I came across with the following solution from Sandler's book: $$\begin{eqnarray}\left( \frac{\partial G}{\partial T} \right)_U &=& \frac{\partial (G,U)}{\partial (T,U)} \\ &=& \frac{\partial (G,U)}{\partial (T,V)}\cdot \frac{\partial (T,V)}{\partial (T,U)} \\ &=&\frac{(\partial G/\partial T)_V (\partial U/\partial V)_T-(\partial G/\partial V)_T (\partial U/\partial T)_V}{(\partial U/\partial V)_T} \end{eqnarray}$$

Please help me understand these steps. What happens in line 1? I know that line 2 is by using Chain Rule by then how did line 2 becomes line 3?

  • $\begingroup$ anyone care to help? please..thanks $\endgroup$ – james25 Aug 17 '15 at 2:25
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    $\begingroup$ Are we supposed to guess how these variables are related? At least tell us what they are. I would guess that $G$ is the Gibbs free energy, $U$ the internal energy, $T$ the temperature, and $V$ the volume. Even if it is a standard notation, you should at least clarify it. Don't make people guess. $\endgroup$ – Batominovski Aug 17 '15 at 3:20
  • $\begingroup$ Oh Im very sorry Sir. But you guessed it right. $\endgroup$ – james25 Aug 17 '15 at 3:48

I'll start with a "quick and dirty", physicist-style approach. This would start with

$$ \begin{align} dG & =-SdT+VdP \\ dU& =TdS-PdV=0 \\ dT& =\frac{\partial^2 U}{\partial S^2} dS+\frac{\partial^2 U}{\partial S \partial V} dV.\end{align}$$

Heuristically, you want to divide the first equation by the third equation, after enforcing the constraint of the second equation. The key here is to try to use the constraint to make each of the first and third equations only involve one differential (and the same differential in each case). In the second case this is easy enough: for instance we can make it entirely $dS$ by doing this:

$$dV=\frac{T}{P} dS \Rightarrow dT=\left ( \frac{\partial^2 U}{\partial S^2} + \frac{\partial^2 U}{\partial S \partial V} \frac{T}{P} \right ) dS.$$

Doing the same thing with the first equation is not so simple. One thing we can do is think of $T$ and $P$ in the natural variables of $U$, i.e. $S$ and $V$. Then we have

$$dT=\frac{\partial^2 U}{\partial S^2} dS + \frac{\partial^2 U}{\partial S \partial V} dV \\ d(-P)=\frac{\partial^2 U}{\partial V \partial S} dS + \frac{\partial^2 U}{\partial V^2} dV.$$

(Please check my minus signs!)

Now you can rearrange these as before, obtaining something of the form $dT=Q(S,V)dS$ and $d(-P)=R(S,V)dS$, for slightly complicated functions $Q$ and $R$. Then

$$\left ( \frac{\partial G}{\partial T} \right )_U = \frac{-S Q(S,V) - V R(S,V)}{\frac{\partial^2 U}{\partial S^2} + \frac{\partial^2 U}{\partial S \partial V} \frac{T}{P}}.$$

Rigorously speaking, we consider the surface which is described by the equation $U(S,V)=U_0$. Since we have one scalar constraint and $U$ naturally depends on two variables, this surface is actually $2-1=1$ dimensional, and so we can parametrize this surface by one of the two natural variables of $U$. I chose to parametrize it by $S$, but $V$ is just as good.

Next we use the fact that

$$\left ( \frac{\partial G}{\partial T} \right )_U = \frac{\left ( \frac{\partial G}{\partial S} \right )_U}{\left ( \frac{\partial T}{\partial S} \right )_U}.$$

To do this, we identify an implicit function $V(S)$, which arises through the equation $U(S,V)=U_0$, using the implicit function theorem. Then we introduce the functions $S \mapsto G(T(S,V(S)),P(S,V(S)))$ and $S \mapsto T(S,V(S))$. (Note that we do all of this using only one "unnatural" function, namely $V(S)$; $G$, $T$, and $P$ are all in their natural variables here.) Then we differentiate each of these with respect to $S$ and divide as above.

  • $\begingroup$ Thank you Sir for such a detailed solution. I tried solving it using Jacobians and I think I've solved it. $\endgroup$ – james25 Aug 17 '15 at 3:46

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