021 small oscillations - george mason universitycomplex.gmu.edu/ small oscillations.pdf · labeling...

PHYS 705: Classical MechanicsSmall Oscillations

1

Assumptions:

Formulation of the Problem

- A conservative system with depending on position only

- The transformation equation defining does not dep on time explicitly

jV q

Now, consider the situation when a system is near an equilibrium point

where the generalized forces acting on the system @ is ZERO, i.e.,

0

0 0,jj q

VQ q jq

jq

0 jq

(conservative force)

Our goal is to describe the motion of the system when it is slightly deviated

from this equilibrium point:

0j j jq q

0 jq

2

Now, we expand the Lagrangian around :


1. The potential energy term: jV q

For small displacement , we can expand V in a Taylor expansion:

2

1 01 0

0 0

1, , ,2n n j j k

j j k

V VV q q V q qq q q

(E’s sum rule applies here and coefficients evaluated at .)

1j

0 jq

Note: - can be set at zero

- since is an equilibrium

01 0, , nV q q

0

0j

Vq

0 jq

0 jq

3

Note: - is just a number (V evaluated at )

- Since V is assumed to be a smooth function and the order of the

partial derivatives can be switch, is symmetric:

So, near an equilibrium , can be well approximated by

a quadratic form:


jV q

12 jk j kV V where

0 jq

2

0

jkj k

VVq q

jkV

jkV jk kjV V

(E’s sum rule applies to all repeated indices)

0j jq q

4

Recall that we can in general write the KE as (Ch. 1):


2. The kinetic energy term:

2 1 0

2

12

12

i ii j k

i j k j k

ii j

j

ii

i j i

i

t

T T T T m q qq q

m q mq t

r r

r r r

2

1

0

- quadratic in

- linear in

- independent of

j

j

j

T q

T q

T q

With the assumption that transformation equation

does not dependent on time explicitly, i.e.,

we will only have the quadratic term remained…

1, , ,i i nq q tr r

0i

t

r

5

(recall & )

Formulation of the ProblemThis gives,

i ii

ik

jj

k

mq q

m

r r

Note: lq

12 jk j kT qm q where,

12 jk j kT T

0j j jq q

- in general can be a function of

- but we can also Taylor expand it around

- keeping only lowest order term (const term) in , we then have

jk lm q

0lq

01 010

, ,, , jkjk n l

ljk nm q q

mm q q

q

01 0, ,jk jk nT m q q where,

j

j jq

6

- the deviation from the equilibrium is our generalized coords

- and evaluated at are constant square matrices

- dynamic near ANY equilibrium quadratic forms in T and V

- coupling information among diff dofs are encoded in the cross terms

- GOAL for the following analysis is to

Formulation of the Problem: Quadratic Form of LWith V and T given near , we can now form the Lagrangian:

j

12 jk j k jk j kL T V T V

0 jq

j kT n n

To find a coordinate transformation so that in the new coords

j kV0 jq

0 jq

and diagonalize simultaneously

the problem decoupled !

j kTj kV

(E’s sum rule)

7

Formulation of the Problem: EOM near Eq.

Now we use the EL equation to get the EOM for :

12 ik i k ik i kL T V T V

j

(pick out only i=j)

1 12 2k k ji i

jj

L V V

(pick out only k=j)

j k kj

L V

(E’s sum rule)

are symmetric

jk kj

L T

Similarly, we have:

Then, EL eq gives EOM:

jk kj

d L Tdt

0jk k jk kT V (sum rule over k)There are j of these eqs and they are all coupled.

8

j kV

Formulation of the Problem: EOM near Eq.

The EOM for the are a set of coupled 2nd order ODEs with constant

coefficients and the solution has the following general form:

j

(sum over k)

i tj jt Ca e

- is the complex amplitude for the generalized coordinate

0jk k jk kT V

jjCa

- Obviously, only the real part of contributes to the actual motionj

(We have chosen the form of the pre-factor for convenience later.)jCawith being real.ja

9

Formulation of the Problem: Condition on Solution

Plugging the oscillatory solution into the 2nd order ODE, we have the following

matrix equation for the amplitudes :

For nontrivial solutions , we need the determinant of the

coefficients to vanish:

2 0jk jk kV T a

ja

0ja

(sum over k)

2det 0jk jkV T

2 211 11 1 1

2 21 1

0n n

n n nn nn

V T V T

V T V T

or

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- is the eigenvector which will determine the relative relations among

the generalized coords in the different eigenmodes (normal modes)

- are the eigenvalues or resonant frequencies of the problem

Formulation of the Problem: Eigenvalue Equation

One can think of this as the characteristic equation for the following

generalized eigenvalue problem:

a

2det 0 V T

Va Ta2

V and T are real

and symmetric

2 is real

a is real and

“orthogonal”

j

11

Labeling the n eigenmodes by the index r, we can write the matrix

equation for the rth eigenmode as:

