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 [1], 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 [2]) 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!

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[1] H. B. Callen, Thermodynamics and an Introduction to Thermostatistics (2nd ed.), New York: John Wiley & Sons, 1985.

[2] 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.