CHAPTER - I V
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.
Classical Mechanics Page No. 225 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
raijij= ij , ≠ = 1,2,...., N . . . (1)
th th where rij denotes the distance between the i and j particles, and aij are constants.
Equation (1) is symmetric in i and j and i≠ j as the distance of the ith from itself is zero, therefore, the possible number of constraints is N( N −1) NC = . . . . (2) 2 2 N( N −1) We notice that for N > 7, > 3N . 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
A of at least three non-collinear particles whose 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, Classical Mechanics Page No. 226 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.
Classical Mechanics Page No. 227 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 3- degrees 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, x 2 , x 3 ) and ( x1′, x 2′ , x 3′ ) be two co-ordinate systems. The general linear transformation between these two co-ordinate systems is defined by the following set of equations
x1′ = ax 111 + ax 122 + ax 133 ,
x2′ = ax 211 + ax 222 + ax 233 , . . . (1)
x3′ = ax 311 + ax 322 + ax 333 , where a11, a 12 ,..., a 33 are constants. These three equations can be combined in to a single equation as
3
xi′ =∑ ax ij j , i = 1,2,3. . . . (2) j=1 where A= ( a ij ) is called the matrix of transformation. Let a vector
r= xi1 + xj 2 + xk 3 defined in ( x1, x 2 , x 3 ) co-ordinate system be transformed to
( x1′, x 2′ , x 3′ ) co-ordinate systems in the form r= xi1′ + xj 2′ + xk 3′ . Since the magnitude of the vector must be the same in both the co-ordinate system, we must have therefore
3 3 2 2 ∑xi′ = ∑ x i . . . . (3) i=1 i = 1
Classical Mechanics Page No. 228 Using equation (2) in equation (3) we get
33 3 3 2 ∑∑axij j ∑ ax ik k = ∑ x i , ij==11 k = 1 i = 1
333 3 ⇒ 2 ∑∑∑aaij ik xx j k= ∑ x i . jki===111 i = 1 Equating the corresponding coefficients on both the sides of the above equation we get
3
∑ aij a ik= δ jk , . . . (4) i=1 where δ jk is the Kronecker delta symbol and is defined by δ =0when j ≠ k , jk . . . (5) =1when j = k . 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′ =∑ bx kj j , k = 1,2,3, . . . (6) j=1 where the matrix of transformation is B= ( b kj ) . Let the succeeding transformation from x′ to x′′ be defined by the matrix
A= ( a ik ) and is given by
3
xi′′=∑ ax ikk ′ , i = 1,2,3. . . . (7) k=1
Classical Mechanics Page No. 229 Now the transformation from x to x′′ is obtained by combining the two equations (6) and (7) as
3
xi′′ =∑ abx ik kj j , i = 1,2,3. j, k = 1 This may also be written as
3
xi′′ =∑ cx ij j , i = 1,2,3. . . . (8) j=1 where C= ( c ij ) is the matrix of transformation from x to x′′ and the elements of the matrix of transformation are defined as
3
cij= ∑ a ik b kj . . . . (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
Classical Mechanics Page No. 230 cij= ∑ a ik b kj . . . (1) k where Aa=( ik), Bb = ( kj ) are matrices of orthogonal transformations. This implies that
∑ aij a ik= δ jk , . . . (2) i
∑bij b ik= δ jk . . . . (3) i Consider now ∑ccij ik= ∑∑ ab im mj ∑ ab il lk i im l
= ∑ aim b mj ab il lk , i, m , l = ∑ ∑ aim a il b mj b lk , m, l i
= ∑δmlb mj b lk , m, l
= ∑blj b lk , l
∑cij c 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 Classical Mechanics Page No. 231 3
xi′ =∑ ax ij j , i = 1,2,3. . . . (1) j=1 where A= ( a ij ) is the matrix of transformation satisfying the condition
∑ aij a ik= δ jk . . . . (2) i
Let the matrix of inverse transformation from xi′ to xi be described by the inverse
−1 matrix A= ( aij′), a ij′ are the elements of the inverse matrix of transformation satisfying
∑ aij′ a ik′ = δ jk . . . . (3) i Also we have
−1 AA= I⇒ ∑ aki a ij′ = δ kj . . . . (4) i
Now consider the double sum ∑ akl a ki a ij′ which can be evaluated either by summing k, i over k first or over i first. Therefore we evaluate the double sum as ∑aaaklkiij′= ∑ ∑ aaa klki ij′ , ki, ik
= ∑δlia ij′ , i
∑ aaakl ki ij′= a lj′ . . . . (5) k, i Now evaluating the double sum over i first and then over k , we obtain ∑aaakl ki ij′= ∑ ∑ aaa ki ij′ kl , ki, ki
= ∑δkja kl , k
∑ aaakl ki ij′ = a jl . . . . (6) k, i
Classical Mechanics Page No. 232 From equations (5) and (6) we have
aij′ = a ji , ∀ ij , ,
⇒ (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,
⇒ A2 =1 as A = A ′ , ⇒ A = ± 1.
