# Need help on tensor

1. Sep 25, 2007

### R3DH34RT

1. Evaluate Ai = (Epsilon)ijk bj bk

2. Show that this is true: Tij Wij = 0, where Tij = symmetric and Wij = anti-symmetric

3. Show that this is true: Bik = (Epsilon)ijk aj is a anti-symmetric tensor

All i, j, k are indices of the tensor.

Thanks a lot...

2. Sep 25, 2007

### Staff: Mentor

R3DH34RT, you need to show some work on this before we can help you. What are the relevant equations here? What have you done to attempt to solve the problem?

3. Sep 25, 2007

### R3DH34RT

That's why I'm confused...
Epsilon is the alternating tensor (permutation). I've tried to expand the equation, but it still can't solve the problem...
Can u help me? Thanks.

4. Sep 25, 2007

### Staff: Mentor

Hint: What is the relation between $\epsilon_{ikj}$ and $\epsilon_{ijk}$?

5. Sep 25, 2007

### R3DH34RT

(Epsilon)ijk is -(Epsilon)ikj right?
But is there any relationship? :(

6. Sep 25, 2007

### Staff: Mentor

But is there any relationship?? You just named it! What doe $\epsilon_{ikj}=-\epsilon_{ijk}$ when it comes to computing $\sum_{j,k}\epsilon_{ijk}b_jb_k$?

7. Sep 25, 2007

### R3DH34RT

Isn't that (Epsilon)ijk bj bk means BxB?
Does that have something to do with (Epsilon)ikj?
Thanks

8. Sep 25, 2007

### R3DH34RT

Is that mean the result is zero? That's in my calculation... I hope it's right...
And can u help me with the other question please? :(
Thanks...

9. Sep 27, 2007

### R3DH34RT

Hello?

10. Sep 27, 2007

### haushofer

1) Just do the summation. You don't have to do all the terms, because you know that $$\epsilon_{ijk}$$ is antisymmetric in {ijk}.
So you get in 3 dimensions:

$$A_{1}=\epsilon_{1jk}b_{j}b_{k}= \epsilon_{123}b_{2}b_{3}+\epsilon_{132}b_{3}b_{2}$$.

Now you know that $$\epsilon_{123}=1=-\epsilon_{132}$$, so you can write it for general components $$A_{i}$$. That looks pretty much like a cross product in 3 dimensies, right?

2) You know that every rank 2 tensor can be written as
$$T^_{ij} = \frac{1}{2} [T_{ij}+T_{ji}] + \frac{1}{2}[T_{ij}-T_{ji} ] = T_{[ij]}+T_{(ij)}$$ You simply decompose it in an symmetric and antisymmetric part. People use [] around the indices to indicate the symmetric part, and () to indicate the antisymmetric part. So we have:
$$T_{ij}=T_{[ij]}, \ \ \ W^{ij}=W^{(ij)}$$

Let's do the contraction:

$$T_{ij}W^{ij}=\frac{1}{2} (T_{ij}+T_{ji})W^{ij} = \frac{1}{2}(T_{ij}W^{ij}+T_{ji}W^{ij})$$

Take a close look at the last contraction; because W is antisymmetric, we have
$$T_{ji} W^{ij} = -T_{ij} W^{ij}$$
And than you're done. ( remember that the names of the indices don't matter; it matters which indices you contract ! )

3) Again, write out the summation and see what happens if you interchange the indices i and k. Or you could contract it with a symmetric tensor of rank 2 and observe that it becomes 0.

Last edited: Sep 28, 2007
11. Sep 27, 2007

### R3DH34RT

Hi haushofer!
But I am still not very clear with you answer to number 3.
Can you explain a bit more?
Thank you.

12. Sep 28, 2007

### Staff: Mentor

The result a 3x3 tensor; simply build it. It is not hard to compute. Once you build it you should be able to easily show the tensor is anti-symmetric.

13. Sep 28, 2007

### robphy

You can use 1 to prove 3.
One way to show something is antisymmetric is to show that one gets zero when contracting those indices with something symmetric. That is, a 2-index tensor is antisymmetric when its symmetric part is zero.

14. Sep 28, 2007

### R3DH34RT

So, I need to do the tensor manipulation?

15. Sep 29, 2007

### haushofer

Yes. Just do the summation from j=1 to j=3 ( I assume you have to proof it in 3 dimensions, but the generalization is straightforward ) So you get

$$B_{ik} = \epsilon_{i1k}a^{1} + \epsilon_{i2k}a^{2} + \epsilon_{i3k}a^{3}$$

With calculations like this, it is always a good thing to check if we have the same indices on the left and on the right. Now, you know that all the terms with epsilons which contain equal indices are 0, due to the antisymmetric properties of it, so these we can forget.You also know that for an antisymmetric tensor we have

$$B_{ik}= - B_{ki}$$

So now you can check this directly. Our tensor B is a sum of 3 tensors, so every component should be antisymmetric to make the whole thing antisymmetric. You also know what happens if you change 2 indices of your epsilon tensor.

But like mentioned, there are other possibilities to proof the antisymmetric character of tensor. Don't be shy; just write out those summations and play a little with the indices ;)

16. Sep 29, 2007

### R3DH34RT

Thanks, haushofer, you are very helpful!
I'll try and hopefully I'll find the answer...
But there are another problem that seems complicated, I found these on general textbook and I can't understand how to solve it. Can you please also help me?

A.> n = 0.5i + 0.5j + 0.7071k is the unit-normal for plane A. b = 4i + 5j + 2k, c = 2i + 3j + k. Calculate the area of parallelogram project from b x c to plane A. Calculate components of vectors b and c that are parallel to plane A.

B.> New right hand coordinate axes are chosen at the same origin with e1' = (2e1 + 2e2 + e3)/3 and e2' = (e1 - e2) x 1.4. Express e3' in term of e1. If t = 10e1 + 10e2 - 20e3, express t in terms of the new basis ek'. Express the old coordinate xi in term of xk' , xi = f(xk')

Thank you very much...

17. Sep 30, 2007

### R3DH34RT

Hi... can anyone olease help me...? :(

18. Oct 2, 2007

### R3DH34RT

Hello...? :(

19. Oct 3, 2007

### robphy

Your follow up question is less about tensors and differential geometry and more about a homework-type problem in linear algebra, which is probably better posted here:
https://www.physicsforums.com/forumdisplay.php?f=155
where you are expected to show your attempt at the problem.