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Notes on Lagrangian Mechanics

Sergey Frolova∗†

a Hamilton Mathematics Institute and School of Mathematics,Trinity College, Dublin 2, Ireland

Abstract

This is a part of the Advanced Mechanics course MA2341. These notes are partially basedon the textbook “Mechanics” by L.D. Landau and E.M. Lifshitz.

∗Email: [email protected]† Correspondent fellow at Steklov Mathematical Institute, Moscow.

1

Contents

1 THE EQUATIONS OF MOTION 4

1.1 Generalised coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.1.1 Cartesian coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.1.2 Dof and generalised coordinates . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.3 Constraints and dof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 Hamilton’s principle of least action . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2.1 Equations of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2.2 Hamilton’s principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2.3 Equations of motion from Hamilton’s principle . . . . . . . . . . . . . . 7

1.2.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2.5 Lagrange’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.2.6 Properties of Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.2.7 Lagrangian of a constrained system . . . . . . . . . . . . . . . . . . . . . 13

1.3 Lagrangians of various systems . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.3.1 Particle in homogeneous and isotropic space and time . . . . . . . . . . . 16

1.3.2 System of particles subject to Galileo’s principle . . . . . . . . . . . . . . 18

1.3.3 Closed system of particles . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.3.4 System A moving in a given external field due to B . . . . . . . . . . . . 19

2 CONSERVATION LAWS 20

2.1 Noether’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.2 Applications of Noether’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.2.1 Cyclic coordinate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.2.2 Linear momentum ↔ homogeneity of space . . . . . . . . . . . . . . . . 23

2.2.3 Angular momentum ↔ isotropy of space . . . . . . . . . . . . . . . . . . 24

2.2.4 Angular momentum in d-dimensional space . . . . . . . . . . . . . . . . . 26

2.2.5 Particle in a magnetic field . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.2.6 Particle in a gravitational field . . . . . . . . . . . . . . . . . . . . . . . . 27

2.3 Conservation of energy ↔ homogeneity of time . . . . . . . . . . . . . . . . . . . 28

2

3 SMALL OSCILLATIONS 30

3.1 Free oscillations in one dimension . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.2 Forced oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.2.1 Inhomogeneous linear differential equations . . . . . . . . . . . . . . . . 31

3.2.2 Green’s function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.2.3 Periodic external force . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.3 Oscillations of systems with several dof . . . . . . . . . . . . . . . . . . . . . . . 36

3.3.1 Normal frequencies and coordinates from eom . . . . . . . . . . . . . . . 37

3.3.2 Normal frequencies and coordinates from Lagrangian . . . . . . . . . . . 39

3.4 Anharmonic oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

3.4.1 Anharmonic oscillations in one dimension . . . . . . . . . . . . . . . . . . 43

3.4.2 Anharmonic oscillations in several dimensions . . . . . . . . . . . . . . . 46

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Chapter 1

THE EQUATIONS OF MOTION

Classical mechanics studies the motion of material bodies which are not too light and not tooheavy (then we need quantum mechanics and the theory of general relativity, respectively),e.g. planets, cars, rockets, footballs and so on. We refer to a set of these material bodies asto a mechanical system. Sometimes we can neglect the dimensions of the object we study andregard it as a (point) particle. Whether an object may or may not be regarded as a particledepends on the problem concerned. For example, an asteroid moving far enough from Earthmay be regarded as a particle but not in considering its collision with Earth.

1.1 Generalised coordinates

To understand the motion of a mechanical system means to know the position of its bodies(and particles they are made of) at any instant of time.

1.1.1 Cartesian coordinates

The position of a particle in a d-dimensional Euclidian space Rd can be defined by its radiusvector

~r = (x1, . . . , xd) , (1.1)

whose components are its Cartesian coordinates x1, . . . , xd. In 3-space we often identify x ≡ x1,y ≡ x2, z ≡ x3. If we know ~r(t) for t1 ≤ t ≤ t2 then the particles trajectory is a parametriccurve in the space. The first derivative

~v =d~r

dt≡ ~r = (

dr1

dt, . . . ,

drddt

) = (r1, . . . , rd) = (v1, . . . , vd) (1.2)

is the velocity of the particle while the second derivative ~a = d2~r/dt2 ≡ ~r is its acceleration.The velocity vector is tangent to the trajectory. The norm of the velocity

|~v| =√v2

1 + · · ·+ v2d =

√v2i = v (1.3)

is the speed, and the distance travelled by the particle in an infinitesimal time interval dt is

ds = vdt =√dx2

1 + · · ·+ dx2d =

√dx2

i =⇒ ds2 = dx2i . (1.4)

If there are N particles then we need N radius vectors or dN coordinates.

4

1.1.2 Dof and generalised coordinates

Definition. The number of degrees of freedom (dof) of a mechanical system is the number ofindependent quantities which define uniquely the position of the system.

Thus, a system of N freely moving particles in 3-space has 3N dof.

To define the position of a particle one does not have to use the Cartesian coordinates.For example, if a system has axial symmetry, i.e. its properties do not change under rotationsabout a given line then the cylindrical coordinates are more convenient. Similarly, if a systemhas spherical symmetry, i.e. its properties do not change under rotations about a given pointthen one prefers the spherical coordinates.

Definition. Generalised coordinates of a system with s dof are any s quantities q1, q2, . . . , qs

which completely define the position of the system. The first derivatives qi are the generalisedvelocities.1

In what follow we will refer to generalised coordinates and velocities simply as to coordinatesand velocities.

1.1.3 Constraints and dof

In some mechanical systems the interaction between different particles restricts their relativeposition. These restrictions are expressed by means of a set of (holonomic) constraints

Cα(q1, . . . , qs, t) = 0 , α = 1, . . . , r ≤ s . (1.5)

Each constraint reduces the number of dof by 1, so a constrained system with r independentconstraints has s− r dof.

The set of all allowed positions of a mechanical system is called the configuration space ofthe system. In most cases the configuration space of a system is a smooth manifold whosedimension is equal to the number of dof.

Example 1. A simple planar pendulum is a particle moving in 2-space and attached to arod of length l whose other end is fixed. The Cartesian coordinates x and y of the particle aresubject to the constraint2

C(x, y) = x2 + y2 − l2 = 0 , (1.6)

which defines a circle of radius l centred at the origin. This circle, S1, is the configuration spaceof the pendulum. In polar coordinates

x = r cosφ , y = r sinφ , (1.7)

the constraint takes the form r = l which makes obvious that the angle φ is the only generalisedcoordinate of the system.

Example 2. A simple pendulum is a particle moving in 3-space and attached to a rod oflength l whose other end is fixed. The Cartesian coordinates x, y and z of the particle aresubject to the constraint

C(x, y, z) = x2 + y2 + z2 − l2 = 0 . (1.8)

1Throughout the course we mainly use upper indices for generalised coordinates qi unless they are Cartesiancoordinates for which we prefer the lower indices. Cartesian coordinates with lower and upper indices areidentified xi ≡ xi.

2Here and in what follows the origin coincides with the fixed end of the rod.

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which defines a sphere of radius l centred at the origin. This sphere, S2, is the configurationspace of the pendulum. In spherical coordinates

x = r cosϕ sin θ , y = r sinϕ sin θ , z = r cos θ , (1.9)

the constraint takes the form r = l which makes obvious that the angles ϕ and θ are the twogeneralised coordinates of the system.

Example 3. A coplanar double pendulum. The Cartesian coordinates x1, y1 and x2, y2 ofthe two particles are subject to the constraints

C1(x1, y1, x2, y2) = x21 + y2

1 − l21 = 0 ,

C2(x1, y1, x2, y2) = (x2 − x1)2 + (y2 − y1)2 − l22 = 0 ,(1.10)

and the system has two dof. Since each of the constraints defines a circle, the configurationspace of the double pendulum is a torus T 2 = S1 × S1.

Introducing polar coordinates for x1, y1 and x2 − x1, y2 − y1

x1 = r1 cosφ1 , y1 = r1 sinφ1 , x2 − x1 = r2 cosφ2 , y2 − y1 = r2 sinφ2 , (1.11)

one getsC1(r1, φ1, r2, φ2) = r2

1 − l21 = 0 ,

C2(r1, φ1, r2, φ2) = r22 − l22 = 0 ,

(1.12)

and the angles φ1 and φ2 are the two generalised coordinates of the system.

Example 4. Find the number of dof of a system of N particles in d-dimensional spacemoving such that the distance between the particles do not vary.

1. N = 2, d = 2.

2. N = 3, d = 2.

3. N = n ≥ 4, d = 2.

4. N = 2, d = 3.

5. N = 3, d = 3.

6. N = 4, d = 3.

7. N = n ≥ 5, d = 3.

1.2 Hamilton’s principle of least action

1.2.1 Equations of motion

If we know the position of a system at a given instant of time we cannot say anything aboutits position at any other instant because the particles of the system can have any velocities,accelerations, and so on. However, for many systems if all the coordinates and velocities ofa system are known at a given instant of time then the position of the system is completelydetermined, and the coordinates can be found at any instant. This implies in particular that

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the accelerations q i at that instant are uniquely defined by qi and qi, and therefore q i arefunctions of qi and qi. Since the same must hold for any instant of time we conclude that

q i = f i(q1, . . . , qs, q1, . . . , qs, t) , i = 1, . . . , s . (1.13)

Definition. The relations between the coordinates, velocities and accelerations are calledthe equations of motion (eom).

They are second-order differential equations for qi(t).

For brevity, we will often denote by q the set of all the coordinates q1, . . . , qs, and by q andq the sets of all the velocities and accelerations, respectively. Then, the eom are written as

q = f(q, q, t) . (1.14)

It is clear that the eom also define all higher-derivatives of q at any instant of time.

1.2.2 Hamilton’s principle

Hamilton’s principle or the principle of least action:

A mechanical system with s dof is characterised by a definite function

L(q1, . . . , qs, q1, . . . , qs, t) or L(q, q, t) ,

called the Lagrangian of the system. Then, the motion of the system between the position withcoordinates q(1) ≡ q(t1) at the instant t1 and the position with coordinates q(2) ≡ q(t2) at theinstant t2 is such that the integral

S =

∫ t2

t1

L(q, q, t)dt , (1.15)

takes the least possible value (for t2 sufficiently close to t1).

Definition. The integral is called the action of the mechanical system.

L depends on q, q but not on q ,...q , . . . because we assume that the motion of the system is

completely determined by coordinates and velocities.

1.2.3 Equations of motion from Hamilton’s principle

Hamilton’s principle can be used to derive the eom. Let q = q(t) be the function (a set of sfunctions for a system with s dof) for which S is a minimum. Then S is increased when q(t) isreplaced by any function of the form

q(t)→ q(t) + δq(t) , δq(t1) = δq(t2) = 0 , (1.16)

where δq(t) is a function which is small everywhere in the interval [t1, t2], and it vanishes at t1and t2 because the initial and final positions of the system are fixed.

Definition. δq(t) is called a variation of the function q(t).

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Thus, according to Hamilton’s principle the following condition must hold

∆S = ∆

∫ t2

t1

L(q, q, t)dt =

∫ t2

t1

L(q + δq, q + δq, t)dt−∫ t2

t1

L(q, q, t)dt

=

∫ t2

t1

(L(q + δq, q + δq, t)− L(q, q, t)

)dt =

∫ t2

t1

∆L(q, q, t) dt ≥ 0 ,

(1.17)

for any variation δq.

1.2.4 Examples

Example 1. Consider a particle with the action

S =m

2

∫ t2

t1

~r 2 dt , ~r = (x, y, z) , ~r 2 = ~r · ~r = x2 + y2 + z2 . (1.18)

The Lagrangian is

L =m

2~r 2 . (1.19)

Let ~r = ~r(t) be the vector-valued function for which S is a minimum, and δ~r(t) is its variation.Then ∆S is

∆S =m

2

∫ t2

t1

((~r(t) + δ~r(t))2 − ~r 2

)dt = m

∫ t2

t1

~r(t) · δ~r(t) dt+m

2

∫ t2

t1

δ~r(t)2 dt

= −m∫ t2

t1

~r (t) · δ~r(t) dt+m

2

∫ t2

t1

δ~r(t)2 dt ≥ 0 ,

(1.20)

where we integrated by parts and used δ~r(t1) = δ~r(t2) = 0. The part linear in δ~r is denoted byδS

δS = −m∫ t2

t1

~r (t) · δ~r(t) dt , (1.21)

and its integrand must vanish for any δ~r because if it does not vanish then both δS and ∆Scan be made negative by choosing for example δ~r(t) = εm ~r (t) where ε > 0 is small enough sothat |δS| is greater than the second term in (1.20). Thus, one gets the equations

~r = ~0 =⇒ ~r = ~v = const , (1.22)

which are the eom for a freely moving particle, and therefore L = m2~r 2 = m

2~v 2 is its Lagrangian.

