# Interaction of charges in gauge

Let's consider Maxwell theory:

$$\mathcal{L} = -F_{\mu\nu}F^{\mu\nu} = 2 A_\mu (\Box \eta^{\mu\nu} - \partial^\mu \partial^\nu) A_\nu$$

Is it possible to fix gauge $$A_0 = 0$$ and concider Lagrangian:

$$\mathcal{L} = 2 A_i (- \Box \delta^{ij} - \partial^i \partial^j) A_j$$

And do quntization of such theory in Lorentz non-invariant manner? Or there are some other barriers?

More concreatly, is it possible to calculate integration between 2 external charges due to gauge field?

I see obstruction in interaction therm for statical charges $$J^\mu = (q_1\delta(\vec r - \vec r_1)+ q_2\delta(\vec r - \vec r_2), 0 , 0, 0)$$ interaction therm $$A_\mu J^\mu = 0$$. Is possible to do such calculation in this gauge?

Even in classical field theory calculation, one cannot just substitute $$A_0=0$$ in the Lagrangian. Before gauge fixing, $$A_0$$, being non-dynamical Lagrange-multiplier-like variable, imposes with its EOM a (gauge-invariant) constraint like $$div\, \vec{E} = J^0$$ (which is the Gauss law). One should remember about this constraint after fixing the gauge to get sensible answers.
I believe that same applies to the quantum case: one should impose Gauss law constraint by hand as a requirement on physical states of the system. As another example of the same problem, recall how for Polyakov action in string theory after fixing the gauge for the $$g_{\mu\nu}$$ field we get just a system of $$d$$ massless free bosonic fields. Yet there, this is not the full story, as one should remember about EOMs for the metric, which is $$T=0$$; only after imposing this constraint we get correct answers.