Complement 1: Exact Differential

Summary

Attention

One difficulty in thermodynamics is to establish a formal link between what makes sense and the mathematics used to describe it. At a central point of the first and second laws of thermodynamics is the concept of function of states.

A function of state $f(x)$ is a function that only depends on the state the system is in, it does not depend on how it got there.

Mathematically, such a function must be an exact differential such that

$\int_i^f \mathrm{d} f = f(x_f)-f(x_i)$

or, equivalently:

$\oint \mathrm{d} f = 0$

Conversely, a function that is not a function of state is one that depends on how it got there. This is an important distinction as the path taken by a system will necessarily yield different values for such a function. Mathematically, this translates into the fact that the function is an inexact differential. We introduce a notation that reminds us that the equations above do not apply, namely, for a function $g(x)$ , which is not a function of state, we have

$\int_i^f \dbar(g) \neq g(x_f)-g(x_i)$

and of course

$\oint \dbar(g) \neq 0$

Partial differential

Consider a function $x=x(y,z)$ , then we have:

$\mathrm{d} x=\left(\frac{\partial x}{\partial y}\right)_{z} \mathrm{~d} y+\left(\frac{\partial x}{\partial z}\right)_{y} \mathrm{~d} z$

Similarly, for $z=z(x,y)$ , we have:

$\mathrm{d} z=\left(\frac{\partial z}{\partial x}\right)_{y} \mathrm{~d} x+\left(\frac{\partial z}{\partial y}\right)_{x} \mathrm{~d} y$

It follows, from those two equations that:

$\mathrm{d} x=\left(\frac{\partial x}{\partial z}\right)_{y}\left(\frac{\partial z}{\partial x}\right)_{y} \mathrm{~d} x+\left[\left(\frac{\partial x}{\partial y}\right)_{z}+\left(\frac{\partial x}{\partial z}\right)_{y}\left(\frac{\partial z}{\partial y}\right)_{x}\right] \mathrm{d} y$

It follows two important theorems:

Reciprocal theorem:: $\left(\frac{\partial x}{\partial z}\right)_{y}=\frac{1}{\left(\frac{\partial z}{\partial x}\right)_{y}}$
index:Reciprocity theorem:: $\left(\frac{\partial x}{\partial y}\right)_{z}\left(\frac{\partial y}{\partial z}\right)_{x}\left(\frac{\partial z}{\partial x}\right)_{y}=-1$
Corollary: By combining the two theorems, we find a very handy relationship:

$\left(\frac{\partial x}{\partial y}\right)_{z}=-\left(\frac{\partial x}{\partial z}\right)_{y}\left(\frac{\partial z}{\partial y}\right)_{x}$

Exact differential

Consider the function $f(x,y)$ . This function will be an exact differential provided that

$\left(\frac{\partial^{2} f}{\partial x \partial y}\right)=\left(\frac{\partial^{2} f}{\partial y \partial x}\right)$

This relation can be proven using Stokes theorem.

Copy of Slides

The slides for this complement are available in pdf format here: pdf