Do gauge fields violate causality? For example, take $$V(\textbf{r},t)=\frac{1}{4\pi \epsilon}\int\frac{\rho(\textbf{r},t)}{|\textbf{r}-\textbf{r}'|}d^3r'.$$ This integral is purely spatial, the time of the density corresponds to the scalar potential. So the distribution of charge determines $\phi$. This seems to imply a signal traveling faster than light. Is this a violation of causality?
 A: 
the time of the density corresponds to the scalar potential.

This is not correct, precisely because of causality considerations. If the charge distribution is time-dependent, the potential at time $t$ will instead depend on the configuration at some retarded time, $t_r=t-\frac{|\boldsymbol{r}-\boldsymbol{r}'|}{c}$.
A: In the Lorenz gauge $\vec{\nabla}\cdot\vec{A}+\frac{1}{c^2}\frac{\partial V}{\partial t}=0$, the Maxwell equations in terms of the potentials are, for $V$:
$$-\frac{1}{c^2}\frac{\partial^2V}{\partial t^2}+\nabla^2V=-\frac{\rho}{\epsilon_0},\tag{1}$$
where $\rho=\rho(\vec{x},t)$ is the charge distribution. The general solution to this is
$$V(\vec{x},t)=V_0(\vec{x},t)+\frac{1}{\epsilon_0}\int d\vec{x}' dt'\,G(\vec{x},t;\vec{x}',t')\rho(\vec{x}',t'),\tag{2}$$
where $V_0(\vec{x},t)$ is the solution to the homogeneous part of $(1)$ (it gives the potential not produced by the charge distribution, so let me omit it in the future), and $G(\vec{x},t;\vec{x}',t')$ is the Green function, which satisfies
$$-\frac{1}{c^2}\frac{\partial^2G}{\partial t^2}+\nabla^2G=-\delta^3(\vec{x}-\vec{x}')\delta(t-t').\tag{3}$$
If you solve that equation with Fourier's help, you get
$$G(\vec{x},t;\vec{x}',t')=\frac{1}{4\pi|\vec{x}-\vec{x}'|}\delta(t'-t_{ret}),\tag{4}$$
where $$t_{ret}=t-\frac{|\vec{x}-\vec{x}'|}{c}.\tag{5}$$ This means that the Green function "is activated" at $(\vec{x},t)$ $\frac{|\vec{x}-\vec{x}'|}{c}$ seconds after something happens on $(\vec{x}',t')$, i.e. the time that light needs to travel from one point to the other. With this, $(2)$ takes the form
$$V(\vec{x},t)=\frac{1}{4\pi\epsilon_0}\int d\vec{x}'dt'\,\frac{\rho(\vec{x}',t')}{|\vec{x}-\vec{x}'|}\delta(t'-t_{ret})=\frac{1}{4\pi\epsilon_0}\int d\vec{x}'\,\frac{\rho(\vec{x}',t_{ret})}{|\vec{x}-\vec{x}'|}.\tag{6}$$
