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

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?

• 1. Since you haven't really shown us what you are doing, it's not clear what you're asking. 2. Even if you showed us, check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. So you should try to identify the step in your computation where it seems to go wrong, and ask specifically about that. – ACuriousMind Jul 2 '17 at 11:16
• (Presumably) related question by OP: physics.stackexchange.com/q/341326/50583 – ACuriousMind Jul 2 '17 at 11:20
• Well spotted. My two questions are related. In the first question I changed a plus into a minus. In the second I am changing a minus into a plus. The second is more ambitious. The first was only on wikipedia. The second is the world literature. – Stephen Wynn Jul 2 '17 at 15:53
• Your Lorenz gauge condition should have a minus sign. And as a tip, consider asking "where is my mistake?" instead of "why is everyone wrong?". – Javier Jul 3 '17 at 12:04
• Please stop editing this question and thereby unnecessarily bumping it to the front page if you don't have anything substantial to add. This post is at its 23rd revision without substantially differing from its 10th. – ACuriousMind Jul 15 '17 at 14:09

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.

• $$\nabla\cdot\vec A+\frac{\partial \phi}{\partial t}=0$$ is not Lorentz invariant. You have just added derivatives. – Stephen Wynn Jul 6 '17 at 9:00
• @StephenWynn As shown in the answers to the other thread, that combination is Lorentz invariant, but you've shown repeatedly that you're uninterested in understanding why you are mistaken, so I will not engage in a back-and-forth over this. I provided this answer to provide a clear on-the-record throwback to show what the issue is (i.e. the same as in your previous question); take it as-is or leave it, it's up to you. – Emilio Pisanty Jul 6 '17 at 9:49
• The Lorenz gauge condition is that the Lorenz divergence of the 4 potential is zero. $\nabla\cdot\vec A+\frac{\partial \phi}{\partial t}$ is not the Lorenz divergence. – Stephen Wynn Jul 6 '17 at 10:19
• The Lorentz divergence is defined on Wikipedia under divergence at the bottom of the page. – Stephen Wynn Jul 6 '17 at 10:28
• @Stephen that definition is wrong, as far as I can see. See en.wikipedia.org/wiki/Lorenz_gauge_condition. – Javier Jul 11 '17 at 11:40

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 $( A_1,A_2,A_3)=-(A^1,A^2,A^3)$.

• Did you answer your own question expecting approval? – Mauricio Nov 22 '17 at 16:00
• This is less wrong than your previous attempts, but it is still not right. It is correct that $(A_1,A_2,A_3)=-(A^1,A^2,A^3)$, but it is *not* correct that $\vec A$ as a three-vector has two different signs (which is plainly nonsense). Instead, the correspondence works as $A^\mu=(\phi,\vec A)$ and $A_\mu=(\phi,-\vec A)$ (setting $c=1$, with signature $+---$). – Emilio Pisanty Nov 22 '17 at 18:32