CHAPTER - IV
THE KINEMATICS OF RIGID BODY
Unit 1: Rigid Body:
Introduction:
In this chapter we define a rigid body and describe how the number of
degrees of freedom of a rigid body with N particles is determined. There are two
types of motion involved in the case of rigid body viz.; the translation and the
rotation. Various sets of variables have been used to describe the orientation of rigid
body. We will discuss in this chapter how the Eulerian angles and the complex
Cayley-Klein parameters can be used for the description of rigid body with one point
fixed.
Geometrically, matrix represents rotation; we will find the matrix of
transformation in terms of Eulerian angles and Cayley-Klein parameters and
establish the relation between them. This unit is devoted to the study of orthogonal
transformations and its properties.
Rigid Body :
A rigid body is regarded as a system of many (at least three) non-collinear
particles whose positions relative to one another remain fixed. i.e., distance between
any two of them remains constant through out the motion. The internal forces
holding the particles at fixed distances from one another are known as forces of
constraint. These forces of constraint obey the Newton’s third law of motion.
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Worked Examples •
Example 1 : Explain how the generalized co-ordinates of a rigid body with N
particles reduce to six for its description.
Solution :
•
Generalized co-ordinates of rigid body :
A system of N particles free from constraints can have 3N degrees of freedom
and hence 3N generalized co-ordinates. But the constraints involved in rigid body
with N particles are holonomic and scleronomic and are given by
rij = aij , i ≠ j = 1, 2,...., N
. . . (1)
where rij denotes the distance between the i th and j th particles, and aij are constants.
Equation (1) is symmetric in i and j and i ≠ j as the distance of the i th from itself is
zero, therefore, the possible number of constraints is
N
C2 =
N ( N − 1)
.
2
We notice that for N > 7,
. . . (2)
N ( N − 1)
> 3 N . Therefore the actual number of
2
degrees of freedom cannot be obtained simply by subtracting the number of
constraints from 3N. This is simply because all constraints in equation (1) are not
independent.
To show how the generalized co-ordinates of a rigid
C
body with N particles reduce to six for its
P
description, let a rigid body be regarded as a system
of at least three non-collinear particles whose
A
positions relative to one another remain fixed. Thus a
B
system of 3 particles free from constraints has
9 degrees of freedom but there
involves 3 constraints. Hence the number of generalized co-ordinates reduces to six.
Thus the total number of degrees of freedom for three non-collinear particles A, B,
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and C of a rigid body is equal to six. This is because each particle has 3-degrees of
freedom and less three equations of constraints.
The position of each further particle say P requires three more co-ordinates
for its description, but there will be three equations of constraints for this particle,
because the distance of P from A, B, C is fixed. Thus three co-ordinates for P and
less three equations of constraints for P gives zero degrees of freedom. Thus any
other particle apart from A, B, C taken to specify the configuration of the rigid body
will not add any degrees of freedom. Once the positions of three of the particles of
the rigid body are determined the constraints fix the positions of all remaining
particles.
Thus the configuration of the rigid body would be completely specified by
only three particles i.e., by six degrees of freedom, no matter how many particles it
may contain.
Example 2 : Describe the motion of the rigid body.
Solution: A rigid body can have two types of motion
(i)
a translational motion and
(ii)
a rotational motion.
Thus a rigid body in motion can be completely specified if its position and
orientation are known.
However, if one of the points of a rigid body is fixed, the translation motion
of the body is absent and the body rotates about any line through the fixed point.
Again, if we fix up a second point, then the motion of the body is restricted to rotate
about the line joining the two fixed points. Further, if we also fix the third point of
the body non-collinear with other two, the position of the body is fixed and there is
no motion of any kind. The co-ordinates of the third point alone will be able to locate
the rigid body completely in space. It follows that the position of the rigid body is
determined by any three non-collinear points of it that is by six degrees of freedom.
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Of the six generalized co-ordinates, 3 co-ordinates are used to describe translational
motion and other three co-ordinates are used to describe rotational motion. Since a
rigid body with one point fixed has no translational motion and hence it has 3degrees of freedom and three generalized co-ordinates-which are used to describe the
rotational motion.
•
Orthogonal Transformation :
Example 3 : Define orthogonal transformation. Show that finite rotation of a rigid
body about a fixed point of the body is not commutative.
Solution : Consider
( x1 , x2 , x3 )
and
( x1′, x2′ , x3′ )
be two co-ordinate systems. The
general linear transformation between these two co-ordinate systems is defined by
the following set of equations
x1′ = a11 x1 + a12 x2 + a13 x3 ,
x2′ = a21 x1 + a22 x2 + a23 x3 ,
. . . (1)
x3′ = a31 x1 + a32 x2 + a33 x3 ,
where a11 , a12 ,..., a33 are constants. These three equations can be combined in to a
single equation as
3
xi′ = ∑ aij x j , i = 1, 2,3.
. . . (2)
j =1
where
A = ( aij )
is called the matrix
r = x1i + x2 j + x3 k defined in
( x1′, x2′ , x3′ )
( x1 , x2 , x3 )
of transformation.
Let
a
vector
co-ordinate system be transformed to
co-ordinate systems in the form r = x1′i + x2′ j + x3′ k . Since the magnitude
of the vector must be the same in both the co-ordinate system, we must have
therefore
3
3
∑ x′ = ∑ x
2
2
i
i
i =1
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.
. . . (3)
i =1
Page No. 228
Using equation (2) in equation (3) we get
3


3
3
∑  ∑ a x   ∑ a
i =1
⇒

ij
j =1
j

ik
k =1
 3
xk  = ∑ xi2 ,
 i =1
3
 3

2
=
a
a
x
x
∑∑
 ∑ ij ik  j k ∑ xi .
j =1 k =1  i =1
i =1

3
3
Equating the corresponding coefficients on both the sides of the above equation we
get
3
∑a a
ij ik
= δ jk ,
. . . (4)
i =1
where δ jk is the Kronecker delta symbol and is defined by
δ jk = 0 when
= 1 when
j ≠ k,
j = k.
. . . (5)
Thus any transformation (2) satisfying (4) is called as an orthogonal transformation.
Ex. 4. Show that two successive finite rotations of a rigid body about a fixed point of
the body are not commutative.
Solution:
Consider two successive linear transformations described by the
matrices B and A corresponding to two successive displacements of the rigid body.
Let the first transformation from x to x′ be denoted by the matrix B and is defined
by
3
xk′ = ∑ bkj x j , k = 1, 2,3,
. . . (6)
j =1
where the matrix of transformation is B = ( bkj ) .
Let the succeeding transformation from x′ to x′′ be defined by the matrix
A = ( aik ) and is given by
3
xi′′ = ∑ ai k xk′ , i = 1, 2, 3.
. . . (7)
k =1
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Now the transformation from x to x′′ is obtained by combining the two equations
(6) and (7) as
3
xi′′ =
∑a
b x j , i = 1, 2,3.
i k kj
j , k =1
This may also be written as
3
xi′′ = ∑ cij x j , i = 1, 2,3.
. . . (8)
j =1
where C = ( cij ) is the matrix of transformation from x to x′′ and the elements of the
matrix of transformation are defined as
3
cij = ∑ aik bkj .
. . . (9)
k =1
These elements are obtained by multiplying the two matrices A and B. Thus the two
successive linear transformations described by A and B is equivalent to a third linear
transformation described by the matrix C, defined by
C = AB.
. . . (10)
Since the matrix multiplication is not commutative in general, hence the finite
rotations of a rigid body about a fixed point of the body are not commutative.
•
Properties of orthogonal transformation matrix :
Example 5: Prove that the product of two orthogonal transformations is again
orthogonal transformation.
Solution: Consider two successive orthogonal linear transformations of a rigid body
with one point fixed corresponding to two successive displacement of the rigid body
and are described by the matrices B and A respectively. We know that the two
successive orthogonal transformations is equivalent to a third linear transformation
described by the matrix C, where C = AB, and its elements are defined by
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cij = ∑ aik bkj
. . . (1)
k
where A = ( aik ) , B = ( bkj ) are matrices of orthogonal transformations.
This implies that
∑a a
ij ik
= δ jk ,
. . . (2)
= δ jk .
. . . (3)
i
∑b b
ij ik
i
Consider now
∑c c
ij ik
i