We also see that if ~r = ~0 then

∆S =m

2

∫ t2

t1

δ~r(t)2 dt , (1.23)

and it is positive only if m > 0. The constant m is the mass of the particle.

Example 2. Consider a particle with the action

S =

∫ t2

t1

(m2~r 2 + ~F (t) · ~r

)dt , L(~r, ~r, t) =

m

2~r 2 + ~F (t) · ~r , (1.24)

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where ~F (t) is independent of ~r, ~r. Then δS, the part linear in δ~r, is

δS =

∫ t2

t1

(−m~r · δ~r + ~F (t) · δ~r

)dt =

∫ t2

t1

(−m~r + ~F (t)

)· δ~r dt , (1.25)

and it again must vanish. Thus, one gets the equations of motion

δS = 0 =⇒ m~r = ~F (t) this is second Newton′s law , (1.26)

which are the eom for a particle moving under the influence of the force ~F .

Example 3. Consider a system of N particles with the action

S =

∫ t2

t1

( N∑a=1

ma

2~ra

2 − U(~r1, . . . , ~rN , t))dt , L(~r, ~r, t) =

N∑a=1

ma

2~ra

2 − U(~r1, . . . , ~rN , t) .

(1.27)where U(~r1, . . . , ~rN , t) is independent of ~r. Then δS is

δS =

∫ t2

t1

(−

N∑a=1

ma~r a · δ~ra −N∑a=1

∂U

∂~ra· δ~ra

)dt = −

∫ t2

t1

N∑a=1

(ma~r a +

N∑a=1

∂U

∂~ra

)· δ~ra dt ,

(1.28)and its vanishing gives the equations of motion

δS = 0 =⇒ ma~r a = −N∑a=1

∂U

∂~ra= ~Fa . (1.29)

This is second Newton’s law for a system of N particles. The function U is the potential energyof the system. We see that the Lagrangian of this system can be written as

L = T − U , T =N∑a=1

ma

2~ra

2 , U = U(~r1, . . . , ~rN , t) , (1.30)

where T is the kinetic energy.

1.2.5 Lagrange’s equations

The eom for a general system are derived in a similar way. First consider a system with onedof. According to (1.17),

∆S =

∫ t2

t1

∆L(q, q, t) dt ≥ 0 , (1.31)

where

∆L(q, q, t) = L(q + δq, q + δq, t)− L(q, q, t) =∂L

∂qδq +

∂L

∂qδq +O(δq2) . (1.32)

Thus, the variation δS, which is the part of ∆S linear in δq, δq, is

δS =

∫ t2

t1

(∂L∂qδq +

∂L

∂qδq)dt , (1.33)

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and the necessary condition for S to have an extremum is that the variation δS vanishes. Thus,the weak form (S has a critical “point” but not necessarily a minimum) of Hamilton’s principleis

δS =

∫ t2

t1

δL(q, q, t) dt = 0 , δL =∂L

∂qδq +

∂L

∂qδq . (1.34)

Since δq = ddtδq, we get

δS =∂L

∂qδq∣∣∣t2t1

+

∫ t2

t1

(∂L∂q− d

dt

∂L

∂q

)δq dt =

∫ t2

t1

(∂L∂q− d

dt

∂L

∂q

)δq dt , (1.35)

where ∂L∂qδq∣∣∣t2t1

= 0 because δq(t1) = δq(t2) = 0. Thus the eom are

d

dt

∂L

∂q− ∂L

∂q= 0 ⇔ d

dt

∂L

∂v− ∂L

∂q= 0 , v = q . (1.36)

For a system with s dof the s functions qi(t) must be varied independently and we get s eom

d

dt

∂L

∂qi− ∂L

∂qi= 0 ⇔ d

dt

∂L

∂vi− ∂L

∂qi= 0 , vi = qi , i = 1, . . . , s . (1.37)

These are Lagrange’s equations. It is a set of s second-order differential equations for s unknownfunctions qi(t). The general solution contains 2s arbitrary constants. To fix them and find themotion of the system, one specifies the initial conditions, i.e. the coordinates and velocities atsome given instant.

The quantity

pi ≡∂L

∂qi(1.38)

is called the (generalised) momentum conjugate to the coordinate qi. By using momenta, theeom take the form

dpidt

=∂L

∂qi, pi =

∂L

∂qi, i = 1, . . . , s . (1.39)

The equation pi = ∂L∂qi

can be used to express q as functions of q and p. Then, the motion of thesystem is defined by the values of coordinates and momenta at a given instant of time. This isHamiltonian mechanics to be discussed in the second half of the course.

Example. Find eom of a system with s dof and the Lagrangian

L =1

2gij(q)q

iqj , (1.40)

where gij = gji is a second rank tensor which depends only on q’s, and we sum over repeatedindices. Geometrically, gij is the metric tensor on the configuration space of the system, and,according to Hamilton’s principle, it must be positive definite. The distance between twoinfinitesimally close points in the configuration space is called the line element, and is definedby

ds2 = gij(q)dqidqj . (1.41)

To find the eom we calculate

∂L

∂qi= gij q

j ,∂L

∂qi=

1

2

∂gjk∂qi

qj qk . (1.42)

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Thus, the eom ared

dt

(gij q

j)− 1

2

∂gjk∂qi

qj qk = 0 , (1.43)

or after a simplification

gij qj +

1

2

(∂gij∂qk

+∂gik∂qj− ∂gjk

∂qi)qj qk = 0 . (1.44)

Introducing the Christoffel symbols

Γijk =1

2gin(∂gnj∂qk

+∂gnk∂qj− ∂gjk∂qn

), (1.45)

where gij is the inverse of gij: gikgkj = δij, one gets

q i + Γijkqj qk = 0 . (1.46)

These are the equations for geodesics (the shortest curve between two points) in the configura-tion space.

It is of interest to see how the metric tensor gij transforms under a change of generalisedcoordinates. Let

qi → Qi : qi = qi(Q1, . . . , Qs) = qi(Q) , i = 1, . . . , s , (1.47)

be a nondegenerate transformation of coordinates. Then, the velocity changes as

qi → Qi : qi =∂qi

∂QkQk , i = 1, . . . , s . (1.48)

This is the transformation rule under a change of coordinates of a vector.

Since the Lagrangian should not change, we get

L =1

2gij(q)q

iqj =1

2gij(q)

∂qi

∂QkQk ∂q

j

∂QlQl =

1

2Gkl(Q)QkQl . (1.49)

Thus, under a change of coordinates the metric tensor transforms as

gkl(q)→ Gkl(Q) : Gkl(Q) =∂qi

∂Qk

∂qj

∂Qlgij(q) , k, l = 1, . . . , s . (1.50)

This is the transformation rule for a second-rank tensor of type (0, 2).

Notice also that a co-vector (a component of one-form) transforms as

pk → Pk : Pk =∂qi

∂Qkpi , (1.51)

and the scalar productpiq

i (1.52)

is invariant under a change of coordinates.

11

1.2.6 Properties of Lagrangian

Additivity of Lagrangian

If a system consists of two subsystems A and B then in the limit where the interaction betweenthe parts may be neglected the Lagrangian L of the system is equal to the sum of Lagrangiansof the subsystems A and B

limno interaction between A and B

L = LA + LB . (1.53)

Multiplicativity of Lagrangian

Multiplication of L of a system by a constant does not change the eom:

L→ gL ⇒ eom→ eom.

However, the Lagrangians of different isolated mechanical systems cannot be multiplied bydifferent constants due to the additivity property. The possibility of multiplying Lagrangiansof all systems by the same constant reflects the natural arbitrariness in the choice of the unitof measurement.

Total derivative freedom

Two Lagrangians differing by a total derivative with respect to time of some function Λ(q, t)of coordinates and time (no velocities!) lead to the same eom, and therefore describe the samemechanical system. Let

L(q, q, t) = L(q, q, t) +d

dtΛ(q, t) . (1.54)

The actions are related as

S =

∫ t2

t1

L(q, q, t)dt =

∫ t2

t1

L(q, q, t)dt+

∫ t2

t1

d

dtΛ(q, t)dt

= S + Λ(q(2), t2)− Λ(q(1), t1) .

(1.55)

Thus,δS = δS , if δq(1) = δq(2) = 0 , (1.56)

and therefore S and S give the same eom.

Example. Find eom of a system with s dof and the Lagrangian

L =1

2gij q

iqj + bij qiqj − U(q) , (1.57)

where gij = gji and bij are constants, and we sum over repeated indices.

We calculate∂L

∂qi= gij q

j + bijqj ,

∂L

∂qi= bjiq

j − ∂U

∂qi. (1.58)

Thus the eom ared

dt

(gij q

j + bijqj)− bjiq

j +∂U

∂qi= 0 , (1.59)

12

or after a simplification

gij qj + (bij − bji)qj +

∂U

∂qi= 0 . (1.60)

The eom depend only on the anti-symmetric part of bij. The reason is that the symmetric partof bij contributes a total time derivative to L. Indeed,

bij qiqj =

1

2(bij − bji)qiqj +

1

2(bij + bji)q

iqj

=1

2(bij − bji)qiqj +

1

4

d

dt

((bij + bji)q

iqj).

(1.61)

So, without loss of generality one can consider bij to be anti-symmetric. In 3-space the termwith bij describes the interaction of a particle with the magnetic field bij.

1.2.7 Lagrangian of a constrained system

Consider a system A with generalised coordinates Qa, a = 1, . . . , s+ r and the Lagrangian

LA = LA(Q1, . . . , Qs+r, Q1, . . . , Qs+r, t) . (1.62)

Let us now restrict the possible configurations of the system by imposing a set of r independentholonomic constraints

Cα(Q1, . . . , Qs+r, t) = 0 , α = 1, . . . , r . (1.63)

The constraints reduce the number of dof of A to s. The general solution of the constraintsequations (1.63) can be written in the form

Qa = F a(q1, . . . , qs, t) , a = 1, . . . , s+ r . (1.64)

It depends on s independent parameters qi which can be chosen as the generalised coordinatesof the reduced system.

To analyse the motion of the reduced system we need to know the Lagrangian L = L(q, q, t)of the reduced system.

A natural proposal for L is that it is just equal to LA evaluated on the solution (1.64) ofthe constraints equations. Taking into account that

Qa =∂F a(q1, . . . , qs, t)

∂qiqi +

∂F a(q1, . . . , qs, t)

∂t, (1.65)

one finds

L = LA(F (q, t),∂F (q, t)

∂qq +

∂F (q, t)

∂t, t) . (1.66)

where F (q, t) denotes the set of F 1(q, t), . . . , F s+r(q, t), and ∂F (q,t)∂q

q + ∂F (q,t)∂t

denotes the set of∂F 1(q,t)∂qi

qi + ∂F 1(q,t)∂t

, . . . , ∂Fs+r(q,t)∂qi

qi + ∂F s+r(q,t)∂t

.

Example 1. L of a simple planar pendulum in a uniform gravitational field.

According to the consideration in section 1.1.3, LA is equal to

LA =1

2m(x2 + y2) +mgx , (1.67)

13

where the x-axis points in the direction opposite to the direction of the gravitational field. Thesolution of the constraint x2 + y2 − l2 = 0 is

x = l cosφ , y = l sinφ , x = −lφ sinφ , y = lφ cosφ . (1.68)

Substituting the solution into LA, one gets

L =1

2ml2φ2 +mgl cosφ . (1.69)

The eom of the pendulum is

φ+g

lsinφ = 0 . (1.70)

Example 2. L of a simple pendulum in 3-space in a uniform gravitational field. Accordingto the consideration in section 1.1.3, LA is equal to

LA =1

2m(x2 + y2 + z2) +mgz , (1.71)

where the z-axis points in the direction opposite to the direction of the gravitational field. Thesolution of the constraint C(x, y, z) = x2 + y2 + z2 − l2 = 0 is

x = l cosϕ sin θ , y = l sinϕ sin θ , z = l cos θ ,

x = −lϕ sinϕ sin θ + lθ cosϕ cos θ , y = lϕ cosϕ sin θ + lθ sinϕ cos θ , z = −lθ sin θ .(1.72)


L =1

2ml2(θ2 + ϕ2 sin2 θ) +mgl cos θ , (1.73)

and the eomd

dt(sin2 θϕ) = 0 ⇒ sin2 θϕ = const ,

θ − 1

2ϕ2 sin 2θ +

g

lsin θ = 0 .

