Lecture 12 - Harvard University Department of...

MechanicsPhysics 151

Lecture 12Oscillations(Chapter 6)

What We Did Last Time

Analyzed the motion of a heavy topReduced into 1-dimensional problem of θQualitative behavior Precession + nutationInitial condition vs. behavior

Magnetic dipole moment of spinning charged objectM = γL, where γ = q/2m is the gyromagnetic ratioL precesses in magnetic field by ω = –γBγ of elementary particles contains interesting physics

Oscillations are everywhereSmall motion near equilibrium (perturbation)Appears in many QM problems

You’ve learned it in 15c or 16Rather easy

We’ll do more formal treatment of multi-dimensional oscillation

Did this (less formally) in 15c for coupled oscillatorsGeorgi’s book does this You may already know

Oscillations

2x xω= − exp( )x A i tω= −

Problem Definition

Consider a system with n degrees of freedomGeneralized coordinates {q1, …, qn}

Generalized force at the equilibrium

V must be minimum at a stable equilibrium

Taylor expansion of V using

0

0ii

VQq

⎛ ⎞∂= − =⎜ ⎟∂⎝ ⎠

Potential V is at an extremum

2

00 0

12

12i i j

i i jij i j

V VVq q q

VV ηηη ηη⎛ ⎞⎛ ⎞∂ ∂

= + + + ≈⎜ ⎟⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝ ⎠

0i i iq q η= +

constant zero

Constant symmetric

matrix

Problem Definition

Kinetic energy is a 2nd-order homogeneous function of velocities

This requires that the transformation functions do not explicitly depend on time, i.e.

mij generally depend on {qi} Taylor expansion

11 1( , , )2 2ij n i j ij i jT m q q q q m ηη= =…

( )0

0

ijij ij k ij

k

mm m T

qη

∂⎛ ⎞= + + ≈⎜ ⎟∂⎝ ⎠

12 ij i jTT ηη≈

1( , , , )i i Nq q x x t= …

Constant symmetric

matrix

Lagrangian

For small deviation {ηi} from equilibrium

The equations of motion are

This looks similar toDifficulty: Tij and Vij have off-diagonal componentsIf T and V were diagonal

1 1 1 12 2 2 2ij i j ij i jL T V T Vηη ηη= − = − = −ηTη ηVη

0ij j ij jT Vη η+ = (Textbook Eq 6.8 wrong)

0mx kx+ =

0ij j ij j ii i ii iT V T Vη η η η+ → + = no sum over i

Can we find a new set of coordinates that diagonalizes T and V?

Solving Lagrange’s Equations

Assume that solution will be

It’s a slightly odd eigenvalue equationSolution can be found byn-th order polynomial of λ Expect n solutions for λλ must be real and λ = ω2 > 0

0ij j ij jT Vη η+ =

i ti iCa e ωη −=

2 0ij j ij jT a V aω− + = 0λ− =Va Taor

0λ− =V T

Can be proven by a bit of work

2λ ω=

Reality of Eigenvalues

Start fromTake adjoint (complex conjugate + transpose)

Multiplying by a† or a gives

Writing a as α + iβ (α and β are real)

Since is positive for any real

λ=Va Ta

† † * †λ λ= =a Va a Ta a Ta

† * †λ=a V a T

* †( ) 0λ λ− =a Ta

† ( ) ( ) ( )i i i= − + = + + −a Ta α β T α β αTα βTβ αTβ βTα

zero12T = ηTη η

† 0= + >a Ta αTα βTβ λ – λ* = 0, i.e. λ is real

Now that we know λ is real

α and β both satisfy the same eigenvalue equationSuppose that the eigenvalues are not degenerate

Will worry about the degenerate case later…Each eigenvalue corresponds to 1 eigenvector

α and β are proportional to each other

a can be made real by absorbing γ in C of

Reality of Eigenvectors

λ=Va Ta i iλ λ+ = +Vα Vβ Tα Tβ

i ti iaC e ωη −=

0λ− =V T 1 2, , , nλ λ λ λ= … All different

i γ= + =a α β α a complex number

We now have an all-real equation

We now have a guarantee that each solution of the eigenvalue equation gives an oscillating solution

with a definite frequency

Positive Definiteness

λ=aVa aTa

λ =aVaaTa

Already shown to be positive

Positive because for any real ηif V is minimum at the equilibrium

12 0V = ≥ηVη

λ = ω2 is positive definite

i tC e ω−=η a 2λ ω=

Normalization

Eigenvector satisfying Va = λTa has arbitrary scalea Ca can absorb such scale as well as imaginary phaseWe fix the normalization by declaring

Just the sign (±) remains ambiguous

This turns

1=aTa

λ =aVaaTa

λ = aVa

Principal Axis Transformation

There are n eigenvectors Call them aj

Take transpose

If we stack aj to make

j j jλ=Va Ta 1, 2,...,j n=

k k kλ=a V a T ( ) 0j k k jλ λ− =a Ta

k j jkδ=a Ta Assuming λj ≠ λk for j ≠ k

=ATA 1

1 0

0 n

λ

λ

⎡ ⎤⎢ ⎥= = ⎢ ⎥⎢ ⎥⎣ ⎦

AVA λ

T and V are diagonalized by the principal axis

transformation

no sum over j or k

[ ]1 2 n=A a a a

k j j jkλ δ=a Va

Normal Coordinates

Lagrangian wasOnce we have A, we can switch to new coordinates

A-1 does exist becauseLagrangian becomes

Solutions are obvious

1 1 1 12 2 2 2ij i j ij i jL T Vηη ηη= − = −ηTη ηVη

1−≡ζ A η

1 1 1 12 2

1 12 2 22 k k k k kL ζ ζ λ ζ ζ= − = = −−ζζATAζ ζAVA ζ ζλζζ

=ATA 1 0≠A

k k kζ λ ζ= − ki tk kC e ωζ −= 2

k kω λ=

Normal coordinates

No cross terms

Normal coordinates are independent simple harmonic oscillators

Initial Conditions

The coefficients Ck is fixed by the initial conditionsSuppose at t = 0

Using

We need an example now…

ki tk kC e ωζ −=

(0)=η η (0)=η η

(0) (0)=η Aζ

(0) Re( ) Imj jk k k jk k ka i C a Cη ω ω= − =

Remember to take the real part!

=ATA 1Re (0)k lk lj jC a T η=

1Im (0)k lk lj jk

C a T ηω

= no sum over k

(0) Rej jk ka Cη =

(0) (0)=η Aζ

Linear Triatomic Molecule

Consider a molecule like CO2Consider only motion along the axis

We want to solve eigenvalue equation

m mm

2 2 21 2 3( )

2mT η η η= + + 2 2

2 1 3 2( ) ( )2 2k kV η η η η= − + −

0 00 00 0

mm

m

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

T0

20

k kk k k

k k

−⎡ ⎤⎢ ⎥= − −⎢ ⎥⎢ ⎥−⎣ ⎦

V

2( ) 0ω− =V T a

Linear Triatomic Molecule

Solutions are

2

2 2

2

02 0

0

k m kk k m k

k k m

ωω ω

ω

− −− = − − − =

− −V T

2 2 2( )(3 ) 0k m k mω ω ω− − =

1 0ω = 2km

ω = 33km

ω =

1

11 13 1m

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

a 2

11 02 1m

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥−⎣ ⎦

a 3

11 26 1m

⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦

a

Is this OK?

Linear Triatomic Molecule

First solution is linear movement of the molecule

This is not an “oscillation”Consider it as an oscillation with infinitely long periodAlthough V is minimum at the equilibrium, it does not increase when the whole molecule is shifted

Position of the CoM is a cyclic coordinateTotal momentum is conserved

1 0ω = 1

11 13 1m

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥⎣ ⎦

am mm

Linear Triatomic Molecule

Two “normal” oscillation modes exist

CoM does not move Orthogonal to the first solution

2km

ω =

33km

ω =

2

11 02 1m

⎡ ⎤⎢ ⎥= ⎢ ⎥⎢ ⎥−⎣ ⎦

a

3

11 26 1m

⎡ ⎤⎢ ⎥= −⎢ ⎥⎢ ⎥⎣ ⎦

a

m mm

m mm

Linear Triatomic Molecule

Putting together a1, a2 and a3

Normal coordinates are

ζ1 is cyclic as we expect

2 3 11 2 0 26

2 3 1m

⎡ ⎤⎢ ⎥

= −⎢ ⎥⎢ ⎥

−⎢ ⎥⎣ ⎦

A 1

2 2 2

3 0 36

1 2 1

m−

⎡ ⎤⎢ ⎥

= −⎢ ⎥⎢ ⎥−⎢ ⎥⎣ ⎦

A

1 1 2 33 ( )mζ η η η= + + 2 1 32 ( )mζ η η= − 3 1 2 36 ( 2 )mζ η η η= − +

2 2 2 2 21 2 3 2 3

1 ( ) ( 3 )2 2

kLm

ζ ζ ζ ζ ζ= + + − +

Degenerate Solutions

We assumed λj ≠ λk for j ≠ kWhat if the eigenvalue equation has multiple roots?Can we still achieve ?

Quick answer: Don’t worryMultiple root corresponds to multiple eigenvectors

Any linear combination cjaj is also an eigenvectorIt is always possible to find a set of m orthogonal vectors

=ATA 1

( ) ( ) 0m fλ λ κ λ− = − =V T λ = κ is an m-fold root

( ) 0 ( 1, , )j j mκ− = =V T a … m eigenvectors

Recipe

For an m-fold root and m eigenvectorsFirst normalize a1 withNext turn a2 into

This satisfiesNormalize a2’ withNext turn a3 into

This satisfies andContinue…

It is now guaranteed that we can make

1 1 1=a Ta

2 2 1 2 1( )′ = −a a a Ta a

1 2 0′ =a Ta

2 2 1′ ′ =a Ta

3 3 1 3 1 2 3 2( ) ( )′ ′ ′= − −a a a Ta a a Ta a

1 3 0′ =a Ta 2 3 0′ ′ =a Ta

=ATA 1

Summary

Studied oscillationDiscussed general features of multi-dimensional oscillators

Equation of motion Eigenvalue problemShowed that oscillating solutions existEigenvalues ω2 are positive definite

Provided that V is minimum at the equilibrium

Principal axis transformation diagonalizes T and VNormal coordinates behave as independent oscillators

Next: forced oscillation

λ=Va Tai t

i iCa e ωη −=