= ∑  ∑ aimbmj ∑ ail blk 
i  m
l

= ∑ aimbmj ail blk ,
i , m ,l


= ∑  ∑ aim ail bmj blk ,
m ,l  i

= ∑ δ ml bmj blk ,
m ,l
= ∑ blj blk ,
l
∑c c
ij ik
= δ jk .
i
This proves that the product of two orthogonal transformations is again an
orthogonal transformation.
Note: Though the matrix multiplication is not commutative in general, but it is
associative when the product is defined.
i.e.,
( AB ) C = A ( BC ) .
Example 6: Show that in the case of an orthogonal transformation the inverse matrix
is identified by the transpose of the matrix.
Solution: Consider an orthogonal transformation from xi to xi′ described by the
matrix A and is given by
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3
xi′ = ∑ aij x j , i = 1, 2,3.
. . . (1)
j =1
where A = ( aij ) is the matrix of transformation satisfying the condition
∑a a
ij ik
= δ jk .
. . . (2)
i
Let the matrix of inverse transformation from xi′ to xi be described by the inverse
matrix A−1 = ( aij′ ) , aij′ are the elements of the inverse matrix of transformation
satisfying
∑ a′ a′
ij ik
= δ jk .
. . . (3)
i
Also we have
∑a
AA−1 = I ⇒
a′ = δ kj .
ki ij
. . . (4)
i
Now consider the double sum
∑a
kl
aki aij′ which can be evaluated either by summing
k ,i
over k first or over i first. Therefore we evaluate the double sum as
∑a
kl
k ,i


aki aij′ = ∑  ∑ akl aki aij′ ,
i  k

= ∑ δ li aij′ ,
i
∑a
kl
aki aij′ = alj′ .
. . . (5)
k ,i
Now evaluating the double sum over i first and then over k , we obtain
∑a
kl
k ,i


aki aij′ = ∑  ∑ aki aij′ akl ,
k  i

= ∑ δ kj akl ,
k
∑a
kl
aki aij′ = a jl .
. . . (6)
k ,i
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From equations (5) and (6) we have
aij′ = a ji , ∀ i, j ,
⇒ ( aij′ ) = ( a ji ) ,
⇒
A−1 = A′ .
. . . (7)
This proves that the inverse matrix of an orthogonal transformation identifies the
transpose matrix.
Example 7 : Show that the determinant of an orthogonal matrix is ±1 .
Solution : Let A be the matrix of an orthogonal transformation and A−1 be its
inverse matrix. Then we have
AA−1 = I
. . . (1)
However, we know that in the case of an orthogonal matrix, its inverse is identified
by its transpose.
A−1 = A′ .
Hence equation (1) becomes
AA′ = I
. . . (2)
Taking the determinant on both the sides of above equation we get
AA′ = 1
•
⇒
A A′ = 1,
⇒
A = 1 as
⇒
A = ±1.
2
A = A′ ,
Infinitesimal Rotation :
Example 8: Define infinitesimal rotation. Show that infinitesimal rotation of a rigid
body with one point fixed is commutative. Also find the inverse matrix of
infinitesimal rotation.
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Solution: An infinitesimal rotation is an orthogonal transformation of co-ordinate
axes in which the components of a vector are almost the same in both the sets of
axes. The new co-ordinates differ from the old co-ordinates by an infinitesimal
amounts.
Mathematically, an infinitesimal transformation is defined as
x1′ = x1 + ε11 x1 + ε12 x2 + ε13 x3 ,
x2′ = x2 + ε 21 x1 + ε 22 x2 + ε 23 x3 ,
x3′ = x3 + ε 31 x1 + ε 32 x2 + ε 33 x3 ,
xi′ = xi + ε ij x j ,
i.e.,
. . . (1)
where summation is defined over the repeated index j and ε ij are the elements of the
matrix of infinitesimal transformation and are infinitesimal. i.e., second order terms
in ε ij can be neglected. Hence we write equations (1) as
xi′ = δ ij x j + ε ij x j ,
xi′ = (δ ij + ε ij ) x j ,
i.e.,
. . . (2)
In matrix notations we write equation (2) as
X ′ = (I +ε ) X ,
. . . (3)
 x1 
 x1′ 
 
 
X =  x2  , X ′ =  x2′  , I = (δ ij ) , ε = ( ε ij )
x 
 x′ 
 3
 3
where
and I + ε is the matrix of infinitesimal transformation.
Now let
I + ε1
and
I + ε2
be two matrices of successive infinitesimal
transformations.
Consider
( I + ε1 )( I + ε 2 ) = I .I + I ε 2 + ε1I + ε1ε 2
= I + ε1 + ε 2 .
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Similarly consider
( I + ε 2 )( I + ε1 ) = I .I + I ε1 + ε 2 I + ε 2ε1
= I + ε1 + ε 2 .
We see from above equations that
( I + ε1 )( I + ε 2 ) = ( I + ε 2 )( I + ε1 )
This shows that the product of the matrices of two successive infinitesimal
transformations is commutative. Now to find the inverse matrix of an infinitesimal
transformation, consider
( I + ε )( I − ε ) = I − I ε + ε I − εε
( I + ε )( I − ε ) = I
This shows that the inverse matrix of an infinitesimal transformation is I − ε .
i.e.,
(I +ε )
−1
= (I −ε )
.
For orthogonal transformation we know that its transpose matrix identifies the
inverse matrix. Hence we have
(I +ε )
−1
= ( I + ε ) = ( I − ε ) ,
where ε is the transpose of ε . Consequently we have ε = −ε .
This shows that the matrix of infinite transformation is anti-symmetric.
Example 9: A constant vector X is given by X = i + 4 j + 2 3k with respect to a
particular co-ordinate system. Find the form of the vector with respect to co-ordinate
system obtained from the first by rotating it about the x-axis through an angle
π
3
in
the anti-clock wise direction. Determine its magnitude and compare with X .
Solution: The matrix of rotation about x-axis through an angle
π
3
in the anti clock
wise direction is given by
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
1
0

π
π 
A   =  0 cos
3
3 

 0 − sin π
3

 
0
0  1


π
1
sin  =  0
3
2


π
cos   0 − 3
3 
2

0 

3
.
2 

1 

2 
Hence the new vector with respect to the new co-ordinate axes obtained from the first
by rotating through an angle
π
3
about x-axis is given by

1
0

1
X ′ =  0
2

3

0 −
2


0 

3
2 

1 

2 
 1   1 

 

 4 = 5 
2 3 − 3

 

X ′ = i + 5 j − 3k
The magnitude of this new vector is
X ′ = 29 .
This shows that X = X ′ .
••
Euler’s Theorem 1 : Show that the general displacement of a rigid body with one
point fixed is a rotation about some axis passing through the fixed point.
Proof : Consider a rigid body with one point fixed and be taken as the origin of the
body set of axes. Then the displacement of the rigid body involves no translation of
the body axes, the only change is in orientation. Hence the body set of axes at any
time t can always be obtained by a single rotation of the space set of axes.
Thus any vector lying along the axis of rotation must have the same
components in both the initial and final axes. Further, the orthogonality condition
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implies that the magnitude of a vector parallel to the axis of rotation is unaffected. It
means that the vector R has same components in both the system.
R′ = AR = R .
. . . (1)
This is the special case of more general equation
R′ = AR = λ R .
. . . (2)
where λ is called as the eigen (characteristic) value. Equation (2) can be written as
( A − λI ) R = 0 .
. . . (3)
The equation (3) can have solution only when
A − λI = 0 .
. . . (4)
This is known as characteristic equation, and the roots of the equation (4) are known
as characteristic values. Since the matrix of rotation A is orthogonal, and then we
have
A−1 = A
. . . (5)
where A is the transpose of A . This orthogonal matrix satisfies the equation
A = A = 1 .
. . . (6)
Now to prove the Euler’s Theorem, we just prove that eigen value λ = 1 . Thus
consider the expression
( A − I ) A = AA − IA
= AA−1 − A
( A − I ) A = I − A .
. . . (7)
Taking the determinant of equation (7) we get
( A − I ) A =
(A− I)
⇒
I − A
A = I − A
A− I = − A− I
⇒ A− I = 0.
. . . (8)
. . . (9)
Comparing equations (4) and (9) we get λ = 1 . This proves the theorem.
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Unit 2: Eulerian Angles :
We have seen that a rigid body with one point fixed has three degrees of
freedom and hence three generalized co-ordinates. To describe the orientation of a
rigid body about a fixed point we use a matrix of rotation, whose elements are called
the direction cosines, which are not linearly independent, therefore they are not
suitable as generalized co-ordinates. So we cannot use them in the description of
Lagrangian of the system. Therefore three new independent parameters are necessary
for the description of a rigid body with one point fixed. A number of such sets of
parameters have been used in the literature but the most common and found to be
useful is the set of Eulerian angles.
Euler has designed three independent parameters called as Eulerian angles, to
describe the orientation of a rigid body with one point fixed. These can be used to
write Lagrangian and hence the Lagrange’s equations of motion. We shall define the
Eulerian angles and show how these angles can be used for the description of the
orientation of the rigid body.
Theorem 2 : Define Eulerian angles. Obtain the matrix of transformation from space
co-ordinates to body co-ordinates in terms of Eulerian angles. Prove further that this
matrix is orthogonal and hence deduce the matrix of inverse transformation from the
body set of axes to space set of axes.
z = z1
Proof : Eulerian angles φ , θ ,ψ are the three
successive angles of rotation about a specified
axes performed in specific sequence. These
y1
angles can be used as generalized co-ordinates to
φ
O
φ
x
fix the orientation of a rigid body with one point
y
fixed. Thus the orientation of a rotating body
with one point fixed can be completely specified
x1
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by three independent Eulerian angles.
Page No. 238
To discuss the rotation of the rigid body, let one of the points of the body be
fixed. This implies that there is no translational motion but the body rotates about an
axis passing through the fixed point. Consider two co-ordinate systems, one of which
is (x, y, z) fixed in space (called an inertial frame) and the other ( x′, y′, z ′ ) fixed in
the body called the body set of axes (also known as non-inertial frame). It has been
observed that the configuration of the rigid body is completely specified by locating
the body set of axes relative to the co-ordinate axes fixed in space. This is achieved
by finding the matrix of transformation from the space set of axes to the body set of
axes. Therefore we shall carry out the transformation from space set of axes to body
set of axes such that x, y, z coincides with x′, y′, z ′ . This is achieved by three
successive rotations about specified axes.
The sequence will be started by rotating the initial system of axes x, y, z
through an angle φ anti-clock wise direction about z – axis. Let the resulting coordinate system be labeled as x1 , y1 , z1 axes as shown in the fig. In this case xy plane
becomes x1 y1 plane. The rotation is affected by the following transformation
equations.
x1 = x cos φ + y sin φ ,
y1 = − x sin φ + y cos φ ,
. . . (1)
z1 = z.
These equations can be written in matrix form as
 x1 
 x
 
