Does Coulomb gauge imply constant density?

Question

Say we have $$\Box A = J$$

and $$\nabla \cdot A = 0\;.$$

Then $$0 = \Box (\nabla \cdot A) = \nabla \cdot J\;.$$

But, $$\nabla \cdot J - \partial_t \rho = 0\;.$$

So $$ \partial_t \rho = 0\;.$$

Thus, $$\rho = const.$$

But that doesn't make sense to me, especially when you consider a nonlinear system like Dirac-Maxwell. Where is my thinking wrong?

EDIT: As the answers point out the confusion stemmed from the fact that a gauge was already fixed when writing $\Box A = J$. However I do want to point out that orginially I was thinking about Coulomb gauge in which one has

$$\Box A = -\mathbb{P} J$$

where $\mathbb{P} = I - \nabla \Delta^{-1} \nabla$ projects onto divergence free vector fields since in the Coulomb gauge

$$\Box A = -J + \nabla \partial_t V$$

but $\nabla \partial_t V$ is the rotation-free part of $J$ as a quick calculation shows. In this gauge, taking the divergence of $\Box A = -\mathbb{P} J$ is not problematic since one applies $\nabla \cdot$ to the divergence free part of the current only. The rotation-free part of $J$ has nonzero divergence.

Answer 1

You secretly impose 2 gauge fixing conditions:

$$\nabla \cdot \vec{A} = 0$$ $$\nabla \cdot \vec{A} + \mu_0 \epsilon_0\frac{\partial \varphi}{\partial t} = 0$$

The latter coming from that fact you use the wave equation,which is implied by the lorenz gauge fixing condition.

As a result of applying 2 gauge fixing conditions, you are finding conditions on $\rho$ such that the field generated by a charge density $\rho$ satisfies both gauge fixing conditions.

As a result

$$\rho = c$$

When $\rho$ is constant in time, the lorenz gauge satisfies the coulomb gauge.

Is this a general condition? No, in general the lorenz gauge does not satisfy the coulomb gauge and thus $\rho$ in general is not constant.

When solving use 1 gauge fixing condition.

Answer 2

In order for the wave equation to be valid, the vector potential has to satisfy Lorenz gauge condition $$\vec{\nabla}\cdot\vec{A}+\frac{1}{c^2}\frac{\partial\varphi}{\partial t}=0$$ You cannot impose Coulomb gauge condition as well.

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As explained by @jensen paull after fixing Lorenz gauge condition, you can fix a supplementary condition, provided that the generating function of your gauge transformation satisfies the wave equation itself. I naively thought this could never be done with Coulomb gauge but it turns out to be possible, so my answer was wrong.