Equating differentials => equating coefficients

[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



Derivator

 

[Jorriss]

This is not always true. See, exact differentials.

 

[Derivator]

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

 

[Mandelbroth]

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?

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,

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

 

[Jorriss]

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

Maxwell equations or maxwell relations?

 

[Derivator]

Maxwell equations or maxwell relations?

Sorry, i meant maxwell relations, of course.

 

[Jorriss]

Sorry, i meant maxwell relations, of course.

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

 

[Derivator]

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

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?


Derivator

 

[Mandelbroth]

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?


Derivator

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 $A \hat{i} + B \hat{j} = \frac{\partial f}{\partial x}\hat{i} + \frac{\partial f}{\partial y}\hat{j}$.

 

[physwizard]

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

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.