 
 y1  = D  y 
z 
 
 1
z
or
X 1 = DX
where
 coxφ

D =  − sin φ
 0

is the matrix of transformation.
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0

cos φ 0 
0
1 
sin φ
. . . (2)
Page No. 239
z = z1
z2
The second rotation is performed about the new
x1 axis. The axes x1 , y1 , z1 are rotated about x1
axis counter clockwise direction by an angle θ .
y2
Let the resultant set of co-ordinate
y1
θ
θ
φ
y
O
axes be relabeled as x2 , y2 , z2 . Here x2 axis
being the line of intersection of xy plane and
φ
x1 y1 plane is called the line of nodes. The
x
x1 = x2
transformation equations from x1 , y1 , z1 to new
set of axes x2 , y2 , z2 can be represented by the following set of equations:
x2 = x1 ,
y2 = y1 cos θ + z1 sin θ ,
. . . (3)
z1 = − y1 sin θ + z1 cos θ ,
 x2 
 x1 
 
 
 y2  = C  y1 
z 
z 
 2
 1
i.e.,
or
X 2 = CX 1
where
0
1

C =  0 cos θ
 0 − sin θ

0 

sin θ 
cos θ 
. . . (4)
is the matrix of transformation.
Finally, the third rotation is performed
z2 axis.
z2 = z3= z’
about
z = z1
y3 = y’
The x2 , y2 , z2 axes are rotated counter
y2
clockwise direction by an angle ψ about z2 axis to
ψ
produce the third and the final set of axes x3 , y3 , z3 ,
which coincide, with body set of axes x′, y′, z ′ . This
y1
θ
φ
O
φ
ψ
x
x1 = x2
x3 = x’
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y
completes the transformation from space set of axes to body set of axes. This
transformation is represented by
x3 = x′ = x2 cosψ + y2 sinψ ,
y3 = y′ = − x2 sinψ + y2 cosψ ,
. . . (5)
z3 = z ′ = z 2
or
 x′ 
 x2 
 ′
 
 y  = B  y2 
 z′ 
z 
 
 2
or
X ′ = BX 2
where
 coxψ

B =  − sinψ
 0

sinψ
cosψ
0
0

0
1 
. . . (6)
is the matrix of transformation.
Thus the space set of axes x, y, z coincides with body set of axes through
three successive rotations φ , θ ,ψ , which are described by matrices D, C and B. The
angles φ , θ ,ψ are called Eulerian angles. The Eulerian angles completely specify the
orientation of the x′, y′, z ′ system relative to the x, y, z system. Now we can obtain
the complete matrix of transformation from x, y, z to x′, y′, z ′ by writing the matrix
as the triple product of the separate rotations.
X ′ = BX 2
= B ( CX 1 )
= BC ( X 1 )
X ′ = BCDX
X ′ = AX ,
Classical Mechanics
Page No. 241
 x′ 
 ′
y =
 z′ 
 
 x
 
A y 
z
 
. . . (7)
where A = BCD is the complete matrix of transformation from x, y, z to x′, y′, z ′ and
is the product of the successive matrices. Using the equations (2), (4) and (6) the
matrix of transformation from space co-ordinates to the body co-ordinates is then
given by
 cosψ

A =  − sinψ
 0

sinψ
cosψ
0
01
0

0   0 cos θ
1   0 − sin θ
0   cos φ

sin θ   − sin φ
cos θ   0
sin φ
cos φ
0
0

0
1 
cosψ sin φ + sinψ cos θ cos φ , sinψ sin θ 
 cosψ cos φ − cos θ sinψ sin φ ,


A =  − sinψ cos φ − cos θ cosψ sin φ , − sinψ sin φ + cos θ cosψ cos φ , cosψ sin θ  .

sin θ sin φ ,
− sin θ cos φ ,
cos θ 

...(8)
This is the required matrix of transformation.
We will now show that this matrix A is orthogonal. Let the matrix A be
represented by
A = ( aij ) .
The condition for orthogonal transformation is that
∑a a
ij ik
= δ jk ,
i
where δ ij is a Kronecker delta symbol.
i.e.,
a1 j a1k + a2 j a2 k + a3 j a3k = δ jk .
We take the case j = k = 1 , and consider
Classical Mechanics
Page No. 242
2
2
a112 + a21
+ a312 = ( cosψ cos φ − cos θ sinψ sin φ ) +
2
+ ( − sinψ cos φ − cos θ cosψ sin φ ) + ( sin θ sin φ )
2
2
2
a112 + a21
+ a31
= 1.
Similarly, we can show for all value of j = k.
Now for j ≠ k ,j, k=1, 2, 3 we consider the case
a11a12 + a21a22 + a31a32 =
= ( cosψ cos φ − cos θ sinψ sin φ )( cosψ sin φ + cos θ sinψ cos φ ) +
+ ( − sinψ cos φ − cos θ cosψ sin φ )( − sinψ sin φ + cos θ cosψ cos φ ) + ( sin θ sin φ )( − sin θ cos φ )
a11a12 + a21a22 + a31a32 = 0 .
Similarly, we can show for all j ≠ k . that
∑a a
ij
jk
=0.
i
Hence the matrix A is orthogonal. To find the inverse of A, we know that in the case
of orthogonal matrix A−1 is the same as the transpose of A. Thus we have
 cosψ cos φ − cos θ sinψ sin φ , − cos φ sinψ − cos θ cosψ sin φ , sin θ sin φ 


A =  sin φ cosψ + cos θ cos φ sinψ , − sinψ sin φ + cos θ cosψ cos φ , − cos φ sin θ 


sin θ sinψ ,
sin θ cosψ ,
cos θ


−1
The matrix A−1 is the desired matrix, which gives the inverse transformation from
the body set of axes to the space set of axes.
This completes the answer.
Worked Examples •
Example 10 : If the matrix of transformation from space set of axes to body set of
axes is equivalent to a rotation through an angle χ about some axis through the
origin then show that
Classical Mechanics
Page No. 243
χ
 φ +ψ
cos   = cos 
2
 2

θ 
 cos   .

2
Solution : We know the matrix of complete rotation from space set of axes to body
set of axes in terms of Eulerian angles φ , θ ,ψ is given by
cosψ sin φ + sinψ cos θ cos φ , sinψ sin θ 
 cosψ cos φ − cos θ sinψ sin φ ,


A =  − sinψ cos φ − cos θ cosψ sin φ , − sinψ sin φ + cos θ cosψ cos φ , cosψ sin θ  .

sin θ sin φ ,
− sin θ cos φ ,
cos θ 

. . . (1)
It is given that this matrix of rotation is equivalent to the matrix of rotation of
co-ordinate axes through an angle χ about some axis with the same origin.
Equivalently, it means that it is always possible by means of some similar
transformation, to transform the matrix A to the matrix B obtained by rotating the coordinate axes through an angle χ about some axis with the same origin. This matrix
of rotation is given by either
 cos χ

B =  − sin χ
 0

sin χ
cos χ
0
0
0
1


0  or B =  0 cos χ
 0 − sin χ
1 

0 

sin χ  .
cos χ 
. . . (2)
It is well known that under similar transformation trace of the matrix is invariant.
Using this result we have
Trace of A = Trace of B
cosψ cos φ − cos θ sinψ sin φ + cosψ cos φ cos θ − sinψ sin φ + cos θ = 2 cos χ + 1
2 cos χ + 1 = ( cosψ cos φ − sinψ sin φ ) + cos θ ( cosψ cos φ − sinψ sin φ ) + cos θ
= [1 + cos θ ] cos (ψ + φ ) + cos θ
θ 
ψ + φ  
2 θ 
2 θ 
2 cos χ + 1 = 2 cos 2    2 cos 2 
 − 1 + cos   − sin  
 2 
 2  
2
2
Classical Mechanics
Page No. 244
θ 
 φ +ψ 
2 cos χ + 1 = 4 cos 2   cos 2 
 −1
2
 2 
θ 
 φ +ψ 
2 ( cos χ + 1) = 4 cos 2   cos 2 

2
 2 
χ
θ 
 φ +ψ 
cos 2   = cos 2   cos 2 

2
2
 2 
χ
θ 
 φ +ψ
cos   = cos   cos 
2
2
 2
•

.

