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How am I supposed to interpret this notation:

$$F^{uv} = \partial^uA^v-\partial^vA^u$$

I know that $\partial^u = (\frac{1}{c}\frac{\partial}{\partial t},- \vec\nabla)$

So for example for the potential $$A=\left(\begin{matrix} 0 & 0 & 0& E_z\\ 0 & 0 & B_y & 0\\ 0 & -B_x & 0 & 0\\ E_z & 0&0&0\end{matrix}\right)$$

So to compute $F^{23} = \partial^u\left(\begin{matrix}0\\B_y\\0\\0\end{matrix}\right) -\partial^v \left(\begin{matrix}0&0&B_y&0\end{matrix}\right) = 0 - - B = B$.

Is this the correct way to do it? I'm just getting confused by the notation.

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The indices come with ordering $(0,1,2,3)$ so that $\partial^0=\frac{1}{c}\frac{\partial}{\partial t}$, $\partial^1=-\frac{\partial}{\partial x}$ etc. $A^\mu$ is a 4-vector with components $(A^0,A^1,A^2,A^3)$, not a matrix as your notation suggest. Thus, in your specific example, $$ F^{23}=\partial^{2}A^3-\partial^3 A^2= -\frac{\partial}{\partial y}A_z+\frac{\partial}{\partial z}A_y\, . $$

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The 4-potential $A_\mu$ is a four-vector, not a matrix. Set the speed of light $c=1$ and it is defined as $$ A^\mu = (\phi,\vec{A})\,, $$ in which $\phi$ is the electric potential and $\vec{A}$ vector potential. The magnetic field is given by $$ \vec{B} = \nabla \times \vec{A}\,, $$ and the electric field is given by $$ \vec{E} = -\frac{\partial \vec{A}}{\partial t}-\nabla \phi\,. $$ Thus the electromagnetic tensor (as you write in the question description) is $$ F^{\mu\nu} = \partial^\mu A^\nu-\partial^\nu A^\mu = \left(\begin{array}{cccc} 0&-E_1&-E_2&-E_3\\ E_1&0&-B_3&B_2\\ E_2&B_3&0&-B_1\\ E_3&-B_2&B_1&0 \end{array}\right)\,. $$ This tensor is gauge invariant. Under the transformation $A_\mu\rightarrow A_\mu+\partial_\mu\chi$, $F^{\mu\nu}$ will not change.

The Lagrangian of electromagnetic field is $$ \mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + A_\mu J^{\mu} $$ and the corresponding Euler-Lagrange equation is Maxwell equation $$ \frac{\partial \mathcal{L}}{\partial A_\nu}-\partial_\mu\frac{\partial \mathcal{L}}{\partial \partial_\mu A_\nu}=0~~\Rightarrow ~~\partial_\mu F^{\mu\nu} = J^\nu\,. $$

Actually, the 4-vector potential $A_\mu$ can be viewed as the connection on a fiber bundle (something like the Christoffel symbol in general relativity), and thus the tensor $F_{\mu\nu}$ is the curvature. In quantum field theory we have the covariant derivative $D_\mu=\partial_\mu + ieA_\mu$, it is similar to the covariant derivative on curved spacetime $\nabla_\mu=\partial_\mu+\Gamma_{\mu\nu}^\lambda$.

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