Can You Crack This Vector Identity Challenge?

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Here is this week's POTW:

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Prove the vector identity

$$\nabla(\mathbf{A}\cdot \mathbf{B}) = (\mathbf{A}\cdot \nabla)\mathbf{B} + (\mathbf{B}\cdot \nabla)\mathbf{A} + \mathbf{A}\times (\nabla \times \mathbf{B}) + \mathbf{B}\times (\nabla \times \mathbf{A})$$

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!

 

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For a hint, use the epsilon tensor and the epsilon-delta identity.

 

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No one answered this week’s problem. You can read my solution below.

Spoiler

Using Einstein notation, the $j$th component of $\nabla(\mathbf{A}\cdot \mathbf{B})$ is $$\partial_j(A^iB_i) = \delta_j^{\mu} \delta^{\nu}_{i} \partial_{\mu}(A^i B_{\nu}) = \delta_j^{\mu} \delta_i^\nu (\partial_{\mu} A^i)B_\nu + \delta_j^{\mu} \delta_i^{\nu} A_i \partial_{\mu} B^\nu$$
By the epsilon-delta identity, $$\delta_j^{\mu} \delta_i^\nu = \delta_{j}^{\mu} \delta_{\nu}^i = \epsilon^{\mu i k}\epsilon_{j\nu k} + \delta^{\mu}_{\nu} \delta^i_j$$
Thus $$\delta_j^{\mu} \delta_i^{\nu} (\partial_{\mu} A_i)B_{\nu} = \epsilon_{j\nu k}(\epsilon^{\mu i k}\partial_{\mu} A^i)B_{\nu} + \delta^{\mu}_{\nu} \delta^i_j (\partial_{\mu} A^i)B_{\nu} = \epsilon_{j\nu k} B_{\nu}(\nabla \times \mathbf{A})^k + B_{\mu}\partial_{\mu} A^j$$which is the $j$th component of the sum $$\mathbf{B}\times (\nabla \times \mathbf{A}) + (\mathbf{B}\cdot \nabla)\mathbf{A}$$ Similarly, $\delta_j^{\mu}\delta_i^{\nu} A_i \partial_{\mu}B^{\nu}$ is the $j$th component of the sum $\mathbf{A}\times(\nabla \times \mathbf{B}) + (\mathbf{A}\cdot \nabla)\mathbf{B}$ The result now follows.