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I'm trying to derive the entropy of a black hole, given the density of states of a bosonic string (the details are not relevant). The density of states is

$$ \omega(E) = E^\alpha e^{\beta E} $$

The entropy is defined as

$$ S = k\ln \Omega, $$

where $\Omega$ is the number of microstates.

I would suppose that the number of microstates in the energy interval $(E,E+\delta E)$ would be $\delta \Omega = \omega \delta E$. So the increase in entropy is

$$\delta S = k \ln{\omega \delta E} = k \ln (E^\alpha e^{\beta E} \delta E),$$

which can't be right because according to equation $11.9.4$ in http://arxiv.org/abs/1506.07798, it should be (in the high energy limit, which ignores the power contribution in $E$)

$$\delta S = k\beta \delta E.$$

So what is the correct way to derive the entropy from the density of states?

It seems that $\delta S = k\ln \omega(\delta E) $ would work, but it involves a logarithm of a quantity with dimensions.

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  • $\begingroup$ Isn't $\delta S = k \frac{\Omega}{\delta \Omega }$? $\endgroup$
    – Prahar
    Jun 29, 2016 at 18:49
  • $\begingroup$ @Prahar Did you mean $\delta S = k \frac{\delta \Omega}{\Omega}$ ? $\endgroup$
    – valerio
    Jun 29, 2016 at 22:04
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    $\begingroup$ @valerio92 - Yes. I absolutely meant that. Tried to type it out from my phone! $\endgroup$
    – Prahar
    Jun 30, 2016 at 2:39

2 Answers 2

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$$S=k \ln [\Omega(E)] = k \ln [\omega (E) \delta E] = k \ln [\omega(E)] +k \ln (\delta E)$$

Last term is an arbitrary constant, so that we can set

$$S = k \ln[\omega (E)]$$

from which

$$\delta S = k \frac{\delta \omega}{\omega}$$

If we can ignore the power contribution and set $\omega (E) \simeq e^{\beta E}$, we get

$$\delta S = k \frac{\delta(e^{\beta E})}{e^{\beta E}} = k \frac{\beta \ e^{\beta E} \delta E}{e^{\beta E}} = k \ \beta \delta E$$

More about entropy and density of states: here.

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  • $\begingroup$ 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? $\endgroup$
    – nervxxx
    Apr 21, 2017 at 19:48
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    $\begingroup$ @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. $\endgroup$
    – valerio
    May 29, 2020 at 7:31
  • $\begingroup$ @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? $\endgroup$
    – FaDA
    Nov 3, 2022 at 12:20
  • $\begingroup$ @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 ;) $\endgroup$
    – valerio
    Nov 4, 2022 at 13:54
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    $\begingroup$ thanks for your help! i understand what you mean $\endgroup$
    – FaDA
    Nov 5, 2022 at 3:05
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As @valerio92 points out, your mistake is that $S = k \ln (\omega\, \delta E)$, not $\delta S$. To get $\delta S$, you differentiate the right-hand expression to get $\delta S = k \frac{\delta \omega}{\omega}$, and the $\delta E$ drops out and you get an expression with the right dimensions. The notation is a bit misleading, because the $\delta$ in the $\delta E$ is not a differential corresponding to the $\delta$ in the $\delta S$ - it just denotes that we should think of $\delta E$ as a small constant quantity. Once you differentiate the expression for $S$, the "differential" $\delta$ actually ends up on the $\omega$, which is the actual variable quantity.

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