Non-invertible function in the context of Legendre transformations

Question

I'm in my 4th semester of physics and currently visit a course about Thermodynamics.

We currently deal with Legendre Transformations and my textbook gave the following example:

Given the function $$U(S,V,N)=A\frac{S^2}{N}e^{\frac{S}{k_B N}}\tag{1}$$ try to find $U(S,V,\mu)$.

If you do this the standard way, calculating $$(\frac{\partial U}{\partial N})=\mu\tag{2}$$ and then try to find a expression for $N(\mu)$, you arrive at a equation which is not invertible in terms of $N$. So the just described doesn't really work out.

It further says that the missing invertibility of this special relation is a standard case in Thermodynamics (only in the case of an ideal gas you can find an analytical solution) and I will encounter it again at least once in a course about Statistical Mechanics.

I do not fully understand what is meant by that sentence so I wanted to ask here if anyone knows what the book is talking about and give me some further details about it? I would appreciate any comments, thanks!

Answer 1

1. The problem with the standard way (2) for OP's system (1) is that the derivative $U^{\prime}\equiv\frac{\partial U}{\partial N}<0$ is negative, so that eq. (2) has no solution for non-negative chemical potential $\mu\geq 0$.

   However, everything is not lost since one may show that the function (1) is convex $U^{\prime\prime}\equiv\frac{\partial^2 U}{\partial N^2}>0.$ Recall that there are at least 2 definitions of Legendre transform, cf. e.g. this related Phys.SE post.

   Mathematically, the issue can in principle be solved by using the Legendre-Fenchel transform $$ \sup_{N\in\mathbb{R}_+}(\mu N-U(N))~=~\left\{ \begin{array}{rcl} \mu (U^{\prime})^{-1}(\mu) -U((U^{\prime})^{-1}(\mu)) &{\rm for}& \mu<0,\cr 0&{\rm for}& \mu=0, \cr \infty &{\rm for}& \mu>0,\end{array} \right.$$ instead if one allows the Legendre transform to take the value $\infty$.

2. Explicit example: Let the Lagrangian be the convex function $L(v)=e^v>0$. Then the derivative $\frac{\partial L}{\partial v}=e^v>0$ is positive so that the equation $p=\frac{\partial L}{\partial v}$ has no solution for non-positive momentum $p\leq 0$. Nevertheless, one may show that the Legendre-Fenchel transform exists $$ H(p)~:=~\sup_{v\in\mathbb{R}}(pv-L(v))~=~\left\{ \begin{array}{rcl} p(\ln p -1)&{\rm for}& p>0,\cr 0&{\rm for}& p=0, \cr \infty &{\rm for}& p<0.\end{array} \right.$$