2

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 $j_i^{\mu}(x) = (c, \vec{v}_i(t))^T \delta(\vec{x} - \vec{x}_i(t))$.

Níck Aguiar Alves
  • 24,192
Quantumwhisp
  • 7,095

1 Answers1

4

Answering the title question: yes, but the exact Lagrangian depends on the dynamics.

If you want to describe a classical scalar field interacting with the electromagnetic field (for whatever reason), then you want the Lagrangian for scalar electrodynamics. If you want to describe fermions, then you need the Lagrangian usually used in quantum electrodynamics. This does not mean you need to quantize the fields: the action for QED will lead you to the theory in relativistic quantum mechanics of an electron interacting with the electromagnetic field. This is different from the theory of a classical scalar field interacting with the electromagnetic field.

If you want the theory for a charged particle interacting with the electromagnetic field, then you should add to the action the kinetic and mass terms for the particle (and eventual forces that you might want to consider) and the current in Maxwell's action will be the usual expression for the four-current of a point-particle.

In short, yes, you can get a Lagrangian describing both the electromagnetic field and the current. However, it does depend on your model for charged matter, as it should be: a charged fluid is expected to behave very differently from a point particle with finite charge. Different models for the source require different Lagrangians.

Níck Aguiar Alves
  • 24,192