Formulation of the Problem: Eigenvalue Equation

Since V and T are square matrices, there are in general n distinct

eigenvalues and eigenvectors. (We will consider degeneracy later.)

r r rVa Ta

n n

The general solution for will in general be a superposition (linear

combination) of all of these eigenmodes:

r ri t i tj jr r rt a C e C e

j

(sum over r)

1 , , Tr r nra aa where is the jth compoent of jra

12

where and are real parameters and they will be determined by initial

conditions: and

As we mentioned previously, the actual physical motion is given by the real

part of the complex solution,

Formulation of the Problem: Real Physical Solution

r

Re cosj j r jr r rt t f a t

rf 0j 0j

0 cos 2 Rej r jr r jr rf a a C

0 sin 2 Imj r jr r r jr r rf a a C

(Here, we have taken the choice that for .) *

r r rC C C j t

13

To show that are real, we will take the conjugate transpose of the

equation:

Formulation of the Problem: Props of Eigenmodes

r r rVa TaV and T are real

and symmetric† * †r r ra V a T

* †0 s r r s a Ta

†sr s sa Va Ta † * †

sr r r a V a aT--

(*)

r

Then,

14

Consider the case when r = s, we have


* †0 r r r r a Ta

For nontrivial solutions, we have and 0r a † 0r r a Ta

Then, the above condition gives * 0r r

2r r are real

15

Conversely, this means that we can always find a real eigenvector

for if is real !


Note: a real eigenvalue can be associated with a complex eigenvector.

Let call it, then straightforwardly gives

r r rVa Ta

r

Thus, are also two real eigenvector for the real eigenvalue

r r ri a α β

r r rVα Tα r r ri iVβ Tβand

and r rα β r

r

From now on, we will only consider real for real

r r rVa Ta

ra r

16


We look back at our Eq. (*) (eigen-system condition) with and

being real: 0s r r s a Ta

r

With distinct eigenvalues for , we then have r s r s

0r s a Ta r s

Lastly, to show the orthogonality of the eigenvectors,

ra

With , the matrix product in Eq. (**) is indeterminate. We will now show

that it is nonzero and in fact positive definite from a physical argument, i.e.,

r s

(**)

0r r a Ta

17

To do that, we will exploit the fact that the Kinetic energy is a positive

definite quantity, i.e.,


0T

jNow, we put in the expansion for in terms of its eigenmodes:

0q q

sum over r

12 jk j kT 1

2 jk j kT m q q

1 sin sin 02 jk r r jr r r s s ks s sT T f a t f a t

sum over s

sum over j and k

cosj r jr r rf a t

18

We have just shown that for , we have the orthogonality condition:


So, the previous Q-triple sum for the kinetic energy T collapses with only

terms surviving:

2 2 21 sin 02 jk jr kr r r r rT T a a f t

, ,sum over r j and k

r s

0r s a Ta or 0jk jr ksT a a

r s

Notice that the [] term are all squares and it will generically be positive

The matrix product or

when we are considering nontrivial solutions .

0jk jr krT a a 0r a

0r r a Ta

19

So, we have the following orthogonality condition for the matrix product:


Recall that the eigenvalue equation, ,determines the

direction of the eigenvector and we do have a freedom in choosing

its magnitude.

r s00r s

a Tar s

We now impose the normalization of the eigenvectors such that:

r r rVa Tara

r s10r s

a Tar s

or jk jr ks rsT a a

20

ra

Now, we will define a new set of coordinates (normal modes) so that the

oscillations decoupled !

Normal Modes

Recall we have the general expansion for our generalized coords:

We will define the normal coordinates as the term inside the ( ),

*r ri t i tj jr r rt a C e C e (sum over r)

1 , , Tr r nra aa where is the jth component of jra

*r ri t i tr r rt C e C e

r t

and the complex amplitude will be determined by IC.rC

21

- This is a linear transformation between two coordinate systems

and this particular linear transformation is specified by the set of

eigenvectors: or .

Then, we can write,

Normal Modes

- Each of the jth generalized coordinate is now written as a linear

combination of the normal coordinates .

- The rth normal coordinates is purely periodic depending on

the rth eigenfrequency only.

(sum over r)

jra

r t

- This linear transformation can be inverted to give in terms of

r

j jr rt a t

j

r

1 | | rA a a

r j

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Now, we are going to explicitly show that the system decoupled in the

normal coordinates !

