CONTINUUM MECHANICS for ENGINEERS

and, furthermore, that for repeated subscripts ε is zero as in
ijk
Therefore, now the vector cross product above becomes
u× v= ueˆ × v eˆ = uv ( eˆ × eˆ i i j j i j i j)= ε uv eˆ
(2.2-10)
ijk i j k
Again, notice how the scalar components pass through the vector cross
product operator.
(d) triple scalar product (box product):
or
(2.2-11)
where in the final step we have used both the substitution property of δ and ij
the sign-change property of .
(e) triple cross product:
ε113 = ε212 = ε133 = ε222
= 0
u⋅ v× w= u× v⋅ w=[ u, v,w]
[ uv,w , ]= ueˆ ⋅ ( v eˆ × w eˆ )= ueˆ ⋅εv
w eˆ
i i j j k k i i jkq j k q
= ε uvw δ = ε uvw
jkq i j k iq ijk i j k
ε ijk
( )= ×( )
u× ( v× w)= ueˆ × veˆ × w eˆ ueˆ ε v w eˆ
i i j j k k i i jkq j k q
= ε ε uvw eˆ = ε ε uvw eˆ
iqm jkq i j k m miq jkq i j k m
(2.2-12)
εε −δδ
IDENTITY
The product of permutation symbols εmiqεjkq in Eq 2.2-12 may be expressed
in terms of Kronecker deltas by the identity
εmiqε = δ δ δ δ
jkq mj − ik mk ij
(2.2-13)
as may be proven by direct expansion. This is a most important formula used
throughout this text and is worth memorizing. Also, by the sign-change
property of ,
ε ijk
ε ε = ε ε = ε ε =
ε ε
miq jkq miq qjk qmi qjk qmi jkq