# Fundamental thermodynamic relation: derivation of “non-natural” variables

Maybe this could be case where a question, other than an answer, could be wrong, but whatever...

Let's start with the fundamental thermodynamic relation in differential form for a hydrostatic system, $dE=TdS-PdV+\mu dN$.

Via the Euler's homogeneous functions theorem we could write too $E=TS-PV+\mu N$.

So what could stop us from writing, i.e., $V$ as $V=-\frac{\partial E}{\partial P}$? Why we have to switch to enthalpy?

• have you forgot that the other variables $T, S, \mu, N, etc.$ may also depend on $p$? – hyportnex Jul 17 '17 at 21:02
• Also, how much time in the calculations of TD processes, especially as you get further into it, does the enthalpy concept save us? Schroeder in "Thermal Physics" covers this pretty well, imo – user163104 Jul 17 '17 at 21:08
• Please @hyportnex could you elaborate more, or offer a reference? From the textbooks I'm reading I know just that $S=S(E,V,N)$ or $T=T(E,V,N)$... – Lo Scrondo Jul 17 '17 at 21:17
• So @JohnKennedy there's just an experimental motivation? – Lo Scrondo Jul 17 '17 at 21:17
• First law of thermodynamics asserts the relationship: $dE=TdS=PdV+\mu dN$. It's an axiom. Beginning from this point no amount of mathematical manipulation can get you the result $V=-\partial E/\partial p$. – Deep Jul 18 '17 at 5:24

It doesn't work because $T,p$ and $\mu$ are not extensive variables.

The natural variables for the internal energy $U$, $(S,N,V)$, are extensive, i.e. they are additive for subsystems: if I put together two systems of entropy $S_1$ and $S_2$, the total entropy is $S_1+S_2$. The same is valid for $N$ and $V$, but it is not valid for $T,P$ and $\mu$.

So it is true that $U$ is homogeneous of degree $1$ in $S,N$ and $V$:

$$U(\lambda S, \lambda N, \lambda V) = \lambda U(S,N,V)$$

and therefore we can use Euler's theorem to write*

$$U = \frac{\partial U}{\partial S}S+ \frac{\partial U}{\partial V}V + \frac{\partial U}{\partial N}N = TS-pV+\mu N$$

but $U$ is not homogeneous in $T,p$ and $\mu$. Therefore we cannot write

$$U = \frac{\partial U}{\partial T}T+ \frac{\partial U}{\partial p}p + \frac{\partial U}{\partial \mu}\mu$$

$^*$ More details here.

• I think my mistake stemmed from a geometric perspective, i.e. trying to think of the thermodynamic space as a 7-dimensional Euclidean one. – Lo Scrondo Jul 21 '17 at 16:15