Verifying a vector identity using Levi-Civita | Quantum Science Philippines

## Verifying a vector identity using Levi-Civita

VERIFYING A VECTOR IDENTITY USING LEVI-CIVITA

Bianca Rae B. Sambo

Hello physics enthusiast! I am BR, a graduate student in Physics in Mindanao State University – Iligan Institute of Technology. Hope my solution will be of use to you. Keep visiting this site!

The vector identity to be verified using Levi-Civita is, $\vec{\nabla}(\vec{a}\bullet\vec{b})=(\vec{a}\bullet\vec{\nabla})\vec{b}+(\vec{b}\bullet\vec{\nabla})\vec{a}+\vec{a}\times(\vec{\nabla}\times\vec{b})+\vec{b}\times(\vec{\nabla}\times\vec{a})$

where we define the following vectors as, $\vec{a} = a_{i} \hat{i}$ $\vec{b} = b_{j} \hat{j}$ $\vec{\nabla} = \partial{k } \hat{k}$

We start by expanding the left hand side of the identity. $\vec{\nabla}(\vec{a}\bullet\vec{b})=\vec{\nabla} (\delta_{ij} a_{i} b_{j})=\partial{k} \hat{k} (\delta_{ij} a_{i} b_{j})$

Then we set i = j and apply the distributive property of the partial derivative.

Equation 1: $\vec{\nabla}(\vec{a}\bullet\vec{b})=\partial{k}(a_{j} b_{j}) \hat{k} =a_{j} \partial{k} (b_{j})\hat{k} + b_{j}\partial{k}(a_{j}) \hat{k}$

Notice in the right hand side of the vector identity, we have the equations $(\vec{a}\bullet\vec{\nabla})\vec{b}$ and $(\vec{b}\bullet\vec{\nabla})\vec{a}$. These two equations can be written as, $(\vec{a}\bullet\vec{\nabla})\vec{b}= a_{k} \partial{k}(b_{j})\hat{j}$

and $(\vec{b}\bullet\vec{\nabla})\vec{a}= b_{k}\partial{k}(a_{i}) \hat{i}$

So we incorporate this to Equation 1 above. $\vec{\nabla}(\vec{a}\bullet\vec{b})= a_{k} \partial{k}(b_{j}) \hat{j}+ b_{k}\partial{k}(a_{i}) \hat{i}+ a_{j}\partial{k} (b_{j}) \hat{k}+b_{j}\partial{k}(a_{j})\hat{k} - a_{k}\partial{k} (b_{j}) \hat{j}- b_{k} \partial{k} (a_{i}) \hat{i}$

Equation 2: $\vec{\nabla}(\vec{a}\bullet\vec{b})=(\vec{a}\bullet\vec{\nabla})\vec{b}+(\vec{b}\bullet\vec{\nabla})\vec{a}+[a_{j}\partial{k}(b_{j})\hat{k}-a_{k}\partial{k}(b_{j})\hat{j}]+[(b_{j} \partial{k}(a_{j}) \hat{k}- b_{k}\partial{k}(a_{i})\hat{i})]$

Notice also that: $\vec{a}\times(\vec{\nabla}\times\vec{b})=a_{i}\hat{i}\times(\epsilon_{kjm}\partial{k}(b_{j})\hat{m})=\epsilon_{iml}\epsilon_{kjm}a_{i} \partial{k}(b_{j}) \hat{l}=\epsilon_{lim}\epsilon_{kjm}a_{i}\partial{k}(b_{j})\hat{l}=(\delta_{lk}\delta_{ij}-\delta_{lj}\delta_{ik})a_{i}\partial{k}(b_{j})\hat{l}$ $\vec{a}\times (\vec{\nabla}\times\vec{b})=[a_{j}\partial{k}(b_{j})\hat{k}-a_{k}\partial{k}(b_{j})\hat{j}]$

Similarly, $\vec{b}\times(\vec{\nabla}\times\vec{a})=b_{j}\hat{j}\times(\epsilon_{kin}\partial{k}(a_{i})\hat{n})=\epsilon_{jnp}\epsilon_{kin}b_{j}\partial{k}(a_{i})\hat{p}=\epsilon_{pjn}\epsilon_{kin}b_{j}\partial{k}(a_{i})\hat{p}=(\delta_{pk}\delta_{ji}-\delta_{pi}\delta_{jk}) b_{j}\partial{k}(a_{i})\hat{p}$ $\vec{b}\times(\vec{\nabla}\times\vec{a})=[b_{j}\partial{k}(a_{j})\hat{k}- b_{k}\partial{k}(a_{i})\hat{i}]$
So plugging these equations to equation 2, we finally verified the vector identity because we get: $\vec{\nabla}(\vec{a}\bullet\vec{b})=(\vec{a}\bullet\vec{\nabla})\vec{b}+(\vec{b}\bullet\vec{\nabla})\vec{a}+\vec{a}\times(\vec{\nabla}\times\vec{b})+\vec{b}\times(\vec{\nabla}\times\vec{a})$