• 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.
Classical Mechanics Page No. 233 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
xx1′ =+ 1ε 111 x + ε 122 x + ε 133 x ,
xx2′ =+ 2ε 211 x + ε 222 x + ε 233 x ,
xx3′ =+ 3ε 311 x + ε 322 x + ε 333 x , i.e., xi′ = x i + ε ij x j , . . . (1) where summation is defined over the repeated index j and εij are the elements of the matrix of infinitesimal transformation and are infinitesimal. i.e., second order terms in εij can be neglected. Hence we write equations (1) as
xi′ =δ ijj x + ε ijj x , i.e., xi′ =(δ ij + ε ij) x j , . . . (2) In matrix notations we write equation (2) as X′ =( I + ε ) X , . . . (3)
x1 x1′ ′ ′ where X= x 2 , X= x 2 , I =(δij), ε = ( ε ij ) x3 x3′ and I + ε is the matrix of infinitesimal transformation.
Now let I + ε1 and I + ε 2 be two matrices of successive infinitesimal transformations. Consider (I+ε)( I +=+++ ε) III. εεεε I 1 2 2112 =I +ε1 + ε 2 .
Classical Mechanics Page No. 234 Similarly consider (I+ε)( I +=+++ ε) III. εεεε I 2 1 1221 =I +ε1 + ε 2 . We see from above equations that
(I+ε1)( I +=+ ε 2) ( I ε 2)( I + ε 1 ) 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) II εεεε I (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 Xi= +4 j + 2 3 k with respect to a particular co-ordinate system. Find the form of the vector with respect to co-ordinate π system obtained from the first by rotating it about the x-axis through an angle in 3 the anti-clock wise direction. Determine its magnitude and compare with X .
π Solution : The matrix of rotation about x-axis through an angle in the anti clock 3 wise direction is given by Classical Mechanics Page No. 235 1 0 0 1 0 0 π π π 1 3 A =0 cos sin = 0 . 3 33 22 π π 3 1 0− sin cos 0 − 3 3 2 2 Hence the new vector with respect to the new co-ordinate axes obtained from the first π by rotating through an angle about x-axis is given by 3 1 0 0 1 1 1 3 X ′ =0 4 = 5 2 2 2 3− 3 3 1 0 − 2 2
Xi′ = +5 j − 3 k
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
Classical Mechanics Page No. 236 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−λ I) R = 0 . . . . (3) The equation (3) can have solution only when A−λ I = 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− IA) = I − A . . . . (7) Taking the determinant of equation (7) we get ( AIA−) = I − A
()AI− A = I − A ⇒ AI− =− AI − . . . (8)
⇒ A− I = 0 . . . . (9) Comparing equations (4) and (9) we get λ =1. This proves the theorem. Classical Mechanics Page No. 237 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 = z 1 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 fix the orientation of a rigid body with one point φ y O fixed. Thus the orientation of a rotating body φ x with one point fixed can be completely specified by three independent Eulerian angles. x1
Classical Mechanics 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 co- ordinate system be labeled as x1, y 1 , z 1 axes as shown in the fig. In this case xy plane becomes x1 y 1 plane. The rotation is affected by the following transformation equations.
x1 = xcosφ + y sin φ ,
y1 = − xsinφ + y cos φ , . . . (1)
z1 = z . These equations can be written in matrix form as
x1 x y1 = D y or X1 = DX z1 z where cox φsin φ 0 D = − sinφ cos φ 0 . . . (2) 0 0 1 is the matrix of transformation.
Classical Mechanics Page No. 239 z = z 1 z2 The second rotation is performed about the new
x1 axis. The axes x1, y 1 , z 1 are rotated about x1 axis counter clockwise direction by an angle θ . y2
θ y1 Let the resultant set of co-ordinate θ axes be relabeled as x2, y 2 , z 2 . Here x2 axis φ y O φ being the line of intersection of xy plane and
x x1 y 1 plane is called the line of nodes. The
x1 = x 2 transformation equations from x1, y 1 , z 1 to new set of axes x2, y 2 , z 2 can be represented by the following set of equations:
x2= x 1 ,
y2= y 1cosθ + z 1 sin θ , . . . (3)
z1= − y 1sinθ + z 1 cos θ ,
x2 x 1 i.e., y2= C y 1 or X2= CX 1 z2 z 1 where 1 0 0 C = 0 cosθ sin θ . . . (4) 0− sinθ cos θ is the matrix of transformation. z = z z = z = z’ 1 Finally, the third rotation is performed about 2 3
