# Equating differentials => equating coefficients

1. May 25, 2013

### Derivator

Hi all,

In thermodynamics one often has equations like

A dx + B dy = ∂f/∂x dx + ∂f/∂y dy

From which follows

A = ∂f/∂x
B = ∂f/∂y

Can anyone explain to me why this conclusion is necessary from a mathematical point of view, please?

Here is my try:
A dx + B dy = ∂f/∂x dx + ∂f/∂y dy
=>
(A-∂f/∂x )dx + (B- ∂f/∂y)dy = 0
since the terms in front of the differentials are independent of each other:
(A-∂f/∂x ) = const_1
(B- ∂f/∂y) = const_2
However, I cannot justify why const_1 = const_2 = 0

Best,
Derivator

2. May 25, 2013

### Jorriss

This is not always true. See, exact differentials.

3. May 25, 2013

### Derivator

But on this the derivation of the maxwellequations is based!?

4. May 25, 2013

### Mandelbroth

Let's suppose that dx and dy are linearly independent of each other. This may not always be true, especially if y is a function of x. However, we'll work under the assumption that dx and dy are linearly independent.

Then, from this point of view, there is no scalar you can multiply dx by to get dy. Thus, we can split the equation into two parts, one for each independent form. That is,

$$\begin{matrix} A \, dx = \frac{\partial f}{\partial x} \, dx \\ B \, dy = \frac{\partial f}{\partial y} \, dy\end{matrix}$$

5. May 25, 2013

### Jorriss

Maxwell equations or maxwell relations?

6. May 25, 2013

### Derivator

Sorry, i meant maxwell relations, of course.

7. May 25, 2013

### Jorriss

The differentials used for Maxwells relations are exact - dU, dS, dP, etc. Work and heat, for example, are inexact though.

8. May 25, 2013

### Derivator

As I showed in my first post, I only can see that
$$\begin{matrix} A \, dx = \frac{\partial f}{\partial x} \, dx + const_1 \\ B \, dy = \frac{\partial f}{\partial y} \, dy + const_2 \end{matrix}$$

Why do the both constant have to vanish?

Best,
Derivator

9. May 25, 2013

### Mandelbroth

Think of dy and dx as orthogonal vectors. They are actually covectors, but that's more math and less physics.

For example, $\begin{bmatrix} A \\ B \end{bmatrix} = \begin{bmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{bmatrix}$ is the same as saying $A \hat{i} + B \hat{j} = \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j}$.

10. May 25, 2013

### physwizard

Set dx = 0
you'll get Bdy = $\partial f/\partial y$ dy .
So B = $\partial f/\partial y$
Similarly you'll get the value of A by setting dy = 0.