(1.74)

Example 3. L of a coplanar double pendulum. According to the consideration in section1.1.3, LA is equal to

LA =1

2m1(x2

1 + y21) +

1

2m2(x2

2 + y22) +m1gx1 +m2gx2 , (1.75)

where the x-axis points in the direction of the gravitational field. The solution of the constraints

x21 + y2

1 − l21 = 0 , (x2 − x1)2 + (y2 − y1)2 − l22 = 0 , (1.76)

isx1 = l1 cosφ1 , y1 = l1 sinφ1 , x2 − x1 = l2 cosφ2 , y2 − y1 = l2 sinφ2 . (1.77)


L =1

2(m1 +m2)l21φ

21 +

1

2m2l

22φ

22 +m2l1l2φ1φ2 cos(φ1 − φ2)

+ (m1 +m2)gl1 cosφ1 +m2gl2 cosφ2 .(1.78)

14

Lagrange multipliers

Sometimes it is more convenient to analyse the motion of a constrained system in terms of theoriginal coordinates Qa. Then one uses a different approach called the method of Lagrange’smultipliers. This is a generalisation of the standard Lagrange’s method for finding an extremumof a function whose coordinates are subject to a set of constraints.

For each constraint Cα one introduces an auxiliary coordinate λα, and considers the followingLagrangian

Laux(Q, Q, λ, t) = LA(Q, Q, t) + λαCα(Q, t) . (1.79)

Then, the motion of the constrained system between Q(1) ≡ Q(t1) at the instant t1 and Q(2) ≡Q(t2) at the instant t2 (both satisfying the constraints) is such that the action

Saux =

∫ t2

t1

Laux(Q, Q, λ, t)dt , (1.80)

takes the least possible value.

Clearly, one gets the usual Lagrange’s eom from Saux

d

dt

∂Laux

∂Qa=∂Laux

∂Qa⇒ d

dt

∂LA

∂Qa=∂LA∂Qa

+ λα∂Cα(Q, t)

∂Qα, a = 1, . . . , s+ r ,

d

dt

∂Laux

∂λα=∂Laux

∂λa⇒ Cα(Q, t) = 0 , α = 1, . . . , r .

(1.81)

Thus, the eom which follow from a variation of the Lagrange multipliers are just the constraints.One can show that if one solves the constraints and substitutes the solution to the equations forQa in (1.81) then the resulting eom for qi coincide with the eom derived from the Lagrangian(1.66) for the reduced system.

1.3 Lagrangians of various systems

The Lagrangian describing a mechanical system depends on the choice of a frame of reference.Usually, a frame of reference can be chosen in which space is homogeneous and isotropic andtime is homogeneous. Such a frame of reference is called inertial.

Let us discuss how to describe the properties mathematically. Consider a system of Nparticles with radius vectors ~ra, a = 1, . . . , N , and the Lagrangian L(~r,~v, t) where ~r and ~v = ~rdenote the sets of radius vectors and velocities.

The homogeneity of space

It means that the eom have the same form at any point of space, or, in other words, theeom are invariant under a simultaneous shift of all the radius vectors ~ra by a constant vector

~ra → ~ra + ~ε . (1.82)

Thus, if space is homogeneous then the Lagrangian L may depend only on the differences~rab = ~ra − ~rb and the velocities which remain unchanged under the shift of space coordinates:L = L(~rab, ~v, t).

The homogeneity of time

15

Similarly, it means that the eom have the same form at any point of time, or, in other words,the eom are invariant under a shift of time by a constant

t→ t+ ε . (1.83)

Thus, if time is homogeneous then the Lagrangian L may not explicitly depend on time: L =L(~r,~v).

The isotropy of space

It means that the eom have the same form in any direction of space, or, in other words, theeom are invariant under a simultaneous rotation of all the radius vectors ~ra about the origin

xai → Aijxaj , i, j = 1, . . . , d , ~ra = (xa1, . . . xad) , (1.84)

where A is an orthogonal d×d matrix with the unit determinant: AikAjk = δij, detA = 1. Thevelocities obviously transform in the same way. Since rotations preserve distances and angles,if space is isotropic then the Lagrangian L may depend only on the norms of radius vectorsand velocities |~ra|, |~va|, the norms of the differences |~rab| = |~ra−~rb|, |~vab| = |~va−~vb|, and scalarproducts ~ra · ~vb: L = L(|~ra|, |~rab|, ~ra · ~vb, |~va|, |~vab|, t).

Combining the results obtained we see that if space is homogeneous and isotropic and timeis homogeneous then the Lagrangian of a system of N particles must have the form

L = L(|~rab|, |~va|, |~vab|) . (1.85)

1.3.1 Particle in homogeneous and isotropic space and time

As the first application of the consideration above, let us consider just a single particle. Then,the Lagrangian is

L = L(v2) , (1.86)

where instead of |~v| we assumed that L depends on its square: v2 = |~v|2. The eom which followfrom this Lagrangian are

d

dt

(∂L∂~v

)= 0 ⇒ ∂L

∂~v= const ⇒ ~v = const . (1.87)

Thus, a single particle always moves with a constant velocity in homogeneous and isotropicspace and time, or equivalently in an inertial frame of reference.

To determine the functional dependence of L on v one has to use a relativity principle whichstates that all inertial frames of reference are equivalent. This implies that the eom derivedfrom the Lagrangian describing the system in one inertial frame are the same as the eom derivedfrom the Lagrangian describing the same system in another inertial frame. Mathematically,this means that any relativity principle must be supplemented with transformation rules whichrelate coordinates and times in two different frames of references K and K ′.

In the case of Galileo’s relativity principle the rules are called Galilean transformations, andthey take the form3

t = t′ , ~r = ~r ′ + ~V t , ⇒ ~v = ~v ′ + ~V , (1.88)

3In the case of Einstein’s relativity principle the rules are called Lorentz transformations, and they will bediscussed in the second part of the course.

16

where ~V is the velocity of K ′ measured in K. The eom will be the same if the Lagrangians L′

and L differ by the total time derivative of a function of ~r and t. Considering an infinitesimalGalilean transformation

~v = ~v ′ + ~ε , (1.89)

and expanding L′ = L(v′2) in powers of ε keeping only the linear term in ε, one gets

L(v′2) = L(v2 − 2~v · ~ε+ ~ε 2) = L(v2)− ∂L

∂v22~v · ~ε+O(ε2)

= L(v2)− 2∂L

∂v2

d

dt(~r · ~ε) +O(ε2) .

(1.90)

It is obvious that the second term on the second line is a total time derivative only if ∂L∂v2

=const.Thus,

L =1

2mv2 , (1.91)

where the quantity m is the mass of the particle which, as was discussed earlier, must bepositive. It is easy to check that L′ and L given by (1.91) differ by a total time derivative fora finite Galilean transformation.

For a system of noninteracting particles the additivity property of the Lagrangian leads to

L =∑a

1

2mav

2a . (1.92)

It is equal to the kinetic energy of the system. Due to the multiplicativity property of L onlyratios of the masses are physically meaningful.

To find the Lagrangian of a particle in other coordinate systems it is useful to notice that

v2 =(dsdt

)2

=ds2

dt2, (1.93)

where ds is the element of arc or the line element in a given coordinate system.

In Cartesian coordinates

ds2 = dx2 + dy2 + dz2 = gijdqidqj ,

q1 = x , q2 = y , q3 = z , (gij) = diag (1, 1, 1) ,(1.94)

where gij are the components of the metric tensor of R3 in Cartesian coordinates.

In cylindrical coordinates

x = r cosφ , y = r sinφ , z = z , (1.95)

one findsds2 = dr2 + r2dφ2 + dz2 = gijdq

idqj ,

q1 = r , q2 = φ , q3 = z , (gij) = diag (1, r2, 1) ,(1.96)

where gij are the components of the metric tensor of R3 in cylindrical coordinates.

In spherical coordinates

x = r cosϕ sin θ , y = r sinϕ sin θ , z = r cos θ , (1.97)

17

one findsds2 = dr2 + r2dθ2 + r2 sin2 θdϕ2 = gijdq

idqj ,

q1 = r , q2 = θ , q3 = φ , (gij) = diag (1, r2, r2 sin2 θ) ,(1.98)

where gij are the components of the metric tensor of R3 in spherical coordinates.

Finally, for arbitrary generalised coordinates qi

x = f1(q1, q2, q3) , y = f2(q1, q2, q3) , z = f3(q1, q2, q3) , (1.99)

one finds

ds2 = gij(q)dqidqj , gij(q) =

∂fa∂qi

∂fa∂qj

, (1.100)

and

L =1

2mgij(q)q

iqj , (1.101)

where we sum over repeated indices, and gij are the components of the metric tensor of R3

expressed in terms of qi. The Lagrangian of a system of N noninteracting particles in d-dimensional space has the same form in terms of generalised coordinates qi, i = 1, . . . , dN , andthe quantity m is fixed by a choice of the unit of mass for the system.

1.3.2 System of particles subject to Galileo’s principle

Consider a system of N particles in homogeneous and isotropic space and time, and assumethat Galileo’s relativity principle is valid. Then, the Lagrangian of the system in Cartesiancoordinates has the form

L =N∑a=1

1

2mav

2a − U(|~rab|) , (1.102)

where the function U(|~rab|) ≡ U(|~r12|, |~r13| , . . . , |~rN−1,N |) is called the potential energy of thesystem, and it may depend on the differences |~rab|, a, b = 1, . . . , N . The Lagrangian eom canbe easily derived

ma~r a = −∂U∂~ra

= ~Fa . (1.103)

These are Newton’s equations, and ~Fa is the force on the ath particle which depends only ondistances between particles. Since the force depends on the particles coordinates any changein the position of any particle instantaneously effects all other particles, and therefore theinteractions are instantaneously propagated. Note also that the Lagrangian (1.102) is invariantunder the time reversal: if t is replaced by −t the Lagrangian and therefore the eom areunchanged. This means that all motions in homogeneous and isotropic space and time whichobey Galileo’s relativity principle are reversible.

1.3.3 Closed system of particles

Consider a system of N particles with the following Lagrangian

L =N∑a=1

1

2mav

2a − U(~r) , (1.104)

18

where the function U(~r) ≡ U(~r1, ~r2 , . . . , ~rN) is again the potential energy of the system, butnow it may depend not only on |~rab| but also on the radius vectors ~ra. Obviously, this system ofparticles does not obey Galileo’s relativity principle, and the Lagrangian describes the systemin a special reference frame. Still, the Lagrangian can be thought of as a limiting case of theLagrangian (1.102) for a system of N + M particles where the N + 1, N + 2, . . . , N + M -thparticles are infinitely massive. According to (1.103), if ma = ∞ then the velocity of athparticle must be constant. Assuming that the velocities of all the infinitely massive particlesare the same, one can choose the inertial frame so that their common velocity is zero. Then,the Lagrangian (1.102) reduces to

L =N∑a=1

1

2mav

2a − U(|~rab|, |~raβ|) , (1.105)

where ~raβ = ~ra − ~rβ, β = N + 1, . . . , N + M , and ~rβ are the constant radius vectors of theinfinitely massive particles. By choosing properly ~rβ and the number of infinitely massiveparticles M (which may be infinite too), one can represent any function U(~r1, ~r2 , . . . , ~rN) inthe form U(|~rab|, |~raβ|).

In particular if N = 1 and M = 1, and ~r2 = ~0, one gets

L =1

2mv2 − U(|~r|) , (1.106)

which is the Lagrangian of a particle moving in a central field.

Another interesting case is described by the Lagrangian

L =1

2mv2 + ~F · ~r , (1.107)

where ~F is a constant force. This Lagrangian is for a particle moving in a uniform field. It canbe obtained from (1.102) if the infinitely heavy particles form an infinite wall and interact withthe “light” particle with a Coulomb potential.

1.3.4 System A moving in a given external field due to B

This is a generalisation of a closed system of particles. Consider systems A and B interactingwith each other. Then, according to (1.102), the Lagrangian for the combined system A + Bcan be written in the form

LA+B = LA + LB + Linteraction ,

LA = TA(vA)− U(~rA) , LB = TB(vB)− U(~rB) , Linteraction = −U(~rA, ~rB) .(1.108)

If the particles of the system B are much heavier than those of A then one can neglect theback-reaction of A on B. Then, the trajectories of the particles of B follow from its LagrangianLB. Fixing some initial conditions for the particles of B one can solve their Lagrange’s eomand find ~rB = ~rB(t) as some definite functions of time. Substituting these functions into LA+B,we see that it is equivalent to

L = TA(vA)− U(~rA)− U(~rA, ~rB(t)) , (1.109)

because they differ by a function of time only which can be always written as a time derivativeof its anti-derivative. The Lagrangian L shows that we can think of A as being in an externaltime-dependent field created by B. Note, that the potential energy of the external field maydepend explicitly on time.

