You're welcome. It works in 3d, but too much to post.

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- Apr 18th 2013, 12:57 PMHartlwRe: Einstein indicial notation problem
- Apr 18th 2013, 01:16 PMmiladRe: Einstein indicial notation problem
A

^{2}_{ik}means each member powered to 2 . A_{ij}*A_{jk}=A^{2}_{ik}is true only if A is a diagonal tensor .which I think you didn't notice :). thank you so much for your time and effort . - Apr 18th 2013, 01:38 PMHartlwRe: Einstein indicial notation problem
- Apr 29th 2013, 03:18 AMmiladRe: Einstein indicial notation problem
I showed the answer to my professor but he says I must do all the work in index notation .

- Apr 30th 2013, 10:56 AMHartlwRe: Einstein indicial notation problem
With principal directions in 3d

(s_{11}^{2}+s_{22}^{2}+s_{33}^{2})^{2}= s_{11}^{4}+s_{22}^{4}+s_{33}^{4}+ 2(s_{11}^{2}s_{22}^{2}+ s_{11}^{2}s_{33}^{2}+ s_{22}^{2}s_{33}^{2})

(s_{11}+s_{22}+s_{33})^{2}= s_{11}^{2}+s_{22}^{2}+s_{33}^{2}+ 2(s_{11}s_{22}+ s_{11}s_{33}+ s_{22}s_{33}) = 0 because s_{ii}=0. So

s_{11}^{2}+s_{22}^{2}+s_{33}^{2}= -2(s_{11}s_{22}+ s_{11}s_{33}+ s_{22}s_{33})

(s_{11}^{2}+s_{22}^{2}+s_{33}^{2})^{2}= 4[s_{11}^{2}s_{22}^{2}+s_{11}^{2}s_{33}^{2}+s_{22}^{2}s_{33}^{2 }+ 2(s_{11}+s_{22}+s_{33})(s_{11}s_{22}s_{33})] and s_{ii}=0. So

s_{11}^{2}s_{22}^{2}+s_{11}^{2}s_{33}^{2}+s_{22}^{2}s_{33}^{2}= ¼(s_{11}^{2}+s_{22}^{2}+s_{33}^{2})^{2}and substitute this into first eq to get:

s_{11}^{4}+s_{22}^{4}+s_{33}^{4 }= ½(s_{11}^{2}+s_{22}^{2}+s_{33}^{2})^{2}, or in indicial notation:

s_{ii}^{4}= s_{ii}^{2}s_{jj}^{2}

The above in indicial notation is:

(s_{ii}^{2})^{2 = }s_{ii}^{4}+ 2A

(s_{ii})^{2}= s_{ii}^{2 }+ 2B = 0

s_{ii}^{2 }= -2B

(s_{ii}^{2})^{2}= 4B^{2}= 4A,__if you can show B__^{2}= A

__Then__

s_{ii}^{4}= s_{ii}^{2}s_{jj}^{2}

I can’t express things like s_{11}s_{22}+ s_{11}s_{33}+ s_{22}s_{33}in indicial notation.

Surely your Prof at some point has to show you how to do this problem in indicial notation.

Please let us know.

EDIT: Had to juggle with Latex help again to get my Word sub-super script version to post.