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I have the following expression that i would like to simplify further. $\mathbf{e}$ is a unit vector. I have used the BAC-CAB Rule to obtain: $$ \begin{aligned} (\phi \mathbf{e}) \times (\nabla \times \psi \mathbf{e}) & = \phi \mathbf{e} \times (\nabla \psi \times \mathbf{e} + \psi\nabla \times \mathbf{e}) \\ & = (\phi \mathbf{e} \times \nabla \psi \times \mathbf{e}) + (\phi \mathbf{e} \times \psi \nabla \times \mathbf{e}) \\ & = \nabla \psi(\phi \mathbf{e}.\mathbf{e}) - \mathbf{e}(\phi \mathbf{e}. \nabla \psi) + (\phi \psi \mathbf{e} \times \nabla \times \mathbf{e}) \\ & = \phi \nabla \psi - \mathbf{e}(\phi \mathbf{e}. \nabla \psi) + (\phi \psi \mathbf{e} \times \nabla \times \mathbf{e}) \end{aligned} $$

I feel $\mathbf{e} \times \nabla \times \mathbf{e}$ should be further reducible. Ideally I would like to have a scalar expression multiplied into one vector that is perhaps a composite expression of $\mathbf{e}$ and/or unit normals to $\mathbf{e}$. I am not interested in higher order tensors.

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  • $\begingroup$ It is only a notation device. I was hoping for more $\endgroup$
    – Edgar
    Jun 17, 2021 at 14:56

1 Answer 1

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Are you familiar with Levi-Civita symbol? It is the perfect tool for dealing with such problems.

$$ [(\phi \, \mathbf e) \times (\nabla \times\psi \, \mathbf e)]_i = \varepsilon^{ijk}\phi \, \mathbf e_j (\nabla \times\psi \, \mathbf e)_k = \varepsilon^{ijk}\phi \, \mathbf e_j \varepsilon^{kmn} \partial_m\psi\,\mathbf e_n = \varepsilon^{kij} \varepsilon^{kmn} \phi\,\mathbf e_j\partial_m\psi \,\mathbf e_n. $$

Then using the property $\varepsilon^{kij} \varepsilon^{kmn} =\delta_{im}\delta_{jn} - \delta_{in}\delta_{jm} $, where $\delta_{ij}$ is Kronecker delta

$$ [(\phi \, \mathbf e) \times (\nabla \times\psi \, \mathbf e)]_i = (\delta_{im}\delta_{jn} - \delta_{in}\delta_{jm})\phi\,\mathbf e_j\partial_m\psi \,\mathbf e_n = \delta_{im}\delta_{jn} \phi\,\mathbf e_j\partial_m\psi \,\mathbf e_n - \delta_{in}\delta_{jm} \phi\,\mathbf e_j\partial_m\psi \,\mathbf e_n $$

Then using the delta property $\delta_{ij} x_i = x_j$

$$ [(\phi \, \mathbf e) \times (\nabla \times\psi \, \mathbf e)]_i = \phi \, \mathbf e_j \partial_i \psi \,\mathbf e_j - \phi \,\mathbf e_j \partial_j\psi\,\mathbf e_i $$

Simplification can be done further by calculating derivatives and using the $\|\mathbf e\| = 1$, but I will leave it for now. Please note that in my derivation I use Einstein notation.

I hope it helps.

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  • $\begingroup$ \ Thanks. It helps a ton. But how do I go about calculating the derivatives and further simplifying with $|\mathbf e\| = 1$ I am not well versed with Einstein Summation convention but willing to grok further. $\endgroup$
    – Edgar
    Jun 18, 2021 at 6:50
  • $\begingroup$ I hope that the following hints will remove any doubts. 1. Regarding derivative use product rule $(fg)' = f'g+fg'$, i.e. $ \partial_i f g \equiv \partial_i(fg) = \partial_i f g + f \partial_i g$. 2. Regarding $\| \mathbf e\|=1$ please note that I use Einstein notation, so $ \mathbf e_i \mathbf e_i \equiv \sum_i \mathbf e_i \mathbf e_i = \mathbf e \cdot \mathbf e = 1$. $\endgroup$
    – andywiecko
    Jun 19, 2021 at 8:00

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