. . . (3)
Moments of Inertia and Products of Inertia :
Moment of Inertia :
A uniformly rotating body possesses the tendency to oppose any change in its
state of rotation motion. This quantity is called the moment of inertia.
Theorem 3 : Obtain the angular momentum of a rigid body about a fixed point of the
body when the body rotates instantaneously with angular velocity ω in terms of
inertia tensor.
Proof: Consider a rigid body composed of N particles having masses mi and position
vectors ri with respect to the fixed point of the body. Since the translational motion
is absent and the body rotates about an axis passes through the fixed point. Let ω be
the instantaneous angular velocity of the body. If vi is the linear velocity of the
i th particle, then we know it is given by
vi = ω × ri .
. . . (1)
If L is the total angular momentum then it is equal to the sum of the angular
momenta of an individual particle.
Therefore we have
N
L = ∑ I i = ∑ ri × mi vi .
i
Classical Mechanics
i =1
Page No. 245
N
L = ∑ ri × mi (ω × ri ) ,
i =1
N
L = ∑ mi  ri × (ω × ri )  .
i =1
Using the vector identity
a × ( b × c ) = ( a .c ) b − ( a .b ) c
we obtain
N
L = ∑ mi ri 2ω − ∑ mi ( ri .ω ) ri .
i =1
. . . (2)
i
Let the components of the position vector ri , the angular velocity ω and the angular
momentum vector L be denoted by
ri = ixi + jyi + kzi ,
ω = iω x + jω y + kω z ,
. . . (3)
L = iLx + jLy + kLz .
Using equations (3) we write
iLx + jLy + kLz =
∑ m ( iω
i
i
i.e.,
x
+ jω y + kω z ) ri 2 − ( xiωx + yiω y + ziω z ) ( ixi + jyi + kzi ) 


iLx + jLy + kLz = i ω x ∑ mi ( ri 2 − xi2 ) − ω y ∑ mi xi yi − ω z ∑ mi xi zi  +
i
i
i




+ j  −ω x ∑ mi xi yi + ω y ∑ mi ( ri 2 − yi2 ) − ω z ∑ mi yi zi  +
i
i
i




+ k  −ω x ∑ mi xi zi − ω y ∑ mi yi zi + ω z ∑ mi ( ri 2 − zi2 ) 
i
i
i


Equating the corresponding coefficients on both the sides of the equation we get
Lx = ω x ∑ mi ( ri 2 − xi2 ) − ω y ∑ mi xi yi − ωz ∑ mi xi zi ,
i
Classical Mechanics
i
. . . (4a)
i
Page No. 246
Ly = −ω x ∑ mi xi yi + ω y ∑ mi ( ri 2 − yi2 ) − ω z ∑ mi yi zi ,
i
i
Lz = −ω x ∑ mi xi zi − ω y ∑ mi yi zi + ω z ∑ mi ( ri 2 − zi2 ) .
i
. . . (4b)
i
i
. . . (4c)
i
We write the components of angular momentum as
Lx = I xxω x + I xyω y + I xzω z ,
Ly = I yxω x + I yyω y + I yzω z ,
. . . (5)
Lz = I zxω x + I zyω y + I zzω z ,
where the coefficients of ω x , ω y , ω z are respectively defined by
I xx = ∑ mi ( ri 2 − xi2 ) = ∑ mi ( yi2 + zi2 ),
i
i
I yy = ∑ mi ( ri − y
2
2
i
) = ∑m (x
i
i
2
i
+ zi2 ),
. . . (6)
i
I zz = ∑ mi ( ri 2 − zi2 ) = ∑ mi ( xi2 + yi2 ).
i
i
and are called the moment of inertia about x, y, and z axes respectively. Equations
(6) show that the moment of inertia is the sum over the particles in the system of the
product of masses and the square of its perpendicular distance from the axis of
rotation. Also the quantities I xy , I xz and I yz are called the product of inertia and are
defined by
I xy = −∑ mi xi yi = I yx ,
i
I xz = − ∑ mi xi zi = I zx ,
. . . (7)
i
I yz = −∑ mi yi zi = I zy .
i
Now the equation (5) can be written in the matrix form as
 Lx   I xx
  
 Ly  =  I yx
L  I
 z   zx
or
Classical Mechanics
I xy
I yy
I zy
L = Iω
I xz   ω x 
 
I yz   ω y 
I zz   ω z 
. . . (8)
. . . (9)
Page No. 247
where I is called the moment of inertia tensor or inertia tensor. Note that moment
of inertia tensor is symmetric and hence it has only six independent components. The
moment of inertia tensor depends only on the mass distribution in the body. It is
given by
I
 xx
I =  I yx
I
 zx
I xy
I yy
I zy
I xz 

I yz  .
I zz 
. . . (10)
The diagonal elements are called the moments of inertia of the body about the
given point and the given set of body set of axes. The off diagonal components of
moment of inertia tensor are called the product of inertia of the body about the given
point and the given set of body axes.
Note that it is always possible to find a set of axes with respect to which all
the products of inertia tensor vanish leaving off diagonal terms and the axis is called
Principal axis of the body.
•
Kinetic Energy of a rigid body with one point fixed :
Theorem 4 : Find the kinetic energy of a rigid body rotating about a fixed point of
the body when the moments and products of inertia of the body relative to the set of
axes through fixed point are known.
Proof: Consider a rigid body composed of N particles having masses mi and rotating
with instantaneous angular velocity ω . If one of the points of the rigid body is fixed
then the translational motion is absent and the body rotates about an axis passes
through the fixed point. If vi is the linear velocity of the i th particle and position
vector ri with respect to the fixed point, then we know it is given by
vi = ω × ri .
. . . (1)
We know the kinetic energy of the body is given by
T=
Classical Mechanics
1
∑ mi vi2 .
2 i
. . . (2)
Page No. 248
Using equation (1) we write (2) as
T=
1
∑ mi vi (ω × ri ) .
2 i
On using the vector identity
a.(b × c ) = b .( c × a ) ,
we write the expression for the kinetic energy as
1
∑ miω. ( ri × vi ),
2 i
1
= ω ∑ ( ri × mi vi ),
2 i
1
= ω ∑ ri × pi
2 i
T=
1
T = ω .L
2
where
. . . (3)
L = ∑ ri × pi
i
is the total angular momentum. Now to express kinetic energy in terms of moment of
inertia and product of inertia, we know the angular momentum of the rigid body is
given by
L = Iω
. . . (4)
where I is the moment of inertia tensor and is given by
I
 xx
I =  I yx
I
 zx
I xy
I yy
I zy
I xz 

I yz  .
I zz 
. . . (5)
Hence equation (3) becomes
1 T = ω.I .ω .
2
. . . (6)
If the components of the angular velocity ω and the angular momentum vector L are
Classical Mechanics
Page No. 249
ω = iω x + jω y + kω z ,
L = iLx + jLy + kLz ,
. . . (7)
then we write equation (4) as
 Lx   I xx
  
 Ly  =  I yx
L  I
 z   zx
I xy
I yy
I zy
I xz   ω x 
 
I yz   ω y 
I zz   ω z 
iLx + jLy + kLz = i ( I xxω x + I xyω y + I xzω z ) + j ( I yxω x + I yyω y + I yzω z ) +
+ k ( I zxω x + I zyω y + I zzω z ) .
Hence equation (6) becomes
T=
1
( iωx + jω y + kωz ) i ( I xxωx + I xyω y + I xzωz ) + j ( I yxωx + I yyω y + I yzωz ) +
2
+ k ( I zxωx + I zyω y + I zzω z )  .
This equation becomes
T=
1
I xxω x2 + I yyω y2 + I zzω z2 + 2 I xyω xω y + 2 I yzω yω z + 2 I zxω zω x )
(
2
. . . (8)
If the body rotates about z-axis with angular velocity ω , we have
ω z = ω , ω x = ω y = 0 . In this case equation (8) becomes
T=
1
I .ω 2
2
. . . (9)
where I is the moment of inertia of the body about z axis.
Worked Examples •
Example 11 : Prove the statement that “ a change in time dt of the components of a
vector r as seen by an observer in the body system of axes will differ from the
corresponding change as seen by an observer in the space system.
Classical Mechanics
Page No. 250
consider two co-ordinate systems S and S ′ , where S ′ is rotating
Solution:
uniformly with angular velocity ω , with respect to the frame S with the same origin
O. Let S be the space set of axes and S ′ the body set of axes. Let i, j, k be the unit
vectors associated with the co-ordinate axes of S frame and i′, j ′, k ′ be the unit
z
z’
k’
vectors associated with
S ′ frame.
k
P
r
Consider the position vector r of a particle
y’
j’
O
i
j
y
in a rigid body with respect to the body set of axes.
It is represented by
r = i′x′ + j ′y′ + k ′z ′
i’
x
x′, y′, z ′ axes of the
. . . (1)
x’
Clearly such a vector appears constant when measured in the body set of
axes. However, to an observer fixed in space set of axes, the components of the
vector will vary in time. Let the components of the vector with respect to the space
set of axes be given by
r = ix + jy + kz .
. . . (2)
The time derivatives of r however will be different in the two systems. For
the space (fixed) system S, we have
dx
dy
dz
 dr 
  =i + j +k .
dt
dt
dt
 dt  fix
. . . (3)
Similarly, the time derivative of the vector r defined in S ′ with respect to
the body set of axes is given by
dx′
dy′
dz ′
 dr 
′
′
′
.
=
i
+
j
+
k
 
dt
dt
dt
 dt body
. . . (4)
However, as body rotates, the unit vectors of the body set of axes will be seen
changing relative to the observer in the space set of axes. Hence we find the time
derivative of the vector r in S ′ with respect to the fixed co-ordinate system as
Classical Mechanics
Page No. 251
dx′
dy′
dz
di′
dj ′
dk ′
 dr 
+ j′
+ k ′ + x′ + y ′ + z ′
.
  = i′
dt
dt
dt
dt
dt
dt
 dt  fix
. . . (5)
The first three terms on the R. H. S. of equation (5) are the time derivative of the
vector in the rotating system, while the remaining three terms arise as a result of
rotation of the system.
Hence we write the equation (5) as
di′
dj ′
dk ′
 dr 
 dr 
.
  =   + x′ + y ′ + z ′
dt
dt
dt
 dt  fix  dt body
. . . (6)
We know the linear velocity of a particle having the position vector r and rotating
with angular ω is given by
v=
dr
=ω×r .
dt
. . . (7)
This formula can be applied to the unit vectors as a special case. Thus we write
di′
dj ′
dk ′
= ω × i′,
= ω × j ′,
= ω × k′ .
dt
dt
dt
. . . (8)
Hence equation (6) reduces to
 dr 
 dr 
  =   + x′ ( i′ω x + j ′ω y + k ′ω z ) × i′ + y′ ( i′ω x + j ′ω y + k ′ω z ) × j ′ +
 dt  fix  dt body
.
+ z ′ ( i′ω x + j ′ω y + k ′ω z ) × k ′
 dr 
 dr 
  =   + i′ ( z ′ω y − y′ω z ) − j ′ ( z ′ω x − x′ω z ) + k ′ ( y′ω x − x′ω y ) . . . (9)
 dt  fix  dt body
This is equivalent to
i′
 dr 
 dr 
  =   + ωx
 dt  fix  dt body
x′
Classical Mechanics
j′
k′
ω y ωz .
y′
z′
Page No. 252
 dr 
 dr 
  =   +ω ×r .
 dt  fix  dt body
. . .(10)
This is the required relation between the two time derivatives of a vector with respect
to two frames of references.
••
Note : The formula (10) can also be represented as
 dr 
 dr 
=   +ω×r .
 
 dt  space  dt  rot
•
. . .(11)
Euler’s Equations of Motion :
A set of equations governing the rotation of a rigid body referred to its own
axis are known as Euler’s equations of motion of a rigid body with one point fixed.
Theorem 6 : Obtain the Euler’s equations of motion of a rigid body when one point
of the body remains fixed.
Proof : Consider a rigid of which one point is fixed. Hence translational motion of
the body is absent and the body rotates about an axis passes through the fixed point.
The rotation of the body takes place under the action of torque acting on it. Thus the
equation of the rotational motion of the body in a fixed frame is given by
Torque = rate of change of angular momentum.
 dL 
N =  .
 dt  fix
However, we know
 dr 
 dr 
  =   +ω ×r .
 dt  fix  dt body
Classical Mechanics
Page No. 253
Therefore the equation of motion of the rigid body becomes
 dL 
 dL 
N =   =   +ω× L .
 dt  fix  dt body
. . . (1)
where ω is the angular velocity of the body and L is the angular momentum and is
given by
L = Iω ,
. . . (2)
I is the moment of inertia tensor and is constant with respect to the body frame of
reference. We choose the principal axis of the body with respect to which the off
diagonal elements of the moment of inertia tensor vanish and only the diagonal
elements remain in the expression for I . If I1 , I 2 , I 3 are the principal moments of
inertia then we have
I
 1
I =0
0

0
I2
0
0

0.
I 3 
In this case the expression for angular momentum (2) becomes
L = I1ω x i + I 2ω y j + I 3ω z k ,
Differentiating this with respect to time t in the body frame we get
 dL 
  = I1ω x i + I 2ω y j + I 3ω z k .
 dt body
Also we find the value of
i
ω × L = ωx
I1ω x
j
k
ωy
I 2ω y
ωz
I 3ωz
⇒ ω × L = i ( I 3 − I 2 ) ω yω z − j ( I 3 − I1 ) ω xω z + k ( I 2 − I1 ) ω yω x .
Classical Mechanics
Page No. 254
Hence the equation of rotational motion of the rigid body becomes
 dL 
N =   + ω × L = i  I1ω x + ( I 3 − I 2 ) ω yωz  + j  I 2ω y + ( I1 − I 3 ) ωxω z  +
 dt body
+ k  I 3ω z + ( I 2 − I1 ) ω xω y 
If torque is expressed in term of its components as
N = iN x + jN y + kN z
then on equating the corresponding components on both sides of the above equations
we obtain
N x = I1ω x + ( I 3 − I 2 ) ω yωz ,
N y = I 2ω y + ( I1 − I 3 ) ω xω z ,
. . . (3)
N z = I 3ω z + ( I 2 − I1 ) ω xω y .
These are the required Euler’s equations of motion of the rigid body with one point
fixed.
Note : In the case of torque free motion of a rigid body, equations (3) reduce to
I1ω x = ( I 2 − I 3 ) ω yω z ,
I 2ω y = ( I 3 − I1 ) ω xωz ,
. . . (4)
I 3ω z = ( I1 − I 2 ) ω xω y .
Worked Examples •
Example 12 : If the rigid body with one point fixed rotates about the principal axis
of the body, then show that
(1)
kinetic energy of the body and
(2)
the magnitude of the angular momentum are constants throughout the motion.
Solution : The kinetic energy of a rigid body with one point fixed is given by
T=
1
I xxω x2 + I yyω y2 + I zzω z2 + 2 I xyω xω y + 2 I yzω yω z + 2 I zxω zω x )
(
2
Classical Mechanics
. . . (1)
Page No. 255
where I xx , I yy , I zz and I xy , I yz , I zx are the moments of inertia and product of inertia
about the co-ordinate axes respectively. If the body rotates about the principal axis of
the body then the products of inertia tensors are zero and the moments of inertia
tensors are constants. In this case the kinetic energy becomes
T=
1
I1ω x2 + I 2ω y2 + I 3ω z2 )
(
2
. . . (2)
I1 = I xx , I 2 = I yy , I 3 = I zz .
where
In the absence of torque the Euler’s equations of motion of the rigid body are given
by
I1ω x + ( I 3 − I 2 ) ω yω z = 0,
I 2ω y + ( I1 − I 3 ) ωxω z = 0,
. . . (3)
I 3ω z + ( I 2 − I1 ) ω xω y = 0.
i)
Multiply each equation in (3) by ω x , ω y , ω z respectively and adding we get
I1ω xω x + I 2ω yω y + I 3ωzω z = 0
. . . (4)
We write this as
1 d
( I1ωx2 + I 2ω y2 + I3ωz2 ) = 0
2 dt
dT
⇒
= 0,
dt
1
⇒ T = ( I1ω x2 + I 2ω y2 + I 3ω z2 ) = const.
2
ii)
Now we claim that the magnitude of the angular momentum is constant.
The moment of inertia tensor with respect to the principal axis of the body, is given
by
I
 1
I =0
0

Classical Mechanics
0
I2
0
0

0,
I 3 
Page No. 256
where I1 , I 2, I 3 are constant with respect to the principal axis of the body. Hence the
expression for angular momentum becomes
L = I1ω x i + I 2ω y j + I 3ω z k ,
⇒ L2 = L.L = I12ω x2 + I 22ω y2 + I 32ωz2 .
. . . (5)
Multiply each equation of (3) by I1ω x , I 2ω y , I 3ω z respectively and adding we get
I12ωxω x + I 22ω yω y + I 32ω zω z = 0 .
This we write as
1 d 2 2
( I1 ωx + I 22ω y2 + I32ωz2 ) = 0,
2 dt
d 2
⇒
( L ) = 0,
dt
⇒ L2 = const.
This shows that the magnitude of the angular momentum of the rigid body is
constant.
•
Components of angular velocity vector along body set of axes :
Example 13 : Show that the components of angular velocity vector along the body
set of axes are given by
ω x′ = φ sin θ sinψ + θ cosψ ,
ω y′ = φ sin θ cosψ − θ sinψ ,
ω z′ = φ cos θ +ψ .
Solution : Let (x, y, z) and ( x′, y′, z ′ ) be the space (fixed) set of axes and body
(rotating) set of axes respectively. Let a rigid body with one point fixed rotate
instantaneously with angular velocity ω . We shall obtain the components of ω
along the body set of axes. If φ , θ ,ψ are the Eulerian angles, then their time
derivatives φ,θ,ψ represent the angular speeds about the space z-axis, the line of
nodes and the body z-axis respectively. We denote these three angular speeds by
Classical Mechanics
Page No. 257
ωφ , ωθ , ωψ which are the three components of the angular velocity ω . Note that these
three components of ω are not all either along the space set of axes or the body set of
axes. Let ω = (ω x′ , ω y′ , ω z′ ) be the components of ω with respect to the body set of
axes x′, y′, z ′ .
If φ (ωφ ) represents the angular speed about space z-axis then its
components along the body set of axes are found by applying orthogonal
transformations C through an angle θ about new x -axis and B through an angle ψ
about new z -axis, to come to the body axes, as two orthogonal transformations are
required to come to body axes. If
( ω ) , (ω ) , ( ω )
φ x′
φ
y′
φ z′
are the components of
φ (ωφ ) along body set of axes, then we have
 (ω ) 
0
 φ x′ 
 (ω )  = BC  0  ,
 
