# Effect of gauge-fixing via Lagrange multipliers on Euler-Lagrange equations

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?

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.

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.