Partition function

The partition function of a system in contact with a thermal bath at temperature $T$ is the normalization constant of the Boltzmann distribution function, and therefore its expression is given by

$Z(T)=\int \Omega (E)\exp(-E/k_{B}T)\,dE$,

where $\Omega (E)$ is the density of states with energy $E$ and $k_{B}$ the Boltzmann constant.

The partition function of a system is related to the Helmholtz energy function through the formula

$\left.A\right.=-k_{B}T\log Z.$

This connection can be derived from the fact that $k_{B}\log \Omega (E)$ is the entropy of a system with total energy $E$. This is an extensive magnitude in the sense that, for large systems (i.e. in the thermodynamic limit, when the number of particles Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle N\to \infty } or the volume $V\to \infty$), it is proportional to $N$ or $V$. In other words, if we assume $N$ large, then

$\left.k_{B}\right.\log \Omega (E)=Ns(e),$

where Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle s(e)} is the entropy per particle in the thermodynamic limit, which is a function of the energy per particle Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle e=E/N} . We can therefore write

$\left.Z(T)\right.=N\int \exp\{N(s(e)-e/T)/k_{B}\}\,de.$

Since $N$ is large, this integral can be performed through steepest descent, and we obtain

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \left.Z(T)\right.=N\exp\{N(s(e_{0})-e_{0}/k_{B}T)\}} ,

where Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle e_{0}} is the value that maximizes the argument in the exponential; in other words, the solution to

$\left.s'(e_{0})\right.=1/T.$

This is the thermodynamic formula for the inverse temperature provided Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e_0} is the mean energy per particle of the system. On the other hand, the argument in the exponential is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{k_BT}(TS(E_0)-E_0)=-\frac{A}{k_BT}}

the thermodynamic definition of the Helmholtz energy function. Thus, when Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N} is large,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left.A\right.=-k_BT\log Z(T).}

Connection with thermodynamics

We have the aforementioned Helmholtz energy function,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left.A\right.=-k_BT\log Z(T)}

we also have the internal energy, which is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U=k_B T^{2} \left. \frac{\partial \log Z(T)}{\partial T} \right\vert_{N,V}}

and the pressure, which is given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p=k_B T \left. \frac{\partial \log Z(T)}{\partial V} \right\vert_{N,T}} .

These equations provide a link between classical thermodynamics and statistical mechanics