 φ y′ 
 φ 


 
ω
(
)
 φ z′ 
 cosψ

=  − sinψ
 0

 (ω ) 
 φ x′   cosψ
 (ω )  =  − sinψ
 φ y′  

  0
ω
(
)
φ
z′ 

⇒
(ω )
(ω )
(ω )
φ x′
φ
y′
φ z′
Classical Mechanics
sinψ
cosψ
0
01
0

0   0 cos θ
1   0 − sin θ
sinψ cos θ
cosψ cos θ
− sin θ
0  0
 
sin θ   0 
cos θ   φ 
sinψ sin θ   0 
 
sin θ cosψ   0  ,
cos θ   φ 
= φ sinψ sin θ ,
= φ cosψ sin θ ,
. . . (1)
= φ cos θ .
Page No. 258
( )
Similarly, if ωθ θ represents the angular speed along the line of nodes
( )
(new x′ -axis), and (ωθ ) x′ , (ωθ ) y ′ , (ωθ ) z′ are the components of ωθ θ about the
body set of axes x′, y′, z ′, then to find these components, we apply orthogonal
transformation B-through an angle ψ about new z -axis to come to the body axes;
after θ rotation has been performed. Thus we have
 (ω ) 
 θ 
 θ x′ 
 
 (ωθ ) y′  = B  0  ,


0
 (ωθ ) 
 

z′ 
 cosψ

=  − sinψ
 0

⇒
sinψ
cosψ
0
(ωθ ) x′ = θ cosψ ,
(ωθ ) y′ = −θ sinψ ,
(ωθ ) z′ = 0.
0   θ 
 
0 0 
1   0 
. . . (2)
Now ωψ (ψ ) is already parallel to z ′ axis, no transformation is necessary.
Hence, if (ωψ ) x′ , (ωψ ) y′ , (ωψ ) z′ are components of ωψ with respect to the body set of
axes, then they are given by
(ω )
(ω )
(ω )
ψ
x′
ψ
y′
ψ
z′
= 0,
= 0,
. . . (3)
= ψ .
Thus the components of angular velocity ω ,
(ω
x′
, ω y′ω z′ ) about body set of axes are
defined by
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Page No. 259
ω x′ = (ωφ ) x′ + (ωθ ) x′ + (ωψ ) x′ ,
ω y′ = (ωφ ) y′ + (ωθ ) y′ + (ωψ ) y′ ,
ω z′ = (ωφ ) z′ + (ωθ ) z′ + (ωψ ) z′ .
Using equations (1), (2) and (3), we readily obtain the components of angular
velocity ω about body set of axes in the form
ω x′ = φ sin θ sinψ + θ cosψ ,
ω y′ = φ sin θ cosψ − θ sinψ ,
. . . (4)
ω z′ = φ cos θ +ψ .
Example 14 : If a rectangular parallelepiped with its edges 2a, 2a, 2b rotates about
its center of gravity under no force, prove that, its angular velocity about one
principal axis is constant and about the other axis it is periodic.
Solution: It is given that the rigid body rotates under the action of no forces. Hence
the Euler’s equation of motion, in the absence of no forces are given by
I1ω x = ( I 2 − I 3 ) ω yω z ,
I 2ω y = ( I 3 − I1 ) ω xωz ,
. . . (1)
I 3ω z = ( I1 − I 2 ) ω xω y .
where I1 , I 2 , I 3 are called the principal moments of inertia about the center of gravity
of the body. Since the rigid body is parallelepiped with its edges 2a, 2a, 2b. Hence
the moments of inertia about the principal axes OX, OY, and OZ are given by
 a2 + b2 
I1 = I 2 = m 

 3 
2
I 3 = ma 2
3
. . . (2)
where m is the mass of the parallelepiped. Substituting these values in equation (1)
we get
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Page No. 260
(a
(a
2
+ b 2 ) ω x = ( b 2 − a 2 ) ω yω z ,
2
+ b 2 ) ω y = ( a 2 − b 2 ) ωxω z ,
. . . (3)
ω z = 0 ⇒ ω z = n = const.
The last equation in (3) shows that the angular velocity about one principal axis is
constant. Consequently, the other two equations give
(a
(a
2
+ b 2 ) ω x = ( b 2 − a 2 ) nω y ,
2
+ b 2 ) ω y = ( a 2 − b 2 ) nωx .
. . . (4)
Eliminating ω y between (4) we obtain
2
 a 2 − b2 
ωx = −  n 2 2  ω x .
 a +b 
. . . (5)
This is a second order differential equation of simple harmonic motion. This shows
that ω x is periodic.
••
Unit 3: Caley-Klein Parameters:
Introduction:
We have seen that the Eulerian angles are used to describe an orientation of a
rigid body. However, it is found that these angles are difficult to use in the numerical
computation, because of the large number of trigonometric functions involved.
Various other groups of variables have been used to describe the orientation of a
rigid body. Klien’s set of four complex parameters is one of them. He introduced the
set of four parameters bearing his name to facilitate the integration of complicated
gyroscopic problems. These parameters are much better adapted for use on
computers. Furthermore these four parameters are of great theoretical interest in
modern branches of physics. Cayley-Klein parameters are the set of four complex
numbers used to describe the orientation of a rigid body in space.
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Page No. 261
We begin this unit with some basic definitions and see how these CayleyKlein parameters define the orientation of the rigid body.
•
Some Definitions:
1.
Conjugate matrix: The matrix obtained from any given matrix A by
replacing its elements by the corresponding conjugate complex numbers is
called the conjugate of A and it is denoted by A* .
2.
Trace of a matrix : Let A be a square matrix of order n. The sum of the
elements of A lying along the principal diagonal is called the trace of A.
3.
Transposed conjugate of a matrix : The transpose of the conjugate of a
matrix Q is called transposed conjugate of Q and it is denoted by Q† . It is
T
*
also called as adjoint of Q. Thus Q † = ( Q* ) = ( QT ) .
4.
Unitary Matrix : A square matrix Q with complex elements is said to be
unitary if Q†Q = I = QQ † . Unitary condition expect that Q = 1 ⇒ Q is
invertible and it is given by Q −1 = Q † . This also gives that Q −1 = Q † = adjQ .
5.
Self adjoint : A linear operator which is identical with its adjoint operator is
called self-adjoint. If P is self- adjoint then P = adjP .
6.
Hermitian Matrix : A square matrix A = ( aij ) is said to be Hermitian if
aij = a*ji for every i, j . ⇒ If
7.
A = A* ⇒
A is Hermitian.
Similar matrices : Let A and B two square matrices of the same order. Then
A and B are said to be similar if there exists a non-singular matrix P such that
AP = PB. ⇒
A = PBP −1 .
Property : Under similar transformation the self-adjoint (Hermitian) property of the
matrix and the trace of the matrix are invariant.
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Page No. 262
Worked Examples •
Example 15 : Explain how the eight quantities of Cayley-Klein parameters can be
reduced to only three independent quantities to describe the orientation of a rigid
body?
Solution: Consider a two –dimensional complex space with u and v as complex
axis. A general linear transformation in such a space is given by
u ′ = α u + β v,
v′ = γ u + δ v ,
. . . (1)
α
Q=
γ
. . . (2)
where
β
δ 
is the rotation matrix in 2-dimensional complex plane, and α , β , γ , δ are four
complex parameters are known as Cayley-Klein parameters. There are eight
quantities in four complex parameters. However, we know that the minimum number
of independent quantities needed to specify the orientation of a rigid body is three.
Thus to reduce eight quantities in equation (2) into three independent quantities, the
matrix Q is restricted by imposing an additional condition that it is unitary. The
unitary condition implies that
Q†Q = I = QQ †
. . . (3)
where Q† is transposed conjugate of Q known as the adjoint matrix. As Q is unitary,
we have
Q = 1 ⇒ αδ − βγ = 1 .
. . . (4)
Also expanding equation (3) we get
α

γ
β  α * γ * 

=I
δ   β * δ * 
 αα * + ββ * αγ * + βδ *   1 0 
 *
=

*
γγ * + δδ *   0 1 
 γα + δβ
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Page No. 263
αα * + ββ * = 1,
⇒
γγ * + δδ * = 1,
. . . (5)
αγ * + βδ * = 0.
We notice that the first two equations of (5) are real while the third is complex.
Therefore equation (5) gives 4 conditions. These 4 conditions plus one condition
given in (4) are totally five conditions on eight quantities. Therefore we are left with
only 3 independent quantities, which are used to describe the orientation of the body.
To calculate those three independent quantities dividing equation (4) by αγ we get
δ β
1
− =
.
γ α αγ
. . . (6)
Now from the last equation in (5) we have
δ
α*
=− * .
γ
β
. . . (7)
Thus from (6) and (7) we have
−
1
α* β
− =
*
β α αγ
⇒
− (αα * + ββ * )
β
*
=
1
γ
⇒ γ = −β * , δ = α * .
. . . (9)
As a result of (9), the matrix Q takes the form
β
 α
Q= *
*
 −β α 
. . . (10)
with the unitary condition
αα * + ββ * = 1 .
. . . (11)
Hence Q involves only 3 independent quantities, which are used to describe the
orientation of a rigid body.
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Page No. 264
•
Matrix of transformation in terms of Cayley-Klein Parameters :
Theorem 7 : Obtain the matrix of transformation in Cayley-Klein parameters, which
specify the orientation of a rigid body.
OR
Show that to each unitary matrix Q in the 2-dimensional complex space there
is associated some real orthogonal matrix of transformation in ordinary 3dimensional space.
Proof : To find the matrix of transformation in terms of Cayley-Klein parameter, let
P be a matrix operator in a specialized u, v complex co-ordinate system in a
particular form
 z
P=
 x + iy
x − iy 
.
−z 
. . (1)
We notice that the matrix P is trace free and is self -adjont. i.e., P = P † , and x, y, z
are real quantities taken as co-ordinates of a point in space. Suppose the matrix P is
transformed to the matrix P′ by means of the unitary matrix Q in the following way
P′ = QPQ†
. . . (2)
α
Q=
γ
. . . (3)
where
β
δ 
is the matrix of transformation in 2- dimensional complex space and Q† is a complex
*
transposed conjugate of Q. i.e., Q † = ( QT ) . Since Q is unitary, from the unitary
property of Q, we have
QQ † = I .
This implies that adjoint of Q is same as its inverse. i.e.,
Q −1 = Q † = adj ( Q ) .
Classical Mechanics
. . . (4)
Page No. 265
Therefore equation (2) becomes
P′ = QPQ −1 .
. . . (5)
This shows that P′ is the similarity transformation of P. It is well known that the
self-adjoint and the trace free property of the matrix are invariant under similarity
transformation. As P is self-adjoint and trace-free, therefore P′ must be like wise
self-adjoint and trace free. Thus P′ can have the form
x′ − iy′ 
 z′
P′ = 

− z′ 
 x′ + iy′
where x′, y′, z ′ are to be determined.
Let us denote
x − iy = x− ,
x + iy = x+ .
Therefore equation (5) with the help of equations (3) and (4) becomes.
 z′
 ′
 x+
x−′   α
=
− z ′   γ
β  z
δ   x+
x−   α * γ * 
.

