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Tensor multiplication 3 dimesnsions

dwsmith

Well-known member
Feb 1, 2012
1,673
\begin{alignat*}{3}
A_{ij}B_{ij} & = & (A_{(ij)} + A_{[ij]})(B_{(ij)} + B_{[ij]})\\
& = & A_{(ij)}B_{(ij)} + A_{(ij)}B_{[ij]} + A_{[ij]}B_{(ij)} + A_{[ij]}B_{[ij]}
\end{alignat*}
$$
A_{(ij)}B_{[ij]} + A_{[ij]}B_{(ij)} = \frac{1}{2}(A_{ji}B_{ij} - A_{ij}B_{ji})
$$
Can I then say $A_{ji}B_{ij} = C_{jj} = C_{3\times 3}$ and $A_{ij}B_{ji} = C_{ii} = C_{3\times 3}$?
Therefore, $A_{(ij)}B_{[ij]} + A_{[ij]}B_{(ij)} = 0$.
 

wmccunes

New member
Jan 30, 2013
8
Dustin, I think you basically solved it in line 1, since
$$
A_{ij} = A_{(ij)} + A_{[ij]}
$$
(decomposition of the tensor into symmetric $(A_{(ij)}$ and antisymmetric $(A_{[ij]})$ parts), so
$$
A_{ij}B_{ij} = (A_{(ij)} + A_{[ij]})(B_{(ij)} + B_{[ij]})
$$

But how does
$$
A_{(ij)}B_{(ij)} + A_{[ij]}B_{[ij]} = (A_{(ij)} + A_{[ij]})(B_{(ij)} + B_{[ij]})
$$
? Thanks
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Dustin, I think you basically solved it in line 1, since
$$
A_{ij} = A_{(ij)} + A_{[ij]}
$$
(decomposition of the tensor into symmetric $(A_{(ij)}$ and antisymmetric $(A_{[ij]})$ parts), so
$$
A_{ij}B_{ij} = (A_{(ij)} + A_{[ij]})(B_{(ij)} + B_{[ij]})
$$

But how does
$$
A_{(ij)}B_{(ij)} + A_{[ij]}B_{[ij]} = (A_{(ij)} + A_{[ij]})(B_{(ij)} + B_{[ij]})
$$
? Thanks
If $A_{(ij)}B_{[ij]} + A_{[ij]}B_{(ij)} = 0$ which I am not sure how to show.
 

wmccunes

New member
Jan 30, 2013
8
If $A_{(ij)}B_{[ij]} + A_{[ij]}B_{(ij)} = 0$ which I am not sure how to show.
Yes nevermind I was looking at your solution backwards. Breaking up $A_{(ij)}B_{[ij]} + A_{[ij]}B_{(ij)}$ into elements it all cancelled out except
$$
\frac{1}{2}(A_{ij}B_{ij} - A_{ji}B_{ji})
$$
so if that equals zero we are good...
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Yes nevermind I was looking at your solution backwards. Breaking up $A_{(ij)}B_{[ij]} + A_{[ij]}B_{(ij)}$ into elements it all cancelled out except
$$
\frac{1}{2}(A_{ij}B_{ij} - A_{ji}B_{ji})
$$
so if that equals zero we are good...
Yup