# Can you check my proof?

by learningphysics
Tags: check, proof
 HW Helper P: 4,124 An antisymmetric tensor of rank two $$E_{ij}$$ has $$E_{ij}=-E_{ji}$$ for all i and j right? Meaning for i=j, $$E_{ij}=0$$... Just wanted to make sure of this. Here's what I'm trying to prove. Given $$E_{ij}$$ is an antisymmetric tensor. Is $$E_{ij,k}$$ a tensor? I get yes. Here's my proof: $$E_{i'j',k'} =\sum_i \sum_j \sum_k (\frac{\partial}{x^k}(E_{ij}\frac{\partial x^i}{\partial x^{i'}}\frac{\partial x^j}{\partial x^{j'}}))\frac{\partial x^k}{\partial x^{k'}}$$ This reduces to: $$\sum_i \sum_j \sum_k E_{ij,k}\frac{\partial x^i}{\partial x^{i'}}\frac{\partial x^j}{\partial x^{j'}}\frac{\partial x^k}{\partial x^{k'}} + \sum_i \sum_j \sum_k E_{ij}(\frac{\partial}{x^k}(\frac{\partial x^i}{\partial x^{i'}}\frac{\partial x^j}{\partial x^{j'}}))\frac{\partial x^k}{\partial x^{k'}}$$ Finally after rewriting the second part: $$\sum_i \sum_j \sum_k E_{ij,k}\frac{\partial x^i}{\partial x^{i'}}\frac{\partial x^j}{\partial x^{j'}}\frac{\partial x^k}{\partial x^{k'}} + \sum_i \sum_j \sum_k E_{ij}(\frac{{\partial}^2 x^i}{\partial x^k \partial x^{i'}}\frac{\partial x^j}{\partial x^{j'}} + \frac{\partial x^i}{\partial x^{i'}}\frac{{\partial}^2 x^j}{\partial x^k \partial x^{j'}})\frac{\partial x^k}{\partial x^{k'}}$$ The expression $$E_{ij}$$ is multiplied by in the second triple summation is symmetric about i and j... so if I write one element of the second summation as: $$E_{ij}X_{ijk}$$... so we add two elements of this sum $$i_1$$ and $$j_1$$ are not equal: $$E_{i_1 j_1}X_{i_1 j_1 k_1} + E_{j_1 i_1}X_{j_1 i_1 k_1} = E_{i_1 j_1}X_{i_1 j_1 k_1} - E_{i_1 j_1}X_{i_1 j_1 k_1} =0$$ since E is antisymmetric, and X is symmetric about i and j... So in this manner each element with indices $$i_1,j_1,k_1$$ where the i and j elements are different, cancels with the element with indices $$j_1,i_1,k_1$$. And when $$i_1 = j_1$$ we have $$E_{i_1,j_1} =0$$ so that element in the summation is 0 anyway. So the second summation goes to zero. And we have: $$E_{i'j',k'} = \sum_i \sum_j \sum_k E_{ij,k}\frac{\partial x^i}{\partial x^{i'}}\frac{\partial x^j}{\partial x^{j'}}\frac{\partial x^k}{\partial x^{k'}}$$ so $$E_{ij,k}$$ is a tensor? Does this look right? Thanks a bunch.
 Sci Advisor HW Helper P: 11,915 Of course it's incorrect. This part $$\frac{\partial^{2}x^{i}}{\partial x^{k'}\partial x^{i'}}\frac{\partial x^{j}}{\partial x^{j'}}+\frac{\partial x^{i}}{\partial x^{i'}}\frac{\partial^{2}x^{j}}{\partial x^{k'}\partial x^{j'}}$$ is obviously nonsymmetric in "i" and "j" and nonantisymmetric in <> and <>. Daniel.
 P: 701 what could you do to make it a tensor? Hint: $$A_{i,j}$$ is not a tensor but $$A_{i,j} - A_{j,i}$$ is a tensor
HW Helper
P: 4,124
Can you check my proof?

 Quote by dextercioby Of course it's incorrect. This part $$\frac{\partial^{2}x^{i}}{\partial x^{k'}\partial x^{i'}}\frac{\partial x^{j}}{\partial x^{j'}}+\frac{\partial x^{i}}{\partial x^{i'}}\frac{\partial^{2}x^{j}}{\partial x^{k'}\partial x^{j'}}$$ is obviously nonsymmetric in "i" and "j" and nonantisymmetric in <> and <>. Daniel.
Thanks dexter.
HW Helper
P: 4,124
 Quote by quetzalcoatl9 what could you do to make it a tensor? Hint: $$A_{i,j}$$ is not a tensor but $$A_{i,j} - A_{j,i}$$ is a tensor

I just proved that $$E_{ij,k} + E_{jk,i} + E_{ki,j}$$ is a tensor where $$E_{ij}$$ is antisymmetric. Is that what you were referring to?
P: 701
 Quote by learningphysics I just proved that $$E_{ij,k} + E_{jk,i} + E_{ki,j}$$ is a tensor where $$E_{ij}$$ is antisymmetric. Is that what you were referring to?
yes

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