19

Chapter 2

CONSERVATION LAWS

For any mechanical system there are functions of coordinates, velocities and time which re-main constant during the motion. Such a function, I(q, q, t), is called an integral of motion:ddtI(q, q, t) = 0. Solving the eom one finds

qi = qi(t, c1, . . . , c2s) and qi = qi(t, c1, . . . , c2s) ,

where the constants ci can be fixed by the initial conditions. One can think of the solution as asystem of 2s equations on ci. Solving these equations, one finds ci as some functions of q, q andt. They obviously do not change during the motion, and therefore they are integrals of motion.Since any integral of motion is a function of ci and t, there are 2s independent integrals.

If the system is closed, i.e. its eom are invariant under shifts of time (homogeneity of time),then the solution of eom depends on t− t0 and 2s− 1 other constants

qi = qi(t− t0, c1, . . . , c2s−1) and qi = qi(t− t0, c1, . . . , c2s−1) ,

One can use one of the equations to express t− t0 as a function of q, q and c. Substituting t− t0into the remaining expressions one can then express the 2s− 1 constants ci as some functionsof q and q without any explicit dependence of t. Thus, for a closed system there are 2s − 1integrals which depend on q and q only.

The most important integrals are additive: if in some limit the system is composed of severalnoninteracting parts then the value of an additive integral is equal to the sum of its values forthese parts. For example consider a system which at t = −∞ is composed of noninteractingsubsystems A and B, while at t = +∞ it is composed of noninteracting subsystems A′ and B′.Then, an additive integral I for the system satisfies the relation

I = IA + IB = IA′ + IB′ , (2.1)

which imposes essential restrictions on possible motions of the system. The quantities repre-sented by additive integrals are said to be conserved. It appears that each conserved quantityis related to a particular continuous symmetry of a mechanical system, e.g. energy is related tohomogeneity of time, momentum to homogeneity of space, and angular momentum to isotropyof space. This relation is made manifest by Emmy Noether’s theorem, formulated in 1918.

20

2.1 Noether’s theorem

Consider an arbitrary transformation of coordinates qi → qi = qi(q, q, t). The coordinates qi

satisfy their Lagrange’s eom while the transformed coordinates qi in general satisfy differenteom because the form of the Lagrangian may not be preserved by the transformation. Atransformation which maps a solution of the eom to another solution of the same eom is calleda symmetry of the mechanical system. Since an inverse symmetry transformation is a symmetry,and a composition of two symmetry transformations is a symmetry too, the set of all symmetrytransformations of a mechanical system forms a group.

A transformation of coordinates is called continuous if it depends on a parameter whichcan take any value in some interval. For example a shift of a coordinate by a constant, or arotation of a radius vector through an angle are continuous transformations. The zero value ofthe parameter usually corresponds to the identity transformation.

Any known continuous symmetry transformation is analytic with respect to its parameter(can be expanded in power series in the parameter). Thus, the set of all continuous symme-try transformations forms a Lie group. According to Noether’s theorem, continuous symmetrytransformations are in one-to-one correspondence with conserved quantities. Here, we onlydiscuss how a continuous symmetry transformation leads to a conservation law. It is suffi-cient to consider infinitesimal transformations close to the identity transformation so that theirparameters are infinitesimally small.

As we discussed, if the Lagrangian under a transformation changes by a total time derivativeof a function of coordinates and time then the eom for the original and transformed coordinatesare the same. Therefore, such a transformation is a symmetry. Moreover, it is clear that itmust only depend on coordinates and time. We will formulate Noether’s theorem only fortransformations which only depend on coordinates and time.

Noether’s theorem. Consider a mechanical system with s dof. Let an infinitesimaltransformation of coordinates

qi → qi = qi + δqi , δqi = ε ζ i(q, t) , i = 1, . . . , s , (2.2)

depend only on q and t, and let the Lagrangian of the system transform under this transfor-mation as

L(q, q, t)→ L(q, ˙q, t) = L(q, q, t) + εd

dtΛ(q, t) . (2.3)

Then, the transformation is a symmetry, and the quantity

J =∂L

∂qiζ i − Λ (2.4)

is conserved, i.e.d

dtJ = 0 , (2.5)

if qi satisfy the eom.

21

Proof. We just take J and differentiate it with respect to time

d

dtJ =

d

dt

(∂L∂qi)ζ i +

∂L

∂qid

dtζ i − d

dtΛ

=d

dt

(∂L∂qi)ζ i +

∂L

∂qiζ i −

(∂L∂qi

ζ i +∂L

∂qiζ i)

=( ddt

(∂L∂qi)− ∂L

∂qi

)ζ i = 0 ,

(2.6)

where the last line vanishes due to Lagrange’s eom, and in the second line we used that underany variation of q

L(q, ˙q, t)− L(q, q, t) =∂L

∂qiδqi +

∂L

∂qiδqi , (2.7)

and therefored

dtΛ =

∂L

∂qiζ i +

∂L

∂qiζ i . (2.8)

Remark 1. Since the momentum is pi = ∂L/∂qi, the conserved integral can be written inthe form

J = piζi − Λ . (2.9)

Thus, if δΛ = 0 then the conserved integral in terms of p and q has a universal form independentof the form of the Lagrangian.

Remark 2. Multiplying J by ε, one gets

Jε =∂L

∂qiδqi − Λε . (2.10)

This form admits a straightforward generalisation to the case where a symmetry transformationqi → qi + δqi , δqi = εα ζ iα(q, t) depends on several parameters εα, so it is a composition ofone-parameter transformations. Then, for each α there is a conserved quantity which can beextracted from the formula

Jαεα =

∂L

∂qiδqi − Λαε

α , (2.11)

where as usual we sum over repeated indices.

Remark 3. One can easily see that the quantity J = ∂L∂qiζ i−Λ is conserved even if ζ i and Λ

have any dependence on q, q, and so on. The difficult part is then to show that the correspondingtransformation is a symmetry, i.e. it indeed maps a a solution of the eom to another solutionof the same eom. We will discuss it later on the example of the energy conservation.

2.2 Applications of Noether’s theorem

2.2.1 Cyclic coordinate

Consider a system with s dof described by a Lagrangian which has no explicit dependence onone of the coordinates, say, the last one qs

L = L(q1, . . . , qs−1, q1 . . . , qs−1, qs, t) . (2.12)

22

Such a coordinate is called cyclic. Obviously, the Lagrangian is invariant under a shift of thecyclic coordinate by a constant

qs → qs + ε , q1 → q1 , . . . , qs−1 → qs−1 ⇒ L→ L . (2.13)

Therefore,δqs = ε , δq1 = 0 , . . . , δqs−1 = 0 , Λ = 0 , (2.14)

and the following quantity is conserved

Jε =∂L

∂qiδqi =

∂L

∂qsε ⇒ J =

∂L

∂qs= ps . (2.15)

Thus, the momentum conjugate to a cyclic coordinate is conserved.

2.2.2 Linear momentum ↔ homogeneity of space

If the space is homogeneous then the Lagrangian of a system is invariant under a simultaneousshift of all radius-vectors by a constant vector

~ra → ~ra + ~ε , δ~ra = ~ε , L→ L ⇒∑a

∂L

∂~ra= 0 , (2.16)

and the following vector quantity is conserved

~P · ~ε =∂L

∂~ra· δ~ra =

∑a

∂L

∂~ra· ~ε ⇒ ~P =

∑a

∂L

∂~ra=∑a

~pa . (2.17)

The total (linear) momentum is called the momentum of the system, and it is conserved ifthe space is homogeneous. The momentum is obviously additive, and moreover, it is equal tothe sum of momenta ~pa of individual particles even if the interaction between them cannot beneglected.

The condition∑

a∂L∂~ra

= 0 has a simple physical meaning. Since the force ~Fa acting on the

ath particle is equal to ∂L∂~ra

, the condition just states that in homogeneous space the sum of theforces on all the particles is zero ∑

a

~Fa = 0 . (2.18)

For a system of two particles one gets ~F1 + ~F2 = 0 which is Newton’s third law: the equality ofaction and reaction.

Centre of mass

Consider a system with the Lagrangian

L =1

2ma~ra

2 − U(~rab, t) . (2.19)

The momentum ~P =∑

ama~va of this system is conserved. It is measured in a frame K, and

in a frame K ′ moving with the velicity ~V relative to K, the momentum ~P ′ is

~P ′ =∑a

ma~v′a =

∑a

ma~va −∑a

ma~V = ~P − µ~V , (2.20)

23

where µ =∑

ama is the total mass of the system. Thus, if

~V =~P

µ=

∑ama~va∑ama

, (2.21)

then the total momentum of the system in K ′ is zero. This is the velocity of the system as awhole. It can be also written in the form

~V =d~R

dt, ~R =

∑ama~ra∑ama

, (2.22)

where ~R is the radius-vector of the centre of mass of the system. The conservation of momentumimplies that the centre of mass moves uniformly in a straight line.

Let us introduce new coordinates ~R, ~ρa by the formula

~ra = ~ρa + ~R . (2.23)

The coordinates ~ρa are not independent. They are subject to the constraint∑a

ma~ρa = 0 ⇒∑a

ma~ρa = 0 . (2.24)

Expressing L in terms of the new coordinates, one finds

L =1

2

∑a

ma(~ρa + ~R)2 − U(~ρab, t)

=1

2

∑a

ma~ρa2 +

∑a

ma~ρa · ~R +1

2

∑a

ma~R2 − U(~ρab, t)

=1

2µ ~R2 +

1

2

∑a

ma~ρa2 − U(~ρab, t) .

(2.25)

We see that L is independent of ~R, and therefore the radius-vector of the centre of mass iscyclic. Therefore, its momentum, which is the momentum of the system, is conserved.

2.2.3 Angular momentum ↔ isotropy of space

If the space is isotropic then the Lagrangian of a system is invariant under a simultaneousrotation of all radius-vectors about any line through the origin. Any rotation in 3-space canbe characterised by a vector ~φ whose direction coincides with the direction of rotation (being

that of a right-handed screw driven along ~φ) and whose norm φ = |~φ| is equal to the angle ofrotation.

Any radius-vector ~ra = (xa1, xa2, xa3) transforms as follows under an infinitesimal rotation

δ~φ

~ra → ~ra + δ~ra , δ~ra = δ~φ× ~ra , (2.26)

In terms of the coordinates xai, i = 1, 2, 3 the formula takes the form

δxai = εijkεjxak , εi ≡ δφi , (2.27)

24

where we sum over repeated indices, and εijk is the anti-symmetric tensor, also called theLevi-Cevita symbol: ε123 = 1, εijk = −εjik = −εikj = −εkji.

If the Lagrangian is invariant under this transformation then

Miεi =∑a

∂L

∂xaiδxai =

∑a

∂L

∂xaiεijkεjxak =

∑a

εijkxajpakεi . (2.28)

ThusMi =

∑a

εijkxajpak =∑a

(~ra × ~pa)i ⇒ ~M =∑a

~ra × ~pa . (2.29)

is conserved. It is called the angular momentum of the system.

Properties of the angular momentum

1. Angular momentum is additive even if the interactions are not negligible (same the linearmomentum).

2. Its value depends on the choice of origin

~ra = ~r′a +~b ,

~M =∑a

~ra × ~pa =∑a

~ra × ~p′a +~b×∑a

~pa = ~M ′ +~b× ~P , (2.30)

except when the system is at rest as a whole, ~P = 0.

3. ~M and ~M ′ in frames K and K ′ are related as follows. Assume the origins coincide at agiven instant, and therefore

~ra = ~r′a , ~va = ~v′a + ~V ,

~M =∑a

~ra ×ma~va =∑a

~ra ×ma~v′a +

∑a

ma~ra × ~V = ~M ′ + µ~R× ~V ,

If K ′ is that in which the system is at rest as a whole, then ~V is the velocity of its centreof mass relative to K, and µ~V = ~P , and therefore ~M = ~M ′ + ~R × ~P . Thus, ~M consistsof its “intrinsic angular momentum” in a frame in which it is at rest, and ~R × ~P due toits motion as a whole.

4. If an external field is symmetric about an axis then the angular momentum along the axisis conserved. The angular momentum must be defined relative to an origin lying on theaxis.

(a) In the case of a centrally symmetric field or central field the potential depends only onthe distance from some particular point (the centre of the field). Then, the system isinvariant under a rotation about any axis passing through the centre, and therefore~M is conserved provided it is defined with respect to the centre of the field.

(b) If the field pointing in the z-direction is homogeneous then Mz is conserved whicheverpoint is taken as the origin.

25

5. The component of ~M along any axis, say the x3-axis, can be found as

M3 =∑a

∂L

∂φa=∑a

pφa , (2.31)

where φa is the polar angle of the ath particle in cylindrical coordinates

xa1 = ra cosφa , xa2 = ra sinφa .

