Tensors for Physics

412 Appendix: Exercises...

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 temperature T∗bykBT∗=ε/A 0 .Furthermore, approximate the
entropy by the single particle contribution

Sa=−NkB〈ln(f/f 0 )〉 0 ,

cf. Sect.12.2.6where the entropy per particle sawas considered, Notice thatSa=
Nsa.Use f =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 infer A 0 ,B,C.Finally, use these values to calculate aniandδ=(Tni−
T∗)/Tni,cf.(15.17)and(15.18).

Due to (12.39), the alignment entropy per particle is given by

sa=−kB

(

1

2

〈Φ^2 〉 0 −

1

6

〈Φ^3 〉 0 +

1

12

〈Φ^4 〉 0 ±...

)

,

where hereΦ=aμνφμνis used. The normalization ofφμνimplies that the sec-
ond order contribution is^12 aμνaμν. The third order term involves the triple product
aμνaμ′ν′aμ′′ν′′〈φμνφμ′ν′φμ′′ν′′〉 0. Due to the integral relation (12.7), this expression
becomes^27

√

30 aμνaνκaκμand the third order contribution is− 211

√

30 aμνaνκaκμ.

The fourth order term involves the quadruple product

aμ 1 ν 1 aμ 2 ν 2 aμ 3 ν 3 aμ 4 ν 4 〈φμ 1 ν 1 φμ 2 ν 2 ′φμ 3 ν 3 φμ 4 ν 4 〉 0.

Due to the integral relation (12.9) and the use of (12.10), this expression becomes
15
7 (aμνaνμ)

(^2) and the fourth order contribution is^5
28 (aμνaνμ)
(^2).
As a consequence, the Landau-de GennesΦLdG, defined viaFa=Ua−TSa=
NkBTΦLdGis given by
ΦLdG=

1

2

Aaμνaμν−

1

3

√

6 Baμνaνκaκμ+

1

4

C(aμνaμν)^2 ,

with

A=A 0 ( 1 −T∗/T), A 0 = 1 , kBT∗=ε, B=

√

5

7

, C=

5

7

.

These specific values, based on the single particle contribution only for the entropy,
yield for the order parameterani= 2 B/( 3 C)at the transition temperatureTnithe
value 152

√

5, corresponding to a Maier-Saupe order parameter of 152 ≈ 0 .133. The

quantityδ=(Tni−T∗)/Tni,asgivenin(15.18), assumes the value 632 ≈ 0 .032.