Transfer rank2 tensor to a new basis

In summary, transferring a rank-2 tensor to a new basis involves applying a transformation matrix that corresponds to the change of basis. This process requires the tensor's components to be multiplied by the transformation matrix and its transpose, ensuring that the tensor's properties are preserved in the new coordinate system. The transformation can be expressed mathematically, often resulting in a new set of components that accurately represent the tensor in the altered basis.
  • #1
GGGGc
The trace of the sigma should be the same in both new and old basis. But I get a different one. Really appreciate for the help.
I’ll put the screen shot in the comment part
 
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  • #2
IMG_9040.jpeg
 
  • #3
I have no idea about your calculations since I haven't really learned the rules of manipulations of tensors. However, I am surprised that you don't get off-diagonal entries although the new basis looks as if there should be several of them.

I see that
$$
\sigma_{ij}=\underline{\hat{e}}^{(1)} \otimes \underline{\hat{e}}^{(1)}+\underline{\hat{e}}^{(2)}\otimes \underline{\hat{e}}^{(2)}+2\cdot\underline{\hat{e}}^{(3)}\otimes \underline{\hat{e}}^{(3)}
$$
Hence, I would write ##\underline{\hat{e}}^{(k)}=\alpha_k \underline{\hat{e}}'^{(1)}+\beta_k \underline{\hat{e}}'^{(2)}+\gamma_k \underline{\hat{e}}'^{(3)},## determine the ## \alpha_k\, , \,\beta_k\, , \,\gamma _k,## substitute all of them in the first equation and rearrange everything with the distributive law to obtain an equation
$$
\sigma'_{ij}=\sum_{p,q,r=1}^3 s_{pqr} \cdot \underline{\hat{e}}'^{(p)}\otimes \underline{\hat{e}}'^{(q)}\otimes \underline{\hat{e}}'^{(r)}
$$
 
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