1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equating differentials => equating coefficients

  1. May 25, 2013 #1
    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

  2. jcsd
  3. May 25, 2013 #2
    This is not always true. See, exact differentials.
  4. May 25, 2013 #3
    But on this the derivation of the maxwellequations is based!?
  5. May 25, 2013 #4
    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,

    $$\begin{matrix} A \, dx = \frac{\partial f}{\partial x} \, dx \\ B \, dy = \frac{\partial f}{\partial y} \, dy\end{matrix}$$
  6. May 25, 2013 #5
    Maxwell equations or maxwell relations?
  7. May 25, 2013 #6
    Sorry, i meant maxwell relations, of course.
  8. May 25, 2013 #7
    The differentials used for Maxwells relations are exact - dU, dS, dP, etc. Work and heat, for example, are inexact though.
  9. May 25, 2013 #8
    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?

  10. May 25, 2013 #9
    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}##.
  11. May 25, 2013 #10
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook