Partial derivative notation in thermodynamics

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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 $$ \left(\frac{\partial}{\partial x_1}\left(f\circ (P,S)^{-1}\right)\right)\circ (P,S). $$