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For a second-order rank, we know that the isotropic and deviatoric parts is given by $$ \frac{1}{2} \left( A_{ij} + A_{ji} \right) - \frac{1}{3} A_{ss} \delta_{ij} \, , $$ whereas for a third-rank tensor, it is given by $$ \frac{1}{6} \left( A_{ijk} + A_{ikj} + A_{jik} + A_{jki} + A_{kij} + A_{kji} \right) - \frac{1}{15} \left[ \left( A_{ssi} + A_{sis} + A_{iss} \right) \delta_{kj} + \left( A_{ssj} + A_{sjs} + A_{jss} \right) \delta_{ik} + \left( A_{ssk} + A_{sks} + A_{kss} \right) \delta_{ij} \right] \, . $$

Then, how can this be generalized for fourth- and fifth-rank tensors?

Thank you!

keynes
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    Could you define deviatoric? Also, in what dimension are you working? – J.G. Oct 22 '21 at 09:17
  • @J.G. This apparently has to do with subtracting the isotropic part of a given tensor. i found this article in Wiki in German (https://de.wikipedia.org/wiki/Deviator) which apparently does not exist in another language. – keynes Oct 22 '21 at 12:11
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    There's an obvious problem with rank $4$ or higher. For rank $4$ we want$$\frac{1}{4!}A_{(ijkl)}-B[A_{ssij}\delta_{kl}+\cdots]-C[A_{sstt}\delta_{ij}\delta_{kl}+\cdots],$$but $B,,C$ won't be unique subject to whatever contraction-based criterion being "deviatoric" imposes. – J.G. Oct 22 '21 at 12:37
  • @J.G. The factorial term is indeed guessable but the second term remains a mystery – keynes Oct 22 '21 at 12:57
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    The decomposition is not symmetric + deviatoric but rather isotropic + deviatoric. In fact, the deviator is defined as $${\rm dev}(A)=A-{\rm iso}(A)$$ Here is a good post about calculating isotropic tensors. And here is another. – greg Oct 22 '21 at 13:27
  • @greg Thanks for the references. The 2nd one is rather very useful – keynes Oct 22 '21 at 13:36

1 Answers1

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$ \def\tss{\mathop{\bullet}\limits} \def\dss{\mathop{\odot}\limits} \def\bbR#1{{\mathbb R}^{#1}} \def\a{\alpha}\def\b{\beta}\def\d{\delta}\def\t{\theta} \def\A{{\cal A}}\def\E{{\cal I}}\def\B{{\cal B}} \def\L{\left}\def\R{\right} \def\LR#1{\L(#1\R)} \def\BR#1{\Big(#1\Big)} \def\iso#1{\operatorname{iso}\LR{#1}} \def\dev#1{\operatorname{dev}\LR{#1}} \def\Diag#1{\operatorname{Diag}\LR{#1}} \def\sym#1{\operatorname{sym}\LR{#1}} \def\skew#1{\operatorname{skew}\LR{#1}} \def\trace#1{\operatorname{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\c#1{\color{red}{#1}} $Let $\A$ denote an arbitrary $k^{th}$ order tensor, $\E$ the totally symmetric isotropic tensor of the same order, and $(\tss)$ the $k^{th}$ order contraction product, i.e. $$\A\tss\B = \sum_{i=1}^n\sum_{j=1}^n\ldots\sum_{\ell=1}^n \;{\A_{ij\ldots\ell}\;\B_{ij\ldots\ell}}$$

Then the isotropic part of $\A$ is defined as $$\eqalign{ \iso{\A} = \LR{\frac{\A\tss\E}{\E\tss\E}} \E \\ }$$ and the deviatoric part of the tensor is what's left over $$\eqalign{ \dev{\A} &= \A - \iso{\A}\qquad \\ }$$ Note that $\LR{\A\tss\E}$ and $\LR{\E\tss\E}\,$ are fully contracted scalar quantities.

Update

It occurs to me that you might be studying irreducible tensor decompositions, in which case you should probably study this post.

greg
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