega$ (and $T$), then the argument holds. You can easily see this from inspection, but you could also say e.g. that for small $\omega$ (specifically, $ \omega \ll kT/\hbar$), the expression on the right becomes inversely proportional to $\omega$, while the expression on the left becomes inversely proportional to $\omega + \frac{kT}{\hbar}(1-B_{mn}/B_{nm})$; demanding the same low-frequency (or high-temperature) behavior requires that $B_{nm}-B_{mn}$ vanishes.

  • $\begingroup$ Ok. I guess it depends on a variation of $T$. I didn't consider that. The reason I balked at a variation in $\omega$ is that in this case $\omega$ is a discrete value determined by the transition energy. $\endgroup$ – Steven Thomas Hatton Jun 3 at 3:37
  • $\begingroup$ @StevenThomasHatton Ah, okay. That's not quite right. The idea is that we are taking $I(\omega)=\hbar\omega^3/\pi^2c^2(e^{\hbar\omega/kT}-1)$ as an input. We say that we know that this is the correct form for the radiation intensity profile for all $\omega$. On the other hand, by balancing the emission/absorption rates, we can express it in terms of the $A$'s and $B$'s. By comparing this to the formula we know, we can read off the fact that $B_{nm}=B_{mn}$ by inspection. $\endgroup$ – J. Murray Jun 3 at 17:36
  • $\begingroup$ "by inspection" is a bit more rigorous than argumentum footnoteum. To me "by inspection" means "the result is obvious", but it still requires a step of reasoning. The $B_{mn}$ and $B_{nm}$ are specific to a species of molecule, and to a particular state transition. The blackbody equation involving them really (as I see it) only applies to the resonant peak near the transition frequency. On the other hand, it is temperature agnostic. Holding everything but the temperature constant requires $B_{nm}=B_{mn}.$ See feynmanlectures.caltech.edu/I_41.html#Ch41-F3 or ... $\endgroup$ – Steven Thomas Hatton Jun 4 at 3:17
  • $\begingroup$ ... [continued] Figure 3 in dx.doi.org/10.1155/2013/503727 $\endgroup$ – Steven Thomas Hatton Jun 4 at 3:20

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