Normal Modes

We have and (sum over r) j jr rt a j jr rt a

Now, calculate T :

2

1212121 12 2

jk j k

jk jr r k

jk jr ks

s s

r

rs

s

r s r

T

T T

a

a

T

a

a

r s rsa Ta

-the are

orthogonal !

'r sa

23

Normal Modes

To simplify the matrix product (blue term), we will use the eigenvalue

equation for a particular rth eigenmode,

Now, calculate V:

121212

jk j k

j

jk jr k

k jr r s

ss

ks

r

V V

V a a

V a a

jk jr r jk jrV a T a (sum over j)

24

Normal Modes

Substituting this back into our equation for the potential energy, we have,

Multiply a different eigenvector on both sides, we have,

2

121 12 2

r s

r s r

jk jr ks

rs rr

V aV a

jk jr r jk jrV a T a

sa

jk jr ks r jk jr ksV a a T a a

Using the orthogonality condition again, we have

jk jr ks r rsV a a ( also diagonalizes !)ra jkV

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Normal Modes

Forming the Lagrangian,

2 212 r r rL T V

Calculating the EOM using EL equation for each rth normal mode ,

(NO sum over r)

0r r

d L Ldt

2 0r r r

r

-EOM for each of the normal mode is decoupled !

-Each normal mode evolutes in time as a simple harmonic oscillator with a

single eigenfrequency .r

2r r

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the general solution will be a linear combination of these normal modes .

Normal Modes

Note:

In general, for a system with n generalized coordinates, we will have n

normal modes and using the linear coordinate transformation

r j jr rt a

*r ri t i tj jr r jr r rt a a C e C e

(sum over r)

(sum over r)

Recall that the complex coefficients are determined by IC.rC

For typical IC, all normal modes will present, i.e., 0rC r

so that ALL normal modes will participate in the motion.

r

27

- Let say, we have all except ,

- Then, the motion is very simple with all generalize coordinates proportional to

ONE non-zero normal mode

Normal Modes- However, one can imagine situations in which the IC will only lead to

the excitation of a certain normal mode.

1

1 1j jt a (for all j)

in proportions given by .

0rC 1 0C

1ja

- As an example in 2D: if , then (symmetric motion)a aa a

A 1 2 1a

if , then (anti-sym motion)a aa a

A 1 2 1a

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We start with our Lagrangian:

To ease in visualization later, let consider a 2 dofs case:

Rotation, Squeeze, and Simultaneous Diagonalization of V and T

The reduction of the small oscillation problem into a set of equations

for its normal modes has an elegant geometric interpretation…

(matrix notation)

12 jk j k jk j kL T V

12

L ηTη ηVη

2 2 2 211 1 12 1 2 22 2 11 1 12 1 2 22 2

1 12 22 2

L T T T V V V

both T and V are symmetric

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The reduction to the normal modes equation corresponds to a linear

transformation A which simultaneously diagonalizes the two real-

symmetric quadratic forms: T and V.

Now, we will try to visualize this geometrically in 2D.


2 2 2 211 1 12 1 2 22 2 11 1 12 1 2 22 2

1 12 22 2

L T T T V V V

2 2 2 21 2 1 1 2 2

1 12 2

L

j jr rt a η Aζ

(no crossed terms)

(in going to the

normal coords)

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For any positive definite quadratic forms:

Note:


2 2,F x y ax bxy cy

, 1F x y

x

y

- If and the ellipse is a circle

a c

- If , the ellipse will be rotated

0b

Matrix

Diagonalization

semi-major &

semi-minor axes

coordinate

axes

x

y

Geometrically:

One can visualize F as the contour curve of the function: in

2D as an ellipse.

31

0b

Finding the normal modes corresponds to the simultaneous

diagonalization of both T and V. Let see now, how this is done

geometrically in 3 steps:


Step 0: in original generalized coordinates: and forT forVη η

Both quadratic forms

are NOT the same and

the ellipses are

oriented differently1

2T

1

2V

32


' η R η

Step 1: Rotate T so that it is aligned with coord axes

1'

2'T

1'

2'V

' η R η

22 21 1 2

222 1 1 2

1 ' ' ' ' '' ' ' '2

12

a cL b

aligns T in space

and rotates V in space

but V is still tilted.

R 'η'η

cos sinsin cos

R

33

2 21 1 2 2

1 ' ' ' ' ' ' '2

a b c


Step 2: Squeeze (expand and contract) T so that it becomes a circle with

unit radius

1''

2''T

1''

2''V

1,2'' 'η G η

2 21

2 21 2 1 2 2

1 '' '' '' '1 '' ' ' '' '' ''2

'2

L a b c

G expands and contracts

T so that it becomes a

circle. G applies the

same scaling to V.

