How to derive entropy from density of states?

Question

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.

Answer 1

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

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.

Answer 2

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.