Exact and Inexact Differentials

In our investigation of heat and work, we have come across various infinitesimal objects, such as $d \overline{E}$ and ${\mathchar'26\mkern-11mud}W$ . It is instructive to examine these infinitesimals more closely.

Consider the purely mathematical problem where $F(x, y)$ is some general function of two independent variables, $x$ and $y$ . Consider the change in $F$ in going from the point $(x$ , $y)$ in the $x$ -$y$ plane to the neighboring point ($x+dx$ , $y+dy$ ). This is given by

$$dF = F(x + dx, y+dy)- F(x,y),$$ (4.22)

which can also be written

$$dF = X(x,y) dx + Y(x,y) dy,$$ (4.23)

where $X= \partial F/\partial x$ and $Y=\partial F/\partial y$ . Clearly, $dF$ is simply the infinitesimal difference between two adjacent values of the function $F$ . This type of infinitesimal quantity is termed an exact differential to distinguish it from another type to be discussed presently. If we move in the $x$ -$y$ plane from an initial point $i\equiv (x_i$ , $y_i)$ to a final point $f \equiv (x_f$ , $y_f)$ then the corresponding change in $F$ is given by

$${\mit\Delta} F = F_f - F_i = \int_i^f dF= \int_i^f(X dx + Y dy).$$ (4.24)

Note that because the difference on the left-hand side depends only on the initial and final points, the integral on the right-hand side can only depend on these points as well. In other words, the value of the integral is independent of the path taken in going from the initial to the final point. This is the distinguishing feature of an exact differential. Consider an integral taken around a closed circuit in the $x$ -$y$ plane. In this case, the initial and final points correspond to the same point, so the difference $F_f- F_i$ is clearly zero. It follows that the integral of an exact differential over a closed circuit is always zero:

$$\oint dF \equiv 0.$$ (4.25)

Of course, not every infinitesimal quantity is an exact differential. Consider the infinitesimal object

$${\mathchar'26\mkern-11mud}G \equiv X'(x,y) dx + Y'(x,y) dz,$$ (4.26)

where $X'$ and $Y'$ are two general functions of $x$ and $y$ . It is easy to test whether or not an infinitesimal quantity is an exact differential. Consider the expression (4.23). It is clear that because $X= \partial F/\partial x$ and $Y=\partial F/\partial y$ then

$$\frac{\partial X}{\partial y} = \frac{\partial Y}{\partial x}= \frac{\partial^{ 2} F} {\partial x \partial y}.$$ (4.27)

Thus, if

$$\frac{\partial X'}{\partial y} \neq \frac{\partial Y'}{\partial x}$$ (4.28)

(as is assumed to be the case), then ${\mathchar'26\mkern-11mud}G$ cannot be an exact differential, and is instead termed an inexact differential. The special symbol ${\mathchar'26\mkern-11mud}$ is used to denote an inexact differential. Consider the integral of ${\mathchar'26\mkern-11mud}G$ over some path in the $x$ -$y$ plane. In general, it is not true that

$$\int_i^f {\mathchar'26\mkern-11mud}G = \int_i^f (X' dx + Y' dy)$$ (4.29)

is independent of the path taken between the initial and final points. This is the distinguishing feature of an inexact differential. In particular, the integral of an inexact differential around a closed circuit is not necessarily zero, so

$$\oint {\mathchar'26\mkern-11mud}G \neq 0.$$ (4.30)

Consider, for the moment, the solution of

$${\mathchar'26\mkern-11mud}G = 0,$$ (4.31)

which reduces to the ordinary differential equation

$$\frac{dy}{dx} = -\frac{X'}{Y'}.$$ (4.32)

Because the right-hand side is a known function of $x$ and $y$ , the previous equation defines a definite direction (i.e., gradient) at each point in the $x$ -$y$ plane. The solution simply consists of drawing a system of curves in the $x$ -$y$ plane such that, at any point, the tangent to the curve is as specified by Equation (4.32). This defines a set of curves that can be written $\sigma (x, y) = c$ , where $c$ is a labeling parameter. It follows that, on a particular curve,

$$\frac{d\sigma}{d x}\equiv \frac{\partial \sigma}{\partial x} +\frac{\partial\sigma}{\partial y}\frac{dy}{dx} = 0.$$ (4.33)

The elimination of $dy/dx$ between Equations (4.32) and (4.33) yields

$$Y' \frac{\partial \sigma}{\partial x} = X' \frac{\partial\sigma}{\partial y} =\frac{X' Y'}{\tau},$$ (4.34)

where $\tau(x, y)$ is function of $x$ and $y$ . The previous equation could equally well be written

$$X' = \tau \frac{\partial \sigma}{\partial x},     Y' = \tau \frac{\partial \sigma} {\partial y}.$$ (4.35)

Inserting Equation (4.35) into Equation (4.26) gives

$${\mathchar'26\mkern-11mud}G = \tau \left( \frac{\partial \sigma}{\partial x} dx + \frac{\partial\sigma} {\partial y} dy\right)= \tau d\sigma,$$ (4.36)

or

$$\frac{{\mathchar'26\mkern-11mud}G}{\tau} = d\sigma.$$ (4.37)

Thus, dividing the inexact differential ${\mathchar'26\mkern-11mud}G$ by $\tau$ yields the exact differential $d\sigma$ . A factor $\tau$ that possesses this property is termed an integrating factor. Because the previous analysis is quite general, it is clear that an inexact differential involving two independent variables always admits of an integrating factor. Note, however, this is not generally the case for inexact differentials involving more than two variables.

