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Vector calculus identities proof using suffix notation
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[QUOTE="kumusta, post: 6547240, member: 693563"] Show that $$(1){~~~~~~~~~~~~~~~~~~~~~~~~~}\vec \nabla\cdot(\vec a \times \vec b) = \vec b \cdot (\vec \nabla \times \vec a) - \vec a \cdot (\vec \nabla \times \vec b){~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}$$ Using suffix notation, we get for the left hand side of eq. (1) $$(2){~~~~~~~~~~~~~~~~~~~~~}[~\vec \nabla\cdot(\vec a \times \vec b)~]_i =\begin{cases} \begin{align}&~\partial_i (\vec a \times \vec b)_i = \partial_i (π_{ijk} a_j b_k) = π_{ijk} \partial_i (a_j b_k) \nonumber \\ & {~~~~~} π_{ijk} [~b_k (\partial_i a_j) + a_j ( \partial_i b_k)~]~\dots\nonumber\end{align} \end{cases}$$For the right hand side of eq. (1), we obtain $$(3){~~~~~~~~~~~~}[~\vec b \cdot (\vec \nabla \times \vec a) - \vec a \cdot (\vec \nabla \times \vec b)~]_i =\begin{cases} \begin{align}& ~b_i ( π_{ijk} \partial_j a_k) - a_i ( π_{ijk} \partial_j b_k) \nonumber \\ &~π_{ijk} [~b_i (\partial_j a_k)~] - π_{ijk} [~a_i (\partial_j b_k)~] \nonumber\end{align} \end{cases}$$According to eqs. (1), (2), and (3) $$(4){~~~~~~~~~~~~~~~~~}π_{ijk} [~b_k (\partial_i a_j) + a_j ( \partial_i b_k)~] = π_{ijk} [~b_i (\partial_j a_k)~] - π_{ijk} [~a_i (\partial_j b_k)~]{~~~~~~~~}$$ For the left hand side of eq. (4), we get after summing over ##k## ##{~~~~~~}(5){~~~~~~~~~~}π_{ijk} [~b_k (\partial_i a_j) + a_j ( \partial_i b_k)~] =\begin{cases}\begin{align}& ~π_{ij1} [~b_1 (\partial_i a_j) + a_j ( \partial_i b_1)~]\nonumber \\ & + π_{ij2} [~b_2 (\partial_i a_j) + a_j ( \partial_i b_2)~]\nonumber \\ & + π_{ij3} [~b_3 (\partial_i a_j) + a_j ( \partial_i b_3)~] \nonumber\end{align}\end{cases}## Now ... 1st line of (5) ##j\neq1##, ##j## = 2, 3 ... 2nd line ##j\neq2##, ##j## = 1, 3 ... 3rd line ##j\neq3##, ##j## = 1, 2 ##{~~~~~~}(6){~~~~~~~~~~}π_{ijk} [~b_k (\partial_i a_j) + a_j ( \partial_i b_k)~] =\begin{cases}\begin{align}& ~π_{i21} [~b_1 (\partial_i a_2) + a_2 ( \partial_i b_1)~] \nonumber \\ &{~~~~~}+ π_{i31} [~b_1 (\partial_i a_3) + a_3 ( \partial_i b_1)\nonumber \\ & + π_{i12} [~b_2 (\partial_i a_1) + a_1 ( \partial_i b_2)~] \nonumber \\ &{~~~~~} + π_{i32} [~b_2 (\partial_i a_3) + a_3 ( \partial_i b_2)~] \nonumber \\ & + π_{i13} [~b_3 (\partial_i a_1) + a_1 ( \partial_i b_3)~] \nonumber \\ &{~~~~~} + π_{i23} [~b_3 (\partial_i a_2) + a_2 ( \partial_i b_3)~] \nonumber\end{align}\end{cases}## Summing over ##i## as in the preceding, we find that ##{~~~~~~~~~~~~~~~~~~~~~}π_{ijk} [~b_k (\partial_i a_j) + a_j ( \partial_i b_k)~] =\begin{cases}\begin{align}& ~π_{321} [~b_1 (\partial_3 a_2) + a_2 ( \partial_3 b_1)~] \nonumber \\ &{~~~~~}+ π_{231} [~b_1 (\partial_2 a_3) + a_3 ( \partial_2 b_1)\nonumber \\ & + π_{312} [~b_2 (\partial_3 a_1) + a_1 ( \partial_3 b_2)~] \nonumber \\ &{~~~~~} + π_{132} [~b_2 (\partial_1 a_3) + a_3 ( \partial_1 b_2)~] \nonumber \\ & + π_{213} [~b_3 (\partial_2 a_1) + a_1 ( \partial_2 b_3)~] \nonumber \\ &{~~~~~} + π_{123} [~b_3 (\partial_1 a_2) + a_2 ( \partial_1 b_3)~] \nonumber\end{align}\end{cases}## Using the fact that ##~π_{ijk} =\begin{cases}\begin{align} &~0~\text {if any index is equal to} \nonumber \\&~\text {any other index}~\dots\nonumber \\ &+1~\text {if}~i ,j, k~\text {form an} \nonumber \\ &~\text {even permutation (cylic}\nonumber \\ &~\text {permutation) of 1, 2, 3}~\dots\nonumber \\ &-1~\text {if}~i ,j, k~\text {form an odd}\nonumber \\ &~\text {permutation of 1, 2, 3}~\dots\nonumber \end{align}\end{cases}## ##{~~~~~~~~~~~~~~~~~~~~~}π_{ijk} [~b_k (\partial_i a_j) + a_j ( \partial_i b_k)~] =\begin{cases}\begin{align}& ~(-1) [~b_1 (\partial_3 a_2) + a_2 ( \partial_3 b_1)~] \nonumber \\ &{~~~~~}+ (+1) [~b_1 (\partial_2 a_3) + a_3 ( \partial_2 b_1)~] \nonumber \\ & + (+1) [~b_2 (\partial_3 a_1) + a_1 ( \partial_3 b_2)~] \nonumber \\ &{~~~~~} + (-1) [~b_2 (\partial_1 a_3) + a_3 ( \partial_1 b_2)~] \nonumber \\ & + (-1) [~b_3 (\partial_2 a_1) + a_1 ( \partial_2 b_3)~] \nonumber \\ &{~~~~~} +(+1) [~b_3 (\partial_1 a_2) + a_2 ( \partial_1 b_3)~] \nonumber\end{align}\end{cases}## ##{~~~~~~~~~~~~~~~~~~~~~}π_{ijk} [~b_k (\partial_i a_j) + a_j ( \partial_i b_k)~] =\begin{cases}\begin{align}& ~ - b_1 (\partial_3 a_2)^* - a_2 ( \partial_3 b_1){\uparrow\uparrow} \nonumber \\ &{~~~~~}+ b_1 (\partial_2 a_3)^* + a_3 ( \partial_2 b_1){\uparrow\uparrow\uparrow}\nonumber \\ & + b_2 (\partial_3 a_1)^{**} + a_1 ( \partial_3 b_2)^{\uparrow} \nonumber \\ &{~~~~~} - b_2 (\partial_1 a_3)^{**} - a_3 ( \partial_1 b_2){\uparrow\uparrow\uparrow} \nonumber \\ & - b_3 (\partial_2 a_1)^{***} - a_1 ( \partial_2 b_3)^{\uparrow} \nonumber \\ &{~~~~~} + b_3 (\partial_1 a_2)^{***} + a_2 ( \partial_1 b_3)^{\uparrow\uparrow} \nonumber \end{align}\end{cases}## ##{~~~~~~}(7){~~~~~~~~~~}π_{ijk} [~b_k (\partial_i a_j) + a_j ( \partial_i b_k)~] =\begin{cases}\begin{align}& ~b_1 [~(\partial_2 a_3) - (\partial_3 a_2)~] \nonumber \\ &{~~~~~} + b_2 [~(\partial_3 a_1) - (\partial_1 a_3)~] \nonumber \\ & + b_3 [~(\partial_1 a_2) - (\partial_2 a_1)~] \nonumber \\ &{~~~~~}+ a_1 [~( \partial_3 b_2) - (\partial_2 b_3)~] \nonumber \\ & + a_2 [~( \partial_1 b_3) - ( \partial_3 b_1)~] \nonumber \\ &{~~~~~}+ a_3 [~( \partial_2b_1) - ( \partial_1 b_2)~] \nonumber\end{align}\end{cases}## Going back to the right side of eq. (4), we get after summing over ##k##: $$π_{ijk} [~b_i (\partial_j a_k)~] - π_{ijk} [~a_i (\partial_j b_k)~] =\begin{cases}\begin{align}&~π_{ij1} [~b_i (\partial_j a_1)~] - π_{ij1} [~a_i (\partial_j b_1)~]\nonumber \\ &~+ π_{ij2} [~b_i (\partial_j a_2)~] - π_{ij2} [~a_i (\partial_j b_2)~]\nonumber \\ &~+ π_{ij3} [~b_i (\partial_j a_3)~] - π_{ij3} [~a_i (\partial_j b_3)~]\nonumber\end{align}\end{cases}$$ Summing over ##j##: ##π_{ijk} [~b_i (\partial_j a_k)~] - π_{ijk} [~a_i (\partial_j b_k)~] =\begin{cases}\begin{align}&~π_{i21} [~b_i (\partial_2 a_1)~] - π_{i21} [~a_i (\partial_2 b_1)~]\nonumber \\ &~+ π_{i31} [~b_i (\partial_3 a_1)~] - π_{i31} [~a_i (\partial_3 b_1)~]\nonumber \\ &~+ π_{i12} [~b_i (\partial_1 a_2)~] - π_{i12} [~a_i (\partial_1 b_2)~]\nonumber \\ &~+ π_{i32} [~b_i (\partial_3 a_2)~] - π_{i32} [~a_i (\partial_3 b_2)~]\nonumber \\ &~+ π_{i13} [~b_i (\partial_1 a_3)~] - π_{i13} [~a_i (\partial_1 b_3)~]\nonumber \\ &~+ π_{i23} [~b_i (\partial_2 a_3)~] - π_{i23} [~a_i (\partial_2 b_3)~]\nonumber\end{align}\end{cases}## Summing over ##i##: ##π_{ijk} [~b_i (\partial_j a_k)~] - π_{ijk} [~a_i (\partial_j b_k)~] =\begin{cases}\begin{align}&~π_{321} [~b_3 (\partial_2 a_1)~] - π_{321} [~a_3 (\partial_2 b_1)~]\nonumber \\ &~+ π_{231} [~b_2 (\partial_3 a_1)~] - π_{231} [~a_2 (\partial_3 b_1)~]\nonumber \\ &~+ π_{312} [~b_3 (\partial_1 a_2)~] - π_{312} [~a_3 (\partial_1 b_2)~]\nonumber \\ &~+ π_{132} [~b_1 (\partial_3 a_2)~] - π_{132} [~a_1 (\partial_3 b_2)~]\nonumber \\ &~+ π_{213} [~b_2 (\partial_1 a_3)~] - π_{213} [~a_2 (\partial_1 b_3)~]\nonumber \\ &~+ π_{123} [~b_1 (\partial_2 a_3)~] - π_{123} [~a_1 (\partial_2 b_3)~]\nonumber\end{align}\end{cases}## ##π_{ijk} [~b_i (\partial_j a_k)~] - π_{ijk} [~a_i (\partial_j b_k)~] =\begin{cases} \begin{align}&~(-1) b_3 (\partial_2 a_1) - (-1) a_3 (\partial_2 b_1) \nonumber \\ &~+ (+1) b_2 (\partial_3 a_1) - (+1) a_2 (\partial_3 b_1) \nonumber \\ &~+ (+1) b_3 (\partial_1 a_2) - (+1) a_3 (\partial_1 b_2) \nonumber \\ &~+ (-1) b_1 (\partial_3 a_2) - (-1) a_1 (\partial_3 b_2) \nonumber \\ &~+ (-1) b_2 (\partial_1 a_3) - (-1) a_2 (\partial_1 b_3) \nonumber \\ &~+ (+1) b_1 (\partial_2 a_3) - (+1) a_1 (\partial_2 b_3) \nonumber\end{align}\end{cases}## ##π_{ijk} [~b_i (\partial_j a_k)~] - π_{ijk} [~a_i (\partial_j b_k)~] =\begin{cases} \begin{align}&~- b_3 (\partial_2 a_1)^{***} + a_3 (\partial_2 b_1)^{\uparrow\uparrow \uparrow} \nonumber \\ &~+ b_2 (\partial_3 a_1)^{**} - a_2 (\partial_3 b_1)^{\uparrow\uparrow} \nonumber \\ &~+ b_3 (\partial_1 a_2)^{***} - a_3 (\partial_1 b_2){\uparrow\uparrow \uparrow} \nonumber \\ &~- b_1 (\partial_3 a_2)^* + a_1 (\partial_3 b_2)^{\uparrow} \nonumber \\ &~- b_2 (\partial_1 a_3)^{**} + a_2 (\partial_1 b_3)^{\uparrow\uparrow} \nonumber \\ &~+ b_1 (\partial_2 a_3)^* - a_1 (\partial_2 b_3)^{\uparrow} \nonumber\end{align}\end{cases}## ##(8){~~~~~}π_{ijk} [~b_i (\partial_j a_k)~] - π_{ijk} [~a_i (\partial_j b_k)~] =\begin{cases} \begin{align}&~b_1 [~(\partial_2 a_3)- (\partial_3 a_2)~] \nonumber \\ &~+ b_2 [~(\partial_3 a_1) - (\partial_1 a_3)~] \nonumber \\ &~+ b_3 [~(\partial_1 a_2) - (\partial_2 a_1)~] \nonumber \\ &~+ a_1 [~(\partial_3 b_2) - (\partial_2 b_3)~] \nonumber \\ &~+ a_2 [~(\partial_1 b_3) - (\partial_3 b_1)~] \nonumber \\ &~+ a_3 [~(\partial_2 b_1) - (\partial_1 b_2)~] \nonumber\end{align}\end{cases}## Comparing eqs. (7) and (8), we find that $$π_{ijk} [~b_k (\partial_i a_j) + a_j ( \partial_i b_k)~] = π_{ijk} [~b_i (\partial_j a_k)~] - π_{ijk} [~a_i (\partial_j b_k)~]$$ $$ \Rightarrow{~~~~~~~~~}\vec \nabla\cdot(\vec a \times \vec b) = \vec b \cdot (\vec \nabla \times \vec a) - \vec a \cdot (\vec \nabla \times \vec b)$$ [/QUOTE]
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Vector calculus identities proof using suffix notation
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