# Expressing Maxwell's equations in tensor notation

I've been teaching myself relativity by reading Sean Carroll's intro to General Relativity textbook, and in the first chapter he discusses special relativity and introduces the concept of tensors, which I am still very new to. Near the end of the chapter, he introduces the electromagnetic field strength tensor: $$F_{\mu\upsilon} = \left( \begin{matrix} 0 & -E_1 & -E_2 & -E_3\\ E_1 & 0 & B_3 & -B_2\\ E_2 & -B_3 & 0 & B_1\\ E_3 & B_2 & -B_1 & 0\\ \end{matrix} \right) = -F_{\upsilon\mu}$$

He then writes the four Maxwell equations in tensor notation, using the elements of the above field tensor: $$\bar{\epsilon}^{ijk}\partial_jB_k - \partial_0E^i = J^i\\ \partial_iE^i = J^0\\ \bar{\epsilon}^{ijk}\partial_jE_k + \partial_0B^i = 0\\ \partial_iB^i = 0.$$ Next, by showing that the field tensor can be represented by the two tensor equations $$F^{0i} = E^i$$ and $$F^{ij} = \bar{\epsilon}^{ijk}B_k$$, he proposes that the first two of Maxwell's equations can be written as: $$\partial_jF^{ij} - \partial_0F^{0i} = J^i\\ \partial_iF^{0i} = J^0$$ Finally, he proposes that by using the antisymmetry of $$F_{\mu\upsilon}$$, the above two equations can be reduced to the single equation: $$\partial_\mu F^{\upsilon\mu} = J^{\upsilon}$$ My question is, can someone show me how to use the antisymmetry of $$F_{\mu\upsilon}$$ to derive the last equation from the penultimate pair of equations? Note that in this context, $$J$$ is the current 4-vector in Gaussian form, $$J = (\rho, J^x, J^y, J^z)$$, $$\bar{\epsilon}^{ijk}$$ is the Levi-Civita symbol in spatial coordinates, and Latin subscripts and superscripts refer to spatial coordinates while Greek subscripts and superscripts refer to spacetime coordinates.

• Start with the last expression and write $\mu$ as $0$ and $i$, and $\nu$ as $0$ and $j$. Then, the antisymmetry implies that $F^{00}=0$. Commented May 5, 2022 at 3:26
• @flippiefanus I understand that the last equation holds; what I don't understand is how to derive it from the last pair of equations involving the partial derivatives of $F$ and the 4-current. In other words, if you did not know that they could be reduced to the final equation, how would you go about simplifying them? Commented May 5, 2022 at 3:34
• $\partial_jF^{ij}-\partial_0F^{0i}=\partial_jF^{ij}+\partial_0F^{i0}=\partial_\mu F^{i\mu}=J^i$ and $\partial_i F^{0i}=\partial_i F^{0i}+\partial_0F^{00}=\partial_\mu F^{0\mu}=J^0$ and thus $\partial_\mu F^{\nu\mu}=J^\nu$ where we used $F^{\mu\nu}=-F^{\nu\mu}$, sum convention and Latin indices running from $1$ to $3$ and Greek indices running from $0$ to $3$?!
– N0va
Commented May 5, 2022 at 5:10
• @N0va Do you want to put down your comment as an answer? Because I think you just solved it for me Commented May 5, 2022 at 5:26

The equation $$\partial_\mu F^{\nu\mu} = J^\nu$$ is to hold for all $$\nu$$, i.e. for $$\nu=0$$ and for all $$\nu=i$$. For $$\nu=0$$ this reads $$\partial_\mu F^{0\mu} = J^0 = \partial_i F^{0 i}$$ because the term with $$\mu=0$$ vanishes due to antisymmetry of $$F$$. For $$\nu=i$$ you have $$\partial_\mu F^{i\mu} = J^i = \partial_j F^{ij}+\partial_0 F^{i0} = \partial_j F^{ij}-\partial_0 F^{0i}$$ where in the last step again we have used the antisymmetry.
• @Chidi $$\forall\upsilon\in\{0,1,2,3\}:\sum_{\mu=0}^3\partial_\mu F^{\upsilon\mu} = J^{\upsilon}\Leftrightarrow\begin{cases}\displaystyle\sum_{\mu=0}^3\partial_\mu F^{0\mu} = J^0\\\forall i\in\{1,2,3\}:\displaystyle\sum_{\mu=0}^3\partial_\mu F^{i\mu} = J^i\end{cases}\Leftrightarrow\begin{cases}\displaystyle\sum_{j=1}^3\partial_j F^{0j} = J^0\\\forall i\in\{1,2,3\}:\displaystyle\sum_{j=1}^3\partial_jF^{ij} - \partial_0F^{0i} = J^i\end{cases}$$ The first equivalence is obvious and kricheli has proven the second equivalence. Commented May 5, 2022 at 6:22