Answer to Question #115212 in Vector Calculus for mohammed

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 "\\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)\\\\\n\\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 \\\\\n\\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)\\\\\n\\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\\\\\n\\overrightarrow{\\nabla} \\cdot (\\overrightarrow{\\nabla} \\times \\overrightarrow{a})=-\\overrightarrow{\\nabla} \\cdot (\\overrightarrow{\\nabla} \\times \\overrightarrow{a})\\\\\n \\implies \\overrightarrow{\\nabla} \\cdot (\\overrightarrow{\\nabla} \\times \\overrightarrow{a})= 0"

Hence, we are done.


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