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## Can you check my proof?

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.
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 Blog Entries: 9 Recognitions: Homework Help Science Advisor 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.
 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

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## 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.

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 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?

 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