Indeed, the Lagrangian is

L =1

2

∑a

ma(r2a + r2

aφ2a + z2

a)− U , (2.32)

andM3 =

∑a

ma(xa1xa2 − xa2xa1) =∑a

mar2aφa . (2.33)

Thus, the M3 component of the angular momentum of the ath particle is the (generalised)momentum pφa conjugate to the angle coordinate φa.

2.2.4 Angular momentum in d-dimensional space

Rotations in d-dimensional space are characterised by real orthogonal matrices A with the unitdeterminant1 which form the special orthogonal group SO(d):

A · AT = AT · A = I , detA = 1 .

The components xi, i = 1, . . . , d of any radius-vector ~r = (x1, . . . , xd) transform as follows

xi → x′i = Aijxj . (2.34)

To find an infinitesimal transformation we represent

A = I + ε , (2.35)

where the entries εij of the matrix ε are infinitesimally small. Since A is orthogonal the matrixε is anti-symmetric

(I + ε)(I + εT ) = I + ε+ εT +O(ε2) = I ⇒ ε+ εT = 0 . (2.36)

We say that the vector space of anti-symmetric matrices is the Lie algebra of the specialorthogonal group SO(d).

Thus, infinitesimal rotations in d-space have the form

xi → x′i = xi + εijxj ⇒ δxi = εijxj , εji = −εij , (2.37)

where we sum over repeated indices. The parameters εij have simple meaning – they are theangles of rotation in the xixj-plane. Indeed, let us assume that only ε12 and ε21 = −ε12 do notvanish. Then, one gets

δx1 = ε1jxj = ε12x2 , δx2 = ε2jxj = −ε12x1 , δxi = 0 , i = 2, 3, ..., d . (2.38)

1An orthogonal matrix with detA = −1 describes a composition of a rotation with an odd number ofreflections, e.g. x1 → −x1.

26

These formulae describe the infinitesimal rotation in the x1x2-plane through the angle ε21.

Assuming now that the Lagrangian is invariant under rotations, one gets the angular mo-mentum

1

2Jijεij =

∑a

∂L

∂xaiδxai =

∑a

paiεijxaj =1

2

∑a

(paixaj − pajxai)εij , (2.39)

where Jij are anti-symmetric, and we multiplied the lhs by 1/2 to avoid the double counting.Thus, in d-dimensional space the angular momentum is given by

Jij =∑a

(paixaj − pajxai) . (2.40)

Finally, let us mention that in 3-dimensional space the relation of εij with εi follows from (2.27)

εij = εikjεk = −εijkεk ⇒ εi = −1

2εijkεjk , (2.41)

and the relation between Jij and Mi is similar

Jij = −εijkMk ⇒ Mi = −1

2εijkJjk . (2.42)

2.2.5 Particle in a magnetic field

Consider a particle in a magnetic field described by the Lagrangian

L =1

2mx2

i + bijxixj , (2.43)

where bij = −bji are constants, and we sum over repeated indices.

The eom are invariant under a constant shift of coordinates

xi → xi + εi , δxi = εi , (2.44)

because L changes by a total time derivative

L→ L+ bijxiεj , δL =d

dtbijxiεj ⇒ Λiεi = bijxiεj . (2.45)

Thus, the corresponding conserved quantities are

Jiεi =∂L

∂xiδxi − bijxiεj = (mxi + bijxj)εi − bijxiεj = (mxi + 2bijxj)εi , (2.46)

orJi = mxi + 2bijxj = const . (2.47)

2.2.6 Particle in a gravitational field

Consider a particle in a gravitational field described by the Lagrangian

L =1

2mx2 + mgx , (2.48)

27

Under a shift of x: x→ x′ = x+ ε, the Lagrangian changes by a total time derivative

L→ L+mgε , δL = mgε =d

dt(mgtε) ⇒ Λε = mgtε . (2.49)

Thus, the shift of x is a symmetry, and the corresponding conserved quantity is

Jε =∂L

∂xδx−mgtε = (mx−mgt)ε ⇒ J = mx−mgt . (2.50)

2.3 Conservation of energy ↔ homogeneity of time

In all the examples we’ve considered an infinitesimal symmetry transformation depended onlyon q. There are, however, situations where it depends also on q, and, therefore, the trans-formed eom naively are third-order differential equations. In reality for a finite transformationthe transformed equations are second-order differential equations and the appearance of thirdderivative terms is an artefact of the expansion in the infinitesimal parameter.

As an example, let the time be homogeneous, and let us therefore consider a system withthe Lagrangian which has no explicit time dependence. Then, the eom are obviously invariantunder a shift of time: t→ t′ = t+ ε

d

dt

∂L

∂qi=∂L

∂qi↔ d

dt

∂L

∂ ˙qi=∂L

∂qi, (2.51)

whereqi(t) = qi(t+ ε) , L(q, ˙q) = L(q, ˙q) . (2.52)

On the other hand, infinitesimal transformations of q and L are given by

qi(t)→ qi(t) = qi(t) + qi(t)ε ⇒ δqi(t) = qi(t)ε ,

L→ L = L(q + qε, q + q ε) = L(q, q) +∂L

∂qiqiε+

∂L

∂qiq iε

= L(q, q) +d

dtL(q, q)ε ⇒ Λ = L .

(2.53)

Now, according to Remark 3 in section 2.1, the corresponding conserved quantity is

E =∂L

∂qiqi − L = piq

i − L , (2.54)

which is the energy of the system. Mechanical system whose energy is conserved are calledconservative systems.

In particular if

L = T (q, q)− U(q) =1

2gij q

iqj − U(q) , (2.55)

then

E =∂L

∂qiqi − L = gij q

iqj − 1

2gij q

iqj + U(q) =1

2gij q

iqj + U(q) = T + U , (2.56)

Thus, it is the sum of the kinetic and potential energies.

28

If we have a system of particles in homogeneous and isotropic space-time, then, as wasdiscussed,

L =∑a

1

2mav

2a − U(|~rab|) , (2.57)

and

E =∑a

1

2mav

2a + U(|~rab|) , (2.58)

The value of the energy obviously depends on the choice of a frame. The relation between Ein K and E ′ in K ′ can be easily found

E =∑a

1

2mav

2a + U =

∑a

1

2ma(~v

′a + ~V )2 + U

=∑a

1

2mav

′2a +

∑a

ma~v′a · ~V +

1

2µV 2 + U

= E ′ + ~V · ~P ′ + 1

2µV 2 .

(2.59)

We see that if the centre of mass is at rest in K ′ then ~P ′ = 0 and E ′ = Einternal, where Einternal

is the internal energy, that is the energy of the system in K ′ where it is at rest as a whole.Thus, the total energy of a system moving as a whole with speed V is given by the sum of theinternal energy and the kinetic energy of its centre of mass

E = Einternal +1

2µV 2 . (2.60)

29

Chapter 3

SMALL OSCILLATIONS

Small or linear or harmonic oscillations of a system about a position of stable equilibrium arevery common. This is the most important type of motion because it allows for a completeanalytic description and serves as a starting point of perturbative study of anharmonic ornonlinear oscillations.

3.1 Free oscillations in one dimension

Consider a system with one degree of freedom and the following Lagrangian

L =1

2a(q)q2 − U(q) . (3.1)

Let us assume that at q = q0 the system is at stable equilibrium, and that U ′′(q0) 6= 0. ThenU ′(q0) = 0, U ′′(q0) > 0, and a(q0) > 0. Expanding L up to quadratic order in q and q, we find

L =1

2a(q0)q2 − U(q0)− 1

2U ′′(q0)(q − q0)2 + · · · . (3.2)

Denotingx = q − q0 , m = a(q0) > 0 , k = U ′′(q0) > 0 , (3.3)

and omitting the constant U(q0), we get the Lagrangian

L =1

2mx2 − 1

2kx2 , (3.4)

which describes a system called a one-dimensional oscillator. It executes harmonic or linearoscillations. The eom is the following linear homogeneous differential equation

mx+ kx = 0 =⇒ x+ ω2x = 0 . (3.5)

Here

ω =

√k

m, (3.6)

is called the frequency of the oscillations because the general solution of this equation is asuperposition of the two independent solutions cosωt and sinωt

x = c1 cosωt+ c2 sinωt = a cos(ωt+ α) , (3.7)

30

wherea =√c1 + c2 , tanα = −c2

c1

. (3.8)

Here a is the amplitude of the harmonic oscillations, ωt+ α is their phase, and α is the initialphase. The frequency ω is determined b the properties of the system, and it is a fundamentalcharacteristic of the harmonic oscillations independent of the initial conditions of the motion.

The energy is conserved and is equal to

E =1

2mx2 +

1

2kx2 =

1

2m(x2 + ω2x2) =

1

2mω2a2 , (3.9)

and it is proportional to the square of the amplitude.

Let us note that the general solution (3.7) can be written as the real part of Aeiωt

x = <(Aeiωt) , A = aeiα , (3.10)

where A is the complex amplitude: |A| = a, arg(A) = α.

3.2 Forced oscillations

Let us now place a harmonic oscillator in a time-dependent external field

L =1

2mx2 − 1

2kx2 − Ue(x, t) , (3.11)

and let us assume that the motion of the system under the influence of the external field isfinite and its maximum displacement from x = 0 is small. Oscillations of such a system arecalled forced. Expanding the external potential in powers of x, we get

Ue(x, t) = Ue(0, t) + U ′e(0, t)x+1

2U ′′e (0, t)x2 + · · · . (3.12)

The first term is a function of time and therefore can be omitted, then U ′e(0, t) = −F (t)where F (t) is the external force acting on the system in the equilibrium position, and, finally,U ′′e (0, t) ≡ ke(t) can be thought of as a time-dependent spring constant which leads to a time-dependent frequency. We assume that ke(t) can be neglected.1 Then, the Lagrangian takes theform

L =1

2mx2 − 1

2kx2 + F (t)x . (3.13)

Since we do not want the external force to change the equilibrium position we should assumethat F (t) is small, and that either F (t) → 0 as t → ±∞, or F (t) is periodic and its averageover the period is 0.

3.2.1 Inhomogeneous linear differential equations

The eom is the following linear inhomogeneous differential equation

x+ ω2x = f(t) , f(t) =F (t)

m, ω =

√k

m. (3.14)

1Oscillatory systems with time-dependent parameters (mass and spring constant) exhibit variuos interestingphenomena such as parametric resonance.

31

Let us assume that we know a particular solution x1 of this inhomogeneous equation. Then, itis easy to see that if x is another solution then the difference x− x1 satisfies the correspondinghomogeneous equation. Thus, the general solution of this inhomogeneous equation is

x = x0 + x1 , (3.15)

where x0 is the general solution of the homogeneous equation, and x1 is any particular solutionof the inhomogeneous equation. We know that x0 = a cos(ωt+ α), and let us discuss how onecan find a particular solution.

Let us introduce an auxiliary complex variable

ξ = x+ iωx . (3.16)

Since x is real, it is equal to the imaginary part of ξ divided by ω

x =1

ω=(ξ) . (3.17)

It is easy to check that if x satisfies eom (3.14), then ξ satisfies the following first-order lineardifferential equation

ξ − iωξ = f(t) . (3.18)

To solve the equation let us use the following ansatz

ξ = C(t)eiωt , (3.19)

where C(t) is a unknown function. Substituting the ansatz in the equation (3.18), we find

dC

dt= e−iωtf(t) , (3.20)

and therefore

C(t) = C0 +

∫ t

t0

e−iωτf(τ)dτ =⇒ ξ(t) = C0eiωt +

∫ t

t0

eiω(t−τ)f(τ)dτ , (3.21)

where t0 is any constant, and C0 = C(t0). Thus, setting C0 = 0, we find the following particularsolution x1 of the inhomogeneous equation (3.14)2

x1(t) =1

ω=(

∫ t

t0

eiω(t−τ)f(τ)dτ) =1

ω

∫ t

t0

sin(ω(t− τ))f(τ)dτ , (3.22)

where we used that f and ω are real. Note that the particular solution satisfies x1(t0) = 0. Thegeneral solution of the homogeneous equation (3.14), therefore, is

x(t) = a cos(ωt+ α) +1

ω

∫ t

t0

sin(ω(t− τ))f(τ)dτ . (3.23)

The constants a, α, t0 are fixed by the initial conditions. For example, since the energy is notconserved, one can ask how much energy is transmitted to the system during all time assuming

2Keeping C0 arbitrary gives the general solution to (3.14).

