Sma Oscillations127

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Lecture 4 : THEORY OF SMALL

OSCILLATIONS

The lowest order Lagrangian

L ≡ T − V = 1

2(T ij ηi η j − V ijηiη j) (1)

which leads via the Lagrange’s equations

to the following n equations of motion of a

mechanical system near the equilibrium.

T kj η j + V kj η j = 0 (k = 1,...,n). (2)

Eigenvalue Problem

These equation of motion are linear dif-

ferential equations with constant coefficients,

and we thus try an oscillatory solution of the form

ηi = Caie−iωt (3)

where Cai is the complex amplitude of the

oscillations for coordinate ηi, the scaling

factor C and the frequency ω being thesame for all coordinates.



n linear algebraic equations for the am-

plitudes ai:

(V ij

−ω2T ij)a j = 0. (4)

These are homogeneous equations, and hence

the nontrivial solution exists only if the de-

terminant of the coefficients vanishes:

V 11 − ω2T 11 . . . V 1n − ω2T 1n

. . . . .

. . . . .

. . . . .

V n1 − ω2T n1 . . . V nn − ω2T nn

= 0. (5)



The simplified version of the general prob-

lem: the appropriate generalized coordi-

nates are the Cartesian coordinates of the

system of point particles multiplied by thesquare root of the particle mass (some-

times referred to as mass-weighted coordinates ).

The kinetic energy takes the form:

T = 1

2

η2

i = 1

2

(η)2, (6)

In this case T ij = δ ij and

V ija j = λai, (7)

where ω2 = λ. These equations forms the

well-known eigenvalue problem of the linear

algebra. It is convenient to put this prob-lem in matrix form as

V a = λa, (8)

where V is n × n matrix with elements V ij

and a is n-dimentional vector with compo-

nents ai. This equation is referred to asthe eigenvalue equations .



The values of λ for which the equation

is solvable are known as the characteristic

values, or eigenvalues of V , and the vector

solutions a are called the eigenvectors of V .Two facts from the linear algebra:

(i) the eigenvectors of a given matrix are

orthogonal to each other;

(ii) the eigenvalues of V are real if V is

symmetrical and real.



For arbitrary symmetric and real matrix

T ij the eigenvalue equation is rewritten in

matrix form as,

V a = λT a, (9)

also stands the eigenvalue problem for a

given matrix V . Note, that here the effect

of V on the eigenvector a is not merely

the same vector times the factor λ, but

a multipe of the result of the other given

matrix T acting on a. Nevertheless, even

in this general case we are in a position

to show that the eigenvectors a are still

orthogonal, and the eigenvalues λ are all

real under some conditions imposed on V and T .



We can form all the eigenvectors ak into

a square matrix A with components a jk.

Then we get one matrix equation:

AT A = I, (10)

the transposed matrix A is obtained from A

by interchanging rows and columns. This

condition expresses in the matrix form or-

thonormality property of the eigenvectors

ak, i.e. these vectors are orthogonal and

of unit magnitude - but in Riemann’s space !

To show this, we must recall some facts

concerning to this space. Riemann’s space,

or Riemann geometry is based on the quadratic

differential form in the variables x1,...,xn

ds2

= (g11dx1 + g12dx2 + g13dx3 + ... + g1ndxn)dx1

+(g21dx1 + g22dx2 + g23dx3 + ... + g2ndxn)dx2

................................................................

+(gn1dx1 + gn2dx2 + gn3dx3 + ... + gnndxn)dxn(11)

whose coefficients, g11,...,gnn, are generally

not constant but functions of the n vari-

ables x1,...,xn. These coefficients form the

metric tensor G of the n-dimensional space.



Clearly for Cartesian coordinates the met-

ric tensor is the unit matrix I . In more

general case of the curvilinear orthogonal

coordinates (such as polar coordinates ) G willbe not unit but still diagonal matrix. The

dot (or scalar ) product of two vectors a and

b in Riemann’s space is given by

a · b ≡ aigikbk ≡ aGb. (12)

If a · b = aGb = 0, (13)

the vectorsa and b are called orthogonal .

Furthermore, vector a is said to be nor-

malized to unity if

a · a = aGa = 1. (14)The eigenvectors ak are orthogonal and

that they are normalized to have unit mag-

nitude in a particular Riemann’s space for

which T is the metric tensor . Note, that the

kinetic energy (8) can be represented inthis space as

T = 1

2T ij ηi η j =

1

2˜ηT η =

1

2˜ηGη ≡ 1

2η · η ≡ 1

2(η)2,

(15)



. General Solution of Equations of Mo-

tion

The trial function of the oscillatory typesatisfies the equations of motion not for a

single frequency but in general for a set of

n frequencies ωk, k = 1,...,n. Mathemati-

cally, it means that the complete solution

of the equations of motion may be writtenas a sum over index k,

ηi = C kaik exp−iωkt, (16)

where C k is a complex scale factor. Strictly

speaking this solution must be generalized

to

ηi = aik(C +k exp+iωkt +C −k exp−iωkt), (17)

since for each eigenvalue λk (and thus for

each eigenvector ak) there are two frequences

+ωk and

−ωk and accordingly two different

scale factors C +k and C −k . Nevertheless thereal parts of these equations can be writ-

ten in the same form

ηi = f kaik cos (ωkt + δ k) (18)



For the determination of the constants f k

and δ k we put

ηi(0) = Re C kaik (19)

Similarly, the initial value of the velocities

is

ηi(0) = Im C kaikωk (20)

These equations form a set of 2n equations

from which the real and imaginary parts of

the scale factors C k, ReC k and ImC k, may

be evaluated.

