360 17 Tensor Dynamics
The quantityQi=−
√
6 Tμνiaνλaλμis explicitly given by
Q 0 =− 3 a 02 + 3 (a^21 +a^22 )−
3
2
(a^23 +a^24 ), (17.29)
Q 1 = 6 a 0 a 1 −
3
2
√
3 (a^23 −a 42 ), Q 2 = 6 a 0 a 2 − 3
√
3 a 3 a 4 ,
Q 3 =− 3 a 0 a 3 − 3
√
3 (a 1 a 3 +a 2 a 4 ), Q 4 =− 3 a 0 a 4 + 3
√
3 (a 1 a 4 −a 2 a 3 ).
TheΦioccurring in the relaxation equation (17.28) are the derivatives of the potential
functionΦwith respect to the componentsai,viz.Φi=∂Φ/∂a 1 , where
Φ=
1
2
θa^2 +Q+
1
2
(a^2 )^2 , Q=−
√
6 aμνaνλaλμ. (17.30)
Apart from the sign and a numerical factor,Qis the determinant of the alignment
tensor, cf. (5.44). In terms of theai, it is given by
Q=−a^30 + 3 a 0
(
a^21 +a^22 −
1
2
a^23 −
1
2
a^24
)
−
3
2
√
3 a 1
(
a 32 −a 42
)
− 3
√
3 a 2 a 3 a 4.
(17.31)
SinceQ, obviously, is not a function ofa^2 , the potentialΦis highly anisotropic in
the 5-dimensional space of theaicomponents.
17.2.4 Stability of Stationary Solutions
Letaμνst be a stationary solution of the inhomogeneous relaxation equation (16.149)
withΦμνgiven by (16.150). Insertion ofaμν=aμνst +δaμνinto the equation and
disregard of terms nonlinear in the small deviationδaμνfrom the stationary state
yields
∂
∂t
δaμν− 2 εμλκΩλδaκν− 2 κΓμκδaκν+Φμν,λκδaλκ= 0 , (17.32)
with the second derivative of the potential, viz.
Φμν,λκ=
∂
∂aλκ
Φμν=(θ+ 2 a^2 )Δμν,λκ+ 4 aμνaλκ− 6
√
6 Δ(μν,αβ,λκ^2 ,^2 ,^2 ) aαβ,(17.33)
evaluated at the stationary value for the alignment tensor. For the isotropic coupling
tensorΔ(···^2 ,^2 ,^2 )see (11.36).