Sum of exact differentials is inexact?

The definition of work is $$\delta W=-\sum\limits_i f_i \, \text{d}X_i,$$ where $f_i$ are the generalized forces ($p$, $\mu$, etc.), and $X_i$ their corresponding extensive quantities ($V$, $N$, etc.).

Is the sum of exact differentials not necessarily exact? Why is $\delta W$ an inexact differential here?

Perhaps this or this is related.

• Inexact differential just means that there is not a function $W(X_i)$, even though changes in $W$ are computable from changes in the $X_i$. It's basically just due to path dependence, varying $X_1$ before $X_2$ is not always the same as varying $X_2$ before $X_1$. The most basic example is the first law of thermo. Only in special cases is there an actual function $U(V, T, N)$. But of course $dU = p dV + S dT + \mu dN$ – Bobak Hashemi Feb 2 '17 at 21:28
• Ok thank you, I get it: the sum of exact differentials can be an inexact differential. – Carucel Feb 2 '17 at 21:42
• @Carucel I believe this is due to the dependence of $f_i$ on $X_j$. For example, even though $dX_i$ is exact, $f_i dX_i$ does not need to be exact. So the sum of exact differentials is exact, but the product of an exact differential with an arbitrary scalar function is not necessarily exact. – user12029 Feb 2 '17 at 21:46
• @NeuroFuzzy, are you sure? $f_i\text{d}X_i$ can be inexact? I thought it was the path dependence that made the sum inexact. – Carucel Feb 2 '17 at 21:48
• @Carucel It is path dependence that makes it inexact. In two dimensions, $dx=d(x)$ and $dy=d(y)$ are exact, and so is $dx+dy=d(x+y)$, but $-y dx+x dy$ is inexact. ie, there is no $f(x,y)$ with $df=-y dx+x dy$. – user12029 Feb 2 '17 at 22:03

Even if $dX_i$ is exact, $f_i~dX_i$ might not be if $f_i$ is not constant with respect to the $X_i$.

This is one of those cases where I think we need a sort of "food pyramid" for physics, with examples at the bottom, equations somewhere in the middle, and physical theories on top, or so. Here's a great example of an inexact differential made of exact ones:$$\delta s = -y ~dx + x~dy.$$Remember what this means?

If not, that's OK, we'll get there. We're talking about a state space $\mathcal S$, in this case $(x, y)$ pairs are here, and there exist some "state functions" which take points in $\mathcal S$ (i.e. states) as input and give us back outputs, let's say points in $\mathbb R$ (i.e. real numbers). And because we're doing physics we often like those functions to be "nice", and one of the ways they can be nice is if, say, the response to a very small deviation in the state coordinates has a linear approximation: $$s(x + dx, y + dy) = s(x, y) + c_1 ~dx + c_2 ~ dy.$$If you chart $s$ as a $z$-component of a 3D space then $s(x,y)$ is some sort of 2D curved surface, and the above tries to approximate it with a tangent plane at some point $(x, y).$ These "partial derivatives" $c_1 = \left(\frac{\partial s}{\partial x}\right)_y$ and $c_2 = \left(\frac{\partial s}{\partial y}\right)_x$ therefore vary as functions of $\mathcal S$ as well, because you can make a tangent plane around any point and then measure the slopes of the lines which make it up.

In this case we would write $s(x + dx, y+dy) - s(x, y)$ as the exact differential $ds$.

Conversely, when I say that the above is an inexact differential I am saying that there exists no such nice state function $s(x,y)$ which leads to this choice of $c_{1,2}(x, y).$ I might also saying that there might be no such nice state function and I am not sure, because you can always pretend an exact differential is inexact (just don't derive the $s(x, y)$ function!), but you can't pretend that an inexact one is exact (because there is no $s(x, y)$ function and pretending that one exists will lead you into trouble).

How am I so sure that the above is inexact? Well call it $\delta s = c_1~dx + c_2 ~ dy$ and let's just integrate those components: for the first one we get $\int dx~c_1(x,y) = -xy + C_1(y)$ and for the second one we get $\int dy~c_2(x,y) = +xy + C_2(x)$ and it doesn't matter how you choose $C_1(y), C_2(x)$, you cannot come out of this rut with a consistent $s(x, y),$ because that $xy$ term has different signs and refuses to go gently.

