44 Classical mechanics
ly=−ωx∑ni=1miyixi+ωy∑ni=1mi(r^2 i−y^2 i)−ωz∑ni=1miyizilz=−ωx∑ni=1mizixi−ωy∑ni=1miziyi+ωz∑ni=1mi(r^2 i−zi^2 ). (1.11.29)Eqn. (1.11.29) can be written in matrix form as
lx
ly
lz
=
Ixx Ixy Ixz
Iyx Iyy Iyz
Izx Izy Izz
ωx
ωy
ωz
. (1.11.30)
The matrix elements are given by
Iαβ=∑ni=1mi(
ri^2 δαβ−ri,αri,β,)
, (1.11.31)
whereα,β= (x,y,z) andri,αis theαth component of theith position vector in the
body-fixed frame. The matrixIαβis known as theinertia tensorand is the general-
ization of the moment of inertia defined previously. The inertia tensor is symmetric
(Iαβ=Iβα), and can therefore be diagonalized via an orthogonal transformation. Con-
sequently, it has real eigenvalues denotedI 1 ,I 2 andI 3. The eigenvectors of the inertia
tensor define a new set of mutually orthogonal axes about which wemay describe the
motion. When these axes are used, the inertia tensor is diagonal. Since the angular
momentum is obtained by acting with the inertia tensor on the angularvelocity vector,
it follows that, in general,lis not parallel toωas it was in the two-dimensional prob-
lem considered above. Thus,ω×l 6 = 0 so that the time derivative oflin a space-fixed
frame obeys eqn. (1.11.19):
(
dl
dt)
space=
(
dl
dt)
body+ω×l. (1.11.32)Accordingly, the rate of change oflin the space-fixed frame will be determined simply
by the torque according to (
dl
dt
)
space=τ. (1.11.33)Expressing this in terms of the body-fixed frame (and dropping the“body” subscript)
eqn. (1.11.32) yields
dl
dt
+ω×l=τ. (1.11.34)Finally, using the fact thatl= Iωand working with a set of axes in terms of which
I is diagonal, the equations of motion for the three componentsω 1 ,ω 2 , andω 3 of the
angular velocity vector become
I 1 ω ̇ 1 −ω 2 ω 3 (I 2 −I 3 ) =τ 1