In physics courses, one is taught to think of $dU$ as "a small change in the Potential Energy". I wonder what a precise mathematical definition of these objects might be. For example, in the first law of thermodynamics $$ dQ=dU+dW $$ what is this $d$? Is it the exterior derivative? I don't know how to make sense of it.

Also, setting $dQ=TdS=dU+dW=dU+pdV-\mu dN$, how can one justify that $\frac{\partial S}{\partial U}|_{V,N}=\frac{1}{T}$ ?

  • $\begingroup$ You are looking for the terms total differential/exact differential $\endgroup$ – Triatticus Aug 12 '17 at 20:47
  • $\begingroup$ @Triatticus All I find about those is based on the assumption that $df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$ (for example) is well-defined. But, again, I do not know what this $dx$ is ("infinitesimal" is usually not a good answer, I think). Maybe it is the differential of the projection $\pi_x$? I don't know. $\endgroup$ – Soap Aug 12 '17 at 20:56
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    $\begingroup$ Generally speaking, $Q$, $U$ and $W$ are all scalar valued functions with a domain some vector space $\mathcal{V}$ of parameters; $\partial$ in this context is an operator looking at the behavior of the output of these functions with respect to a 'differential change' in their arguments. This can be formalized by looking at the limiting behavior of sequences $\{v_i\}_{i=0}^\infty$ in $\mathcal{V}$ under the image of our functions for total differentials, or sequences in individual coordinates of the vector space with other coordinates fixed for partial differentials. $\endgroup$ – Alec Rhea Aug 12 '17 at 21:09
  • $\begingroup$ @AlecRhea Could you please elaborate on that or give a reference? $\endgroup$ – Soap Aug 12 '17 at 21:11
  • $\begingroup$ It's just my understanding of a thermodynamics course I took a few semesters back, but I will think some on it and post a more detailed answer. $\endgroup$ – Alec Rhea Aug 12 '17 at 21:13

For a different example, we can look at


This is a compact notation saying that $E$ depends on $S$ and $V$, $\frac{\partial E}{\partial S}=T,\frac{\partial E}{\partial V}=-p$. One does not need to really make sense of the differential symbols themselves, instead you can think of this as just a shorthand. Indeed I would suggest this approach, because without this you will tend to make mistakes such as expecting $\frac{\frac{\partial E}{\partial S}}{\frac{\partial E}{\partial V}}=\frac{\partial V}{\partial S}$ when in fact there is a minus sign involved.

The dQ example is similar except that Q is not a state function, so it does not make sense to say "Q depends on U and W". Still, given a path between two configurations you can determine the heat change by integrating dQ along that path.

Part of the reason that this seemingly peculiar notation is used in thermodynamics is that really every state function can be thought of as depending on some other set of state functions. For example, one can describe the equilibrium behavior of some gas through $E(S,V)$ or $S(E,V)$; both are equally valid, it just depends on your preferences. This differential notation puts the two presentations on more similar footing than the usual mathematical notation does.

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    $\begingroup$ Yes, a good analogy to keep in mind is $d\theta$ in the plane, which it makes perfectly good sense to integrate along a curve, but there is—notation notwithstanding—no everywhere defined differentiable angle function $\theta$ on $\Bbb R^2-\{0\}$. $\endgroup$ – Ted Shifrin Aug 12 '17 at 23:31
  • $\begingroup$ @TedShifrin Good point -- would it not make sense to perhaps define $\theta(x,y)=\arctan(\frac{y}{x})$, then view $\mathbb{R}^2$ as isomorphic to $\mathbb{C}$ and project $\theta$ onto the Riemann sphere to make the discontinuities meet at 'infinity' on top of the sphere? This is not my area of expertise by any stretch, I was just curious. $\endgroup$ – Alec Rhea Aug 13 '17 at 19:59
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    $\begingroup$ @AlecRhea Your function is undefined on a line. The best you can do is a "branch cut" along a ray emanating from the origin. $\endgroup$ – Ted Shifrin Aug 13 '17 at 20:25
  • $\begingroup$ @TedShifrin Thanks for the clarification -- this fascinates me, particularly when integrating $d\theta$ around a closed curve containing the origin! $\endgroup$ – Alec Rhea Aug 13 '17 at 20:35
  • $\begingroup$ That's why I brought this up as a simple analogy with the inexact differentials in thermodynamics, @AlecRhea. :) $\endgroup$ – Ted Shifrin Aug 13 '17 at 21:44

