Proving Vector Identity Involving the Unit Vector Using the Levi-Civita and the Kronecker Delta | Quantum Science Philippines

## Proving Vector Identity Involving the Unit Vector Using the Levi-Civita and the Kronecker Delta

*author: Michelle R. Fudot

Prove: $(\vec{a}.\vec{\nabla)}\hat{n}) =\frac{1}{r}[\vec{a}-\hat{n}(\vec{a}.\hat{n}]=\frac{a_\perp}{r}$ ___________________________________________________________

Proof: First, we define the following vectors as:

$\vec{a}=a_i \widehat{e_i}$;

$\hat{n}=\frac{\vec{X}}{|X|}=\frac{X_j\widehat{e_j}}{|X|}$; and

$\vec{\nabla} = \partial_k\widehat{e_k}$

Now, $(\vec{a}.\vec{\nabla})\hat{n}=(a_i \widehat{e_i}\bullet\partial_k\widehat{e_k})\frac{X_j\widehat{e_j}}{|X|}$

$=\delta_{ik}(a_i\partial_k)\frac{X_j\widehat{e_j}}{|X|}$

if we let i=k, then $\delta_{ik} = \delta_{ii} = 1$. Furthermore,

$= a_i\partial_i\frac{X_j\widehat{e_j}}{|X|}$ $= a_i\lgroup\frac{1}{|X|}\partial_iX_j\widehat{e_j} + X_j\widehat{e_j}\partial_i(\frac{1}{|X|})\rgroup$ $= a_i\lgroup\frac{1}{|X|}\lgroup\widehat{e_j}\partial_iX_j+X_j\partial_i\widehat{e_j}\rgroup+X_j\widehat{e_j}\partial_i(\frac{1}{|X|})\rgroup$

Now, the derivative of orthonormal basis $\widehat{e_j}$, that is, $\partial_i\widehat{e_j}=0$ and the derivative of a coordinate X, $\partial_iX_j = \delta_{ij}$. Also, $\partial_i(\frac{1}{|X|})=-\frac{X_j\widehat{e_j}}{|X|^3}$, thus

=$a_i\lgroup\frac{1}{|X|}\lgroup\widehat{e_j}\delta_{ij}-X_j\widehat{e_j}(\frac{X_j\widehat{e_j}}{|X|^3})\rgroup\rgroup$

=$\frac{1}{|X|}a_i-\frac{1}{|X|}\frac{X_j\widehat{e_j}}{|X|}a_i\frac{X_j\widehat{e_j}}{|X|}$

=$\frac{1}{|X|}(\vec{a}-\frac{\vec{X}}{|X|}a_i\frac{\vec{X}}{|X|})$

=$\frac{1}{|X|}(\vec{a}-\hat{n}(\vec{a}.\hat{n}))$

It is noted that $|X| = r$. Therefore,

=$\frac{1}{r}(\vec{a}-\hat{n}(\vec{a}.\hat{n}))$

This is further equivalent to the ratio of the component of $a$ perpendicular to $\vec{X}$, that is

= $\frac{a_\perp}{r}$

since $a_\perp = \hat{n}\times(\hat{n}\times\vec{a})$.

To show their equivalence, we use the BAC-CAB Rule in the definition of $a_\perp$. So,

$\frac{a_\perp}{r} = \frac{\hat{n}(-\hat{n}.\vec{a}-\vec{a}(-\hat{n}.\hat{n})}{r}$

=$\frac{\hat{n}(-\vec{a}.\hat{n})-\vec{a}(-1)}{r}$

=$\frac{\vec{a}-\hat{n}(\vec{a}.\hat{n})}{r}$

Thus, we conclude that

$(\vec{a}.\vec{\nabla})\hat{n} =\frac{1}{r}(\vec{a}-\hat{n}(\vec{a}.\hat{n})) =\frac{a_\perp}{r}$

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