# Isotropic tensor functions

1. Nov 4, 2006

### Andrea

Hello,

I consider only Cartesian tensors in the following. The definition of
isotropic tensor function I know is

1) T = F ( G )

such that, for any rotation ( ' = transpose),

2) O F( G ) O' = F( O G O' )

But, if I change to component notation, it seem to me that any tensor
function is isotropic, which cannot obviously be. Denoting the
components in the new basis with ^*, I have

3a) T_ij^* = O_ir T_rs (O_sj)'

3b) G_ij^* = O_ir G_rs (O_sj)'

since T and G are tensors. Then, by 1),

4a) T_rs = F_rs ( G_mn )

4b) T_rs^* = F_rs^* ( G_mn^* )

Then, substituting 3a) and 3b) into 4b), I get

5) O_ir T_rs (O_sj)' = F_rs^* ( O_mk G_kl (O_ln)' )

Finally, substituting 4a) into 5), I have

6) O_ir F_rs ( G_mn ) (O_sj)' = F_rs^* ( O_mk G_kl (O_ln)' )

that is,

O F( G ) O' = F (O G O' )

So any tensor function would be isotropic. Clearly that's false, but I
don't see where the error is. Can you help me find it? Thanks,

Andrea

2. Feb 11, 2009

### edmi0002

Here's an example of a tensor valued, anisotropic function

$$H = g(T) = (T\cdot{n\otimes{n}})n\otimes{n}$$
where n is a fixed unit vector.

Then
$$g(QTQ^T) = (QTQ^T\cdot{n\otimes{n}})n\otimes{n} = (T \cdot (Q^Tn \otimes Q^Tn)) n\otimes{n}$$
which is not the same as
$$Qg(T)Q^T = (T\cdot{n\otimes{n}})Qn\otimes{Qn}$$
for general orthogonal Q, only if n is a proper vector of Q.
Im basically extrapolating from Jaunzemis' Continuum Mechanics book, pg 287.