5
Rotations
5.1
Rotation matrix
The rotation operator
ˆ
R
(
t
) has a matrix
R
ij
(
t
) defined so that if a vector
a
=
a
i
e
i
(where
e
i
are
the basis vectors in the lab frame) is rotated into vector
b
=
b
i
e
i
, then
b
=
ˆ
R
a
,
b
i
=
R
ij
a
j
The rotated basis vectors
e
′
i
(
t
) =
ˆ
R
e
i
satisfy
e
′
i
(
t
)
=
e
j
R
ji
(
t
)
(2)
e
i
=
e
′
j
(
t
)
(
R
−
1
(
t
)
)
ji
(3)
To memorize the above, recall that when
a
→
b
by the same rotation which takes
e
i
→
e
′
i
b
=
e
i
b
i
=
e
′
j
a
j
e
i
b
i
=
e
i
R
ij
a
j
=
e
′
j
a
j
The requirement of length conservation defines the rotation. It can be expressed as (
⃗e
i
(
t
)
·
e
j
(
t
)) =
δ
ij
.
Of course, the labframe vectors also satisfy (
e
i
0
·
e
j
0
) =
δ
ij
. From (2) we have
δ
ij
= (
e
′
i
(
t
)
·
e
′
j
(
t
)) =
R
mi
R
nj
(
e
m
·
e
n
) =
R
mi
R
nj
δ
mn
=
R
mi
R
mj
In matrix notation that means
R
T
R
=
E,
or alternatively
R
−
1
=
R
T
.
5.2
Instantaneous angular velocity
Let us consider any vector
r
(
t
) =
c
′
i
e
′
i
(
t
) =
c
i
(
t
)
e
i
that is fixed in the rotating body (
c
′
i
are time
independent). At
t
= 0 the body has not yet rotated, so
e
′
i
(0) =
e
i
and
c
i
(0) =
c
′
i
. Rotation
R
(
t
)
transforms vector
r
(0) =
c
′
i
e
i
into
r
(
t
) =
c
i
e
i
, so
c
i
(
t
) =
R
ij
(
t
)
c
′
j
Differentiating w.r.t. time we get
˙
c
i
=
˙
R
ij
c
′
j
=
˙
R
ij
R
(
−
1)
jk
c
k
=
X
ik
c
k
,
where we defined
X
jk
=
˙
R
ji
(
R
−
1
)
ik
=
˙
R
ji
(
R
T
)
ik
, so
X
=
˙
RR
T
.
First, we prove that
X
is an
antisymmetric matrix. From
RR
T
=
E
one gets
˙
RR
T
+
R
˙
R
T
= 0
,
˙
RR
T
=
−
(
˙
RR
T
)
T
,
X
=
−
X
T
.
Since
X
is antisymmetric, it has a form
X
=
0
x
12
x
13
−
x
12
0
x
23
−
x
13
−
x
23
0
with only three independent elements.
7
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The velocity of the point described by
r
can be determined as
˙
r
= ˙
e
′
i
c
′
i
=
e
j
˙
R
ji
c
′
i
=
e
′
j
˙
R
ji
(
R
−
1
)
ik
c
k
=
e
j
X
jk
c
k
Note
that we made a substitution ˙
e
′
i
=
e
j
˙
R
ji
. This is only true for constant
e
j
. Later we will study
the subject of rotation in rotating reference frame, where
e
j
are timedependent and expressions
become more complicated.
Now we want to prove that
v
can be given by the formula
v
= [
⃗ω
×
r
]
(4)
with
some
vector
⃗ω
. This equation can be translated into a matrix form. Expand
⃗ω
=
ω
i
e
i
(
t
). Then
v
= [
ω
i
e
i
×
c
j
e
j
] =
ω
i
c
j
[
e
i
×
e
j
] =
ω
i
c
j
ϵ
ijk
e
k
= (
ϵ
ijk
ω
j
c
k
)
e
i
which gives
v
i
=
ϵ
ijk
ω
j
c
k
=
Y
ik
c
k
.
In matrix form
Y
=
0
ω
1
−
ω
2
−
ω
2
0
ω
3
ω
2
−
ω
3
0
We see that by choosing
ω
1
=
x
12
,
ω
2
=
−
x
13
, and
ω
3
=
x
23
we can set
Y
=
X
. With this choice of
components
ω
i
we have
X
ik
=
ϵ
ijk
ω
j
The equivalence between
X
and
Y
proves formula (4).
Let us return to the issue of timedependent
e
i
(
t
). Formula
˙
c
i
=
ϵ
ijk
ω
j
c
k
holds regardless time dependence of
e
i
.
Components
ω
i
are given by the elements of matrix
X
that describes the rotation of the
{
e
′
i
}
frame relative to the
{
e
i
}
frame. Introduction of the vector
⃗ω
=
ω
i
e
i
is a convenience which allows one to write down formula (4). The latter formula is only
true when the frame
{
e
i
}
is at rest. But we can rewrite it in a form
e
i
˙
c
i
= [
ω
j
e
j
×
e
k
c
k
]
,
(5)
which is true even if
e
i
(
t
) are timedependent. It’s just that in this case the l.h.s. is not equal to ˙
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 Fall '11
 Bazaliy
 mechanics, Rotation, Angular velocity, ej Rji

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