I Transfer rank2 tensor to a new basis

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
 
Physics news on Phys.org
IMG_9040.jpeg
 
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
$$
\sigma'_{ij}=\sum_{p,q,r=1}^3 s_{pqr} \cdot \underline{\hat{e}}'^{(p)}\otimes \underline{\hat{e}}'^{(q)}\otimes \underline{\hat{e}}'^{(r)}
$$
 
Thread 'How to define vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 19 ·
Replies
19
Views
5K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
2
Views
4K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
18K
  • · Replies 16 ·
Replies
16
Views
5K