This document discusses principal axes and Eulerian angles for describing the orientation and motion of rigid bodies. It contains the following key points: 1) The principal moments and axes of an inertia tensor can always be found for real, symmetric tensors like the inertia tensor, and they will always be real values. 2) Under translation, the inertia tensor about a point not at the center of mass can be expressed in terms of the inertia tensor at the center of mass using Steiner's theorem. 3) Eulerian angles define a rotation as three successive rotations about fixed axes and can be used to relate an inertial frame to a body-fixed frame for a rigid body. 4) Euler's equations describe

1. Lecture 35: Principal Axes,
Translations, and Eulerian Angles

2. When Can We Find Principal Axes?
• We can always write down the cubic equation that one
must solve to determine the principal moments
• But if we want to interpret these as physically meaningful
quantities, the roots of that cubic have to be real
– Recall that in general, cubics can have two complex roots
• Fortunately, we’re not in the general case here
• The inertia tensor is both real and symmetric – in
particular, it satisfies:
• Matrices that satisfy this restriction are called Hermitian
• For such matrices, the principal moments can always be
found, and they are always real (see proof in text)
*
ij ij
I I
=
This mathematics will come up again in Quantum Mechanics
Principal Moments ↔ Eigenvalues
Principal Axes ↔ Eigenfunctions

3. Inertia Tensor Under Translation
• We’ve already seen how the inertia tensor transforms
under a rotation of the coordinate system
• Now we’ll see how it behaves under translation – that is,
keeping the direction of the the coordinate axes fixed, but
changing the origin
• In particular, we’ll compare I for a coordinate system at
the center of mass of an object with that for a different
origin:
x1
x2
x3
O
X1
X2
X3
Q
rCM
O is at the center of
mass; Q can be
anywhere

4. • In the X reference frame (the one with origin at Q), the
position of any point in the rigid body can be written as:
which means the inertia tensor becomes:
α
α = +
CM
R r
r Position in center-of-
mass frame
( )
( ) ( )( )
( )
( )
(
)
( )
( )
2
, ,
2
, , ,
2
, , , , , ,
2 2
, , , ,
, , , , , , , ,
,
2
, , , ,
2
2
ij ij k i j
k
ij k CM k i CM i j CM j
k
ij k CM k k CM k
k
i j CM i j i CM j CM i CM j
ij CM k k
ij k i j
k
CM k CM
k
I m X X X
m x x x x x x
m x x x x
x x x x x x x x
m x x x
m x x x
x
α α α α
α
α α α α
α
α α α
α
α α α α
α
α α
α
α α
α
δ
δ
δ
δ
δ
= −
= + − + +
= + +
− − − −
=
+ + −
−
( )
, , , , ,
i j i CM j CM i CM j
x x x x x
α α
α
− −
Inertia tensor about the center of mass

5. • So the inertia tensor about Q is:
• Each term in the second line has a factor that looks like:
• For a general reference frame, this quantity equals the mass of
the system times the position of the ith component of the center
of mass
– But the xi and in the center of mass frame, so all these terms are
zero
• Therefore we have the result (Steiner’s Parallel Axis Theorem):
( )
( )
2
, , , ,
, , , , , ,
2
ij CM ij ij CM k CM i CM j
k
ij CM k k CM i j i CM j
k
I I x x x m
m x x x x x x
α
α
α α α α
α
δ
δ
= + −
+ − −
,i
m x
α α
α
( )
2
, , , ,
ij CM ij ij CM k CM i CM j
k
I I M x x x
δ
= + −

6. Eulerian Angles
• In analyzing the motion of a rigid body, it is most
convenient to use a reference frame that is fixed to the
body
– For example, a symmetry axis of the body could be the z
axis in this “body frame”
– But that’s usually not an inertial frame
• So we need to define a translation between some inertial
“fixed” frame and the body frame
• Three angles are needed to specify how one coordinate
system is rotated with respect to another (in three
dimensions)
– These angles determine the elements of the rotation matrix λ
λ
λ
λ

