Tensors for Physics

17.2 Nonlinear Relaxation, Component Notation 359

The scalar constructed from the triple product of these tensors is determined by

√

6 Tμλi TλνjTνμk ≡C(i,j,k). (17.25)

The coupling coefficientC(i,j,k)is symmetric under the interchange of any two
of the labelsi,j,k,e.g.C(i,j,k)=C(i,k,j)=C(k,j,i).From(17.24) and
the orthogonality relation, one infers: apart from interchanges, the only nonzero
coefficients are

C( 0 , 0 , 0 )= 1 , C( 0 , 1 , 1 )=C( 0 , 2 , 2 )=− 1 , C( 0 , 3 , 3 )=C( 0 , 4 , 4 )=

1

2

,

C( 1 , 3 , 3 )=C( 2 , 3 , 4 )=

1

2

√

3 , C( 1 , 4 , 4 )=−

1

2

√

3. (17.26)

17.1 Exercise: Components of a Uniaxial Alignment
Consider a uniaxial alignment given byaμν=

√

3 / 2 anμnν. Determine the com-
ponentsaiin terms of the polar coordinatesθandφ.Usenx = sinθcosφ,
ny=sinθsinφ,nz=cosθ.
Consider the special casesθ= 0 , 45 , 90 ◦and cos^2 θ= 1 /3.

17.2.2 Third-Order Scalar Invariant and Biaxiality Parameter

The third-order scalar invariant is defined byI 3 =

√

6 aμνaνκaκμ,cf.(15.4). Due to
(17.25) with (17.26),I 3 =

√

6 aμνaνκaκμis expressed in terms of the components
aiby

I 3 =a 0

[

a^20 − 3 (a^21 +a^22 )+

3

2

(a 32 +a 42 )

]

+

3

2

√

3 a 1 (a 32 −a 42 )+ 3

√

3 a 2 a 3 a 4.
(17.27)
The square of the biaxiality parameterb, cf. Sect.5.5.2, in particular (15.5), is

b^2 = 1 −I 32 /I 23.

17.2.3 Component Equations

The relaxation equations for the 5 components, which correspond to (16.149), with
(16.150), in the absence of a flow, are

∂

∂t

ai+Φi= 0 ,Φi=(θ+ 2 a^2 )ai+Qi, i= 0 , 1 , 2 , 3 , 4. (17.28)