3.3 Antisymmetric Part, Vector Product 41
Use of (3.31)fora′ 1 and of the corresponding expressions fora′ 2 ,a′ 3 (obtained by the
cyclic permutations of 1, 2 ,3) leads to
b 1 =(U 11 U 2 μU 3 ν+U 21 U 3 μU 2 ν+U 31 U 1 μU 2 ν)(Aμν−Aνμ). (3.34)Now we consider the special casea 1 =A 23 −A 32 =0 anda 2 =A 31 −A 13 =0,
a 3 =A 12 −A 21 =0. Then (3.34) yields, with the summation over double indices
written explicitly
b 1 =(U 11 (U 22 U 33 −U 23 U 32 )+U 21 (U 32 U 13 −U 33 U 12 )
+U 31 (U 12 U 23 −U 13 U 22 ))a 1 =det(U)a 1 , (3.35)where det(U)is the determinant of the transformation matrix. For a proper rotation
det(U)= 1 holds true, and consequentlyb 1 =a 1. The proof for the equality
of the other componentsbandarequires just the cyclic permutation of 1, 2 ,3.
This then completes the proof that the three-component quantityalinked with the
antisymmetric part of a tensor by the duality relation (3.29) is a vector in the sense
used here.
3.3.2 Vector Product
In the case of a dyadicA=aband witha,in(3.29), replaced byc, the relation
(3.29) corresponds to the usualcross productorvector productc=a×bof the two
vectorsaandb. More specifically one has
c 1 =(a×b) 1 =a 2 b 3 −a 3 b 2 ,
c 2 =(a×b) 2 =a 3 b 1 −a 1 b 3 ,
c 3 =(a×b) 3 =a 1 b 2 −a 2 b 1. (3.36)By analogy to (3.30), an antisymmetric tensor is linked with the cross product. In
particular, one has for the 12-component:
2 (ab)asy 12 :=a 1 b 2 −a 2 b 1 =(a×b) 3. (3.37)The other components are obtained by a cyclic interchange of 1, 2 ,3.
The antisymmetric part of the dyadicab, as well as the cross productc=a×b
vanish, when the vectorsaandbare parallel to each other.
As an alternative to (3.36), the components of the cross product can be expressed
with the help of a determinant according to
cμ:=∣ ∣ ∣ ∣ ∣ ∣
δ 1 μa 1 b 1
δ 2 μa 2 b 2
δ 3 μa 3 b 3∣ ∣ ∣ ∣ ∣ ∣
=
∣ ∣ ∣ ∣ ∣ ∣
δ 1 μδ 2 μδ 3 μ
a 1 a 2 a 3
b 1 b 2 b 3