I am trying hard to understand the solution to a problem from David Chandler's Introduction to Modern Statistical Mechanics. The equation of state for a rubber band is given in terms of S(E,L,n) where E is the internal energy, L is the length of the rubber band, and n (I believe) is the number of molecules making up the rubber band. The task is to find an expression for the tension, $f$, as a function of L/n and the temperature T.
Earlier in the chapter (pg. 10), it is shown that for generalized displacements X and generalized forces f $$ \frac{\partial S}{\partial \textbf{X}}|_E = -\frac{\textbf{f}}{T} $$
Additionally, earlier in the chapter (pg. 9). It is explained that $ \frac{\partial S}{\partial \textbf{X}}|_E \textrm{ }d\textbf{X} = \frac{\partial S}{\partial X_1}|_E \textrm{ }dX_1 + \frac{\partial S}{\partial X_2}|_E \textrm{ }dX_2 … $
Therefore, when I tried to solve this problem, taking that general equation above and applying it to this system I arrived at:
$$ \frac{\partial S}{\partial \textbf{X}}|_E = \frac{\partial S}{\partial L}|_{E,n} + \frac{\partial S}{\partial n}|_{E,L} = -\frac{f}{T} $$
However, in the solutions manual, it is stated that:
$$ \frac{\partial S}{\partial L}|_{E,n} = -\frac{f}{T} $$
Why is the $ \frac{\partial S}{\partial n}|_{E,L} $ not included? At first I was thinking that the variables for $n$ and $L$ get some fundamentally different treatment, but in fact on page 6 he explicitly states that: " We choose to ignore the distinction [between the composition variables (n) and mechanical variables] because one can …design experiments in which transfer of moles occurs with the consumption… of work".
Here is the specific form of the equation of state if it is relevant:
$$ S = nl_0\gamma \sqrt{\frac{\theta E}{nl_0}} – nl_0\gamma[\frac{1}{2}(\frac{L}{nl_o})^2 + \frac{nl_o}{L} – \frac{3}{2}]$$
Another post here many years ago asked about the same question (problem 1.2 from David Chandler's Intro. to Modern Stat. Mech.) , but the answer to that post does not answer my specific question about the problem.
Best Answer
This is not statistical mechanics, but standard thermo-statics. Starting from the Gibbs equation for the internal energy with variables $S, X_1, X_2, ..$ you have $$dE=TdS+\sum_k Y_k dX_k \tag{1}$$ or expressed for $S$ is $$dS=\frac{1}{T}dE - \sum_k \frac{Y_k}{T}dX_k \tag{2}.$$ Here $E$ and the other extensives $X_k$ are the independent variables therefore you have $$\frac{\partial S}{\partial E} = \frac{1}{T} \tag{3}$$and $$\frac{\partial S}{\partial X_k} = -\frac{Y_k}{T} \tag{4}.$$ Eq. (4) is the formula you have been asking about when $X_1=L$, the length, and $Y_1 = f$ the corresponding conjugate intensive, i.e., the tension, $\frac{\partial S}{\partial L} = -\frac{f}{T}$. The number of molecules $n$ does not show up there. When it is included in the energy (or entropy) form, $dU=TdS+fdL+\mu n$, then it shows up with its conjugate intensive, the so-called chemical potential, usually denoted by $\mu$, therefore $\frac{\partial S}{\partial n} = -\frac{\mu}{T}$.