# Deriving all of electrodynamics from one single action

Preamble

The action for a relativistic particle of charge $m$ and charge $q$ moving in an external electromagnetic 4-potential $A^\mu$ is $$\mathcal S_p[y]=-\int_{a}^{b}\left(mc+\frac q c A_\mu(y)\frac {dy^\mu} {ds} \right)ds$$ where $a,b\in\mathbb R^{1,3}$, and $y$ is the path of the particle in spacetime. From $\delta \mathcal S_p=0$ one derives the Lorentz equation: $$mc\frac {d^2y^\mu}{ds^2}=\frac q c F_{\mu \nu} \frac {dy^\nu}{ds}$$ where $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ is the electromagnetic field.

The action for the electromagnetic field generated by a 4-current $j^\mu$ is $$\mathcal S_f[A]=-\int_{\Sigma_1}^{\Sigma_2}\left (\frac 1 4 F_{\mu\nu}F^{\mu\nu}+A_\mu j^\mu\right)d^4x$$ where $\Sigma_1$ and $\Sigma_2$ are spacelike surfaces in spacetime. From $\delta \mathcal S_f=0$ one derives the Maxwell equations: $$\partial_\mu F^{\mu\nu}=j^\nu$$

Question

Is there a way to write one action, functional of both particle path and electromagnetic 4-potential, whose extrema are all and only $A$ and $y$ that satisfy both Lorentz and Maxwell equations? (In the latter, the 4-current should be replaced by $j^\mu(x)=\int q \frac {dy^\mu} {ds}\delta^4(x-y(s))ds$.)

Partial answer

In Lechner's book, the author writes the following action: \begin{align}\mathcal S[A,y]&=\mathcal S_1[A]+\mathcal S_2[y]+\mathcal S_3[A,y]=\\&=-\int_{\Sigma_1}^{\Sigma_2}\frac 1 4 F_{\mu\nu}F^{\mu\nu}d^4x-\int_a^bmcds-\int_{\Sigma_1}^{\Sigma_2}A_\mu j^\mu d^4x\end{align} and, since $\mathcal S_2[y]$ doesn't depend on $A$, he notices that the field $A$ minimizes $\mathcal S_1[A]+\mathcal S_3[A,y]=\mathcal S_f[A]$ iff Maxwell equations are satisfied. In a similar way, since $\mathcal S_1[A]$ doesn't depend on $y$, the path $y$ minimizes $\mathcal S_2[y]+\mathcal S_3[A,y]=\mathcal S_p[y]$ iff the Lorentz equation is satisfied.

But in this way, isn't he excluding the possibility that $A$ and $y$ could simultaneously minimize the action which is the sum of the three terms?

## 1 Answer

But in this way, isn't he excluding the possibility that $A$ and $y$ could simultaneously minimize the action which is the sum of the three terms?

No. For $A$ and $y$ to be a solution of the combined equations of motion, it is necessary that their values are an extremum of the combined action. But this is equivalent to saying that "$\delta S/\delta A = 0$ when $y$ is held fixed" and "$\delta S/\delta y = 0$ when $A$ is held fixed", which are exactly the conditions he outlines.

This is a bit easier to see in the context of multivariable calculus. If $f(x,y) = g_1(x) + g_2(y) + g_3(x,y)$, then $f$ is extremized at a particular value of $x$ and $y$ if and only if $$\left( \frac{\partial f}{\partial x} \right)_y = \frac{\partial g_1}{\partial x} + \left( \frac{\partial g_3}{\partial x} \right)_y = 0$$ and $$\left( \frac{\partial f}{\partial y} \right)_x = \frac{\partial g_2}{\partial y} + \left( \frac{\partial g_3}{\partial y} \right)_x = 0$$ simultaneously. But these conditions are exactly analogous to the conditions outlined by Lechner.