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I want to show the following thermodynamic identity (Pathria, 3rd Edition, Appendix H, Pg 677): $$\left(\frac{\partial x}{\partial y}\right)_w = \left(\frac{\partial x}{\partial y}\right)_z + \left(\frac{\partial x}{\partial z}\right)_y \left(\frac{\partial z}{\partial y}\right)_w$$

I start with

$$ dx = \left(\frac{\partial x}{\partial y}\right)_z dy + \left(\frac{\partial x}{\partial z}\right)_y dz $$ $$ dx = \left(\frac{\partial x}{\partial y}\right)_w dy + \left(\frac{\partial x}{\partial w}\right)_y dw $$

Yielding $$\left[\left(\frac{\partial x}{\partial y}\right)_w - \left(\frac{\partial x}{\partial y}\right)_z \right] dy = \left(\frac{\partial x}{\partial z}\right)_y dz - \left(\frac{\partial x}{\partial w}\right)_y dw $$ Now consider $$dz= \left(\frac{\partial z}{\partial y}\right)_w dy + \left(\frac{\partial z}{\partial w}\right)_y dw$$ and substitute it in: $$\left[\left(\frac{\partial x}{\partial y}\right)_w - \left(\frac{\partial x}{\partial y}\right)_z \right] dy = \left(\frac{\partial x}{\partial z}\right)_y \left[\left(\frac{\partial z}{\partial y}\right)_w dy + \left(\frac{\partial z}{\partial w}\right)_y dw \right] - \left(\frac{\partial x}{\partial w}\right)_y dw $$ Giving: $$\left[\left(\frac{\partial x}{\partial y}\right)_w - \left(\frac{\partial x}{\partial y}\right)_z - \left(\frac{\partial x}{\partial z}\right)_y \left(\frac{\partial z}{\partial y}\right)_w \right] dy = \left[\left(\frac{\partial z}{\partial w}\right)_y - \left(\frac{\partial x}{\partial w}\right)_y\right] dw $$

This is strange though. If the first identity is true, this implies that $$ \left(\frac{\partial z}{\partial w}\right)_y = \left(\frac{\partial x}{\partial w}\right)_y$$ Something which can easily be shown to not be true, for $x=yz^2$ where $w=yz,$ so $x=w^2 /y,$ and $z=w/y$.

What has gone wrong?

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  • $\begingroup$ Try to have a look at this notes. You should find something useful basics.altervista.org/test/Physics/TD/partial_derivatives.html $\endgroup$
    – basics
    Commented Oct 26, 2022 at 22:05
  • $\begingroup$ LOL. I'm answering your question. Give me a while. Are you 100% sure? I'm using chrome on android and everything works fine $\endgroup$
    – basics
    Commented Oct 26, 2022 at 22:18
  • $\begingroup$ I'm on android, and the link works with chrome and firefox. try to explicitly add https:// in front of the link $\endgroup$
    – basics
    Commented Oct 26, 2022 at 22:45
  • $\begingroup$ @basics It seems like a Safari specific problem $\endgroup$
    – Jbag1212
    Commented Oct 26, 2022 at 22:47

1 Answer 1

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I left you a reference about partial derivatives and some useful notation often used in thermodynamics in the comment.

Anyway, for this identity it's enough to

  • write $x(y, z)$ and $z(y, w)$,
  • define the composite function $\tilde{x}(y, w) = x(y, z(y, w))$
  • evaluate the partial derivative of $\tilde{x}(y, w)$

$\left(\dfrac{\partial \tilde{x}}{\partial y}\right)_w (y,w) = \left(\dfrac{\partial x}{\partial y}\right)_w (y, z(y, w)) =\left(\dfrac{\partial x}{\partial y}\right)_z (y, z(y, w)) +\left(\dfrac{\partial x}{\partial z}\right)_y (y, z(y, w))\left(\dfrac{\partial z}{\partial y}\right)_w (y, z(y, w)) $

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  • $\begingroup$ I understand this method, but it still leaves the curious question in the OP: "what went wrong?" $\endgroup$
    – Jbag1212
    Commented Oct 26, 2022 at 22:47
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    $\begingroup$ you missed the term $(\partial x /\partial z)_w$ in front of the first term in the square bracket on the rhs. With this missing term, the content of the square brackets reads $(\partial x /\partial w)_y - (\partial x /\partial w)_y \equiv 0$ $\endgroup$
    – basics
    Commented Oct 26, 2022 at 22:58
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    $\begingroup$ Ah yes, I've been starting at a screen all day evidently, thank you! $\endgroup$
    – Jbag1212
    Commented Oct 26, 2022 at 22:59
  • $\begingroup$ do we solve the mystery? $\endgroup$
    – basics
    Commented Oct 26, 2022 at 23:00
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    $\begingroup$ Yes we definitely do. $\endgroup$
    – Jbag1212
    Commented Oct 26, 2022 at 23:02

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