Timeline for How to derive entropy from density of states?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 5, 2022 at 3:05 | comment | added | FaDA | thanks for your help! i understand what you mean | |
| Nov 4, 2022 at 13:54 | comment | added | valerio | @FaDA Taking the limit $\delta E \to 0$ is meaningless in this situation because $\delta E$ is just a notation for an "infinitesimal", and if you want to be rigorous you can rewrite all the derivation using derivatives and series expansions. The limit $\delta E/E\to 0$, on the other hand, makes sense and it's typically taken in derivations related to this one. Also note that a logarithmic divergence is a "very slow" one so in any case you shouldn't worry too much about it ;) | |
| Nov 3, 2022 at 12:20 | comment | added | FaDA | @valerio sorry for response to this early post. your reply is important to me. my question is why does $k \ln (\delta E)$ an arbitrary constant. should $k \ln (\delta E)$ goes to infinity as $\delta E$ approaches zero? | |
| May 29, 2020 at 7:31 | comment | added | valerio | @nervxxx Sorry for the very late reply, I had missed this comment. You can always throw in some energy scale to make the log dimensionless. For example, $\ln[\omega(E)\delta E] = \ln[\omega(E) E_0] + \ln (\delta E/E_0)$. This doesn't change the result since the expression for $\delta S$ is unchanged. | |
| Apr 21, 2017 at 19:48 | comment | added | nervxxx | I know this is what people do all the time, but how should I think of the units in $S = k \ln[\Omega(E)]$ and $S = k \ln[\omega(E)]$? $\Omega(E)$ is the number of microstates which is dimensionless, but $\omega(E)$ is the density of states which has units of inverse energy. What does it mean to take the logarithm of something with dimensions? | |
| Jun 29, 2016 at 22:20 | history | answered | valerio | CC BY-SA 3.0 |