# Any difference between thermodynamic double-derivative and derivative “at constant” value?

Reading about the Maxwell relations has left me confused, and I want a basic sanity check regarding the notation. The Wikipedia article breezes over the following switch of notation without really describing it:

$$\left(\frac{\partial T}{\partial V}\right)_S = \frac{\partial^2 U }{\partial S \partial V}$$

Probably, it is also true to say:

$$\left(\frac{\partial T}{\partial V}\right)_S = \frac{\partial^2 T }{\partial S \partial V}$$

There are many examples, this is just the first case-in-point. I'm familiar with the subscripts being presented as "at constant x". In the above case, that would be saying, "dT/dV at constant entropy".

Firstly, is that a correct interpretation?

Secondly (and if it is), how are the above two things the same? For another example, how would the following two be the same things:

• differentiating pressure with respect to temperature at constant volume vs.
• differentiating temperature with respect to volume and then pressure?

While I ask this, I can already see that my mental picture is incomplete. It's nonsense to ask to differentiate temperature with respect with volume with no other specifiers because that doesn't respect the total degrees of freedom. So could you introduce another auxiliary variable, and then and then explicitly show the above equality? I'm sure this is often taken as trivial, but I find it non-trivial for myself.

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Your second expression probably has a typo, it's incorrect (by simple dimensional analysis) –  ScroogeMcDuck Jul 11 '14 at 16:07
@ScroogeMcDuck Actually, that looks to be tied to my error in reasoning in the first place. I could try to salvage that, but it doesn't look good. –  Alan Rominger Jul 11 '14 at 16:09

They are not the same $-$ those are the Maxwell relations, which are a consequence of $S$, $U$ and other thermodynamical potentials being functions of state. The thermodynamical notation of parital derivatives can be defined as follows:

$$\left( \frac{\partial U}{\partial S} \right)_V := \frac{\partial U(S,V)}{\partial S}$$

This means it is only a first order partial derivative. The subscript "at constant $x$" only reminds us, what are the independent thermodynamical parameters we consider, so we don't have to write $U(S,V)$ all the time (the other parameter is the one with respect to which we are differentiating). And because of the existence of the equation of state we know, that there are only 2 independent parameters.

As for the Maxwell relations, they tell us about second order derivatives (mixed derivatives) of $S$, $U$ etc. and how changing the order of differentiation does not matter ($\frac{\partial^2 U}{\partial S \partial V} = \frac{\partial^2 U}{\partial V \partial S}$). However we also know, that e.g.:

$$\left( \frac{\partial U}{\partial S} \right)_V = T$$

Putting these two equation together we get, what you wrote as your first equation above. All the Maxwell relations are obtained in the same fashion.

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The fundamental equation of thermodynamics says that $$\mathrm{d}U = T\mathrm{d}S-P\mathrm{d}V$$ It is a mathematical identity that $$\mathrm{d}U = \left(\frac{\partial U}{\partial S}\right)_V\mathrm{d}S+\left(\frac{\partial U}{\partial V}\right)_S\mathrm{d}V$$ Comparing the two we find that $$T = \left(\frac{\partial U}{\partial S}\right)_V \qquad P= -\left(\frac{\partial U}{\partial V}\right)_S$$ your first equation comes from simply differentiating these identities (the "things being held constant" can disappear in the second derivative. Clearly the derivative with respect to volume at a constant volume has to be 0) and Maxwell's Relations then follow from the fact that partial derivatives commute.

The intuition for your second relation comes from assuming that $U$ is a proportional to $T$ or that it is a function of $T$ only. This is only true for an ideal gas and even then I think there is a constant factor missing. The equation is certainly not true in general.Similarly with regard to your second question, in general \begin{align}\left(\frac{\partial P}{\partial T}\right)_V &\ne \frac{\partial^2 T}{\partial P\partial V}\\ &\ne \frac{\partial^2 P}{\partial V\partial T}\end{align} (I'm guessing the second one is what you were going for)

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