3
$\begingroup$

In thermodynamics one works with state functions, e.g., the energy function $U(Y_1,\dotsc,Y_n)$, where $Y_i>0$ are the so called extensive variables. This function is $1$st order homogeneous. Sometimes one uses the specific energy function $$u(y_1,\dots,y_{n-1})=U(y_1,\dots,y_{n-1},1),$$ where $y_i=\frac{Y_i}{Y_n}$. The relation between $u$ and $U$ is $$Y_n\cdot u(y_1,\dotsc,y_n)=U(Y_1,\dots,Y_n).$$ I want to understand how to express second order partial derivatives $\frac{\partial^2u}{\partial y_i\partial y_j}$ in terms of $\frac{\partial^2U}{\partial Y_i\partial Y_j}$ and $Y_i$.

Let's start with the first order derivatives. Using the chain rule we can write $$\frac{\partial u}{\partial y_i}=\sum_{k=1}^n \frac{\partial U}{\partial Y_k}\frac{\partial Y_k}{\partial y_i}.$$ But now I get stuck. What is $\frac{\partial Y_k}{\partial y_i}$? Can we simply write it as $\frac{\partial Y_k}{\partial y_i}=1/\frac{\partial y_i}{\partial Y_k}$? Apparently not as otherwise, e.g., $\frac{\partial y_1}{\partial Y_2}=0$ and $\frac{\partial Y_2}{\partial y_1}=\infty$. How can we deal with that?

Technically, it must hold that $\frac{\partial U}{\partial Y_i}=\frac{\partial u}{\partial y_i}$ because the first order partial derivatives of a $1$st order homogeneous function are $0$th order homogeneous. But I do not see how to show that formally using mathematical arguments.

$\endgroup$

1 Answer 1

2
$\begingroup$

There are a couple of typos. First, the relation between $u$ and $U$ is

$$Y_n \cdot u(y_1, ..., y_{n-1}) = U(Y_1, ..., Y_n).$$

Next, you lost a factor of $Y_n$ when you wrote the chain rule.

Perhaps the best way to show that

$$\frac{\partial U}{\partial Y_i} = \frac{\partial u}{\partial y_i}$$

(for $i \neq n$, because $u$ only depends on $y_1, ..., y_{n-1}$), is to use the chain rule the other way around, i.e.

$$\frac{\partial U}{\partial Y_i} = Y_n \cdot \sum_{k=1}^{n-1} \frac{\partial u}{\partial y_k} \frac{\partial y_k}{\partial Y_i}$$

for $i \neq n$. Since the $y$'s are defined as $y_k = \frac{Y_k}{Y_n}$ ($k = 1, ..., n-1$), we have

$$\frac{\partial y_k}{\partial Y_i} = \frac{\delta_{ik}}{Y_n}$$

where $\delta_{ik} = 1$ if $i = k$ and $0$ otherwise, you can conclude that

$$\frac{\partial U}{\partial Y_i} = \frac{\partial u}{\partial y_i}.$$

$\endgroup$
2
  • 2
    $\begingroup$ (+1) Now continue, letting $z_i:={\partial u\over\partial y_i}$: $${\partial^2 U\over\partial Y_j\partial Y_i} ={\partial z_i\over\partial Y_j} =\sum_{k=1}^{n-1}{\partial z_i\over\partial y_k}{\partial y_k\over\partial Y_j} ={\partial z_i\over\partial y_j}{1\over Y_n}={\partial^2 u\over\partial y_j\partial y_i}{1\over Y_n}\quad(i,j\neq n)$$ $\endgroup$
    – r.e.s.
    Jul 22, 2020 at 14:11
  • $\begingroup$ Thank you! Apparently I got confused when deriving this relation. It's clear now. $\endgroup$
    – orthxx
    Jul 22, 2020 at 16:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .