Tensors for Physics

Appendix: Exercises... 415

e 1 =Pref

(

1 +

√

2

3

c 2 aeq

)− 1 √

3

2

c 1 aeq,

e 3 =Pref

(

1 −

√

1

6

c 2 aeq

)− 1 √

3

2

c 1 aeq.

Forc 2 =0 one hase 1 =e 3.

Exercises Chapter 16

16.1 Nonlinear Electric Susceptibility in a Polar Material(p.302)
In a medium without hysteresis, the electric polarizationPcan be expanded in powers
of the electric fieldE,cf.(2.59),thus

Pμ=ε 0

(

χμν(^1 )Eν+χμνλ(^2 )EνEλ+...

)

.

The second rank tensorχμν(^1 )≡χμνcharacterizes the linear susceptibility. The third

rank tensorχμνλ(^2 ) describes the next higher order contributions toP.Consider a
material whose isotropy is broken by a polar unit vectord.Formulate the expressions
for these tensors which are in accord with the symmetry and with parity conservation.
Treat the casesEparallel and perpendicular tod.

To conserve parity, even rank tensors must contain even powers ofd, the third rank
tensor must be an odd function ofd. The ansatz which fulfills the required proper-
ties is

χμν(^1 )=χ^10 δμν+χ^12 dμdν,

χμνλ(^2 ) =χ^20 dμδνλ+χ^21

1

2

(δμνdλ+δμλdν)+χ^23 dμdνdλ,

with scalar coefficientsχ^10 ,...,χ^23. The resulting expression forP=P(^1 )+P(^2 )+
...is

Pμ(^1 )=χ^10 Eμ+χ^12 dμdνEν,

Pμ(^2 )=χ^20 dμE^2 +χ^21 EμdνEν+χ^23 dμdνdλEνEλ.

ForE‖done hasP(^1 )=(χ^10 +^23 χ^12 )EandP(^2 )=(χ^20 +χ^21 +^25 χ^23 )E^2 d.
Similarly,theresultforE⊥disP(^1 )=(χ^10 −^13 χ^12 )EandP(^2 )=(χ^20 −^15 χ^23 )E^2 d.
Here