An antisymmetric tensor of rank two [tex]E_{ij}[/tex] has [tex]E_{ij}=-E_{ji}[/tex] for all i and j right? Meaning for i=j, [tex]E_{ij}=0 [/tex]... Just wanted to make sure of this.(adsbygoogle = window.adsbygoogle || []).push({});

Here's what I'm trying to prove. Given [tex]E_{ij}[/tex] is an antisymmetric tensor. Is [tex]E_{ij,k}[/tex] a tensor?

I get yes. Here's my proof:

[tex]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'}}[/tex]

This reduces to:

[tex]\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'}}[/tex]

Finally after rewriting the second part:

[tex]\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'}}[/tex]

The expression [tex]E_{ij}[/tex] 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:

[tex]E_{ij}X_{ijk}[/tex]... so we add two elements of this sum [tex]i_1[/tex] and [tex]j_1[/tex] are not equal:

[tex]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[/tex] since E is antisymmetric, and X is symmetric about i and j...

So in this manner each element with indices [tex]i_1,j_1,k_1[/tex] where the i and j elements are different, cancels with the element with indices [tex]j_1,i_1,k_1[/tex].

And when [tex]i_1 = j_1[/tex] we have [tex]E_{i_1,j_1} =0 [/tex] so that element in the summation is 0 anyway.

So the second summation goes to zero. And we have:

[tex]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'}}[/tex] so [tex]E_{ij,k}[/tex] is a tensor?

Does this look right? Thanks a bunch.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Can you check my proof?

**Physics Forums | Science Articles, Homework Help, Discussion**