# Boltzmann factor and ratio of number of states

When using Boltzmann factor for two states, we see that $$N_1/N_2 = \exp\left(\frac{h\omega}{k_BT}\right)\;.$$ Where $N_1$ is the number of atoms in state with lower energy $E_1$ than atoms in state $N_2$ with energy $E_2$. The Boltzmann factor shows that there will always be more atoms in state $N_1$ and $N_2$, no matter how much we turn up the temperature. But if we turn up the temperature more and more, wouldn't we expect at some point that all atoms would want to be in the state with highest energy $E_2$? Since otherwise, where would that energy go from turning up the temperature? Thank you.

Your intuition is correct: even though $N_2$ can never exceed $N_1$ as $T \to \infty$, something has to happen if we keep putting in more energy. What happens is that the temperature "overflows" and goes to $-\infty$. It then increases as we put more energy in, finally reaching $-0$ when $N_1 = 0$.

The reason this looks unnatural is because temperature $T$ is not the right variable; the inverse temperature $\beta \propto 1/T$ is more fundamental. In this case, the inverse temperature always decreases when we put more energy in; it changes continuously from $+\infty$ to $-\infty$.

• No, you can't do that simply by raising the temperature. You can invert the population only in non-equilibrium situations. If the temperature approaches $\infty$, then you simply cannot add any more energy by turning up the temperature. – garyp Jun 19 '16 at 0:33
• @garyp I never said anything about raising the temperature, I only talked about "putting more energy in". – knzhou Jun 19 '16 at 3:03
• Ok you are right about that. But then you haven't answered the question that the OP asked. In three places the OP mentions raising the temperature. – garyp Jun 19 '16 at 3:33
• @garyp Huh, you're right, I autocorrected it mentally. I'll fix this in an edit soon. – knzhou Jun 19 '16 at 3:34

The reason why $N_2$ can never exceed $N_1$ in your expression is that a state in which $N_2>N_1$ is a non-equilibrium state, and your expression is not going to allow that case because it is derived from the canonical probability density, which is only valid at equilibrium.

A state in which $N_1>N_2$ is a non-equilibrium state because a system cannot normally remain in an excited state for an infinite time: after a certain finite amount of time, a fluctuation in the electromagnetic field will cause spontaneous emission and the excited state will decay to the lower energy level (until eventually the ground state is reached).

$N_1=N_2$ means that the number of transition from level 1 to level 2 matches the number of transitions from level 2 to level 1, so it is a dynamical kind of equilibrium.

Non-equilibrium situations in which $N_2>N_1$ can be achieved in many ways, a process known as population inversion.

• It makes sense from an excited atom viewpoint in that it will do spontaneous emission as more and more atoms go into state N2. However, when Boltzmann did this a long time ago, I don't think he knew about spontaneous emission. Maybe he thought about a molecule flying in the air (E1), and a molecule flying in the air AND vibrating (E2)at the same time. So as T increases, you would expect more more and molecules to be both flying and vibrating, and eventually N2 > N1. However this seems impossible from the Boltzmann factor. – Bootleg Jun 19 '16 at 9:31