lecture 12 - harvard university department of...
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MechanicsPhysics 151
Lecture 12Oscillations(Chapter 6)
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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
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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= −
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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
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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
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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?
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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=
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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
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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
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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=
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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
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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
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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
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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
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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
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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?
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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
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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
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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= + + − +
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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
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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
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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 −=