32

its initial energy to be 0. Then, it is natural to choose t0 = −∞, and therefore a = 0 becausethe system was at rest initially. The solution of eom (3.14), therefore, is

x(t) =1

ω

∫ t

−∞sin(ω(t− τ))f(τ)dτ , (3.24)

the corresponding ξ is

ξ(t) =

∫ t

−∞eiω(t−τ)f(τ)dτ , (3.25)

and the energy transferred is

E(∞) =1

2m(x2 + ω2x2) =

1

2m|ξ(∞)|2 =

1

2m

∣∣∣∣∫ ∞−∞

e−iωtf(t)dt

∣∣∣∣2 =1

2m∣∣∣f(ω)

∣∣∣2 =1

2m

∣∣∣F (ω)∣∣∣2 ,

(3.26)where F (ω) is the Fourier component of the force F (t). If the force acts only during someperiod of time t0 −∆ ≤ t ≤ t0 −∆ the energy transferred can be written as

E(∞) =1

2m

∣∣∣∣∫ t0+∆

t0−∆

e−iωtF (t)dt

∣∣∣∣2 =1

2m

∣∣∣∣∫ ∆

−∆

e−iωτF (τ + t0)dτ

∣∣∣∣2 . (3.27)

Thus if the time interval 2∆ is short in comparison with 1/ω, i.e. ω∆ << 1 then the formulacan be approximated by

E(∞) ≈ 1

2m

(∫ ∆

−∆

F (τ + t0)dτ

)2

=1

2m

(∫ ∞−∞

F (t)dt

)2

=p2

2m. (3.28)

This formula shows that a force of short duration gives the system a momentum p =∫∞−∞ F (t)dt

without causing a visible displacement.

3.2.2 Green’s function

Sometimes one is interested in a particular solution of (3.14) which satisfies the vanishingboundary conditions

x(t0) = x(t1) = 0 , t0 < t1 , (3.29)

at some given instants of time t0, t1. The corresponding solution can be written in the form

x(t) =

∫ t1

t0

G(t, τ)f(τ)dτ , (3.30)

where G(t, τ) is called Green’s function, and it satisfies the following conditions

∂2G(t, τ)

∂t2+ ω2G(t, τ) = δ(t− τ) , G(t0, τ) = G(t1, τ) = 0 . (3.31)

Here δ(t− τ) is Dirac’s delta-function which satisfies the defining relation∫ t1

t0

δ(t− τ)h(τ)dτ = h(t) , t0 < t < t1 , (3.32)

33

for any function h continuous on the interval [t0, t1]. Clearly, x(t) given by (3.30) satisfies (3.14)if G(t, τ) satisfies (3.31).

To find G(t, τ) we can regard τ as a parameter. Then, for t0 ≤ t < τ and for τ < t ≤ t1Green’s function satisfies the homogeneous equation, and therefore one can look for a solutionof the form

G(t, τ) = Gt<τ (t, τ)θ(τ − t) +Gt>τ (t, τ)θ(t− τ) , (3.33)

where θ is the Heaviside θ-function

θ(t) =

0 if t < 01 if t > 0

. (3.34)

The two functions Gt<τ (t, τ) and Gt>τ (t, τ) must satisfy

∂2Gt<τ (t, τ)

∂t2+ ω2Gt<τ (t, τ) = 0 , Gt<τ (t0, τ) = 0 ,

∂2Gt>τ (t, τ)

∂t2+ ω2Gt>τ (t, τ) = 0 , Gt>τ (t1, τ) = 0 ,

Gt<τ (τ, τ) = Gt>τ (τ, τ) ,(∂Gt>τ (t, τ)

∂t− ∂Gt<τ (t, τ)

∂t

)∣∣∣t=τ

= 1 ,

(3.35)

where the third and fourth conditions guarantee that Green’s function satisfy (3.31).

The first two equations give

Gt<τ (t, τ) = A sinω(t− t0) , Gt>τ (t, τ) = B sinω(t− t1) . (3.36)

The third and fourth conditions then give

A sinω(τ − t0)−B sinω(τ − t1) = 0 ,

B cosω(τ − t1)− A cosω(τ − t0) =1

ω.

(3.37)

Solving these equations one finds

A =1

ω

sinω(τ − t1)

sinω(t1 − t0), B =

1

ω

sinω(τ − t0)

sinω(t1 − t0), (3.38)

and therefore

G(t, τ) =sinω(t− t0) sinω(τ − t1)

ω sinω(t1 − t0)θ(τ − t) +

sinω(t− t1) sinω(τ − t0)

ω sinω(t1 − t0)θ(t− τ) . (3.39)

3.2.3 Periodic external force

Let us analyse forced oscillations under the action of a periodic external force. Since anyperiodic function can be expanded in a Fourier series it is sufficient to consider the simplestforce of this type

F (t) = F0 cos(γt+ β) . (3.40)

The eom is

x+ ω2x = f cos(γt+ β) , f =F0

m, ω =

√k

m. (3.41)

34

We choose t0 = 0 in (3.42). Then, the particular solution is

x1(t) =f

ω

∫ t

0

sin(ω(t− τ)) cos(γτ + β)dτ

= − f

2ω

∫ t

0

(sin((ω − γ)τ − ωt− β) + sin((γ + ω)τ − ωt+ β)

)dτ .

(3.42)

To compute this integral we need to consider the two cases: i) γ 6= ω, and i) γ = ω.

If γ 6= ω we get

x1(t) =f

2ω

( 1

ω − γcos((ω − γ)τ − ωt− β) +

1

ω + γcos(ω((γ + ω)τ − ωt+ β)

)∣∣∣t0

=f

ω2 − γ2cos(γt+ β)− f

2ω

( 1

ω − γcos(ωt+ β) +

1

ω + γcos(ωt− β)

).

(3.43)

Since the last term in (3.43) is a solution to the homogeneous equation, we can choose the firstterm as a particular solution to (3.41), and therefore the general solution to (3.41) is

x(t) = a cos(ωt+ α) +f

ω2 − γ2cos(γt+ β) . (3.44)

Thus the motion of the system is a superposition of two oscillations. It is not a periodic motionunless the frequencies ω and γ are commensurable, i.e. ω/γ is a rational number, see thepictures below

-60 -40 -20 20 40 60

-2

-1

1

2

-60 -40 -20 20 40 60

-1.5

-1.0

-0.5

0.5

1.0

1.5

The solution (3.44) cannot be used if γ = ω when resonance occurs. In this case computingthe integral in (3.42), we get

x1(t) =f

2ω

∫ t

0

(sin(ωt+ β)− sin(2ωτ − ωt+ β)

)dτ =

f

2ωt sin(ωt+ β) , (3.45)

and therefore the general solution is

x(t) = a cos(ωt+ α) +f

2ωt sin(ωt+ β) = a(t) cos(ωt+ α(t)) . (3.46)

Obviously, in resonance the amplitude a(t) of oscillations increases linearly with the time (atlarge t), and therefore the approximation of small oscillations will be broken eventually.

Let us analyse the motion of the system near resonance when

γ = ω + ε , ε << ω . (3.47)

35

To this end it is convenient to write the solution (3.44) in the complex form

x(t) = <(Aeiωt +Beiγt

)= <

((A+Beiεt)eiωt

), (3.48)

where

A = aeiα , B = beiβ , b =f

ω2 − γ2. (3.49)

Since ε/ω << 1, the quantityC(t) = A+Beiεt , (3.50)

which may be regarded as a time-dependent complex amplitude of oscillations, changes slowlyover the period 2π/ω of eiωt, and the motion near resonance may be regarded as small oscillationsof the variable amplitude. The absolute value of C is

|C(t)|2 = a2 + b2 + 2ab cos(εt+ β − α) , (3.51)

and therefore it varies with frequency ε between3∣∣a− |b|∣∣ ≤ |C(t)| ≤ a+ |b| . (3.52)

This phenomenon is called beats, see the pictures below

-150 -100 -50 50 100 150

-6

-4

-2

2

4

6

-150 -100 -50 50 100 150

-10

-5

5

10

3.3 Oscillations of systems with several dof

Consider a system with s degrees of freedom and the following Lagrangian

L =1

2gjn(q)qj qn − U(q) . (3.53)

Let us assume that at qj = qj0, j = 1, . . . , s the system is at stable equilibrium, and that the

matrix of second-partial derivatives(∂2U(q0)∂qj∂qn

)is nondegenerate. Then ∂U(q0)

∂qj= 0, j = 1, . . . , s,

and(∂2U(q0)∂qj∂qn

)and

(gjn(q0)

)are positive-definite. Expanding L up to quadratic order in q and

q, we find

L =1

2gjn(q0)qj qn − U(q0)− 1

2

∂2U(q0)

∂qj∂qn(qj − qj0)(qn − qn0 ) + · · · . (3.54)

3Note that b may be negative.

36

Denoting

xj = qj − qj0 , mjn = gjn(q0) , kjn =∂2U(q0)

∂qj∂qn, (3.55)

and omitting the constant U(q0), we get the Lagrangian

L =1

2mjnxjxn −

1

2kjnxjxn , (3.56)

which describes a system of coupled oscillators. Its coordinates execute superpositions of har-monic oscillations with different frequencies as we shall see in a moment.

3.3.1 Normal frequencies and coordinates from eom

The eom is the following set of s linear homogeneous differential equations

mjnxn + kjnxn = 0 , j = 1, 2, . . . , s , (3.57)

and, as usual, we sum over the repeated index n. To solve the equations let us use the ansatz4

xj = Ajeiωt , j = 1, 2, . . . , s , (3.58)

where Aj and ω are some constants to be determined. Substituting (3.58) in the eom (3.57),we get a set of linear homogeneous algebraic equations(

− ω2mjn + kjn)An = 0 , j = 1, 2, . . . , s . (3.59)

These equations have a nontrivial solution only if the determinant of the matrix(−ω2mjn+kjn

)vanishes

det(− ω2mjn + kjn

)= 0 . (3.60)

The equation (3.60) is called the characteristic equation. Its lhs is a polynomial of degree sin ω2 which is called the characteristic polynomial, and its roots ω2

α, α = 1, . . . , s are calledthe characteristic frequencies or eigenfrequencies of the mechanical system.5 Since we haveassumed that at xj = 0 the system is in stable equilibrium all eigenfrequencies must be real,and therefore ω2

α must be positive. To prove it we multiply (3.59) by Aj, and take a sum over j

∑j,n

(− ω2mjn + kjn

)AnAj = 0 =⇒ ω2 =

∑j,n kjnAnAj∑j,nmjnAnAj

. (3.61)

Since the matrices (kjn) and (mjn) are positive-definite6 both the numerator and denominatorin (3.61) are positive as can be seen by representing Aj as the sum of its real and imaginaryparts: Aj = aj + ibj. Then∑

j,n

kjnAnAj =∑j,n

kjn(an + ibn)(aj − ibj) =∑j,n

kjn(anaj + bnbj) > 0 , (3.62)

4To be precise xj are given by the real part of (3.58).5For systems with additional symmetries some of these characteristic frequencies may coincide.6Let us mention that if (kjn) is not positive-definite but it is nondegenerate then some ω2

α < 0 and thesystem is unstable. The corresponding modes are called tachyonic. If (kjn) is degenerate then some ω2

α = 0and the centre of mass of the system can move with constant velocities in the corresponding directions whichare called flat.

37

where we used that (kjn) is symmetric: knj = kjn.

Having found ωα we substitute each of them in (3.59) and find the corresponding A(α)n . If

all ωα are different then the rank of(−ω2

αmjn + kjn)

is equal to s− 1, and the solution A(α)n is

proportional to the minors ∆nα of(− ω2

αmjn + kjn): A

(α)n ∼ ∆nα. The general solution is then

the superposition of all these solutions

xn = <s∑

α=1

∆nαCαeiωαt =

s∑α=1

∆nαΘα , n = 1, 2, . . . , s , (3.63)

whereΘα = <

(Cαe

iωαt), α = 1, 2, . . . , s . (3.64)

It is a superposition of s simple periodic oscillations Θ1,Θ2, . . . ,Θs with arbitrary amplitudesand phases but definite frequencies determined by the properties of the mechanical system.

The equations xn =∑s

α=1 ∆nαΘα can be considered as a set of equations for s new gen-eralised coordinates Θα which are called the normal coordinates. According to (3.64), theyexecute simple periodic oscillations called normal oscillations of the system, and satisfy theequations

Θα + ω2αΘα = 0 , α = 1, 2, . . . , s . (3.65)

Thus, in terms of Θα the eom become a set of decoupled harmonic oscillator equations.

It is clear that the Lagrangian which leads to eom (3.65) must be a sum of the harmonicoscillators Lagrangians

L =1

2

∑α

mα(Θ2α − ω2

αΘ2α) , mα > 0 . (3.66)

This form of L means that the change of coordinates xn =∑s

α=1 ∆nαΘα simultaneously diag-onalises both the kinetic and potential quadratic forms.

