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Most thermodynamics textbooks introduce a notation for partial derivatives that seems redundant to students who have already studied multivariable calculus. Moreover, the authors do not dwell on the explanation of the notation, which leads to a poor conceptual intuition of the subject.

For example, in maths, given a sufficiently well-behaved function $f: \mathbb{R}^3 \to \mathbb{R}$, we can define its partial derivatives unambiguously by: $$\frac{\partial f}{\partial x}(x,y,z) \qquad\text{ or simply}\qquad\frac{\partial f}{\partial x} \text{ (in shorthand notation)}$$ Writing $$\left(\frac{\partial f}{\partial x} \right)_{y,z}$$ would be verbose.

In thermodynamics, why do we have to specify the variables which are held constant by writing subscripts?

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    $\begingroup$ Schroeder's Introduction to Thermal Physics, at least, provides an exercise to show that holding different quantities constant does affect the answer. (It's something along the line of defining $z = xy$ and $w = yz$, taking various partial derivatives with different variables held constant, and showing that they don't agree.) I always make sure to assign it to my students when I teach that class, or to do it as an in-class activity. $\endgroup$ – Michael Seifert Mar 24 at 12:43
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    $\begingroup$ Thermodynamics does a thing where quantities aren't really defined as functions a priori: instead we have an implicit relationship that holds between all the thermodynamic variables together, and then we think about this implicitly defined surface using a wide variety of explicitly defined functions. $\endgroup$ – Ian Mar 25 at 14:23
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That's because in thermodynamics we sometimes use the same letter to represent different functions. For example, one can write the volume of a system as $V=f_1(P,T)$ (a function of the pressure and the temperature) or as $V=f_2(P,S)$ (a function of the pressure and the entropy). The functions $f_1$ and $f_2$ are distinct in the mathematical sense, since they take different inputs. However, they return the same value (the volume of the system). Thus, in thermodynamics it is convenient to symbolize $f_1$ and $f_2$ by the same letter (simply $V=V(P,T)$ or $V=V(P,S)$).

The subtlety here is that there can be more than one rule that associates pressure (and other variable) to volume. Therefore, the notation $$\frac{\partial V}{\partial P}$$ is ambiguous, since it could represent either $$\frac{\partial V}{\partial P}(P,T)=\frac{\partial f_1}{\partial P} \qquad\text{or}\qquad\frac{\partial V}{\partial P}(P,S)=\frac{\partial f_2}{\partial P}$$

(Here, I am supposing a single component system. Due to Gibbs' phase rule, we need $F=C-P+2$ independent variables to completely specify the state of a system.)

However, if we write $$\left(\frac{\partial V}{\partial P}\right)_{T}\qquad\text{or}\qquad\left(\frac{\partial V}{\partial P}\right)_{S}$$ there is no doubt about what we mean, hence the importance of the subscripts.

You can indeed check that for a single component system, $\left(\frac{\partial V}{\partial P}\right)_{T}\neq\left(\frac{\partial V}{\partial P}\right)_{S}$. $$\left(\frac{\partial V}{\partial P}\right)_{S}-\left(\frac{\partial V}{\partial P}\right)_{T}=\frac{TV^2 \alpha^2}{Nc_p}$$

If you want to read more about this, I suggest Representations of Partial Derivatives in Thermodynamics.

