Tensors for Physics

52 4 Epsilon-Tensor

When the center of the circle is chosen as the originr=0, the position vectorr
is perpendicular tow, thusr·w=0 andL=mr^2 w=mR^2 w, whereRis the
radius of the circle. For a single particle, such a special choice of the origin, wherer
has only components perpendicular to the rotation axis, can always be made. This,
however, is not possible for a rotating solid body which is composed of many mass
points.

4.3.2 Moment of Inertia Tensor

Asolid bodyis composed ofNmass points, i.e. atoms, molecules or small parts
of the body, with massesm 1 ,m 2 ,...,mNlocated at positionsr(^1 ),r(^2 ),...,r(N).
The total mass isM=

∑N

i= 1 mi.Solidmeans: the distances between the constituent
parts of the body do not change, the body moves as a whole. In the case of a rotation
with the angular velocityw, each mass point has the velocityv(i)=w×r(i),the
linear momentump(i)=miv(i), and the angular momentumL(i)=r(i)×p(i),for
i= 1 , 2 ,...,N. Again, the origin of the position vectors is a point on the rotation
axis,wis parallel to this axis. By analogy to (4.21), the total angular momentum is
found to be

Lμ=

∑N

i= 1

L(μi)=Θμνwν, (4.22)

with themoment of inertia tensorΘμνgiven by

Θμν=

∑N

i= 1

mi[(r(i))^2 δμν−rμ(i)rν(i)]. (4.23)

By definition, the moment of inertia tensor is symmetric:Θμν=Θνμ.
The equation (4.22) is an example for a linear relation between two vectors gov-
erned by a second rank tensor. Here both vectorswandLhave positive parity, i.e.
they are axial or pseudo vectors. The moment of inertia tensor has also positive parity,
i.e. it is a proper tensor of rank 2.
Themoment of inertiafor a rotation about a fixed axis is defined via the linear
relation between the component of the angular momentum parallel to this axis and
the magnitudewof the angular velocity. Scalar multiplication of (4.22) with the unit
vector̂wμ=wμ/wleads to

̂wμLμ=̂wμΘμνwν=Θw,

with the moment of inertia

Θ=̂wμΘμν̂wν=

∑N

i= 1

mi

(

r⊥(i)

) 2

. (4.24)