## Classical canonical partition function

### Set

 context $\langle \mathcal M, \mathcal H,\pi,\pi_0,{\hat\rho},{\hat\rho}_0\rangle$ … classical canonical ensemble context $\mathrm{dim}(\mathcal M) = 3N$ context $\hbar$ … Reduced Planck's constant definiendum $Z(\beta):=\frac{1}{h^{3N}N!}\int_{\Gamma_{\mathcal M}}\ \hat\rho(\beta;{\bf q},{\bf p}) \ \mathrm d\Gamma$

#### Discussion

This mirrors the classical microcanonical phase volume.

We have

$Z(\beta):=\frac{1}{h^{3N}N!}\int_{\Gamma_{\mathcal M}} \mathrm{e}^{-\beta\ H({\bf q},{\bf p})}\ \mathrm d\Gamma$

$U\equiv\langle H\rangle=-\dfrac{\partial}{\partial\beta}\log\,Z(\beta)$

##### Usage as moment generating function for $H$

The usage of $Z(\beta)$ is very similar to that of a characteristic function/moment generating function for a probability density $f(x)$:

$E[e^{-ikX}]:=\int f(x)\ \mathrm e^{-ikx}\ \mathrm dx$

in probability theory, where

$E[X^n]=\left(i\frac{\partial}{\partial k}\right)_{k=0}^nE[e^{-ikX}].$

In fact, to compute expectation values with the partition function is even a little more straight forward, since in that case the probability function itself has exponential form and the temperature acts like the otherwise auxiliary parameter $k$. Notice a second difference: The parameter $\beta$ is multiplied by the energy $H(q,p)$, not directly by the variables of integration $q,p$.

#### In terms of the density of states

$Z(\beta)=\int{\mathrm e}^{-\beta\,E}{\mathrm d}\varphi(E)=\int D(E)\ \mathrm{e}^{-\beta\ E}\ \mathrm d E$

$U\equiv\langle H\rangle=-\dfrac{\partial}{\partial\beta}\log\,Z(\beta)=\int E\,\dfrac{{\mathrm e}^{-\beta\,E}}{Z(\beta)}\,D(E)\,{\mathrm d}E$

Note that in QM, there is no phase spaces to smoothly integrate over.

Experimental observation says that certain objects (the ones involved in experiments measuring the energy spectrum from a box, the sun, etc.) the itegrand $E\,\dfrac{{\mathrm e}^{-\beta\,E}}{Z(\beta)}\,D(E)$ ought to be replaces by $\propto E\,\dfrac{1}{\mathrm e^{\beta\hbar\omega}-1}\,E^2$.
For a single particle and $E\propto \omega\propto |k|$, it can indeed by argued that $D(E)\propto E^2$, see density of states. The temperature dependency of the energy distribution however isn't observed to follow a Boltzmann Distribution but instead a statistics known from the Grand canonical partition function.

Planck's ad hoc step is to say that any radiation ray of energy $E$ is actually partitioned into $n$ parts of $(\hbar\omega)$. Here $n$ isn't fixed but partioning is instead also thermailzed (that's basically the Definition of a black body) and thus follows a Boltzmann distribution in $(n\hbar\omega)$. His derivation amounts to raplacing certain functions of energy (they are energy densities, if we introduce spatial variables into the model)

If we compare the classical expression with the empirical result, the replacement must be

$E\,\dfrac{{\mathrm e}^{-\beta\,E}}{Z(\beta)}\mapsto\sum_{n=0}^\infty\dfrac{(n(\hbar\omega))\,{\mathrm e}^{-\beta\,n\,(\hbar\omega)}}{\sum_{n=0}^\infty{\mathrm e}^{-\beta\,n\,(\hbar\omega)}}=(\hbar\omega)\dfrac{1}{\mathrm e^{\beta\hbar\omega}-1}$

The common derivation of Plancks law makes the above step and then neglects spatial inhomogenies and just introduces a characteristic length $L$. Then we can use the characteristic speed (of light) $c$ to get a characteristic frequency $c/L$. Thus we can intorduce powers of $\omega$ via the unitless expression $\omega\left/\right.\dfrac{c}{L}$. If $D$ is quadratic (the case of photon gas), we get

$U=\int_0^\infty \dfrac{1}{\pi^2}\dfrac{1}{\mathrm e^{\beta\hbar\omega}-1} \left(\omega\left/\right.\dfrac{c}{L}\right)^3 \mathrm d(\hbar\omega)$

#### Context 