Mechanics Physics 151

Transcription

1 Mechanics Physics 5 Lecture Oscillations (Chapter 6)

2 What We Did Last Tie Analyzed the otion of a heavy top Reduced into -diensional proble of θ Qualitative behavior Precession + nutation Initial condition vs. behavior Magnetic dipole oent of spinning charged object M = γl, where γ = q/ is the gyroagnetic ratio L precesses in agnetic field by ω = γb γ of eleentary particles contains interesting physics

3 Oscillations Oscillations are everywhere Sall otion near equilibriu (perturbation) Appears in any QM probles You ve learned it in 5c or 6 Rather easy x= ω x x= Aexp( iωt) We ll do ore foral treatent of ulti-diensional oscillation Did this (less forally) in 5c for coupled oscillators Georgi s book does this You ay already know

4 Proble Definition Consider a syste with n degrees of freedo Generalized coordinates {q,, q n } Generalized force at the equilibriu Q i V = = qi V ust be iniu at a stable equilibriu Taylor expansion of V using constant V Potential V is at an extreu V V = V + η + ηη + q zero q = q + η i i i i i j i q i qj V ηη ij i j Constant syetric atrix

5 Proble Definition Kinetic energy is a nd -order hoogeneous function of velocities T = ij ( q,, qn) qq i j = ijηη i j This requires that the transforation functions do not explicitly depend on tie, i.e. q = q ( x,, x, t ) i i N ij generally depend on {q i } Taylor expansion = + η + T ( ) ij qk ij ij k ij T T ηη ij i j Constant syetric atrix

6 Lagrangian For sall deviation {η i } fro equilibriu L= T V = T ijηη i j Vijηη i j = ηtη ηvη The equations of otion are T η + Vη = ij j ij j This looks siilar to (Textbook Eq 6.8 wrong) x + kx = Difficulty: T ij and V ij have off-diagonal coponents If T and V were diagonal T η + Vη T η + Vη = no su over i ij j ij j ii i ii i Can we find a new set of coordinates that diagonalizes T and V?

7 Solving Lagrange s Equations Assue that solution will be T η + Vη = ij j ij j It s a slightly odd eigenvalue equation Solution can be found by n-th order polynoial of λ Expect n solutions for λ λ ust be real and λ = ω > Can be proven by a bit of work η = V λt = i i t Ca e ω i ω Ta ij j + Va ij j = or Va λta = λ = ω

8 Reality of Eigenvalues Start fro Take adjoint (coplex conjugate + transpose) Multiplying by a or a gives Writing a as α + iβ (α and β are real) Since Va = λta * av= λ at * ( λ λ ) ata= * ava= λata= λ ata = ( ) ( + ) = + + ( ) ata α iβ T α iβ αtα βtβ i αtβ βtα T = ata αtα βtβ ηtη is positive for any real η = + > zero λ λ * =, i.e. λ is real

9 Reality of Eigenvectors Now that we know λ is real Va = λta Vα + ivβ = λtα + iλtβ α and β both satisfy the sae eigenvalue equation Suppose that the eigenvalues are not degenerate V λt = λ = λ, λ,, λn All different Will worry about the degenerate case later Each eigenvalue corresponds to eigenvector α and β are proportional to each other a = α+ iβ= γ α a can be ade real by absorbing γ in C of a coplex nuber η = i i t C e ω ai

10 Positive Definiteness We now have an all-real equation λ = ava ata Already shown to be positive ava = λata Positive because V = ηvη if V is iniu at the equilibriu for any real η λ = ω is positive definite We now have a guarantee that each solution of the eigenvalue equation gives an oscillating solution i t C e ω η = a with a definite frequency λ = ω

11 Noralization Eigenvector satisfying Va = λta has arbitrary scale a Ca can absorb such scale as well as iaginary phase We fix the noralization by declaring Just the sign (±) reains abiguous This turns λ = ava ata ata = λ = ava

12 Principal Axis Transforation There are n eigenvectors Call the a j Va j = λ jta j j =,,..., Take transpose ata k j = δ jk If we stack a j to ake ATA = ava k j = λ jδ jk λ AVA = λ = λ n n av = λ at ( λ λ ) ata = k k k no su over j or k Assuing λ j λ k for j k [ ] A= a a a n j k k j T and V are diagonalized by the principal axis transforation

13 Noral Coordinates L= T ijηη i j Vijηη i j = ηtη ηvη Once we have A, we can switch to new coordinates ζ A η Noral coordinates A - does exist because ATA = A Lagrangian was Lagrangian becoes L = ζataζ ζava ζ = ζ ζ ζλζ = ζζ k k λkζζ k k Solutions are obvious i k ζ = λ ζ ζ = Ce ω k k k k k t ω k = λ k No cross ters Noral coordinates are independent siple haronic oscillators

14 Initial Conditions ζ k = i kt Ce ω k The coefficients C k is fixed by the initial conditions Suppose at t = η() = Aζ() η () = Aζ () Using ATA = Re C = a Tη () k lk lj j η= η () η = η () η j () = ajk ReCk η () = a Re( iω C ) = a ω IC j jk k k jk k k I Ck = alktη lj j() no su over k ω k We need an exaple now Reeber to take the real part!

15 Linear Triatoic Molecule Consider a olecule like CO Consider only otion along the axis T = ( η + η + η3) T = k k V = ( η η ) + ( η η ) 3 k k V = k k k k k We want to solve eigenvalue equation ( V ω T) a=

16 Linear Triatoic Molecule V k ω k ω T = k k ω k = k k ω ω ( k ω )(3 k ω ) = Solutions are Is this OK? k ω = ω = ω 3 = a = a = 3 3 a 3k = 6

17 Linear Triatoic Molecule First solution is linear oveent of the olecule ω = a = This is not an oscillation 3 Consider it as an oscillation with infinitely long period Although V is iniu at the equilibriu, it does not increase when the whole olecule is shifted Position of the CoM is a cyclic coordinate Total oentu is conserved

18 Linear Triatoic Molecule Two noral oscillation odes exist CoM does not ove Orthogonal to the first solution k ω = 3 3k ω = = a 3 6 = a

19 Linear Triatoic Molecule Putting together a, a and a 3 3 A = 6 3 Noral coordinates are A = ( ) ζ = 3 η+ η + η ( ) 3 ζ = η η3 ζ 3 = ( 6 η η + η ) 3 k L = ( ζ + ζ + ζ3 ) ( ζ + 3 ζ3 ) ζ is cyclic as we expect

20 Degenerate Solutions We assued λ j λ k for j k What if the eigenvalue equation has ultiple roots? Can we still achieve ATA =? Quick answer: Don t worry Multiple root corresponds to ultiple eigenvectors V λt = ( λ κ) f ( λ) = λ = κ is an -fold root ( V κ T) a j = ( j =,, ) eigenvectors Any linear cobination c j a j is also an eigenvector It is always possible to find a set of orthogonal vectors

21 Recipe For an -fold root and eigenvectors First noralize a with ata = Next turn a into a = a ( a Ta ) a This satisfies ata = Noralize a with ata = Next turn a 3 into a 3 = a3 ( a Ta3) a ( a Ta3) a This satisfies ata 3 = and ata 3 = Continue ATA = It is now guaranteed that we can ake

22 Suary Studied oscillation Discussed general features of ulti-diensional oscillators Equation of otion Eigenvalue proble Showed that oscillating solutions exist Eigenvalues ω are positive definite Provided that V is iniu at the equilibriu Principal axis transforation diagonalizes T and V Noral coordinates behave as independent oscillators Next: forced oscillation i t η = Ca e ω i i Va = λta