Proving E_{ij,k} as a Tensor: A Comprehensive Explanation

[learningphysics]

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.

 

[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 <<j'>> and <<k'>>.

Daniel.

 

[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

 

[learningphysics]

Of course it's incorrect.

This part

\frac{\partial^{2}x^{i}}{\partial x^{k&#039;}\partial x^{i&#039;}}\frac{\partial x^{j}}{\partial x^{j&#039;}}+\frac{\partial x^{i}}{\partial x^{i&#039;}}\frac{\partial^{2}x^{j}}{\partial x^{k&#039;}\partial x^{j&#039;}}

is obviously nonsymmetric in "i" and "j" and nonantisymmetric in <<j'>> and <<k'>>.

Daniel.

Thanks dexter.

 

[learningphysics]

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?

 

[quetzalcoatl9]

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