Quantum Science Philippines
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}

 

 

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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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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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