2
$\begingroup$

Preamble

Consider the Lagrangian density for electrodynamics: $$L=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}-A_\mu J^\mu\tag{1}$$ With the usual definition of $F_{\mu\nu}=\partial_\mu A_\nu - \partial_\nu A_\mu$. This results in the Euler-Lagrange equations: $$\partial_\rho F^{\rho\mu}=J^\mu\tag{2}$$ From which we can determine that $$\partial_\mu J^\mu=0,\tag{3}$$ so the vector field can only be consistently coupled to a conserved current. Now let's say we want to fix the gauge via a Lagrange multiplier, using Coulomb gauge. Thus: $$L=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}-A_\mu J^\mu+\lambda\partial_i A_i.\tag{4}$$ The Euler-Lagrange equations now are: $$\partial_i F^{i0}=J^0,\tag{5}$$ $$-\partial_0 F^{k0}+\partial_i F^{ik}-\partial_k \lambda=J^k,\tag{6}$$ $$\partial_i A_i=0.\tag{7}$$ We have Coulomb gauge, but it looks like the equations of motion have changed with the appearance of the Lagrange multiplier field $\lambda$ in equation (6).

In fact though, using the condition that $\partial_\mu J^\mu=0$, one can show, by adding the time derivative of (5) to the divergence of (6), that: $$\partial_i \partial_i \lambda =0.\tag{8}$$ So that (at least with appropriate boundary conditions?) $\lambda$ is constant and the equations of motion are not changed after all.

Question

My question is, does this hold generally? Given a gauge theory (e.g., $SU(N)$ Yang-Mills), if one adds the gauge condition as a constraint with a Lagrange multiplier, are the resulting Euler-Lagrange equations always equivalent to the original ones, just supplemented with the gauge condition?

$\endgroup$
0

2 Answers 2

2
$\begingroup$

OP has correctly identified the mechanism in their example: In order for OP's action (4) to be gauge invariant, we must assume the continuity equation (3). Eq. (3) in turn implies that the Lagrange multiplier $\lambda$ is a harmonic function (8), so that with appropriate boundary conditions, $\lambda$ disappears from the Euler-Lagrange (EL) eq. (6).

More generally, one may show that in any gauge theory, if a gauge fixing is complete, i.e., the gauge functions are determined uniquely by the gauge conditions, the EL equations derived from the gauge-fixed action are equivalent to those derived from the original action supplemented with the gauge conditions, cf. Ref. 1.

For other examples, see e.g. my Phys.SE answers here & here.

References:

  1. H. Motohashi, T. Suyama & K. Takahashi, Phys. Rev. D 94 (2016) 124021, arXiv:1608.00071.
$\endgroup$
0
$\begingroup$

The answer is 'yes' under the condition that $\lambda$ is independent of $x^\mu$. If $\lambda$ depends on $x^\mu$ it contributes in the equations of motion and the conservation laws.

Note that $\lambda$ is not constant. If it were dL/d$\lambda$ would be meaningless and Eq. (7) would not follow.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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