Thermodynamics potential problem How to show that $U(T, V, N)$ is explicitly not a fundamental relation because there is an infinite number of functions $A(T, V, N)$ that can lead to same function $U(T, V, N)$. Where $U$ is the internal energy and $A$ is the Helmholtz free energy.
 A: It is an axiom of thermodynamics that the internal energy function expressed in terms of $S,V,N$, viz. $U=f(S,V,N)$, is a fundamental relation. To say that $f$ is the fundamental relation means that, if for a thermodynamic system, $S,V,N$ are given, then its absolute $U$ is determined. Therefore you may plot $U$ versus $S,V,N$.
Given this, and the definition of temperature, $T\equiv 1 \Bigr /\frac{\partial f}{\partial S}\Bigr |_{V,N}$, it is easy to see why $U=g(T,V,N)$ cannot be a fundamental relation. To say that $g$ is not a fundamental relation means that if $T,V,N$ are given then $U$ is determined only up to a constant. In this case, it becomes meaningful to speak of only difference in internal energy $U$.
Now, $U=g(T,V,N)=g(1 \Bigr /\frac{\partial f}{\partial S}\Bigr |_{V,N},V,N)$. Suppose $T,V,N$ for a particular state is given. This state must have a definite value of $U$ and $S$. This may be plotted as a point in the graph of $U$ vs $S$ (for a given $V,N$), but how would you locate it given only the slope $\frac{\partial f}{\partial S}=1/T$, instead of the value of $S$ itself? For a given $V,N$, there are many curves in $U$-$S$ plane, all shifted parallel to each other (along $S$-axis), that have the same slope (for a particular $U$). To pick any particular curve then is to fix your reference internal energy $U_0$, and therefore only changes in internal energy become meaningful in this case. That is why $g$ is not a fundamental relation.
Reference: Thermodynamics by Callen.
