# Natural log introduced in microstates derivative with respect to energy in equilibrium equation

In Pathria and Beale's Statistical Mechanics, 3rd ed, Chapter 1.2 (Contact between statistics and thermodynamics: physical significance of the number $$Ω(N, V, E)$$ )

The equation to maximize $$Ω^{(0)}$$ is:

$$\left(\frac{∂Ω_1(E_1)}{∂E_1}\right)_{E_1 = \bar{E_1}} Ω_2(\bar{E_2}) \space + Ω_1(\bar{E_1}) \left(\frac{∂Ω_2(E_2)}{∂E_2}\right)_{E_2 = \bar{E_2}} \cdot {\frac{∂E_2}{∂E_1}} = 0$$

where $$\frac{∂E_2}{∂E_1} = -1$$

but then in the next equation a natural log term $$\ln$$ is slid into the partial derivatives becoming:

$$\left( \frac{∂ \ln Ω_1(E_1)}{∂E_1}\right)_{E_1 = \bar{E_1}} = \left( \frac{∂ \ln Ω_2(E_2)}{∂E_2}\right)_{E_2 = \bar{E_2}}$$

My confusion is why is the must the natural log of the number of microstates be taken instead of just the partial derivative as it was in the first equation?

As well as what happened to the two multiplicative terms $$Ω_2(\bar{E_2})$$ and $$Ω_1(\bar{E_1})$$, did these terms cancel out or something?

It is not clear in the steps presented in the book. A thorough explanation is much appreciated.

• Expand out the derivatives of the natural log terms, and you'll see that this equation is equivalent to the first. Jun 8, 2022 at 15:26

Recall the chain rule: $$\left(f\circ g \right)' =\left(f'\circ g\right) \cdot g',\tag{1}$$ where $$f\circ g\equiv f\left(g\left(x\right)\right)$$. In your case, $$f(\cdots)\leftarrow\ln(\cdots)$$ and $$g(\cdot)\leftarrow\Omega(\cdot)$$. These can be directly fed into (1): $$\frac{\partial}{\partial E}\left(\ln(\Omega(E))\right)=\frac{1}{\Omega(E)}\cdot\frac{\partial\Omega(E)}{\partial E}.$$
This allows us to write the entropy as the logarithm of the number of microstates: $$S\propto\ln\Omega$$.