2. Sect. 4.5: Cayley-Klein Parameters 3 independent quantities are needed to specify a rigid body orientation. Most often, we choose them to be the Euler Angles: ,θ,ψ. Sometimes, it’s convenient to use variable sets which contain more than the minimum number of 3, even though these can’t be used as indep generalized coords. One set of 4 parameters, due to Klein (& Cayley) & originally Euler  Cayley-Klein Parameters is often convenient. –These are very useful in theoretical physics. –Also much easier to deal with than Euler Angles when obtaining numerical solutions to rigid body problems!

3. Cayley-Klein Parameters: 4 complex variables denoted as: α,β,γ,δ. Constraints: β = - γ*, δ = α* DEFINE general transformation matrix A in terms of these as: Of course, this is the SAME A that we wrote in terms of Euler Angles!  There MUST be a connection between α,β,γ,δ & angles ,θ,ψ. A in this form looks complex. But A is real! PHYSICAL INTERPRETATION of α,β,γ,δ ? *  complex conjugate

4. To see that A is real, define the real quantities e 0, e 1, e 2, e 3 (  Euler parameters) as: α  e 0 +i e 3 β  e 2 +i e 1 Also, since β = - γ* & δ = α* γ  -e 2 +i e 1 δ  e 0 - i e 3 In terms of these real parameters, A looks like: Clearly, A in this form is real. Orthogonality of a ij  (e 0 ) 2 + (e 1 ) 2 + (e 2 ) 2 + (e 3 ) 2 = 1 PHYSICAL INTERPRETATION of e 0, e 1, e 2, e 3 ?

5. Sect. 4.6: Euler’s Theorem Now have the complete math formalism to describe the motion of any rigid body. –At any time t, the body orientation with respect to some external set of axes can be specified by an orthogonal transformation A. –Can express elements of A (a ij ) with a convenient choice of parameters: Euler angles: ,θ,ψ. Cayley-Klein Parameters: α,β,γ,δ. Euler Parameters: e 0,e 1,e 2,e 3. Each is time dependent.  As time progresses, the orientation of the body changes: A = A(t) Time dependence is obtained by solving Lagrange’s Eqtns!

6. Assume initial conditions so that body axes are the same as the external axes at t = 0. Initial condition: A(0) = 1 Later as orientation changes, A = A(t)  1 Physics: Motion must be continuous  The transformation matrix A must evolve continuously from the identity transformation 1.  Euler’s Theorem: The general displacement of a rigid body with one point fixed is a rotation about some axis. Proof: As we proceed.

7. Euler’s Theorem: The general displacement of a rigid body with one point fixed is a rotation about some axis. Physical meaning: For every such rotation it is always possible to find an axis through the fixed point, oriented at particular polar angles θ,  such that a rotation about this axis by a particular angle ψ duplicates the rotation.  3 parameters θ, ,ψ characterize the rotation & these ARE the Euler angles The fixed point is often (but not necessarily!) the CM of the body. Take the fixed point as origin of body axes  “Displacement” of body involves no translation of body axes, but only a change in ORIENTATION

8. Euler’s Theorem restated: The body axes at time t can be obtained by a SINGLE rotation (about an appropriate axis, to be determined!) of the initial axes. In other words: The OPERATION implied by the general orthogonal transformation A describing the motion of a rigid body IS A ROTATION! Given the Euler angles θ, ,ψ, the rotation axis is to be determined! 1 st Characteristic of a rotation: The magnitudes of all vectors are unchanged on rotation! –This results automatically from the orthogonality conditions on the a ij !

9. 2 nd Characteristic of a rotation: The direction of the rotation axis is unchanged on rotation.  Any vector lying on this axis has the same components in both the initial & the final axis systems!  If we can show that there exists a vector R having the same components in both systems, will have proven Euler’s Theorem. This proof follows: In general, for vector R, under a rotation characterized by A: R = AR If R = rotation axis, then R = R For generality, write R = λR. To prove Euler’s theorem, look for solutions where λ = +1 Combining gives: (A - λ1)R = 0

10. To prove Euler’s theorem, we need to solve: (A - λ1)R = 0 (1)  The Eigenvalue Problem Values of λ which satisfy (1)  Eigenvalues In general, λ might be real or complex Vectors R which satisfy (1)  Eigenvectors Eigenvalue  German for characteristic value. Euler’s theorem restated: The real, orthogonal matrix specifying the physical motion of a rigid body with one fixed point always has the eigenvalue λ = + 1

