Classical Mechanics

LECTURE NOTES ON CLASSICAL MECHANICS

Jos Thijssen

Spring 2012

Copyright © 2012 by TU DelftAn electronic version of these notes is available athttp://blackboard.tudelft.nl/.

http://blackboard.tudelft.nl/

PREFACE

These notes have been developed over several years for use with the courses Classical andQuantum Mechanics A and B, which for several years have been part of the third year ap-plied physics degree program at Delft University of Technology. Part of these notes stem fromcourses which I taught at Cardiff University of Wales, UK. Nowadays, in the Delft programme,the subjects of classical and quantum mechanics have been separated again, and I have ex-tracted the part dealing with classical mechanics into this set of lecture notes.

These notes are intended to be used alongside standard textbooks. Several texts can beused, such as the books by Morin (Introduction to Classical Mechanics, Cambridge UniversityPress, 2008), Hand and Finch (Analytical Mechanics, Cambridge University Press, 1999) andGoldstein (Classical Mechanics, third edition, Addison Wesley, 2004), the older book by Cor-ben and Stehle (Classical Mechanics, second edition, Dover, 1994, reprint of 1960 edition),and the textbook by Kibble and Berkshire, (Classical Mechanics, 5th edition, World Scientific,2004). You should be aware that consultation of one or more of the texts mentioned is essen-tial for a thorough understanding of this field.

In the appendix there is a large problem set, which is more essential than the notes them-selves. There are many things in life which you can only learn by doing it yourself. Nobodywould seriously believe you can master any sport or playing a musical instrument by readingbooks. For physics, the situation is exactly the same. You have to learn the subject by doingit yourself – even by failing to solve a difficult problem you learn a lot, since in that situationyou start thinking about the structure of the subject.

In writing these notes I had numerous discussions with and advice from Herre van derZant and Miriam Blaauboer. Part of the problems were chosen and developed or formulatedby Miriam, Wim Bouwman and Ad Verbrugge. Wim Bouwman has contributed to the finaledit. Jos Seldenthuis has provided the LATEX style files with which these notes were formatted.I hope the resulting set of notes and problems will help students learn and appreciate thebeautiful theory of classical mechanics.

Delft, January 2012

iii

CONTENTS

Preface iii

1 Introduction: Newtonian mechanics and conservation laws 11.1 Newton’s laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Systems of point particles – symmetries and conservation laws . . . . . . . . . . 3

2 Lagrange and Hamilton formulations 92.1 Generalised coordinates and virtual displacements . . . . . . . . . . . . . . . . . 102.2 d’Alembert’s principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.3.1 The pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.3.2 The block on the inclined plane . . . . . . . . . . . . . . . . . . . . . . . . 142.3.3 Heavy bead on a rotating wire . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.4 d’Alembert’s principle in generalised coordinates . . . . . . . . . . . . . . . . . . 172.5 Conservative systems – the mechanical path . . . . . . . . . . . . . . . . . . . . . 182.6 Summary and examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.6.1 Overview of the theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.6.2 A system of pulleys . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.6.3 Example: the spinning top . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.7 Non-conservative forces – charged particle in an electromagnetic field . . . . . 262.7.1 Charged particle in an electromagnetic field . . . . . . . . . . . . . . . . . 26

2.8 Hamilton mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.9 Applications of the Hamiltonian formalism . . . . . . . . . . . . . . . . . . . . . . 32

2.9.1 The three-pulley system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.9.2 The spinning top . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.9.3 Charged particle in an electromagnetic field . . . . . . . . . . . . . . . . . 34

3 The two-body problem 353.1 Formulation and analysis of the two-body problem . . . . . . . . . . . . . . . . . 363.2 Solution of the Kepler problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4 Examples of variational calculus, constraints 434.1 Variational problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444.2 The brachistochrone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444.3 Fermat’s principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464.4 The minimal area problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474.5 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4.5.1 Constraint forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474.5.2 Global constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

5 Symmetry and conservation laws 535.1 Noether’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 545.2 Liouville’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

v

vi CONTENTS

6 Systems close to equilibrium 596.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 606.2 Analysis of a system close to equilibrium . . . . . . . . . . . . . . . . . . . . . . . 61

6.2.1 Examples: a spring system and the double pendulum . . . . . . . . . . . 626.3 Normal modes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

6.3.1 Normal coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 656.4 Vibrational analysis in molecules . . . . . . . . . . . . . . . . . . . . . . . . . . . . 676.5 The chain of particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 706.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

A Problems and Exercises Classical Mechanics 73

B Antwoorden 85

1INTRODUCTION: NEWTONIAN

MECHANICS AND CONSERVATION LAWS

In this lecture course, we shall introduce some mathematical techniques for studying prob-lems in classical mechanics and apply them to several systems. In a previous course, youhave already met Newton’s laws and some of its applications. In this chapter, we briefly re-view the basic theory, and consider Newton’s laws in some detail. Furthermore, we considerconservation laws of classical mechanics which are connected to symmetries of forces, andderive these conservation laws from Newton’s laws.

1

1

2 1. INTRODUCTION: NEWTONIAN MECHANICS AND CONSERVATION LAWS

1.1 NEWTON’S LAWSThe aim of a mechanical theory is to predict the motion of objects. It is convenient to startwith point particles which have no dimensions. The trajectory of such a point particle isdescribed by its position at each time. Denoting the spatial position vector by r, the trajectoryof the particle is given as r(t ), a three-dimensional function depending on a one-dimensionalcoordinate: the time. The velocity is defined as the time-derivative of the vector r(t ), and byconvention it is denoted as r(t ):

r(t ) = d

d tr(t ), (1.1)

and the acceleration a is defined as the second derivative of the position vector with respectto time:

a(t ) = r(t ). (1.2)

The last concept we must introduce is that of momentum p: it is defined as

p = mr(t ), (1.3)

where m is the mass. Although we have an intuitive idea about the meaning of mass, this isalso a rather subtle physical concept, as is clear from the frequent confusion of mass with theconcept of weight (see below).

Now let us state Newton’s laws:

1. A body not influenced by any other matter will move at constant velocity

2. The rate of change of momentum of a body is equal to the force, F:

dp

d t= F(r, t ). (1.4)

3. When a particle exerts a force F on another particle, then the other particle exerts aforce on the first particle which is equal in magnitude but opposite in direction to theforce F – these forces are directed along the line connecting the two particles. Denotingthe particle by indices 1 and 2, and the force exerted on 1 by 2 by F1,2 and the forceexerted on 2 by 1 by F2,1, we have:

F1,2 =−F2,1 =±F1,2r1,2. (1.5)

where r1,2 is a unit vector pointing from r1 to r2. The ± denotes whether the force isrepulsive (−) or attractive (+).

Some remarks about these laws are in place. It is questionable whether the second lawis really a statement, as a new vector quantity, called ‘force’, is introduced, which is not yetdefined. Only if we know the force, we can predict how a particle will move. In that sense,a real ‘law’ is only formed by combining Newton’s second law together with an explicit ex-pression for the force. In table 1.1, known forces are given for several systems. Note that theforce generally depends on the position r, on the velocity r, and also explicitly on time (e.g.when an external, time-varying field is present). An implicit dependence on time is furtherprovided by the time dependence of the position vector r(t ).

In most cases, the mass is taken to be constant, although this is not always true: you maythink of a rocket burning its fuel, or disposing of its launching system, or bodies moving ata speed of the order of the speed of light, where the mass deviates from the rest mass. Withconstant mass, the second law reads:

mr(t ) = F(r, t ). (1.6)

In fact, the second law disentangles two ingredients of the motion. One is the mass m,which is a property of the moving particle which is acted upon by the force, and the other the

1.2. SYSTEMS OF POINT PARTICLES – SYMMETRIES AND CONSERVATION LAWS 3

1TABLE 1.1: Forces for various systems. The symbol mi stand for the mass of point particle i , qi stands for electriccharge of particle i , B is a magnetic, and E an electric field. G , ε0 and g are known constants. The gravitationaland the electrostatic forces are directed along the line connecting the two particles i = 1,2.

Forces in natureSystem ForceGravity FG =GmM 1

r 2

Gravity near earth’s surface Fg =−mg zElectrostatics FC = 1

4πε0q1q2

1r 2

Particle in an electromagnetic field FEM = q (E+ r×B)Air friction Ffr =−γr

force, which itself arises from some external origin. In the case of gravitational interaction,the force depends on the mass, which drops out of the equation of motion. Generally, masscan be described as the resistance to velocity change, as the second law states that the largerthe mass, the smaller the change in velocity (for the same force). It is an experimental fact thatthe mass which enters the expression for the gravitational force is the same as this universalquantity mass, which occurs for any force in the equation of motion. The weight is the gravityforce acting on a body.

Usually, the first law is phrased as follows: ‘when there is no force acting on a point par-ticle, the particle moves at constant velocity’. This statement obviously follows from the sec-ond law by taking F = 0. The formulation adopted above (page 2) emphasises that force hasa material origin. It is impossible to fulfill the requirements of this law, as everywhere in theuniverse gravitational forces are present: the first law is an idealisation. The first law is notobvious from everyday life, where it is never possible to switch friction off completely: ineveryday life, motion requires a force in order to be maintained.

The third law is a statement about forces. It turns out that this statement does not hold ex-actly, as the forces of this statement should act simultaneously. In quantum field theory, parti-cles travelling between the interacting particles are held responsible for the interactions, andthese particles cannot travel at a speed faster than that of light in vacuum (about 3 ·108m/s).However, for everyday life mechanics, the third law holds to sufficient precision, unless themoving particles carry a charge and interact through electromagnetic interactions. In thatcase, the force acts no longer along the line connecting the two particles.

1.2 SYSTEMS OF POINT PARTICLES – SYMMETRIES AND CONSERVATION

LAWS

Real objects which we describe in mechanics are not point particles, but to very good accu-racy they can be considered as large collections of interacting point particles – in this sectionwe consider systems consisting of N point particles. It is possible to disentangle the mutualforces acting between these particles from the external ones. The mutual forces satisfy New-ton’s third law: for every force Fi , j , which is the force exerted by particle j on particle i , theforce F j ,i is equal in magnitude but opposite in direction to Fi , j . For a particle i , we considerall the mutual forces Fi , j for j 6= i – the remaining forces on i must then be due to externalsources (i.e., not depending on the other particles in our system), and we lump these forcestogether in one external force, FExt

i :

Fi =N∑

j=1; j 6=iFi , j +FExt

i , i = 1, . . . , N . (1.7)

1


The equations of motion read:

mi ri =N∑

j=1; j 6=iFi , j +FExt

i . (1.8)

The total momentum of the system is the sum of the momenta of all the particles:

p =N∑

i=1pi =

N∑i=1

mi ri . (1.9)

We can view the total momentum of the system as the momentum of a single particle with amass equal to the total mass M of the system, and position vector rC. This position vector isthen defined through:

p = M rC =N∑

i=1mi ri ; M =

N∑i=1

mi . (1.10)

This is equivalent to

rC = 1

M

N∑i=1

mi ri (1.11)

up to an integration constant which is always taken to be zero. The vector rC is called centreof mass of the system. A particle of mass M at the centre of mass (which obviously changes intime) represents the same momentum as the total momentum of the system.

Let us find an equation of motion for the centre of mass. We do this by summing Eq. (1.8)over i :

N∑i=1

mi ri =N∑

i , j=1,i 6= jFi , j +

N∑i=1

FExti . (1.12)

In the first term on the right hand side, for every term Fi , j , there will also be a term F j ,i , butthis is equal in magnitude and opposite to Fi , j ! So the first term vanishes, and we are left with

N∑i=1

mi ri = p =N∑

i=1FExt

i ≡ FExt. (1.13)

We see that the centre of mass behaves as a point particle with mass M subject to the totalexternal force acting on the system.

Conservation of physical quantities, such as energy, momentum etcetera, is always theresult of some symmetry. This deep relation is borne out in a beautiful theorem, formulatedby E. Noether, which we shall consider in the next semester. In this section we shall derivethree conservation properties from Newton’s laws and the appropriate symmetries.

The first symmetry we consider is that of a system of particles experiencing only mutualforces, and no external ones. We then see immediately from Eq. (1.13) with FExt = 0 that p = 0,in other words, the total momentum is conserved.

Conservation of momentumIn a system consisting of interacting particles, not subject to an external force, thetotal momentum is always conserved.

Next, let us consider the angular momentum L. This is a vector quantity, which for par-ticle i is defined as Li = ri ×pi . The total angular momentum L is the sum of the vectors Li :

L =N∑

i=1Li . (1.14)

To see how L varies in time, we calculate the time derivative of Li :

Li = ri ×pi + ri × pi . (1.15)


1r

r Frd

2

1

FIGURE 1.1: Path from r1 to r2. The force at some point along the path is shown, together with the contributionto the work of a small segment dr along the path.

The first term of the right hand side vanishes because pi is parallel to ri , so we are left with

Li = ri × pi = Ni ; (1.16)

Ni is the torque acting on particle i .Now we calculate the torque on the total system by summing over i and replacing pi by

the force (according to the second law):

L =N∑

i=1ri ×Fi =

N∑i=1

ri ×(

N∑j=1, j 6=i

Fi , j +FExti

). (1.17)

The first term in right hand side vanishes, again as a result of the third law:

ri ×Fi , j + r j ×F j ,i = Fi , j ×(ri − r j

)= 0, (1.18)

where the last equality is a result of the direction of Fi , j coinciding with that of the line con-necting ri and r j (which excludes electromagnetic interactions between moving particlesfrom the discussion). We therefore have

L =N∑

i=1ri ×FExt

i . (1.19)

We see that if the external forces vanish, the angular momentum does not change:

Conservation of angular momentumIn a system consisting of interacting particles (not electromagnetic), not subject toan external force, the angular momentum is always conserved.

Finally, we consider the energy. Let us evaluate the work W done by moving a singleparticle from r1 to r2 along some path Γ (see figure 1.1). This is by definition the inner productof the force and the infinitesimal displacements, summed over the path:

W =∫Γ

F ·dr(t ) =∫ t2

t1

F · r d t . (1.20)

Using Newton’s second law, we can write:

W =∫ t2

t1

mrrd t =∫ t2

t1

m

2

d

d t

(r2)d t = m

2

(r2

2 − r21

), (1.21)

where r1 is the velocity at time t1 and similar for r2. We see that from Newton’s second lawit follows that the work done along the path Γ is equal to the change in the kinetic energyT = mr2/2.

1


A conservation law can be derived for the case where F is a conservative and time-independentforce. This means that F can be written as the negative gradient of some scalar function,called the potential:1

F(r) =−∇V (r). (1.22)

In that case we can write the work in a different way:

W =−∫Γ∇V (r)dr(t ) =−

∫ t2

t1

dV (r)

d td t =V (r1)−V (r2). (1.23)

From this and from Eq. (1.21) it follows that

T1 +V1 = T2 +V2, (1.24)

where T1 is the kinetic energy in the point r1 (or at the time t1) etcetera. Thus T +V is aconserved quantity, which we call the energy E .