y3 = y’ z2 axis. The x2, y 2 , z 2 axes are rotated counter
y2 clockwise direction by an angle ψ about z2 axis to y ψ 1 produce the third and the final set of axes x, y , z , θ 3 3 3 φ y O which coincide, with body set of axes x′, y ′ , z ′ . This φ x ψ
x1 = x 2
x3 = x’ Classical Mechanics Page No. 240 completes the transformation from space set of axes to body set of axes. This transformation is represented by
xxx3=′ = 2cosψ + y 2 sin ψ ,
yyx3==−′ 2sinψ + y 2 cos ψ , . . . (5)
z3= z′ = z 2 or
x′ x 2 ′ y= B y 2 z′ z 2 or X′ = BX 2 where cox ψsin ψ 0 B = − sinψ cos ψ 0 . . . (6) 0 0 1 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′ x y′ = A y . . . (7) z′ z 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 cosψψ sin 01 0 0 cos φφ sin 0 A = −sinψψ cos 0 0 cos θθ sin − sin φφ cos 0 0 0 10 − sinθ cos θ 0 0 1
cosψφ cos− cos θψφ sin sin , cos ψφψθφ sin + sin cos cos , sin ψθ sin A =− sinψφ cos − cos θψφ cos sin , − sin ψφ sin + cos θψφ cos cos , cos ψθ sin . sinsin,θφ− sincos, θφ 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= ( a ij ) . The condition for orthogonal transformation is that
∑ aij a ik= δ jk , i where δij is a Kronecker delta symbol. i.e., aa11jk+ aa 22 jk + aa 33 jk = δ jk . We take the case j= k = 1, and consider
Classical Mechanics Page No. 242 a2++= a 2 a 2 ()cosψ cos φ − cos θψφ sin sin 2 + 11 21 31 +−()()sinψφ cos − cos θψφ cos sin2 + sin θφ sin 2
2 2 2 a11+ a 21 + a 31 = 1. Similarly, we can show for all value of j = k. Now for j≠ k ,j, k=1, 2, 3 we consider the case aa+ aa + aa = 1112 21 22 31 32 =−(cosψφ cos cos θψφ sin sin)( cos ψφ sin + cos θψφ sin cos ) + +−−()()()()sinψφ cos cos θψφ cos sin −+ sin ψφ sin cos θψφ cos cos + sin θφ sin − sin θφ c os
aa1112+ aa 21 22 + aa 31 32 = 0 . Similarly, we can show for all j≠ k . that
∑ aij a 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 cosψφ cos− cos θψφ sin sin , − cos φψ sin − cos θψφ cos sin , sin θφ sin −1 A =+sinφψ cos cos θφψ cos sin , −+ sin ψφ sin cos θψφ cos cos , − cos φθ sin sinsin,θψ sincos, θψ cos θ
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 φψ+ θ cos= cos cos . 2 2 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
cosψφ cos− cos θψφ sin sin , cos ψφψθφ sin + sin cos cos , sin ψθ sin A =− sinψφ cos − cos θψφ cos sin , − sin ψφ sin + cos θψφ cos cos , cos ψθ sin . sinsin,θφ− sincos, θφ 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 co- ordinate axes through an angle χ about some axis with the same origin. This matrix of rotation is given by either cosχ sin χ 0 1 0 0 B = − sinχ cos χ 0 or B = 0 cosχ sin χ . . . . (2) 0 0 1 0− sinχ cos χ 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 cosψφ cos− cos θψφ sin sin + cos ψφθψφ cos cos −+=+ sin sin cos θ 2cos χ 1 2cosχ+= 1( cos ψφψφ cos − sin sin) + cos θψφψφ( cos cos − sin sin) + cos θ
=+[]1 cosθ cos() ψφ ++ cos θ
22θ ψφ+ 22 θ θ 2cosχ += 1 2cos 2cos −+ 1 cos − sin 2 2 22
Classical Mechanics Page No. 244 θ φ+ ψ 2cosχ + 14cos =2 cos 2 − 1 2 2 θ φ+ ψ 2() cosχ + 1 = 4cos2 cos 2 2 2 χ θ φψ+ cos2= cos 2 cos 2 2 2 2
χ θ φψ+ cos= cos cos . . . . (3) 2 2 2
• 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 ith particle, then we know it is given by
vi=ω × r i . . . . (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=∑ Ii = ∑ rmv i × ii . i i =1 Classical Mechanics Page No. 245 N
L=∑ rmi × i()ω × r i , i=1
N L=∑ mri i ×()ω × r i . i=1 Using the vector identity a××=( bc) ( acb.) − ( abc . ) we obtain
N 2 L=∑ mriiω − ∑ mrr ii(). ω i . . . . (2) i=1 i
Let the components of the position vector ri , the angular velocity ω and the angular momentum vector L be denoted by
ri= ix i + jy i + kz i ,
ωω=ix + j ω y + k ω z , . . . (3)
L= iLx + jL y + kL z . Using equations (3) we write
iLx+ jL y + kL z =
2 ∑ mii( ωωω x++ j y k zi) r −++( x ixiyizi ωωω y z)() ixjykz ++ i i i
2 2 iLx++= jLkLi y zω x∑ mr ii() −− x i ω y ∑ mxy iiiz − ω ∑ mxz iii + i i i
2 2 i.e., +−j ωx∑ mxy iii + ω y ∑ mry ii() −− i ω z ∑ myz iii + i i i
2 2 +−k ωx∑ mxz iii − ω y ∑ myz iii + ω z ∑ mrz ii() − i i i i Equating the corresponding coefficients on both the sides of the equation we get
2 2 Lx=ω x∑ mrx ii( −− i) ω y ∑ mxy iii − ω z ∑ mxz iii , . . . (4a) i i i
Classical Mechanics Page No. 246 2 2 Ly=−ω x∑ mxy iii + ω y ∑ mry ii( −− i) ω z ∑ myz iii , . . . (4b) i i i
2 2 Lz=−ω x∑ mxz iii − ω y ∑ myz iii + ω z ∑ mrz ii( − i ) . . . . (4c) i i i We write the components of angular momentum as
LIx= xxxω + I xyy ω + I xzz ω ,
LIy= yxxω + I yyy ω + I yzz ω , . . . (5)
LIz= zxω x + I zy ω y + I zz ω z , where the coefficients of ωx, ω y , ω z are respectively defined by
22 22 Ixx=∑ mrx iii( −=) ∑ myz iii( + ), i i 22 22 Iyy=∑ mry ii()() −= i ∑ mxz iii + , . . . (6) i i 22 22 Izz=∑ mrz iii()() −= ∑ mxy iii + . 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 Ixy, I xz andI yz are called the product of inertia and are defined by
Ixy= −∑ mxy i i i = I yx , i
Ixz= −∑ mxz i i i = I zx , . . . (7) i
Iyz= −∑ myz i i i = I zy . i Now the equation (5) can be written in the matrix form as Lx II xx xy I xz ω x Ly= I yx I yy I yz ω y . . . (8) Lz I zx I zy I zz ω z or L= I ω . . . (9)
Classical Mechanics 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 I I I xx xy xz given by I= Iyx I yy I yz . . . . (10) Izx I zy I zz 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
th through the fixed point. If vi is the linear velocity of the i particle and position vector ri with respect to the fixed point, then we know it is given by
vi=ω × r i . . . . (1) We know the kinetic energy of the body is given by
1 2 T= ∑ mi v i . . . . (2) 2 i
Classical Mechanics Page No. 248 Using equation (1) we write (2) as 1 T=∑ mvi i()ω × r i . 2 i On using the vector identity abc.( ×) = bca . ( × ) , we write the expression for the kinetic energy as 1 T=∑ mrviω.() i × i , 2 i 1 =ω∑()ri × m i v i , 2 i 1 =ω∑ ri × p i 2 i 1 T= ω. L . . . (3) 2 where L=∑ ri × p i 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= I ω . . . (4) where I is the moment of inertia tensor and is given by I I I xx xy xz I= Iyx I yy I yz . . . . (5) Izx I zy I zz Hence equation (3) becomes 1 T= ω. I . ω . . . . (6) 2 If the components of the angular velocity ω and the angular momentum vector L are
Classical Mechanics Page No. 249 ωω=ix + j ω y + k ω z , . . . (7) L= iLx + jL y + kL z , then we write equation (4) as Lx II xx xy I xz ω x Ly= I yx I yy I yz ω y Lz I zx I zy I zz ω z
iLjLkLiIx++= y z( xxxωωω ++ I xyy I xzz) + jI( yxx ωωω ++ I yyy I yzz ) +
+kI()zxω x + I zy ω y + I zz ω z . Hence equation (6) becomes
1 TijkiI=++()()()ωωωx y z xxx ωωω +++ I xyy I xzz jI yxx ωωω +++ I yyy I yzz 2 +kI()zxω x + I zy ω y + I zz ω z . This equation becomes
1 2 2 2 TIIII=( xxxω +++ yyy ω zzz ω2 xyxy ωω + 2 I yzyz ωω + 2 I zxzx ωω ) . . . (8) 2 If the body rotates about z-axis with angular velocity ω , we have
ωz= ω, ω x = ω y = 0 . In this case equation (8) becomes 1 T= I .ω 2 . . . (9) 2 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 Solution : consider two co-ordinate systems S and S′ , where S′ is rotating 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 vectors associated with x′, y ′ , z ′ axes of the S′ frame. k’ k P Consider the position vector r of a particle r y’ j’ in a rigid body with respect to the body set of axes.
y j i O It is represented by
i’ x r= ix′′ + jy ′′ + kz ′′ . . . (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 dr dx dy dz =i + j + k . . . . (3) dt fix dt dt dt Similarly, the time derivative of the vector r defined in S′ with respect to the body set of axes is given by
dr dx′ dy ′ dz ′ =i′ + j ′ + k ′ . . . . (4) dt body dt dt dt 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 dr dx′′ dy dz di ′′′ dj dk =i′ + j ′ ++ kxyz ′′ + ′′ + . . . . (5) dt fix dt dt dt dt dt dt 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 dr dr di′ dj ′ dk ′ = +++x′ y ′ z ′ . . . . (6) dt fix dt body dt dt dt We know the linear velocity of a particle having the position vector r and rotating with angular ω is given by dr v= =ω × r . . . . (7) dt This formula can be applied to the unit vectors as a special case. Thus we write di′ dj ′ dk ′ =×ωi′, =× ω j ′ , =× ω k ′ . . . . (8) dt dt dt Hence equation (6) reduces to dr dr = +xi′′()()ωωωxyz ++×+ j ′ k ′ iyi ′′′ ωωω xyz ++×+ j ′ k ′ j ′ dt fix dt body .
+zi′′()ωx ++ j ′ ω y k ′ ω z × k ′
dr dr = +iz′′()ωωyz −− y ′ jz ′′() ωω xz −+ x ′ ky ′′() ωω xy − x ′ . . . (9) dt fix dt body
This is equivalent to i′ j ′ k ′ dr dr = + ω ω ω . dt dt x y z fix body x′ y ′ z ′
Classical Mechanics Page No. 252 dr dr = +ω × r . . . .(10) dt fix dt body This is the required relation between the two time derivatives of a vector with respect to two frames of references.
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Note : The formula (10) can also be represented as dr dr = +ω × r . . . .(11) dtspace dt rot
• 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 . . . . (1) dt fix dt body where ω is the angular velocity of the body and L is the angular momentum and is given by L= I ω , . . . (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
I1 0 0 I= 0 I 2 0 . 0 0 I3 In this case the expression for angular momentum (2) becomes
LI=1ωx iI + 2 ω y jI + 3 ω z k , Differentiating this with respect to time t in the body frame we get
dL =I1ωx iI + 2 ω y jI + 3 ω z k . dt body Also we find the value of i j k
ω×L = ωx ω y ω z
I1ωx I 2 ω y I 3 ω z
⇒ ω×=−LiII( 32) ωωyz −− jII( 31) ωω xz +− kII( 21 ) ωω yx .
Classical Mechanics Page No. 254 Hence the equation of rotational motion of the rigid body becomes dL N= +×=ω LiI1 ωx +−()() II 32 ωω yz + jI 2 ω y +− II 13 ωω xz + dt body +kI3ωz +() II 2 − 1 ω x ω y If torque is expressed in term of its components as
N= iNx + jN y + kN z then on equating the corresponding components on both sides of the above equations we obtain NIx=1ω x +( II 3 − 2 ) ω yz ω , NIy=2ω y +() II 1 − 3 ω xz ω , . . . (3) NIz=3ω z +() II 2 − 1 ω xy ω . 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 I1ωx=( I 2 − I 3 ) ω y ω z , I2ωy=() I 3 − I 1 ω x ω z , . . . (4) I3ωz=() I 1 − I 2 ω x ω 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
1 2 2 2 TIIII=( xxxω +++ yyy ω zzz ω2 xyxy ωω + 2 I yzyz ωω + 2 I zxzx ωω ) . . . (1) 2
Classical Mechanics Page No. 255 where Ixx,, I yy I zz andI 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
1 2 2 2 T=() I1ωx + I 2 ω y + I 3 ω z . . . (2) 2 where I1= IIxx, 2 = II yy , 3 = I zz . In the absence of torque the Euler’s equations of motion of the rigid body are given by I1ωx+( I 3 − I 2 ) ω y ω z = 0, I2ωy+() I 1 − I 3 ω x ω z = 0, . . . (3) I3ωz+() I 2 − I 1 ω x ω y = 0. i) Multiply each equation in (3) by ωx, ω y , ω z respectively and adding we get I1ωωxx+ I 2 ωω yy + I 3 ωω zz = 0 . . . (4) We write this as
1 d 2 2 2 ()I1ωx+ I 2 ω y + I 3 ω z = 0 2 dt dT ⇒ = 0, dt 1 ⇒ 2 2 2 T=() I1ωx ++ I 2 ω y I 3 ω z = 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
I1 0 0 I= 0 I 2 0 , 0 0 I3
Classical Mechanics 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
LI=1ωx iI + 2 ω y jI + 3 ω z k ,
2 222222 ⇒ LLLI=. =1ωx + I 2 ω y + I 3 ω z . . . . (5)
Multiply each equation of (3) by I1ωx, I 2 ω y , I 3 ω z respectively and adding we get
2 2 2 I1ωωxx+ I 2 ωω yy + I 3 ωω zz = 0. This we write as
1 d 22 22 22 ()I1ωx+ I 2 ω y + I 3 ω z = 0, 2 dt d ⇒ ()L2 = 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 ωx′ = φsin θ sin ψθ + cos ψ , ωy′ = φsin θ cos ψθ − sin ψ , ωz′ = φcos θψ + . 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, ω θ , ω ψ 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= ( ωx′, ω y ′ , ω z ′ ) be the components of ω with respect to the body set of axes x′, y ′ , z ′ . If φ( ω φ ) 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 are the components of ( φ)x′( φ) y ′( φ ) z ′ φ( ω φ ) along body set of axes, then we have
ω ( φ )x′ 0 ω = BC 0 , ()φ y′ φ ω ()φ z′ cosψ sin ψ 01 0 0 0 = − sinψψ cos 0 0 cos θθ sin 0 0 0 10 − sinθ cos θ φ
ω ( φ )x′ cosψ sin ψθ cos sin ψθ sin 0 ω = − sin ψψθθψ cos cos sin cos 0 , ()φ y′ 0− sinθ cos θ φ ω ()φ z′
⇒ ω= φψsin sin θ , ( φ )x′ ω= φψcos sin θ , . . . (1) ()φ y′ ω= φcos θ . ()φ z′
Classical Mechanics Page No. 258 Similarly, if ωθ ( θ ) represents the angular speed along the line of nodes
(new x′ -axis), and , , are the components of about the (ωθ)x′( ω θ) y ′( ω θ ) z ′ ωθ ( θ ) 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′ B 0 , ()ωθ y′ = ()ω 0 θ z′ cosψ sin ψ 0 θ = − sinψ cos ψ 0 0 0 0 10 ⇒ (ωθ )x′ = θcos ψ , . . . (2) ()ωθ y′ = − θsin ψ ,
()ωθ z′ = 0. Now ωψ ( ψ ) is already parallel to z′ axis, no transformation is necessary.
Hence, if ω, ω , ω are components of ω with respect to the body set of ( ψ)x′( ψ) y ′( ψ ) z ′ ψ axes, then they are given by ω = 0, ( ψ )x′ ω = 0, . . . (3) ()ψ y′ ω= ψ . ()ψ z′
Thus the components of angular velocity ω,( ωx′ , ωω y ′ z ′ ) about body set of axes are defined by
Classical Mechanics Page No. 259 ωω= + ω + ω , x′ ( φ)x′( θ)x′ ( ψ ) x ′ ωω= + ω + ω , y′ ()φy′() θy′ () ψ y ′ ωω= + ω + ω . z′ ()φz′() θz′ () ψ z ′ Using equations (1), (2) and (3), we readily obtain the components of angular velocity ω about body set of axes in the form ωx′ = φsin θ sin ψθ + cos ψ , ωy′ = φsin θ cos ψθ − sin ψ , . . . (4) ωz′ = φcos θψ + . 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 I1ωx=( I 2 − I 3 ) ω y ω z , I2ωy=() I 3 − I 1 ω x ω z , . . . (1) I3ωz=() I 1 − I 2 ω x ω 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+ b 2 I= I = m 1 2 3 . . . (2) 2 I= ma 2 3 3 where m is the mass of the parallelepiped. Substituting these values in equation (1) we get
Classical Mechanics Page No. 260 22 22 (ab+)ωx =( ba − ) ω y ω z , 22 22 ()()ab+ωy = ab − ω x ω z , . . . (3) ωz= 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 22 22 (ab+)ωx =( ban − ) ω y , . . . (4) 22 22 ()()ab+ωy = abn − ω x .
Eliminating ωy between (4) we obtain
2 2 2 a− b ωx= − n ω x . . . . (5) a2+ b 2 This is a second order differential equation of simple harmonic motion. This shows that ωx is periodic.
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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.
Classical Mechanics Page No. 261 We begin this unit with some basic definitions and see how these Cayley- Klein 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 * ) = ( Q T ) . 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= ( a ij ) is said to be Hermitian if
* * aij= a ji for every i, j .⇒ If A= A ⇒ A is Hermitian. 7. 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.
Classical Mechanics 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 , . . . (1) v′ =γ u + δ v , where α β Q = . . . (2) γ 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 β α* γ * = I γ δ β* δ *
αα*+ ββ * αγ * + βδ * 1 0 = γα*+ δβ * γγ * + δδ * 0 1
Classical Mechanics 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 − = . . . . (6) γ α αγ Now from the last equation in (5) we have δ α * = − . . . . (7) γ β * Thus from (6) and (7) we have α* β 1 − − = β* α αγ * * −()αα + ββ 1 ⇒ = β* γ
⇒ γ= − βδ*, = α * . . . . (9) As a result of (9), the matrix Q takes the form α β Q = . . . (10) −β* α * 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.
Classical Mechanics 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 3- dimensional 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 x− iy P = . . . (1) xiy+ − z
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) where α β Q = . . . (3) γ δ is the matrix of transformation in 2- dimensional complex space and Q† is a complex
* transposed conjugate of Q. i.e., Q† = ( Q T ) . 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)
Classical Mechanics 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 z′ x ′− iy ′ P′ = x′+ iy ′ − z ′ 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.
