What Does an Isotropic Cartesian Tensor Look Like in Higher Dimensions?

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Discussion Overview

The discussion revolves around the characteristics and proofs related to isotropic Cartesian tensors, specifically focusing on their forms in three and four dimensions, as well as higher ranks. Participants explore the implications of isotropy under rotation and seek to establish general forms for rank 3 and rank 4 tensors.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant requests a proof of the form of isotropic Cartesian tensors in three or four dimensions.
  • Another participant provides a series of equations derived from the isotropy condition for a rank 2 tensor in three dimensions, leading to specific relations among tensor components.
  • A later post clarifies that the original inquiry was about rank 3 and rank 4 tensors, not the dimensions themselves.
  • One participant references external sources for isotropic tensors and permutation symbols, suggesting that the derivation for rank 4 tensors may follow similar lines to rank 2 but is more complex.
  • Another participant notes the existence of three types of solutions for rank 4 tensors and mentions a sequence related to the number of isotropic tensors of various ranks, but expresses uncertainty about the derivation of these solutions.
  • There is a suggestion that the recurrence relation for higher ranks may require a more sophisticated approach than that used for lower ranks.

Areas of Agreement / Disagreement

Participants express uncertainty and differing views regarding the proofs and forms of isotropic tensors, particularly for ranks 3 and 4. No consensus is reached on the specific forms or derivations.

Contextual Notes

Participants acknowledge the complexity of deriving isotropic tensors of higher ranks and the potential need for advanced methods beyond elementary reasoning. Specific assumptions and definitions related to isotropy and tensor ranks are not fully explored.

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Does anyone have a proof of what a isotropic cartesian tensor should look like in three or four dimensions?
 
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An isotropic tensor must remain identical under any rotation:

t' = R.t.R-1 = t

For three dimensions, for a rotation around the first axis this leads to:

<br /> \text{t&#039;}(1,1)=t(1,1) \\<br /> \text{t&#039;}(1,2)=t(1,2) \cos (\alpha )-t(1,3) \sin (\alpha ) \\<br /> \text{t&#039;}(1,3)=t(1,2) \sin (\alpha )+t(1,3) \cos (\alpha ) \\<br /> \text{t&#039;}(2,1)=t(2,1) \cos (\alpha )-t(3,1) \sin (\alpha ) \\<br /> \text{t&#039;}(2,2)=t(3,3) \sin ^2(\alpha )+t(2,2) \cos ^2(\alpha )-(t(2,3)+t(3,2)) \sin (\alpha ) \cos (\alpha ) \\<br /> \text{t&#039;}(2,3)=-t(3,2) \sin ^2(\alpha )+t(2,3) \cos ^2(\alpha )+(t(2,2)-t(3,3)) \sin (\alpha ) \cos (\alpha ) \\<br /> \text{t&#039;}(3,1)=t(2,1) \sin (\alpha )+t(3,1) \cos (\alpha ) \\<br /> \text{t&#039;}(3,2)=-t(2,3) \sin ^2(\alpha )+t(3,2) \cos ^2(\alpha )+(t(2,2)-t(3,3)) \sin (\alpha ) \cos (\alpha ) \\<br /> \text{t&#039;}(3,3)=t(2,2) \sin ^2(\alpha )+t(3,3) \cos ^2(\alpha )+(t(2,3)+t(3,2)) \sin (\alpha ) \cos (\alpha ) \\
The isotropy condition then leads you to a bunch of relations:

t(1,2) = 0
t(1,3) = 0
t(3,1) = 0
t(2,1) = 0
t(2,2) = t(3,3)
t(2,3) = 0
t(3,2) = 0

The rotations around another axis yields only one additional conclusion:

t(1,1) = t(2,2)
 
Last edited:
I must apologize.
What I meant is rank 3 and rank 4, and not 3rd and 4th dimension. So sorry.
You have got a proof for the general form of rank 3 or rank 4 isotropic cartesian tensor?
 
No proof provided :(
 
For the permutation symbol, it should be similar to the rank-2 derivation above, just a bit longer.
See: http://mathworld.wolfram.com/PermutationSymbol.html .

For rank-4, there are 3 types of solutions:
"The number of isotropic tensors of rank 0, 1, 2, ... are 1, 0, 1, 1, 3, 6, 15, 36, 91, 232, ... (Sloane's A005043). These numbers are called the Motzkin sum numbers and are given by the recurrence relation ..."
See: http://mathworld.wolfram.com/IsotropicTensor.html

I have no idea what these 3 independent solutions would be.
However, in principles the derivation should go along the sames lines.

I have no idea how the recurrence relation for higher ranks can be derived.
It must be interesting to find out, because it is unlikely that it would be based on the same elementary reasoning.
There must be a more powerful approach.

See also p87and p88 in
 
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