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 - The Fusion of Science and Community**

# Can you check my proof?

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

- Similar discussions for: Can you check my proof?

Loading...

**Physics Forums - The Fusion of Science and Community**