Non-invertible function in the context of Legendre transformations

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!

• Related: physics.stackexchange.com/q/336818/2451 , physics.stackexchange.com/q/546495/2451 and links therein. Jun 11, 2021 at 17:21
• Thank you for linking these two! So the top answer in the second post talks about how if the mapping between the derivative with respect to velocity and the momentum is neither injective nor surjective we are dealing with so called constrained Hamiltonian theories and to solve such problems we can use "Dirac bracket". Would that be correct so far? Is this procedure and reasoning directly convertible to Thermodynamics aswell? Because all posts that I read so far were dealing with Lagrange/Hamiltonian mechanics. Jun 12, 2021 at 8:00

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 $$H(p)~:=~\sup_{v\in\mathbb{R}}(pv-L(v))~=~\left\{ \begin{array}{rcl} p(\ln p -1)&{\rm for}& p>0,\cr 0&{\rm for}& p=0, \cr \infty &{\rm for}& p<0.\end{array} \right.$$
• Thanks again for taking the time to answer this! I would have a follow up question reading this: Why is it a problem that there exists no solution for $\mu\geq 0$? Wouldn't a system that only has negative chemical potential also be totally valid? Isn't the real problem that $\frac{\partial U}{\partial N}=\mu$ is not invertible for $N(\mu)$? I guess also with the Legendre Fenchel transformation it's not possible to do a legendre transform in my example because also the toy model example has an invertible derivative. If that's the case, is there even a way to solve it? Jun 13, 2021 at 7:18