# Coulomb gauge choice: Does $A_0=0$ imply that we also need to choose $\nabla \cdot \vec{A} =0$ from the EOM of $A_0$?

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)?

• 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. Jul 13, 2022 at 9:26