Coulomb gauge Lorenz invariant?

[Sebas4]

TL;DR

What is meant by Coulomb gauge not being Lorenz invariant?

Hey,

What is meant by Coulomb gauge not being Lorenz invariant?

The Coulomb gauge is just a constraint on \mathbf{A} and \phi and thus it is independent of inertial frame.

I posted the question in the wrong section. This question is in the context of QFT. The notes says:
A disadvantage of working in Coulomb gauge is that it breaks Lorentz invariance.

 

[Dale]

TL;DR Summary: What is meant by Coulomb gauge not being Lorenz invariant?

What is meant by Coulomb gauge not being Lorenz invariant?

It means that if you have some four-potential $A^\mu=(\phi,\vec A)$ where $A^\mu$ satisfies the Coulomb gauge condition in some unprimed inertial frame, then $A^{\mu'}=\Lambda^{\mu'}_\mu A^\mu$ generally will not satisfy the Coulomb gauge condition in the primed inertial frame. It will still be a perfectly valid four-potential in the primed frame, but just not in the Coulomb gauge.

 

[vanhees71]

Of course you can write the Coulomb gauge in a manifestly covariant way.

The point is that usually you take an arbitrary inertial frame $\Sigma^{*}$ and write the Coulomb-gauge condition in (1+3)-notation as
$$\vec{\nabla}^* \cdot \vec{A}^*=0.$$
You can make this manifestly covariant by introducing the four-vector with components $U^*=(1,0,0,0)$ in this frame.

Then the Coulomb-gauge condition in a general frame reads
$$\partial_{\mu} (A^{\mu}-U^{\mu} U^{\nu} A_{\nu})=0.$$
The point is that you now have introduced a preferred inertial reference frame to define your gauge constraint.

Using this covariant notation, you get a manifestly covariant photon propagator, containing the $U^{\mu}$ of course. Now the important point is that due to gauge-invariance for any physically observable quantities like S-matrix elements for scatterings between photons and electrons+positrons in standard spinor QED the frame-dependent terms, i.e., those containing $U^{\mu}$ cancel thanks to the Ward identities.

The advantage of the Coulomb gauge is that you have a complete gauge fixing and no unphysical degrees of freedom. The disadvantage is this "fictitious breaking of Lorentz symmetry" due to the introduction of an arbitrary reference frame, i.e., the vector $U^{\mu}$.

You can also use a manifestly covariant gauge like the Landau gauge, demanding $\partial_{\mu} A^{\mu}=0$, but this fixes the gauge only partially, and you have to deal with unphysical degrees of freedom like longitudinal and timelike photons, which however also cancel using the Gupta-Bleuler formalism. The advantage is that there's no arbitrary preferred frame and the Feynman rules lead to simpler expressions.