Tensors for Physics

60 5 Symmetric Second Rank Tensors

Now the case is considered, wherevis not parallel tou. By symmetry, one of the
principal directions is parallel to the vector productu×v. The pertaining principal
eigenvalueis0.Thetwootherprincipaldirectionsareparalleltothebisectorsbetween
the vectorsuandv, i.e. they are parallel tou±v. Withh=u±v, one hash·u= 1 ±c,
h·v=c±1, andh·h= 2 ( 1 ±c), withc=u·v=cosθ, whereθis the angle
between the vectorsuandv. Thus the two other principal values are found to be
(c± 1 )/2. In summary, the principle values are

{

1

2

(c+ 1 ),

1

2

(c− 1 ), 0

}

.

Clearly, forθ=0, corresponding tou=v, one hasc=1, and the uniaxial case
is recovered. Foru⊥v,c=0 applies and the principal values are{ 1 / 2 ,− 1 / 2 , 0 }.
This corresponds to a symmetric traceless planar biaxial tensor.
The trace of the tensor given by (5.15) is the scalar productc=u·v. Thus the
principal values of the symmetric traceless tensoruμvν are

{

1

6

c+

1

2

,

1

6

c−

1

2

,−

1

3

c

}

,c=u·v=cosθ. (5.17)

Comparison with (5.14) shows that the coefficientssandq, introduced in Sect.5.2.4,
are here given bys=−c/2 andq= 1 /2.

5.3 Applications.

5.3.1 Moment of Inertia Tensor of Molecules.

The moment of inertia tensorΘμν, as defined in (4.23), is symmetric. When the
origin of the position vectors of the constituent parts of a solid body coincides with
the center of mass of the body, the moment of inertia tensor reflects and characterizes
the shape and symmetry of the body. Molecules in their vibrational ground state can
be looked upon as “solid bodies”. Some simple examples are considered next.

(i) Linear Molecules

A linear molecule is composed of two atoms, with massesm 1 andm 2 , separated by
the distanced. The unit vector parallel to the axis joining the nuclei of the two atoms
is denoted byu. Their position vectors, with respect to the center of mass of the
molecule, arer(^1 )=d 1 uandr(^2 )=−d 2 u, with the distancesd 1 andd 2 determined
byd 1 +d 2 =0 andm 1 d 1 =m 2 d 2 .Useof(4.23) leads to the moment of inertia tensor

Θμν=

(

m 1 d 12 +m 2 d 22

)(

δμν−uμuν

)

=Θ

(

δμν−uμuν

)

, (5.18)