4
$\begingroup$

This question already has an answer here:

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?

$\endgroup$

marked as duplicate by Qmechanic lagrangian-formalism Jan 7 at 7:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ I am a little confused as to why didgeridoo92's answer and yours do not match with regards to the sign. By the Euler-Lagrange equation, $-\partial_\mu\left(\frac{\partial L}{\partial A_{\nu,\mu}}\right)+\frac{\partial L}{\partial A_\nu} = 0$ followed by substitution of $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$, wouldn't one get $\partial^\mu \partial_\mu A^\nu - \partial^\nu\partial^\mu A_\mu-m^2 A_\mu=-\beta J_\mu$, which then gives us: $(\square-m^2)A_\nu=-\beta J_\nu$. Apologies for playing around with the indices, I'm quite new to this kind of notation.Thank you for helping! $\endgroup$ – dnninja May 3 '16 at 7:45
  • $\begingroup$ yes, that corresponds to the signs present in "my" equations of motion. And depending on your definition of $\square$, I would rather write it in the more familiar form $$(\square + m^2) A_\mu = \beta j_\mu$$ which then looks like a generalization of the Klein-Gordon equation. But again: Don't forget the constraint! Otherwise it is not equivalent to Proca's equation. $\endgroup$ – schmui May 3 '16 at 7:55
  • $\begingroup$ How would one, then, go about finding the stress tensor for such a field? $\endgroup$ – dnninja May 3 '16 at 8:24
  • $\begingroup$ That is a bit of a tricky task. In principle you'd have to do a variation of the Lagrangian w.r.t. to the metric. If you google stress-energy tensor or energy moment tensor in classical field theories, you will find detailed information. For example arxiv.org/ftp/hep-th/papers/0307/0307199.pdf or physics.umd.edu/courses/Phys624/agashe/F10/solutions/HW1.pdf for an application in electromagnetism $\endgroup$ – schmui May 3 '16 at 9:05
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.