Let us start from $\textbf{B}=\nabla \times \textbf{A}$ and write its components $B_k=\epsilon_{ijk}\partial_i A_j$.

I want to show that $\partial_i A_j - \partial_j A_i = \epsilon_{ijk}B_k$. I can sense that it works, but I want to see it directly. How should I start?

  • 2
    $\begingroup$ Contract with $\epsilon_{ijk}$ and figure out what the contraction of two $\epsilon$-symbols is. $\endgroup$ – Nephente Sep 2 at 7:58

Using the identity $\epsilon_{kij}\epsilon_{klm}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$,$$\epsilon_{ijk}B_k=\epsilon_{ijk}\epsilon_{klm}\partial_lA_m=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})\partial_lA_m=\partial_iA_j-\partial_jA_i=F_{ij}.$$


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