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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.

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  • $\begingroup$ Start with the last expression and write $\mu$ as $0$ and $i$, and $\nu$ as $0$ and $j$. Then, the antisymmetry implies that $F^{00}=0$. $\endgroup$
    – flippiefanus
    Commented May 5, 2022 at 3:26
  • $\begingroup$ @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? $\endgroup$
    – Chidi
    Commented May 5, 2022 at 3:34
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    $\begingroup$ $\partial_jF^{ij}-\partial_0F^{0i}=\partial_jF^{ij}+\partial_0F^{i0}=\partial_\mu F^{i\mu}=J^i$ 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 $3$?! $\endgroup$
    – N0va
    Commented May 5, 2022 at 5:10
  • $\begingroup$ @N0va Do you want to put down your comment as an answer? Because I think you just solved it for me $\endgroup$
    – Chidi
    Commented May 5, 2022 at 5:26

1 Answer 1

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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.

I.e. the equation at the top is comprises of/is equivalent to the two below.

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  • $\begingroup$ I think you're working backwords from the last expression to the penultimate pair. Like I mentioned in the comments, I know the last equation holds; what I want to figure out is how to get to that last equation from the pair of equations above it. $\endgroup$
    – Chidi
    Commented May 5, 2022 at 5:32
  • $\begingroup$ You can easily make your way backwards through the calculation... $\endgroup$
    – kricheli
    Commented May 5, 2022 at 5:34
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    $\begingroup$ @Chidi \begin{equation} \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} \end{equation} The first equivalence is obvious and kricheli has proven the second equivalence. $\endgroup$
    – Filippo
    Commented May 5, 2022 at 6:22
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    $\begingroup$ @Filippo Alright, I think with the answer and all the comments I am able to piece together the reasoning behind the derivation. Thank you $\endgroup$
    – Chidi
    Commented May 5, 2022 at 7:12