# Independence of thermodynamic variables

A given thermodynamical system has a number of state variables, not all of which are independent. Suppose that we have a system which can be specified by $k+1$ extensive variables: $U,X_1,\cdots,X_k$. We also have $k$ intensive variables which are conjugate to the $k$ extensive variables $X_i$. To what extent are these $2k+1$ variables independent?

For example, we can always express the internal energy in terms of its natural variables: $$U=U(S,V,N_1,\cdots, N_{k-2})$$ But what about in terms of other variables? How about exchanging an extensive variable for its conjugate internal variable, say $T$ in place of $S$: $$U=U(T,V,N_1,\cdots,N_{k-2})?$$ Or perhaps replacing $T$ by some unrelated intensive variable, say $p$ in place of $S$: $$U=U(p,V,N_1,\cdots,N_{k-2})?$$ Will any combination of the remaining thermodynamic variables serve to functionally determine $U$? If not, then what exactly relates these $2k+1$ variables?

• No answers yet? The variables must not be conjugate, and you need to switch from internal energy to the thermodynamic potential associated with your choice of variables... – Christoph Dec 23 '14 at 16:35
• @Christoph Thank you for your response. My current understanding is this: Let $\{X_i\}_{i=1}^k$ denote the extensive parameters of the system and let $\{Y_j\}$ denote the conjugate intensive parameters. Let $I\cup J = {1,2,\cdots,k}$ denote a partition of the indices. Then $\{X_i\}_{i\in I} \cup \{Y_j\}_{j\in J}$ will serve as a valid coordinate system for the thermodynamic states so long as we include at least one extensive parameter. – EuYu Dec 23 '14 at 18:01
• cont.. This is because for any partition of extensive/intensive variables that we choose, there is some thermodynamic potential for which the variables serve as natural variables and hence is enough to determine the entire thermodynamic state. So I guess this is a sufficient requirement. Am I correct in assuming that this is also sufficient? I.e. the inclusion of a conjugate pair of variables means that the state will not be uniquely determined. So for a simple system with variables $\{U,S,V,T,p\}$, we cannot use $\{S,T,p\}$ or $\{V,p,T\}$ as coordinates for example. – EuYu Dec 23 '14 at 18:01
• sounds good to me (I forgot to mention that at least one variable has to be extensive); also note that you can define conjugate pairs with respect to state functions other than energy - eg the entropy representation – Christoph Dec 23 '14 at 19:03
• @Christoph Thank you Christoph. If you convert your comments to an answer, I'd be happy to award you the bounty. – EuYu Dec 23 '14 at 19:17

Perhaps you are looking for the notion of Legendre transform. Here is the general idea: suppose you have a function of some sets of variables, say $\mathbf x$ and $\mathbf v$, and so $f = f(\mathbf x,\mathbf v)$. You usually get a new set of variables, conjugate to $\mathbf v$ by taking the Legendre transform of $f$ w.r.t. $\mathbf v$. So what you do is to define said conjugate variables, say $\mathbf p = \nabla_{\mathbf v}f$, which depend on $\mathbf x$ and $\mathbf v$ through $$\mathbf p(\mathbf x,\mathbf v)=\nabla_{\mathbf v}f(\mathbf x,\mathbf v).$$ Now one has to require $f$ to have suitable concavity property, and in this case we could require $f$ to be such that the Hessian $D^2_{\mathbf v}f$ to be non-zero on the relevant domain. If this is the case we can then invert the relation to get $$\mathbf v(\mathbf x,\mathbf p) = g(\mathbf x,\mathbf p),$$ for some function $g$.
Suppose that $\mathbf x\in\mathbb R^n$ and that $\mathbf v\in\mathbb R^m$; then among the $n+2m$ variables $\mathbf x,\mathbf v,\mathbf p$ there are $m+1$ constraints: $m$ coming from $g$ and 1 coming from $f$. Therefore you only have $n-1$ independent variables. The ones that you can then take as independent among the set $\mathbf x,\mathbf v,\mathbf p$ usually depends on how well-behaved the functions you get are. In general you get a (possibly smooth) submanifold of $\mathbb R^{n+2m}$, and the choice correspond to the fact that your atlas should have well-behaved charts (think of the sphere in $\mathbb R^3$, where you can't give a global chart, but you have to patch at least two charts together).