Of course, now that we know the expression for the energy, we can verify that it is a con-served quantity by calculating its time derivative, using Newton’s second law:

E = mr · r+∇V (r) · r = F · r−F · r = 0. (1.25)

For a many-particle system, the derivation is similar – the condition on the force is thenthat there exists a potential function V (r1,r2, . . . ,rN ), such that the force Fi on particle i isgiven by

Fi =−∇i V (r1,r2, . . . ,rN ). (1.26)

Note that V depends on 3N coordinates – the gradient ∇i acting on V gives a 3-dimensionalvector

∇i V =(∂V

∂xi,∂V

∂yi,∂V

∂zi

). (1.27)

The kinetic energy is the sum of the one-particle kinetic energies. Now the energy conserva-tion is derived as follows:

E =∑i

mi ri · ri +∑

i∇i V · ri =

∑i

(Fi · ri −Fi · ri ) = 0. (1.28)

The function V above depends on time only through the time-dependence of the argu-ments ri . If we consider a charged particle in a time-dependent electric field, this is no longerthe case: then t occurs as an additional, explicit argument in V . If V would depend explicitlyon time, the energy would change at a rate

E = ∂

∂tV (r1,r2, . . . ,rN , t ), (1.29)

where the arguments ri also depend on time (but do not take part in the differentiation withrespect to t ). If V does not depend explicitly on time, we can define the zero of time (i.e. thetime when we set our clock to zero) arbitrarily. This time translation invariance is essentialfor having conservation of energy.

Similarly, the conservation of momentum is related to space translation invariance of thepotential, i.e. this potential should not change when we translate all particles all over thesame vector. Finally, angular momentum is related to rotational symmetry of the potential.In quantum mechanics, all these symmetries lead to the same conserved quantities (or rathertheir quantum mechanical analogues).

1From vector calculus it is known that a necessary and sufficient condition for this to be possible is that the forceis curl-free, i.e. ∇×F = 0.


1There exists a correspondence between symmetries and conservation laws. We haveconsidered the following examples of this correspondence:

• Spatial translation symmetry implies conservation of the total momentum P.

• Rational symmetry implies conservation of angular momentum L.

• If the forces in a system are conservative, time translation symmetry impliesconservation of total energy E .

A final remark concerns the evaluation of the kinetic energy of a many-particle system.As we have seen above, the motion of the centre of mass can be split off from the analysis in asuitable way. This procedure also works for the kinetic energy. Let us decompose the positionvector ri of particle i into two parts: the centre of mass position vector rC and the positionrelative to the centre of mass, which we call r′i :

ri = rC + r′i . (1.30)

As, by definition, rC =∑i mi ri /M , we have∑

imi r′i =

∑i

mi ri −MrC = 0. (1.31)

We can use this decomposition to rewrite the kinetic energy:

T =∑i

mi

2

(rC + r′i

)2 = M

2r2

C + rC ·∑i

mi r′i +∑

i

mi

2r′2i . (1.32)

The second term vanishes as a result of (1.31) and therefore we have succeeded in writing thekinetic energy of the many-particle system as the kinetic energy of the centre of mass plus thekinetic energy of the relative coordinates:

T = TCM +∑i

mi

2r′2i . (1.33)

This formula is a convenient device for calculating the kinetic energy in many applications.

In a many-particle system, it is often useful to split the velocities into two compo-nents: the centre of mass velocity, and the velocities relative to the centre of mass.The kinetic energy can be written as the sum of the centre of mass kinetic energyand the internal kinetic energy, i.e. the kinetic energy associated with the relativevelocities.

2LAGRANGE AND HAMILTON

FORMULATIONS OF CLASSICAL

MECHANICS

9

2

10 2. LAGRANGE AND HAMILTON FORMULATIONS

ϕ

θ

x

y

z

FIGURE 2.1: The pendulum in three dimensions. The position of the mass is described by the two angles ϕ andϑ.

The laws of classical mechanics, formulated by Newton, and the various laws for theforces (see table 1.1) supply sufficient ingredients for predicting the motion of mechanicalsystems in the classical limit. Working out the solution for particular cases is not always easy,however. In this chapter we shall develop an alternative formulation of the laws of classicalmechanics, which renders the analysis of many systems easier than the traditional Newto-nian formulation, in particular when the moving particles are subject to constraints. The newformulation will not only enable us to analyse new applications more easily than using New-ton’s laws, but it also leads to an important example of a variational formulation of a phys-ical theory. Broadly speaking, in a variational formulation, a physical solution is found byminimising a mathematical expression involving a function by varying that function. Manyphysical theories can be formulated in a variational way, in particular quantum mechanicsand electrodynamics.

2.1 GENERALISED COORDINATES AND VIRTUAL DISPLACEMENTSWhen observing motion in everyday life, we often encounter systems in which the movingparticles are subject to constraints. For example, when a car moves on the road, the roadsurface withholds the car from moving downward, as is the case with the balls on a billiardtable. Another example is a particle suspended on a rigid rod (i.e. the pendulum), whichcan only move on the the sphere around the suspension point with radius equal to the rodlength. The constraints are realised by forces, which we call the forces of constraint. The forcesof constraint guarantee that the constraints are met – they often do not influence the motionwithin the subspace.1 The main object of the next few sections is to show that it is possible toeliminate these constraint forces from the description of the mechanical problem.

As the presence of constraints reduces the actual degrees of freedom of the system, itis useful to use a smaller set of degrees of freedom to describe the system. As an example,consider a ball on a billiard table. In that case, the z-coordinate drops out of the description,and we are left with the x and y coordinates only. This is obviously a very simple example,in which one of the Cartesian coordinates is simply left out of the description of the system.More interesting is a ball suspended on a rod. In that case we can use the angular coordinatesϑ andϕ to describe the system – that is, we replace the coordinates x, y and z by the anglesϕand ϑ – see figure 2.1. In this case, we see that the coordinates no longer represent distances,that is, they do not have the dimension of length, but rather they are values of angles, andtherefore dimensionless. This is the reason why we speak of generalised coordinates. Thesecoordinates form a reduced representation of a system subject to constraints. In chapter2 of the Schaum book you find many examples of constraints and generalised coordinates.Generalised constraints are denoted by q j , where j is an index which runs over the degreesof freedom of the constrained system.

1The subspace on which the particle is allowed to move is not necessarily a linear subspace, e.g. the sphericalsubspace in the case of a pendulum. Mathematicians would use the term ‘submanifold’ rather than subspace.

2.1. GENERALISED COORDINATES AND VIRTUAL DISPLACEMENTS 11

2

We now shall look at constraints and generalised coordinates from a more formal view-point. Let us consider a system consisting of N particles in 3 dimensions, so that the totalnumber of coordinates is 3N . The system is subject to a number of constraints, which are ofthe form

g (k)(r1, . . . ,rN , t ) = 0, k = 1, . . . ,K . (2.1)

Constraints of this form (i.e. independent of the velocities) are called holonomic. Usually, it isthen possible to transform the 3N degrees of freedom to a reduced set of 3N −K generalisedcoordinates q = q j , j = 1, . . . ,3N −K . It is now possible to express the position vectors interms of these new coordinates:

ri = ri (q, t ). (2.2)

As an example, consider the particle suspended on a rod; see figure 2.2. The Cartesian coor-dinates are x, y and z and they can be written in terms of the generalised coordinates ϑ andϕ as:

x = l sinϑcosϕ; (2.3)

y = l sinϑsinϕ; (2.4)

z =−l cosθ, (2.5)

where l is the length of the rod (and therefore fixed). These equations are a particular exampleof Eqs. (2.2).

The velocity can be expressed in terms of the q j using the chain rule:

ri =3N−K∑

j=1

∂ri

∂q jq j + ∂ri

∂t. (2.6)

If we now take the partical derivative with respect to q j in the left and right hand side, wefind:

∂ri

∂q j= ∂ri

∂q j, (2.7)

a result which will be very useful further on.Newton’s laws predict the evolution of a mechanical system without ambiguity from a

given initial state (if that state is not on an unstable point, such as zero velocity at the topof a hill). However, we are sometimes interested in a variation of the path of a system, i.e.a displacement of one or more particles in some direction. Such displacements are calledvirtual displacements in order to distinguish them from the actual displacement, which isalways governed by the Newton equations of motion. If we now generalise the definition ofwork, Eq. (1.20) to include virtual displacements δri rather than the mechanical displace-ments which actually take place, then the work done due to this displacement is defined as

δW =N∑

i=1F ·δri . (2.8)

The notion of virtual work is very important in the following section.Summarizing this section, we have

• A system consisting of N point particles has 3N coordinates, usually given asri , i = 1, . . . , N . If the system is subject to S independen constraints, it can bedescribed by K = 3N −K new variables, called generalised coordinates.

• We may consider instantaneous displacements δri of the particles in the sys-tem and assign a virtual work δW to these displacements as in Eq. (2.8).

2


2.2 D’ALEMBERT ’S PRINCIPLEWe start from Newton’s law of motion for an N -particle system.

pi = mri = Fi , i = 1, . . . , N . (2.9)

It is always possible to decompose the total force on a particle into a force of constraint FC

and the remaining force, which we call the applied force FA:

F = FC +FA. (2.10)

If you consider any system consisting of a single particle (or nonrotating rigid body), subjectto constraints, you will find that the work forces of constraint are always perpendicular tothe space in which the particle is allowed to move. For example, if a particle is attached to arigid rod which is suspended such that it can rotate freely, the particle can only rotate on aspherical surface. The force of constraint, which is the tension in the rod, is always normalto that surface. Similarly, the force of the billiard table on the balls is always vertical, i.e. per-pendicular to the plane of motion. This notion provides a way to eliminate these forces fromthe description. Consider an arbitrary but small virtual displacement δr within the subspaceallowed by the constraint. Because the force of constraint is perpendicular to this subspace,we have:

p ·δr = (FC +FA) ·δr = FA ·δr. (2.11)

We see that the force of constraint drops out of the system, and we are left with a motiondetermined by the applied force only. Because (2.11) holds for every small δr, we have

p = FA (2.12)

if we restrict both p and FA to be tangential to the constraint subspace. The principle we haveformulated in Eq. (2.11) is called d’Alembert’s principle. For systems consisting of a single rigidbody, it expresses the fact that the forces of constraint are perpendicular to the subspace ofthe constraint. This is a nontrivial fact: it can be verified for particular systems and can beused as a general rule for handling constraints. The expression F ·δr is the virtual work doneas a result of the virtual displacement.

It is important to note that the virtual displacements are always considered to be spatial– the time is not changed. This is particularly important in cases where the constraints aretime-dependent. In the next section we shall consider an example of this.

For more than one object, the contributions to the virtual work must be added, so that weobtain:

N∑i

pi ·δri =∑

iFA

i ·δri . (2.13)

In this form, the contributions of the constraint forces to the virtual work do not all vanish foreach individual object, but the total virtual work due to the constraint forces vanishes:∑

iFC

i ·δri = 0. (2.14)

In summary, we can formulate d’Alembert’s principle in the following, concise form:

According to d’Alembert’s principlem the virtual work due to the forces of constraintis always zero for virtual displacements which do not violate the constraint.

The use of d’Alembert’s principle can simplify the analysis of systems subject to con-straints, although we often use this principle tacitly in tackling problems in the ‘Newtonian’approach. In that approach we usually demand that the forces of constraint balance the com-ponents of the applied force perpendicular to the constraint subspace. Nevertheless, it isconvenient to skip this step, using d’Alembert’s principle, especially in complicated problems(many applied forces and constraints).

2.3. EXAMPLES 13

2

Fg

l

FTϕ

ϕ

ϕ

m

FIGURE 2.2: The pendulum moving in a plane. The rod of length is rigid, massless, and is suspended frictionless.

2.3 EXAMPLES

2.3.1 THE PENDULUM

As a simple example, let us consider a pendulum moving in a plane. This system is shownin figure 2.2. Using Newton’s mechanics, we say that the ball of mass m is kept on the circleby the tension in the suspension rod. This tension is directed along the rod, and it preciselycompensates the component of the gravitational force along the same line. The componentof the gravitational force tangential to the circle of motion determines the motion. The dis-tance traveled by the pendulum is given by r (t ) = lϕ(t ), where ϕ is the angle shown in thefigure. The speed is therefore given by r (t ) = l ϕ(t ), and the equation of motion is

l ϕ(t ) =−g sinϕ(t ). (2.15)

Using d’Alembert’s principle simplifies the first part of this analysis. We can simply saythat the motion is determined by the component of the applied force (i.e. gravity) lying inthe subspace of the motion (i.e. the circle) and this leads to the same equation of motion. Al-though in this simple case the difference between the approaches with and without d’Alembert’sprinciple is minute, in more complicated systems, the possibility to avoid analysing the forcesof constraint is a real gain.

Let us redo this calculation in terms of the Cartesian components x, y and z. These aregiven by

x(t ) = l sin[ϕ(t )

](2.16a)

y(t ) =−l cos[ϕ(t )

](2.16b)

z(t ) = 0 (2.16c)

where the last equation defines the plane of the pendulum motion as the x y-plane. Theseequations clearly show how the r(t ) as 3 Cartesian coordinates can be written in terms of asingle generalised coordinateϕ, as a result of the two constraints: motion in one plane (z = 0)and the length l =

√x2 + y2 fixed.

We can now calculate the velocities:

x(t ) = l ϕ(t )cos[ϕ(t )

](2.17a)

y(t ) = l ϕ(t )sin[ϕ(t )

](2.17b)

z(t ) = 0. (2.17c)

From this we immediately see that

v =√

x2 + y2 + z2 = l ϕ. (2.18)

2


FF

F

FF

1

z

Z

1-

2

IP

SB

d^

x

y

n

α

FIGURE 2.3: Small block on an inclined plane.

Similarly, we find for the acceleration a = l ϕ, a result we used above.

2.3.2 THE BLOCK ON THE INCLINED PLANE

Now we consider a more complicated example: that of a block sliding on a wedge. We shalldenote the block by SB (small block) and the wedge by IP (Inclined plane). The setup is shownin figure 2.3. It consists of the wedge (inclined plane) of mass M which can move freely (i.e.without friction) over a horizontal table, and the small block off mass m, which can slide overthe inclined plane (also frictionless). The aim is to find expressions for the accelerations of IPand SB. The Cartesian unit vectors are x and y, and the unit vector along the inclined plane,pointing to the right, is d, and the upward normal unit vector to the plane is called n. Let ussolve this problem using the standard approach. The acceleration of IP is called A, and that ofthe small block is A+a, i.e., a is the acceleration of the small block with respect to the inclinedplane.

Newton’s second law for the two bodies reads:

MA =−M g y+F2y−F1n, (2.19a)

m(A+a) =−mg y+F1n. (2.19b)

As we know that the motion of IP is horizontal, we know that all y components of the forcesacting on it will cancel, and A is directed along x. Similarly, we know that a is zero along n.This allows us to simplify the equations:

M A =−F1 sinα; (2.20a)

m(Ax+a‖d) =−mg y+F1n (2.20b)

where a‖ is the component of a directed along d. The first of these equations is a scalar equa-tion. The second equation represents in fact two equations, one for the x and one for the ycomponent. We have three unknowns: A, a‖ and F1. Translating d and n into the x- and y-components is straightforward, and (2.20b) becomes:

m(A+a‖ cosα) = F1 sinα, (2.21a)

−ma‖ sinα= F1 cosα−mg . (2.21b)

2.3. EXAMPLES 15

2

Now we can solve for the accelerations by eliminating F1 from our equations, and we find:

a‖ = g(M +m)sinα

M +m sin2α; (2.22a)

A =−gm sinαcosα

M +m sin2α. (2.22b)

The solution of this problem involved one nontrivial step: the fact that we have split the ac-celeration of SB into the acceleration of IP plus the acceleration of the SB with respect to theIP has enabled us to remove the latter’s component along n. This is not so easy a step when adifferent representation is used (e.g. when the acceleration is not split into these parts).

Now we turn to the solution using d’Alembert’s principle:

pSB ·δrSB + pIP ·δrIP = FASB ·δrSB +FA

IP ·δrIP. (2.23)

We identify two natural candidates to be used as generalized coordinates: the coordinate Xof the IP along the horizontal direction, and the distance d from the top of the IP to the SB.The total virtual work done as a result of displacements δX and δd is the sum of the virtualwork done by each body:

δrSB = δd d+δX x and δrIP = δX x. (2.24)

The applied forces are the gravity forces – we do not care about constraint forces any longer– and we find

FAIP ·δrIP = 0, (2.25)

as the displacement is perpendicular to the applied (gravity) force. Furthermore

FASB ·δrSB = mg sinαδd . (2.26)

On the other hand:pSB = m(X x+ d d) (2.27)

andpIP = M X x, (2.28)

so that pIP = M Ax and pSB = m(Ax+a‖d). Taking time derivatives of (2.27) and (2.28) and us-ing d’Alembert’s equations (2.23) for this problem, together with (2.25) and (2.26), we obtain

m AδX +ma‖δd +m A cosαδd +ma‖ cosαδX +M AδX = mg sinαδd . (2.29)

This equation should hold for any pair of virtual displacements δX and δd . We may thereforetake δX = 0 or δd = 0. From this we see that the coefficients of both δX and δd should eachvanish individually, giving the equations:

(m +M)A+ma‖ cosα= 0. (2.30a)

m(a‖+ A cosα) = mg sinα. (2.30b)

Not surprisingly, these equations lead to the same result (2.22) as obtained before. Althoughthe second approach using d’Alembert’s principle does not seem simpler, it is less prone toerrors since the constraint forces do not have to be taken into account explicitly. This mani-fests itself explicitly in the fact that we do not have to eliminate the constraint force F1 as inthe direct approach.

An important remark concerns this neglect of the constraint force: the inclined planemoves along the x-axis – therefore, the constraint force −F1 is not perpendicular to the dis-placement of the inclined plane. This seems to be in contradiction with what we said aboveabout all constrained forces being perpendicular to the allowed motion. However, when the

2


z

x

y

ω

α

tq

^

FIGURE 2.4: Bead on a rotating wire.

inclined plane moves over a distance δx in the horizontal direction, the displacement of theblock can be decomposed into (i) the same displacement δx in the horizontal direction plus(ii) a distance δd over which the block slides over the wedge’s surface. The second displace-ment is perpendicular to the constraint force F1, but as the constraint force on the blockis opposite to the constraint force −F1 on the inclined plane, the first displacement gives acontribution to the virtual work which is opposite to that on the inclined plane. Thereforethe total work done by the constraint forces is zero: note every displacement in the systemis perpendicular to the constraint force, but the sum of all contributions to the virtual workvanishes.

2.3.3 HEAVY BEAD ON A ROTATING WIRE

In this section, we consider a system with a time-dependent constraint. A bead slides with-out friction along a straight wire which rotates along a vertical axis, under an angle α (seefigure 2.4). The position of the bead along the wire is denoted by q , which is the distance ofthe bead from the origin. The momentum of the bead is given by

p = mqωsinα t+mq q (2.31)

It should however be noted that the unit vectors t and q rotate themselves, and hence theirtime derivatives occur in p. The latter occurs in d’Alembert’s equation, in which gravity entersas the applied force FA. Instead of working out p explicitly, we can use the following trick:

p ·δr = d

d t(p ·δr)−p ·δr. (2.32)

At first sight, you might think that the second term on the right hand side is zero as δr = δq qand δq does not involve any time dependence: virtual displacements are always assumedto be instantaneous and do not involve any time dependence. However, even with a time-independent δq , the displacement δr is time-dependent as the displacement is carried outin a rotating frame. This can also be seen from the fact that q is time-dependent. In fact, inour system the displacement along the wire will cause a change in the rotational velocity, andit is this velocity change which gives δr. If the bead is moved upward, for example, the beadwill move along a circle which has a larger radius, but still at the same angular velocity, so thatthe orbital speed increases. The orbital speed is given as qωsinα, so that we have:

δr =ωsinαδq t. (2.33)

As δr is given by δq q, we find

p ·δr = mq δq −mω2 sin2αq δq = Fa ·δr =−mg cosαδq (2.34)

2.4. D’ALEMBERT’S PRINCIPLE IN GENERALISED COORDINATES 17

2

and we find the equation of motion:

q −ω2 sin2αq =−g cosα. (2.35)

The solution to this equation can be found straightforwardly:

q(t ) = q0 + AeΩt +Be−Ωt (2.36)

with q0 = g cotα/(ω2 sinα), A and B arbitrary constants and Ω = ωsinα. This is quite a pe-culiar behaviour: as soon as we release the bead a bit away from the position where the cen-trifugal and gravity forces are in balance, it will run away exponentially fast from this positionas the first term in the solution grows with time. This is reasonable if you think of it: displac-ing the bead to a higher position increases the centrifugal force; therefore the bead will bepushed even further outward. The same holds for a downward displacement, which causesthe gravity force to become dominant and pull the bead further downward. Later we shallencounter more powerful techniques which enable us to solve such a problem more easily.

2.4 D’ALEMBERT ’S PRINCIPLE IN GENERALISED COORDINATESIn the previous section we have encountered a few examples of systems subject to constraints,and analysed them using d’Alembert’s principle. In this section we shall do the same for anunspecified system and derive the equations of motion for a general constrained system us-ing d’Alembert’s principle.

We start from d’Alembert’s equation for N objects:

N∑i=1

pi ·δri =N∑

i=1FA

i ·δri . (2.37)

Now let us consider a small displacement. This can be described in terms of a small changein the generalised coordinates

q j → q j +δq j . (2.38)

The Cartesian coordinates ri are functions of the q j . Therefore we can express the changesin these ri via a first order Taylo expansion in the q j :

δri =3N−K∑

j=1

∂ri

∂q jδq j . (2.39)

Now we realise ourselves that the δq j can be chosen independently (provided they are small).Choosing all but one of them zero, we see that the coefficients of each δq j in d’Alembert’sprinciple must be identical on the left and right hand side in equation (2.37). This leads to

N∑i=1

pi · ∂ri

∂q j=

N∑i=1

FAi · ∂ri

∂q j. (2.40)

This is just a new formulation of ’dAlembert’s principle.In order to make progress, we use the sum rule for differentiation, similar to the one we

applied already to the bead sliding along the wire. Using the product rule for differentiation:

N∑i=1

pi · ∂ri

∂q j= d

d t

(N∑

i=1pi · ∂ri

∂q j

)−

N∑i=1

pi · d

d t

∂ri

∂q j. (2.41)

We note furthermore that in time derivative in the second term on the right hand side, we canswap the derivative with respect to time with the derivative with respect to q j to obtain:

d

d t

(∂ri

∂q j

)= ∂ri

∂q j. (2.42)

2


We see that we can formulate Eq. (2.40) in the form

d

d t

(N∑

i=1pi · ∂ri

∂q j

)−

N∑i=1

pi · ∂ri

∂q j=

N∑i=1

FAi · ∂ri

∂q j. (2.43)

In section 1.2 we have seen that the work done equals the change in kinetic energy. Thissuggests that the kinetic energy might be a convenient device for expressing d’Alembert’sequation in generalised coordinates. To see that this is indeed the case, we first calculate thederivative of the kinetic energy with respect to q j :

∂T

∂q j=

N∑i=1

mi ri · ∂ri

∂q j. (2.44)

Then we calculate its derivative with respect to q j :

∂T

∂q j=

N∑i=1

mri · ∂ri

∂q j=

N∑i=1

pi · ∂ri

∂q j, (2.45)

where we have used (2.7). If we stare long enough at these derivatives and at (2.43), we seethat the left hand side of d’Alembert’s equation as in (2.43) can be written as

d

d t

(∂T

∂q j

)− ∂T

∂q j. (2.46)

Let’s now tackle the right hand side of Eq. (2.43). Defining

N∑i=1

FAi ·

∂ri

∂q j=F j , (2.47)

where F j is the generalised force, we have the following

Formulation for d’Alembert’s principle in generalised coordinates:

d

d t

(∂T

∂q j

)− ∂T

∂q j=F j . (2.48)

There is no sum over j in this equation because the variations δq j are arbitrary and indepen-dent. It is then possible to obtain the form (2.48) from d’Alembert’s principle by taking onlyone particular δq j to be nonzero.

2.5 CONSERVATIVE SYSTEMS – THE MECHANICAL PATHConsider now a particle which moves in a constrained subspace under the influence of apotential. As an example you can imagine a non-flat surface on which a ball is moving fromr1 to r2. If the ball is not forced to obey the laws of mechanics, it can move from r1 at timet1 to r2 at time t2 along many different paths. Instead of approaching the problem of findingthe motion of the ball from a time-evolution point of view, where we update the position andthe velocity of a particle at each infinitesimal time step, we consider the path allowed for bythe laws of mechanics2 as a special one among all the available paths from r1 at t1 to r2 at t2.

We thus try to find a condition on the path as a whole rather than for each of its infinites-imal segments. To this end, we start from d’Alembert’s principle, and apply it to two paths,ra(t ) and rb(t ), which are close together for all times. The difference between the two paths atany time t between t1 and t2 is δr(t ) = rb(t )−ra(t ), and we write down d’Alembert’s principleat time t using this δr(t ):

mr(t ) ·δr(t ) = F ·δr(t ), (2.49)

2This path is not always, but nearly always, unique.

2.5. CONSERVATIVE SYSTEMS – THE MECHANICAL PATH 19

2

where it is understood that F is the applied force only, as δr lies in the constrained subspace.3

This equation holds for every t between t1 and t2, and we can formulate a global conditionon the path by integrating it over time from t1 to t2:∫ t2

t1

mr(t ) ·δr(t )d t =∫ t2

t1

F ·δr(t )d t . (2.50)

The analysis which follows resembles that of the previous chapter when we derived the con-servation property of the energy. Indeed, the right hand side looks like an expression for thework, but it should be kept in mind that δr is not a real displacement of the particle, but adifference between two possible paths.

Via partial integration, and using the fact that the begin and end point of the path arefixed, we can transform the left hand side of (2.50), using partial integration:

∫ t2

t1

mr(t ) ·δr(t )d t =−∫ t2

t1

mr(t ) ·δr(t )d t =−∫ t2

t1

m

2

∂r2

∂rδrd t =

−∫ t2

t1

∂T

∂rδrd t ≈−

∫ t2

t1

[T (rb)−T (ra)]d t , (2.51)

where the approximation holds to first order in δr as we have performed a first-order Taylorexpansion in δr. The resulting expression is the difference in kinetic energy between the twopaths, integrated over time.

If we are dealing with a conservative force field, the right hand side of (2.50) can also betransformed to a difference between two global quantities:∫ t2

t1

F ·δr(t )d t =−∫ t2

t1

∇V ·δr(t )d t ≈−∫ t2

t1

[V (rb)−V (ra)]d t . (2.52)

Combining (2.51) and (2.52) we obtain:

δ

∫ t2

t1

(T −V )d t = 0, (2.53)

where the notation with the δ means that we calculate the integral for two paths that differonly slightly from each other and then calculate the difference between these two integrals.We conclude that d’Alembert’s principle for a conservative force can be transformed to thecondition that the linear variation (2.53) vanishes. This global condition distinguishes themechanical path from all other ones.

The quantity T −V is called the Lagrangian, L. The integral over time of this quantity∫ t2t1

L d t is called the action, denoted by S:

S =∫ t2

t1

d t (T −V ) =∫ t2

t1

d t L. (2.54)

We have derived a new principle:

The mechanical path of a particle moving in a conservative potential field from aposition r1 at time t1 to a position r2 at t2 is a stationary solution of the action, i.e.the linear variation of the action with respect to an allowed variation of the patharound the mechanical path, vanishes.

This principle is called Hamilton’s principle. Note that the variations of the path are re-stricted to lie within the constrained subspace. The advantage of this new formulation ofmechanics with conservative force fields over the Newtonian formulation is that it holds for

3We suppose that the constrained subspace is smooth and that ra(t ) is close to rb(t ) for all t .

2


any system subject to constraints, and that it holds independently of the coordinates whichwere chosen to represent the motion. This is clear from the fact that we search for the min-imum of the action within the subspace allowed for by the constraint, and this subspace isproperly described by the generalised coordinates q j . When solving the motion of some par-ticular mechanical system our task is therefore to properly express T and V in terms of thesegeneralised coordinates, plug the Lagrangian L = T −V into the action, and minimise the lat-ter with respect to the generalised coordinates (which are functions of time). Although thismight seem a complicated way of solving a simple problem, it should be realised that thetransformation of forces and accelerations to generalised coordinates is usually more com-plicated than writing the kinetic energy and the potential in terms of these new coordinates.Furthermore we shall see below that the problem of finding the stationary solution for a givenaction leads straightforwardly to a second-order differential equation, which is the correctform of the Newtonian equation of motion in terms of the chosen generalised coordinates.

As an example, consider the simple pendulum (see figure 2.1). The position of the massm is given by the 2 coordinates x and y (we neglect the third coordinate z). The constraintobeyed by these coordinates is x2 + y2 = l 2. This constraint allows us to use only a singlegeneralised coordinate ϕ: x = l sinϕ and y = −l cosϕ. The velocity is given by vϕ = l ϕ. Thisexample shows that the generalised coordinate q = ϕ does not necessarily have to have thedimension of length (ϕ is an angle and is therefore dimensionless), and likewise q = ϕ doesnot necessarily have the dimension of velocity (with dimension 1/time rather than displace-ment/time). The kinetic energy is now given as T = ml 2ϕ2/2, and the potential energy byV =−mg l cosϕ. The Lagrangian of the pendulum is therefore

L = T −V = m

(l 2ϕ2

2+ g l cosϕ

). (2.55)

We now turn to the problem of determining the stationary solution for an action with such aLagrangian.

The Lagrangian can have many different forms, depending on the particular set of gen-eralised coordinates chosen; therefore we shall now work out a general prescription for de-termining the stationary solution of the action without making any assumptions concerningthe form of the Lagrangian, except that it may depend on the q j and on their time derivativesq j :

S[q] =∫ t2

t1

L(q, q, t )d t . (2.56)

Here q(t ) is any vector-valued function, q(t ) = (q1(t ), . . . , qN (t )

). We now consider an arbi-

trary, but small variation δq(t ) of the path q(t ), and calculate the change δS in S as a result ofthis variation:

δS[q] = S[q+δq]−S[q] =∫ t2

t1

L(q+δq, q+δq, t

)d t −

∫ t2

t1

L(q, q, t

)d t ≈∫ t2

t1

[∂L

(q, q, t

)∂q

δq+ ∂L(q, q, t

)∂q

δq]

d t . (2.57)

In the last line of this equation, we have performed a Taylor expansion in δq and in δq. Notethat both q and q depend on time. Note further that ∂/∂q is a vector – the derivative mustbe interpreted as a gradient with respect to all the components of q. The use of the partialderivative ∂ and not the full derivatives d in the derivatives indicates that when calculatingthe gradient with respect to q, q is considered as a constant, and vice-versa.

