Verifying vector formulas using Levi-Civita: (Divergence & Curl of normal unit vector n) | Quantum Science Philippines

## Verifying vector formulas using Levi-Civita: (Divergence & Curl of normal unit vector n)

By Sim P. Bantayan, MS Physics I, MSU-IIT

Let $\Vec{\bigtriangledown} = \partial_i\hat{e}_i$,

and $\hat{n}= \frac{\Vec{x}}{r}$ where $\Vec{x}= {x_j}\hat{e}_j$ and $r=|\Vec{x}|$.

1. Prove that $\Vec{\bigtriangledown}\cdot\hat{n} = \frac{2}{r}$.

Proof:

Now, $\Vec{\bigtriangledown}\cdot\hat{n} = \partial_i\hat{e}_i\cdot\frac{{x_j}\hat{e}_j}{r}$. Since i=j for the divergence of normal unit vector n,

$\begin{array}{lcl}\Vec{\bigtriangledown}\cdot\hat{n} &=&\partial_i\frac{x_i}{r} \hat{e}_i\cdot\hat{e}_i\\& =& \partial_i\frac{x_i}{r} \delta_{ii}\end{array}$

but $\delta_{ii}=1$ (i=j). Moreover, for three dimensions, $r=\sqrt{x{^2_1}+x{^2_2}+x{^2_3}}$, so

$\begin{array}{lcl}\Vec{\bigtriangledown}\cdot\hat{n} & =& \partial_i\frac{x_i}{r}\\&=&\frac{2(x{^2_1}+x{^2_2}+x{^2_3})^2}{(x{^2_1}+x{^2_2}+x{^2_3})^3} \\&=& \frac{2}{(x{^2_1}+x{^2_2}+x{^2_3})}\\&=&\frac{2}{r}\end{array}$

Therefore, $\Vec{\bigtriangledown}\cdot\hat{n} = \frac{2}{r}$.

2. Prove that $\Vec{\bigtriangledown}\times\hat{n} = 0$.

Proof:

$\Vec{\bigtriangledown}\times\hat{n} = \partial_i\hat{e}_i\times\frac{{x_j}\hat{e}_j}{r}$. Since i=j for the curl of normal unit vector n,

$\begin{array}{lcl}\Vec{\bigtriangledown}\times\hat{n}&=&\partial_i\frac{x_i}{r} \hat{e}_i\times\hat{e}_i \\&=&\epsilon_{iik}\partial_i\frac{x_i}{r}\hat{e}_k\end{array}$

but $\epsilon_{iik} = 0$ (index i is repeated).

Therefore, $\Vec{\bigtriangledown}\times\hat{n} = 0$.