# On writing the entropy of a system as function of other parameters; Reif exercise 5.14

I am having trouble understanding the proposed solution for a problem taken from "Fundamentals of Statistical And Thermal Physics" by F. Reif. The problem is taken from the chapter 5 - Simple applications of macroscopic thermodynamics and reads as follows:

"In a temperature range near absolute temperature $T$, the tension force $F$ of a stretched plastic rod is related to its length $L$ by the expression $$F = aT^2(L-L_0)$$ where $a$ and $L_0$ are positive constants, $L_0$ being the unstretched length of the rod. When $L = L_0$, the heat capacity $C_L$ of the rod (measured at constant length) is given by the relation $C_L = bT$, where $b$ is a constant."

I understand what is said here. Then the part which is bogging me:

"a) Write down the fundamental thermodynamic relation for this system, expressing $dS$ in terms of $dE$ and $dL$."

My train of thought was as follows:

I am familiar with the expressions $\Delta E = \Delta Q + \Delta W$ and $dQ = TdS$. From those one can write $dE = TdS + dW$ For the usual case of a gas (which is the one most spoken of in previous sections) one gets

$$dE = TdS - pdV$$

where $p$ is the pressure; Having that, and because the $-pdV$ part has the differential of an external parameter multiplied by its generalized force, I thought I could infer that I would have

$$dE = TdS - FdL \iff\\ \iff TdS = dE + FdL \tag{1}$$

however the proposed solution I found here, on page 27, writes instead

$$TdS = dE - FdL$$

While I was thinking about why the signs differ, I also remembered another possible path that could help me understand this. In the section, the author wrote plenty of thoughts analogous to this one:

If we want to write $S$ as a function of $E, L$: $S \equiv S(E, L)$, then one has

$$dS = \left(\frac{\partial S}{\partial E}\right)_L dE + \left(\frac{\partial S}{\partial L}\right)_E dL \iff\\ TdS = T\left(\frac{\partial S}{\partial E}\right)_L dE + T\left(\frac{\partial S}{\partial L}\right)_E dL$$

Then I would only have to be able to understand why is it that

$$\begin{cases}T\left(\frac{\partial S}{\partial E}\right)_L = 1 \\ T\left(\frac{\partial S}{\partial L}\right)_E = -F \end{cases}\tag2$$

If any one is able to explain me why my extrapolation in $(1)$ is incorrect or why the system in $(2)$ holds by mathematical reasoning / physical intuition, it would be great.

• when you compress a gas you work on it hence the work is positive when $dV<0$ and increase the internal energy; this is the opposite as to what happens when $dV>0$ (expand the gas). When an elastic rod is stretched, $dL>0$, relative to its zero stress position you again increase its internal energy, hence the sign change and so the work is $\delta W = FdL$ – hyportnex Jan 16 '17 at 13:59
• @hyportnex thanks for that! I tried to reason about that, but I had no idea whether the internal energy of the rod would increase OR decrease if I stretched it. – RGS Jan 16 '17 at 14:16

The sign of the generalized forces can change according to the used convention. Using the same sign rule employed for the pressure:

$$\frac{\partial E}{\partial X_k} := -P_k$$

where in your example: $P_k = F$ and thus : $$dE = TdS - FdL$$ Like you found.

With this sign convention is also easy to see how your equations in $(2)$ hold.

In thermodynamics, the system equilibrium state is macroscopically characterized by its internal energy E, which is a function of some extensive parameters ($S$ and $X_1, X_2, \ldots$): $$E(S, X_1, X_2, \ldots)$$

For the system of your interest, the significant extensive parameters are $S$ and the rod length $L$ (there could be other extensive parameters, like the number of the various atoms which form the rod, the rod volume and so on. We can indicate all of them simply as $X_2, X_3, \ldots$).

In an equivalent way, the macroscopic state of the system can be completely characterized by its entropy, which is a function of $E$ and all its extensive parameters:

$$S(E, L, X_2, \ldots)$$

The differential of $S$ is therefore:

$$dS = \frac{\partial S}{\partial E} dE + \frac{\partial S}{\partial L} dL + \sum_{k=2} \frac{\partial S}{\partial X_k} dX_k$$

Applying the differentiation rules for composite functions we get:

$$\frac{\partial S}{\partial X_k} = \frac{\partial S}{\partial E} \frac{\partial E}{\partial X_k} = - \frac{P_k}{T}$$ Where the definition of absolute temperature was used: $$\frac{\partial E}{\partial S} := T \rightarrow \frac{\partial S}{\partial E} = \frac{1}{T}$$
For $X_k = L$: $$\frac{\partial S}{\partial L} = -\frac{F}{T}.$$