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How to justify the Coulomb gauge fixing condition choice with $$ A_0=0, \quad \nabla \cdot \vec{A} =0? $$

Below in the text image, I find a text explaining that imposing $A_0=0$ is always possible because I think we set $$ A_\mu \mapsto A_\mu' = A_\mu + \partial_\mu \omega. $$

We can choose $A_0 \mapsto A_0' = A_0 + \partial_0 \omega =0$. This means finding $\omega$ such that $A_0 + \partial_0 \omega =0$.

Then we impose the equation of motion of $A_0$, which seems to be, at least for pure Maxwell theory, to $$ \partial_j(\partial_j A^0 - \partial_0 A^j )=0. $$

Even if $A^0 =0$, we have still $$ \partial_j( \partial_0 A^j )=0. \tag{1} $$

This seems not implying that the Coulomb gauge condition requires $$ \partial_j A^j =\nabla \cdot \vec{A} =0. \tag{2} $$

How to show that (1) can deduce (2)?

p.s. See also a different condition is discussed in https://physics.stackexchange.com/a/33140/310987

enter image description here

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  • $\begingroup$ Related: physics.stackexchange.com/q/544133 $\endgroup$
    – Buzz
    Jul 13, 2022 at 1:56
  • $\begingroup$ I don't quite understand the question: 1. No one is saying that the equation of motion implies the gauge choice (i.e. $(1)\implies (2)$). If it did, it would not be a choice, would it? Why do you think you can deduce the gauge choice from an equation of motion? Why do we need to "justify" a gauge choice? 2. The answer you link discusses the exact same equation of motion, because $E^i = F^{i0} = \partial_i A^0 - \partial_0 A^i$. 3. Where is the screenshot at the end from? Please always cite your sources and type out text you want to quote so that it can be indexed by search engines. $\endgroup$
    – ACuriousMind
    Jul 13, 2022 at 9:26
  • $\begingroup$ It is from a lecture note - for example p.2 of damtp.cam.ac.uk/user/tong/gaugetheory/72d.pdf $\endgroup$ Jul 14, 2022 at 1:10

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