Tensors for Physics

414 Appendix: Exercises...

SinceΦμνa =Aaaμν−

√

6 Baaμλaλν+Caaμνaλκaλκ,

∂Φ

∂aμν

=Aaμν−

√

6 Baμλaλν +Caμνaλκaλκ

is obtained, where

A=Aa−c^21 , B=Ba+ 3 c 1 c 2 , C=Ca− 2 c^21 ,

in accord with (15.53).

15.4 Flexo-electric Coefficients(p.297)

Start from equation(15.56)for the vector dμ,use aμν=

√

3
2 aeqnμnνand Pμ=
Prefdμin order to derive an expression of the form(15.57)and express the flexo-
electric coefficients e 1 ande 3 toc 1 ,c 2 and aeq=

√

5 S,where S is the Maier-Saupe
order parameter. Furthermore, compute the contribution to the electric polarization
which is proportional to the spatial derivative of aeq=

√

5 S.

Hint: treat the components ofPparallel and perpendicular tonseparately.
With the recommended ansatz foraμν, the right hand side of (15.56) becomes

c 1 ∇νaνμ=

√

3

2

c 1

[

aeqnμ∇νnν+aeqnν∇νnμ+nμnν∇νaeq

]

.

Likewise, the left hand side of (15.56) becomes

(

δμν+

√

3

2

c 2 aeqnμnν

)

dν=

(

1 +

√

2

3

c 2 aeq

)

dμ‖+

(

1 −

√

1

6

c 2 aeq

)

d⊥μ,

wheredμ‖=nμnνdνanddμ⊥=dμ−dμ‖are the components parallel and perpendic-
ular to the director. The solution of (15.56)fordis

dμ‖=

(

1 +

√

2

3

c 2 aeq

)− 1 √

3

2

c 1 nμ

(

aeq∇νnν+

2

3

nν∇νaeq

)

,

dμ⊥=

(

1 −

√

1

6

c 2 aeq

)− 1 √

3

2

c 1

(

aeqnν∇νnμ−

1

3

∇μ⊥aeq

)

.

The resulting flexo-electric coefficients are