Of course, δq and δq are not independent: if we know q(t ) for all t in the interval underconsideration, we also know the time derivative q. We can remove δq by partial integration:∫ t2

t1

[∂L

(q, q, t

)∂q

δq+ ∂L(q, q, t

)∂q

δq]

d t =∫ t2

t1

(∂L

∂q− d

d t

∂L

∂q

)δqd t . (2.58)

2.6. SUMMARY AND EXAMPLES 21

2

Because δq is small but arbitrary, this variation can only vanish when the term in brackets onthe right hand side vanishes. Consider for example a δq which is zero except for a very smallrange of t-values around some t0 in the interval between t1 and t2. Then the term betweenthe square brackets must vanish in that small range, which is means that it should vanish att0. We can do this for any small interval on the time axis, and we conclude that the term inbrackets vanishes for all t in the integration interval. So our conclusion reads

The action S[q] is stationary, that is, its variation with respect to q vanishes to firstorder, if the following equations are satisfied:

∂L

∂q j= d

d t

∂L

∂q j, for j = 1, . . . , N . (2.59)

The equations (2.59) are called Euler equations. In the case where L is the Lagrangian ofclassical mechanics, L = T −V , the equations are called Euler–Lagrange equations (note thatin the above derivation, no assumption has been made with respect to the form of L nor whatit means – the only assumption is that L depends at most on q, q and t ). The Euler equationshave many applications outside mechanics.

Often the following notation is used:

δL =N∑

j=1

(∂L

∂q j− d

d t

∂L

∂q j

)δq j (2.60)

andδL

δq=

(∂L

∂q− d

d t

∂L

∂q

), (2.61)

or, written in another way:δL

δq j=

(∂L

∂q j− d

d t

∂L

∂q j

). (2.62)

Note that (2.61) is an equality between (N -dimensional) vector quantities.Solving problems using this formalism usually proceeds along a number of fixed steps.

We will outline those at the beginning of the next subsection.Turning again to our simple example of a pendulum, we use theLagrangian found in (2.55) and write down the Euler–Lagrange equation for this:

∂L

∂ϕ=−mg l sinϕ= d

d t

∂L

∂ϕ= ml 2ϕ. (2.63)

The solution to this equation can be found through numerical integration. In the next sectionwe shall encounter some more complicated examples which show the advantages of the newapproach more clearly.

2.6 SUMMARY AND EXAMPLESWe start by summarising the general results we have obtained so far in this chapter. Then weshall consider several nontrivial examples where these results are used to find the equationsof motion for several nontrivial mechanical systems.

2.6.1 OVERVIEW OF THE THEORY

• When dealing with a system of M degrees of freedom subject to K constraints, there areconstraint forces present in order for these constraints to be realized. The constraintsleave M − K degrees of freedom. These degrees of freedom can be represented us-ing M −K generalised coordinates q j . These coordinates may be distances, or anglesetcetera.

2


• In addition to the constraint forces, we usually have applied forces, FA. In order to getrid of the constraint forces, we use d’Alembert’s principle:∑

ipi ·δri =

∑i

FAi ·δri , (2.64)

where FAi is the applied force on particle i .

• d’Alembert’s principle can be reformulated in terms of the generalised coordinates q j :

d

d t

(∂T

∂q j

)− ∂T

∂q j=F j , (2.65)

where T is the total kinetic energy, and F j is the generalised force:

F j =N∑

i=1FA

i ·∂ri

∂q j. (2.66)

• If the forces are all conservative, we can reformulate d’Alembert’s principle as Hamil-ton’s principle, which says that the mechanical path q(t ) running from a fixed initialposition at t1 to a fixed final position at t2, is given by the condition that the action

S =∫ t2

t1

L(q, q, t )d t (2.67)

is stationary, i.e. S varies to second order in the deviation from the mechanical path. Lis the Lagrangian which has the form

L = T −V (2.68)

where T is the kinetic, and V is the potential energy.

The path q j (t ) for which S is stationary obeys the Euler-Lagrange equations:

d

d t

(∂L

∂q j

)= ∂L

∂q j. (2.69)

The art of solving problems in analytical mechanics usually involves the following steps.

• Find the number of degrees of freedom without constraints. For N point particles, thisis 3N .

• Identify the constraints (by carefully reading the description of the system).

• Find a convenient set of generalised coordinates. This is the step where experience andintuition are convenient.

• Find the kinetic energy. This is sometimes difficult. If so, you write the particle coor-dinates ri as functions of the generalised coordinates q j and, from this, the velocitiesri as functions of the q j and q j . Then the kinetic energy T =∑

imi2 r2

i can be written asa function of the q j and q j . The resulting expression must be quadratic in the q j ! It isalways recommended to check this.

• Is the force is non-conservative, solve d’Alembert’s equation in generalised coordi-nates. If it is conservative, write V in terms of the q j and write L = T −V . You cannow write down the Euler-Lagrange equation and these may be solvable or not.

Often, the first few steps are not necessary as it is usually possible to directly identify thenumber of degrees of freedom, together with a convenient set of q j . If also the kinetic energycan be written easily in terms of the q j and q j , there is no need to involve the full set ofcoordinates ri .


2

2.6.2 A SYSTEM OF PULLEYS

We consider a system of massless pulleys as in the figure below.

mb

ma

mc

ll l

l1

2 34

The string is also massless and furthermore inextensible. It is quite complicated to find outwhat the forces on the system are when taking all the forces on the pulleys and the tension ofthe string into account. However, it turns out that using Hamilton’s principle makes it an easyproblem. The total string length is l = l1+l2+l3+l4 and this is fixed. This is the first constraint.Note that we have omitted the motion in the horizontal direction and perpendicular to theplane, which eliminates 6 degrees of freedom from the problem. From the 9 original degreesof freedom, there are therefore only 9−1−6 = 2 left. From the coordinates l1, l2, l3 and l4,one is redundant as l2 = l3. The constraint on the total string length fixes l2 = l3 when l1 andl4 are given:

l2 = l3 = 1

2(l − l1 − l4). (2.70)

Therefore we use only l1 and l4 as generalised coordinates.We can calculate the total potential energy now entirely in terms of l1 and l4. This en-

ergy is composed of three contributions, corresponding to the masses ma , mb and mc . Theheights of the outer two are directly given by l1 and l4, and the height of the central mass isgiven by that of the central pulley, which is given by l2 (or l3), and the total potential energyis therefore given as:

V =−g[

ma l1 + mb

2(l − l1 − l4)+mc l4

]. (2.71)

The speed of the left mass ma is given by l1, and that of the right one, mc , by l4. Using (2.70)we find that the speed of the central pulley is given by 1

2 (−l1− l4). The Lagrangian is thereforegiven as

L = 1

2ma l 2

1 +1

2mc l 2

4 +1

8mb(l 2

1 + l 24 +2l1 l4)+ g

[ma l1 + mb

2(l − l1 − l4)+mc l4

]. (2.72)

The Euler-Lagrange equations can be derived straightforwardly:

d

d t

∂L

∂l1= (ma + 1

4mb)l1 + 1

4mb l4 = ∂L

∂l1=

(ma − 1

2mb

)g ; (2.73a)

d

d t

∂L

∂l4= (mc + 1

4mb)l4 + 1

4mb l1 = ∂L

∂l4=

(mc − 1

2mb

)g . (2.73b)

2


The two equations can be solved for l1 and l4 and the result is

l1 = 4mamc +mamb −3mc mb

mc mb +4mamc +mambg ; (2.74a)

l4 = 4mamc +mc mb −3mamb

mamb +4mamc +mbmcg . (2.74b)

To check whether the answer is reasonable we verify that a stationary motion (i.e. a motionwith constant velocity) is possible if mb = 2ma = 2mc (convince yourself of this). The solutionis now trivial, since the right hand sides of (2.74) vanish as should indeed be the case. We seethat the Lagrange equations provide a framework which enables us to find the equations ofmotion quite easily.

2.6.3 EXAMPLE: THE SPINNING TOP

??Consider a top with cylindrical symmetry. In this section, we shall formulate and analyze

the equations of motion of the top. This is material that you have already covered in your firstyear classical mechanics. If you have forgotten most of this, refer back to that course! Theposition of the top is defined by its two polar angles ϑ and ϕ and a third angle, ψ, defines therotation of the top around its symmetry axis. The angular velocity is given in terms of thesethree polar angles as:

ωωω= ϕz+ ϑe+ ψd (2.75)

where z is a unit vector along the z-axis; e is a unit vector in the x y plane which is perpendic-ular to the axis of the top, and d is a unit vector along the axis of the top. The axis of the top isshown in the figure:

ϑ

ϕ

z

x

y

e

f

ϕ

d

ψ

ϑ

From this figure, it is clear that

e = (−sinϕ,cosϕ,0) and (2.76a)

d = (cosϕsinϑ, sinϕsinϑ,cosϑ). (2.76b)

And it follows that

f = e× d = (cosϕcosϑ, sinϕcosϑ,−sinϑ). (2.77)

The rotational kinetic energy of the top is given by

T = 1

2ωωωT Iωωω (2.78)


2

(the superscript T denotes transpose in a linear algebra sense: it turns the column vector ωωωinto a row vector). Diagonalizing the 3× 3 matrix I, it is always possible to find some axeswith respect to which this moment of inertia tensor is diagonal, and as a result of the axialsymmetry of the top one diagonal element, which we shall denote by I3, corresponds to thesymmetry axis d, and two other diagonal elements correspond to axes in the plane perpen-dicular to the body axis, such as e and f – we call these elements I1 (the axial symmetry forcesthem to be equal).

The kinetic energy is then given by

T = 1

2I1(ωωω · e)2 + 1

2I1(ωωω · f)2 + 1

2I3(ωωω · d)2 = 1

2I1ϕ

2 sin2ϑ+ I1

2ϑ2 + 1

2I3(ψ+ ϕcosϑ)2. (2.79)

The gravitational force results in a potential V = M g R cosϑ, where M is the top’s massand R the distance from the point where it rests on the ground to the centre of mass – clearly,the height of the centre of mass is then given as R cosϑ. The Lagrangian therefore reads:

L = 1

2I1ϕ

2 sin2ϑ+ I1

2ϑ2 + 1

2I3(ψ+ ϕcosϑ)2 −M g R cosϑ. (2.80)

The Lagrange equations for ϑ, ϕ and ψ are then given by:

d

d t

∂L

∂ϑ= I1ϑ= ∂L

∂ϑ= I1ϕ

2 sinϑcosϑ− I3(ψ+ ϕcosϑ)ϕsinϑ+M g R sinϑ; (2.81a)

d

d t

∂L

∂ϕ= d

d t

[I1ϕsin2ϑ+ I3(ψ+ ϕcosϑ)cosϑ

]= ∂L

∂ϕ= 0; (2.81b)

d

d t

∂L

∂ψ= I3

d

d t(ψ+ ϕcosϑ) = ∂L

∂ψ= 0. (2.81c)

We immediately see from the last equation that ψ+ ϕcosϑ is a constant of the motion – weshall call this ω3:

ω3 = ψ+ ϕcosϑ= Constant. (2.82)

ω3 denotes the component of angular velocity along the spining axis.Let us search for solutions of constant precession: ϑ= constant, or ϑ= 0. We furthermore

set ϕ=Ω. The first Euler-Lagrange equation then gives:

I1Ω2 cosϑ− I3ω3Ω+M g R = 0. (2.83)

If ω3 is large, we find the two solutions

Ω= M g R

I3ω3(2.84)

for whichΩ is inversely proportional to ω3 and

Ω= I3ω3

I1 cosϑ(2.85)

i.e. Ω is proportional to ω3. The first solution corresponds to slow precession and fast spin-ning around the spinning axis; the second solution corresponds to rapid precession in whichthe gravitational force is negligible.

For general ω3, the quadratic equation (2.83) with ϑ= 0 has two real solutions forΩ if

I 23ω

23 > 4I1 cosϑM g R. (2.86)

For smaller values of ω3, a wobbling motion sets in (“nutation”).

2


2.7 NON-CONSERVATIVE FORCES – CHARGED PARTICLE IN AN ELEC-TROMAGNETIC FIELD

In this section we consider one particular type of force which is not conservative, but whichcan still be analysed fully within the Lagrangian approach. This is the very important exampleof a charged particle in an electromagnetic field. Before we turn to this specific example ofa particle in an electromagnetic field, we discuss how the Euler-Lagrange technique can beextended to a particular type of nonconservative forces.

Suppose we have a collection of N particles which experience a non-conservative forcewhich can be derived from a generalised potential W (ri , ri ) in the following way:

F =−∂W

∂ri+ d

d t

∂W

∂ri. (2.87)

Analogous to the previous section we can then derive a variational condition, starting fromd’Alembert’s principle:∫ t2

t1

mriδri d t =−∫ t2

t1

δT d t =∫ t2

t1

[−∂W

∂ri+ d

d t

∂W

∂ri

]δri d t . (2.88)

The left hand side has been transformed as in (2.51). To rewrite the the right hand side, weneed a special partial integration step for the second term:

−∫ t2

t1

δT d t =∫ t2

t1

[−∂W

∂riδri − ∂W

∂riδri

]d t =−

∫ t2

t1

δW d t . (2.89)

So we see that the variation of the action

S[q] =∫ t2

t1

[T −W ]d t (2.90)

vanishes. We can also work out the Euler-Lagrange equations for the ‘Lagrangian’ L = T −W ,which, for

T =N∑

i=1mi r2

i , (2.91)

directly leads to the classical equation of motion:

d

d t

dL

d ri= mi ri + d

d t

dW

d ri= dW

dri, (2.92)

which is equivalent tomi ri = Fi . (2.93)

2.7.1 CHARGED PARTICLE IN AN ELECTROMAGNETIC FIELD

A point particle with charge q moving in an electromagnetic field experiences a force

F = q (E+v×B) . (2.94)

The charge q of the particle should not be confused with the generalised coordinates qi in-troduced before. E is the electric field, B is the magnetic field. Note that the force depends onthe velocity, which shows that it cannot be conservative. Therefore we try to cast it into theform (2.87).

The fields are not independent, but they are related through the Maxwell equations. Weuse the following two Maxwell equations

∇·B = 0 and (2.95a)

∇×E+ ∂B

∂t= 0. (2.95b)

2.8. HAMILTON MECHANICS 27

2

We know from vector calculus that a vector field whose divergence is zero, can always bewritten as the curl of a vector function depending on space (and, in our case, time); applyingthis to (2.95a) we see that we can write B in the form B =∇×A, where A is a vector function,called the vector potential, depending on space and time. Substituting this expression for Bin Eq. (2.95b) leads to

∇×(

E+ ∂A

∂t

)= 0. (2.96)

Now we use another result from vector calculus, which says that any function whose curl iszero can be written as the gradient of a scalar function, which in this case we call the potential,φ(r, t ). This results in the following representations of the electromagnetic field:

E(r, t ) =−∇φ(r, t )− ∂A

∂t(r, t ); (2.97a)

B(r, t ) =∇×A(r, t ). (2.97b)

In fact, by using two Maxwell equations, we have reduced the set of 6 field values (3 for E and3 for B) to 4 (3 for A and 1 for φ).

We are after a function W (r, r) which, when used in an action of the usual form, yields thecorrect equation of motion with the force (2.94). The potential which does the job is

W (r, r) = qφ(r, t )−q r ·A(r, t ) = qφ−q(x Ax + y Ay + z Az ). (2.98)

Note that Ax denotes the x-component and not the partial derivative with respect to x. TheLagrangian occurring in the action is therefore:

L = 1

2mr2 +q r ·A(r, t )−qφ(r, t ). (2.99)

To see that this Lagrangian is indeed correct we work out the force component in thex-direction. First we calculate the derivative of the potential W with respect to x:

−∂W

∂x=−q

∂φ

∂x+q

(x∂Ax

∂x+ y

∂Ay

∂x+ z

∂Az

∂x

). (2.100)

Furthermore

d

d t

(∂W

∂x

)=−q

d Ax

d t=−q

(∂Ax

∂t+ ∂Ax

∂xx + ∂Ax

∂yy + ∂Ax

∂zz

). (2.101)

The Euler-Lagrange equations for the action contain the two contributions resulting from thepotential. We have

mx =−∂W

∂x+ d

d t

(∂W

∂x

)=−q

(∂φ

∂x+ ∂Ax

∂t

)+

q

[y

(∂Ay

∂x− ∂Ax

∂y

)+ z

(∂Az

∂x− ∂Ax

∂z

)]= qEx +q(yBz − zBy ) (2.102)

i.e. precisely the equation of motion (for the x-component) with the force given in (2.94)! Theother components follow in a similar way.

