- #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
I’ll put the screen shot in the comment part
Transfer rank2 tensor to a new basis
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- Thread starter GGGGc
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- #2
GGGGc
- #3
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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
$$
\sigma'_{ij}=\sum_{p,q,r=1}^3 s_{pqr} \cdot \underline{\hat{e}}'^{(p)}\otimes \underline{\hat{e}}'^{(q)}\otimes \underline{\hat{e}}'^{(r)}
$$
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
$$
\sigma'_{ij}=\sum_{p,q,r=1}^3 s_{pqr} \cdot \underline{\hat{e}}'^{(p)}\otimes \underline{\hat{e}}'^{(q)}\otimes \underline{\hat{e}}'^{(r)}
$$
1. How do you transfer a rank 2 tensor to a new basis?
To transfer a rank 2 tensor to a new basis, you can use the transformation matrix. Multiply the original tensor by the inverse of the transformation matrix to obtain the tensor in the new basis.
2. What is a rank 2 tensor?
A rank 2 tensor is a mathematical object that takes two vectors as input and produces a scalar as output. It can be represented as a matrix of components.
3. Why is it important to transfer a tensor to a new basis?
Transferring a tensor to a new basis allows us to analyze the same physical quantity from different perspectives. It helps simplify calculations and can reveal underlying patterns or symmetries.
4. What is a basis in linear algebra?
In linear algebra, a basis is a set of linearly independent vectors that can be used to represent any vector in a vector space. Changing the basis can make calculations easier or reveal different properties of the vectors.
5. How do you calculate the transformation matrix for a new basis?
To calculate the transformation matrix for a new basis, you can use the old basis vectors and the new basis vectors. Arrange the new basis vectors as columns in a matrix and multiply it by the inverse of the matrix formed by the old basis vectors. This will give you the transformation matrix.
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