15.5 Energetic Coupling of Order Parameter Tensors 297
the number density and to the permanent electric dipole moment of the molecules.
The result is derived in the following exercise. Both flexo-electric coefficients are
proportional toPrefand toc 1 ,c 2.
The description of non-equilibrium phenomena dealing with the dynamics of
the second rank tensora, as presented in Chap. 17 , can be extended to include the
coupling with an electric dipole moment, cf. [97]. The differential equations contain
the derivatives (15.55) of the potential function. Flexo-electric effects in cholesteric
liquid crystals are treated in [98].
15.4 Exercise: Flexo-electric Coefficients
Start from (15.56) for the vectordμ,useaμν=
√
32 aeqnμnν andPμ=Prefdμ
in order to derive an expression of the form (15.57) and express the flexo-electric
coefficientse 1 ande 3 toc 1 ,c 2 andaeq=
√
5 S, whereSis the Maier-Saupe order
parameter. Furthermore, compute the contribution to electric polarization which is
proportional to the spatial derivative ofaeq=
√
5 S.
Hint: treat the components ofPparallel and perpendicular tonseparately.
15.5.3 Second- and Third-Rank Tensors
By analogy to the coupling between a second rank tensor with a vector, as treated
in Sect.15.5.2, the dimensionless free energyΦ=Φ(a,T)underlying the cou-
pling between the second rank tensoraand third rank tensorT, cf. Sect.15.4.3,is
written as
Φ=Φa+ΦT+ΦaT,ΦaT =−c 1 Tμνλ∇μaνλ+1
2
c 2 TμκλTκλνaμν,(15.58)whereΦa=Φa(a)is a Landau-de Gennes potential function,ΦT =ΦT(T)is
a similar expression for the third rank tensorT, andΦaT, with the coefficients
c 1 ,c 2 , characterizes the coupling between the second and third rank tensors. Terms
of higher order are possible, but not included here, for simplicity. A scalar linear
in bothaandT must involve an additional vector, here it is the nabla-vector. The
coupling coefficientc 1 is a true scalar whenT has negative parity. The coefficient
c 2 is a true scalar, in any case.
Thederivativesofthepotentialwithrespecttothesecondandthirdranktensorsare
∂Φ
∂aμν=Φμνa +c 1 ∇λTλμν+1
2
c 2 TμλκTλκν,∂Φ
∂Tμνλ=ΦμνλT −c 1 ∇μaνλ+c 2 Tμνκaκλ. (15.59)