Tensor identity - Help

[jvicens]

I'm having some trouble to prove the following tensor identity shown below in Einstein's summation convention:

(a_{ij}+a_{ji})x_{i}x_{j}=2a_{ij}x_{i}x_{j}

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

a_{ij}=a_{ji}

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?

[Lonewolf]

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

[robphy]

To elaborate on Lonewolf's reply...

\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*}


You'll learn that \frac{1}{2}(a_{ij} + a_{ji})
is called "the symmetric part of a_{ij} ", and is written as a_{(ij)} .
Hence, your identity can be written
2a_{(ij)}x_{i}x_{j} =2a_{ij}x_{i}x_{j}.

Here is an "index gymnastics" proof, starting with half of your right-hand-side:

\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*}

[jvicens]

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
a_{ij}=a_{ji}
I was thinking of it as when we say
a=b
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 :smile: