Tensors for Physics

282 15 Liquid Crystals and Other Anisotropic Fluids

Clearly,aeq∗ =1forθ=1. The nematic state is metastable in the range 1<θ<^98
of the reduced temperature. The isotropic state corresponding toa=0 is metastable
for 0<θ<1.
The scaled variables can be denoted by the original symbols without the star, when
no confusion arises. Notice that the scaled potential function (15.21) corresponds to
(15.12) withA=θand the universal coefficientsB=3,C=2.
By definition, the Maier-Saupe order parameterS 2 =〈P 2 〉lies within the range
−^12 ≤ S 2 ≤1. Consequently the order parametera =

√

5 S 2 is bounded by

−

√

5 / 2 ≈ 1 .12 and

√

5 ≈ 2 .24. The bounds for the corresponding scaled vari-
ablea∗=a/aniinvolve the factor 1/ani. For typical thermotropic nematics,

− 1. 25 <a∗< 2. 5.

is the range of the scaled order parameter. The Landau-de Gennes free energy is
well suited to study the isotropic state and the nematic phase in the vicinity of the
transition temperature. The bounds on the order parameter just discussed, however,
are not taken care of. An amended version of a Landau-de Gennes type potential
function which implies an upper bound of the magnitude of the order parameter was
considered in [86].

15.1 Exercise: Derivation of the Landau-de Gennes Potential
In general, the free energyFis related to the internal energyUand the entropySby
F=U−TS. Thus the contributions to these thermodynamic functions which are
associated with the alignment obey the relation

Fa=Ua−TSa.

Assume that the relevant internal energy is equal to

Ua=−N

1

2

εaμνaμν,

whereε>0 is a characteristic energy, per particle, associated with the alignment.
It is related to the temperatureT∗bykBT∗=ε/A 0. Furthermore, approximate the
entropy by the single particle contribution

Sa=−NkB〈ln(f/f 0 )〉 0 ,

cf. Sect.12.2.6, where the entropy per particlesa was considered. Notice that
Sa = Nsa.Usef = f 0 ( 1 +aμνφμν)and (12.39) to compute the entropy and
consequently the free energy up to fourth order in the alignment tensor. Compare
with the expression (15.12) to inferA 0 ,B,C. Finally, use these values to calculate
aniandδ=(Tni−T∗)/Tni,cf.(15.17) and (15.18).