418 Appendix: Exercises...
Exercises Chapter 17
17.1 Components of a Uniaxial Alignment(p.359)
Consider a uniaxial alignment given by aμν=
√
3 / 2 anμnν.Determine the com-
ponents aiin terms of the polar coordinatesθandφ.Use nx = sinθcosφ,
ny =sinθsinφ,nz =cosθ.Consider the special casesθ = 0 , 45 , 90 ◦and
cos^2 θ= 1 /3.
By definition, the components areai=Tμνi aμν. Thus, due to (17.20), one finds
a 0 =a
3
2
(cos^2 θ−1
3
), a 1 =1
2
√
3 asin^2 θcos( 2 φ), a 2 =1
2
√
3 asin^2 θsin( 2 φ),a 3 =
1
2
√
3 asin( 2 θ)cosφ, a 4 =1
2
√
3 asin( 2 θ)sinφ.Forθ=0, one hasa 0 =a,a 1 =a 2 =a 3 =a 4 =0, as expected. The result for
θ= 90 ◦isa 0 =−a/2,a 1 =^12
√
3 acos( 2 φ),a 2 =^12√
3 asin( 2 φ),a 3 =a 4 =0.
Forθ = 45 ◦, one has cos^2 θ = sin^2 θ = 1 /2, anda 0 = a/4,a 1 =
1
4√
3 acos( 2 φ),a 2 =^14√
3 asin( 2 φ),a 3 =^12√
3 acosφ,a 4 =^12√
3 asinφ.
For the case cos^2 θ= 1 /3, corresponding toθ≈ 55 ◦, the components area 0 =0,
a 1 =^13
√
3 acos( 2 φ),a 2 =^13√
3 asin( 2 φ),a 3 =^13√
6 acosφ,a 4 =^13√
6 asinφ.17.2 Stability against Biaxial Distortions(p.362)
Compute the relaxation frequenciesν(^1 )andν(^3 )for biaxial distortionsδaμν=
Tμν^1 δa 1 andδaμν=Tμν^3 δa 3 from the relevant relations given in Sect.17.2.4.
Solve the full nonlinear relaxation equation for a 3 with a 1 =a 2 =a 4 = 0 and
a 0 =aeq.
The expressions given in Sect.17.2.4yield
ν(^1 )=θ+ 6 aeq+ 2 aeq^2 = 9 aeq.For all temperaturesθ< 9 /8, this relaxation frequency is positive and the stationary
nematic state is stable against biaxial distortions of this type described byδa 1 ,the
same applies forδa 2. On the other hand, one finds
ν(^3 )=θ− 3 aeq+ 2 a^2 eq= 0.The last equality follows from the equilibrium condition. Thus the nematic state has
just marginal stability against biaxial distortions of the typeδa 3 and alsoδa 4 .In
this case the nonlinear equation fora 3 has to be studied. From (17.28) with (17.29)
follows
da 3
dt+ 2 a^33 = 0.