Let us recall the propertied of Levi-Civata's symbol ϵ i j k \epsilon_{ijk} ϵ ijk i , j , k ∈ { 1 , 2 , 3 } i,j,k \in \{1,2,3\} i , j , k ∈ { 1 , 2 , 3 }
Properties,
ϵ 123 = ϵ 231 = ϵ 312 = 1 ( ⋆ ) \epsilon_{123}=\epsilon_{231}=\epsilon_{312}=1 \hspace{1cm} (\star) ϵ 123 = ϵ 231 = ϵ 312 = 1 ( ⋆ ) ϵ 321 = ϵ 213 = ϵ 132 = − 1 ( ⋆ ⋆ ) \epsilon_{321}=\epsilon_{213}=\epsilon_{132}=-1 \hspace{1cm} (\star \star) ϵ 321 = ϵ 213 = ϵ 132 = − 1 ( ⋆ ⋆ )
ϵ i j k = 0 ( † ) \epsilon_{ijk}=0 \hspace{1cm} (\dag) ϵ ijk = 0 ( † )
if at least two of i , j , k i,j,k i , j , k
If we exchange any two consecutive indices like i ↔ j i \leftrightarrow j i ↔ j ϵ i j k = − ϵ j i k ( ♠ ) \epsilon_{ijk}=-\epsilon_{jik} \:(\spades) ϵ ijk = − ϵ jik ( ♠ )
The cross product of any two vector in terms of Levi-Civata's symbol is defined as bellow,
[ A → × B → ] i = ∑ j = 1 3 ∑ k = 1 3 ϵ i j k A j B k ( † † ) [\overrightarrow{A} \times \overrightarrow{B}]_i =\sum_{j=1}^{3} \sum_{k=1}^{3}\epsilon_{ijk} A_j B_k \hspace{1cm} (\dag \dag) [ A × B ] i = j = 1 ∑ 3 k = 1 ∑ 3 ϵ ijk A j B k ( †† )
where,subscript i , j , k i ,j,k i , j , k i t h , j t h , k t h i^{th},j^{th},k^{th} i t h , j t h , k t h
Now, denote
∂ i = ∂ ∂ x i \partial_i = \frac{\partial}{\partial x^{i}} ∂ i = ∂ x i ∂ Let the vector a → = a 1 e 1 ^ + a 2 e 2 ^ + a 3 e 3 ^ \overrightarrow{a} = a_1 \hat{e_1}+a_2 \hat{e_2}+a_3 \hat{e_3} a = a 1 e 1 ^ + a 2 e 2 ^ + a 3 e 3 ^ e i ^ \hat{e_i} e i ^ i ∈ { 1 , 2 , 3 } i \in \{1,2,3\} i ∈ { 1 , 2 , 3 }
Thus,from ( † ) , ( † † ) , ( ⋆ ) , ( ⋆ ⋆ ) (\dag),(\dag \dag) ,(\star),(\star \star) ( † ) , ( †† ) , ( ⋆ ) , ( ⋆ ⋆ )
∇ → ⋅ ( ∇ → × a → ) = ∑ i = 1 3 ∂ i ( ∑ j = 1 3 ∑ k = 1 3 ϵ i j k ∂ j a k ) ∇ → ⋅ ( ∇ → × a → ) = ∑ i = 1 3 ∑ j = 1 3 ∑ k = 1 3 ϵ i j k ∂ i ∂ j a k ∇ → ⋅ ( ∇ → × a → ) = ∑ i = 1 3 ∑ j = 1 3 ∑ k = 1 3 − ϵ j i k ∂ j ∂ i a k ( f r o m , ♠ ) ∇ → ⋅ ( ∇ → × a → ) = ∑ i = 1 3 ∑ j = 1 3 ∑ k = 1 3 − ϵ i j k ∂ i ∂ j a k ∇ → ⋅ ( ∇ → × 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 ∇ ⋅ ( ∇ × a ) = i = 1 ∑ 3 ∂ i ( j = 1 ∑ 3 k = 1 ∑ 3 ϵ ijk ∂ j a k ) ∇ ⋅ ( ∇ × a ) = i = 1 ∑ 3 j = 1 ∑ 3 k = 1 ∑ 3 ϵ ijk ∂ i ∂ j a k ∇ ⋅ ( ∇ × a ) = i = 1 ∑ 3 j = 1 ∑ 3 k = 1 ∑ 3 − ϵ jik ∂ j ∂ i a k ( f ro m , ♠ ) ∇ ⋅ ( ∇ × a ) = i = 1 ∑ 3 j = 1 ∑ 3 k = 1 ∑ 3 − ϵ ijk ∂ i ∂ j a k ∇ ⋅ ( ∇ × a ) = − ∇ ⋅ ( ∇ × a ) ⟹ ∇ ⋅ ( ∇ × a ) = 0 Hence, we are done.
#340153 on Dec 2023
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