# Derive the Boltzmann factor in classical statistical mechanics

In both quantum and classical statistical mechanics, the probability of an NVT system having an energy $E$ is proportional to $$p(E)\propto e^{-E/T}$$

However, all of the derivations (that I can find) for this are implicitly built on quantum mechanics (i.e. discrete sums and the traditional $S=\log\Omega$ definition of entropy) and then merely generalise the result to the continuum case at the end. But is there a way of deriving this result entirely within a continuous framework (i.e. integrals and differential entropy $S=-\int p\log p$)?

I may have misunderstood what you're looking for exactly, but the usual way of deriving the Boltzmann factor for a canonical ensemble does not involve any discretization. Here's a short sketch, if you are interested in additional details I will add them. (Or simply look into Vol. 5 of Landau & Lifshitz, or Stat. Mech. by J. Woods Halley chapter 1.)

Start from Liouville's theorem, which tells us that the density distribution function $f$ is a constant of motion defined by Hamilton's equations. Then we make two simplifying assumptions:

1. Suppose the system has a large number of degrees of freedom.
2. All interactions are short range.

Next we partition the considered system into two parts each containing a large number of particles, and use the second assumption in order to be able to neglect the finite size effects due to the partition in the limit of $L\to \infty$ ($L$ being the partition size). Now any average property can be calculated equally well in either partition of the divided system (of course here there are underlying ergodic considerations). Thus in the limit of large systems the original density distribution function can be written as a product of the that of the separated parts: $$f(\mathbf{q},\mathbf{p})=f_1(\mathbf{q_1},\mathbf{p_1})f_2(\mathbf{q_2},\mathbf{p_2}) \tag{1}$$ Taking the $\ln:$ $$\ln{f(\mathbf{q},\mathbf{p})}=\ln{f_1(\mathbf{q_1},\mathbf{p_1})}+\ln{f_2(\mathbf{q_2},\mathbf{p_2})} \tag{2}$$ Now each of the density distributions $f_1$ and $f_2$ still obey Liouville's theorem and are thus constants of motion. More importantly, from $(2)$ we see now that $\ln{f}$ is an additive constant of motion. Knowing that there are only $7$ additive constants of motion, the energy, the three momenta and the three angular momenta, it follows that $\ln{f}$ can also be written as a linear combination of the other additive constants of motion:

$$\ln{f(\mathbf{q},\mathbf{p})}=\alpha - \beta H(\mathbf{q},\mathbf{p})+\mathbf{\gamma} \cdot \mathbf{p} + \mathbf{\delta} \cdot \mathbf{L} \tag{3}$$ One last simplification is to consider the system being enclosed in a rigid box (or simply confined in a wall), which means that momentum and angular momentum are not conserved anymore, thus the only remaining constant of motion is the energy, so $(3)$ becomes: $$\ln{f(\mathbf{q},\mathbf{p})}=\alpha - \beta H(\mathbf{q},\mathbf{p}) \tag{4}$$

From which, knowing that $\alpha$ and $\beta$ are constants independent of $(\mathbf{q},\mathbf{p}),$ we can straightforwardly write (ignoring normalization factors):

$$f(\mathbf{q},\mathbf{p}) \propto e^{-\beta H(\mathbf{q},\mathbf{p})} \tag{5}$$ Equation $(5)$ is just the canonical distribution function.