Evaluate Ai = (Epsilon)ijk bj bk

[R3DH34RT]

Hi guys, can u please help me in this problem?

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.

Please help me guys, I'm so confused... :(
Thanks a lot...

 

[D H]

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?

 

[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.

 

[D H]

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

 

[R3DH34RT]

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

 

[D H]

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?

 

[R3DH34RT]

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

 

[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...

 

[R3DH34RT]

Hello?

 

[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} ]<br /> = 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.

 

[R3DH34RT]

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

 

[D H]

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.

 

[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.

 

[R3DH34RT]

So, I need to do the tensor manipulation?

 

[haushofer]

So, I need to do the tensor manipulation?

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}<br />

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 ;)

 

[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...

 

[R3DH34RT]

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

 

[R3DH34RT]

Hello...? :(

 

[robphy]

Hello...? :(

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.