
#1
Jul3104, 01:45 AM

P: 18

I'm having some trouble to prove the following tensor identity shown below in Einstein's summation convention:
[tex](a_{ij}+a_{ji})x_{i}x_{j}=2a_{ij}x_{i}x_{j}[/tex] I expanded the terms but when I did group them I didn't get the identity. The only way I could get the identity is if [tex]a_{ij}=a_{ji}[/tex] and I don't see a reason why this would be so. Obviously I'm missing something. Can somebody tell what is it that I'm doing wrong? 



#2
Jul3104, 05:03 AM

P: 333

It's true that [tex]a_{ij}x_{i}x_{j} = a_{ji}x_{i}x_{j}[/tex]. Putting in the summation signs may help you to see this.




#3
Jul3104, 09:47 AM

Sci Advisor
HW Helper
PF Gold
P: 4,108

To elaborate on Lonewolf's reply...
[tex] \begin{align*} a_{ij}x_{i}x_{j} &= a_{ji}x_{j}x_{i}&&\text{relabel repeated [dummy] indices}\\ &= a_{ji}x_{i}x_{j}&&\text{reorder the writing of the x factors}\\ \end{align*} [/tex] You'll learn that [itex]\frac{1}{2}(a_{ij} + a_{ji})[/itex] is called "the symmetric part of [itex] a_{ij} [/itex]", and is written as [itex] a_{(ij)} [/itex]. Hence, your identity can be written [tex] 2a_{(ij)}x_{i}x_{j} =2a_{ij}x_{i}x_{j}[/tex]. Here is an "index gymnastics" proof, starting with half of your righthandside: [tex] \begin{align*} a_{ij}x_{i}x_{j} &= a_{ij}x_{(i}x_{j)}&&\text{since } x_{i}x_{j} = x_{(i}x_{j)} \\ &= a_{(ij)}x_{i}x_{j}&&\text{since i and j are being symmetrized} \end{align*} [/tex] 



#4
Jul3104, 11:10 PM

P: 18

Tensor identity  Help
Guys, thanks to you help I got it right this time after some thinking and some calculations. I realized one of my problems was the fact that when I was looking at
[tex]a_{ij}=a_{ji}[/tex] I was thinking of it as when we say [tex]a=b[/tex] The way I should look at that expression is not isolated from the rest of the other terms. The important thing, I think, is not the term itself isolated but how the summation terms comes up after the entire summation is expanded. I can continue now with the next page of my book 


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