# Isotropic Cartesian Tensors

1. Jun 30, 2014

### TSC

Does anyone have a proof of what a isotropic cartesian tensor should look like in three or four dimensions?

2. Jul 1, 2014

### maajdl

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:

$$\text{t'}(1,1)=t(1,1) \\ \text{t'}(1,2)=t(1,2) \cos (\alpha )-t(1,3) \sin (\alpha ) \\ \text{t'}(1,3)=t(1,2) \sin (\alpha )+t(1,3) \cos (\alpha ) \\ \text{t'}(2,1)=t(2,1) \cos (\alpha )-t(3,1) \sin (\alpha ) \\ \text{t'}(2,2)=t(3,3) \sin ^2(\alpha )+t(2,2) \cos ^2(\alpha )-(t(2,3)+t(3,2)) \sin (\alpha ) \cos (\alpha ) \\ \text{t'}(2,3)=-t(3,2) \sin ^2(\alpha )+t(2,3) \cos ^2(\alpha )+(t(2,2)-t(3,3)) \sin (\alpha ) \cos (\alpha ) \\ \text{t'}(3,1)=t(2,1) \sin (\alpha )+t(3,1) \cos (\alpha ) \\ \text{t'}(3,2)=-t(2,3) \sin ^2(\alpha )+t(3,2) \cos ^2(\alpha )+(t(2,2)-t(3,3)) \sin (\alpha ) \cos (\alpha ) \\ \text{t'}(3,3)=t(2,2) \sin ^2(\alpha )+t(3,3) \cos ^2(\alpha )+(t(2,3)+t(3,2)) \sin (\alpha ) \cos (\alpha ) \\$$
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: Jul 1, 2014
3. Jul 1, 2014

### TSC

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?

4. Jul 1, 2014

5. Jul 1, 2014

### TSC

No proof provided :(

6. Jul 1, 2014

### maajdl

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.