* * zx′− ′ α β zx − α γ = * * . xz+′− ′ γ δ xz + − β δ We know that α, β , γ , δ are not independent but are related by the equations
α*= δβ, * = − γ , Therefore we have
zx′− ′ αβ zx − δβ− P′ = = xz+′− ′ γδ xz + − − γα
zx′− ′ α β zxδγ− −, − βα zx + − P′ = = xz+′− ′ γ δ δγ xzxz + +−−, + βγ
z′ x ′ z(αδβγ+−+) αγ xx βδ, α2 xzx −− 2 αβ β 2 − = −+ − + 2 2 x+′− z ′ 2γδzxx−+ γ−+ δ , αγ x − −+−() αδγβ zx βδ + 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
2 2 x+′ =δ x + − γ x − + 2 γδ z , 2 2 ⇒ x−′ =−β xx + + α − − 2 αβ z , . . . (6)
zxx′ =βδ+ − αγ − ++() αδ βγ z . In matrix notation, we write these equations as 2 2 x+′ δ− γ2 γδ x + ′ 2 2 x− = −β α − 2 αβ x − . . . . (7) z′ βδ− αγ αδ + βγ z Explicitly, we write equations (6) as ⇒ xiy′+=− ′ (δ22 γ) xi +( δ 22 + γ) y + 2 γδ z , xiy′−=− ′ ()()αβ22 xi − αβ 22 + y − 2 αβ z , z′ =−()()()βδ αγ xi + αγ + βδ y ++ αδ βγ z . Solving these equations for x′, y ′ , z ′ we ready obtain
1 2222i 2222 ()αβδγ−+−() −−++ αβδγ γδαβ − x′ 2 2 x i 22221 2222 y′ =()αβδγ −−+() αβδγ +++ −+i() αβγδ y . . . . (8) 2 2 z′ z βδαγ−i () αγβδ + αδβγ + Thus we have
1 2222i 2222 ()αβδγ−+−,() −−++ αβδγ , γδαβ − 2 2 i 22221 2222 A =()αβδγ −−+,() αβδγ +++ , −+i () αβγδ . . . . (9) 2 2 βδαγ−,i() αγβδ + , αδβγ + 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, Q θ , Q ψ 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, Q θ , Q ψ corresponding to the separate successive rotations φ, θ , ψ and then combine them to form the complete matrix Q= QQQψ θ φ of rotation.
First Rotation : α β Let Q = be the matrix in 2-complex plane corresponding to the first φ γ 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′ = xcosφ + y sin φ , y′ = − xsinφ + y cos φ , z′ = z We write these equations as
Classical Mechanics Page No. 268 −iφ − i φ x+′=+ x ′ iy ′ = xe + iye ⇒ x′ = e−iφ x , + + . . . (1) iφ x−′ = e x − , z′ = z . But we know that
2 2 x+′ =δ x + − γ x − + 2 γδ z , 2 2 x−′ =−β xx + + α − − 2 αβ z , . . . (2)
zxx′ =βδ+ − αγ − ++() αδ βγ z . Now comparing equations (1) and (2) we get α2=eiφ, δ 2 = e − i φ , βγ == 0.
Therefore the Qφ matrix corresponding to the rotation through an angle φ becomes
iφ e 2 0 Q = . . . . (3) φ iφ − 0 e 2 Second Rotation : α β Let Q = be the matrix in 2-complex plane corresponding to the θ γ 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′ = ycosθ + z sin θ , . . . (4) z′ = − ysinθ + z cos θ . From equations (4) we obtain
2θ 2 θ xxiyx+′=+= ′ ′ cos + sin + iy cosθ + iz sin θ 2 2
2θ 2 θ xxiyx−′=−= ′ ′ cos + sin − iy cosθ − iz sin θ 2 2 z′ = − ysinθ + z cos θ . Classical Mechanics Page No. 269 2θθ 2 22 θθ θθ xx+′ =cos ++ sin iy cos −+ sin 2 iz sin cos . 22 22 22 This can be written as θ θ θθ xx′ =cos2 + x sin 2 + 2sin iz cos . + +2 − 2 22 Similarly, we write θ θ θθ xx′ =sin2 + x cos 2 − 2 iz sin cos , − +2 − 2 22 . . . (5) θθ θθ2 θ 2 θ zix′ =+sin cos − ix − sin cos +− z cos sin . 22 22 2 2 Comparing equations (5) with (2) we get θ θ αδ222==cos , βγ 22 ==− sin 2 . 2 2
Hence Qθ matrix becomes
θ θ cos ,i sin 2 2 Q = . . . . (6) θ θ θ i sin , cos 2 2
α β Third Rotation: Let Q = be the matrix in 2-complex plane corresponding ψ γ 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′ = xcosψ + y sin ψ , y′ = − xsinψ + y cos ψ , z′ = z
Classical Mechanics Page No. 270 We write these equations in the form
−iψ x+′ = e x + , iψ x−′ = e x − , . . . (7) z′ = z .
