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.


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

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

| cite | improve this answer | |
  • 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 at 14:11
  • $\begingroup$ Thank you! Apparently I got confused when deriving this relation. It's clear now. $\endgroup$ – orthxx Jul 22 at 16:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.