Relation between Legendre and Laplace transforms in statistical mechanics

Question

I am working off of these lecture notes by Stephen Teitel for his spring 2025 Statistical Mechanics course to study the canonical ensemble in statistical mechanics. In doing so, I have come across the following claim

When two thermodynamic potentials are related by a Legendre transform, then their corresponding partition functions are related by a Laplace transform.

Is there a way that I can see the logic behind why this statement is true? To clarify this question, consider functions f and g, that are related by a Legendre transform. Then, I send them to some new functions by a mapping $$\phi: f \rightarrow h$$ and $$\psi: g \rightarrow k.$$ Consider the further restriction that $h$ and $k$ are related by Laplace transforms. What must be true about the mappings $\phi$ and $\psi$ in order for the claim "if $f$ and $h$ are related by Legendre transforms, then $g$ and $k$ are related by Laplace transforms" to be true, and how is this criteria satisfied in the context of statistical mechanics?

Answer 1

TL;DR: The Legendre transformation can be e.g. seen as the leading classical tree-level formula of a formal semiclassical Fourier transformation; which in turn is a Wick-rotated version of the Laplace transform, cf. my Phys.SE answer here.

Superficially, in Stephen Teitel's statistical physics example with Helmholtz free energy $$\begin{align} \exp&\left\{ -\frac{F(T,V,N;k_B)T^{-1}}{k_B}\right\}\cr~=~&\int\!\frac{dE}{\Delta E} \exp\left\{ -\frac{S(E,V,N)}{k_B}\right\}\exp\left\{ \frac{ET^{-1}}{k_B}\right\}, \end{align}\tag{2.8.23'}$$ the conjugate variables are energy $E$ and inverse temperature $T^{-1}$, with the Boltzmann constant $k_B$ playing the role of the semiclassical parameter.

However, a physically better motivated expansion in the thermodynamic limit $N\gg 1$ is to work with intrinsic variables $$\begin{align} \exp&\left\{ -N f(\beta,v,N^{-1})\beta\right\}\cr~=~&\int\!\frac{d\epsilon}{\Delta\epsilon} \exp\left\{ -N s(\epsilon,v,N^{-1})\right\}\exp\left\{ N\varepsilon\beta\right\}, \end{align}\tag{2.8.23"}$$ so that the conjugate variables are energy $\epsilon=E/N$ per particle and inverse temperature $\beta=1/k_BT$, with $N^{-1}$ playing the role of the semiclassical parameter, cf. LPZ's answer.

The analogy in QM would be energy $E$ and time $t$ as conjugate variables, with the Planck constant $\hbar$ playing the role of the semiclassical parameter.

Answer 2

The general correspondence is captured by large deviation theory. You can check out a previous answer of mine where I illustrate the general relation on specific examples. To recap, using the notation of statistical mechanics, say you start with a function $W(E)$ (e.g. partition function of the microcanonical ensemble), and consider its Laplace transform (e.g. partition function of canonical ensemble): $$ Z(\beta) = \int W(E)e^{-\beta E}dE. $$ In the thermodynamic limit, $N\to\infty$, you expect the scaling: $$ W(E)\asymp e^{Ns(E/N)}\quad Z(\beta)\asymp e^{-N\beta f(\beta)} $$ with $s,f$ the thermodynamic potentials (e.g. entropy and free energy). By Laplace's method, you can estimate: $$ Z(\beta)\asymp e^{N\sup_\epsilon[s(\epsilon)-\beta \epsilon]} $$ so $s,\beta f$ are related by a Legendre transform (maximum entropy principle): $$ \beta f(\beta) = \inf_\epsilon[\beta \epsilon-s(\epsilon)] $$ The correspondence is therefore not general, you need the specific thermodynamic scaling. More generally, this is the large deviation principle and the correspondence is given by Varadhan's lemma.

Hope this helps.

Answer 3

Quote: "A mathematician may say anything, but a physicist must be partially sane" (JW Gibbs, 1839-1903) For Legendre-transforms in thermodynamics, see: 'Going around in circles - Legendre transformations' (P Mander, carnotcycle.worldpress)