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

## 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

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

1. Alexandrine Says:

Alexandrine…

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

2. energy7.snappages.com Says:

I am regular visitor, how are you everybody? This piece of writing posted at this web site is genuinely nice.

3. Caio Fernando Says:

one thing got me curious:

it is said that the levi-cevita symbol is coordinate independent, however, the way you wrote the del operator represents del in cartesian-like coordinates.

The final result is, of course, correct, but I can’t see why we don’t need to change our levi-cevita symbol (when using polar, spherical coordinates, for example)

Thanks!