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