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Next: Statistical thermodynamics Up: Heat and work Previous: Quasi-static processes

Exact and inexact differentials

In our investigation of heat and work we have come across various infinitesimal objects such as $d \bar{E}$ and $\,{\mathchar'26\mskip-12mud}W$. 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 neighbouring point ($x+dx$, $y+dy$). This is given by

dF = F(x + dx, y+dy)- F(x,y),
\end{displaymath} (135)

which can also be written
dF = X(x,y) \,dx + Y(x,y)\, dy,
\end{displaymath} (136)

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).
\end{displaymath} (137)

Note that since 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.
\end{displaymath} (138)

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

{\mathchar'26\mskip-12mud}G \equiv X'(x,y)\, dx + Y'(x,y)\, dz,
\end{displaymath} (139)

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 (136). It is clear that since $X= \partial F/\partial x$ 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}.
\end{displaymath} (140)

Thus, if
\frac{\partial X'}{\partial y} \neq \frac{\partial Y'}{\partial x}
\end{displaymath} (141)

(as is assumed to be the case), then $\,{\mathchar'26\mskip-12mud}G$ cannot be an exact differential, and is instead termed an inexact differential. The special symbol $\,{\mathchar'26\mskip-12mud}$ is used to denote an inexact differential. Consider the integral of $\,{\mathchar'26\mskip-12mud}G$ over some path in the $x$-$y$ plane. In general, it is not true that
\int_i^f {\mathchar'26\mskip-12mud}G = \int_i^f (X'\, dx + Y' \,dy)
\end{displaymath} (142)

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\mskip-12mud}G \neq 0.
\end{displaymath} (143)

Consider, for the moment, the solution of

{\mathchar'26\mskip-12mud}G = 0,
\end{displaymath} (144)

which reduces to the ordinary differential equation
\frac{dy}{dx} = -\frac{X'}{Y'}.
\end{displaymath} (145)

Since the right-hand side is a known function of $x$ and $y$, the above 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 in Eq. (145). This defines a set of curves which can be written $\sigma (x, y) = c$, where $c$ is a labeling parameter. It follows that
\frac{d\sigma}{d x}\equiv \frac{\partial \sigma}{\partial x}
+\frac{\partial\sigma}{\partial y}\frac{dy}{dx} = 0.
\end{displaymath} (146)

The elimination of $dy/dx$ between Eqs. (145) and (146) yields
Y'\, \frac{\partial \sigma}{\partial x} = X'\, \frac{\partial\sigma}{\partial y}
=\frac{X' \,Y'}{\tau},
\end{displaymath} (147)

where $\tau(x, y)$ is function of $x$ and $y$. The above equation could equally well be written
X' = \tau\, \frac{\partial \sigma}{\partial x},~~~~~Y' = \tau\,
\frac{\partial \sigma}
{\partial y}.
\end{displaymath} (148)

Inserting Eq. (148) into Eq. (139) gives
{\mathchar'26\mskip-12mud}G = \tau \left( \frac{\partial \si...
... \frac{\partial\sigma}
{\partial y}\,dy\right)= \tau\,d\sigma,
\end{displaymath} (149)

\frac{{\mathchar'26\mskip-12mud}G}{\tau} = d\sigma.
\end{displaymath} (150)

Thus, dividing the inexact differential $\,{\mathchar'26\mskip-12mud}G$ by $\tau$ yields the exact differential $d\sigma$. A factor $\tau$ which possesses this property is termed an integrating factor. Since the above 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 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 $\bar{E}$. 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 $d \bar{E}$ are infinitesimal differences between well-defined quantities: i.e., they are exact differentials. For example, $d\bar{E} = \bar{E}_f - \bar{E}_i$ 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 $f$ the mean energy change is given by

{\mit\Delta} \bar{E} = \bar{E}_f - \bar{E}_i = \int_{i}^f d\bar{E}.
\end{displaymath} (151)

However, since the mean energy is just a function of the macrostate under consideration, $\bar{E}_f$ and $\bar{E}_i$ depend only on the initial and final states, respectively. Thus, the integral $\int d\bar{E}$ 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, $\,{\mathchar'26\mskip-12mud}W = \sum \bar{X}_\alpha\,dx_\alpha$ is not the difference between two numbers referring to the properties of two neighbouring 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 $\,{\mathchar'26\mskip-12mud}W$ 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\mskip-12mud}W,
\end{displaymath} (152)

where the integral represents the sum of the infinitesimal amounts of work $\,{\mathchar'26\mskip-12mud}W$ 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 ${\mit\Delta} \bar{E}$ does not depend on the process used whereas the work $W$, in general, does. Thus, it follows from the first law of thermodynamics, Eq. (123), that the heat $Q$, in general, also depends on the process used. It follows that

{\mathchar'26\mskip-12mud}Q \equiv d \bar{E} + {\mathchar'26\mskip-12mud}W
\end{displaymath} (153)

is an inexact differential. However, by analogy with the mathematical example discussed previously, there must exist some integrating factor, $T$, say, which converts the inexact differential $\,{\mathchar'26\mskip-12mud}Q$ into an exact differential. So,
\frac{{\mathchar'26\mskip-12mud}Q}{T} \equiv dS.
\end{displaymath} (154)

It will be interesting to find out what physical quantities correspond to the functions $T$ and $S$.

Suppose that the system is thermally insulated, so that $Q=0$. In this case, the first law of thermodynamics implies that

W_{if} = - {\mit\Delta} \bar{E}.
\end{displaymath} (155)

Thus, in this special case, the work done depends only on the energy difference between in the initial and final states, and is independent of the process. In fact, when Clausius first formulated the first law in 1850 this is how he expressed it:
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 $\,{\mathchar'26\mskip-12mud}W=0$, Eq. (124) reduces to

{\mathchar'26\mskip-12mud}Q = d \bar{E},
\end{displaymath} (156)

and $\,{\mathchar'26\mskip-12mud}Q$ 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.
next up previous
Next: Statistical thermodynamics Up: Heat and work Previous: Quasi-static processes
Richard Fitzpatrick 2006-02-02