4

Provided an action: $$S[A_\nu] = \int\left(\frac{1}{4\mu_0}(A_{\gamma,\mu}-A_{\mu,\gamma})(A_{\zeta,\alpha}-A_{\alpha,\zeta})\eta^{\gamma\zeta}\eta^{\mu\alpha}+\frac{1}{2}\nu^2A_\mu A_\gamma -\beta A_\mu J^\mu\right)\sqrt{-\eta}~dtd^3x,$$

How would one go about finding the field equations for the same? I do understand that using the Euler-Lagrange method is how one should start out.

Does it tell us what kind of field we're looking at, at a glance?

dnninja
  • 43

3 Answers3

3

As already mentioned, the action corresponds to massive electrondynamics, including external sources, in Minkowski spacetime. This is also known under the term Proca action.
As you mention, the corresponding equations of motion can be found using Euler-Lagrange equation, that is $$0 = \frac{\partial \mathcal{L}}{\partial A_\mu} - \partial_\nu \frac{\partial \mathcal{L}}{\partial (\partial_\nu A_\mu)}$$ Both terms are calculated rather straightfowardly and you should and up with the equations of motion $$0=\partial^\nu F_{\nu \mu} - m^2 A_\mu + \beta j_\mu$$ or $$0= \partial^\mu \partial_{[\mu} A_{\nu]} - m^2A_\mu + \beta j_\mu$$ I have put in the field strenght tensor $F = dA$, or $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ for simplicity and substituted $m^2$ for $\nu^2$ since I think $\nu$ is kind of reserved for coordinate indices (in that sense, your definition is also a bit formally incorrect as on the right hand side you write $S[A_\nu]$). Writing down the Lagrangian with the field strenght tensor, I think, also allows for a more direct identification of the underlying theory.
Proca fields are the most straightforward massive generalizations of vector fields and can be studied also in a quantum mechanical, or quantum field, context. But note that, unlike Maxwells equation, Proca's equation are not gauge invariant, as the symmetry is broken by the mass term!

EDIT: Note, that in another answer it is stated that the equations of motions are $(\square - m^2)A_\mu = \beta j_\mu$, but this is a priori incorrect. However, if you try to solve Proca's equation you find that it is actually equivalent to the mentioned wave equation PLUS a constraint, $m^2 \partial^\nu A_\nu = \beta \partial^\nu j_\nu$. This is in a sense analogue to the case in electrodynamics where you find that Maxwell's equations can be solved by solving a wave equation (of the four-vector potential) and a constraint (the Lorenz constraint).

1

by the Euler-Lagrange method you would simply get the following field equation:

$(\square-\nu^2)A_\alpha=-\beta J_\alpha$,

which is the Proca equation. You can read up about the Proca action online. I'm new to field theory, and someone should correct me if I am wrong.

0

This is just the lagrangian for electromagnetism. The A is the vector potential and the expressions in parenthesis are the F tensor. You can read about it here

magma
  • 1,552