Question #115212

Prove that the divergence of a curl is always zero:

∇⃗⃗. (∇⃗⃗

×a)=0


1
Expert's answer
2020-05-11T17:17:30-0400

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

Properties,


ϵ123=ϵ231=ϵ312=1()\epsilon_{123}=\epsilon_{231}=\epsilon_{312}=1 \hspace{1cm} (\star)ϵ321=ϵ213=ϵ132=1()\epsilon_{321}=\epsilon_{213}=\epsilon_{132}=-1 \hspace{1cm} (\star \star)

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


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




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



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


[A×B]i=j=13k=13ϵijkAjBk()[\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,ki ,j,k of any vector denotes respectively ith,jth,kthi^{th},j^{th},k^{th} component of that vector.

Now, denote

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

Let the vector a=a1e1^+a2e2^+a3e3^\overrightarrow{a} = a_1 \hat{e_1}+a_2 \hat{e_2}+a_3 \hat{e_3} ,where ei^\hat{e_i} is the orthogonal unit vector for all i{1,2,3}i \in \{1,2,3\} .

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

(×a)=i=13i(j=13k=13ϵijkjak)(×a)=i=13j=13k=13ϵijkijak(×a)=i=13j=13k=13ϵjikjiak(from,)(×a)=i=13j=13k=13ϵijkijak(×a)=(×a)    (×a)=0\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.


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