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Aug 20, 2021 at 0:08 answer added Han Xu timeline score: 0
Dec 5, 2015 at 21:42 vote accept Zane Dufour
Dec 3, 2015 at 21:01 answer added anonymous67 timeline score: 3
Oct 27, 2015 at 3:09 comment added Zane Dufour Also, I'm not trying to "break" statistical mechanics or something stupid like that. The first result was something I was asked to prove on a homework assignment and the second result was used by my prof to demonstrate the Stefan-Boltzmann law. My assumption is that the two situations are fundamentally different, but I still don't understand why the overall partition function for Blackbody Radiation is a sum instead of a product (it doesn't seem like we're counting states with photons with different frequencies).
Oct 27, 2015 at 3:04 history edited Zane Dufour CC BY-SA 3.0
Tidied up notation
Oct 27, 2015 at 2:59 comment added Zane Dufour I tidied up my notation a bit so that there shouldn't be any equivocation.
Oct 27, 2015 at 2:55 history edited Zane Dufour CC BY-SA 3.0
Tidied up notation
Oct 27, 2015 at 2:32 comment added CR Drost My gut feeling is that you are equivocating (using the same term to mean two different things), and I think that the term you're equivocating over is $s$. We should probably say "state $i$ contains $n_{ij}$ photons at frequency $\omega_j$ leading to $E_i = \sum_j n_{ij} \hbar \omega_j.$ This causes your "derivation" to look instead like $\sum_i \Pi_j \exp(-\beta~n_{ij}~\hbar~\omega_j).$
Oct 27, 2015 at 2:12 history asked Zane Dufour CC BY-SA 3.0