11. Solve the eigenvalue problem: (A - λ1)R = 0 (1) Note that: X a 11 a 12 a 13 1 0 0 R  Y A  a 21 a 22 a 23 1  0 1 0 Z a 31 a 32 a 33 0 0 1  (1) becomes 3 simultaneous, homogeneous, linear algebraic eqtns for the components X, Y, Z (a 11 - λ)X + a 12 Y + a 13 Z = 0 a 21 X + (a 22 - λ)Y + a 23 Z = 0 (1) a 31 X + a 32 Y + (a 33 - λ) Z = 0 Solutions to (1): (in general 3) eigenvalues λ. For each λ, ratios of components of corresponding eigenvector R. Physics: Only the direction of R, not the magnitude, can be determined.

12. Eigenvalue problem: (A - λ1)R = 0 (1) Or: (a 11 - λ)X + a 12 Y + a 13 Z = 0 a 21 X + (a 22 - λ)Y + a 23 Z = 0 (1) a 31 X + a 32 Y + (a 33 - λ) Z = 0 Has a solution only when the determinant of the coefficients of X,Y, & Z vanishes.  Solution requires: |A - λ1| = 0 or: (a 11 - λ) a 12 a 13 a 21 (a 22 - λ) a 23 = 0 (2) a 31 a 32 (a 33 - λ) (2)  Characteristic or secular eqtn of matrix A. Euler’s theorem restated again: For real, orthogonal matrices A, the secular eqtn must have the root λ = +1

13. Eigenvalue problem, (slightly) alternate formulation: (A - λ1)R = 0 (1)  Solution requires: |A - λ1| =0 (2) Notation: 3 eigenvalues  λ k (k =1,2,3) 3 eigenvectors R  X k (k =1,2,3) Each eigenvector X k has 3 components labeled as X ik (Change of notation from X,Y,Z!) 1 st subscript (i) labels the component 2 nd subscript (k) labels the eigenvector to which the component belongs.  Eqtn resulting from (1) for k th eigenvalue (summation convention not used!): ∑ j a ij X jk = λ k X ik (3)

14. Eigenvalue problem: ∑ j a ij X jk = λ k X ik (3) Rewrite using δ j,k notation:  ∑ j a ij X jk = ∑ j X ij δ j,k λ k (3) (3): Both sides have the form of matrix products: Define the (diagonal) eigenvalue matrix: λ 1 0 0 λ  0 λ 2 0  (3) becomes: AX = Xλ (4) 0 0 λ 3 Multiply (4) from left by X -1 : X -1 AX = λ (5) (5): A similarity transformation operating on A.  Can diagonalize A by performing a suitable similarity transformation. If an appropriate X can be found, the elements of diagonal A  are the eigenvalues sought & the X’s which do this are the eigenvectors.

15. A proof of Euler’s Theorem in form: “For real, orthogonal matrices A, the secular eqtn must have the root λ = +1” Diagonalize A & find eigenvalue λ = +1. Another proof: Use property of transpose Ã. Recall for orthogonal matrices, the reciprocal is equal to the transpose: A -1 = Ã Consider the expression (A - 1)Ã = 1 - Ã. Take the determinant of both sides: |A - 1||Ã| = |1 - Ã| To describe rigid body motion, A(t) must correspond to a proper rotation  |A| = |Ã| =1  |A - 1| = |1 - Ã|. Determinant of a matrix = determinant of its transpose:  |A - 1| = |1 - A|  Determinant of matrix A - 1 = determinant of matrix 1 - A = -(A - 1)

16. |A - 1| = |1 - A| = -|A -1|  |A -1| = 0. Compare to secular eqtn |A - λ1| = 0  A must always have at least one eigenvalue λ = + 1.  Euler’s Theorem is proven! What about the other 2 eigenvalues? Determinant of any matrix is unaffected by similarity transformation. We had: AX = Xλ (4) and X -1 AX = λ (5) Take determinant of (5), noting that |A| = 1 (previous result) & that determinant is invariant under similarity transformation:  |A| = 1 = |λ| = λ 1 λ 2 λ 3

