Hey guys im having a little trouble with this exercice, which is supposed to be quite simple QQ



Thank you!

Choose i, j and k as 3 integers among 1, 2 or 3.

If at least 2 among i, j and k are equals, then Epsilon(i,j,k) = 0, then Epsilon(i,jk,)*Delta(i,j) = [ 0 ]

If (i, j,k) = (1,2,3) or (2,3,1) or (3,1,2), then Epsilon(i,j,k) = 1, then Epsilon(i,jk,)*Delta(i,j) = Delta(i,j)

If (i,j,k) = (1,3,2) or (2,1,3) or (3,2,1) then Epsilon(i,j,k) = -1, then Epsilon(i,jk,)*Delta(i,j) = - Delta(i,j)

But I think there's a mistake in your exercice, since Delta(i,j) doesn't depend neither on i nor on j.

General knowledge :

http://en.wikipedia.org/wiki/Cartesian_tensor

ty for reply!
i know the answer must be a vector ( one dimension - > k )
so that would be [ 0 , 0, 0 ] because delta (i,j) = 1 only when i=j
e: delta(i,j) = 1 if i=j
delta (i,j) = 0 if i=/j
ee: delta(i,j is kroneker tensor

This post was edited by Toilet on Nov 10 2013 10:14am

Write it out component by component. It will seem long and tedious but to be honest its the best way to learn tensors.

The best part about this tensor is that there is no summation. So you can just pick indices and write out the value.

For example,

choose (i,j,k) to be (0,1,0)

That gives you e(0,1,0)delta(0,1). You should know that for both of the tensors with these indices are zero, so your value is zero. Continue with the rest of the possible combinations (and you can eliminate a LOT by inspection. Epsilon depends on cyclic rotation of the indices, while the Kronecker delta depends on i,j being the same.

Then you can write your new tensor, say X(i,j,k) with the components found by writing out the value corresponding to the ith, jth, and kth component.

Now that I think about it, every value should be zero, but I havnt done tensors for a few months.

Edit: Take this with a grain of salt. Its early and I might be mistaken. My professor did this in class but I recently threw away my notes because I finished college ><

This post was edited by khemist on Nov 10 2013 10:34am

thank you for reply
when you do a product, the answer's dimension is the indice that isnt repeted
the only indice not repeted here is k, so that would make the asnwer dimension one, being a vector
and i assume it would be vector 0 since every value should be zero.
sorry english isnt my first language.

The indices on a tensor are NOT a label of the dimension. Its a label of the RANK of the tensor. A rank zero tensor is a scalar (or an ordinary number), rank 1 a vector, and everything else is called a tensor. Technically, scalars and vectors are tensors too, but they are very special types. You will eventually learn the difference between the upstairs and downstairs index, but even a tensor with 2 upstairs and 3 downstairs is still a rank 5 tensor (with 4x4x4x4x4=1024 components, yikes!).

We can eliminate all components with indices that have i not equal to j (because of the delta). But we already know that the epsilon has to have a cyclic permutation of (1,2,3) as the ith, jth, and kth component (by definition). Thus, every value is zero.

You can try writing it out, and I am guessing your professor would like you to write at least a few of the values to show you know how to compute the component of a tensor.

This post was edited by khemist on Nov 10 2013 10:41am

thank you,
sorry meant rank (traduction from french lol)
i agree with every value being zero, but
the rank ( as i called it dimension in last post) should be 1 ( vector ) because there is one indice left and not scalar right ?
ye ive been writing some of them

No.

Epsilon is not TECHNICALLY a tensor. Taken from wiki, "the Levi-Civita symbol represents a collection of numbers; defined from the sign of a permutation of the natural numbers 1, 2, …, n, for some positive integer n." So it is a number (either -1, 0 or 1).

So you will actually get a rank 2 tensor (not a scalar like I previously said). There is no contraction here because all the indices are "downstairs" and are not summed over (definition of contraction). Rather, you are multiplying a number by a rank 2 tensor, which yields a rank 2 tensor.

You can write the new tensor as n(i,j) = epsilon(i,j,k)delta(i,j)

And you know this tensor will be the zero tensor.

This post was edited by khemist on Nov 10 2013 11:02am

gotcha!
thanks a lot for help,
very appreciated