2.8 HAMILTON MECHANICSIn the previous sections of this chapter, we have seen that Hamilton principle, formultedas the Lagrangian equations of motion, enable us to solve complicated systems relativelyquickly. It is possible to formulate Lagrangian mechanics in a different way. At first sight thisdoes not add anything new to the formalism which was constructed in the previous sections,but we shall see that this new formalism provides us with a conserved quantity which is the

2


energy or some analogous object. More importantly, this formalism is essential for setting upquantum mechanics in a structured way, as will be shown in a later course.

Let us again consider a system described by a Lagrangian formulated in terms of gener-alised coordinates, with the equations of motion given by:

d

d t

∂L

∂q j= ∂L

∂q j. (2.103)

This is a second order differential equation, which we shall transform into two first orderones.

We define the conjugate momentum p j as

p j = ∂L

∂q j. (2.104)

The conjugate momentum should not be confused with the mechanical momentum, whichis simply

∑i mri , although the two coincide when the generalised coordinates are simply the

ri . Using the conjugate momenta, the equations of motion can be formulated as:

p j = ∂L

∂q j. (2.105)

In the particular example of a conservative system formulated in terms of the position coor-dinates ri :

L =N∑

i=1

mi

2r2

i −V (r1, . . . ,rN ), (2.106)

the momenta are given as

pi = mri (2.107)

and the equations of motion are

pi =−∂V

∂ri. (2.108)

We see that in the case of a particle moving in a conservative force field, the conjugate mo-mentum corresponds to the mechanical momentum.

We have reformulated the Euler-Lagrange equation as two first-order differential equa-tions. The Euler-Lagrange equations were derived from a variational principle, the Hamiltonprinciple, which requires the action to be stationary for the mechanical path. This led to arelation between the derivatives of the Lagrangia, ∂L/∂q j and ∂L/∂q j . We may ask ourselvesif it is possible to define our two new equations also as relations of a derivative of some otherfunction. This means that the Lagrangian, which is formulated as a function of q j and q j

(and perhaps the time t ) is traded in for a new function which depends on the coordinates q j

and p j . We call this new function the Hamiltonian, H(p j , q j , t ). We shall for now not discusshow this function is constructed (see the box below), but just give its form:

H(p j , q j , t ) =N∑

j=1p j q j −L

[q j , q j (qk , pk ), t

]. (2.109)

Note that we can indeed express q j in terms of the pk and qk , as indicated in the secondargument of L by inversion of Eq. (2.104).4 This function can be used to formulate the twoequations (2.104) and (2.105) in an elegant way in terms of the Hamiltonian.

4For this inversion to be possible, the Lagrangian should be convex, but we shall not go into details concerningthis point.


2

To see this, let us calculate the derivatives of H with respect to q j and p j :

∂H

∂p j= q j +

∑k

pk∂qk

∂p j−∑

k

∂L

∂qk

∂qk

∂p j. (2.110)

Note that it follows from (2.104) that the second and third terms on the right hand side cancel,so that we are left with

∂H

∂p j= q j . (2.111)

Now let us calculate the derivative with respect to q j :

∂H

∂q j=− ∂L

∂q j+∑

kpk∂qk

∂q j−∑

k

∂L

∂qk

∂qk

∂q j. (2.112)

Again using (2.104) we see that the second and third term on the right hand side cancel –furthermore the first term on the right hand side is equal to −pi and we are left with:

∂H

∂q j=−p j . (2.113)

Eqs. (2.111) and (2.113), together with the definition of the Hamiltonian (2.109) and of themomentum (2.104) are equivalent to the equations of motion. Eqs (2.111) and (2.113) arecalled Hamilton’s equations. Note that we must consider the generalised coordinates and thecanonical momenta as independent coordinates.

2


** Legendre transformationsIt seems that the Hamiltonian has been found by some magic. Now we shall explain howthis magic works. Let us assume that we have a function f (x, y) depending on two variablesx and y . A variation of these variables induces a variation of f :

d f = ∂ f

∂xd x + ∂ f

∂yd y. (2.114)

Callingu = ∂ f /∂x and v = ∂ f /∂y, (2.115)

we haved f = ud x + vd y. (2.116)

In mathematical terms, we say that f is a exact differential. Most ‘well-behaved’ functionsare exact differentials, in particular the Lagrangian is an exact differential of the q j and q j .Now we search for a function g (u, y) which depends on u and y such that

d g = ∂g

∂udu + ∂g

∂yd y. (2.117)

It turns out that such a function is given by

g = xu − f . (2.118)

Let us check this. We haved g = xdu +ud x −d f . (2.119)

Using (2.116), we can work this out as

d g = xdu +ud x −ud x − vd y = xdu − vd y. (2.120)

We see that indeed the new function g does the job! The procedure outlined here is knownas Legendre transformation. It occurs in many area of physics.Now let us turn back to L and assume that it is an exact differential of q j and q j (we forgetthe explicit time dependence for simplicity):

dL = ∂L

∂q jd q j + ∂L

∂q jd q j = ∂L

∂q jd q j +

∑j

p j d q j , (2.121)

where we have used the definition of p j , (2.104). Now we find the exact differential H interms of q j and p j analogous to our recipe for finding g from f (applied to each pair ofcoordinates q j , q j ):

H(p j , q j ) =∑j

p j q j −L(q j , q j ), (2.122)

which is the form we have given above.

If the system does not depend explicitly on time, the Hamiltonian is the analogue of theenergy. The simplest case is a conservative system with the positions ri as coordinates. Inthat case it is easy to see that

H =N∑

i=1

p2i

2m+V (r1, . . . ,rN ). (2.123)

More generally, let us consider a conservative system formulated in terms of generalisedcoordinates q1, . . . , qs . Note the difference with Eq. (2.2), where ri may contain an explicittime-dependence – in the present case we assume that the constraints have no explicit time-dependence. In that case it is possible to express the position coordinates ri in terms of the sgeneralised coordinates q j , j = 1, . . . , s:

ri = ri (q1, . . . , qs) (2.124)


2

and therefore the velocities can be calculated as

ri =s∑

j=1

∂ri

∂q jq j . (2.125)

Therefore, if we formulate the kinetic energy∑

i12 mr2

i in terms of the generalised coordinates,we obtain an expression which is quadratic in the q j :

T =s∑

k, j=1Mk j (q1, . . . , qs)q j qk (2.126)

where

M j k = Mk j =N∑

i=1

mi

2

∂ri

∂q j

∂ri

∂qk. (2.127)

If we calculate the contribution to the momenta arising from the kinetic energy, we find thatthey depend linearly on the q j :

p j = ∂T

∂q j=

s∑k=1

(M j k +Mk j

)qk = 2

s∑k=1

Mk j qk . (2.128)

Hence

s∑j=1

q j p j = 2T (2.129)

and

H = 2T − (T −V ) = T +V = Energy. (2.130)

For a general system Hamilton’s equations of motion can be used to derive the time deriva-tive of the Hamiltonian:

d H

d t=

s∑j=1

∂H

∂q jq j +

s∑j=1

∂H

∂p jp j + ∂H

∂t. (2.131)

Using Hamilton’s equation of motion (2.111) and (2.113) we see that the first two terms onthe right hand side cancel and we are left with:

d H

d t= ∂H

∂t. (2.132)

We see therefore that if H (or L) does not depend explicitly on time, then H is a conservedquantity. If the potential does not contain a q j dependence, this implies conservation of en-ergy. If the potential on the other hand does contain such a dependence, then (2.132) impliesconservation of some quantity which plays a role more or less equivalent to energy.

2


Let us summarize the formalism which we have developed in this section. Westarted with the Euler Lagrange equation

d

d t

∂L

∂q j= ∂L

∂q j. (2.133)

Defining

p j = ∂L

∂q j, (2.134)

we could write the Euler Lagrange equation in the form

p j = ∂L

∂q j. (2.135)

It is possible to formulate these equations using a new functional depending on q j

and p j , the Hamiltonian:

H(p j , q j ) =∑j

p j q j −L. (2.136)

The new equations are called Hamilton equations:

q j = ∂H

∂p j, p j =− ∂H

∂q j. (2.137)

The Hamiltonian is preserved in time. In the case where the Lagrangian (and there-fore the Hamiltonian) have no explicit time dependence, the Hamiltonian coincideswith the total mechanical energy.

2.9 APPLICATIONS OF THE HAMILTONIAN FORMALISMIn this section we shall reconsider the systems studied before in the Lagrange framework andpoint out which features are different when these systems are considered within the Hamil-tonian framework. From the derivation of the Hamiltonian and Hamilton’s equations, it isseen that the latter can be viewed as a different way of writing Lagrange’s equations. The rea-son for introducing the Hamiltonian and Hamilton’s equations is that they are often used inquantum mechanics and because the Hamilton formalism is more convenient for discover-ing some conserved quantities.

2.9.1 THE THREE-PULLEY SYSTEM

From the Lagrangian (2.72):

L = 1

2ma l 2

1 +1

2mc l 2

4 +1

8mb(l 2

1 + l 24 +2l1 l4)+ g

[ma l1 + mb

2(l − l1 − l4)+mc l4

], (2.138)

the momenta p1 and p4 associated with the degrees of freedom l1 and l4 are found as:

p1 = ∂L

dl1=

(ma + mb

4

)l1 + mb

4l4; (2.139a)

p4 = ∂L

dl4=

(mc + mb

4

)l4 + mb

4l1; (2.139b)

We should express l1 and l4 in terms of the momenta:

l1 = 1

∆

[(mc + mb

4

)p1 − mb

4p4

], (2.140a)

l4 = 1

∆

[−mb

4p1 +

(ma + mb

4

)p4

], (2.140b)

2.9. APPLICATIONS OF THE HAMILTONIAN FORMALISM 33

2

with∆= (ma +mc )mb/4+mamc . (2.141)

After some calculation, we find for the Hamiltonian:

H = 1

2∆

[mc p2

1 +ma p24 +

mb

4(p1 −p4)2

]− g

[ma l1 + mb

2(l − l1 − l4)+mc l4

]. (2.142)

Note that this Hamiltonian only depends on the p j and q j (which are in this particular prob-lem the l j ’s) – there is no dependence of this Hamiltonian on the q j = l j here).

The Hamilton equations read:

p1 = (ma − mb

2)g ; (2.143a)

p4 = (mc − mb

2)g . (2.143b)

The solution is simple since the right hand sides of these equations are constants:

p1 = (ma − mb

2)g t ; (2.144a)

p4 = (mc − mb

2)g t , (2.144b)

where the initial conditions are that the system is standing still at t = 0. Together with Eqs. (2.139),we obtain the same solution as in the Lagrangian case. We see that the difference betweenthe two approaches are not very dramatic in this case. Note that it is now easy to see that forma = mc = 2mb the system is in equilibrium.

2.9.2 THE SPINNING TOP

L = 1

2I1ϕ

2 sin2ϑ+ I1

2ϑ2 + 1

2I3(ψ+ ϕcosϑ)2 −M g R cosϑ. (2.145)

From this Lagrangian, we can derive the momenta associated with the three degrees of free-dom ϕ, ϑ and ψ:

pϕ = I1ϕsin2ϑ+ I3(ψ+ ϕcosϑ)cosϑ; (2.146a)

pϑ = I1ϑ; (2.146b)

pψ = I3(ψ+ ϕcosϑ). (2.146c)

If we want to express the kinetic energy in terms of these momenta, we need to solve for thetime derivatives of the angular coordinates ϑ, ϕ and ψ in terms of these momenta:

ϕ= pϕ−pψ cosϑ

I1 sin2ϑ; (2.147a)

ϑ= pϑI1

; (2.147b)

ψ= pψI3

− pϕ−pψ cosϑ

I1 sin2ϑcosϑ. (2.147c)

After some calculation, the Hamiltonian is then found to be

H = (pϕ−pψ cosϑ)2

2I1 sin2ϑ+ p2

ϑ

2I1+

p2ψ

2I3+M g r cosϑ. (2.148)

As the Hamiltonian does not depend on ψ and ϕ, we see immediately that pψ and pϕmust be constant. Coordinates of which only the momentum does appear in the Hamiltonianare called ignorable: these momenta are constant in time – they represent constants of themotion. We have seen that both pψ and pϕ are constants of motion.

2


The Hamiltonian now reduces to a simple form:

H = p2ϑ

2I1+U (ϑ), (2.149)

where

U (ϑ) = (pϕ−pψ cosϑ)2

2I1 sin2ϑ+

p2ψ

2I3+M g r cosϑ. (2.150)

The Hamilton equations yield

−I1ϑ=−pϑ =dU

dϑ. (2.151)

This equation is difficult to solve analytically. Note that apart from the ignorable coordinates,we have an additional constant of the motion, the energy:

p2ϑ

I1+U (ϑ) = E = constant. (2.152)

2.9.3 CHARGED PARTICLE IN AN ELECTROMAGNETIC FIELD

Finally we consider again the charged particle in an electromagnetic field. The momentumcan be found as usual from the Lagrangian – we obtain

p = mr+qA. (2.153)

The Hamiltonian is

H = pr− m

2r2 −q r ·A+qφ= m

2r2 +qφ(r) = (p−qA)2

2m+qφ(r). (2.154)

You might already know that this Hamiltonian is used in quantum mechanics for a particle inan electromagnetic field.

3THE TWO-BODY PROBLEM

35

3

36 3. THE TWO-BODY PROBLEM

In this chapter we consider the two-body problem within the framework of Lagrangianmechanics. One of the most impressive results of classical mechanics is the correct descrip-tion of the planetary motion around the sun, which is equivalent to electric charges movingin each other’s field. With the analytic solution of this problem, we shall recover the famousKepler laws. The problem is also important in quantum mechanics: the hydrogen atom is aquantum version of the Kepler problem.

3.1 FORMULATION AND ANALYSIS OF THE TWO-BODY PROBLEMThe two-body problem describes two point particles with masses m1 and m2. We denotetheir positions by r1 and r2 respectively, and their relative position, r2 − r1, by r. Finding theLagrangian is quite simple. The kinetic energy is the sum of the kinetic energies of the twoparticles, and the potential energy is the interaction, which depends only on the separationr = |r| of the two particles, and is directed along the line connecting them (note that this lastrestriction excludes magnetic interactions). We therefore have:

L = m1

2r 2

1 + m2

2r 2

2 −V (r ). (3.1)

Before deriving the equations of motion, we note that instead of writing the kinetic energyas the sum of the kinetic energies of the two particles, it can also be separated into the kineticenergy of the centre of mass and that of the relative motion, as in Eq. (1.33):

T = TCM +2∑

i=1

mi

2r′2i , (3.2)

where

r′i = ri − rCM; (3.3)

rCM = m1r1 +m2r2

M, (3.4)

and

TCM = M

2r2

CM, M = m1 +m2. (3.5)

As there are only two particles, we can work out the coordinates r′i relative to the centreof mass rC, and we find, using Eq. (3.4):

r′1 = r1 − rCM = m2

M(r1 − r2), (3.6)

andr′2 = r2 − rCM = m1

M(r2 − r1). (3.7)

We can take time derivatives by simply putting a dot on each r in these equations and then,after some calculation, we find for the kinetic energy:

T = M

2r2

CM + m1m2

2Mr2. (3.8)

The Lagrangian is therefore

L = T −V = M

2r2

CM + m1m2

2Mr2 −V (r ). (3.9)

We see that the kinetic energy of the relative motion has the form of the kinetic energy of a sin-gle particle of mass m1m2/(m1+m2) and position vector r(t ). The mass termµ= m1m2/(m1+m2) is called reduced mass.

