Why is $\vec{E}$ defined as $-\nabla\phi - \partial\vec{A}/\partial t$?

Question

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?

Answer 1

Why does the literature use the identity $$ \vec E = - \nabla\phi - \frac {\partial \vec A} {\partial t}? $$ Because it's the only way to make $\vec B=\nabla\times\vec A$ consistent with the Faraday induction law: if you know that $\vec B=\nabla\times \vec A$ (which you can always set, given the magnetic Gauss law) then the Faraday law reads \begin{align} \nabla \times \vec E &=-\frac{\partial}{\partial t}\vec B =-\frac{\partial}{\partial t}\nabla\times \vec A \\& =-\nabla\times \frac{\partial\vec A}{\partial t} \\ \implies \nabla\times\left(\vec E+ \frac{\partial\vec A}{\partial t}\right)&=0, \end{align} which implies that $\vec E+ \frac{\partial\vec A}{\partial t}$ is irrotational and you can therefore find some function $\phi$ (whose sign is arbitrary but chosen here as negative because conventions) which satisfies $$ \vec E+ \frac{\partial\vec A}{\partial t} = -\nabla\phi, $$ or in other words $$ \vec E = -\nabla\phi - \frac{\partial\vec A}{\partial t} $$ as used in the literature.

Now, your version disagrees with the literature on the sign of $\phi$, which is thus far arbitrary and unconnected to anything else, so the missing joint needs to be on the first nontrivial connection of $\phi$ with the rest of the formalism.

This can only be the Lorenz gauge, and a closer examination shows that you're misapplying it: the condition reads $$ \nabla\cdot\vec A+\frac{\partial \phi}{\partial t}=0 $$ instead of your version with a flipped sign on $\phi$. This is the only possible covariant combination of the spatial and temporal derivatives, and the proof goes along the lines of Qmechanic's comment:

Hint: Distinguish between upper and lower Lorentz indices.

If what you want is help understanding why the Lorenz gauge is as in the literature and not as in your version, I would suggest actually asking that instead of steadfastly insisting that you're right and everyone else is wrong.

Answer 2

Both

$$ E = - \frac {\partial \vec A} {\partial t} - \nabla\phi ,$$

and

$$ E = + \frac {\partial \vec A} {\partial t} - \nabla\phi $$

are true. The difference is that in the first equation we are using upper indices for $\vec A$ and in the second lower, and $ ( A_1,A_2,A_3)=-(A^1,A^2,A^3) $.