−z   β * δ * 
We know that α , β , γ , δ are not independent but are related by the equations
α * = δ , β * = −γ ,
Therefore we have
 z′
 ′
 x+
 z′
P′ = 
 x+′
x−′   α
=
− z ′   γ
β  z
δ   x+
x−   δ

− z   −γ
 z′
P′ = 
 x+′
x−′   α
=
− z ′   γ
β   zδ − γ x− , − β z + α x− 
δ   δ x+ + γ z, − x+ β − zγ 
−β 
α 
 z (αδ + βγ ) − αγ x− + βδ x+ ,
x−′ 
α 2 x− − 2αβ z − β 2 x+ 
=



− z′ 
2γδ z − γ 2 x− + δ 2 x+ ,
αγ x− − (αδ + γβ ) z − βδ x+ 

This is the matrix transformation equation in complex 2-plane. Obviously the matrix
on the r. h. s. is hermitean, proving that the hermitean property is invariant under any
unitary similar transformation.
Classical Mechanics
Page No. 266
Solving these equations, we obtain
x+′ = δ 2 x+ − γ 2 x− + 2γδ z ,
⇒
x−′ = − β 2 x+ + α 2 x− − 2αβ z ,
. . . (6)
z ′ = βδ x+ − αγ x− + (αδ + βγ ) z.
In matrix notation, we write these equations as
2
 x+′   δ
 ′  2
 x−  =  − β
 z ′   βδ
  
−γ 2
2
α
−αγ
  x+ 
 
−2αβ   x−  .
αδ + βγ   z 
2γδ
. . . (7)
Explicitly, we write equations (6) as
⇒ x′ + iy′ = (δ 2 − γ 2 ) x + i (δ 2 + γ 2 ) y + 2γδ z ,
x′ − iy′ = (α 2 − β 2 ) x − i (α 2 + β 2 ) y − 2αβ z ,
z ′ = ( βδ − αγ ) x + i (αγ + βδ ) y + (αδ + βγ ) z.
Solving these equations for x′, y′, z ′ we ready obtain
1 2
2
2
2
 2 (α − β + δ − γ )
 x′  
 ′  i 2
2
2
2
 y  =  2 (α − β − δ + γ )
 z′  
 
βδ − αγ


i

γδ − αβ 
−α 2 − β 2 + δ 2 + γ 2 )
(
2
 x 
1 2
2
2
2
α + β + δ + γ ) −i (αβ + γδ )   y  . . . . (8)
(

2
z
i (αγ + βδ )
αδ + βγ   


Thus we have
1 2
2
2
2
 2 (α − β + δ − γ ) ,

i
A =  (α 2 − β 2 − δ 2 + γ 2 ) ,
2

βδ − αγ ,


i

−α 2 − β 2 + δ 2 + γ 2 ) ,
γδ − αβ 
(
2

1 2
2
2
2
α + β + δ + γ ) , −i (αβ + γδ )  .
(

2
i (αγ + βδ ) ,
αδ + βγ 


. . . (9)
This is the required matrix of transformation in terms of Cayley-Klein parameters.
This matrix specifies the orientation of a rigid body. Hence the Cayley-Klein
parameters specify the orientation of a rigid body.
Classical Mechanics
Page No. 267
•
Relation between the Cayley-Klein Parameters and Eulerian Angles :
Theorem 8 : Establish the relation between the Eulerian angles φ , θ ,ψ and the
Cayley-Klein parameters α , β , γ , δ .
OR
Obtain Q matrices Qφ , Qθ , Qψ in complex 2-plane corresponding to the separate
successive rotations through an angle φ , θ ,ψ in 3 × 3 real space. Hence obtain the
orthogonal matrix of complete rotation.
Proof : The Eulerian angles φ , θ ,ψ are the three successive angles of rotations about
the specified axes, such that the space set of co-ordinates (x, y, z) coincide with the
body set of co-ordinates ( x′, y′, z ′ ) . These angles are used to describe the orientation
of a rigid body. Similarly, Cayley-Klein parameters α , β , γ , δ are also used to
describe the orientation of rigid body. Now to find the relation between φ , θ ,ψ and
α , β , γ , δ , we first construct Q matrices say Qφ , Qθ , Qψ corresponding to the
separate successive rotations φ , θ ,ψ and then combine them to form the complete
matrix Q = Qψ Qθ Qφ of rotation.
First Rotation :
α
Let Qφ = 
γ
β
be the matrix in 2-complex plane corresponding to the first
δ 
rotation through an angle φ in 3 × 3 real space. This rotation through an angle φ is
performed about z-axis. Hence the transformation equations are given by
x′ = x cos φ + y sin φ ,
y′ = − x sin φ + y cos φ ,
z′ = z
We write these equations as
Classical Mechanics
Page No. 268
x+′ = x′ + iy′ = xe −iφ + iye− iφ
⇒
x+′ = e− iφ x+ ,
x−′ = eiφ x− ,
z ′ = z.
. . . (1)
But we know that
x+′ = δ 2 x+ − γ 2 x− + 2γδ z ,
x−′ = − β 2 x+ + α 2 x− − 2αβ z ,
. . . (2)
z ′ = βδ x+ − αγ x− + (αδ + βγ ) z.
Now comparing equations (1) and (2) we get
α 2 = eiφ , δ 2 = e − iφ , β = 0 = γ .
Therefore the Qφ matrix corresponding to the rotation through an angle φ becomes
 i2φ
e
Qφ = 
 0


0 
.
iφ 
−

e 2
. . . (3)
Second Rotation :
α
Let Qθ = 
γ
β
be the matrix in 2-complex plane corresponding to the
δ 
second rotation through an angle θ in 3 × 3 real space. This rotation through an angle
θ is performed about new x-axis. Hence the transformation equations are given by
x′ = x,
y′ = y cos θ + z sin θ ,
. . . (4)
z ′ = − y sin θ + z cos θ .
From equations (4) we obtain

θ 
θ 
x+′ = x′ + iy′ = x  cos 2   + sin 2    + iy cos θ + iz sin θ
2
 2 


θ 
θ 
x−′ = x′ − iy′ = x  cos 2   + sin 2    − iy cos θ − iz sin θ
2
 2 

z ′ = − y sin θ + z cos θ .
Classical Mechanics
Page No. 269


θ 
θ 
θ 
θ 
θ 
θ 
x+′ = x  cos 2   + sin 2    + iy  cos 2   − sin 2    + 2iz sin   cos   .
2
 2 
2
 2 
2
2


This can be written as
θ 
θ 
θ 
θ 
x+′ = x+ cos 2   + x− sin 2   + 2iz sin   cos   .
2
2
2
2
Similarly, we write
θ 
θ 
θ 
θ 
x−′ = x+ sin 2   + x− cos 2   − 2iz sin   cos   ,
2
2
2
2
θ 
θ 
θ 
θ  
θ 
θ 
z ′ = ix+ sin   cos   − ix− sin   cos   + z  cos 2   − sin 2    .
2
2
2
2 
2
 2 
. . . (5)
Comparing equations (5) with (2) we get
θ 
θ 
α 2 = δ 2 = cos 2   , β 2 = γ 2 = − sin 2   .
2
2
 
 
Hence Qθ matrix becomes

θ 
θ 
 cos  2  , i sin  2  
 
 
Qθ = 
.

θ 
θ  
 i sin   , cos   
2
2

α
Third Rotation: Let Qψ = 
γ
. . . (6)
β
be the matrix in 2-complex plane corresponding
δ 
to the third rotation through an angle ψ in 3 × 3 real space. This rotation through an
angle ψ
is performed about new z-axis. This rotation is affected by the
transformation equations
x′ = x cosψ + y sinψ ,
y′ = − x sinψ + y cosψ ,
z′ = z
Classical Mechanics
Page No. 270
We write these equations in the form
x+′ = e − iψ x+ ,
x−′ = eiψ x− ,
z ′ = z.
. . . (7)
Hence comparing equation (7) with (2) we obtain the matrix Qψ corresponding to
the third rotation through an angle ψ about new z- axis as
 iψ2
e , 0
Qψ = 
iψ
 0, e − 2



.