3.1. FORMULATION AND ANALYSIS OF THE TWO-BODY PROBLEM 37

3

Of course we could write down the Euler–Lagrange equations for this Lagrangian as be-fore, but it is convenient to perform a further separation: that of the kinetic energy of therelative coordinate into a radial and a tangential part. First we must realise that the planethrough the origin and the initial velocity vector r of the relative position will always remainthe plane of the motion, as the force acts only within that plane. In this plane, we choose anx and a y axis. Then we can conveniently introduce polar coordinates r and ϕ, in which thex and y coordinate can be expressed as follows:

x = r cosϕ; (3.10)

y = r sinϕ. (3.11)

It then immediately follows that the kinetic energy of the relative motion can be rewritten as

µ

2

(x2 + y2)= µ

2

(r 2 + r 2ϕ2) . (3.12)

The Lagrange equations given in Eq. (2.59) then take the form:

M rCM = 0; (3.13a)

d

d t

(µr 2ϕ

)= 0; (3.13b)

µr −µr ϕ2 =−dV (r )

dr. (3.13c)

The first equation tells us that the centre of mass moves at constant speed: it does not feela net force. This follows from the fact that it does not appear in the potential, and is in ac-cordance with the conservation of total momentum in the absence of external forces. Co-ordinates such as rCM with constant conjugate momentum, are called ignorable – see sec-tion 2.9.2. The second and third equation do not depend on rCM – therefore we see that therelative motion can entirely be understood in terms of a single particle with mass µ and mov-ing in a plane under the influence of a potential V (r ).

We now use the second equation to eliminate ϕ. First note that the term in bracketsoccurring in this equation must be a constant, – we call this constant ` (`=µr 2ϕ is preciselythe angular momentum, and we see that it is conserved); the third equation then transformsinto

µr − `2

µr 3 =−dV (r )

dr, (3.14)

Note that this equation can be viewed as that of a one-dimensional particle subject to a forceF = `2

µr 3 − dV (r )dr . Such a force can in turn be derived from a conservative potential:

F (r ) =− d

drVEff(r ) =− d

dr

[`2

2µr 2 +V (r )

]. (3.15)

The subscript Eff is used to distinguish between this ‘effective’ potential and the original,bare attraction potential V (r ). The potential VEff is represented in figure 3.1 for the caseV (r ) = −1/r . From Eqs. (3.13b) and (3.13c) and from figure 3.1, we can infer the qualitativebehaviour of the motion.

We have seen [Eq. (3.13b)] that the angular momentum is constant. This implies thatthe motion will always keep the same orientation (i.e. clockwise or anti-clockwise). If theparticles move apart, the speed at which they orbit around each other will be slower (sinceincreasing r implies decreasing r ϕ).

The motion in the radial direction can be understood qualitatively as follows. Note thatwe can interpret Eq. (3.13c) as the motion of a particle in one dimension. This particle has amechanical energy which is the sum of its kinetic energy and the effective potential, and thisenergy should remain constant. Furthermore, the energy cannot be lower than the lowest

3


VE

ff(r

)-1

0

1

2

3

0 1 2 3 4

rminrmax

rmin

E<

E>

0

0

r

FIGURE 3.1: Effective potential for a two-particle system.

value of the effective potential shown in figure 3.1. If it lies between this value and 0, thenthen r will vary between some minimum and maximum value as shown in this figure. If E ison the other hand positive, r will vary between some minimum value and infinity.

We have seen that the r -component of the two-body motion can be described in termsof a single particle in one dimension. The energy of this particle is the sum of its kinetic andpotential energy – the latter is the effective potential [see Eq. (3.15)]. It turns out that thisenergy is equal to the total energy of the two-particle system (neglecting the contributionof the centre of mass motion to the latter). As we have already worked out the kinetic andpotential energy of the two-body problem above, we immediately see that

E = T +V = µ

2

(r 2 + r 2ϕ2)+V (r ), (3.16)

which can easily be identified as the kinetic energy µr 2/2 of the one-dimensional particleplus the effective potential.

3.2 SOLUTION OF THE KEPLER PROBLEMThe special case V (r ) =−A/r is very important as it describes the gravitational and the Coulombattraction. Also, in this special case, the motion can be studied further by analytical means.Finding the solution in the form r (t ),ϕ(t ) is not convenient – rather, we search for r (ϕ), whichcontains explicit information about the shape of the orbit.

We use the fact that the angular momentum `= µr 2ϕ is constant and combine this withthe fact that the energy is constant and given by (3.16):

ϕ= `

µr 2 ; (3.17)

r 2 = 2

µ(E −V )− `2

µ2r 2 . (3.18)

Eliminating the d t of the time derivatives by dividing (3.17) by the square root of (3.18) leadsto

dϕ

dr= ±`

r 2[2µ(E −V (r ))−`2/r 2

]1/2. (3.19)

With V (r ) =−A/r this can directly be integrated to give

ϕ−C = arcsin

(µAr −`2

εµAr

). (3.20)

3.2. SOLUTION OF THE KEPLER PROBLEM 39

3

In addition to the integration constant C on the left hand side, we see a constant ε, called theeccentricity, which is given in terms of the problem parameters as

ε=√

1+ 2E`2

µA2 . (3.21)

Inverting Eq. (3.20) to find r as a function of the polar angle ϕ gives:

r = `2

µA[1−εsin(ϕ−C )

] . (3.22)

We have some freedom in choosing C – it changes the definition of the angle ϕ. If we takeϕ= 0 as the angle for which the two particles are closest (perihelion), we see that C =π/2.

The motion can now be classified according to the value of ε. We take ε positive – chang-ing the sign of ε does not change the shape of the orbit (putting ϕ→ϕ+π compensates thissign change). For ε= 0, r does not depend on φ. This corresponds to a circle. If 0 < ε< 1, wehave an ellipse (r varies between some maximum and minimum value). For ε= 1, we have aparabola (r →∞ for ϕ=π), and for ε> 1 we have an hyperbola (r →∞ for cosϕ= 1/ε).

Defining

r0 = `2

µA, (3.23)

We see that the radius of the orbit varies between r0/(1+ ε) and r0/(1− ε). In figure 3.2, thefocal point F1 must be viewed as the origin where r = 0. We see that the shortest and longestdistance lie in this picture along the horizontal axis. If we add the two, we get

r0

1+ε +r0

1−ε = 2r0

1−ε2 . (3.24)

From the figure, we see that this sum must be equal to twice the long axis, 2a. Therefore

a = r0

1−ε2 . (3.25)

Usually, the notation

λ= `2

µA

1

1+ε (3.26)

is used for the perihelion distance, so that the equation relating the two polar coordinates onthe curve of the motion reads:

r = λ(1+ε)

1+εcosϕ. (3.27)

In figure 3.2, we indicate the semi-major and semi-minor axis a and b respectively and thefocal points. From the definitions for λ and the value of the semi-major axis, we immediatelysee that

a = λ

1−ε . (3.28)

The area of an ellipse in terms of its semi-major axis is πa2p

1−ε2. This can be related to theangular momentum by realising that the infinitesimal area swept by a line from the originto the point of the motion is given by r 2dϕ/2. This tells us that the rate at which this areachanges is given as `2/(2µ), so that the total area, which is swept in one revolution of periodT is equal to T`/(2µ), so that we have:

T`

2µ=πa2

√1−ε2. (3.29)

The quantities a and ε are not independent – remember a = λ/(1− ε); furthermore ` relatedto λ and ε [see Eq. (3.26)]. Using this to eliminate ε finally leads to

T 2 = 4π2µ

Aa3. (3.30)

We have now recovered all three laws of Kepler:

3


a εFF

1 2

a

b

FIGURE 3.2: Ellipse with various parameters indicated.

• All planets move around the sun in elliptical paths. In fact, most planets haveeccentricities very close to zero.

• A line drawn from the sun to a planet sweeps out equal areas in equal times.The rate at which this area increases is given by `2/(2µ) as we have seen above.

• The squares of the periods of revolution of the planets about the sun are pro-portional to the cubes of the semimajor axes of the ellipses. See Eq. (3.30).

3.3 SUMMARYIn this chapter, we have analysed the two-body problem, with a focus on the Kepler case,where the interaction potential of the two bodies is V (r ) = −A/r . Very generally, we haveseen that we can decompose any two-particle system with an ineraction potential betweenthe two particles (and no external field), into the centre of mass motion with mass M andcoordinate

R = m1r1 +m2r2

M. (3.31)

This coordinate experiences a uniform motion.The relative motion is described by the coordinate r = r2 − r1. Its motion is described by

the Lagrangian

L = µ

2r2 −V (r ), (3.32)

where µ is the reduced mass

µ= m1m2

m1 +m2. (3.33)

This motion takes place in a plane spanned by the velocity and the force at any time and istherefore essentially two-dimensional.

For the particular case of the Kepler problem, where V (r ) = −A/r , with positive A, weexpect the particles to either orbit around each other (this happens when their total energyis negative) or they scatter off each other (positive energy). We have solved this problemanalytically for the negative energy solutions. We have seen that the shape of the orbit wasdetermined by

r = r0

1+εcosϕ, (3.34)

where r,ϕ are polar coordinates.

3.3. SUMMARY 41

3

From the solution, the Kepler laws follow. They are summarized at the end of the previoussection.

4EXAMPLES OF VARIATIONAL CALCULUS,

CONSTRAINTS

Suspended chain; this shape is called ‘catenary’, from the latin word ‘catena’.

43

4

44 4. EXAMPLES OF VARIATIONAL CALCULUS, CONSTRAINTS

4.1 VARIATIONAL PROBLEMSIn the previous chapters, we have considered a reformulation of classical mechanics in termsof a variational principle. This will lead the way to formulating quantum mechanics – thisis the subject of the next chapter. In this chapter we make an excursion which is still in thefield of classical problems, though not classical dynamics as in the previous chapters. In fact,variational calculus is not only useful for mechanics. Many physical problems which occurin every day life can be formulated as variational problems. In the next sections we shallconsider a few examples.

We shall first introduce some further analysis concerning the problems we are about totreat in this chapter. Consider an expression of the form

J =∫

d x F (y, y ′, x). (4.1)

We have used a notation which differs from that used in previous chapter in order to empha-sise that J is not always the action and F not always the Lagrangian. J assigns a real value toevery function y(x) – it is called a functional. There is a whole branch of mathematics, calledfunctional analysis, dedicated to such objects. Here we shall only consider finding the sta-tionary solutions (minima, maxima or saddle points) of J ; they are given as the solutions tothe Euler equations

∂F

∂y− d

d x

∂F

∂y ′ = 0 (4.2)

In the case where F does not depend explicitly on x, i.e.

F = F (y, y ′), (4.3)

we can directly integrate the Euler equation(s) once: by multiplying the Euler equation by y ′

we findd

d x

[F (y, y ′)− y ′ ∂F (y, y ′)

∂y ′

]= 0. (4.4)

From this it follows that the solution must obey

F (y, y ′)− y ′ ∂F (y, y ′)∂y ′ = Constant. (4.5)

This is a first order differential equation: we have integrated the second order Euler equationsonce.

This is a quick way of solving the Euler equations. Note that it only works for functionalswhich do not depend explicitly on x!

4.2 THE BRACHISTOCHRONENear the end of the 18-th century, Jean Bernouilli was studying a problem, which we formu-late as follows. Suppose you are to design a monorail in an amusement park. There is a trackin your monorail where the trolleys, which arrive at some high point A with low (approxi-mately zero) speed should move to another place B under the influence of gravity (no motoris used and friction is neglected) in the shortest possible time. The problem is to design theshape of the track in order to achieve this goal. It will be clear that the track lies in a plane.Let us first consider the possible solutions heuristically. One could argue that a straight linewould be the best solution because it is the shortest path between A and B . On the otherhand, it would seem favourable to increase the particle’s velocity as much as possible in thebeginning of the motion. This would call for a steep slope near the starting point A, followedby a more or less horizontal path to B , but the resulting curve is considerably longer than thestraight line, which is the shortest path between A and B . We must therefore find the opti-mum between the shortness of the path and the earliest increase of the velocity by a steeperslope.

4.2. THE BRACHISTOCHRONE 45

4

We can solve this problem using the techniques of the previous section. We must min-imise the time for a curve which can be parametrised as x(s), y(s). Obviously, there are manyways to parametrise a curve – we shall use for s the distance along the curve, measured fromthe point A. The infinitesimal segment d s is given by

d s =√

d x2 +d y2 = d x

√1+

(d y

d x

)2

= d x√

1+ y ′2 (4.6)

s can be expressed as a function of t – the relation between the two is given by

d s = vd t (4.7)

where v =√

v2x + v2

y is the particle speed. The time needed to go from A to B is given by

t =∫ B

A

d s

v. (4.8)

We need an equation for v in terms of the path length. As the gravitational force is responsiblefor the change in velocity it is useful to consider the x- and y-components of the path. In fact,we have the following relation between v and y as a result of conservation of energy:

1

2v2 = g y, (4.9)

where the height y is measured from the point A. This means that we have put in the bound-ary condition that when y = 0, then v = 0, which is correct since the particle is released fromA with zero velocity. Therefore, using (4.9), we arrive at

t [y] =∫ x0

0d x

√1+ y ′2√

2g y(4.10)

where x0 is the horizontal distance between A and B . We have to find the stationary functiony(x) for the functional t [y]. The Euler-Lagrange equations have the solution [see Eq. (4.5)]:√

1+ y ′2

2g y− y ′2

√1(

1+ y ′2)2g y= Constant. (4.11)

Rewriting the first term on the left hand side as√1+ y ′2

2g y= 1+ y ′2

(√

1+ y ′2)2g y, (4.12)

this can be simplified toy(1+ y ′2) =C = Constant. (4.13)

In order to solve this equation we substitute

y ′ = tanφ (4.14)

so that we have:

y =C cos2φ=C

[1

2+ 1

2cos(2φ)

]. (4.15)

Andd x

dφ= 1

y ′d y

dφ= C sin(2φ)

tanφ= 2C cos2φ. (4.16)

4


The solution is therefore

x =−C

[φ+ 1

2sin(2φ)

]+D (4.17a)

y =C

[1

2+ 1

2cos(2φ)

]. (4.17b)

D and C are integration constants – if we identify the point A with (0,0), the curve starts atφ= π/2, and D/C =−π/2. The two coordinates of B are used to fix the value of φ at point Band the constants D and C . Note that the boundary condition y ′ = 0 at point A was alreadyrealised in Eq. 4.9). The resulting curve is called the cycloid – it is the curve described by apoint on a horizontally moving wheel.

4.3 FERMAT ’S PRINCIPLEThe path traversed by a ray of light in a medium with a varying refractive index is not a straightline. According to Fermat’s principle this path is determined by the requirement that the lightray follows the path which allows it to go from one point to another in the shortest possibletime. The time needed to traverse a path is determined by the speed of light along that path,and this quantity is given as

c(n) = c

n(4.18)

where c is the speed of light in vacuum and n is the refractive index. The latter might varywith position.

