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

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SUMMARY

An isotropic Cartesian tensor remains unchanged under any rotation, as defined by the transformation t' = R.t.R-1 = t. In three dimensions, specific relations emerge, such as t(1,2) = 0 and t(1,3) = 0, leading to the conclusion that t(1,1) = t(2,2). For rank 4 isotropic tensors, there are three independent solutions, and the derivation of the recurrence relation for higher ranks remains an open question. Resources such as MathWorld provide foundational insights into isotropic tensors and their properties.

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  • Understanding of tensor mathematics
  • Familiarity with rotation matrices
  • Knowledge of isotropic properties in physics
  • Basic grasp of mathematical proofs and derivations
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  • Research the general form of rank 3 and rank 4 isotropic Cartesian tensors
  • Explore the derivation of the recurrence relation for higher rank isotropic tensors
  • Study the properties of permutation symbols in tensor calculus
  • Examine the Motzkin sum numbers and their applications in tensor theory
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Mathematicians, physicists, and engineers interested in advanced tensor analysis, particularly those working with isotropic properties in multi-dimensional spaces.

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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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