# Definition of entropy for canonical vs. microcanonical ensembles

The two following definitions of entropy are often used in the microcanonical and canonical ensembles, respectively:

1. $$S=k_b\ln(\Omega)$$
2. $$S=-k_b\sum_ip_i\ln(p_i)$$

I am curious how the second can be derived from the first, and if they can both be applied to the canonical ensemble. Specifically, how could one apply equation 1 to the canonical ensemble?

As well, this is somewhat unrelated, but if you work backwards from the canonical ensemble, and use $$F=-kT\ln(Z)$$ and $$S=-(\partial F/ \partial T)$$, you can arrive at $$S/k_b =\ln(Z)+\beta \bar E$$. If you then apply definition 1, you can rearrange this to get $$\Omega = Ze^{\beta \bar E}$$. Is this application of definition 1 valid, and is the relation between $$\Omega$$ and $$Z$$ true for the canonical ensemble?

Definition $$1$$ is specific to the microcanonical ensemble, because the quantity $$\Omega$$ is the number of microstates (mutually orthogonal microstates in the quantum case) compatible with the specified conditions. The logarithm of $$\Omega$$ is used for mathematical convenience, and the factor of Boltzmann's constant $$k_b$$ is included only because old traditions are hard to break, despite the minor inconvenience it causes.
Definition $$2$$ can be applied to any ensemble. It is a natural measure of how "non-presumptous" a given probability distribution is. The larger the value of $$S$$, the less presumptuous the distribution. The previous comment about the factor of $$k_b$$ applies here, too.
When definition $$2$$ is applied to the microcanonical ensemble, all of the $$p_i$$ are equal to each other for all states that are compatible with the specified conditions, and are zero otherwise. That is, $$p_i=1/\Omega$$. Using this in definition $$2$$ gives $$S=-k_b\sum_i p_i\ln(p_i)=-k_b \frac{1}{\Omega}\sum\ln(1/\Omega) =k_b\ln(\Omega)$$ because $$\Omega$$ is also the number of terms in the sum. In this sense, definition $$1$$ is a special case of definition $$2$$.
Definition $$1$$ does not apply to the canonical ensemble, but it can be used to deduce the canonical ensemble for a subsystem that is part of a larger system described by the microcanonical ensemble. This derivation only works if $$\Omega(E)$$ does not increase exponentially (or faster) as a function of $$E$$. If it does, then the canonical ensemble doesn't exist. The microcanonical ensemble is more generally applicable than the canonical ensemble, but the canonical ensemble (when it exists) is usually more convenient.
Definition $$2$$ can also be used to deduce the canonical ensemble: among all probability distributions with a given average value of the total energy, the canonical ensemble is the least presumptuous distribution — it maximizes the entropy in definition $$2$$.
Given the canonical ensemble $$p_i\propto \exp(-\beta E_i)$$, the partition function can be written as $$Z=\sum_i \exp(-\beta E_i) \sim \sum_E\Omega(E) e^{-\beta E}$$ where $$\Omega(E)$$ is the number of states with total energy $$E$$. (I'm writing "$$\sim$$" so that I don't have to define $$\Omega(E)$$ more carefully, which would probably not be helpful here.) Typically, $$\Omega(E)$$ is a rapidly increasing function of $$E$$, but not exponentially increasing, so the summand is sharply peaked at a particular value $$\bar E$$ of $$E$$ (which is typically essentially the average value), and we might as well have $$Z\sim \Omega(\bar E)\exp(-\beta \bar E)$$. Using this in the equation $$S/k_b=\ln(Z)+\beta\bar E$$ that was shown in the OP gives $$S\sim k_b\ln\Omega(\bar E)$$, in agreement with definition $$1$$. The agreement is only approximate, as expected, because the canonical ensemble with the given temperature is only approximately equivalent to a microcanonical ensemble with the specific total energy $$\bar E$$. The approximation is typically very good for a macroscopic system, and a more-careful version of this derivation can be used to quantify just how good the approximation is.