We start by defining partial differentiation versus total differentiation for a scalar valued function $f:\mathbb{R}^n\rightarrow\mathbb{R}$ in a vector field $\mathcal{V}:\mathbb{R}^n\rightarrow\mathbb{R}^n$. For each $v\in\mathbb{R}^n$, we will write $\mathcal{V}(v)$ for the vector assigned to $v$ under $\mathcal{V}$ and $$f(v)=f(v_0,\dots,v_{n-1}),$$ where the indexing is a relic of the fact that I am mostly a set theorist by training.

Let $\mathcal{F}(\mathbb{R}^n)$ be the set of all functions from $\mathbb{R}^n$ to $\mathbb{R}$. We define an operator $\partial_i:\mathcal{F}(\mathbb{R}^n)\rightarrow\mathcal{F}(\mathbb{R}^n)$ such that for each function $f:\mathbb{R}^n\rightarrow\mathbb{R}$, $\partial_i(f):\mathbb{R}^n\rightarrow\mathbb{R}$ is the unique function obtained by defining $$\partial_i(f(v))=\partial_if(v)=\lim_{h\rightarrow0}\frac{f(v_0,\dots,v_{i}+h,\dots,v_{n-1})-f(v_0,\dots,v_i,\dots,v_{n-1})}{h}$$ for all $v\in\mathbb{R}^n$. We will refer to $\partial_i$ as the $i^{th}$ partial derivative operator on $\mathbb{R}^n$. Viewing a function as a set of ordered pairs, we have $$\partial_i=\{\big(f,\partial_i(f)\big):f\in\mathcal{F}(\mathbb{R}^n)\}.$$

This is a pretty rough definition of partial differentiation, but it gets the point across and is fully correct for strictly Euclidean space. Computing the $i^{th}$ partial derivative of some scalar valued function amounts to taking a limit of the 'change quotient' of the function as we change the $i^{th}$ input coordinate by a smaller and smaller amount at each vector in the domain of our function.

For total differentiation, I beieve we can use the following:

Let $\mathcal{F}(\mathbb{R}^n)$ be as above. We define an operator $\mathcal{d}_i:\mathcal{F}(\mathbb{R}^n)\rightarrow\mathcal{F}(\mathbb{R}^n)$ such that for each function $f:\mathbb{R}^n\rightarrow\mathbb{R}$, $\mathcal{d}_i(f):\mathbb{R}^n\rightarrow\mathbb{R}$ is the unique function obtained by defining $$\mathcal{d}_i(f(v))=\mathcal{d}_if(v)=\partial_if(v)+\sum_{j\neq i}\partial_jf(v)\frac{dv_j}{dx_i}$$ for all $v\in\mathbb{R}^n$. We will refer to $\mathcal{d}_i$ as the $i^{th}$ total derivative operator on $\mathbb{R}^n$. Viewing a function as a set of ordered pairs, we have $$\mathcal{d}_i=\{\big(f,\mathcal{d}_i(f)\big):f\in\mathcal{F}(\mathbb{R}^n)\}.$$

I have cheated a little bit by using the Liebniz notation $\frac{dv_j}{dx_i}$ to 'bake in' all of the one-dimensional calculus implicitly taking place here, but essentially the idea is that if $v=\mathcal{V}(r)$ for $r=(x_0,\dots,x_{n-1})\in\mathbb{R}^n$ (if it is a vector under the image of the vector field) then each $v_i$ is a function in up to $n$ variables $x_0,\dots,x_{n-1}$. If the coordinates of $v$ depend on more than one coordinate of $r$, the total derivative takes this mutual dependence into account using the above process.

I will have to think some more on how exactly to get a nice view of these operations in a physical parameter space, but these should be the definitions you're looking for.


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