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 −=
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