Calculating with tensors and simplifying

[EsmeeDijk]

Homework Statement

I have a tensor which is given by $t_{ij} = -3bx_i x_j + b \delta_{ij} x^2 + c \epsilon_{ijk} x_k$
And now I am asked to calculate $(t^2)_{ij} : = t_{ik} t_{kj}$

Homework Equations

The Attempt at a Solution

At first I thought I had to calculate the square of the original tensor and set it equal to the product of the same tensors only with the indices i,j replaced by i, k and k,j.
Later I thought that because of the " : " sign in front of the "= " that I only have to calculate the product of the tensor with the replaced indices because t² is defined by this. So with these different indices I got 9 different terms of which none have same indices so it is not possible to simplify it any further than just taking the product.
Also in the epsilon tensor there are three indices and I gave the last one a different name evertime so m and n giving me epsilon_ikm and epsilon_kjn.
I hope it is clear what my question is.

 

[Fightfish]

So with these different indices I got 9 different terms of which none have same indices so it is not possible to simplify it any further than just taking the product

Could you show us the expression that you got? You might be able to simply it by using the properties of the Kronecker delta as well as certain identities related to the Levi-Civita symbol, such as $\varepsilon_{ijk} \varepsilon^{imn} = \delta_{j}^{m} \delta_{k}^{n} - \delta_{j}^{n} \delta_{k}^{m}$

 

[EsmeeDijk]

Could you show us the expression that you got? You might be able to simply it by using the properties of the Kronecker delta as well as certain identities related to the Levi-Civita symbol, such as $\varepsilon_{ijk} \varepsilon^{imn} = \delta_{j}^{m} \delta_{k}^{n} - \delta_{j}^{n} \delta_{k}^{m}$

Yes of course, I got $9b^2x_i x_j (x_k)^2 - 3b^2 x_i x_k x^2 \delta _{kj} - 3bc x_i x_k x_n \epsilon _{kjn} - 3b^2 x_k x_j x^2 \delta _{ik} + b^2 \delta _{ik} \delta _{kj} x^4 + bc x_n x^2 \delta _{ik} \epsilon _ {kjn} - 3bc x_k x_j x_m \epsilon _{ikm} + bc x_mx^2 \delta _{kj} \epsilon _{ikm} + c^2 x_m x_n \epsilon _{ikm} \epsilon _{kjn}$
We are not making a difference yet between whether the indices are on the top or bottom. Thanks!

 

[Fightfish]

We are not making a difference yet between whether the indices are on the top or bottom. Thanks!

Strictly speaking, the Einstein summation convention only 'works' when one of the indices is up and one is down. But no worries of course - just convert all the up indices on the identities that you see to down indices. In your context, they work in the same way.

You can see that there are many Kronecker deltas $\delta_{kj}$ and whatnot lying around - use them to simplify your expressions. For instance, $x_{i}x_{k}\delta_{kj} = x_{i}x_{j}$. Notice also that because you are using the summation convention, repeated index terms like $x_{k}x_{k}$ work out to be just $x^{2}$. Another useful way to combine terms is to notice that dummy indices can be freely relabelled: for example, $x_{k} \epsilon_{ijk}$ is the same as $x_{m} \epsilon_{ijm}$.

 

[Orodruin]

Apart from what has been mentioned already, you should also note that the Levi-Civita symbol is anti-symmetric. What happens when you contract two indices from an anti-symmetric object with two indices from a symmetric object?