I Coulomb gauge Lorenz invariant?

Sebas4
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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.
 
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Sebas4 said:
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.
 
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.
 
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