280 15 Liquid Crystals and Other Anisotropic Fluids
or direct computation from (15.11) leads to
ΦμνLdG≡∂ΦLdG
∂aμν=Aaμν−B√
6 aμκaκν+Caμνaλκaλκ. (15.13)Before the equilibrium conditionΦLdGμν =0 is discussed further, the Landau de
Gennes potential (15.11) is expressed in terms of the variablesaandαas introduced
in the previous Sect.15.2.1. The result is
ΦLdG=1
2
Aa^2 −1
3
Ba^3 cos 3α+1
4
Ca^4. (15.14)The conditions for an extremum of this function are
∂ΦLdG
∂α=Ba^3 sin 3α= 0 ,and
∂ΦLdG
∂a=aA−Ba^2 cos 3α+Ca^3 = 0.The first of these conditions implies that the biaxiality parameterb=sin 3αvanishes.
Thus the equilibrium state is uniaxial andα=0 is used. Then the second condition is
a(A−Ba+Ca^2 )= 0. (15.15)The solutions area=0 and, provided thatB^2 ≥AC,also
a=a 1 , 2 =B
2 C
±
1
2 C
√
B^2 − 4 AC.
The casea=0 corresponds to an isotropic state. In an uniaxially ordered state
one hasa =0. Notice thatA=A(T)is a function of the temperatureT.Atthe
isotropic-nematic coexistence temperatureTni, the potential has a minimum at the
value ofawhereΦLdG(a)=0 holds true. WithAni=A(Tni)=A 0 ( 1 −T∗/Tni),
one has
1
2
Ani−1
3
Ba+1
4
Ca^2 = 0. (15.16)Since
Ani−Ba+Ca^2 = 0 ,