# change of variables and partial derivatives in thermodynamics

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

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

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