Now sometimes, you can get away with simply augmenting $\mathcal S,$ a great example of this is $$\delta s_2 = \frac{-y}{x^2 + y^2}~dx + \frac{x}{x^2 + y^2}~dy.$$In this case we have indeed that $\frac{\partial c_1}{\partial y} = \frac{\partial c_2}{\partial x}$ where those are defined, but this is not defined at $(0, 0).$ It is the almost-state function $d\theta$ which has a discontinuity at $2\pi$ when you try to represent it as a function $\mathcal S \to \mathbb R.$ Instead the idea is to choose a curve, any curve, from $0$ out to $\infty$ that does not self-intersect. And in addition to your place on $\mathcal S$ we will store an integer (in $\mathbb Z$) saying how many times you passed around the origin, we will increment it when you pass over the curve one way, or decrement it when you pass over the curve in the other way. Then $d\theta$ becomes an exact differential on this space $\mathbb Z \times \big(\mathcal S - \{(0,0)\}\big).$

But very often we can't and we're stuck, especially when we don't have this $\nabla \times [c_1, c_2] = 0$ type of equation to help us.

• I can't quite decipher what your last two paragraphs are doing. How does $\mathrm{d}\theta$ become an exact differential on your "space"? $\mathbb{Z}\times(S - \{0\}$ is a rather strange object, if you want it to be a manifold with differential forms on it you get the countable disjoint union of $S-\{0\}$ with itself, on which $\mathrm{d}\theta$ is no more exact than it was before - the $2\pi$ still obstruct it, since every exact form must give 0 when integrating along closed paths. – ACuriousMind Feb 2 '17 at 22:16
• @ACuriousMind fortunately at the level that you're at I can just tell you "those last two paragraphs are constructing a standard Riemann surface out of a countable supply of branch sheets that share a branch cut." The rule that the topology on $\mathbb Z \times (\mathcal S - \{\vec 0\})$ connects points across the branch cut only when their $z$-index increments/decrements by one is precisely the rule which makes $d\theta$ exact by allowing it to increase to, say, $3\pi$ as your path travels onto a new branch sheet. The function is no longer discontinuous with respect to the new topology. – CR Drost Feb 2 '17 at 22:28
• I think I understand what you're doing (it's an infinite helicoid if $S$ is the puntured disc), but I don't think that object is actually $\mathbb{Z}\times (S-0)$. You probably should take a quotient along the branch cuts. – ACuriousMind Feb 2 '17 at 22:52
• My experience with quotient sets is not very good and I think what you're getting hung up on is that the topology I'm using is not the product topology on $A\times B.$ I don't think you're actually confused on the isomorphism: the fact that $\mathcal S-0$ is isomorphic (via polar coordinates) to $[0,2\pi)\times(0,\infty)$, and I'm proposing to promote that $[0,2\pi)$ to $(-\infty,\infty)$ to make $d\theta$ an exact differential, and there's an isomorphism betwen $(-\infty, \infty)$ and $\mathbb Z \times [0,2\pi),$ and that $\times$ is associative so I can reduce back to $Z \times (S - 0).$ – CR Drost Feb 2 '17 at 23:24

The sum of exact differentials is exact: Given $\mathrm{d}f$ and $\mathrm{d}g$ as exact differentials, their sum is the exact differential $\mathrm{d}(f+g)$ - that is, the sum of differentials of functions is the differential of the sum of the functions. It's a basic consequence of the exterior derivative being linear.

However, you are not taking the "sum of exact differentials" there. A differential $f(x_1,\dots,x_n)\mathrm{d}x_1$ is not exact - if you assume there existed a function $F(x_1,\dots,x_n)$ such that $\mathrm{d}F = f\mathrm{d}x_1$ you quickly find a contradiction unless $F$ (and thereby $f$) does not depend on the $x_2,\dots,x_n$ at all, which the $f_i$ in your differential do. So $\delta W$ is the sum of inexact differentials an can therefore be inexact (note that this doesn't prove $\delta W$ is inexact, since the sum of inexact differentials can easily be exact, consider the trivial case $\omega + (-\omega)$ for some inexact $\omega$).