# Details in the derivation of the second law starting from the phase space volume

I had a question on one of the details of the derivation of the second law of thermodynamics starting from the phase space volume. I'll type out what I understand so far:

Letting the Hamiltonian depend on some external parameter $a$ so that $H=H(a)$. The phase space volume can be written as

$$\bar{\Omega}(E,a)=\int d\Gamma \Theta(E-H(a)),$$

where $\Theta$ is the heaviside function. The total differential is

$$d\bar{\Omega}(E,a)=\int d\Gamma \delta(E-H(a))(dE-\frac{\partial H}{\partial a}da)=\Omega(E,a)(dE-\langle\frac{\partial H}{\partial a}\rangle da).$$

In the equation above $$\Omega=\int d\Gamma \delta(E-H(a)).$$

Using the logarithmic derivative

$$d \log\bar{\Omega}=\frac{\Omega}{\bar{\Omega}}(dE-\langle\frac{\partial H}{\partial a}\rangle da).$$

Using the definition of entropy $S=k\log\bar{\Omega}$

$$k\ dS=\frac{\Omega}{\bar{\Omega}}(dE-\langle\frac{\partial H}{\partial a}\rangle da).$$

At this point the book jumps to

$$dS=\frac{1}{T}(dE-\langle\frac{\partial H}{\partial a}\rangle da)$$

without much explanation. My question is how the $1/T$ follows from the previous line?

The expression $$k_B \frac{\Omega}{\bar{\Omega}}$$ equals $$k_B\frac{1}{\bar{\Omega}}\frac{d\bar{\Omega}}{dE}$$ which equals $$\frac{dS}{dE}.$$ In thermodynamics, where $S$ is the Clausius entropy, this is equal to $1/T$ where $T$ is the Kelvin temperature. In statistical physics, this expression can be taken as a definition of $1/T$ of a system from the microcanonical ensemble $E,a$.