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ponjavic
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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 =)
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
a)
(u dot v) = uivj
(u cross v) = eijkujvk
(eijk is the permutation tensor)
b)
trace(A) = Aii
(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?
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?
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