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classical_canonical_partition_function [2016/03/09 11:39] nikolaj |
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| >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 $\propto E\,\dfrac{1}{\mathrm e^{\beta\hbar\omega}-1}\,E^2$. | >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 $\propto E\,\dfrac{1}{\mathrm e^{\beta\hbar\omega}-1}\,E^2$. | ||
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| >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]]. | >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]]. | ||
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| - | >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) |
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| + | >If we compare the classical expression with the empirical result, the replacement must be | ||
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| + | >$E\,\dfrac{{\mathrm e}^{-\beta\,E}}{Z(\beta)}\mapsto\sum_{n=0}^\infty\dfrac{(n(\hbar\omega))\,{\mathrm e}^{-\beta\,n\,(\hbar\omega)}}{\sum_{n=0}^\infty{\mathrm e}^{-\beta\,n\,(\hbar\omega)}}=(\hbar\omega)\dfrac{1}{\mathrm e^{\beta\hbar\omega}-1}$ | ||
| >Roughly | >Roughly | ||
| > | > | ||
| >$\int {\mathrm d}E\cdot p_E \, \mapsto \, \sum_{n=0}^\infty\int{\mathrm d}\,(n\hbar\omega)\,p_\omega(n)\,\mapsto\,\sum_{n=0}^\infty\int{\mathrm d}\,(n\hbar|k|)\,|k|^2\,p_k(n)$ | >$\int {\mathrm d}E\cdot p_E \, \mapsto \, \sum_{n=0}^\infty\int{\mathrm d}\,(n\hbar\omega)\,p_\omega(n)\,\mapsto\,\sum_{n=0}^\infty\int{\mathrm d}\,(n\hbar|k|)\,|k|^2\,p_k(n)$ | ||
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| - | >If we compare the classical expression with the empirical result, the replacement must be | ||
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| - | >$E\,\dfrac{{\mathrm e}^{-\beta\,E}}{Z(\beta)}\mapsto\sum_{n=0}^\infty\dfrac{(n(\hbar\omega))\,{\mathrm e}^{-\beta\,n\,(\hbar\omega)}}{\sum_{n=0}^\infty{\mathrm e}^{-\beta\,n\,(\hbar\omega)}}=(\hbar\omega)\dfrac{1}{\mathrm e^{\beta\hbar\omega}-1}$ | ||
| 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 | ||
