I find to obtain $\nabla\cdot \vec E= 0$ where there is no electric charge or current, I need $$\vec E = \frac {\partial \vec A} {\partial t} - \nabla\phi ,$$
($\vec E = \nabla\phi - \frac {\partial \vec A} {\partial t} $ is obviously not suitable) where $\vec A=(A_1, A_2, A_3)$. If $$\vec E = \frac { \partial \vec A } { \partial t} - \nabla\phi = - \left( \frac {\partial \phi }{\partial x }- \frac{\partial A_1 }{\partial t }, \frac {\partial \phi }{\partial y }- \frac{\partial A_2 }{\partial t }, \frac {\partial \phi }{\partial z }- \frac{\partial A_3 }{\partial t }\right) $$
Applying the wave equation for $\phi $ and the Lorenz gauge condition $\nabla\cdot\vec A = \frac {\partial \phi} {\partial t}$ (see my previous question) , we find $\nabla\cdot \vec E= 0$. If there are no complex numbers or minus signs, then time and space have to be on opposite sides of the equation. This is true for example of the heat equation, the wave equation and the Lorenz gauge condition.
In the literature there is the equation
$$\vec E = - \nabla\phi - \frac {\partial \vec A} {\partial t} , $$
which does not seem to work. But
$$\vec E = \frac {\partial \vec A} {\partial t} - \nabla\phi, $$
does work. I am seeking an explanation.
Setting also (in my next question)
\begin{align}
\
\vec B &= - \nabla\times \vec A
\end{align}
Maxwell's equations can be largely derived. So why in the literature is $\vec E$ defined as
$$\vec E = - \nabla\phi - \frac {\partial \vec A} {\partial t}? $$
It seems this is wrong. Is that accurate?