If the path lies in the x y plane, the path length dl of a segment corresponding to a dis-tance d x along the x-axis is given by

dl = d x

√1+

(d y

d x

)2

. (4.19)

The time d t needed to traverse the path dl is given as:

d t = dl

c/n, (4.20)

so that the total time can now be given as an integral over d x:

t =∫ L

0d x

n(y)

c

√1+

(d y

d x

)2 . (4.21)

Here we have assumed that n depends on the coordinate y only. Now take n(y) =√

1+ y2,then we must minimise

ct =∫ L

0d x

√1+ y2

√1+

(d y

d x

)2 . (4.22)

For this case, the Euler-Lagrange equations reduce to the equation [see Eq. (4.5)]

d y

d x= 1

A

√(1− A2)+ y2. (4.23)

The solution is given as

y(x) =±√

1− A2 sinh( x

A+B

). (4.24)

The possible range of A-values is |A| ≤ 1.

4.4. THE MINIMAL AREA PROBLEM 47

4

x

y(x)

4.4 THE MINIMAL AREA PROBLEMConsider a soap film which is suspended between two parallel hoops (see figure). The soapfilm has a finite surface tension, which means that its energy scales linearly with its area. Asthe film tends to minimise its energy, it minimises

its area. The minimal area for a surface of revolution described by a function y(x) is givenby:

2π∫ L

0d x y

√1+ y ′2. (4.25)

Minimising this functional of y leads to the standard Euler-Lagrange solution Eq. (4.5) forfunctionals with no explicit time dependence:

y√1+ y ′2

=C (4.26)

The solution to this equation is given by

y(x) =C cosh

(x + A

C

)(4.27)

We now assume that the hoops have the same diameter, and that they are placed at a distanceL from each other. Let us furthermore choose the x-axis such that the origin is in the middlebetween the two hoops. Using the fact that cosh is an even function, we have:

R =C cosh

(L

2C

), (4.28)

where R is the radius of the hoops. Consider now the graph of C cosh[L/(2C )] as a functionof C for fixed L: It is clear that for R lying in the “gap” of the graph, no solution can be found.What happens is that if the hoops are not too far apart, the soap film will form a nice cosh-formed shape. However, when we pull the hoops apart, there will be moment at which thefilm can no longer be sustained and it collapses.

It can be seen from the graph that usually there are two different solutions. The one withthe smallest surface is to be selected. The surface area is found as

A(y) =π[

LC +2R√

C 2 +R2]

. (4.29)

4.5 CONSTRAINTS

4.5.1 CONSTRAINT FORCES

In d’Alembert’s approach, the forces of constraint are neglected as they are usually of limitedphysical importance. In some cases, however, it might be useful to know what these forces

4


R/L

-4

-2

0

2

4

-4 -2 0 2 4

x*cosh(0.5/x)

C /L

are. For example, a designer of a monorail would like to know the force which is exerted onthat rail by the train in order to certify that the monorail is robust enough. In fact, it is possibleto work out within a Lagrangian analysis what the forces of constraint are.

Let us first recall the solution to the following problem

Find the minimum of the function f (x), where x = (x1, x2, . . . , xN ), under the con-dition g (k)(x) =Ck , where Ck are constants; k = 1, . . . ,K .

Consider a small variation δx such that g (k)(x+δx) =Ck still holds for all k. Then it holds that

g (k)(x)+δx ·∇g (k)(x) = g (k)(x) =Ck (4.30)

hence

δx ·∇g (k)(x) = 0 (4.31)

for all k. On the other hand, for variations δx satisfying (4.31), f should not change to firstorder along δx, so we have

δx ·∇ f (x) = 0. (4.32)

Now we can show that ∇ f (x) must lie in the span of the set ∇g (k)(x). If it would lie outsidethe span, we can write it as the sum of a vector lying in the span of ∇g (k)(x) plus a vectorperpendicular to this space. If take δx to be proportional to the latter, then (4.31) is satisfied,but (4.32) is not. Therefore we conclude that ∇ f can be written as a linear combination of thegradients ∇g (k):

∇ f (x) =K∑

k=1λk∇g (k)(x). (4.33)

This is the well-known Lagrange multiplier theorem.Let us consider a simple example: finding the minimum or maximum of the function

f (x, y) = x y on the unit circle: g (x, y) = x2 + y2 −1 = 0. There is only one Lagrange parameterλ, and Eq. (4.33) for this case reads

(y, x) =λ(2x,2y), (4.34)

whose solution is x = ±y , λ = ±1/2. The constraint x2 + y2 = 1 then fixes the solution tox = ±y = 1/

p2 and x = ±y = −1/

p2. Indeed, the symmetry of the problem allows only the

axes x = ±y and the x = 0 or y = 0 as the possible solutions, and it is easy to identify thestationary points (minima, maxima, saddle points) among these.

4.5. CONSTRAINTS 49

4

Now suppose we have a mechanical N -particle system without constraints. For such asystem we know the Lagrangian L. The combined coordinates of the system are representedas a vector R = (r1, . . . ,rN ). Then we have for any displacement δR that the correspondingchange in Lagrangian vanishes:

M∑i=1

δri · δL(R, R, t )

δri≡ δL(R, R, t ) = 0 (4.35)

for all t , where we have used the notation of (2.60). Now suppose that there are constraintspresent of the form

g (k)(R) = 0. (4.36)

The argument used above for ordinary functions of a single variable can be generalised toshow that we should have

δL(R, R, t )

δR=

K∑k=1

λk∇Rg (k)(R); (4.37)

the reader is invited to verify this. As L is the Lagrangian of a mechanical system withoutconstraints, we know that the left hand side of this equation can be written as p−FA . Theright hand side has the dimension of a force and must therefore coincide with the constraintforce.

Let us analyse the simple example of the pendulum once again. Without constraints wehave

L = m

2

(x2 + y2)−mg y. (4.38)

The constraint is given byl 2 = x2 + y2. (4.39)

So the pendulum equations of motion become

mx = 2λx (4.40a)

my =−mg +2λy. (4.40b)

These equations cannot be solved analytically, as they describe the full pendulum, and notthe small angle limit. In the small angle limit the force in the x-direction dominates, andtherefore mg /(2λ) should be approximately equal to l . We then see that λ is negative, so

that the solution is oscillatory and the frequency is given by ω =∣∣∣p2λ/m

∣∣∣ = √g /l . The λ-

dependent terms in the equation of motion represent indeed a force in the +y direction ofmagnitude mg : this is the tension in the string or rod on which the weight is suspended.

When using polar coordinates, we have

L = m

[r 2

2+ (r ϕ)2

2

]+mg cosϕ, (4.41)

with the constraint r = l . which leads to the Lagrange equations:

mr = mg cosϕ+mr ϕ2 −λ; (4.42)

m(r 2ϕ+ r r ϕ

)=−mg r sinϕ. (4.43)

Filling the constraint is particularly easy. The constraint force is given by

λ= mg cosϕ+mr ϕ2. (4.44)

The constraint force consists of a term which compensates for the gravity force (first term)and an extra term which is necessary for keeping the circular motion going (a centripetalforce, the second term). The equation for ϕ reduces to the usual pendulum equation whenthe constraint is used:

ϕ=−g /l sinϕ. (4.45)

In practice, constraints are seldom used explicitly in the solution of mechanical problems.

4


4.5.2 GLOBAL CONSTRAINTS

In (4.5.1) we have analysed constraints of the form:

g (k)(r1,r2, . . . ,rN ; t ) = 0. (4.46)

This type of constraint is called holonomic, and it frequently allows us to represent the sys-tem using generalised coordinates. This type of constraint imposes conditions on the systemwhich should hold at any moment in time. We may therefore consider this type of constraintsas an infinite set (one constraint for each time). Such constraints are called local, where thisterm refers to the fact that the constraint is local in time.

Sometimes however, we must deal with constraints of a different form. Consider for ex-ample the problem of finding the shape of a chain of homogeneous density ρ (= mass/unitlength) suspended at its two end points. We represent this shape by a function y(x) where xis the coordinate along the line connecting the two end points and y(x) is the height of thechain for coordinate x. The shape is determined by the condition that it minimises the (grav-itational) potential energy, and it is readily seen that this energy is given by the functional

V = gρ∫ X

0d x y

√1+

(d y

d x

)2

. (4.47)

We leave out the constants g and ρ in the following as they do not affect the shape.If we would minimise the potential energy (4.47), we would get divergences, as we have

not yet restricted the total length L of the wire to have a fixed length. This requirement canbe formulated as

L =∫ X

0d x

√1+

(d y

d x

)2

. (4.48)

This is a constraint which is not holonomic, and there is no way to reduce the number ofdegrees of freedom. This type of constraint is called global as it is formulated as a conditionon an integral of the same type as the functional to be minimised, and is not to be satisfiedfor all values of the integration variable. Therefore, we must generalise the derivation of theEuler equations to cases in which a functional constraint is present.

Let us consider two functionals, J and K :

J =∫ b

ad x F (y, y ′, x) (4.49a)

K =∫ b

ad x G(y, y ′, x), (4.49b)

and suppose we want to minimise J under the condition that K has a given value, i.e., foreach variation δy which satisfies: ∫

δG(y, y ′, x) d x = 0; (4.50)

we require that ∫δF (y, y ′, x) d x = 0. (4.51)

Consider now a particular variation which is nonzero only in a small neighbourhood of twovalues x1 and x2 (see figure 4.1). If the areas under these two humps are A1 and A2 respec-tively, we have∫ b

aδG(y, y ′, x) d x = A1

δG[y(x1), y ′(x1), x1]

δy+ A2

δG[y(x2), y ′(x2), x2]

δy, (4.52)

and thereforeδG1/δy

δG2/δy=− A2

A1(4.53)

4.5. CONSTRAINTS 51

4

x x1 2

FIGURE 4.1: Example of a function δy(x) which is nonzero only near x1 and x2.

y(0,0) (X,0)

x

FIGURE 4.2: The cosh solution to the suspended chain problem.

with an obvious shorthand notation.Applying this argument once again we see that for functions y(x) satisfying requirement

(4.53), we should haveδF1/δy

δF2/δy=− A2

A1(4.54)

But this can only be true for arbitrary x1 and x2 when δF /δy and δG/δy are proportional:

δF

δy=λδG

δy. (4.55)

Therefore, we must solve the Euler equations for the combined functional

J (y)−λK (y) (4.56)

where λ is fixed by putting the solution of this minimisation back into the constraint. This isthe Lagrange multiplier theorem for functionals.

We shall now apply this to the suspended chain problem. We have F = y√

1+ y ′2 and

G =√

1+ y ′2. Therefore, the Euler equations read:

(y +λ)

√1+ y ′2 − y ′2√

1+ y ′2

= Constant, (4.57)

which leads to

y +λ=C√

1+ y ′2. (4.58)

The solution is given byy(x) = A cosh[α(x −x0)]+B (4.59)

4


with

A =C = 1/α (4.60a)

B =−λ. (4.60b)

Boundary conditions are y(0) = y(X ) = 0 and the length of the wire must be equal to L. Theseconditions fix x0 and λ: x0 = X /2, λ = −C cosh[X /(2C )] and C = L/sinh[X /(2C )]. In fig-ure 4.2 the solution is shown.

4.6 SUMMARYIn this chapter we have visited a few applications of functional minimization to several everyday-life problems. An expression

J =∫ b

aF (y, y ′, x)d x (4.61)

where y(x) is a function defined between a and b, and y ′ its derivative with respect to x,assumes its extremal values when

d

d x

∂

∂y ′ F (y, y ′, x) = ∂F (y, y ′, x)

∂y. (4.62)

This is the general form of the Euler equation. We have seen that, for the particular case whereF does not depend explicitly on x, the Euler equation can be integrated once, leading to thenew equation

F (y, y ′)− y ′ ∂F (y, y ′)∂y ′ =C , (4.63)

where C is an integration constant.We also considered constraints of two kinds: local and global. A local constraint is a con-

dition on the function y which should hold for any x. If a set of K such constraints of the form

g (k)(r1, . . . ,rN ) = 0, k = 1, . . . ,K (4.64)

holds for an N -particle problem, then the Euler Lagrange equations take the form

d

d t

∂L

∂r j= ∂L

∂r j+

K∑k=1

∇ jλk g (k)(r1, . . . ,rN ). (4.65)

The solutions for r j obviously depend on the parameters λk – these parameters are fixed atthe end by requiring that the solutions r j satisfy the constraints.

Now we turn to the global constraints: they are conditions on the function y as a whole,usually formulated in terms of one or more integrals. In that case, if we want to minimize thefunctional J = ∫

F (y, y ′, x)d x where y is subject to the constraint

K (y) =∫

G(y)d x = constant, (4.66)

This problem is solved by the solutions of the Euler equation of the functional

J −λK =∫ [

F (y, y ′, x)−λG(y, y ′, x)]

d x. (4.67)

The resulting solution depends on the Lagrange parameter λ, which is solved by substitutingthe solution into the constraint equation.

5SYMMETRY AND CONSERVATION LAWS

53

5

54 5. SYMMETRY AND CONSERVATION LAWS

In this chapter, we return to classical mechanics and shall explore the relation betweenthe symmetry of a physical system and the conservation of physical quantities. In the firstchapter, we have already seen that translational symmetry implies momentum conservation,that time translation symmetry implies energy conservation and that rotational symmetryimplies conservation of angular momentum. There exists a fundamental theorem, calledNoether’s theorem, which shows that, indeed, for every spatial continuous symmetry of a sys-tem which can be described by a Lagrangian, some physical quantity is conserved, and thetheorem also allows us to find that quantity.

The special form of the equations of motion for a system described by a Lagrangian (orHamiltonian) leads already to a large number of conserved quantities, called Poincaré invari-ants. We shall consider only one Poincaré invariant here: phase space volume. The associatedconservation law is called Liouville’s theorem.

5.1 NOETHER’S THEOREMSuppose a mechanical system is invariant under symmetry transformations which can beparametrised using some real, continuous parameter. Examples include those mentionedalready above: rotations (parametrised by the rotation angles) or translations in space ortime. The fact that the system is invariant under these transformations is reflected by theLagrangian being invariant under these symmetries. For simplicity we shall restrict ourselvesto a single continuous parameter, s. In the case of rotations one could imagine s to be therotation angle about an axis fixed in space, such as the z-axis. The mechanical path for somesystem, i.e. the solution of the Euler-Lagrange equations of motion, is called q(t ). Now weperform a symmetry transformation. This gives rise to a different path, which we call Q(s, t ),with Q(0, t ) = q(t ). The path Q(s, t ) should have the same value of the Lagrangian L as thepath q(t ), in other words, L should not depend on s:

d

d sL(Q(s, t ),Q(s, t )) = 0. (5.1)

This leads toN∑

j=1

[∂L

∂Q j

∂Q j

∂s+ ∂L

∂Q j

∂Q j

∂s

]= 0. (5.2)

Now we use the Euler-Lagrange equations:

∂L

∂Q j= d

d t

∂L

∂Q j(5.3)

in order to write

dL

d s=

N∑j=1

[∂L

∂Q j

∂Q

∂s+ ∂L

∂Q j

∂Q j

∂s

]=

N∑j=1

[d

d t

(∂L

∂Q j

)∂Q j

∂s+ ∂L

∂Q j

∂Q j

∂s

]= d

d t

[N∑

j=1

∂L

∂Q j

dQ j

d s

]= 0,

(5.4)and we see that the term within brackets in the last expression must be a constant of themotion:

N∑j=1

∂L

∂Q j

dQ j

d s=

N∑j=1

p jdQ j

d s= Constant in time. (5.5)

We see that any continuous symmetry of the Lagrangian leads to a constant of the motion,given by (5.5). This analysis is obviously rather abstract, so let us now consider an example.

