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.


2 Answers 2


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.