After this mathematical excursion, let us return to a physical situation of interest. The macrostate of a macroscopic system can be specified by the values of the external parameters (e.g., the volume), and the mean energy, $\overline{E}$ . This, in turn, fixes other parameters, such as the mean pressure, $\bar{p}$ . Alternatively, we can specify the external parameters and the mean pressure, which fixes the mean energy. Quantities such as $d\bar{p}$ and $d \overline{E}$ are infinitesimal differences between well-defined quantities: that is, they are exact differentials. For example, $d\overline{E} = \overline{E}_f - \overline{E}_i$ is just the difference between the mean energy of the system in the final macrostate $f$ and the initial macrostate $i$ , in the limit where these two states are nearly the same. It follows that if the system is taken from an initial macrostate $i$ to any final macrostate $f$ then the mean energy change is given by

$${\mit\Delta} \overline{E} = \overline{E}_f - \overline{E}_i = \int_{i}^f d\overline{E}.$$ (4.38)

However, because the mean energy is just a function of the macrostate under consideration, $\overline{E}_f$ and $\overline{E}_i$ depend only on the initial and final states, respectively. Thus, the integral $\int d\overline{E}$ depends only on the initial and final states, and not on the particular process used to get between them.

Consider, now, the infinitesimal work done by the system in going from some initial macrostate $i$ to some neighbouring final macrostate $f$ . In general, ${\mathchar'26\mkern-11mud}W = \sum \overline{X}_\alpha dx_\alpha$ is not the difference between two numbers referring to the properties of two neighboring macrostates. Instead, it is merely an infinitesimal quantity characteristic of the process of going from state $i$ to state $f$ . In other words, the work ${\mathchar'26\mkern-11mud}W$ is in general an inexact differential. The total work done by the system in going from any macrostate $i$ to some other macrostate $f$ can be written as

$$W_{if} = \int_i^f {\mathchar'26\mkern-11mud}W,$$ (4.39)

where the integral represents the sum of the infinitesimal amounts of work ${\mathchar'26\mkern-11mud}W$ performed at each stage of the process. In general, the value of the integral does depend on the particular process used in going from macrostate $i$ to macrostate $f$ .

Recall that, in going from macrostate $i$ to macrostate $f$ , the change in energy, ${\mit\Delta} \overline{E}$ , does not depend on the process used, whereas the work done, $W$ , in general, does. Thus, it follows from the first law of thermodynamics, Equation (4.10), that the heat absorbed, $Q$ , in general, also depends on the process used. It follows that

$${\mathchar'26\mkern-11mud}Q \equiv d \overline{E} + {\mathchar'26\mkern-11mud}W$$ (4.40)

is an inexact differential. However, by analogy with the mathematical example discussed previously, there must exist some integrating factor, $T$ (say), that converts the inexact differential ${\mathchar'26\mkern-11mud}Q$ into an exact differential. So,

$$\frac{{\mathchar'26\mkern-11mud}Q}{T} \equiv dS.$$ (4.41)

It will be interesting to discover the physical quantities that correspond to the functions $T$ and $S$ . (See Section 5.5.)

Suppose that the system is thermally insulated, so that ${\mathchar'26\mkern-11mud}Q=0$ . In this case, the first law of thermodynamics implies that

$$W_{if} = - {\mit\Delta} \overline{E}.$$ (4.42)

Thus, in this special case, the work done depends only on the energy difference between the initial and final states, and is independent of the process. In fact, when Clausius first formulated the first law of thermodynamics, in 1850, he wrote:

If a thermally isolated system is brought from some initial to some final state then the work done by the system is independent of the process used.

If the external parameters of the system are kept fixed, so that no work is done, then ${\mathchar'26\mkern-11mud}W=0$ , Equation (4.11) reduces to

$${\mathchar'26\mkern-11mud}Q = d \overline{E},$$ (4.43)

and ${\mathchar'26\mkern-11mud}Q$ becomes an exact differential. The amount of heat, $Q$ , absorbed in going from one macrostate to another depends only on the mean energy difference between them, and is independent of the process used to effect the change. In this situation, heat is a conserved quantity, and acts very much like the invisible indestructible fluid of Lavoisier's calorific theory.