Why does the classical electrodynamics Lagrangian density equation have a "field" term and an "interaction" term? On Wikipedia's page on classical electrodynamics, they state the Lagrangian density equation as follows 
\begin{equation}
\mathcal{L} = \mathcal{L}_{\text{field}} + \mathcal{L}_{\text{int}} = -\frac{1}{4\mu_0} F^{\alpha\beta} F_{\alpha\beta} - A_{\alpha}J^{\alpha}
\end{equation} 
I don't understand what we mean by "field" and "interaction". I have seen this several times before but I still don't know what it means. 
Can someone explain why we have these two terms (ie $\mathcal{L}_{\text{field}}$ and $\mathcal{L}_{\text{int}}$)? 
 A: The "field" term describes electromagnetic waves moving around in space-time, and that's it.
It only describes electromagnetic fields (the $F$ tensor), no charged particles and therefore no sources of electromagnetic waves.
With just the field term, you can describe travelling waves but you can't describe something like an antenna which actually emits those waves.
The "interaction" term describes how electromagnetic fields interact with charges (the $J$ tensor).
This term allows the theory to describe things like oscillating current in an antenna emitting radio waves, screening of elecromagnetic waves in a plasma, and other electromagnetic/matter interactions.
A: I know that the question specifically refers to classical electrodynamics, but I think it is helpful to look at this from a QED perspective. The term $-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$ is the kinetic term, i.e. from it we obtain the propagator. In a course on QFT, you probably derived the general relation
$$Z[J]=\int\mathcal{D}\Psi\,\exp\left(i\int d^4x\,[\mathcal{L}_\text{kinetic}(\Psi)+\sum_mJ^m\Psi_m]\right)\propto \exp(iW[J])$$
$$W[J]=\frac{1}{2}\sum_{mn}\int d^4xd^4y\,J^m(x)\Delta_{mn}(x,y)J^n(x)$$
where $\{\Psi_m\}$ are the fields, $J_m$ are the currents and $\Delta$ is the propagator. (Note that $m,n$ are multi-indices.)
We also have 
$$W[J]=-ET$$
where $T$ is the duration of the interaction. In the case of QED, we can make $J^\mu$ a stationary point charge. We then have
$$E\propto\frac{1}{r}$$
which is precisely Coulomb's potential. 
We need the $A_\mu J^\mu$ term  in the Lagrangian to allow particles to interact (non-vertex interactions of course). It describes the exchange of a boson, which generates the force bosons are famous for.
