# Maxwell relations, confused on how the solution got to this answer

I have a problem where, basically, in part (a) I correctly found a fundamental equation $$TdS = dE - 2\sigma l dx$$

Then the problem goes on to say that the only parameters of interest are $x$ and $T$, so I would assume we can write the relation $$dS = \frac{\partial S}{\partial x} dx + \frac{\partial S}{\partial T}dT$$ by taking the total derivative. From this information, the solution goes on to say "We can read off the Maxwell relation $$\left(\frac{\partial S}{\partial x}\right)_T = \left(\frac{\partial(-2\sigma l)}{\partial T}\right)_x = -2l \frac{d \sigma}{dT} \qquad \sigma = \sigma_0 - \alpha T$$

Where $\sigma$ is given in the problem statement. I'm just not seeing where they find this maxwell relation....been stuck for quite some time. Any help would be appreciated, thanks.

• The maxwell relation you wrote implies a potential, we'll call it $M$, that has differential $dM = S dT - 2\sigma l dx$. – lionelbrits Dec 1 '13 at 20:06

I'll elaborate on my comment. Maxwell's relations are just observations that second partial derivatives of a function commute. So for some potential $\Phi(x,y)$, if you differentiate, $d\Phi = (\partial_x \Phi) dx + (\partial_y \Phi) dy$, then because second partial derivatives commute, i.e., $\partial_y \partial_x \Phi = \partial_x \partial_y \Phi$, we obtain an identity on the coefficients of $dx$ and $dy$ in $d\Phi$.
The Maxwell relation that the solution gives you is reminiscent of a potential $A$ that has differential
$dA = -S dT + 2 \sigma l dx$,
because if $\partial_T A = -S$ and $\partial_x A = 2\sigma l$, then $\partial_T \partial_x A = \partial_x A \partial_T A$ reproduces your identity.
Now, I am going about the problem from the wrong end, because I am given the solution and asked to find how to derive it. However, my suspicion is that you are supposed to recognize from the start that $dx$ looks like a change in volume term ($dV$) and $-2 \sigma l$ a measure of pressure ($P$). Once you have done this, you have many Maxwell relations at your disposal. The mathematics don't care about the exact form of $P dV$, in other words.