# 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

$$\mathcal{L} = \mathcal{L}_{\text{field}} + \mathcal{L}_{\text{int}} = -\frac{1}{4\mu_0} F^{\alpha\beta} F_{\alpha\beta} - A_{\alpha}J^{\alpha}$$ 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 $\mathcal{L}_{\text{int}}$)?

• The field term is analogous to a kinetic energy term for a particle while the interaction term is analogous to a potential energy term. Thus, in the absence of an interaction term, the field is free; the resulting equations of motion are homogeneous differential equations. In this case, the source free Maxwell equations. – Alfred Centauri Jan 31 '15 at 4:24
• Do you know when the first draft/form of this Lagrangian was constructed? Was this conceptualization of field and interaction known to Maxwell in this specific Lagrangian form? – Stan Shunpike Jan 31 '15 at 4:36

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.

• So is the basic idea that everything that's part of a Lagrangian density equation is describing the different forms or states the "energy" can take? In other words, if I have $\mathcal{L}_{\text{field}}$, this is one form of electromagnetic energy. Then the antenna example you gave would be another form which the energy could convert into. Is that the idea? – Stan Shunpike Feb 17 '15 at 22:58
• I would put it differently. Each term in the Lagrangian describes a thing that can happen. The field term describes waves travelling around. The $J$ term describes charges moving around. Any term with both $F$ (or $A$) and $J$ describes the interaction of electromagnetic waves and charges. – DanielSank Feb 18 '15 at 3:19
• Wow! That's a great way of phrasing it. I will have to think about that. I never considered looking at the equations that way. No wonder I am getting confused . – Stan Shunpike Feb 18 '15 at 4:21
• @StanShunpike: Please note that this way of thinking about terms in the Lagrangian is not limited to quantum field theory. Even in something as mundane as an electrical circuit we find that terms in the Lagrangian in which two dynamical variables are multiplied indicates interaction. – DanielSank Feb 18 '15 at 9:11

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.

• Since you know QFT, can you explain what a functional integral is? I asked about it here math.stackexchange.com/questions/1085236/… and here physics.stackexchange.com/questions/161274/… but didn't get an answer. I think if I could grasp that, QFT would come much easier to me. – Stan Shunpike Feb 2 '15 at 1:36
• @StanShunpike: What we really mean when we say functional integration is $$\int\mathcal{D}\phi=\prod_{x}\int_{-\infty}^\infty d\phi(x)$$ In other words, we integrate at each point over all field configurations at that point. You'll never actually do a path integral. You'll use various tricks to derive general results. I recommend Shankar for nonrelativistic path integrals. For QFT ones Zee and Weinberg work well. – Ryan Unger Feb 2 '15 at 1:46
• Ohhhhhhh.....which makes sense because Dick Feynman always said you have to sum over all possible paths. So the integral represents the sum of all the field configurations at the point while the product represents doing that over all $x$. Why do we multiply the integrals? – Stan Shunpike Feb 2 '15 at 2:11
• @StanShunpike: Whoops. It was supposed to read $$\int_{-\infty}^\infty\prod_x d\phi(x)$$ in the sense that $$\int\prod_{i=1}^n dx_i=\underbrace{\int\cdots\int}_n dx_1\cdots dx_n$$ – Ryan Unger Feb 2 '15 at 2:16
• @StanShunpike: I hope you can appreciate why we say this is not well-defined. $\prod_x$ really makes no sense. – Ryan Unger Feb 2 '15 at 2:17