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This term is appeared in the process of deriving continuous eq. from maxwell's eq.

$1 \over 2$$\varepsilon_{i,j,k}$($\partial_i$$\partial_j$+$\partial_j$$\partial_i$)$B_k$

my professor said $\varepsilon_{i,j,k}$ is anti-symmetry, and ($\partial_i$$\partial_j$+$\partial_j$$\partial_i$) is symmetry, so this term is zero because sym and anti-sym's contraction is zero.

first question, I want to know why this is contraction. It seems like just multiplication.

second question, I want to know why sym and anti-sym's contraction is zero.

og zo
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1 Answers1

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The reason it is a contraction and not just multiplication is because there is an implied summation on the indices $i$, $j$, and $k$. As for your second question, let $A_{ij} = A_{ji}$ be symmetric and $B_{ij} = -B_{ji}$ be antisymmetric. Then,

$$ \sum_{ij} A_{ij} B_{ij} = \sum_{ij} A_{ji} (-B_{ji}) = - \sum_{ji} A_{ij} B_{ij} $$

In the first equality I just used the symmetry/antisymmetry, in the second equality I relabeled my summation indices $i \leftrightarrow j$. Since the sum is equal to minus itself, it must be zero.

Zack
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  • It seems like, anti-sym's sum is must be zero. Because there is no A, just there is B, B is also zero. so, is it good understanding follows? : anti-symmetry's summation is zero. – og zo Apr 14 '21 at 03:19
  • And also I wonder that, is it tensor contraction? – og zo Apr 14 '21 at 03:21
  • Of course, trivially if you sum over all the entries of an antisymmetric matrix you will get zero: for every entry $B_{ij}$, there is an entry $B_{ji}$ of same value but opposite sign. – Zack Apr 14 '21 at 05:36