358 17 Tensor Dynamics
The basis tensorsTiare defined by
Tμν^0 =√
3
2
ezμeνz, Tμν^1 =1
2
√
2
(
eμxeνx−eyμeyν)
,
Tμν^2 =√
2 eμxe
y
ν, T
3
μν=√
2 exμezν, Tμν^4 =√
2 e
y
μeνz, (17.20)where theex,ey,ezare unit vectors parallel to the coordinate axes. In a principal
axes system, just the first two of these tensors occur, cf. Sect.15.2.1. The first three
of these tensors have the symmetry of the plane Couette geometry. In general, all 5
components are needed.
In matrix notation, the basis tensors (17.20) read
√
6 T^0 =
⎛
⎝
− 100
0 − 10
002
⎞
⎠,
√
2 T^1 =
⎛
⎝
100
0 − 10
000
⎞
⎠,
√
2 T^2 =
⎛
⎝
010
100
000
⎞
⎠,
√
2 T^3 =
⎛
⎝
001
000
100
⎞
⎠,
√
2 T^4 =
⎛
⎝
000
001
010
⎞
⎠. (17.21)
The basis tensors obey the ortho-normalization relation
Tμνi Tμνk =δik. (17.22)The square of the alignment tensor is equal to the sum of its squared components, viz.
a^2 =aμνaμν=∑^4
i= 0a^2 i. (17.23)Furthermore, as presented in [182], the symmetric traceless part of the product of
two of these tensors is explicitly given by
√
6 Tμλ^0 Tλν^0 =Tμν^0 ,√
6 Tμλ^1 Tλν^1 =√
6 Tμλ^2 Tλν^2 =−Tμν^0 , Tμλ^1 Tλν^2 = 0 ,
√
6 Tμλ^0 Tλν^1 =−Tμν^1 ,√
6 Tμλ^0 Tλν^2 =−Tμν^2 ,√
6 Tμλ^0 Tλν^3 =1
2
Tμν^3 ,
√
6 Tμλ^0 Tλν^4 =1
2
Tμν^4 ,√
6 Tμλ^1 Tλν^3 =√
6 Tμλ^2 Tλν^4 =1
2
√
3 Tμν^3 ,
√
6 Tμλ^1 Tλν^4 =−1
2
√
3 Tμν^4 ,√
6 Tμλ^2 Tλν^3 =1
2
√
3 Tμν^4 ,√
6 Tμλ^3 Tλν^4 =1
2
√
3 Tμν^2 ,
√
6 Tμλ^3 Tλν^3 =1
2
(Tμν^0 +√
3 Tμν^1 ),√
6 Tμλ^4 Tλν^4 =1
2
(Tμν^0 −√
3 Tμν^1 ). (17.24)