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In the study of statistics, a given family of probability densities depending smoothly upon a parameter $\theta$ can be expressed in the form

$p(x,\theta)=exp\left[c(x)+\sum_{r}\theta^{r}S_{r}(x)-\psi(\theta)\right],$

where the variable $x$ ranges over the sample space.

In models of statistical mechanics, we generally deal with Gibbs measures in the form

$p(x,\theta)=exp\left[\sum_{r}\theta^{r}S_{r}(x)-ln(\psi(\theta))\right],$

where the $S_{r}$ determines the form of the action and $\psi(\theta)=lnZ(\theta)$ is gibbs free energy, and $Z$ the partition function.

Based on the above, what is the meaning of $x$ in $Z$?

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  • $\begingroup$ There is no $x$ in your $Z$. $\endgroup$
    – NDewolf
    Commented Feb 15, 2021 at 20:01
  • $\begingroup$ Hi @NDewolf, I have just updated the question. Tell me if you see it $\endgroup$
    – user252965
    Commented Feb 15, 2021 at 20:03
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    $\begingroup$ It's not true you can write any distribution in this form, this is only true for exponential family of distributions en.wikipedia.org/wiki/Exponential_family $\endgroup$
    – Andrew
    Commented Feb 15, 2021 at 20:03
  • $\begingroup$ Hi @Andrew. Ok, I see your point. However, what is the meaning of $x$ in $Z$? $\endgroup$
    – user252965
    Commented Feb 15, 2021 at 20:05
  • $\begingroup$ Can you give an example? In the context of Bayesian inference, normally $\theta$ would refer to some parameters and $x$ to observed data. In statistical mechanics I would guess the $\theta$ are playing the role of Lagrange multipliers (inverse temperature, chemical potential...) and the $x$ are quantities the system could exchange with its environment (energy, number of particles, ...). But it would help to see an example or have more context. $\endgroup$
    – Andrew
    Commented Feb 15, 2021 at 20:23

1 Answer 1

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Eq. 2 of Ref [1] (which is by the same first author as Ref [2], which the OP mentioned in the comments) defines the Gibbs measure for the canonical ensemble \begin{equation} p(H,\beta) = q(x) \exp\left[-\beta H(x)-W(\beta)\right] \end{equation} where $x$ ranges over configuration space, $H(x)$ is the energy, $\beta=1/kT$ is the inverse temperature, and $W(\beta)$ is a normalization constant.

Let's compare this with the normal way to write Boltzman distribution. The Boltzman distribution gives the probability of finding the system in a given microstate, which is determined by the configuration space variables $x$. \begin{equation} p(x,\beta) =\frac{e^{-\beta H(x)}}{\int dx e^{-\beta H(x)}} = \frac{e^{-\beta H(x)}}{Z} \end{equation} where the partition function is $Z\equiv \int dx e^{-\beta H(x)}$.

It should be understood that $x$ here is not necessarily a single variable, but in general could be a vector of parameters. For instance, $x$ could represent the positions and momenta of $10^{23}$ particles, and in this case $\int dx$ really should be understood as a huge integral over each of these $10^{23}$ configuration space variables.

Now this isn't quite the same thing as Eq 2, which gives the probability of the system having a certain energy $H$, rather than being in a given microstate labelled by $x$.

To convert, we need to include a factor of the density of states $\rho(H)$. This function tells us how many states $\rho(H)d H$ have an energy in the interval from $H$ to $H+dH$.

\begin{equation} p(H,\beta) = \frac{\rho(H) e^{-\beta H}}{Z} = \rho(H) \exp\left[-\beta H - \log Z\right] \end{equation}

Comparing this expression with the first one, we can identify the normalization $W(\beta)$ with the log of the partition function $Z$

\begin{equation} W(\beta) = \log Z \end{equation}

Now it appears that the function $q(x)$ in the high temperature $\beta\rightarrow 0$ limit is related to the density of states \begin{equation} q(x) \equiv \rho(H(x)) \end{equation} To be honest I don't understand the notation $q(x)$, because each microstate should be equally likely in the high temperature limit. So I am not sure if the author is intending a distribution of microstates that violates the assumption that each accessible microstate is equally likely, or if $q(x)$ is meant to include some coarse graining over some subset of microstates, or something else.

References:

[1] https://cds.cern.ch/record/332060/files/9708032.pdf

[2] "Geometrical aspects of statistical mechanics" Physical Review E, 1995, volume 51, number 2.

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  • $\begingroup$ Hi @Andrew! Thanks for the clarifying answer!. In the light of the foregoing, what is the meaning of $x$ in $Z$? $\endgroup$
    – user252965
    Commented Feb 16, 2021 at 17:14
  • $\begingroup$ @ViniciusHolmes $x$ ranges over the configuration space, so in a gas it might label the positions and momenta of $10^{23}$ particles. The partition function $Z$ involves an integral over configuration space, so in the gas example $\int dx$ would be a multi-dimensional integral over all the positions and momenta (so integrating over a space with an enormous number of dimensions). $\endgroup$
    – Andrew
    Commented Feb 16, 2021 at 17:57
  • $\begingroup$ Now, I see your point @Andrew. However, the partition function of reference [2] seems to be independent of $x$ in the examples discussed in the paper. Is that correct? $\endgroup$
    – user252965
    Commented Feb 16, 2021 at 18:03
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    $\begingroup$ @ViniciusHolmes The partition function is an integral over $x$ so therefore doesn't depend on $x$. Much like $Z=\int_{-\infty}^{\infty} dx e^{-x^2/2}=\sqrt{2\pi}$ doesn't depend on $x$. $\endgroup$
    – Andrew
    Commented Feb 16, 2021 at 18:04

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