11,2

2

1 0

0 1

G 1,2'' 'η G η

34


Step 3: Rotate again but this time we rotate so that V is in

1

2T

1

2V

''ζ R η

2 21 2

2 21 1 2 2

1122

L

aligns V in space.

Since T is a circle, the

rotation does NOT affect T.

R ζ

1 1,

''ζ R ηcos sinsin cos

R

35


1 1,

2 2, T

1 1,

2 2, V

A R GR

2 2 2 21 2 1 1 2 2

1 12 2

L

ζ Aη

- So the sequence of linear transformations needed to diagonalize both T

and V consists of a rotation followed by a squeeze and by another rotation.

ζ Aη

- The key step is the squeeze so that T becomes a circle before the last

rotation to align V along its axes.

36

Unlike the standard eigenvalue problem, the linear transformation

needed to simultaneously diagonalize both real-symmetry quadratic forms

(T and V) is NOT orthogonal in the traditional sense, i.e.,


A R GR

(the squeeze matrix is not

orthogonal)AA 1

However, it is orthogonal in a more generalized sense

A is orthogonal with respect to the metric tensor T

ATA 1r s rsa Tarecall

37

A metric tensor is just a generalization of our Euclidean distance

measure.


With regular Euclidean space, the distance between two points is,

2 22 2jk j kd d dx dyds d qdz q d s s

2ds

For other non-Euclidean space, distance in general can be measured by a

metric tensor G give by,

2jk j kds g dq dq

where is the matrix element of G.jkg

38

Other familiar examples of metrix tensors:


Mikowski space (Lorentzian metric): (non-positive definite)

2 2 2 2 2ds dr r d dz 2

1 0 00 00 0 1

G r

Cylindrical coord:

Spherical coord:

2 2 2 2 2 2 2sinds dr r d r d 2

2 2

1 0 00 00 0 sin

G rr

2 2 2 2 2 2ds dx dy dz c dt

Posi

tive

def

init

e m

etri

cs39

So, in our discussion for small oscillations, the condition for the kinetic

energy matrix given by


simply means that the set of eigenvectors are orthonormal with

respect to a different metric given by the metric tensor T or .

ATA 1

rajkT

AND, the set of eigenvectors from T diagonalizes simultaneously the

potential energy matrix V,

AVA λ

40

When one or more of the roots from the eigenvalue characteristic equation,

is repeated i.e., , , then our argument in showing that

Degeneracy

from 0s r r s a Ta

r s r s0r s a Ta

does not follow directly.

However, one can still construct a FULL set of orthonormal eigenvectors

from the degenerate set of allowed vectors.

To see how to do that, let consider a simple 3D case when we have 1

distinct eigenvalue and a double root for the other two eigenvalues3

1 2 3

41

There is an infinite set of eigenvectors associated with in the

degenerate plane to

Let be a normalized eigenvector associate with

Degeneracy

r r rVa Ta

3 3 1a Ta 3

3a

1a

2a3a3 1 0a Ta

3 2 0a Ta

1, 2,3r

is orthogonal to both 3a

1 2 and a a

But, might NOT be orthogonal to 1a 2a

3a

42

Check:

Fortunately, one can always construct an orthogonal pair from any

randomly chosen vectors in the plane , e.g.,

Degeneracy

1 2 22 1' c c VaV Vaa

1a

2a3a

1 2 and a a

Let be a new eignenvector given by,

2 1 1 2 2' c c a a a

2'a

2'a- Any linear combination of will

also be an eigenvector for 1 2 and a a

1 2

2

1 1 1 2 2 2

1 1 2 2 'c c

c c

Ta TaT a a Ta

43

Now, we enforce the orthogonality between

Degeneracy

1 2 and 'a a

1 2 1 1 1 2 1 2' 0c c a Ta a Ta a Ta

1a

2a3a 2'a

This new vector must also be properly normalized,

is normalized1 2 1 2 0c c a Ta1a

11 2

2

cc

a Ta

2 2 1 1 2 2 1 1 2 21 ' ' c c c c a Ta a a T a a 2 21 1 1 1 2 1 2 2 2 22c c c c a Ta a Ta a Ta

1 2 2 1a Ta a Ta T is symmetric

2 211 1 2 2

2

1 2 cc c c cc

2 22 1 1c c

44

Thus by solving the two equations

for , we have successful construct a set

of orthonormal eigenvectors for the degenerate

eigenvalue

Degeneracy

1a

2a3a 2'a

11 2

2

cc

a Ta and 2 22 1 1c c

1 2 and c c

1 2

The standard Gram-Schmidt orthogonalization procedure can be used

when the degeneracy space is more than 2 dimensions.

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021 small oscillations - george mason universitycomplex.gmu.edu/ small oscillations.pdf · labeling...

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