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Transformation rule for product of 3rd, 2nd order tensors

  • #1
4
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1. Problem statement:

Assume that u is a vector and A is a 2nd-order tensor. Derive a transformation rule for a 3rd order tensor Zijk such that the relation ui = ZijkAjk remains valid after a coordinate rotation.


Homework Equations

:
[/B]
Transformation rule for 3rd order tensors: Z'ijk = CilCjmCknZlmn. Transformation rule of 2nd order tensors: A'jk = CjmCknAmn. Transformation rule for 1st order tensors: u'i = Cilul.


3. My attempt:

To begin with, I am confused as to the wording of this question. I assume that it means: come up with an expression for Z'ijk such that the relation u'i = Z'ijkA'jk holds, but if I am wrong, I would appreciate an explanation of what we are trying to do! If I am correct, then I don't see why the normal transformation rule for third order tensors does not work here. I have:

Z'ijkA'jk = CilCjmCknZlmnCjmCknAmn = CilZlmnAmn = u'i

I think I've done something very wrong here, but I am unfamiliar with tensors and I don't know how to go about fixing it. Help would be much appreciated, thank you!
 

Answers and Replies

  • #2
Not much wrong with your work. I do have a quibble with your derivation. When writing A'(j,k)= C(j.m)C(k,n)A(m,n) you should use different dummy variables to prevent confusion with the dummy variables in the expression for Z. Thus A'(j,k)= C(j.p)C(k,q)A(p,q) is better. Then you can collapse the C matrices using the relation C(j,p)C(j,m) = delta(p.m). You get to the same answer.
 
  • #3
4
0
Not much wrong with your work. I do have a quibble with your derivation. When writing A'(j,k)= C(j.m)C(k,n)A(m,n) you should use different dummy variables to prevent confusion with the dummy variables in the expression for Z. Thus A'(j,k)= C(j.p)C(k,q)A(p,q) is better. Then you can collapse the C matrices using the relation C(j,p)C(j,m) = delta(p.m). You get to the same answer.
Ah, I see what you mean. Thanks for the input! The new notation helps.
 

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