Timeline for These two definitions in statistical mechanics
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 7, 2020 at 15:59 | history | edited | Qmechanic♦ |
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| Apr 7, 2020 at 14:16 | answer | added | eapovo | timeline score: 0 | |
| Mar 1, 2020 at 23:32 | comment | added | Charlie | I can see the ambiguity of my question given lack of context. Your comment is helpful though, thank you. | |
| Mar 1, 2020 at 23:27 | comment | added | Andrew | It's impossible to give a complete and correct answer without links to the notes for context. However, with the fundamental assumption of statistical mechanics that all allowed microstates are equally probable, it's certainly true that $\Omega(E)$ is (proportional to) the probability that the system has energy $E$. In other words, $\Omega(E)$ counts the degeneracy of states at energy $E$ (sometimes the degeneracy of states is also called $g(E)$). In that sense, you can think of $\Omega(E)$ as a statistical weight for the (sub-)ensemble of states with energy $E$. | |
| Mar 1, 2020 at 21:59 | comment | added | Charlie | Are you talking about the distribution that will include the Boltzmann factor? | |
| Mar 1, 2020 at 21:50 | comment | added | Superfast Jellyfish | Perhaps $\Omega$ refers to the distribution function of $E$. | |
| Mar 1, 2020 at 21:47 | history | asked | Charlie | CC BY-SA 4.0 |