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We know that the EM-Lagrangian \begin{align} -\frac{1}{16 \pi} F_{\mu \nu} F^{\mu \nu} - \frac{1}{c} j_{\mu} A^{\mu} \end{align} correctly describes the evolution of the electromagnetic field in the presence of a given $j(x) = (c \rho(\vec{x}, t), \vec{j}(\vec{x}, t))$, which is externally specified, and not governed by the Lorentz-force (or rather its equivalent action on charge densities). Treating $j^{\mu}$ as an independent field won't work, because then the e.l. equations would yield $A^{\mu} = 0$.

Is there any classical (classical as in: not quantized, no operators) lagrangian that yields the said equations, not only for the E.M. field, but also for the "sources"?

The next best thing I can think of is the EM-field coupled to a scalar field, with mass- and kinetic term for both and $j^{\mu} = \phi^* \partial_{\mu} \phi + \phi \partial^{\mu} \phi^*$ in the interaction term: \begin{align} -\frac{1}{16 \pi} F_{\mu \nu} F^{\mu \nu} - \frac{1}{c}( \phi^* \partial_{\mu} \phi + \phi \partial^{\mu} \phi^*) A_{\mu} + (\partial_{\mu} - A_{\mu})\phi^*(\partial^{\mu} - A^{\mu})\phi + (m^2)\phi^* \phi \end{align}

But is there a lagrangian that is even closer? If not, why isn't this classical lagrangian used to do classical calculations? Doesn't it reproduce the dynamics that would follow from the Lorentz force?

To make more explicit what the lagrangian should yield: It should yield Maxwell's equations, coupled to sources, and additionally the behaviour consistent with a charge density that describes point particles with a trajectory $r_i$ by $j_i^{\mu}(x) = (c, \vec{v}_i(t))^T \delta(\vec{x} - \vec{x}_i(t))$.

Níckolas Alves
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Quantumwhisp
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  • https://en.wikipedia.org/wiki/Quantum_electrodynamics#QED_action and https://en.wikipedia.org/wiki/Scalar_electrodynamics#Lagrangian – Andrew Dec 01 '22 at 15:34
  • @Andrew since I already mentioned the scalar QED lagrangian in my question, I guess the answer to the question "is there anything closer than this" is "no"? – Quantumwhisp Dec 01 '22 at 16:02
  • You didn't write the right scalar lagrangian so I thought maybe you didn't know about minimal coupling. You need a term like $A_\mu A^\mu |\phi|^2$ as well coming from the covariant derivative. – Andrew Dec 01 '22 at 16:04
  • I will write it out more explicit. – Quantumwhisp Dec 01 '22 at 16:05
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    The QED and QED-scalar lagrangians are the answer to your question. You can view what you want to do -- replace the external current with dynamical fields -- in terms of the Noether procedure. This is an iterative procedure to discover consistent, gauge invariant lagrangians by adding terms order-by-order until you get a gauge invariant answer. For fermion-QED, the current is simply the Noether current for the $U(1)$ symmetry in the fermion case. For scalar-QED, there's a correction term proportional to $A^2$, which arises due to the derivative in the current. – Andrew Dec 01 '22 at 16:13
  • The "in matter" in the title suggests you might want to use a displacement tensor and a material law instead of just the $F_{\mu \nu} F^{\mu \nu}$ term.
  • From the Lagrangian you obtain the inhomogeneous Maxwell equations. What more exactly do you want to derive from it? Perhaps you specify the equations you want to obtain, this is unclear to me from what you write currently.
    • – kricheli Dec 01 '22 at 16:57
  • Related : Deriving Lagrangian density for electromagnetic field. – Frobenius Dec 01 '22 at 20:33
  • Similar question and my answer here: https://physics.stackexchange.com/questions/795485/does-classical-electrodynamics-have-a-lagrangian-that-gives-both-the-lorentz-for/795528 – Ján Lalinský Feb 04 '24 at 02:41