Prove the following identities using index notation

  • Thread starter Thread starter ponjavic
  • Start date Start date
  • Tags Tags
    Tensor
Click For Summary
The discussion revolves around proving identities using index notation, specifically for vectors and second-order tensors. Key points include demonstrating that (u cross v) dot w equals u dot (v cross w) and establishing that trace(AB) equals trace(BA) for tensors. Participants also explore the properties of a skew symmetric tensor V and its relationship with a vector w, showing that V can be expressed as a matrix and that Vv equals w cross v. The conversation highlights challenges in understanding tensor operations and the importance of cyclic permutations in tensor notation. Overall, the thread serves as a collaborative effort to clarify tensor identities and their proofs.
ponjavic
Messages
221
Reaction score
0
I hope no one minds but I'm going to keep all my questions and solutions here and then write it down for the report =)

Homework Statement


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

Homework Equations


a)
(u dot v) = uivj
(u cross v) = eijkujvk
(eijk is the permutation tensor)

b)
trace(A) = Aii

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:
Physics news on Phys.org
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.
 
edit: check first post
 
Last edited:
Comon people I'm really struggling with this, no one done tensors?
 
ponjavic said:
Comon people I'm really struggling with this, no one done tensors?

Sorry, I forgot that we were at your beck and call :rolleyes:
 
cristo said:
Sorry, I forgot that we were at your beck and call :rolleyes:
Helpful, I hope you can do tensors and aren't just wasting my time
 
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.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
892
  • · Replies 7 ·
Replies
7
Views
2K