# Feynman's derivation of Saha equation (FLP v. I)

I like to go back to the Feynman lectures on physics because Feynman was so good at cutting through the formalism in order to focus on understanding. Yet I can't understand his explanation of the general form of the Saha equation (see Feynman's lecture volume one - section 42-3) for a hydrogen plasma.

From what he says

Now we [...] say that the number of free electrons per unit volume in the “vapor” is equal to the number of bound electrons per unit volume in the atoms, times e to the minus the energy difference between being bound and being free, over kT .

I would have expected him to write $n_e = n_a e^{-W/kT}$, since the density of bound electrons is equal to the density of bound atoms. Instead, he goes on to write

Now what about the number of electrons per unit volume that are bound to atoms? The total number of places that we could put the electrons is apparently $n_a+n_i$ , and we will suppose that when they are bound each one is bound within a certain volume $V_a$ . So the total amount of volume which is available to electrons which would be bound is $(n_a+n_i)V_a$ , so we might want to write our formula as $n_e=\frac{n_a}{(n_a+n_i)V_a}e^{−W/kT}$.

I don't understand at all why he does that. Why does the total volume available to would-be bound electrons matter? Shouldn't everything be relative to the volume of a hypothetical box which the plasma occupies?

It seems I don't understand either how the different densities are defined, or how Feynman uses the Boltzmann factor.

I'm not looking for a formal derivation of the Saha equation. I would like to understand Feynman's intuition.

• I answered this rather hastily, so I've rewritten my answer after a bit of thought. The key to understanding Feynman's use of the Boltzmann factor is in understanding the fact that the densities in $n_2/n_1 = e^{-W/kT}$ are taken with respect to accessible volume, as it is this which determines the number of available states with energy $E_2$. – J. Murray Jan 21 '18 at 3:35

The general idea of the Boltzmann factor is that if two states $a$ and $b$ are separated by an energy difference $W$, then the relative probability of the higher energy state being occupied is

$$\frac{P(E_b)}{P(E_a)} = e^{-W/kT}$$

For a system of a large number of particles, the number of particles with energy $E$ is equal to the number of states with energy $E$ times the probability that such states are occupied. If we call the number of such states $g(E)$ and the number of particles $N(E)$, then

$$N(E) \propto g(E) \cdot P(E)$$

The degeneracy $g(E)$ is loosely the number of ways we could arrange to give a particle an energy $E$. In particular, it is proportional to the volume $V(E)$ which is accessible to particles with energy $E$. We can therefore say that

$$n(E) \equiv \frac{N(E)}{V(E)} \propto P(E)$$ and so

$$\frac{n_b}{n_a} = e^{-W/kT}$$

From here, Feynman's logic is pretty reasonable. $n_i,n_e,$ and $n_a$ are defined to be the total number of ions, free electrons, and neutral atoms, respectively divided by the total volume of the box $V$.

However, the total volume of the box is not necessarily the total volume accessible to each category of particle. Again, it is the total accessible volume which is important because that is the quantity which determines the degeneracy of thermodynamic states.

From the above, $$\frac{N_e}{V_e} = \frac{N_{\text{atoms}}}{V_{\text{atoms}}} e^{-W/kT}$$

The total volume available to the ions is simply the total volume of the box, as they are free to roam anywhere - therefore,

$$\frac{N_e}{V_e} = \frac{N_e}{V} \equiv n_e$$

On the right hand side, the total volume available for atomic electrons is $(N_i+N_e)V_a$, so we get

$$\frac{N_a}{(N_i+N_a)V_a}$$

Dividing the top and bottom by $V$ gives

$$\frac{n_a}{(n_i+n_a)V_a}$$

and so putting everything together,

$$n_e = \frac{n_a}{(n_i+n_a)V_a}e^{-W/kT}$$