Tensor problem

1. Nov 5, 2007

ponjavic

I hope no one minds but I'm gonna keep all my questions and solutions here and then write it down for the report =)

1. The problem statement, all variables and given/known data
1 Prove the following identities using index notation:
a) For vectors, u, v, w,
(u cross v) dot w = u dot (v cross w) XsolvedX

b)
For two second order tensors, show that trace(AB) = trace(BA) using index notation

c)
For two second order tensors, show that trace(AB) = trace(BA) using index notation
(AB)^t = B^t A^t

2 Consider a tensor V and a vector w related as follows
V_kj = e_ijk*w_i

a) Write out V as a 3 x 3 matrix

b) Show, using index notation, that V is a skew symmetric tensor (V = -V^t)

c) Show, using index notation or otherwise, that for any vector, v,
Vv = w x v

d) Show, using index notation, that the vector w is written in terms of V by,
w_i = 1/2 * e_ijk * V_kj

The vector w is known as the vector dual of the skew symmetric tensor V

Note: For part (d) you may make use of one or the both identities:
e_ijk * e_ipq = delta_jp * delta_kq - delta_kp * delta_jq
e_ijk * e_jkr = e_ijk * e_rjk = 2delta_ir

2. Relevant equations
a)
(u dot v) = uivj
(u cross v) = eijkujvk
(eijk is the permutation tensor)

b)
trace(A) = Aii

3. The attempt at a solution
(u cross v) dot w= eijkujvk * wi =
eijkwiujvk =
ejkiuivjwk =
(as eijk = ejki) number of permutations are even:
eijkvjwk * ui=
u dot (v cross w)

solved

b)
trace(AB) = trace(ABij) = trace(AikBkj) = AikBki = BkiAik = trace(BkiAjk) =
trace(BAij) = trace(BA)

c)
A=Aik, A^t = Aki
B=Bkj, B^t = Bjk
(AB)^t = (AikBkj)^t .. not quite sure how the transpose operation would work on a multiplication

2
a)
V_kj =
w1, w1, w1
w2, w2, w2
w3, w3, w3

No idea what to do with e_ijk...

b)
For A to be skew symmetric, A^t = -A

V_kj^t = ... Do not know how to transpose e_ijk * w_i
possibly: e_ikj * w_k
Then permutation k <-> j:
-e_ijk * w_k = -V_kj

Is this allowed?

Last edited: Nov 5, 2007
2. Nov 5, 2007

Meir Achuz

a) your cyclic permuttation of ijk is correct because e_ijk is invarinat ulnder cyclic permutation.

b) recall (AB)_ij=AikBkj. Now the trace sets i=j and sums.

3. Nov 5, 2007

ponjavic

edit: check first post

Last edited: Nov 5, 2007
4. Nov 7, 2007

ponjavic

Comon people I'm really struggling with this, no one done tensors?

5. Nov 7, 2007

cristo

Staff Emeritus
Sorry, I forgot that we were at your beck and call

6. Nov 7, 2007

ponjavic

Helpful, I hope you can do tensors and aren't just wasting my time

7. Nov 7, 2007

cristo

Staff Emeritus
For (c) you know how to write a product, you know how to write a transpose, and so you should be able to write the transpose of that product.

The parts you've answered from 2 look correct.