What is the (practical) difference between a stationary distribution and an equilibrium distribution of a MC? I know that, for a Markov Chain, a stationary distribution is the (row) vector $\pi$ such that $\pi \cdot P = \pi$, where $P$ is the one-step transition matrix for the MC. Intuitively, I assume that this means that $\pi$ is the collection of probabilities that the process ends up in each of the possible states in the long run.
My understanding of equilibrium distributions is that the (row) vector $P^{*} = (P_{1}^{*}, P_{2}^{*}, \dots, P_{m}^{*})$ is an equilibrium distribution if each $P_{j}^{*} = \lim_{n \rightarrow \infty} P_{ij}^{(n)}$ exists for every $j \in S$ and is not dependent of the initial state $i$.
This implies that $P^{*}$ is the collection of probabilities of the process being in each state after $\infty$ steps. 
Does this not mean that equilibrium distributions are the same as stationary distributions, in practice?
 A: "Does this not mean that equilibrium distributions
are the same as stationary distributions, in practice?"
Stationary Distributions:
Let $\mathbf{P}$ be the transition probability matrix of a
homogeneous Markov chain $\{X_n, n \geq 0\}$. If there exists
a probability vector $\mathbf{\pi}$ such that
$$\mathbf{\pi} \mathbf{P} = \mathbf{\pi} \:\:\:\:\:\:\: (1)$$
then $\mathbf{\pi}$ is called a stationary distribution for the Markov chain.
Equilibrium Distributions:

Thm:
Let $\{X_n, n \geq 0\}$ be a regular homogeneous finite-state Markov chain
with transition matrix $\mathbf{P}$.
Then
$$\lim \limits_{n \to \infty} \mathbf{P}^n = \mathbf{\hat{P}}$$
where $\mathbf{\hat{P}}$ is a matrix whose rows are identical
and equal to the stationary distribution $\mathbf{\pi}$ for the
Markov chain defined by $(1)$.


To express the relationship in another way:
Let $\mathbf{\pi}$ be the stationary distribution. $\mathbf{\pi}$ is related to the expected return time $\mu_j$ by:
$$\mathbf{\pi}_j = \frac{1}{\mu_j}$$
If the chain is both irreducible and aperiodic, it is said to have an equilibrium distribution
$$\lim_{n \to \infty} p_{ij}^n =  \frac{1}{\mu_j}$$
Such $\mathbf{\pi}$ is called the equilibrium distribution.
