Proving Vector Formula with Kronecker Delta Function and Levi-Civita Symbol | Quantum Science Philippines

## Proving Vector Formula with Kronecker Delta Function and Levi-Civita Symbol

Applying $\delta_{ij}$ and $\epsilon_{ijk}$ in Proving the Vector Formula: $\vec{\nabla}\bullet\left(\vec{a}\times\vec{b}\right)=\vec{b}\bullet\left(\vec{\nabla}\times \vec{a}\right)-\vec{a}\bullet\left(\vec{\nabla}\times\vec{b}\right)$

By: Quennie J. Paylaga

Prove:

$\vec{\nabla}\bullet\left(\vec{a}\times\vec{b}\right)=\vec{b}\bullet\left(\vec{\nabla}\times \vec{a}\right)-\vec{a}\bullet\left(\vec{\nabla}\times\vec{b}\right)$

using Kronecker Delta Function and Levi-Civita Symbol.

To prove this, we let

$\begin{array}{rcl} \vec{a} & = & a_{i} \hat{e}_{i} \\ \vec{b} & = & b_{j} \hat{e}_{j} \\ \vec{\nabla} &=& \frac{\partial}{\partial x_{k}} \hat{e}_{k} = \partial_{k} \hat{e}_{k} \end{array}$

We can write the expression for $\vec{a} \times \vec{b}$ in summation form as:

$\begin{array}{rcl} \vec{a} \times \vec{b} & = & a_{i} \hat{e}_{i} \times b_{j} \hat{e}_{j} \\ & = & a_{i} b_{j} \hat{e}_{i} \times \hat{e}_{j} \\ & = & \epsilon_{ijl} a_{i} b_{j} \hat{e}_{l} \\ \end{array}$      where $\hat{e}_{i} \times \hat{e}_{j} = \hat{e}_{l} \\$

where i, j, l are dummy summation variables. Each of which can be any letter (a,b,c) or number (1,2,3).

In the same way, we can write $\vec{\nabla} \bullet \left(\vec{a} \times \vec{b}\right)$ as:

$\begin{array}{rcl} \vec{\nabla} \bullet \left(\vec{a} \times \vec{b}\right) & = & \partial_{k} \hat{e}_{k} \bullet \epsilon_{ijl} a_{i} b_{j} \hat{e}_{l} \\ & = & \epsilon_{ijl} \left(\hat{e}_{k} \bullet \hat{e}_{l}\right) \partial_{k} \left(a_{i} b_{j}\right) \\ & = & \epsilon_{ijl} \delta_{kl} \partial_{k} \left(a_{i} b_{j}\right) \hspace{0.7cm} where \hspace{0.2cm} \delta_{kl} = \hat{e}_{k} \bullet \hat{e}_{l} \\ & & if \hspace{0.2cm} k = l, \hspace{0.2cm} \delta_{kl} = \delta_{kk} = \delta_{ll} = 1, and \hspace{0.2cm} \partial_{k} = \partial_{l} \\ & = & \epsilon_{ijl} \partial_{l} \left(a_{i} b_{j}\right) \\ & = & \epsilon_{ijl} \left(a_{i} \partial_{l} b_{j} + b_{j} \partial_{l} a_{i}\right) \\ & = & \epsilon_{ijl} a_{i} \partial_{l} b_{j} + \epsilon_{ijl} b_{j} \partial_{l} a_{i} \\ & = & \epsilon_{jli} b_{j} \partial_{l} a_{i} - \epsilon_{ilj} a_{i} \partial_{l} b_{j} \\ & & where \\ & & \epsilon_{ijl} = \epsilon_{jli} = \epsilon_{lij} = +1 \hspace{0.2cm} (cyclic) \\ & & \epsilon_{ilj} = \epsilon_{lji} = \epsilon_{jil} = -1 \hspace{0.2cm} (anti-cyclic) \\ & = & \vec{b} \bullet \left(\vec{\nabla} \times \vec{a}\right) - \vec{a} \bullet \left(\vec{\nabla} \times \vec{b}\right). \\ \end{array}$

Thus, we have prove that

$\vec{\nabla} \bullet \left(\vec{a} \times \vec{b}\right) = \vec{b} \bullet \left(\vec{\nabla} \times \vec{a}\right) - \vec{a} \bullet \left(\vec{\nabla} \times \vec{b}\right)$