Quantum Science Philippines

## Curl of the Gradient of a Scalar

proof that the curl of the gradient of a scalar function is equal to zero
$\vec{\nabla}\times\vec{\nabla}\psi=0$
let
$\vec{\nabla}_j= \partial_j\hat{e}_j$
$\vec{\nabla}_k= \partial_k\hat{e}_k$
and
$\triangledown\psi= \partial_k\psi\hat{e}_k$

$\vec{\nabla}_j \times \vec{\nabla}_k \psi=\in_{ijk} \partial_j \partial_k \psi\hat{e}_j \times \hat{e}_k$
$= \in_{ijk}\hat{e}_i\partial_j\partial_k\psi$
$= \hat{e}_1(\partial_2\partial_3\psi - \partial_3\partial_2\psi) + \hat{e}_2(\partial_3\partial_1\psi - \partial_1\partial_3\psi) + \hat{e}_3(\partial_1\partial_2\psi - \partial_2\partial_1\psi)$
$= \hat{e}_1(\partial^2\psi - \partial^2\psi) + \hat{e}_2(\partial^2\psi - \partial^2\psi) + \hat{e}_3(\partial^2\psi - \partial^2\psi)$
$= 0$

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## Proving Vector Identity Using Levi-Civita Symbol

Roel N. Baybayon

MSPhysics1

————————————————————————————————–

We are going to prove the following vector identity using Levi-Civita symbol:

$\left(\vec{a}\times\vec{b}\right)\cdot\left(\vec{c}\times\vec{d}\right)=\left(\vec{a}\cdot\vec{c}\right)\left(\vec{b}\cdot\vec{d}\right)-\left(\vec{a}\cdot\vec{d}\right)\left(\vec{b}\cdot\vec{c}\right)$

Solution:

Let   $\vec{a}=a_i \hat{e_i}$ ,    $\vec{b}=b_j \hat{e_j}$ ,   $\vec{c}=c_k \hat{e_k}$ ,   $\vec{d}=d_l\hat{e_l}$.

Then,

$\begin{array}{rcl} \left(\vec{a} \times \vec{b}\right) \cdot \left(\vec{c} \times\vec{d}\right) & = & \left(a_i \hat{e_i} \times b_j \hat{e_j}\right) \cdot \left(c_k \hat{e_k} \times d_l\hat{e_l}\right) \\ & = & \left(\epsilon_{ijm}a_i b_j \hat{e_m}\right) \cdot \left(\epsilon_{kln} c_k d_l \hat{e_l}\right) \\ & = & \epsilon_{ijm}\epsilon_{kln} a_i b_j c_k d_l \hat{e_m}\cdot\hat{e_n} \\ & = & \epsilon_{ijm}\epsilon_{kln} a_i b_j c_k d_l \delta_{mn}\end{array}$

By definition:

$\delta_{mn} = \begin{cases} 1, & \mbox{if } m=n \\ 0, & \mbox{if } m\neq n \end{cases}$

We have to let m=n so that,

$\left(\vec{a} \times \vec{b}\right) \cdot \left(\vec{c} \times\vec{d}\right)= \epsilon_{ijm}\epsilon_{klm} a_i b_j c_k d_l$

Levi-Civita symbol can be expressed in terms of Kronecker delta given by:

$\epsilon_{ijm}\epsilon_{klm}=\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk}$

Thus,

$\begin{array}{rcl} \left(\vec{a} \times \vec{b}\right) \cdot \left(\vec{c} \times\vec{d}\right) & = & \left(\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk}\right)a_i b_j c_k d_l \\ & = & \delta_{ik}\delta_{jl}a_i b_j c_k d_l-\delta_{il}\delta_{jk}a_i b_j c_k d_l \\ & = & \left(a_ic_k\delta_{ik}\right)\left(b_jd_l\delta_{jl}\right)-\left(a_id_l\delta_{il}\right)\left(b_jc_k\delta_{jk}\right) \\ & = & \left(a_ic_k \hat{e_i}\cdot \hat{e_k}\right)\left(b_jd_l\hat{e_j}\cdot \hat{e_l}\right)-\left(a_id_l\hat{e_i}\cdot \hat{e_l}\right)\left(b_jc_k\hat{e_j}\cdot \hat{e_k}\right) \\ & = & \left(a_i\hat{e_i}\cdot c_k\hat{e_k}\right)\left(b_j\hat{e_j}\cdot d_l\hat{e_l}\right)-\left(a_i\hat{e_i}\cdot d_l\hat{e_l}\right)\left(b_j\hat{e_j}\cdot c_k\hat{e_k}\right) \\ & = & \left(\vec{a}\cdot\vec{c}\right)\left(\vec{b}\cdot\vec{d}\right)-\left(\vec{a}\cdot\vec{d}\right)\left(\vec{b}\cdot\vec{c}\right)\end{array}$

## 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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## Prove that the Divergence of a Curl is Zero by using Levi Civita

Author: Kayrol Ann B. Vacalares

The divergence of a curl is always zero and we can prove this by using Levi-Civita symbol. The Levi-Civita symbol, also called the permutation symbol or alternating symbol, is a mathematical symbol used in particular in tensor calculus.

Prove that:

$\vec{\nabla} \bullet (\vec{\nabla} \times \vec a)$ = 0

Proof:

Let:

$\vec{\nabla} = \partial i \hat{e_i}$ and

$\vec a = (a_j) \hat{e_j}$

To show that:

$\vec{\nabla} \bullet (\vec{\nabla} \times \vec a)$  = 0

First,

$\vec{\nabla} \times \vec{a} = \epsilon_{ijk} \partial_{i} a_{j} \hat{e_i} \times \hat{e_j}$

$\vec{\nabla} \times \vec{a} = \epsilon_{ijk} \partial_{i} a_{j} \hat{e_k}$

Here are the possible values of $\epsilon_{ijk}$ :

$\epsilon_{ijk} = 1$ if i,j,k is cyclic and non-repeating.