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    $\begingroup$ The final link is not working at the moment, but this appears to be an archive of the same document: web.archive.org/web/20210208015005/https://www.compadre.org/per/… $\endgroup$ – StayOnTarget Mar 24 at 11:48
  • $\begingroup$ And of course the beauty of it is that we can change between functions that have intrinsic variables and extensive variables. We are literally following Gallileo's bidding; make measurable that which is not so. $\endgroup$ – Stian Yttervik Mar 25 at 11:12
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    $\begingroup$ +1. There is another aspect to it as well: one could say "well, ok, but what's wrong with just writing $\frac{\partial V}{\partial P}(P,T)$ to make it clear which variables it's a function of?" The problem is then, what happens if you want to evaluate that function at the point, say, $P=\phi, T=T_0$? Then you'd either have to write something like $\frac{\partial V}{\partial P}(P,T)(\phi,T_0)$, in order to indicate both the parameters of the function and their values. That sort of thing ends up looking pretty awkward, and it's not so surprising that people prefer to use a different notation. $\endgroup$ – Nathaniel Mar 26 at 3:15
  • $\begingroup$ I think that to be consistent, one needs to add that in this approach, even expressions like $\left(\frac{\partial V}{\partial P}\right)_T$ can mean different functions depending on the context, otherwise expressions like $\left(\frac{\partial V}{\partial P}\right)_{S}-\left(\frac{\partial V}{\partial P}\right)_{T}$ have no physical sense: $\left(\left(\frac{\partial V}{\partial P}\right)_{S}-\left(\frac{\partial V}{\partial P}\right)_{T}\right)(x,y) = \left(\frac{\partial V}{\partial P}\right)_{S}(P=x,S=y)-\left(\frac{\partial V}{\partial P}\right)_{T}(P=x,T=y)$. $\endgroup$ – Litho Mar 26 at 16:25
  • $\begingroup$ @Litho Technically in that case the point is that for these derivatives to make sense in the traditional sense, you need all three of them included explicitly in the domain to make sense of the difference using functions. In other words you are dealing with $\left ( \left ( \frac{\partial V}{\partial P} \right )_S - \left ( \frac{\partial V}{\partial P} \right )_T \right ) (P,S,T)$. But this is misleading to write in this way, because $P,S,T$ are not bona fide independent variables; for example given $P,T$ and the equation of state I can at least in principle compute $S$. $\endgroup$ – Ian Mar 31 at 0:31
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Your presumption that partial derivatives can be uniquely specified by a single coordinate is false. Consider the function $f : \mathbb{R}^2 \to \mathbb{R}$, $$ f : (x, y) \mapsto x + y \qquad \left(\frac{\partial f}{\partial x}\right)_y = 1 \, . $$ But I can replace the second coordinate by $z = x+y$. Now we have $$ f : (x, z) \mapsto z \qquad \left(\frac{\partial f}{\partial x}\right)_z = 0 \, .$$

There is a simple geometric explanation for this. The derivative $\big( \frac{\partial}{\partial x} \big)_y$ measures the rate of change as $x$ varies along a surface of constant $y$. The derivative $\big( \frac{\partial}{\partial x} \big)$ measures the rate of change as $x$ varies along a surface of constant $z$. Drawing a picture makes the difference obvious. There is also a sophisticated geometric explanation, that you will find in differential geometry textbooks. Nakamura, Geometry, Topology and Physics, is written for and popular with a physicist audience, but any other book should do, too.

The somewhat informal name for this, due to Penrose, is the second fundamental confusion of calculus.

A question in the comments:

If you look at it in the technical sense, aren't the two f's you defined different functions? The way I see it the arguments of the functions are just placeholders

Well, no, because in thermodynamics -- and physics in general -- we deal with functions on some manifold (space time, phase space, ...) and what coordinates we use shouldn't matter. Slots of a function can't have physical meaning. Being pedantic I should write $$ (f \circ x_1)(x,y) = x + y \qquad (f \circ x_2)(x, z) = z $$ where $f : M \to \mathbb{R}$ is a scalar field on spacetime and $x_i ; \mathbb{R}^2 \to M$ are coordinate charts. In this way I distinguish between a point and its coordinates. $f \circ x_i$ are two different functions, but there is only one $f$.

This is the sophisticated geometric explanation I was alluding to before.

