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Question Number 186074 by Mastermind last updated on 31/Jan/23

Prove that div(curlA^− )=0 Help!

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{div}\left(\mathrm{curl}\overset{−} {\mathrm{A}}\right)=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

Answered by aleks041103 last updated on 30/Apr/23

Using einstein convention: div(curl(v))= =∂_i (ε_(ijk) ∂_j v_k )= =ε_(ijk) ∂_i ∂_j v_k = =−ε_(jik) ∂_i ∂_j v_k Schwartz rule : ∂_i ∂_j =∂_j ∂_i ⇒ε_(jik) ∂_i ∂_j v_k =ε_(jik) ∂_j ∂_i v_k renaming i→j and j→i ⇒ε_(jik) ∂_i ∂_j v_k =ε_(ijk) ∂_i ∂_j v_k =div(curl(v)) ⇒div(curl(v))=−div(curl(v)) ⇒div(curl(v))=0 Note: ε_(ijk) is the Levi−Civita symbol where ε_(ijk) = { ((+1, (ijk) is an even perm. of (123))),((−1, (ijk) is an odd perm. of (123))),((0, if any two of i,j or k are equal)) :} ⇒ε_(ijk) is completely antisymmetric, i.e. ε_(ijk) =−ε_(jik) Also, it is true that: a^→ =a_i e_i ^(→) , b^(→) =b_j e_j ^(→) a^→ ×b^(→) = ε_(ijk) a_i b_j e_k ^(→)

$${Using}\:{einstein}\:{convention}: \\ $$$${div}\left({curl}\left({v}\right)\right)= \\ $$$$=\partial_{{i}} \left(\varepsilon_{{ijk}} \partial_{{j}} {v}_{{k}} \right)= \\ $$$$=\varepsilon_{{ijk}} \partial_{{i}} \partial_{{j}} {v}_{{k}} = \\ $$$$=−\varepsilon_{{jik}} \partial_{{i}} \partial_{{j}} {v}_{{k}} \\ $$$${Schwartz}\:{rule}\::\:\partial_{{i}} \partial_{{j}} =\partial_{{j}} \partial_{{i}} \\ $$$$\Rightarrow\varepsilon_{{jik}} \partial_{{i}} \partial_{{j}} {v}_{{k}} =\varepsilon_{{jik}} \partial_{{j}} \partial_{{i}} {v}_{{k}} \\ $$$${renaming}\:{i}\rightarrow{j}\:{and}\:{j}\rightarrow{i} \\ $$$$\Rightarrow\varepsilon_{{jik}} \partial_{{i}} \partial_{{j}} {v}_{{k}} =\varepsilon_{{ijk}} \partial_{{i}} \partial_{{j}} {v}_{{k}} ={div}\left({curl}\left({v}\right)\right) \\ $$$$\Rightarrow{div}\left({curl}\left({v}\right)\right)=−{div}\left({curl}\left({v}\right)\right) \\ $$$$\Rightarrow{div}\left({curl}\left({v}\right)\right)=\mathrm{0} \\ $$$$ \\ $$$${Note}: \\ $$$$\varepsilon_{{ijk}} \:{is}\:{the}\:{Levi}−{Civita}\:{symbol} \\ $$$${where}\:\varepsilon_{{ijk}} =\begin{cases}{+\mathrm{1},\:\left({ijk}\right)\:{is}\:{an}\:{even}\:{perm}.\:{of}\:\left(\mathrm{123}\right)}\\{−\mathrm{1},\:\left({ijk}\right)\:{is}\:{an}\:{odd}\:{perm}.\:{of}\:\left(\mathrm{123}\right)}\\{\mathrm{0},\:{if}\:{any}\:{two}\:{of}\:{i},{j}\:{or}\:{k}\:{are}\:{equal}}\end{cases} \\ $$$$\Rightarrow\varepsilon_{{ijk}} \:{is}\:{completely}\:{antisymmetric},\:{i}.{e}. \\ $$$$\varepsilon_{{ijk}} =−\varepsilon_{{jik}} \\ $$$${Also},\:{it}\:{is}\:{true}\:{that}: \\ $$$$\overset{\rightarrow} {{a}}={a}_{{i}} \overset{\rightarrow} {{e}_{{i}} },\:\overset{\rightarrow} {{b}}={b}_{{j}} \overset{\rightarrow} {{e}_{{j}} } \\ $$$$\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\:=\:\varepsilon_{{ijk}} {a}_{{i}} {b}_{{j}} \:\overset{\rightarrow} {{e}_{{k}} } \\ $$

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