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Need some help evaluating the following 4-gradient, that of the gradient of the field strength tensor

$$F^{\mu\nu}= \begin{bmatrix} 0 & -E_x & -E_y & -E_z\\\ E_x & 0 & -B_z & B_y \\\ E_y & B_z & 0 & -B_x \\\ E_z & -B_y & B_x & 0 \end{bmatrix}$$

such that $$\partial_\mu F^{\mu\nu} = \frac{4\pi}{c}j^\nu.$$

I know that the gradient has the form

$$\partial_\mu = (\frac{\partial}{\partial t}, \nabla).$$

If I am not mistaken, applying the 4-gradient to $j^\mu$ would give a dot product such that

$$\partial_\mu j^\mu = (\frac{\partial}{\partial t}, \nabla) \cdot (c\rho(x),j(x)) = \frac{\partial \rho}{\partial t} + \nabla\cdot j = 0. $$

When applying it to 4-vectors I can understand how the procedure works, nonetheless,when it comes to tensors I am at a complete loss. Any help would be greatly appreciated.

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Note: I will set $c=1$ here. You can restore $c$ by dimensional analysis, or (preferably) by working out the math on your own.

In the expression $\partial_\mu F^{\mu\nu}$, $\mu$ is implicitly summed over, so what we really have is $\sum_{\mu=0}^3 \partial_\mu F^{\mu\nu}$. This has one free index $\nu$, so we will have four components:

$$\nu=0: \quad \partial_0 F^{00} + \partial_1 F^{10} + \partial_2 F^{20} + \partial_3 F^{30}$$

$$\nu=1: \quad \partial_0 F^{01} + \partial_1 F^{11} + \partial_2 F^{21} + \partial_3 F^{31}$$

and so on. The $\nu=0$ component, if you substitute the expressions for the $F^{\mu\nu}$, works out to be $\partial_0 (0) + \partial_1 E_x + \partial_2 E_y + \partial_3 E_z = \nabla \cdot{E}$. For $\nu=1$ we get $-\partial_t E_x + \partial_y B_z - \partial_z B_y$, which we recognize as the $x$-component of $-\partial_t \mathbf{E} + \nabla \times \mathbf{B}$, and similarly for the other two components.

A quicker way to do this is to notice that the expression $\sum_\mu \partial_\mu F^{\mu\nu}$ has exactly the structure of a row vector $\partial$ multiplied on the right by a matrix $F$:

$$(\partial_t, \partial_x, \partial_y, \partial_z) \cdot \begin{pmatrix} 0 & -E_x & -E_y & -E_z\\\ E_x & 0 & -B_z & B_y \\\ E_y & B_z & 0 & -B_x \\\ E_z & -B_y & B_x & 0 \end{pmatrix}$$

so you can just use the rules of matrix multiplication to do it and not have to worry about identifying components and risk making sign mistakes and all that.

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I think you just have to remember the Einstein summation convention : $$\partial_{\mu}F^{\mu\nu} = \sum_{\mu=t,x,y,z}\frac{\partial F^{\mu\nu}}{\partial \mu}$$

Then, for each $\nu$, you just have to write your sum term by term. You will end up with four equations, which you should be able to identify as Maxwell-Gauss and Maxwell-Ampere to conclude.

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