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    $\begingroup$ If you look at it in the technical sense, aren't the two $f$'s you defined different functions? The way I see it the arguments of the functions are just placeholders. So $f:(x,z)\mapsto z$ and $f: f(x,y)\mapsto y$ are the same function. To me it seems you are implicitly defining a new function by saying $\tilde f(x,z)=f(x,y(x,z))$ where $f$ is the first function and $\tilde f$ is the composition of $f$ and $y(x,z)=z-x$. $\endgroup$ – AccidentalTaylorExpansion Mar 25 at 13:18
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    $\begingroup$ @AccidentalTaylorExpansion : With $f$ given as $f(x,y) = \dots$, what's $\frac{\partial f(y,x)}{\partial x}$. This notation is unintelligible unless there is some way to mark which instances of "$x$" refer to slots in the formal parameter list in the definition of $f$ and which refer to independent variables in the specialization "$f(y,x)$". Further, the object under study is a manifold, for instance the pressure-temperature-volume-entropy manifold in 4-space. Fixing one coordinate implicitly restricts to a curve on that manifold, along which the partial derivative is unambiguous. $\endgroup$ – Eric Towers Mar 25 at 17:13
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    $\begingroup$ @AccidentalTaylorExpansion : The underlying problem is that there are two ways to interpret $\frac{\partial f(y,x)}{\partial x}$ -- the way you first described where the "$\partial x$" references a slot in the original definition of $f$ or the way you just described, where it references the variable being substituted into the second slot in the function. If one seriously studies PDEs, one switches to notation to completely remove this ambiguity: $f_1(x,x)$ is the derivative of $f$ with respect to its first slot, then specialize the result at $(x,x)$. Or, better, $f^{(1,0)}(x,x)$. $\endgroup$ – Eric Towers Mar 25 at 23:15
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    $\begingroup$ @AccidentalTaylorExpansion : The manifold viewpoint is necessary because none of the variables in Thermo are independent variables and none are dependent. Around any point, one can pick a couple of variables, declare them independent, then assert the existence of functions for the other, dependent variables on that chart of the atlas. But the underlying problem is that state equations are (1) not solvable as functions for whatever variable you want and (2) are unknowable in detail. They are a relation that picks out a manifold, and you still need to take directional derivatives on it. $\endgroup$ – Eric Towers Mar 25 at 23:18
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    $\begingroup$ @AccidentalTaylorExpansion Everything Eric said is correct, and the key takeaway is that as far as possible we should think in concepts that are coordinate independent. Partial derivatives are not, and the notation is spectacularly unhelpful and confusing. A function in thermodynamics (or say, on spacetime in general relativity) is not $\mathbb{R}^n \to \mathbb{R}$, say, but $M \to \mathbb{R}$ where $M$ is a manifold. It can be precomposed with a coordinate chart to get something $\mathbb{R}^n \to \mathbb{R}$, but this is, subtly, not the same thing. $\endgroup$ – Robin Ekman Mar 26 at 15:32
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While the OP's answer describes a perfectly valid interpretation, I would like to suggest another possible way of looking at it. In this approach, when one writes things like $\left(\frac{\partial V}{\partial P}\right)_T$ and $\left(\frac{\partial V}{\partial P}\right)_S$ in thermodynamics, the symbol $V$ stands for the same function in both cases; however, this function is defined not on $\mathbb{R}^2$, but on the manifold of the system's thermodinamic macrostates, let's denote it $\mathcal{M}$. $P, T$ and $S$ are also functions from $\mathcal{M}$ to $\mathbb{R}$. We can combine them into functions $(P,T)$ and $(P,S)$ from $\mathcal{M}$ to $\mathbb{R}^2$. These functions are injective, so we can define $(P,T)^{-1} : \text{Im}~(P,T)\to\mathcal{M}$ and $(P,S)^{-1} : \text{Im}~(P,S)\to\mathcal{M}$.

Then, when we write $\left(\frac{\partial }{\partial P}\right)_T$, we mean the operator which maps a function $f$ on $\mathcal{M}$ to $$ \left(\frac{\partial}{\partial x_1}\left(f\circ (P,T)^{-1}\right)\right)\circ (P,T), $$ which is also a function on $\mathcal{M}$ (assuming that $f\circ (P,T)^{-1}$ is differentiable on $\text{Im}~(P,T)$). Similarly, $\left(\frac{\partial f}{\partial P}\right)_S$ means $$ \left(\frac{\partial}{\partial x_1}\left(f\circ (P,S)^{-1}\right)\right)\circ (P,S). $$

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