Boltzmann vs Gibbs definition of entropy [duplicate]

Question

I am learning Statistical Mechanics and I have a question regarding different definitions of (statistical) entropy. If we use Boltzmann's definition:

$$\sigma \propto\ln(W)$$

Where $\sigma$ is the entropy of the macrostate associated with the microstates $W$. However, this means that entropy is a property of macrostates, not of microstates. If a macrostate corresponds to a single microstate, then the entropy is zero.

We can instead define the entropy per microstate as:

$$\sigma_i=-\ln(P_i)$$

Where $\sigma_i$ is the entropy of the "i"-th microstate, and $P_i$ is the probability of observing said microstate. Using this definition of entropy for microstates, we can then define the entropy of a macrostate as:

$$\sigma=\langle\sigma_i\rangle$$

Where $\langle\sigma_i\rangle$ is the average entropy over all the microstates of the respective macrostate. Is the latter (Gibbs entropy) simply a more general version of Boltzmann's definition, and hence better? Is it okay to define entropy for a given microstate or does give problems down the line?

Answer 1

Provided that the average of a quantity is defined by $$ \langle A\rangle = \sum_i P_i A_i, $$ the entropy by definition can be viewed as the average of the negative logarithm of the probability of a microstate $$ \sigma = -\sum_iP_i\log P_i = \langle -\log P\rangle. $$ This is just a mathematical fact (which may occasionally turn out to be convenient).

Note however, that here we are dealing with information/Shannon entropy, of which Boltzmann entropy is a particular case (mathematically), when all the state have the same probability ($P_i=1/W$, see also this post and reference therein). Boltzmann entropy has however many special properties related to the nature of the microstates in thermodynamic systems, so one should really distinguish the two.

Gibbs entropy is defined phenomenologically via clausius inequality, but can be shown to be equivalent to microscopically defined Boltzmann entropy.