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Hooke's law for an elastic isotropic 2-d material is a rank 4 tensor with 16 elements $$ \begin{pmatrix} a_{11} & a_{12} & 0 & 0 \\ a_{12} & a_{11} & 0 & 0 \\ 0 & 0 & a_{33} & 0 \\ 0 & 0 & 0 & a_{33} \end{pmatrix} $$ Since the material is isotropic the tensor is the same for all orthogonal frames, hence it is invariant for for all orthonormal coordinate transformations.

But, looking at an orthonormal coordinate transformation for the top left $2\times2$ submatrix, $$ \begin{pmatrix} \cos(1) & \sin(1) \\ -\sin(1) & \cos(1)\end{pmatrix} \begin{pmatrix} .9 & .1 \\ .1 & .9\end{pmatrix} \begin{pmatrix} \cos(1) & -\sin(1) \\ \sin(1) & \cos(1)\end{pmatrix} $$ doesn't equal $\begin{pmatrix} .9 & .1 \\ .1 & .9\end{pmatrix}$.

So, where is the error in the above?

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  • $\begingroup$ Please use MathJax to format your mathematical expressions. I've edited your question this time, to give you an example to follow in future. $\endgroup$ – user197851 Nov 29 '18 at 18:26
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Looks like you've confused a fourth rank tensor with a $4\times4$ matrix. Often a condensed notation (Voigt notation) is used for these tensor elements: each index of $a$ really stands for two indices in the true tensor. You seem to have adopted this. If you want to express the transformation properties of the tensor, you need to express it in the full notation, where each element has four indices. Then the transformation will involve a summation over indices in which the typical term has four contributions from the transformation matrix, like this $$ a_{IJKL} = M_{Ii}M_{Jj}M_{Kk}M_{Ll} \, a_{ijkl} $$ where the $M$ terms describe the coordinate rotation (this corresponds to your $2\times2$ rotation matrix) and we sum over the lowercase indices (Einstein convention).

You should be able to express all the $a_{ijkl}$ in terms of two Lamé constants $\lambda$ and $\mu$ (see here for example), and it will take the form explained in the answer of @JEB ; then you can easily verify that it is isotropic, according to the transformation rule given here.

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There are 3 independent isotropic rank-4 tensors:

$$ \delta_{ij}\delta_{kl} $$

$$ \delta_{ik}\delta_{jl} $$

$$ \delta_{il}\delta_{jk} $$

so if your tensor is not proportional to a linear combination of those, it is not isotropic.

Isotropic tensors, esp. for rank greater than 3, fall under the study of "syzygy", as not all index combinations are linearly independent thanks to Capelli's identity.

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