Mechanics
Physics 151
Lecture 12
Oscillations
(Chapter 6)
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Mechanics Oscillations, Lecture Notes - Physics, Study notes of Mechanics
Mechanics, Physics, Oscillations, Lagrangian, Solving Lagrange’s Equations, Reality of Eigenvectors, Positive Definiteness, Normalization, Principal Axis Transformation, Normal Coordinates, Initial Conditions, Linear Triatomic Molecule, Degenerate Solutions.
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Mechanics
Physics 151
Lecture 12Oscillations(Chapter 6)
What We Did Last Time ^
Analyzed the motion of a heavy top^
Reduced into 1-dimensional problem of
^
Qualitative behavior
Æ
Precession + nutation
^
Initial condition vs. behavior
^
Magnetic dipole moment of spinning charged object^
M
γ L
, where
γ^ =
q
m
is the gyromagnetic ratio
^
L^
precesses in magnetic field by
ω
γ B
γ^ of elementary particles contains interesting physics
Problem Definition ^
Consider a system with
n
degrees of freedom
^
Generalized coordinates {
q^1
qn
^
Generalized force at the equilibrium^
V
must be minimum at a stable equilibrium
^
Taylor expansion of
V
using
0
i
V i
Q
q ⎛^
⎜^
Potential
V is at
an extremum^2
0
0
0
i^
i^
j
i^
i^
j
ij^
i^
j
V^
V
V^
q^
q^
q^
V
V
η^
⎛^
⎛^
∂^
=^
+^
+^
+^
⎜^
⎜^
⎟^
⎜^
∂^
∂^
⎝^
⎠^
⎝^
0 i^
i^
i
q^
q
=^
constant
zero
Constantsymmetricmatrix
Problem Definition ^
Kinetic energy is a 2
nd
-order homogeneous function of
velocities^
This requires that the transformation functions do notexplicitly depend on time, i.e. ^
m
ij^
generally depend on {
q i
}^
Æ
Taylor expansion
1
1
(^
,^
,^
ij^
n^
i^
j^
ij^
i^
j
T^
m
q^
q^
q q
m
η^
=^
(^
)^0
0 ij
ij^
ij^
k^
ij
m k
m
m
T
q
=^
+^
+^
⎜^
⎝^
ij^
i^
j
T
T
η^
≈^
1 (^
,^
,^
,^
i^
i^
N
q^
q^
x^
x^
t
=^
Constantsymmetricmatrix
Solving Lagrange’s Equations ^
Assume that solution will be
^
It’s a slightly odd eigenvalue equation^
Solution can be found by n
-th order polynomial of
λ^
Æ
Expect
n
solutions for
λ^
must be real and
λ^
ij^
j^
ij^
j
T^
V
η^
+^
i^ t
i^
i Ca e
ω
η^
−
=
2
ij^
j^
ij^
j
T a
V a
ω−^
+^
=^
−^
Va
Ta
or 0
λ−
V
T
Can be proven by
a bit of work
2
λ^
Reality of Eigenvalues ^
Start from^
Take adjoint (complex conjugate + transpose) Multiplying by
a
†^ or
a
gives
^
Writing
a
as
α
i β
(α
and
β
are real)
^
Since
is positive for any real
Va
Ta
†^
†^
*^
†
λ^
=^
a Va
a Ta
a Ta
†^
*^
†
a V
a T *^
†
(^
)^
λ^
−^
a Ta
†^
(^
)^
(^
)^
(^
i^
i^
i
=^
−^
+^
=^
+^
+^
a Ta
α^
β^
T^
α^
β^
αT
α^
βT
β^
αT
β^
βT
α
^
^
^
^
zero
1 2 T^
=^
ηT
η ^
^
η
†^
=^
+^
a Ta
αT
α^
βT
β
λ^
= 0, i.e.