It is also convenient to rescale the normal coordinates Θα

Θα =1√mα

Qα , α = 1, 2, . . . , s , (3.67)

so that L takes the form

L =1

2

s∑α=1

(Q2α − ω2

αQ2α) . (3.68)

In particular this form makes obvious that in the degenerate case where some of the normalfrequencies are equal7 the mechanical system has additional rotational symmetry. Indeed as-suming that the first r frequencies are equal, the Lagrangian (3.68) becomes

L =1

2

r∑α=1

(Q2α − ω2

1Q2α) +

1

2

s∑α=r+1

(Q2α − ω2

αQ2α) . (3.69)

It is clear that it is invariant under SO(r) which rotates the coordinates Q1, Q2, . . . , Qr.

7In this case the coefficients ∆nα in (3.63) are not the minors because they vanish. They are independentsolutions to (3.59).

38

The analysis of forced oscillations also simplifies by using the normal coordinates. TheLagrangian

L =1

2mjnxjxn −

1

2kjnxjxn + Fj(t)xj , (sum over repeated indices!) , (3.70)

in the normal coordinates takes the form

L =s∑

α=1

(1

2Q2α −

1

2ω2αQ

2α + fα(t)Qα

), (3.71)

where

fα(t) =s∑j=1

1√mα

Fj(t)∆jα . (3.72)

The eom areQα + ω2

αQα = fα(t) , (3.73)

which are s independent equations for unknown functions Qα(t). Thus, in normal coordi-nates the motion of a system of s coupled oscillators reduces to the motion of one-dimensionaloscillator.

3.3.2 Normal frequencies and coordinates from Lagrangian

It is not necessary to solve eom to find the normal frequencies and coordinates. one can justuse the Lagrangian and a chain of simple coordinate transformations. We begin by rewritingthe Lagrangian L (3.70) as follows

L =1

2mjnxjxn −

1

2kjnxjxn + Fj(t)xj =

1

2XTMX − 1

2XTKX + F TX . (3.74)

Here we used the usual matrix product, and the following columns, rows and matrices

X =

x1

x2...xs

, F =

F1

F2...Fs

, M =

m11 m12 . . . m1s

m21 m22 . . . m2s...

......

...ms1 ms2 . . . mss

, K =

k11 k12 . . . k1s

k21 k22 . . . k2s...

......

...ks1 ks2 . . . kss

,

(3.75)and AT for any A denotes the matrix transposed to A (i.e. XT is the column (x1, . . . , xs)).Both M and K are real symmetric. As we know from Hamilton’s principle, M must be positive-definite. If X = 0 is stable equilibrium K is positive-definite too. For the sake of generality weconsider arbitrary symmetric K and discuss what happens if it is not positive-definite at theend of this subsection.

We want to bring L to the form (3.71). It can be done in the following three steps

1. Diagonalise M by using an orthogonal matrix A: ATA = I

M = AµAT , µ = diag (µ1, . . . , µs) , µi > 0 , i = 1, . . . , s . (3.76)

L takes the form

L =1

2XTAµAT X − 1

2XTKX + F TX , (3.77)

39

and it is obvious that the transformation

ATX = Y ⇐⇒ X = AY , (3.78)

brings it to the form

L =1

2Y TµY − 1

2Y TATKAY + F TAY , (3.79)

2. Rescale Y so that Y TµY = ZT Z

Z =√µY ⇐⇒ Y =

1√µZ ,

√µ = diag (

√µ1, . . . ,

√µs) . (3.80)

In components zi =√µi yi, i = 1, . . . , s. The rescaling brings L to the form

L =1

2ZT Z − 1

2ZT 1√µATKA

1√µZ + F TA

1√µZ

=1

2ZT Z − 1

2ZT KZ + F TA

1√µZ ,

(3.81)

where

K =1√µATKA

1√µ, (3.82)

is symmetric: KT = K.

3. Diagonalise K by using an orthogonal matrix B: BTB = I

K = BκBT , κ = diag (κ1, . . . , κs) . (3.83)

L takes the form

L =1

2ZT Z − 1

2ZTBκBTZ + F TA

1√µZ , (3.84)

and it is obvious that the transformation

BTZ = Q ⇐⇒ Z = BQ , (3.85)

brings it to the desired form

L =1

2QTBTBQ− 1

2QTκQ+ F TA

1√µBQ

=1

2QT Q− 1

2QTκQ+ F TQ

=s∑

α=1

(1

2Q2α −

1

2καQ

2α + fα(t)Qα

),

(3.86)

where

F T = F TA1√µB = (f1(t), . . . , fs(t)) . (3.87)

Combining the transformations we also get he relations between X and the normal coor-dinates Q

X = A1√µBQ ⇐⇒ Q = BT√µATX . (3.88)

40

Few comments are in order

• If K is positive-definite then all κα > 0, and normal frequencies are given by ω2α = κα.

• If K is degenerate, i.e. detK = 0 then some of κα are equal to zero. Let for exampleκ1 = 0. In the absence of the external force f1(t) = 0, the normal coordinate Q1 iscyclic, and therefore the momentum p1 conjugate to Q1 is conserved. Thus the centre ofmass of the system moves with a constant velocity in the Q1 direction which is called flatdirection. If all the other κα are positive then in the reference frame where the velocityis 0 the system undergoes the usual oscillatory motion in the remaining directions.

• If some κα are negative, say κ1 < 0, then the system is unstable, and the correspondingnormal coordinate Q1 is called tachyonic. In the absence of the external force f1(t) = 0,the general solution of eom

Q1 + κ1Q1 = 0 , (3.89)

is

Q1 = C−e−√−κ1 t + C+e

+√−κ1 t , (3.90)

and therefore unless we tune the initial conditions so that C+ = 0, the normal coordinateincreases exponentially fast. The system is still unstable in the past.

Let us now recall how to diagonalise a symmetric matrix. Let S be an s × s symmetricmatrix, and let D = diag (d1, . . . , ds) be the diagonal matrix of its eigenvalues. Then there isan orthogonal matrix A such that S = ADAT . Let us view A as a collection of s columns Aiand AT as a collection of s rows ATi

A =

| | . . . |A1 A2 . . . As| | . . . |

, Ai =

a1i...asi

, AT =

––– AT1 –––––– AT2 –––

......

...––– ATs –––

. (3.91)

Now multiplying the equation S = ADAT by A from the right we get

SA = AD ⇒ SAi = diAi . (3.92)

Thus for each i = 1, . . . , s the column Ai is an eigenvector of S with the eigenvalue di. Since Ais orthogonal the columns Ai are orthonormal: ATA = I ⇒ ATi Aj = δij. Therefore finding Areduces to finding a set of orthonormal eigenvectors of S. Let us also note that by using (3.91)we get the spectral decomposition of S

S = ADAT =

| | . . . |A1 A2 . . . As| | . . . |

d1 0 . . . 00 d2 . . . 0...

......

...0 0 . . . ds

––– AT1 –––––– AT2 –––

......

...––– ATs –––

= d1A1A

T1 + d2A2A

T2 + · · · dsAsATs =

s∑i=1

diPi , Pi = AiATi ,

(3.93)

where Pi are the projection operators which satisfy

s∑i=1

Pi = I , PiPj = δijPj , no summation over j . (3.94)

41

Example

Let S be equal to8

S =

2 −1 0−1 2 −10 −1 2

. (3.95)

First we find the eigenvalues of S by solving the characteristic equation

det(S − dI) =

∣∣∣∣∣∣2− d −1 0−1 2− d −10 −1 2− d

∣∣∣∣∣∣ = (2− d)((2− d)2 − 1)− (2− d)

= (2− d)(d2 − 4d+ 2) = 0 .

(3.96)

Thus, the eigenvalues of S are

d1 = 2−√

2 , d2 = 2 , d3 = 2 +√

2 . (3.97)

1. Consider d1 = 2−√

2

(S − d1I)A1 =

√2 −1 0

−1√

2 −1

0 −1√

2

a11

a21

a31

= 0 ⇒ a21 =√

2 a11 , a31 = a11 .

(3.98)The normalisation condition then gives

AT1A1 = 1 ⇒ 4a211 = 1 ⇒ a11 = ±1

2. (3.99)

Choosing the plus sign we get

AT1 =

(1

2,

1√2,

1

2

). (3.100)

2. Consider d2 = 2

(S − d2I)A2 =

0 −1 0−1 0 −10 −1 0

a12

a22

a32

= 0 ⇒ a22 = 0 , a32 = −a12 . (3.101)

The normalisation condition then gives

AT2A2 = 1 ⇒ 2a212 = 1 ⇒ a12 = ± 1√

2. (3.102)


AT2 =

(1√2, 0 , − 1√

2

). (3.103)

Obviously, it is orthogonal to A1.

8Note that S is the Cartan matrix of the Lie algebra A3 which is the Lie algebra of the Lie group SL(4,C).

42

3. Consider d3 = 2 +√

2

(S − d3I)A3 =

−√2 −1 0

−1 −√

2 −1

0 −1 −√

2

a13

a23

a33

= 0 ⇒ a23 = −√

2 a13 , a33 = a13 .

(3.104)The normalisation condition then gives

AT3A3 = 1 ⇒ 4a213 = 1 ⇒ a13 = ±1

2. (3.105)


AT3 =

(1

2, − 1√

2,

1

2

). (3.106)

Obviously, it is orthogonal to A1 and A2.

Thus,

A =

12

1√2

12

1√2

0 − 1√2

12− 1√

212

, D =

2−√

2 0 00 2 0

0 0 2 +√

2

. (3.107)

3.4 Anharmonic oscillations

Expanding ajn(q) and U(q) in (3.53) in powers of xj = qj − qj0, and keeping higher-order termsin xj, one gets a Lagrangian which describes anharmonic or nonlinear oscillations.

3.4.1 Anharmonic oscillations in one dimension

The method we use is a version of the Poincare-Lindstedt expansion method which can beapplied to some multi-dimensional cases.

Consider a system with the Lagrangian (3.1) which at q = q0 is at stable equilibrium. ThenU ′(q0) = 0, m = a(q0) > 0 , k = U ′′(q0) > 0. Let us introduce a new coordinate x = x(q)(similar to an arc length parameter) such that x = 0 corresponds to q = q0, and

a(q)q2 = mx2 . (3.108)

Then the Lagrangian (3.1) takes the form

L =1

2mx2 − U(x) , (3.109)

where the potential U can be expanded in powers of x

U(x) = U(0) +1

2kx2 +

1

3g3x

3 +1

4g4x

4 + · · · . (3.110)

Here the constants g2, g3, ... are proportional to derivatives of U at x = 0 and are called couplingconstants. Introducing

ω0 =

√k

m, α =

g3

m, β =

g4

m, (3.111)

43

and omitting U(0), one gets the Lagrangian up to fourth order in x4

L =1

2mx2 − 1

2mω2

0x2 − 1

3mαx3 − 1

4mβx4 . (3.112)

The eom isx+ ω2

0x = −αx2 − βx3 . (3.113)

The motion is periodic with the period T = 2π/ω, and therefore x(t) can be expanded in aFourier series

x(t) = x0 +∞∑k=1

(xk cos(kωt) + yk sin(kωt)

). (3.114)

Imposing the initial condition v(0) = x(0) = 0, and taking into account that with this initialcondition if x(t) is a solution then x(−t) is a solution too, we see that all yk = 0. Thus, withoutloss of generality we can look for a solution of the form

x(t) =∞∑k=0

xk cos(kωt) . (3.115)

The frequency ω and the Fourier coefficients xk depend on the energy E of the system

E =1

2mω2

0a2R +

1

3mαa3

R +1

4mβa4

R , (3.116)

where aR = x(0) =∑∞

k=0 xk, which we assume to be positive aR > 0, is one of the amplitudesof the oscillations (the other one is aL = x(T/2)). We want to analyse small corrections toharmonic oscillations at α = β = 0. To give a precise meaning to the notion of smallness weneed to use some dimensionless quantities. Taking into account that the dimension of α is1/cm/sec2, and the dimension of β is 1/cm2/sec2, we see that

aRα

ω20

anda2Rβ

ω20

, (3.117)

are dimensionless, and the approximation of harmonic oscillations is applicable if aRαω20<< 1,

a2Rβ

ω20<< 1. Then, corrections to harmonic oscillations are obtained by expanding ω and xk

in powers of aRαω20

,a2Rβ

ω20

. Computationally, however, it is easier to use the following expansionparameters

ε1 =aα

ω20

, ε2 =a2β

ω20

, (3.118)

where a = x1 is the amplitude of the harmonic oscillations with frequency ω. Obviously, theyare equivalent to (3.117), in particular aR → a as ε1, ε2 → 0. It is also clear that since ε1 ∼ aand ε2 ∼ a2 the order ε21 is the same as the order ε2.