17. Determinant of A = product of its 3 eigenvalues: |A| = 1 = λ 1 λ 2 λ 3 Euler’s theorem: At least one eigenvalue is 1 (say λ 3 = 1)  λ 1 λ 2 = 1 A is real  If λ is an eigenvalue, then it’s complex conjugate λ* is also an eigenvalue.  If the eigenvalue λ 1 is complex, must have λ 2 = λ 1 *. If the eigenvalue λ 1 is complex, the corresponding eigenvector R 1 is also complex. –For complex vectors R, the square of the magnitude is given by RR*

18. The square of the magnitude is invariant under a real orthogonal transformation A  RR* = (AR)AR* = RAAR* = RR* (1) If R is a complex eigenvector corresponding to a complex eigenvalue λ, the first part of (1) becomes: RR* = λλ* RR* (2) (1) & (2) together  λλ* =1 Conclusion: All eigenvalues of a general orthogonal transformation A have unit magnitude.

19. Summary: The 3 eigenvalues λ 1, λ 2, λ 3 of an orthogonal transformation matrix A must satisfy: a. λ 1 λ 2 λ 3 = 1 b. Euler’s Theorem  One of them (say λ 3 ) = 1 c. λ 1 λ 2 = 1 d. λ i λ i * =1 (i =1,2,3)  The λ’s have 3 possible distributions: 1. All are = +1.  A = 1 (trivial) 2. One, say λ 3 = 1 & the other 2, λ 1 = λ 2 = -1  A = rotation by π about some axis 3. One, say λ 3 = 1 & the other 2, λ 1 & λ 2 are complex conjugates of each other.

20. Consider case 3. where λ 3 = 1 & λ 2 = λ 1 *: Still must have λ 1 λ 2 = 1  λ 1 & λ 2 must be of the form λ 1 = e iΦ λ 2 = e -iΦ Direction cosines of axis of rotation (eigenvector R for eigenvalue λ = 1) are obtained by going back to eigenvalue eqtns: (a 11 - λ)X + a 12 Y + a 13 Z = 0 a 21 X + (a 22 - λ)Y + a 23 Z = 0 (1) a 31 X + a 32 Y + (a 33 - λ) Z = 0 setting λ = 1 & solving for X, Y, Z.

21. Can also get angle of rotation for eigenvalue λ = 1. Do this by a similarity transformation BAB -1  A to transform A into a system of coords where z axis is axis of rotation. A  a rotation about the z axis through angle Φ  Can easily write: cosΦ sinΦ 0 A = -sinΦ cosΦ 0 0 0 1 Trace of A = TrA = a ii = 1 + 2 cosΦ The trace of a matrix is invariant under a similarity transformation  TrA = a ii = 1 + 2 cosΦ

22. TrA = a ii = 1 + 2 cosΦ Again using the same property, this  TrA = Trλ = 1 + 2 cosΦ (1) Now, the rotation angle Φ can clearly be seen to be identical to the phase angle of the complex eigenvalues: λ 1 = e iΦ, λ 2 = e -iΦ, λ 3 = 1  TrA = Trλ = 1 + e iΦ + e -iΦ (2) (1) & (2) are the same, since e iΦ + e -iΦ  2 cosΦ Clearly, cases when eigenvalues are all real, are special cases of complex eigenvalues (special choices for rotation angle Φ). Φ = 0  All λ’s =1  A = 1 (as already noted) Φ = π  λ 1 = λ 2 = -1 (as already noted)

23. Note: The prescription for getting the rotation axis direction R & for rotation angle Φ are not unique & unambiguous. –If R is eigenvector, so is -R  Sense of direction of rotation axis is not exactly specified. Could be direction of R or -R –Also, if replace Φ by -Φ all of the formalism is unchanged. (e.g. -Φ satisfies TrA = 1 + 2 cosΦ)  The sense of direction of the rotation angle is not exactly specified. Could be Φ or - Φ. –Can go even further & say that the eigenvalue eqtn does not uniquely specify the orthogonal transformation matrix A. e.g. can show that the inverse or transpose A -1 = Ã has the same eigenvalues & eigenvectors as A.

24. Euler’s Theorem: The general displacement of a rigid body with one point fixed is a rotation about some axis. Corollary  Chasles’ Theorem: The most general displacement of a rigid body is a rotation about some axis plus a translation. –Removing the constraint of the one fixed point introduces 3 more (translational) degrees of freedom (3 more generalized coords, giving a total of 6, as discussed at beginning of the chapter). –A stronger form (footnote, p 161): It is always possible to choose the origin of the body set of coordinates so that the translational motion is in the same direction as the rotation axis  Screw motion. Useful in crystallography!