Prove that the Divergence of a Curl is Zero by using Levi Civita | Quantum Science Philippines
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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

 

2 Responses to “Prove that the Divergence of a Curl is Zero by using Levi Civita”

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