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Can you check my proof?

by learningphysics
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learningphysics
#1
Jun25-05, 06:52 PM
HW Helper
P: 4,124
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.

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.
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dextercioby
#2
Jun25-05, 07:21 PM
Sci Advisor
HW Helper
P: 11,927
Of course it's incorrect.

This part

[tex] \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'}} [/tex]

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

Daniel.
quetzalcoatl9
#3
Jun26-05, 09:13 AM
P: 701
what could you do to make it a tensor?

Hint:

[tex]A_{i,j}[/tex] is not a tensor but
[tex]A_{i,j} - A_{j,i}[/tex] is a tensor

learningphysics
#4
Jun26-05, 09:26 PM
HW Helper
P: 4,124
Can you check my proof?

Quote Quote by dextercioby
Of course it's incorrect.

This part

[tex] \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'}} [/tex]

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

Daniel.
Thanks dexter.
learningphysics
#5
Jun26-05, 09:45 PM
HW Helper
P: 4,124
Quote Quote by quetzalcoatl9
what could you do to make it a tensor?

Hint:

[tex]A_{i,j}[/tex] is not a tensor but
[tex]A_{i,j} - A_{j,i}[/tex] is a tensor

I just proved that [tex]E_{ij,k} + E_{jk,i} + E_{ki,j}[/tex] is a tensor where [tex]E_{ij}[/tex] is antisymmetric. Is that what you were referring to?
quetzalcoatl9
#6
Jun27-05, 01:40 AM
P: 701
Quote Quote by learningphysics
I just proved that [tex]E_{ij,k} + E_{jk,i} + E_{ki,j}[/tex] is a tensor where [tex]E_{ij}[/tex] is antisymmetric. Is that what you were referring to?
yes


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