$\epsilon_{ijk} = -1$ if i,j,k is anti-cyclic or counterclockwise.

$\epsilon_{ijk} = 0$ if there are any repeated index.

Consider i,j,k to be cyclic and non-repeating, so

$\epsilon_{ijk} = 1$ and $\vec{\nabla} \times \vec{a} = \partial_{i} a_{j} \hat{e_k}$

$\begin{array}{rcl} \vec{\nabla} \bullet \left(\vec{\nabla} \times \vec{a}\right) & = & \partial_{i} \hat{e_i} \bullet \partial_{i} a_{j} \hat{e_k} \\ & = & \partial_{i} \left(\partial_{i} a_{j} \right) \hat{e_i} \bullet \hat{e_k} \\ & =& \partial_{i} \left(\partial_{i} a_{j} \right) \delta_{ik} \end{array}$

But $\delta_{ik} = 0$ if i is not equal to  j

and $\delta_{ik} = 1$ if i= k

Since i,j,k is non-repeating and $i \ne k$ , therefore

$\delta_{ik} = 0$

Thus,

$\vec{\nabla} \bullet (\vec{\nabla} \times \vec a)$  = 0

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## Proving Vector Formula with Kronecker Delta Function and Levi-Civita Symbol

Applying $\delta_{ij}$ and $\epsilon_{ijk}$ in Proving the Vector Formula: $\vec{\nabla}\bullet\left(\vec{a}\times\vec{b}\right)=\vec{b}\bullet\left(\vec{\nabla}\times \vec{a}\right)-\vec{a}\bullet\left(\vec{\nabla}\times\vec{b}\right)$

By: Quennie J. Paylaga

Prove:

$\vec{\nabla}\bullet\left(\vec{a}\times\vec{b}\right)=\vec{b}\bullet\left(\vec{\nabla}\times \vec{a}\right)-\vec{a}\bullet\left(\vec{\nabla}\times\vec{b}\right)$

using Kronecker Delta Function and Levi-Civita Symbol.

To prove this, we let

$\begin{array}{rcl} \vec{a} & = & a_{i} \hat{e}_{i} \\ \vec{b} & = & b_{j} \hat{e}_{j} \\ \vec{\nabla} &=& \frac{\partial}{\partial x_{k}} \hat{e}_{k} = \partial_{k} \hat{e}_{k} \end{array}$

We can write the expression for $\vec{a} \times \vec{b}$ in summation form as:

$\begin{array}{rcl} \vec{a} \times \vec{b} & = & a_{i} \hat{e}_{i} \times b_{j} \hat{e}_{j} \\ & = & a_{i} b_{j} \hat{e}_{i} \times \hat{e}_{j} \\ & = & \epsilon_{ijl} a_{i} b_{j} \hat{e}_{l} \\ \end{array}$      where $\hat{e}_{i} \times \hat{e}_{j} = \hat{e}_{l} \\$

where i, j, l are dummy summation variables. Each of which can be any letter (a,b,c) or number (1,2,3).

In the same way, we can write $\vec{\nabla} \bullet \left(\vec{a} \times \vec{b}\right)$ as:

$\begin{array}{rcl} \vec{\nabla} \bullet \left(\vec{a} \times \vec{b}\right) & = & \partial_{k} \hat{e}_{k} \bullet \epsilon_{ijl} a_{i} b_{j} \hat{e}_{l} \\ & = & \epsilon_{ijl} \left(\hat{e}_{k} \bullet \hat{e}_{l}\right) \partial_{k} \left(a_{i} b_{j}\right) \\ & = & \epsilon_{ijl} \delta_{kl} \partial_{k} \left(a_{i} b_{j}\right) \hspace{0.7cm} where \hspace{0.2cm} \delta_{kl} = \hat{e}_{k} \bullet \hat{e}_{l} \\ & & if \hspace{0.2cm} k = l, \hspace{0.2cm} \delta_{kl} = \delta_{kk} = \delta_{ll} = 1, and \hspace{0.2cm} \partial_{k} = \partial_{l} \\ & = & \epsilon_{ijl} \partial_{l} \left(a_{i} b_{j}\right) \\ & = & \epsilon_{ijl} \left(a_{i} \partial_{l} b_{j} + b_{j} \partial_{l} a_{i}\right) \\ & = & \epsilon_{ijl} a_{i} \partial_{l} b_{j} + \epsilon_{ijl} b_{j} \partial_{l} a_{i} \\ & = & \epsilon_{jli} b_{j} \partial_{l} a_{i} - \epsilon_{ilj} a_{i} \partial_{l} b_{j} \\ & & where \\ & & \epsilon_{ijl} = \epsilon_{jli} = \epsilon_{lij} = +1 \hspace{0.2cm} (cyclic) \\ & & \epsilon_{ilj} = \epsilon_{lji} = \epsilon_{jil} = -1 \hspace{0.2cm} (anti-cyclic) \\ & = & \vec{b} \bullet \left(\vec{\nabla} \times \vec{a}\right) - \vec{a} \bullet \left(\vec{\nabla} \times \vec{b}\right). \\ \end{array}$

Thus, we have prove that

$\vec{\nabla} \bullet \left(\vec{a} \times \vec{b}\right) = \vec{b} \bullet \left(\vec{\nabla} \times \vec{a}\right) - \vec{a} \bullet \left(\vec{\nabla} \times \vec{b}\right)$

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