Answer to Question #115212 in Vector Calculus for mohammed

Question

Prove that the divergence of a curl is always zero:

∇⃗⃗. (∇⃗⃗

×a)=0

Answer 1

Let us recall the propertied of Levi-Civata's symbol $\epsilon_{ijk}$ in three dimension,where $i,j,k \in \{1,2,3\}$ .

Properties,

$$\epsilon_{123}=\epsilon_{231}=\epsilon_{312}=1 \hspace{1cm} (\star)$$ $$\epsilon_{321}=\epsilon_{213}=\epsilon_{132}=-1 \hspace{1cm} (\star \star)$$

$$\epsilon_{ijk}=0 \hspace{1cm} (\dag)$$

if at least two of $i,j,k$ are equal.

If we exchange any two consecutive indices like $i \leftrightarrow j$ then $\epsilon_{ijk}=-\epsilon_{jik} \:(\spades)$

The cross product of any two vector in terms of Levi-Civata's symbol is defined as bellow,

$$[\overrightarrow{A} \times \overrightarrow{B}]_i =\sum_{j=1}^{3} \sum_{k=1}^{3}\epsilon_{ijk} A_j B_k \hspace{1cm} (\dag \dag)$$

where,subscript $i ,j,k$ of any vector denotes respectively $i^{th},j^{th},k^{th}$ component of that vector.

Now, denote

$$\partial_i = \frac{\partial}{\partial x^{i}}$$

Let the vector $\overrightarrow{a} = a_1 \hat{e_1}+a_2 \hat{e_2}+a_3 \hat{e_3}$ ,where $\hat{e_i}$ is the orthogonal unit vector for all $i \in \{1,2,3\}$ .

Thus,from $(\dag),(\dag \dag) ,(\star),(\star \star)$ and above notation, we get ,

$$\overrightarrow{\nabla} \cdot (\overrightarrow{\nabla} \times \overrightarrow{a})= \sum_{i=1}^{3} \partial_{i} (\sum_{j=1}^{3} \sum_{k=1}^{3}\epsilon_{ijk} \partial_j a_k)\\ \overrightarrow{\nabla} \cdot (\overrightarrow{\nabla} \times \overrightarrow{a})= \sum_{i=1}^{3} \sum_{j=1}^{3} \sum_{k=1}^{3}\epsilon_{ijk}\partial_i \partial_j a_k \\ \overrightarrow{\nabla} \cdot (\overrightarrow{\nabla} \times \overrightarrow{a})= \sum_{i=1}^{3} \sum_{j=1}^{3} \sum_{k=1}^{3}-\epsilon_{jik}\partial_j \partial_i a_k \hspace{1cm} (from, \spades)\\ \overrightarrow{\nabla} \cdot (\overrightarrow{\nabla} \times \overrightarrow{a})= \sum_{i=1}^{3} \sum_{j=1}^{3} \sum_{k=1}^{3}-\epsilon_{ijk}\partial_i \partial_j a_k\\ \overrightarrow{\nabla} \cdot (\overrightarrow{\nabla} \times \overrightarrow{a})=-\overrightarrow{\nabla} \cdot (\overrightarrow{\nabla} \times \overrightarrow{a})\\ \implies \overrightarrow{\nabla} \cdot (\overrightarrow{\nabla} \times \overrightarrow{a})= 0$$

Hence, we are done.