In the following derivation I am trying to show that the function $Z_C(\beta)$ is obtained from the function $Z_M(E)$ by Laplace transform. Let, \begin{equation} \frac{1}{Z_M}\frac{\partial Z_M}{\partial E}=\frac{\partial \ln Z_M}{\partial E}=\beta \end{equation} Then, \begin{equation} \frac{\partial }{\partial E}(\ln Z_M-\beta E)=0 \end{equation} Let the bracketed term equal $Z_C(\beta)$ be the Legendre transform relationship, \begin{equation} \ln Z_M-\beta E:=\ln Z_C \end{equation} Take the exponential and integrate over $E$ we get, \begin{equation} Z_C(\beta)=\int Z_M(E)e^{-\beta E}dE \end{equation} I have two questions, Firstly why integration over $E$ in the final step doesn't lead to, \begin{equation} \int Z_C(\beta)dE=\int Z_M(E)e^{-\beta E}dE \end{equation} Secondly, is there a general relationship between the Laplace transform of a function and the Legendre transform of the $\ln$ of that function or is this a property of physics rather than the mathematics?

There is also a related question here.

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    $\begingroup$ Crossposting is frowned upon, please choose one site for your question to avoid duplication of answers. If it doesn't recieve answers for a few days on the site you chose, you might consider flagging for migration to the other. $\endgroup$
    – ACuriousMind
    Commented Aug 26, 2015 at 22:23
  • $\begingroup$ Okay, sorry about that haha. In terms of the question itself would you advise deleting here or maths (i.e. where is it more appropriate for?) because on the one hand I'm thinking the question is about the mathematics, but on the other it is a concept in physics? $\endgroup$
    – RedPen
    Commented Aug 26, 2015 at 22:26
  • $\begingroup$ @ACuriousMind I chose here, i think after all it is thermodynamics and on the maths forum that might not be appreciated fully unless there are applied mathematicians etc. $\endgroup$
    – RedPen
    Commented Aug 26, 2015 at 22:29
  • $\begingroup$ Why are you defining $Z_C$ as such in your third equation? I'm not sure this equation is correct. Clearly $\mathrm{ln} Z_M - \beta E$ can be understood as the Legendre transform of $\mathrm{ln} Z_M$ but I don't see how that's useful. $\endgroup$ Commented Aug 27, 2015 at 0:02
  • $\begingroup$ @KyleArean-Raines Hmm, I see what you mean, I agree it might be a bit strong. I just thought that it would have a greater emphasis. Especially since as far as the microcanonical ensemble goes this is the first real time we have seen the function $Z_C(\beta)$. $\endgroup$
    – RedPen
    Commented Aug 27, 2015 at 0:10

2 Answers 2


General Mathematical Result

If you can evaluate the Laplace transform via saddle point method you will get a Legendre transform from a Laplace Transformation. This means the function your transforming should have a significant peak only at one point:

As you can see, the Laplace transformation doesn't necessarily get you a Legendre transformation in the exponential: $\int f(x)e^{-xs}=\int e^{ln(f(x))-xs}$ here $s$ isn't guaranteed to be $\frac{d}{dx}ln(f(x))$ but if f(x) is peaked at one point you can apply the saddle point approximation. In the saddle point approximation you evaluate the integral at the point where the integrand is maximum:

$\frac{d}{dx}[ln(f(x))-xs]|_{x_0}=\frac{d}{dx}ln(f(x))|_{x_0}-s=0$ which implies $\frac{d}{dx}ln(f(x))|_{x_0}=s$

so the evaluation of that integral is $e^{ln(f(x_0))-x_0s}$ where $x_0$ and $s$ have the usual relationship for a Legendre transformation.

Aplication in physics

Of course in physics you want all your thermodynamic potentials to have the have minimum peaks(- sign in $Z=e^{-F\beta}$) so your system settles down into a single well defined state. In this way all the partition functions for the different ensembles can be found by Laplace transformation and you will get the corresponding Legendre transformed thermodynamic potential.

I suggest you read this article which demonstrates this result for the micro-canonical to canonical and gives a very good review of the Legendre transformation. Then from there it is easy to go gibs or grand(which is really just gibs when you consider $\mu$ a force):

So starting from $Z = \int e^{-\beta H}$ you need to add energy from the force,$J$, you are constraining instead of its conjugate $x$: $H-> H' =H+Jx$. Then since you are now constraining $J$ instead of $x$, $x$ now parametrizes your probability and you need to integrate over it to get the correct normalization($Z$).

so $Z->Z'=\int dx \int e^{-\beta(H+Jx)}= \int dx e^{Jx}\int e^{-\beta(H)}= \int dx e^{Jx}Z$

Thus you have your Laplace transform which by following the steps above will get you the desired Legendre transform of the Helmholtz free energy; the Gibs free energy.


The relationship between the Laplace and Legendre transforms is through the method of steepest descent, which is usually exact in the thermodynamic limit (since the configuration space typically goes with the system size as $C(N)\Omega_0(E)^N$). For finite $N$, the microcanonical entropy (density) is slightly different from the canonical entropy since it does not include fluctuations about the average energy. Here $\Omega_0(E)=e^{\log(S_M(E))}$ is the same as your $Z_M(E)$.


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