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classical_canonical_partition_function [2016/03/09 11:27]
nikolaj
classical_canonical_partition_function [2016/03/09 11:40]
nikolaj
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 Note that in QM, there is no phase spaces to smoothly integrate over.  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 [[Classical density of states|density of states]].+>​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 [[Classical density of states|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 $(n\hbar\omega)$. ​This amounts to raplacing certain functions of energy (they are energy densities, if we introduce spatial variables into the model)+>​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
 +>
 +>​$E\,​\dfrac{{\mathrm e}^{-\beta\,​E}}{Z(\beta)}\mapsto\sum_{n=0}^\infty\dfrac{(n(\hbar\omega))\,​{\mathrm e}^{-\beta\,​n\,​(\hbar\omega)}}{\sum_{n=0}^\infty{\mathrm e}^{-\beta\,​n\,​(\hbar\omega)}}=(\hbar\omega)\dfrac{1}{\mathrm e^{\beta\hbar\omega}-1}$
 >Roughly >Roughly
 > >
 >$\int {\mathrm d}E\cdot p_E \, \mapsto \, \sum_{n=0}^\infty\int{\mathrm d}\,​(n\hbar\omega)\,​p_\omega(n)\,​\mapsto\,​\sum_{n=0}^\infty\int{\mathrm d}\,​(n\hbar|k|)\,​|k|^2\,​p_k(n)$ >$\int {\mathrm d}E\cdot p_E \, \mapsto \, \sum_{n=0}^\infty\int{\mathrm d}\,​(n\hbar\omega)\,​p_\omega(n)\,​\mapsto\,​\sum_{n=0}^\infty\int{\mathrm d}\,​(n\hbar|k|)\,​|k|^2\,​p_k(n)$
-> 
->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 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
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