To find the solution it is convenient to rescale x and t as follows

t =τ

ω0

, x(t) = a z(τ) , (3.119)

so that both z and τ are dimensionless. Then, it is easy to check that the eom (3.113) takesthe form

d2z

dτ 2+ z = −ε1z2 − ε2z3 . (3.120)

44

It is also useful to represent a solution in the complex form

z(τ) =∞∑

k=−∞

zkeikγτ = z0 + 2

∞∑k=1

zk cos(kγτ) , ∀k z−k = zk , γ =ω

ω0

, (3.121)

where the solution we are looking for has z1 = 12

= z−1. Then, the Fourier series of z(τ)2 andz(τ)3 are given by

z(τ)2 =∞∑

l,m=−∞

zlzmei(l+m)γτ , z(τ)3 =

∞∑l,m,n=−∞

zlzmznei(l+m+n)γτ . (3.122)

Substituting these formulae in (3.120), we get

∞∑k=−∞

(1− k2γ2)zkeikγτ = −ε1

∞∑l,m=−∞

zlzmei(l+m)γτ − ε2

∞∑l,m,n=−∞

zlzmznei(l+m+n)γτ , (3.123)

and therefore the Fourier coefficients satisfy the equations

(1− k2γ2)zk = −ε1∑l+m=k

zlzm − ε2∑

l+m+n=k

zlzmzn , k ∈ Z . (3.124)

Since z−k = zk it is sufficient to consider nonnegative k only. Let us analyse the equations.

• k = 0z0 = −ε1

∑l+m=0

zlzm − ε2∑

l+m+n=0

zlzmzn . (3.125)

Since z−1 = z1 = 1/2 while all the other zk’s are at least of first order in ε1 we get thatthe leading term in z0 is obtained if l = −m = ±1, and therefore

z0 = −1

2ε1 +O(ε31) . (3.126)

• k = 2(1− 4γ2)z2 = −ε1

∑l+m=2

zlzm − ε2∑

l+m+n=2

zlzmzn . (3.127)

Since z−1 = z1 = 1/2 while all the other zk’s are at least of first order in ε1 we get thatthe leading term in z2 is obtained if l = m = 1, and therefore

z2 =1

12ε1 +O(ε31) . (3.128)

• k = 3(1− 9γ2)z3 = −ε1

∑l+m=3

zlzm − ε2∑

l+m+n=3

zlzmzn . (3.129)

It is clear that the rhs is of order ε21 ∼ ε2. The leading term in z3 is obtained if in the firstterm on the rhs l = 1,m = 2 or l = 2,m = 1, and if in the second term l = m = n = 1,and therefore

z3 =1

8(2ε1z1z2 + ε2z

31) +O(ε41) =

1

32(1

3ε21 +

1

2ε2) +O(ε41) . (3.130)

45

• k ≥ 4

(1− k2γ2)zk = −ε1∑l+m=k

zlzm − ε2∑

l+m+n=k

zlzmzn . (3.131)

The consideration above can be now easily generalised, and It is clear that the rhs, andtherefore zk, is of order εk−1

1 . Thus, for all k ∈ Z but k = 0 the Fourier coefficients zk are

of order ε|k|−11 : zk ∼ ε

|k|−11 , k 6= 0, z0 ∼ ε1. Thus, the leading term in zk is obtained by

taking sums over positive l,m, n, and can be written explicitly as

zk =1

k2 − 1

(ε1

k−1∑l=1

zlzk−l + ε2

k−2∑l,m=1l+m<k

zlzmzk−l−m)

+O(εk+11 ) . (3.132)

• Let us now use the zk’s to find the leading correction to γ2. It is found from the equationfor k = 1

(1− γ2)1

2= −ε1

∑l+m=1

zlzm − ε2∑

l+m+n=1

zlzmzn . (3.133)

The rhs is of order ε21 the leading term in γ2 − 1 is obtained only if l,m, n = −1, 0, 1, 2.Thus, we get

γ2 = 1 + 2(ε1(2z0z1 + 2z−1z2) + 3ε2z−1z1z1

)+O(ε41)

= 1 + 2(− 5

12ε21 +

3

8ε2)

+O(ε41) ,(3.134)

and therefore

γ = 1− 5

12ε21 +

3

8ε2 +O(ε41) . (3.135)

The formulae above for z0, z2, z3 and γ provide us with the leading corrections up to order ε21to the harmonic oscillations. Collecting all the formulae, we get

x(t) = a cosωt+a

2ε1(−1 +

1

3cos 2ωt) +

a

16(1

3ε21 +

1

2ε2) cos 3ωt ,

ω = ω0γ = ω0(1− 5

12ε21 +

3

8ε2) .

(3.136)

It is straightforward to calculate subleading corrections with the help of a computer.

3.4.2 Anharmonic oscillations in several dimensions

Anharmonic oscillations in several dimensions are much more complicated then the ones in onedimension, and they can lead to a chaotic behaviour of the mechanical system. In what followswe’ll discuss only the case where the chaotic motion is absent.

Consider again a system with s degrees of freedom and the Lagrangian

L =1

2gjn(q)qj qn − U(q) . (3.137)

Let the system be at stable equilibrium at qj = qj0, j = 1, . . . , s. Then, ∂U(q0)∂qj

= 0, j = 1, . . . , s,

and(∂2U(q0)∂qj∂qn

)and

(gjn(q0)

)are positive-definite. As was mentioned gij can be thought of as a

metric tensor on the configuration space of the mechanical system. In the course of differential

46

geometry it is proven that one can choose new coordinates xj (which are also called normalcoordinates) such that x = 0 corresponds to q = q0 and

gjn(q)qj qn = δjnxjxn +

1

3Rjklnx

kxlxjxn +O(x3x2) , (3.138)

where Rjkln are constants equal to the components of the Riemann curvature tensor in xj

coordinates at x = 0.

Thus, expanding L up to quartic order in x and x, and switching to normal coordinates Qα,we find

L =∑α

(1

2Q2α −

1

2ω2αQ

2α

)− 1

3

∑α,β,γ

g(3)αβγQαQβQγ

− 1

4

∑α,β,γ,ρ

g(4)αβγρQαQβQγQρ +

1

6

∑α,β,γ,ρ

RαβγρQαQβQγQρ ,(3.139)

where the constants g(3)αβγ and g

(4)αβγρ are symmetric in their indices.

To simplify the formulae let us keep only the cubic term. Then the eom are

Qα + ω2αQα = −g(3)

αβγQβQγ , α = 1, . . . , s , (3.140)

where we sum over β and γ. The small expansion parameters now are

εαβγ = εαγβ =g

(3)αβγaβaγ

ω2αaα

, (3.141)

where aα are amplitudes of the harmonic oscillations Qα = aα cos(ωαt+ ϕα).

To solve these equations perturbatively we again assume that the normal frequencies ωα ofharmonic oscillations are replaced by the exact frequencies Ωα

Ωα = ωα(1 + εβγργα,βγρ + · · · ) , (3.142)

where we sum over β, γ and ρ. We also assume that the only solution of ~k · ~Ω = 0 is ~k = 0where ~k = (k1, . . . , ks) is a vector of integers , and ~Ω = (Ω1, . . . ,Ωs) is a vector of frequencies.Then, we look for a solution of eom (3.140) of the form

Qα =∑

k1,...,ks

Aα,k1k2···kseik1Ω1t+ik2Ω2t+···+iksΩst =

∑~k

Aα,~kei~k·~Ω t . (3.143)

The reality of Qα then leads toAα,−~k = Aα,~k . (3.144)

Substituting (3.143) in the eom (3.140), one gets the equations for the coefficients Aα,~k

(ω2α − (~k · ~Ω)2)Aα,~k = −

∑β,γ

∑~l+~m=~k

g(3)αβγAβ,~lAγ,~m . (3.145)

The analysis of these equations in principle follows the one-dimensional case. However, thesolution depends on assumptions made about the frequencies ωα, in particular in what follows weassume that the only solution of ~k·~ω = 0 is ~k = 0. Note that this condition is independent of thesame condition on Ωα because even if the frequencies Ωα satisfy the condition the unperturbedfrequencies ωα may not. If these conditions are satisfied then as we will see the only amplitudesAα,~k of order 1 are Aα,±~eα ≡ A

(±)α = 1

2aαe

±iϕα where the only nonvanishing component of thebasis vector ~eα is (~eα)α = 1, e.g. ~e3 = (0, 0, 1, 0, . . . , 0).

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• ~k = ~0ω2αAα,~0 = −

∑β,γ

∑~l+~m=~0

g(3)αβγAβ,~lAγ,~m . (3.146)

Since Aα,~eα ≡ Aα while all the other amplitudes are at least of first order in εαβγ we get

that the leading term in Aα,~0 is obtained if γ = β and ~l = −~m = ±~eβ, and therefore

ω2αAα,~0 = −1

2

∑β

g(3)αββa

2β +O(ε2αβγ) ⇒ Aα,~0 = −1

2

∑β

g(3)αββa

2β

ω2α

+O(ε2αβγ) ⇒

Aα,~0 = −1

2aα∑β

εαββ +O(ε2αβγ) .

(3.147)

Note that Aα,~0 is of order εαβγ because ~0 can be written as a linear combination of twobasis vectors ~eρ with coefficients ±1. Consider then the general case of this type

• ~k = ερ~eρ + εµ~eµ, ε2ρ = 1 ∀ρ, ρ, µ = 1, . . . , s

(ω2α − (ερΩρ + εµΩµ)2)Aα,ερ~eρ+εµ~eµ = −

∑β,γ

∑~l+~m=ερ~eρ+εµ~eµ

g(3)αβγAβ,~lAγ,~m .

(3.148)

There are two cases to consider

1. ρ 6= µ

We can have β = ρ,~l = ερ~eρ, γ = µ, ~m = εµ~eµ or γ = ρ, ~m = ερ~eρ, β = µ,~l = εµ~eµ.Thus

(ω2α − (ερωρ + εµωµ)2)Aα,ερ~eρ+εµ~eµ = −1

2g(3)αρµaρaµe

iερϕρ+iεµϕµ +O(ε2αβγ)⇒

Aα,ερ~eρ+εµ~eµ = −1

2aα

ω2αεαρµ

ω2α − (ερωρ + εµωµ)2

eiερϕρ+iεµϕµ +O(ε2αβγ) .(3.149)

Note that according to our assumption, ωα 6= ερωρ + εµωµ for any α, ρ, µ, ερ, εµ.

2. ρ = µ

We can have β = γ = ρ,~l = ~m = ερ~eρ. Thus

(ω2α − 4ω2

ρ)Aα,2ερ~eρ = −1

4g(3)αρρaρaρe

2iερϕρ +O(ε2αβγ)⇒

Aα,2ερ~eρ = −1

4aα

ω2αεαρρ

ω2α − 4ω2

ρ

e2iερϕρ +O(ε2αβγ) .(3.150)

Next we consider amplitudes of order e2αβγ. They are given by ~k which are of the form

~k =∑3

i=1 ερi~eρi .

• ~k = ερ~eρ, ρ = 1, . . . , s

(ω2α − Ω2

ρ)Aα,ερ~eρ = −∑β,γ

∑~l+~m=ερ~eρ

g(3)αβγAβ,~lAγ,~m . (3.151)

There are two cases to consider

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1. α 6= ρ

Then we take ~l = εβ~eβ, ~m = ερ~eρ − εβ~eβ or ~m = εγ~eγ, ~l = ερ~eρ − εγ~eγ, and get

(ω2α − ω2

ρ)Aα,ερ~eρ = −∑β,γ

g(3)αβγaβAγ,ερ~eρ−εβ~eβe

iεβϕβ +O(ε3αβγ)⇒

Aα,ερ~eρ = −∑εβ ,β,γ

ω2αεαβγ

ω2α − ω2

ρ

1

aγAγ,ερ~eρ−εβ~eβe

iεβϕβ +O(ε3αβγ) ,(3.152)

where Aγ,ερ~eρ−εβ~eβ are given by (3.147), (3.149), (3.150).

2. α = ρ

This allows one to determine the leading correction to Ωα

(ω2α − Ω2

α)Aα,εα~eα = −∑β,γ

∑~l+~m=εα~eα

g(3)αβγAβ,~lAγ,~m . (3.153)

We take ~l = εβ~eβ, ~m = εα~eα − εβ~eβ or ~m = εγ~eγ, ~l = εα~eα − εγ~eγ, and get

(1− Ω2α

ω2α

)1

2eiεαϕα = −

∑β,γ

εαβγ1

aγAγ,εα~eα−εβ~eβe

iεβϕβ ⇒

Ω2α

ω2α

= 1 + 2∑εβ ,β,γ

εαβγ1

aγAγ,εα~eα−εβ~eβe

iεβϕβ−iεαϕα +O(ε3αβγ) .

(3.154)

It is clear that one can proceed and determine leading and subleading corrections to any am-plitude and frequency. The resulting series may be divergent. The conditions for it to beconvergent are discussed in the framework of the Kolmogorov-Arnold-Moser theory.

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