# 2.3: Mathematical Interlude - Exact and Inexact Differentials

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$$\newcommand\ncdot{\!\cdot\!}$$
$$\newcommand\NS{N\ns_{\textsf S}}$$

The differential $dF=\sum_{i=1}^k A\ns_i\,dx\ns_i \label{dFeqn}$ is called exact if there is a function $$F(x\ns_1,\ldots,x\ns_k)$$ whose differential gives the right hand side of Equation \ref{dFeqn}. In this case, we have $A\ns_i={\pz F\over\pz x\ns_i} \qquad\Longleftrightarrow\qquad {\pz A_i\over\pz x\ns_j} = {\pz A_j\over\pz x\ns_i} \quad \forall\ i,j\ .$ For exact differentials, the integral between fixed endpoints is path-independent: $\int\limits_{\RA}^{\RB}\!\!dF = F(x^{\ssr{B}}_1,\ldots,x^{\ssr{B}}_k)-F(x^{\ssr{A}}_1,\ldots,x^{\ssr{A}}_k)\ ,$ from which it follows that the integral of $$dF$$ around any closed path must vanish: $\oint\!dF=0\ .$

When the cross derivatives are not identical, when $$\pz A\ns_i/\pz x\ns_j\ne \pz A\ns_j/\pz x\ns_i$$, the differential is inexact. In this case, the integral of $$dF$$ is path dependent, and does not depend solely on the endpoints.

As an example, consider the differential $dF=K\ns_1\,y\,dx + K\ns_2\,x\,dy\ . \label{dFe}$ Let’s evaluate the integral of $$dF$$, which is the work done, along each of the two paths in Fig. [work_path]: \begin{aligned} W^\ssr{(I)}&=K\ns_1\!\int\limits_{x\ns_\RA}^{x\nd_\RB}\!\!dx\>y\subA + K\ns_2\!\int\limits_{y\ns_\RA}^{y\nd_\RB}\!\!dy\>x\subB= K\ns_1\,y\subA \,(x\subB-x\subA) + K\ns_2\,x\subB\,(y\subB-y\subA )\\ W^\ssr{(II)}&=K\ns_1\!\int\limits_{x\ns_\RA}^{x\nd_\RB}\!\!dx\>y\subB + K\ns_2\!\int\limits_{y\ns_\RA}^{y\nd_\RB}\!\!dy\>x\subA = K\ns_1\,y\subB\,(x\subB-x\subA ) + K\ns_2\,x\subA \,(y\subB-y\subA )\ .\end{aligned} Note that in general $$W^\ssr{(I)}\ne W^\ssr{(II)}$$. Thus, if we start at point A, the kinetic energy at point B will depend on the path taken, since the work done is path-dependent.

The difference between the work done along the two paths is $W^\ssr{(I)}-W^\ssr{(II)}=\oint\!dF=(K\ns_2-K\ns_1)\,(x\subB-x\subA)\,(y\subB-y\subA)\ . \label{Wdiff}$ Thus, we see that if $$K\ns_1=K\ns_2$$, the work is the same for the two paths. In fact, if $$K\ns_1=K\ns_2$$, the work would be path-independent, and would depend only on the endpoints. This is true for any path, and not just piecewise linear paths of the type depicted in Fig. [work_path]. Thus, if $$K\ns_1=K\ns_2$$, we are justified in using the notation $$dF$$ for the differential in Equation [dFe]; explicitly, we then have $$F=K\ns_1\,xy$$. However, if $$K\ns_1\ne K\ns_2$$, the differential is inexact, and we will henceforth write $$\dbar F$$ in such cases.

This page titled 2.3: Mathematical Interlude - Exact and Inexact Differentials is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Daniel Arovas.