Wolfgang,
> > Is there any reason why SymmetricTensor <3, dim> does not exist?
>
> Symmetric tensors are of the form
> A_{ij} = A_{ji}
> or
> A{(ij)(kl)} = A_{(kl),(ij)}
> or similar if you had 6 or 8 indices.
My odd-order tensor rule works in this way:
A_{ijk} = A_{ikj}, or, A_{i(jk)} = A_{i(kj)}.
Which is the same as the rank-two (++even-order) tensor rule if we
blissfully ignore the first index. Fifth order I am *very* unsure about.
> I think what I'm unsure about is what sort of symmetry relationship you
> would impose on rank-3 symmetric tensors? If you explain what exactly such
> a tensor is, then we can think about implementing it.
I am unsure if this is a unique (ie. proper) definition... that's what
makes me unsure about imposing this symmetry relation. Nevertheless, this
is the only definition I have seen in literature in the Physical Sciences,
which is one reason it never occured to me to think deeper about it.
Of course I have not yet read everything ;-)
The real problem, I guess, is that I can not multiple a SymmetricTensor
with a Tensor (apparently). Otherwise I may be happy enough to do:
SymmetricTensor<1,dim> * Tensor<3,dim>.
What do you think?
I will have a look through my text-books again to see if there is a "real"
definition of a "symmetric third-order tensor". It's not straightforward
by any entirely gracious means; is it?
Best,
Toby
-----
Toby D. Young
Assistant Professor
Philosophy-Physics
Polish Academy of Sciences
Warszawa, Polska
www: http://www.ippt.gov.pl/~tyoung
skype: stenografia
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