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

 

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