Curl of the product of a scalar and a vector using Levi-Civita | Quantum Science Philippines

## Curl of the product of a scalar and a vector using Levi-Civita

### $\vec{\nabla} \times f \vec{A}=f \vec{\nabla} \times \vec{A} \ + \ \vec{\nabla} f \times \vec{A}$

By Eliezer Estrecho

To prove this formula, we use the following:

$\vec{\nabla} \times \vec{A}=\epsilon _{ijk} \hat{e}_{i} \nabla_{j} A_{k}$

Where: $\vec{\nabla}=\nabla_{j} \hat{e}_{j}$ and $\vec{A}=A_{k} \hat{e}_{k}$

Using the equation above:

$\vec{\nabla} \times f \vec{A} =\epsilon _{ijk} \hat{e}_{i} \nabla_{j} f A_{k} \\ \vec{\nabla} \times f \vec{A} =\epsilon _{ijk} \hat{e}_{i} \ (f \nabla_{j} A_{k} + A_{k}\nabla_{j}f) \ \ \ \ \ \ \ product \ rule\\ \vec{\nabla} \times f \vec{A} =\epsilon_{ijk}\hat{e}_{i}f \nabla_{j} A_{k} + \epsilon_{ijk}\hat{e}_{i} A_{k} \nabla_{j} f$

We can factor out $f$ in the first term to give:

$f \epsilon_{ijk}\hat{e}_{i} \nabla_{j} A_{k}=f \vec{\nabla} \times \vec{A}$

Note that for the second term, the permutation of indices are odd, rearranging them to ijk will give the negative:

$\epsilon_{ijk}\hat{e}_{i} A_{k} \nabla_{j} f = -\epsilon_{ijk}\hat{e}_{i} A_{j} \nabla_{k} f = -\vec{A} \times \vec{\nabla}f = \vec{\nabla}f \times \vec{A}$

Thus,

$\vec{\nabla} \times f \vec{A}=f \vec{\nabla} \times \vec{A} \ + \ \vec{\nabla} f \times \vec{A}$

About the author: Eliezer Estrecho is currently a MS Physics student of MSU-IIT.