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
iψ e 2 , 0 Q = . . . . (8) ψ iψ − 0, e 2 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= QQQψ θ φ ,
θ θ iψ cos ,i sin i φ e2 ,02 2 e 2 ,0 Q = , iψ i φ − θ θ − 0,e2 isin , cos 0, e 2 2 2
i i ()ψφ+θ() ψφ − θ e2cos , ie 2 sin α β 2 2 Q = = . . . . (9) i i γ δ ()φψ−θ −() ψφ + θ ie2sin , e 2 cos 2 2
i ()ψ+ φ θ α = e 2 cos , 2
i ()ψ− φ θ β = ie 2 sin , 2
Classical Mechanics Page No. 271 i ()φ− ψ θ γ = ie 2 sin , 2 . . . (10) i −()ψ + φ θ δ = e 2 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 2cos . 2
Worked Examples •
Example 16 : (Aliter) With usual notations show that χ φψ+ θ cos= cos cos . 2 2 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 β We know the Q matrix in terms of Eulerian angles is given by γ δ
i i ()ψφ+θ() ψφ − θ e2cos , ie 2 sin α β 2 2 Q = = . . . (1) i i γ δ ()φψ−θ −() ψφ + θ ie2sin , e 2 cos 2 2 where α, β , γ , δ are Cayley-Klein parameters such that α*= δβ, * = − 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 cosχ sin χ 0 1 0 0 B = − sinχ cos χ 0 or B = 0 cosχ sin χ . . . . (2) 0 0 1 0− sinχ cos χ Then the Q matrix in complex 2-dimensional plane corresponding to the matrix B is similarly obtain in the form χ χ iχ cos ,i sin e 2 0 2 2 Q= or Q = . . . . (3) χiχ χ − χ χ 0 e 2 isin , cos 2 2 From equation (1) we have
i ()ψ+ φ θ α =e + ie = e 2 cos , 0 3 2 i −()ψ + φ θ δ =e − ie = e 2 cos , 0 3 2 . . . (4) i ()ψ− φ θ β =e + ie = ie 2 sin , 2 1 2 i −()ψ − φ θ γ =−e + ie = ie 2 sin . 2 1 2 Solving these equations, we find αδ+ θ φψ + e = = cos cos , 0 2 2 2 i θ ψ− φ e =−()β + γ = sin cos , 1 2 2 2 . . . (5) βγ− θ ψφ − e = = − sin sin , 2 2 2 2 αδ− θ φψ + e = = cos sin . 3 2 2 2 We see that θ φ+ ψ trace of Q =α + δ = 2cos cos . . . . (6) 2 2
Classical Mechanics Page No. 273 Similarly, from equation (3) we have χ trace of Q = 2cos . . . (7) χ 2 Since trace is invariant, therefore, from (6) and (7) we have χ φψ+ θ cos= cos cos . 2 2 2
Example 17 : Find a real matrix of orthogonal transformation in the 3-dimensional space corresponding to the unitary matrix θ θ cos ,i sin 2 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 ,i sin 2 2 Q = . . . (1) θ θ i sin , cos 2 2 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 x− iy P = . . . (2) xiy+ − z
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 † . . . (3)
Classical Mechanics Page No. 274 * where Q† is a complex transposed conjugate of Q. i.e., Q† = ( Q T ) 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 z′ x ′− iy ′ P′ = . . . (6) x′+ iy ′ − z ′ where x′, y ′ , z ′ are to be determined. Let us define x− iy = x , − . . . (7) x+ iy = x + . Therefore equation (5) with the help of equations (1), (2), (6) and (7) becomes. θθ θθ cosi sin cos − i sin zx′ ′ 22 zx 22 − = − . xz′− ′ θθ xz − θθ + isin cos + − i sin cos 22 22
Classical Mechanics Page No. 275 Therefore we have
z′ x − ′ = x+′− z ′ 2θθ 2 θθθθ 2 θ 2 θθθ zcos−+− sin ix cos sin ix cos sin , x sin +− x cos 2 iz cos sin 22 + 2222 − +− 2 222 = . 2θ 2 θ θ θ2 θθ 2 θθ θθ xcos+ x sin + 2 iz cos sin ,z sin − cos + ix cos sin − ix cos sin +2 − 2 2 2 22 − 22 + 22
On equating the corresponding elements of the matrix we get θθ θ θ xxiyiz′=+= ′ ′ 2cos sin ++() xiy cos2 +−() xiy sin 2 , + 22 2 2
θθ θ θ xxiy′=−=− ′ ′ 2cos iz sin +−() xiy cos2 ++() xiy sin 2 , − 22 2 2
2θθ 2 θθ θθ zz′ =cos − sin −− ixiy() cos sin ++ ixiy() cos sin . 22 22 22 ⇒ x′+=+ iy ′ xiycosθ + iz sin θ , x′−=− iy ′ xiycosθ − iz sin θ , z′ = zcosθ − y sin θ . ⇒ x′ = x y′ = ycosθ + z sin θ , z′ = − ysinθ + z cos θ .
x′ 1 0 0 x iey..,′ = 0 cosθ sin θ y z′ 0− sinθ cos θ z This shows that corresponding to the matrix θ θ cosi sin 2 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 1 0 0 A = 0 cosθ sin θ 0− sinθ cos θ in 3- dimensional real space.
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Exercise: 1. Show that the components of angular velocity vector along the space set of axes are given by ωθx =cos φψ + sin θ sin φ , ωy = θsin φψ − sin θ cos φ , ωz = ψcos θφ + . 2. Find a real matrix of orthogonal transformation in the 3-dimensional space corresponding to the unitary matrix iφ e 2 0 Q = iφ − 0 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 iψ e 2 0 Q = iψ − 0 e 2 in 2- dimensional complex plane.
Classical Mechanics Page No. 277