In matrix relation,

Re C l = a jlT jkηk(0). (21)

A similar procedure with velocities gives

Im C l = 1

ωl

a jlT jk ηk(0). (22)



Normal Coordinates

The general solution for each coordinate

ηi being a sum of simple harmonic oscilla-tions of the frequencies ωk is not itself a

periodic function of time. It is possible,

however, to transform ηi to a new set of

generalized coordinates that are periodic

function of time. Such set of variables isknown as the normal coordinates .

The potential energy can be written as

can be written as

V = 1

2ζ AV Aζ =

1

2ζ Λζ =

1

2ζ lΛlkζ k =

1

2ζ lδ lkλkζ k =

1

2ω2

kζ

(23)In analogous way we obtain for the ki-

netic energy

T = 1

2˜ηT η =

1

2˜ζ AT A ζ , (24)

which can be reduced to

T = 12

˜ζ · ζ = 12

ζ 2k . (25)



We notice that (i) both the potential and

kinetic are sums of squares only (any cross

terms vanish), and (ii) both have been di-

agonalized by A which defines the princi-pal axes transformation. It follows then

that the principal axes transformation which

we employ here is the well-known alge-

braic process of simultaneous diagonalization

of two quadratic forms .The Lagrangian in a new reference sys-

tem

L = 1

2( ζ k ζ k − ω2

kζ 2k ), (26)

which results in the Lagrange’s equations

for ζ k

ζ k + ω2

kζ k = 0, (27)

with an obvious solution

ζ k = C k exp (−iωkt). (28)

This equation demonstrates that each co-

ordinate ζ k is a periodic function of one of the frequences ωk and amplitudes aik. Such

states are said to define the normal modes

of vibration , and ζ k are properly called the

normal coordinates of the system.



Example: Triatomic Molecule

We approximate the actual complicated

interatomic potential by two strings of forceconstant k, and thus take the potential en-

ergy in the form

V = k

2(x2 − x1 − b)2 +

k

2(x3 − x2 − b)2. (29)

Denoting the equilibrium position of the

atoms by x0i, i = 1, 2, 3, we have

x2 − x1 − b = x2 − x02 − (x1 − x01) − b + x02 − x01 ≡x3 − x2 − b = x3 − x03 − (x2 − x02) − b + x03 − x02 ≡where we introduce new coordinates

ηi = xi − x0i (31)

and took into account that

x02 − x01 = b, x03 − x02 = b. (32)

In terms of these new variables the po-

tential is rewritten as

V = k

2(η2 − η1)2 +

k

2(η3 − η2)2 =

k

2(η2

1 + 2η2

2 + η2

3 − 2η1η2 − 2η2η3)

≡ 1

2V ijηiη j ≡ 1

2ηV η. (33)



where matrix V has the form

V =

k −k 0−k 2k −k

0 −k k

. (34)

The kinetic energy has an obvious form

T = m

2 ( η2

1 + η2

3) + M

2 η2

2, (35)

or, in matrix form

T =

1

2˜ηT η, (36)

where T is the diagonal matrix

T =

m 0 00 M 00 0 m

(37)

The characteristic equation , we obtain

|V −ω2T | =

k − ω2m −k 0−k 2k − ω2M −k

0 −k k − ω2m

= 0.

(38)

Direct evaluation of this determinant yields

ω2(k − ω2m)[ω2M m − k(M + 2m)] = 0, (39)

which immediately yields the following so-

lutions:

ω1 = 0, ω2 =

k

m, ω3 =

k

m

1 +

2m

M

. (40)



The first solution represents an exam-

ple of the neutral equilibrium mentioned

above: the molecule under consideration

is translated along the line without anychange in its potential energy - in full agree-

ment with the corresponding equation for

the first normal coordinate

η1 = 0, (41)

which produces a uniform translational mo-

tion.

The second solution, ω2, is recognized as

the frequency of oscillations for a mass m

without any reference to the central mass

M which remains stationary in this modeof motion. And only the third solution, ω3

depends on the value of the central mass

which points out the participation of this

mass in the motion. All these predictions

are completely verified by examining the

eigenvectors on the basis of the eigenvalue

equation.



For the components aij with the frequency

ω j, we have

(k

−ω2m)a1 j

−ka2 j = 0,

−ka1 j + (2k − ω2M )a2 j − ka3 j = 0,−ka2 j + (k − ω2m)a3 j = 0.

(42)

The normalization condition

m(a2

1 j + a2

3 j) + M a2

2 j = 1 (43)

For the first mode ω1 = 0, and the system

immediately gives

a11 = a21 = a31, (44)

i.e. all three amplitudes are the same as

it should be in the case of a pure transla-

tional motion . The values of these ampli-tudes are fixed by the normalization con-

ditions which gives

a11 = 1√ 2m + M

, a21 = 1√ 2m + M

(45)

a31 =

1

√ 2m + M , (46)



For the second mode the amplitudes are

a22 = 0 and a12 = −a32, i.e. the center atom

is at rest, whereas the two edge atoms vi-

brate exactly out of phase. The normal-ized values of the amplitudes are

a12 = 1√

2m, a22 = 0, (47)

a32 = − 1√ 2m

, (48)

For the third mode when (k − ω2m) = 0,

a13 = a33. Combining this with the normal-

ization condition, we obtain

a13 = 1

2m(1 + 2mM

), a23 = − 2

2M (2 + M m

),

(49)

a33 = 1

2m(1 + 2mM

), (50)

In this case the two edge atoms oscillates

with the same amplitudes, while the cen-

tral atom vibrates out of phase with themand has a different amplitude .

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Sma Oscillations127