I am studying thermodynamics and in the first chapter the concept of exact and inexact differentials were used to talk about the differences between internal energy, work and heat.

From Blundell and Blundell: $\Delta f=\int_{x_i}^{x_f}\mathrm df=f(x_f)-f(x_i)$.

My issue is I can not imagine a function where this is not the case.

Can someone explain what I am not understanding?

  • 2
    $\begingroup$ The definition of an inexact different is that it's not the derivative of a function. $\endgroup$
    – DanielSank
    Oct 3, 2019 at 17:04
  • $\begingroup$ $\oint df =0 \implies$ exact differential - or a state function. Otherwise, it's an inexact differential. In general, you can convert an inexact differential into an exact differential by using an integrating factor. For instance, $\oint dQ \neq 0$ but $\oint dQ/T = 0$ where $T$ is the integration factor. Typically if you know the differential equation you can calculate the integration factor. $\endgroup$ Oct 12, 2019 at 0:55
  • 1
    $\begingroup$ it is bad notation to denote inexact differentials as $df$ since hitting $d$ in a function $f$ is usually denoted by $df$, and inexact differentials cannot be obtained by hitting $d$ in functions. Since there is a physical quantity, say $U[C]\equiv \int _{C} \omega$, associated with the differential $\omega$ I prefer to use the notation $\omega=\delta U$ instead of $\omega=dU$. $\endgroup$
    – Nogueira
    Jan 12, 2020 at 17:54

3 Answers 3


There is no such function. It just doesn't exist.

We use the term "exact differential" to refer to the differentials of functions, i.e. if $g = g(x,y)$ is a function of $x$ and $y$, the differential $$ \mathrm d g = \frac{\partial g}{\partial x}\mathrm dx + \frac{\partial g}{\partial y}\mathrm dy \tag 1 $$ is by definition an exact differential. We define inexact differentials as those differentials $\mathrm df$ which are not the differentials of an underlying function, so $f$ is not, and cannot be, a function. It can't "look like" anything.

That said, though, what I've done basically amounts to kicking the can down the road, and it's then perfectly legitimate for you to turn around and ask "but what is an inexact differential, if it's not the differential of a function?" ─ and there, to be honest, I'll bounce that to you: what do you understand a differential to be?

This is an honest question, as there are multiple ways to understand the concept, and they form a continuum in terms of mathematical rigour as well as a continuum in their level of physical intuition. (Unfortunately, those two gradients point, essentially, in completely opposite directions.) Thus, you can just use your physical intuition of "a small change" (and then work hard to make sure that you keep your mathematics in line), but you can also use rigorous mathematics (specifically, the language of differential forms) and then work to make sure that the physical intuition corresponds to the concepts you're using.

Either way, the notion of a general differential boils down to objects that can be expressed in the form $$ \mathrm d h = \alpha(x,y)\mathrm dx + \beta(x,y)\mathrm dy, \tag 2 $$ where the basis differentials $\mathrm dx$ and $\mathrm dy$ obey some set of algebraic rules (and one may or may not follow up on what exactly those two objects are). We also institute a way (the $d$ operator) to go from functions to forms, which I've written above in $(1)$.

This then establishes a set of differentials which are differentials (techincally, 'exterior derivatives') of a function, and those are what we call 'exact' differentials. The inexact differentials are all the rest, which cannot be written in that form, i.e. those for which $\alpha$ and $\beta$ are not partial derivatives as in $(1)$ of a common function. (The cleanest test for this, at least in simply-connected 2D space, is to simply ask whether $$ \frac{\partial\alpha}{\partial y}= \frac{\partial\beta}{\partial x}, $$ because if that's not the case, then no underlying function can exist.)

For those inexact differentials, there simply isn't an quantity $f$ that you can look at. And that's OK.

And finally, to follow up on one of your specific questions:

From Blundell and Blundell: $\Delta f=\int_{x_i}^{x_f}\mathrm df=f(x_f)-f(x_i)$

This may or may not be the case, depending on the dimensionality you're working in.

  • If you're working in 1D, i.e. if $x$ is a single real variable, then you're right. There are no inexact differentials if the underlying space is one-dimensional.
  • If you're working in more than one dimension, i.e. if $x$ is really $\vec x = (x_1,\ldots,x_n)$, and $n$-dimensional tuple of real numbers, then the notation $\int_{x_i}^{x_f}\mathrm df$ is ill-defined, and cannot be used to go from $\mathrm df$ into $\Delta f$, because in the integral you have not specified which path should be taken for the integration.

    In an exact differential, the integration path does not affect the results, so you can go from $\mathrm df$ to $f(x) = \int^{x}_{x_\mathrm{ref}}\mathrm df+C$. However, this path independence is not guaranteed and it can fail; for those differentials there is no way to get $f$, and we call those 'inexact'.

  • $\begingroup$ This is a thoroughly good answer that highlights many important aspects of this concept. $\endgroup$ Sep 1, 2023 at 15:12

An inexact differential is a 1-form $\omega =\sum_{i=1}^n f_i \mathrm{d}x^i$ that is not exact, i.e. the component tuple $(f_1,\ldots, f_n)$ is not a gradient $\left(\frac{\partial f}{\partial x^1},\ldots, \frac{\partial f}{\partial x^n}\right)$ of some function $f$.


The definition of $df$ being an inexact differential is that $f$ is not a function of any sort. It's a small quantity of something. The classic example of this is the first law of thermodynamics $$\mathrm{d} U = dQ - dW.$$ In this law, the left hand side is an exact differential, it's the difference in internal energy of a system. Neither of the quantities on the right hand side are exact differentials - they're just small quantities.

You can imagine that there are some other systems supplying the work in the system that might have $dQ =\mathrm{d}U_{\mathrm{heater}}$ and $dW = -\mathrm{d}U_{\mathrm{worker}}$, but that is not necessarily the case. In thermodynamics we often deal with things called "heat baths" What makes a heat bath a heat bath is that it can supply or absorb an infinite amount of heat without changing temperature. In that case, treating a heat bath as having a finite amount of internal energy $U_{\mathrm{hb}}$ is mathematically questionable/tricky.


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