λ^
is real
^
We now have an all-real equation
^
We now have a guarantee that each solution of theeigenvalue equation gives an oscillating solution
with a definite frequency
Positive Definiteness
aVa
aTa
^
aVa aTa
Positive because^ Already shown to be positive
for any real
η
if V is minimum at the equilibrium
1 2
V^
=^
ηV
η
λ^
ω
2 is positive definite
i^ t C e
ω−
η^
a^
2
λ^
Normalization ^
Eigenvector satisfying
Va
λ Ta
has arbitrary scale
^
a^
Æ
C
a^
can absorb such scale as well as imaginary phase
^
We fix the normalization by declaring^
Just the sign (±) remains ambiguous ^
This turns
aTa
λ^
=^
aVa aTa
λ^
=^
aVa
Normal Coordinates ^
Lagrangian was^
Once we have
A
, we can switch to new coordinates
^
A
-^
does exist because
^
Lagrangian becomes ^
Solutions are obvious
ij^
i^
j^
ij^
i^
j
L^
T^
V
η η
η η
=^
−^
=^
ηT
η^
ηV
η
^
^
(^1) − ≡ ζ
A
η 1
k^
k^
k^
k^
k
L
ζ ζ
λ ζ ζ
=^
−^
=^
=^
ζ
ζATA
ζ^
ζAVA
ζ^
ζλζ
ζ
^
^
^
^
ATA
^
A
k^
k^
k
ζ^
λ ζ = − ^
i^ tk
k^
C ek
ω
ζ^
−
=^
(^2) k^
k
ω^
λ
Normal coordinates
No cross terms
Normal coordinates are independent simple harmonic oscillators
Initial Conditions ^
The coefficients
C
is fixed by the initial conditions k
^
Suppose at
t^
^
Using
^
We need an example now…
i^ t^ k
k^
k C e
ω
ζ^
−
=
η^
η^
η^
η ^
η^
A
ζ
Re(
)^
Im
j^
jk^
k^
k^
jk^
k^
k
a^
i^
C^
a^
C
=^
−^
Rememberto take thereal part!
ATA
Re
k^
lk^
lj^
j
C^
a T
Im
k^
lk^
lj^
j
k
C^
a T
=^
^
no sum over
k
Re
j^
jk^
k
a^
C
η^
η^
A
ζ
Linear Triatomic Molecule
^
Solutions are
2
2
2
2 0
k^
m
k
k^
k^
m
k
k^
k^
m
−^
−^
=^
−^
−^
−^
−^
V
T^2
2
2
(^
)^
k^
m
k^
m
ω^
−^
−^
1
ω^
=^
2
k m
ω^
=^
3
3 k m
ω^
1
m
⎡^
=^
a^
2
m
⎡^
⎢^
=^
⎢^
⎢^
a^
3
m
⎡^
⎢^
=^
⎢^
⎣^
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 period Although
V
is minimum at the equilibrium, it does not
increase when the whole molecule is shifted^
Position of the CoM is a cyclic coordinate Total momentum is conserved 1
ω^
=^
1
m
⎡^
=^
a^
m
m
m
Linear Triatomic Molecule ^
Putting together
a
,^1
a^2
and
a
3
^
Normal coordinates are
ζ^1
is cyclic as we expect
m
⎡^
⎢^
=^
⎢^
⎢^
⎢^
⎣^
A
1
m
−
⎡^
⎢^
=^
⎢^
⎢^
⎢^
⎣^
A
1
1
2
3
m
ζ^
η^
η^
=^
+^
+^
2
1
3
( 2
m
η^
=^
−^
3
1
2
3
m
η^
η^
=^
−^
2
2
2
2
2
1
2
3
2
3
(^
)^
(^
k 2
L^
m
ζ^
ζ^
ζ
ζ^
ζ
=^
+^
+^
−^
^
^
Degenerate Solutions ^
We assumed
λ j
λ k
for
j^
≠^
k
^
What if the eigenvalue equation has multiple roots? ^
Can we still achieve
^
Quick answer: Don’t worry^
Multiple root corresponds to multiple eigenvectors Any linear combination
c^ j
a j^
is also an eigenvector
^
It is always possible to find a set of
m
orthogonal vectors
ATA
(^
)^
(^
)^
m^ f
λ^
λ^
κ^
λ
−^
=^
−^
V
T
λ^ =
κ^ is an
m
-fold root
(^
)^
(^
1,^
,^
j^
j^
m
=^
V
T a
m^
eigenvectors