Equating differentials => equating coefficients

In summary, the equations A dx + B dy = ∂f/∂x dx + ∂f/∂y dy in thermodynamics can be used to solve for A and B, where A = ∂f/∂x and B = ∂f/∂y. This is possible by setting either dx or dy to 0 and solving for the respective variable.
  • #1
Derivator
149
0
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
 
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  • #2
This is not always true. See, exact differentials.
 
  • #3
But on this the derivation of the maxwellequations is based!?
 
  • #4
Derivator said:
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,

$$\begin{matrix} A \, dx = \frac{\partial f}{\partial x} \, dx \\ B \, dy = \frac{\partial f}{\partial y} \, dy\end{matrix}$$
 
  • #5
Derivator said:
But on this the derivation of the maxwellequations is based!?
Maxwell equations or maxwell relations?
 
  • #6
Jorriss said:
Maxwell equations or maxwell relations?

Sorry, i meant maxwell relations, of course.
 
  • #7
Derivator said:
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.
 
  • #8
Mandelbroth said:
$$\begin{matrix} A \, dx = \frac{\partial f}{\partial x} \, dx \\ B \, dy = \frac{\partial f}{\partial y} \, dy\end{matrix}$$

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
 
  • #9
Derivator said:
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}##.
 
  • #10
Derivator said:
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 = [itex]\partial f/\partial y[/itex] dy .
So B = [itex]\partial f/\partial y[/itex]
Similarly you'll get the value of A by setting dy = 0.
 

1. What is the purpose of equating differentials?

The purpose of equating differentials is to find the relationships between different variables in a given system. This allows scientists to understand how changes in one variable affect the others, and to make predictions and calculations based on these relationships.

2. How is equating differentials different from equating coefficients?

Equating differentials involves setting two derivatives or differentials equal to each other, while equating coefficients involves setting two numerical coefficients equal to each other. Equating differentials is used to analyze changes in a system, while equating coefficients is used to solve equations.

3. Can equating differentials be used in any scientific field?

Yes, equating differentials can be used in any scientific field that deals with systems and relationships between variables. This includes fields such as physics, chemistry, biology, and economics.

4. What are some applications of equating differentials?

Some applications of equating differentials include predicting the motion of objects in physics, analyzing chemical reactions in chemistry, and modeling population growth in biology. It can also be used in engineering to optimize designs and in economics to predict market trends.

5. Is there a specific method for equating differentials?

There are various methods for equating differentials, depending on the specific system and variables involved. Some common methods include using the chain rule, implicit differentiation, and separation of variables. It is important to choose the appropriate method based on the problem at hand.

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