Question Number 186074 by Mastermind last updated on 31/Jan/23
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{div}\left(\mathrm{curl}\overset{−} {\mathrm{A}}\right)=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$
Answered by aleks041103 last updated on 30/Apr/23
$${Using}\:{einstein}\:{convention}: \\ $$$${div}\left({curl}\left({v}\right)\right)= \\ $$$$=\partial_{{i}} \left(\varepsilon_{{ijk}} \partial_{{j}} {v}_{{k}} \right)= \\ $$$$=\varepsilon_{{ijk}} \partial_{{i}} \partial_{{j}} {v}_{{k}} = \\ $$$$=−\varepsilon_{{jik}} \partial_{{i}} \partial_{{j}} {v}_{{k}} \\ $$$${Schwartz}\:{rule}\::\:\partial_{{i}} \partial_{{j}} =\partial_{{j}} \partial_{{i}} \\ $$$$\Rightarrow\varepsilon_{{jik}} \partial_{{i}} \partial_{{j}} {v}_{{k}} =\varepsilon_{{jik}} \partial_{{j}} \partial_{{i}} {v}_{{k}} \\ $$$${renaming}\:{i}\rightarrow{j}\:{and}\:{j}\rightarrow{i} \\ $$$$\Rightarrow\varepsilon_{{jik}} \partial_{{i}} \partial_{{j}} {v}_{{k}} =\varepsilon_{{ijk}} \partial_{{i}} \partial_{{j}} {v}_{{k}} ={div}\left({curl}\left({v}\right)\right) \\ $$$$\Rightarrow{div}\left({curl}\left({v}\right)\right)=−{div}\left({curl}\left({v}\right)\right) \\ $$$$\Rightarrow{div}\left({curl}\left({v}\right)\right)=\mathrm{0} \\ $$$$ \\ $$$${Note}: \\ $$$$\varepsilon_{{ijk}} \:{is}\:{the}\:{Levi}−{Civita}\:{symbol} \\ $$$${where}\:\varepsilon_{{ijk}} =\begin{cases}{+\mathrm{1},\:\left({ijk}\right)\:{is}\:{an}\:{even}\:{perm}.\:{of}\:\left(\mathrm{123}\right)}\\{−\mathrm{1},\:\left({ijk}\right)\:{is}\:{an}\:{odd}\:{perm}.\:{of}\:\left(\mathrm{123}\right)}\\{\mathrm{0},\:{if}\:{any}\:{two}\:{of}\:{i},{j}\:{or}\:{k}\:{are}\:{equal}}\end{cases} \\ $$$$\Rightarrow\varepsilon_{{ijk}} \:{is}\:{completely}\:{antisymmetric},\:{i}.{e}. \\ $$$$\varepsilon_{{ijk}} =−\varepsilon_{{jik}} \\ $$$${Also},\:{it}\:{is}\:{true}\:{that}: \\ $$$$\overset{\rightarrow} {{a}}={a}_{{i}} \overset{\rightarrow} {{e}_{{i}} },\:\overset{\rightarrow} {{b}}={b}_{{j}} \overset{\rightarrow} {{e}_{{j}} } \\ $$$$\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\:=\:\varepsilon_{{ijk}} {a}_{{i}} {b}_{{j}} \:\overset{\rightarrow} {{e}_{{k}} } \\ $$