Tensor identity - Help

by jvicens
Tags: identity, tensor
jvicens is offline
Jul31-04, 01:45 AM
P: 18
I'm having some trouble to prove the following tensor identity shown below in Einstein's summation convention:


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


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?
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Lonewolf is offline
Jul31-04, 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.
robphy is offline
Jul31-04, 09:47 AM
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To elaborate on Lonewolf's reply...
&= 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}\\

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 right-hand-side:
&= 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}

jvicens is offline
Jul31-04, 11:10 PM
P: 18
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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
I was thinking of it as when we say
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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