# Variation in Entropy in Einstein's Brownian Motion Paper

In Einstein's Brownian motion paper, he derives a formula for the diffusion coefficient of suspended particles by assuming the system is in dynamic equilibrium and thus, for a variation $\delta x$ (which may vary with the position $x$) in the particle positions, the variation in free energy vanishes: $\delta F = \delta U - T \delta S = 0$. It is also assumed that a force $K$ (also a function of $x$) acts in the $x$-direction, so that the variation in internal energy, $\delta U$, ends up being the virtual work done by $K$.

What I don't understand and am hoping someone can help clarify, is the formula for $\delta S$.

The setup is that there are $n$ particles suspended in a subvolume $V^*$ which lies between $x=0$ and $x=l$. Using the earlier-derived formula $S = \frac{\bar{E}}{T} + \frac{R}{N}(\ln J + n \ln V^*)$, where $J$ is independent of the particle coordinates, Fürth's footnote derives the $\delta S$ formula as follows: $$\delta S = \frac{R}{N}\frac{n}{V^*}\delta v^* = \int_0^l \frac{R}{N} \nu \delta dx = \int_0^l \frac{R}{N} \nu d \delta x = \int_0^l \frac{R}{N} \nu \frac{\partial \delta x}{\partial x} dx = -\int_0^l \frac{R}{N} \frac{\partial \nu}{\partial x} \delta x dx$$ where $\nu$ (also a function of $x$) is the particle density. The last equality is integration by parts and the fact that $\delta x(0) = \delta x(l)=0$, but I was hoping someone could clarify the following points:

• how should one interpret $\delta v^*$? I thought the volume $V^*$ is just a fixed constant, so why would it vary at all with $\delta x$?
• is the second equality just the definition of the functional derivative?
• where does the third equality come from? And what does $d \delta x$ here mean? $d (\delta x (x))$? Or something else? The notation is a little tough since he leaves out the arguments when things are a function of $x$.
• finally, why is the $\bar{E}$ from the entropy formula not included in the variation? Apparently, the total energy does have a variation given that there was a $\delta U$? Obviously that's not correct, but can anyone explain why?

Thanks very much for your responses!