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?

"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.
