7—Operators and Matrices 148u 1
u 2
u 3
=
f 11 f 12 f 13
f 21 f 22 f 23
f 31 f 32 f 33
v 1
v 2
v 3
is u 1 =f 11 v 1 +f 12 v 2 +f 13 v 3 etc.
And this is the reason behind the definition of how to multiply a matrix and a column matrix. The
order in which the indices appear is the conventional one, and the indices appear in the matrix as they
do because I chose the order of the indices in a (seemingly) backwards way in Eq. (7.6).
Components of Rotations
Apply this to the first example, rotate all vectors in the plane through the angleα. I don’t want to
keep using the same symbolfforeveryfunction, so I’ll call this functionRinstead, or better yetRα.
Rα(~v)is the rotated vector. Pick two perpendicular unit vectors for a basis. You may call themxˆand
ˆy, but again I’ll call them~e 1 and~e 2. Use the definition of components to get
Rα(~e 2 )
~e 2
cosα
α
Rα(~e 1 )
sinα
~e 1
Rα(~e 1 ) =
∑
kRk 1 ~ek
Rα(~e 2 ) =
∑
kRk 2 ~ek
(7.10)
The rotated~e 1 has two components, so
Rα(~e 1 ) =~e 1 cosα+~e 2 sinα=R 11 ~e 1 +R 21 ~e 2 (7.11)
This determines the first column of the matrix of components,
R 11 = cosα, and R 21 = sinα
Similarly the effect on the other basis vector determines the second column:
Rα(~e 2 ) =~e 2 cosα−~e 1 sinα=R 12 ~e 1 +R 22 ~e 2 (7.12)
Check:Rα(~e 1 ).Rα(~e 2 ) = 0.
R 12 =−sinα, and R 22 = cosα
The component matrix is then
(
Rα
)
=
(
cosα −sinα
sinα cosα
)
(7.13)
Components of Inertia
The definition, Eq. (7.3), and the figure preceding it specify the inertia tensor as the function that
relates the angular momentum of a rigid body to its angular velocity.
~L=
∫
dm~r×
(
~ω×~r
)
=I(~ω) (7.14)
Use the vector identity,
A~×(B~×C~) =B~(A~.C~)−C~(A~.B~) (7.15)
then the integral is
~L=
∫
dm
[
~ω(~r.~r)−~r(~ω.~r)
]
=I(~ω) (7.16)
Pick the common rectangular, orthogonal basis and evaluate the components of this function. Equa-