7. • There are many ways the three angles can be specified
• One convenient choice is the Eulerian Angles, in which the
3D rotation is the result of three 1D rotations:
1. Rotate CCW by φ around axis
2. Rotate CCW by θ around axis
3. Rotate CCW by ψ around axis
3 3
x x
′ ′′
=
1
x′
2
x′
φ
θ
1
x
ψ
1 2 3
( , , )
x x x
′ ′ ′
1 2 3
( , , )
x x x
′′ ′′ ′′
1 2 3
( , , )
x x x
′′′ ′′′ ′′′
2
x′′
1
x′′
1
x′
2
x′
3 3
x x
′ ′′
=
2
x′′′
3
x′′′
1 1
x x
′′ ′′′
=
3 3
x x
′′′=
1
x′
3
x′
2
x′
2
x
Line of
nodes
1 2 3
( , , )
x x x
3
x′
1
x′′
3
x′′′

8. • Note that in each rotation, one axis (from the previous
rotation) is held fixed. That means the rotation matrices
have the form:
• The complete rotation is given by:
• i.e., the elements of λ
λ
λ
λ are obtained by multiplying the three
individual rotation matrices (see Equation 11.99 in the text
for their values)
cos sin 0 1 0 0
sin cos 0 ; 0 cos sin
0 0 1 0 sin cos
cos sin 0
sin cos 0
0 0 1
φ φ
φ φ θ θ
θ θ
ψ ψ
ψ ψ
= − =
−
= −
′ ′
= ≡
x x x

9. Angular Velocity in the Body Frame
• It’s going to be easiest to find the equations of motion for
the object in the body frame
• In general, we can write:
• To write the components of this vector in the body frame
we need to do some geometry:
= + +
1
x
ψ
3
x
1
x′
3
x′
2
x′
2
x
φ
θ
• is along the x3 direction:
• is in the x1-x2 plane:
• has components in all three
directions:
ψ
= 3
e
2
cos sin
θ ψ θ ψ
= −
1
e e
2 3
sin sin sin cos cos
φ θ ψ φ θ ψ φ θ
= + +
1
e e e

10. • Adding up these vectors, we find that the components of ω
ω
ω
ω
in the body frame are:
1
2
3
sin sin cos
sin cos sin
cos
ω φ θ ψ θ ψ
ω φ θ ψ θ ψ
ω φ θ ψ
= +
= −
= +

11. Euler’s Equations for Rigid Body
Motion
• Now that we’ve defined the geometry, we can set up the
Lagrangian for rigid body motion
• To make things easy, we’ll first consider the case where no
external forces act on the body, so:
• In that case, we can put the origin of both the fixed
(meaning inertial) and body frames at the center of mass
• Furthermore, we can cleverly choose the body frame to
coincide with the principal axes of the body, so that:
L T
=
2
1
2
i i
i
I
T ω
=
Principal moment of inertia
associated with the ith axis

12. • Since the Eulerian angles fully specify the orientation of
the rigid body in the inertial frame, we can take them to be
our generalized coordinates
• The equation of motion for ψ is:
which becomes (using the chain rule):
• One piece that enters everywhere is which is simply:
0
T d T
dt
ψ ψ
∂ ∂
− =
∂ ∂
0
i i
i i
i i
T d T
dt
ω ω
ω ψ ω ψ
∂ ∂
∂ ∂
− =
∂ ∂ ∂ ∂
,
i
T
ω
∂
∂
i i
i
T
I ω
ω
∂
=
∂

13. • The other relations we need are:
• Putting these back into the Lagrange equation gives:
1
2
2
1
3
sin cos sin
sin sin cos
0
ω
φ θ ψ θ ψ ω
ψ
ω
φ θ ψ θ ψ ω
ψ
ω
ψ
∂
= − =
∂
∂
= − = −
∂
∂
=
∂
1
2
3
0
0
1
ω
ψ
ω
ψ
ω
ψ
∂
=
∂
∂
=
∂
∂
=
∂
( )
( )
1 1 2 2 2 1 3 3
1 2 1 2 3 3
0
0
d
I I I
dt
I I I
ω ω ω ω ω
ω ω ω
+ − − =
− − =

14. • The choice of which axis to call x3 was completely
arbitrary
– That means we can find similar relations for the other
components of the angular velocity
( )
( )
( )
1 2 1 2 3 3
3 1 3 1 2 2
2 3 2 3 1 1
0
0
0
I I I
I I I
I I I
ω ω ω
ω ω ω
ω ω ω
− − =
− − =
− − =
Euler’s Equations
for force-free motion