March 4, 2013

The Euler equation in thermodynamics

The internal energy $$U$$ and its natural variables $$S, V, N_i$$ are extensive quantities. It is then -mathematically- very easy (see Callen , section 3.1 for the canonical derivation) to prove the Euler equation:
\begin{equation}
U = TS -pV + \sum_i \mu_i N_i
\label{eq:Euler}
\end{equation}Briefly, one only needs to write the definition of first-order homogeneity:
\begin{equation}
U(\lambda S, \lambda V, \lambda N_i) = \lambda U(S, V, N_i)
\label{eq:homog}
\end{equation}take the derivative with respect to $$\lambda$$ and set $$\lambda = 1$$.

This demonstration is very elegant but it can hide the physical meaning of the relation \eqref{eq:Euler}. It is always a good idea to consider a system undergoing a precise transformation. Unless we can clearly identify the latter, we have not really understood the problem.

In our case, the transformation can be described as follows: all variables increase at the same rate. Since they are extensive, we can consider a system with length $$L$$ (see Figure 1) and a cursor that can slide along the $$x$$ axis.
Figure 1

At the position of the cursor we can introduce in the medium a wall that defines a new system, of length $$\lambda L$$. Clearly, all extensive variables $$U, S, V, N_i$$ have been multiplied by $$\lambda$$, while the intensive parameters $$T, p, \mu_i$$ remain unchanged. We are thus moving along the "diagonal" of parameter space, i.e. the line connecting the current point to the origin, as shown in Figure 2 for a one-component system.
Figure 2
As on any path, we can of course write the fundamental relation for an infinitesimal displacement:
\begin{equation}
\text{d}U = T\text{d}S -p\text{d}V + \sum_i \mu_i  \text{d}N_i
\label{eq:fund}
\end{equation} On this particular path (and only on this one ) the derivatives $$T, p, \mu_i$$ are all constant so we can extend \eqref{eq:fund} to arbitrary displacements, yielding precisely \eqref{eq:Euler}.

In a future post I will try to show how the Gibbs-Duhem relation fits into this geometrical picture. UPDATE: here it is!

 H. B. Callen, Thermodynamics and an Introduction to Thermostatistics (2nd ed.), New York: John Wiley & Sons, 1985.
 Since we derive \eqref{eq:Euler} by moving along one particular path, one might wonder why it holds for all parameter values, even outside the given path. It should be noted that \eqref{eq:Euler} applies to a given state (a point in parameter space), and for each point we can draw its "diagonal". In contrast, \eqref{eq:fund} is written for a transformation.

1 comment:

1. Wow, great post.