5

In the context of Maxwell's relations, there are two frequently used expressions relating the partial derivatives of different thermodynamic quantities:

$$\left(\frac{\partial x}{\partial y}\right)_{z} \left(\frac{\partial y}{\partial z}\right)_{x} \left(\frac{\partial z}{\partial x}\right)_{y}= -1 \tag{1}$$

$$\left(\frac{\partial y}{\partial x}\right)_{z}=1 \bigg{/} \left(\frac{\partial x}{\partial y}\right)_{z} \tag{2}$$

What is the mathematical background of these two expressions? Is there a simple demonstration for them?

Invenietis
  • 851
  • 6
  • 17

3 Answers3

6

These are known as the reciprocal and reciprocity theorems. I feel these are perhaps not stated as clearly as they should be in introductory courses. I always had trouble with them until I read Andrew Steane's fantastic book on Thermodynamics (pretty much the only book on the subject that I like), which I will be following in this answer.

The idea is to start with the well known theorem of partial derivatives. If $f$ is a function of two variables $x$ and $y$, then we can relate a change in $f$ to changes in $x$ and $y$ using the partial derivatives: $$\text{d}f = \left(\frac{\partial f}{\partial x}\right)_{y} \text{d}x + \left(\frac{\partial f}{\partial y}\right)_{x} \text{d}y.\label{1}\tag{1}$$

In the above formula, the notation $(\cdot)_x$ means that the partial derivative is taken while keeping the subscript (here, $x$) constant. Note that in general (I encourage you to try to come up with an example for this)$$\left(\frac{\partial f}{\partial x}\right)_{y} \neq \left(\frac{\partial f}{\partial x}\right)_{z}.$$

We can now use the above relation carefully, while realising that we can't quite be as cavalier with partial derivatives as we could with total derivatives. For example, we can't talk about quantities like $\text{d}f/\text{d}x$, since $f$ is a function of two variables, $x$ and $y$. You might naively call the quantity in question the "rate of change of $f$ with $x$", but since it depends on $y$ as well, we need to specify along which path we choose to describe this variation completely.

The Reciprocal Theorem:

let's start by dividing Equation (\ref{1}) by $\text{d}x$, while keeping $y$ constant. The second term in the equation would trivially be zero, since $\text{d}y=0$, and we'd get the trivial result: $$\left(\frac{\partial f}{\partial x}\right)_{y} = \left(\frac{\partial f}{\partial x}\right)_{y}.$$

A more interesting result appears if we carefully divide Equation (\ref{1}) by $\text{d}f$, but we still keep $y$ constant. In this case, it reduces to: $$1 = \left(\frac{\partial f}{\partial x}\right)_{y} \left(\frac{\partial x}{\partial f}\right)_{y},$$ from which you get the relation

$$\boxed{\left(\frac{\partial f}{\partial x}\right)_{y} = \left(\frac{\partial x}{\partial f}\right)_{y}^{-1}}.$$

The Reciprocity Theorem:

Let's go back to Equation (\ref{1}) again, and this time we divide throughout by $\text{d}x$ again, except we don't keep $y$ constant, but rather we do it while keeping $f$ constant. It should be easy to show that $$0 = \left(\frac{\partial f}{\partial x}\right)_{y} + \left(\frac{\partial f}{\partial y}\right)_{x} \left(\frac{\partial y}{\partial x}\right)_{f}.$$

Using the reciprocal theorem above, we can easily see that this means that $$\boxed{\left(\frac{\partial f}{\partial y}\right)_{x} \left(\frac{\partial y}{\partial x}\right)_{f} \left(\frac{\partial x}{\partial f}\right)_{y} = -1.}$$

Philip
  • 522
2

The second one, which mathematicians usually call the "inverse function theorem", should actually be rather geometrically obvious: if you have the same set of constraints, then $\frac{\Delta y}{\Delta x}$ will be just the reciprocal of $\frac{\Delta x}{\Delta y}$ even for finite changes. It is only nontrivial when you write it down in terms of functions instead of relations between quantities. Indeed, thinking in terms of functions, the equation seems rather strange, since the two sides don't appear to depend on the same thing (the left side seems to depend on $x$ and not on $y$ while the right side seems to depend on $y$ and not on $x$). The fix at the level of functions is that you have to evaluate $\frac{\partial x}{\partial y}$ at $y^{-1}(x)$, or vice versa.

The first one is more mysterious, especially when written in this nice symmetrical "triple product rule" form. It is a lot more intuitive to me when you distinguish one quantity from the other two by rewriting it in the form

$$\left ( \frac{\partial y}{\partial x} \right )_f = -\frac{\left ( \frac{\partial f}{\partial x} \right )_y}{\left ( \frac{\partial f}{\partial y} \right )_x}.$$

This obviously has the right units, so the only remaining mysterious element is why the coefficient is just $-1$ all the time. The intuition for why the coefficient is negative is that if $\left ( \frac{\partial f}{\partial x} \right )_y$ and $\left ( \frac{\partial f}{\partial y} \right )_x$ have the same sign, then $f$ will change if you change both $x$ and $y$ in the same direction. As for why it is actually $-1$, the intuitive explanation is to write out the total differential

$$df=\left ( \frac{\partial f}{\partial x} \right )_y dx + \left ( \frac{\partial f}{\partial y} \right )_x dy = 0$$

since $f$ is held constant, and then formally rearrange to get $\frac{dy}{dx}$, which gets interpreted as $\left ( \frac{\partial y}{\partial x} \right )_f$ since you held $f$ constant in the derivation. In math notation, we formalize this by considering an arbitrary path $(x(t),y(t))$ along which $f$ is constant, and then

$$\frac{df}{dt} = \left ( \frac{\partial f}{\partial x} \right )_y \frac{dx}{dt} + \left ( \frac{\partial f}{\partial y} \right )_x \frac{dy}{dt} = 0$$

from the chain rule. Then you apply the property $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ to conclude.

Ian
  • 104,572
1

They follow from the mathematical definition of the derivative, which is:

$\left(\frac{\partial x}{\partial y} \right)_z=\lim_{h \to 0}\frac{x(y+h,z)-x(y,z)}{h}$,

$\left(\frac{\partial y}{\partial z} \right)_x=\lim_{h \to 0}\frac{y(z+h,x)-y(z,x)}{h}$,

$\left(\frac{\partial z}{\partial x} \right)_y=\lim_{h \to 0}\frac{z(x+h,y)-z(x,y)}{h}$,

$dx=\left(\frac{\partial x}{\partial y} \right)_z dy+\left(\frac{\partial x}{\partial z} \right)_y dz$.

Some Student
  • 177