Complement 1: Exact Differential

Summary

Attention

One difficulty in thermodynamics is establishing a formal link between what makes sense and the mathematics used to describe it. At the core of the first and second laws of thermodynamics is the concept of a function of state.

A function of state $f(x)$ depends only 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 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 to the fact that the function is an inexact differential. To remind us that the equations above do not apply, we introduce a different notation. For a function $g(x)$ that is not a function of state, we write:

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

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:

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

From this we derive two important theorems:

Reciprocal theorem:: $\left(\frac{\partial x}{\partial z}\right)_{y} = \frac{1}{\left(\frac{\partial z}{\partial x}\right)_{y}}$
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 is 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