. . . (8)
Hence the orthogonal matrix for the complete transformation from space set of axes
to the body set of axes is obtained by taking the product of separate Q matrices for
each of the three successive rotations φ , θ ,ψ . Thus we obtain
Q = Qψ Qθ Qφ ,
 iψ2
e , 0
Q=
iψ
 0, e− 2

α
Q=
γ
i
α = e2
i
Classical Mechanics


,


i
 2i (ψ +φ )
(ψ −φ )
θ 
θ  
2
e
cos
,
ie
sin   



β 
2
2
=
.

i
δ   2i (φ −ψ )  θ 
− (ψ +φ )
θ 
2
sin   , e
cos   
 ie
2
 2 

(ψ +φ )
β = ie 2

θ 
θ 
  cos   , i sin     i2φ
2
 2 e , 0



iφ
 i sin  θ  , cos  θ    0, e− 2
 
  

2
2

. . . (9)
θ 
cos   ,
2
(ψ −φ )
θ 
sin   ,
2
Page No. 271
i
γ = ie 2
δ =e
(φ −ψ )
i
− (ψ +φ )
2
θ 
sin   ,
2
. . . (10)
θ 
cos   .
2
These are the required relations between the Eulerian angles and the Cayley-Klein
parameters.
Note: Note from equations (3), (6) and (8) that the trace of any Q matrix through an
χ
angle say χ about some axis is 2 cos   .
2
Worked Examples •
Example 16 : (Aliter) With usual notations show that
χ
 φ +ψ
cos   = cos 
2
 2

θ 
 cos   .

2
Solution: This example is solved in Example (9). However, we will attempt this
problem by using the relation between Eulerian angles and Cayley-Klein parameters.
α
We know the Q matrix 
γ
α
Q=
γ
β
in terms of Eulerian angles is given by
δ 
i
 2i (ψ +φ )
(ψ −φ )
θ 
θ  
2
e
cos
,
ie
sin   



β 
2
2
=

i
δ   2i (φ −ψ )  θ 
− (ψ +φ )
θ 
sin   , e 2
cos   
 ie
2
 2 

. . . (1)
where α , β , γ , δ are Cayley-Klein parameters such that α * = δ , β * = −γ .
This is the matrix of rotation, which describes the orientation of the rigid
body with one point fixed. Let this matrix be equivalent to the matrix B obtained by
rotating the co-ordinate axes through an angle χ about any axis with the same
origin. Then the matrix B is given by
Classical Mechanics
Page No. 272
 cos χ

B =  − sin χ
 0

0
0
1


0  or B =  0 cos χ
 0 − sin χ
1 

sin χ
cos χ
0
0 

sin χ  .
cos χ 
. . . (2)
Then the Q matrix in complex 2-dimensional plane corresponding to the matrix B is
similarly obtain in the form

χ
 χ 
cos   , i sin   


0 
2
 2 
or Qχ = 
.
iχ 

−
χ
 χ 
2 
e 
i sin   , cos   
2
 2 

 i2χ
e
Qχ = 
 0
. . . (3)
From equation (1) we have
i
α = e0 + ie3 = e 2
δ = e0 − ie3 = e
(ψ +φ )
θ 
cos   ,
2
i
− (ψ +φ )
2
θ 
cos   ,
2
i
(ψ −φ )
2
θ 
sin   ,
2
β = e2 + ie1 = ie
γ = −e2 + ie1 = ie
i
− (ψ −φ )
2
. . . (4)
θ 
sin   .
2
Solving these equations, we find

,
2

i
θ 
ψ − φ 
e1 = − ( β + γ ) = sin   cos 
,
2
2
 2 
e0 =
α +δ
θ 
 φ +ψ
= cos   cos 
2
 2
 θ  ψ − φ 
e2 =
= − sin   sin 
,
2
2  2 
α −δ
 θ   φ +ψ 
e3 =
= cos   sin 
.
2
2  2 
β −γ
. . . (5)
We see that
θ 
 φ +ψ 
trace of Q = α + δ = 2 cos   cos 
.
2
 2 
Classical Mechanics
. . . (6)
Page No. 273
Similarly, from equation (3) we have
χ
trace of Qχ = 2 cos  
2
. . . (7)
Since trace is invariant, therefore, from (6) and (7) we have
χ
 φ +ψ
cos   = cos 
2
 2

θ 
 cos   .

2
Example 17 : Find a real matrix of orthogonal transformation in the 3-dimensional
space corresponding to the unitary matrix

θ 
 θ 
 cos  2  , i sin  2  
 
 
Q=

θ 
θ  
i sin   , cos   
2
 2

in two dimensional complex plane.
Solution: To find the matrix of orthogonal transformation in 3-dimensional space
corresponding to the matrix

θ 
 θ 
 cos  2  , i sin  2  
 
 
Q=

θ 
θ  
i sin   , cos   
2
 2

. . . (1)
in 2-dimensional complex plane, let P be a required matrix operator in a specialized
u, v complex co-ordinate system in a particular form
 z
P=
 x + iy
x − iy 

−z 
. . . (2)
We notice that the matrix P is trace free and is self -adjont. i.e., P = P † , and x, y, z
are real quantities taken as co-ordinates of a point in space. Suppose the matrix P is
transformed to the matrix P′ by means of the unitary matrix Q in the following way
P′ = QPQ†
Classical Mechanics
. . . (3)
Page No. 274
*
where Q† is a complex transposed conjugate of Q. i.e., Q † = ( QT ) called adjoint of
Q. Since Q is unitary, from the unitary property of Q, we have QQ † = I .
This implies that adjoint of Q is same as its inverse. i.e.,
Q −1 = Q † = adj ( Q ) .
. . . (4)
Therefore equation (3) becomes
P′ = QPQ −1 .
. . . (5)
This is the similarity transformation of P. Under similar transformation, we know
that the trace-free and self-adjoint property of the matrix are invariant. Since P is
self-adjoint and trace-free, therefore P′ must be like wise self-adjoint and trace free.
Thus P′ can have the form
x′ − iy′ 
 z′
P′ = 

− z′ 
 x′ + iy′
. . . (6)
where x′, y′, z ′ are to be determined.
Let us define
x − iy = x− ,
x + iy = x+ .
. . . (7)
Therefore equation (5) with the help of equations (1), (2), (6) and (7) becomes.
 z′
 ′
 x+

θ 
 θ 
cos   i sin   

x−′ 
2
 2   z
=


− z′  
θ 
 θ    x+
i sin   cos   
2
 2

Classical Mechanics

θ 
 θ 
cos   −i sin   

x− 
2
 2 
.

−z  
θ 
θ  
 −i sin   cos   
2
2 

Page No. 275
Therefore we have
 z′ x−′ 
′
=
 x+ −z′
  2θ  2θ 

θ  θ 
θ  θ 
θ 
θ 
θ  θ 
x sin2 +x cos2 −2izcos sin 
zcos  −sin  +ix+cos sin −ix−cos sin ,

+  2 −  2 
 2
 2  2
 2  2
 2  2
  2
.
=

θ

θ
θ

θ
θ
θ
θ
θ
θ
θ
 
 
   
 
 
   
   
x cos2 +x sin2 +2izcos sin ,
zsin2 −cos2 +ix cos sin −ix cos sin 

+  2 −  2

 2  2
 2  −  2  2  +  2   2
  2
On equating the corresponding elements of the matrix we get
θ  θ 
θ 
θ 
x+′ = x′ + iy′ = 2iz cos   sin   + ( x + iy ) cos 2   + ( x − iy ) sin 2   ,
2 2
2
2
θ  θ 
θ 
θ 
x−′ = x′ − iy′ = −2iz cos   sin   + ( x − iy ) cos 2   + ( x + iy ) sin 2   ,
2 2
2
2

θ 
θ 
θ  θ 
θ  θ 
z ′ = z  cos 2   − sin 2    − i ( x − iy ) cos   sin   + i ( x + iy ) cos   sin   .
2
 2 
2 2
2 2

⇒ x′ + iy′ = x + iy cos θ + iz sin θ ,
x′ − iy′ = x − iy cos θ − iz sin θ ,
z ′ = z cos θ − y sin θ .
⇒ x′ = x
y′ = y cos θ + z sin θ ,
z ′ = − y sin θ + z cos θ .
0
 x′   1
 ′ 
i.e.,  y  =  0 cos θ
 z ′   0 − sin θ
  
0  x 
 
sin θ  y 

cos θ 
 z 
This shows that corresponding to the matrix

θ 
 θ 
 cos  2  i sin  2  
 
 
Q=

θ 
θ  
i sin   cos   
2
 2

Classical Mechanics
Page No. 276
in 2-dimensional complex plane, there exits 3 × 3 real matrix
0
1

A =  0 cos θ
 0 − sin θ

0 

sin θ 
cos θ 
in 3- dimensional real space.
••
Exercise:
1.
Show that the components of angular velocity vector along the space set of
axes are given by
ω x = θ cos φ +ψ sin θ sin φ ,
ω y = θ sin φ −ψ sin θ cos φ ,
ω z = ψ cos θ + φ.
2.
Find a real matrix of orthogonal transformation in the 3-dimensional space
corresponding to the unitary matrix
 i2φ
e
Q=
 0


0 
iφ 
−
e 2 
in 2- dimensional complex plane.
3.
Find a real matrix of orthogonal transformation in the 3-dimensional space
corresponding to the unitary matrix
 iψ2
e
Q=
 0


0 
iψ 
−
e 2 
in 2- dimensional complex plane.
Classical Mechanics
Page No. 277