Derivation of the Boltzmann distribution (in a specific form) I am trying to derive a version of the Boltzmann distribution from statistical mechanics:
$$n(\mathbf{x}) = n_{0}e^{-q\phi(\mathbf{x})/kT}.$$
This expression is commonly used in the derivation of the Debye length in plasma physics (see Wikipedia).
I have found this article helpful. I have also looked at the discussion around equations (7.29)-(7.31) in Thermal Physics by Daniel Schroeder. However, the results in these sources don't exactly match the equation as given above. Both sources give:
$$P_i = \frac{\left< N_i \right>}{N} = \frac{1}{Z}e^{-\varepsilon_i/kT},$$
where $\left< N_i \right>$ is the average number of particles in the $i$th state. Obviously, this is close to the expression above.
I have tried the following with energies that are not discrete. I assume a canonical ensemble and a single particle of charge $q$ in a potential $\phi(\mathbf{x})$. The probability density is
$$P(\mathbf{x}, \mathbf{p}) = \frac{\exp\left[-q \phi(\mathbf{x})/kT\right] \exp\left[-p^2 / 2mkT\right]}{\int d\mathbf{x} \, d\mathbf{p} \, \exp\left[-q \phi(\mathbf{x})/kT\right] \exp\left[-p^2 / 2mkT\right]}$$
Integrating over momentum space gives the probability density over configuration space:
$$\int d\mathbf{p} \, P(\mathbf{x}, \mathbf{p}) = P(\mathbf{x}) = \frac{\exp\left[-q \phi(\mathbf{x})/kT\right]}{\int d\mathbf{x} \, \exp\left[-q \phi(\mathbf{x})/kT\right]}$$
I interpret this as the probability per unit volume that particle q will be located at $\mathbf{x}$. If I multiply now by the number of particles, $N$, I have the average number of particles per unit volume at $\mathbf{x}$, which I interpret as $n(\mathbf{x})$:
$$n(\mathbf{x}) = N \frac{\exp\left[-q \phi(\mathbf{x})/kT\right]}{\int d\mathbf{x} \, \exp\left[-q \phi(\mathbf{x})/kT\right]. }$$
Comparing to the equation that I gave above, it seems that I am close. What am I missing?
 A: I believe that the steps in the posted question are (more or less) correct. The final connection between the first and last equations can be seen by integrating the first equation:
$$\int d\mathbf{x} \, n(\mathbf{x}) = \int d\mathbf{x} \, n_0 e^{-q\Phi(\mathbf{x})/kT}$$
$$N = n_0 \int d\mathbf{x} \, e^{-q\Phi(\mathbf{x})/kT}$$
Solving for $n_0$ gives
$$n_0 = \frac{N}{\int d\mathrm{x} \, \exp(-q\Phi(\mathbf{x})/kT)}$$
Comparing this to the last equation in the original question, we see that the last equation is in fact identical to the first equation.
As a side note, for $q\Phi(\mathbf{x}) \ll kT$ the density becomes
$$n(\mathbf{x}) = n_0$$
If the source of $\Phi$ is at the origin, then this approximation is valid for $\mathbf{x} \to \infty$  For this reason, F.F. Chen (Introduction to Plasma Physics and Controlled Fusion) uses $n_\infty$ for $n_0$.
Also, Chapter 4 of Statistical Mechanics by McQuarrie gives a very extensive discussion of how to transition from the probability that a system will be in a given state to the probability that a particle will be in a given state. This helps justify using a single particle in the original question. The assumptions  are that the particles do not interact, temperature is high, and density is low. The first assumption of non-interacting particles is perhaps not so great for charged particles in a plasma. However, if the high temperature and low density conditions are met, then the assumption of non-interaction becomes better.
I would like to thank one of my colleagues for helping me with this. He could have posted the answer, but he doesn't have an account. Pity :)
