How is the index notation for the electromagnetic potentials defined?

Question

I don't understand, how the $\vec{A}$ potential is defined in the covariant formulation of the Maxwell-Equations.

I am using the metric $(+,-,-,-)$.

The electric field (SI-Units) is defined as: $$ \vec{E}=-\vec{\nabla}\phi-\partial_t\vec{A}.\tag{1} $$ In index notation, I would write this as: $$ E^i/c=-(\vec{\nabla}A^0)^i-\partial_0A^i.\tag{2} $$ And with the definition of the 4-Gradient: $$ E^i/c=-E_i/c=-\partial_iA_0-\partial_0A^i=\partial^iA^0-\partial^0A^i=F^{i0}.\tag{3} $$ But this is exactly the opposite of the field tensor in Wikipedia: $$ E_i/c=F_{0i}=-F^{0i}=F^{i0}.\tag{4} $$

What is going wrong here?

Answer 1

You need to be very careful with upper and lower indices.

The 4-position (with upper index) and the 4-gradient (with lower index) are: $$x^\mu=(ct,\vec{x}) \tag{1}$$ $$\partial_\mu=\frac{\partial}{\partial x^\mu} =\left(\frac{\partial}{c\ \partial t},\vec{\nabla}\right) \tag{2}$$

The electromagnetic 4-potential with upper index is defined as $$A^\alpha = \left( \frac{\Phi}{c},\vec{A} \right) \tag{3}$$ and hence with lower index it is (by index lowering with the metric) $$A_\alpha = \left( \frac{\Phi}{c},-\vec{A} \right) \tag{4}$$

The electromagnetic tensor with lower indices is defined as: $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu \tag{5}$$

From this, for $\mu=0$ and $\nu=i\in\{1,2,3\}$, using (2) and (4), and $A_i$ meaning the $i$-component of $\vec{A}$, we get $$\begin{align} F_{0i}&=\frac{\partial}{c\ \partial t} (-A_i) -\nabla_i \frac{\Phi}{c} \\ &= \frac{1}{c} \left(-\frac{\partial\vec{A}}{\partial t}-\vec{\nabla}\Phi \right)_i \\ &= E_i/c \end{align} \tag{6}$$ in agreement with Wikipedia.

_{In the above I have adopted the convention to use greek indices ($\alpha,\mu,\nu\in\{0,1,2,3\}$) for 4 dimensions and latin indices ($i\in\{1,2,3\}$) for 3 dimensions.}