Suppose a one-particle system in three dimensional space is invariant under rotationsaround the z-axis. The rotation angle is calledα. In order to be able to evaluate the derivativesof the coordinates with respect to α, we use cylindrical coordinates (r,ϕ, z) with x = r cosϕand y = r sinϕ. A rotation about the z axis over an angle α then corresponds to

ϕ→ϕ+α. (5.6)

5.2. LIOUVILLE’S THEOREM 55

5

so that we have

pxd x

dα=−px r sin(ϕ+α) =−px y ; (5.7)

pyd y

dα= py r cos(ϕ+α) = py x; (5.8)

pzd z

dα= 0 (5.9)

so that the conserved quantity, from (5.5) is

xpy − y px = Lz , (5.10)

the z-component of the angular momentum. Similarly, we would find Lx and Ly for the con-served quantities associated with rotations around the x- and y- axes respectively. Also, it iseasy to verify that for more than one particle, the total angular momentum is conserved.

The reader is invited to check that space translation symmetry results in momentum con-servation.

5.2 LIOUVILLE’S THEOREMA special conservation law is due to the fact that the equations of motion can be derivedfrom a Hamiltonian (or from a Lagrangian). Such equations of motion are called canonical.The fact that the equations of motion are canonical reflects a symmetry which is called sym-plecticity (or symplecticness), a discussion of which is outside the scope of these notes. Theimportant notion is that this type of symmetry leads to a number of conserved quantities,called Poincaré invariants, of which we shall consider only one, the volume of phase space.

The proof of Liouville’s theorem hinges upon the fact that whenever in a volume integrallike

V =∫Ω

d n x (5.11)

we perform a variable transformation x → y, we must put a correction factor det(J) in theintegral, where J is the Jacobian matrix, given by

Ji j = ∂yi

∂x j. (5.12)

We thus have

V =∫Ω

d n x det(J) =∫Ω′

d n y. (5.13)

whereΩ′ is the volumeΩ transformed to y-space.The state of a mechanical system consisting of N degrees of freedom is represented by

a point in 2N -dimensional phase space (qi , p j ). In the course of time, this point moves inphase space, and forms a trajectory. We now consider not a single mechanical system inphase space, but a set of systems which are initially homogeneously distributed over someregionΩ0, with volume V0. In the course of time, every point inΩ0 will move in phase space,Ω0 will therefore transform into some new regionΩ(t ). The volume of this new space is givenas

V (t ) =∫Ω(t )

d n q d n p (5.14)

We want to show that V (t ) = V0, hence the volume does not change in time. To this end, weconsider a transformation from time t = 0 to ∆t :

q ′i ≡ qi (∆t ) = qi (0)+ ∂H(p,q)

∂pi∆t +O (∆t 2); (5.15)

p ′i ≡ pi (∆t ) = pi (0)− ∂H(p,q)

∂qi∆t +O (∆t 2), (5.16)

5

56 5. SYMMETRY AND CONSERVATION LAWS

where we have used a first order Taylor expansion and replaced time derivatives of qi and pi

using Hamilton’s equations. Now we can evaluate the original volume V0 as follows:

V0 =∫Ω

d n q d n p =∫Ω(∆t )

d n q d n p det[J(∆t )] . (5.17)

The Jacobi determinant can be written in block-form as follows:

det[J(∆t )] =∣∣∣∣∣ 1+∆t ∂2 H

∂qi∂p j−∆t ∂2 H

∂qi∂qi

∆t ∂2 H∂pi∂p j

1−∆t ∂2 H∂q j∂pi

∣∣∣∣∣ (5.18)

Careful consideration of this expression should convince you that det[J(∆t )] = 1+O (∆t 2). Wesee therefore that V (∆t ) =V0 +O (∆t 2), from which it follows that

dV (0)

d t= 0. (5.19)

This argument can be extended for arbitrary times, so that we have proven that V is a constantof the motion. We have found Liouville’s theorem in the form:

The volume of a region phase occupied by a set of systems does not change in time.

Of course, the region can change in shape, but its total volume will remain constant in time.We could have put any density distribution of points in phase space in the integrals, whichdoes not change the derivation.

Liouville’s theorem is important in equilibrium statistical mechanics. So-called ergodicsystems are assumed to move to a time-independent distribution in phase space, that is, anylarge set of systems setting off at time t = 0 from different points in phase space and movingaccording to the Hamiltonian equations of motion will assume the same, invariant distribu-tion after long times. Liouville’s theorem, moreover tells us that the systems will not all endup in the same point in phase space, but spread over a region with a volume equal to theinitial volume.

There exists a more specific theorem concerning the behaviour of systems in time. This isPoincaré’s theorem, which says that a system which is to evolve under the mechanical equa-tions of motion will always return arbitrarily close to its starting point within a finite time.Consider for example a box partitioned into two sub-volumes (figure 5.1). There is a smallhole in the middle, and there are N particles in the right hand volume. Obviously, a fractionof these particles will move to the left hand volume, but the Poincaré recurrence theorem tellsus that after a finite time, all particles will reassemble in the right hand volume! This seems tobe in contradiction with the second law of thermodynamics. This law states that the entropywill not decrease in the course of time. Here we see an increase of entropy when the particlesdistribute themselves over the two volume halves rather than a single one, but come backin a more ordered (less entropic) state after some time. This is only an apparent contradic-tion, as the Poincaré theorem holds for a finite number of particles (finite-dimensional phasespace). What we see here is an example of the inequivalence of interchanging the order inwhich limits are taken: if we first take the system size to infinity (the approach of statisticalmechanics and thermodynamics), the recurrence time will become infinite. If, on the otherhand, we consider a finite system over infinitely large times (the mechanics approach), wesee that it returns arbitrarily close to its initial state infinitely often. Taking then the systemsize to infinity does not alter this conclusion.

5.2. LIOUVILLE’S THEOREM 57

5

FIGURE 5.1: A box divided into two halves by a wall with a hole. Initially, particles will be in the right hand volume,and will move to the left. After large times, they will all come back to the right hand volume.

6SYSTEMS CLOSE TO EQUILIBRIUM

Vibrational modes of carbon molecules and methane play a crucial role in the greenhouse effect –

see section 6.4. The right figure shows the vibrational modes of methane.

59

6

60 6. SYSTEMS CLOSE TO EQUILIBRIUM

S

M

V

FIGURE 6.1: System with a metastable (M) and a stable (S) state. A strong perturbation may kick the ball out of itsmetastable state, and under the influence of dissipation it will then move to the stable state.

6.1 INTRODUCTION

When we prepare a conservative system in a state with a certain energy, it will conserve thisenergy ad infinitum. In practice, such is never the case, as it is impossible to decouple asystem from its environment or from its internal degrees of freedom. This requires someexplanation. We usually describe macroscopic objects in terms of the coordinates of theircentre of mass and their Euler angles. These are the macroscopic degrees of freedom. As theseobjects consist of particles (atoms, molecules), they have very many additional, internal, ormicrosopic degrees of freedom. In fact the heat which is generated during friction is nothingbut a transfer of mechanical energy associated with the macroscopic degrees of freedom toa mechanical energy of the internal (microscopic) degrees of freedom: if we throw a ball atthe wall, the molecules of the ball will vibrate more as a result of the collision, resulting in aloss of the mechanical energy associate with the centre of mass motion of the ball. So heat inthe end is a form of mechanical energy. As a result of friction, macroscopic objects will, whensubject to a conservative, time-independent force (apart from friction), always end up at restin a point where there potential energy is minimal. Any system at a point where its potentialenergy is minimal is said to be in a stable state. A system which looses its kinetic energy viafriction is called a dissipative system. All the macroscopic systems we know are dissipative,although some can approach conservative systems very well.

If the interactions are not all harmonic (‘harmonic’ means that the potential energy is aquadratic functions of the coordinates) then there may be more than one minimum. Localminima correspond to metastable states. A system in a metastable state will return to thatstate under a small perturbation, but, when it is strongly perturbed, it might move to an-other metastable with lower potential energy, or to the stable state. An example is shown infigure 6.1, for a particle in a one-dimensional potential.

Molecular systems, in which we take all degrees of freedom explicitly into account, arebelieved to be non-dissipative. We know from statistical physics that every degree of freedomin a system in thermal equilibrium carries a kinetic energy equal (on average) to kBT , wherekB is Boltzmann’s constant. At low temperatures, the energies of the particles are small as canbe seen from the Boltzmann distribution, which gives the probability of finding a system withenergy E as exp[−E/(kBT )]. Therefore, for low temperatures, the kinetic and the potentialenergy of a system are small. It can therefore be inferred that at low temperatures, a systemis close to a (meta-)stable state.

In this section, we analyse systems close to mechanical equilibrium. The beautiful resultof this analysis is a description in terms of a set of uncoupled harmonic oscillators, whichare themselves trivial to analyse. Moreover we obtain a straightforward recipe for finding theresonance frequencies (related to the coupling strengths) of those oscillators.

6.2. ANALYSIS OF A SYSTEM CLOSE TO EQUILIBRIUM 61

6

6.2 ANALYSIS OF A SYSTEM CLOSE TO EQUILIBRIUMConsider a conservative system characterised by generalised coordinates q j , j = 1, . . . , N . Thesystem is in equilibrium, defined by q1, . . . , qN , if its potential energy is minimal. In that casewe have

∂

∂q jV (q1, . . . , qN ) = 0; j = 1, . . . , N . (6.1)

Now suppose that we perturb the system slightly, i.e. we change the values of the q j slightlywith respect to their equilibrium values. As the first derivative of the potential with respect toeach of the q j vanishes, a Taylor expansion of the potential only contains second and higherorder terms:

V (q1, . . . , qN ) =V (q1, . . . , qN )+ 1

2

N∑j ,k=1

(q j − q j )(qk − qk )∂2

∂q j∂qkV (q1, . . . , qN )+ . . . . (6.2)

The terms of order higher than two will be neglected, as we are interested in systems close toequilibrium (i.e. q j − q j small).

We can represent the resulting expansion using matrix notation. Introduce the matrix

K =

∂2V

∂q1∂q1

∂2V∂q1∂q2

· · · ∂2V∂q1∂qN

∂2V∂q2∂q1

∂2V∂q2∂q2

· · · ∂2V∂q2∂qN

......

......

∂2V∂qN∂q1

∂2V∂qN∂q2

· · · ∂2V∂qN∂qN

. (6.3)

The matrix K is obviously symmetric. We can write

V (q1, . . . , qN ) =V (q1, . . . , qN )+ 1

2δqTKδq (6.4)

where δq is a column vector with components q j − q j , j = 1, . . . , N ; the superscript T denotesthe transpose of the vector. For clarity, we write the last term in this equation explicitly:

1

2δqTKδq = 1

2

(q1, . . . , qN

)

K11 K12 · · · K1N

K21 K22 · · · K2N... · · · ...

...KN 1 KN 2 · · · KN N

q1

q2...

qN

. (6.5)

Now we write down the kinetic energy in terms of the generalised coordinates. We as-sume that the constraints only depend on the generalised coordinates q j and not on theirderivatives or on time. In that case, the kinetic energy can be written in the form (see page30):

T = 1

2

N∑j ,k=1

M j k q j qk , (6.6)

where the matrix M is symmetric: M j k = Mk j . Note that M j k may depend on the q j . In termsof q j − q j , and using vector notation, we can rewrite the kinetic energy as

T = 1

2δqTMδq. (6.7)

The equations of motion now read:

N∑k=1

(M j kδqk +δqk Mk j ) =− ∂V

∂q j+ ∂T

∂q j=−

N∑k=1

(Kk jδqk +δqk Kk j ). (6.8)

We have omitted the dependence of Mi j on q j – this dependence generates terms on the left-and right hand side, which are both of order δq2 and can therefore be neglected. The easiest

6


way of finding the matrix M in practical applications is by writing up the kinetic energy andthen identifying each M j k as the coefficient of the term containing q j qk . Formally, we canalso formulate M j k as the second derivative matrix of the kinetic energy with respect to thevelocity coordinates:

M j k = ∂2T

∂q j∂qk.

Using the symmetry of the matrices M and K , (6.8) reduces to

N∑k=1

M j kδqk =−N∑

k=1K j kδqk . (6.9)

Let us consider the two-dimensional case to clarify the procedure. We consider two gen-eralised coordinates q1 and q2 with the matrix M j k equal to the identity. Then, the kineticenergy has the form:

T = 1

2q2

1 +1

2q2

2 . (6.10)

The potential energy depends on the two coordinates q1 and q2:

V =V (q1, q2). (6.11)

The equations of motion read:

q1 = δq1 =− ∂V

∂q1(6.12)

q2 = δq2 =− ∂V

∂q2(6.13)

Expanding about the point q1, q2, where V is supposed to be minimal, we have

V (q1, q2) =V (q1, q2)+ 1

2(q1 − q1)2 ∂2

∂q21

V (q1, q2)+

(q1 − q1)(q2 − q2)∂2

∂q1∂q2V (q1, q2)+ 1

2(q2 − q2)2 ∂2

∂q22

V (q1, q2). (6.14)

This can be written in the form:

V (q1, q2) =V (q1, q2)+ 1

2(q1 − q1, q2 − q2)

∂2V∂q2

1

∂2V∂q1∂q2

∂2V∂q1∂q2

∂2V∂q2

1

(q1 − q1

q2 − q2

). (6.15)

Defining δq1 = q1 − q1 and similarly for δq2, this equation reads:

V (q1, q2) =V (q1, q2)+ 1

2(δq1,δq2)

∂2V∂q2

1

∂2V∂q1∂q2

∂2V∂q1∂q2

∂2V∂q2

1

(δq1

δq2

). (6.16)

The 2×2 matrix occurring in this expression is our matrix K.

6.2.1 EXAMPLES: A SPRING SYSTEM AND THE DOUBLE PENDULUM

As a trivial example, we consider two masses m1 and m2 and three springs with spring con-stants κ1 and κ2 and κ3 in the arrangement shown in the figure.

κ 1 m 1 κ 3 m 2 κ 2

1x 2x

6.2. ANALYSIS OF A SYSTEM CLOSE TO EQUILIBRIUM 63

6

In terms of the displacements x1 and x2 of the masses with respect to the their equilibriumpositions when the system is at rest, the Lagrangian is

L = m1

2x2

1 +m2

2x2

2 −κ1

2x2

1 −κ2

2x2

2 −κ3

2(x2 +x1)2. (6.17)

We do not have to linearise as the Lagrangian is already in quadratic form. The matrices Mand K are

M =(

m1 00 m2

); K =

(κ1 +κ3 κ3

κ3 κ2 +κ3

). (6.18)

Note the off-diagonal elements which have been chosen such as to make K symmetric.The second, somewhat more interesting example is the double pendulum, consisting of

two rigid massless rods of length l and L, with masses M and m:

l

L

M

m

θ

ϕ

The velocity of the upper mass is Lϑ, that of the lower one is a vector sum of the velocity ofthe upper one and that of the lower one with respect to the upper one. For very small angles ϑandϕ both velocities will be approximately in the horizontal direction so that they can simplybe added: vm = Lϑ+ l ϕ. The kinetic energy therefore reads:

T = M

2

(Lϑ

)2 + m

2

(Lϑ+ l ϕ

)2. (6.19)

Let us perform a transformation to more convenient variables

x = Lϑ (6.20a)

y = Lϑ+ lϕ. (6.20b)

Note that x and y do not denote cartesian coordinates. In that case the kinetic energy cansimply be written as

T = M

2x2 + m

2y2. (6.21)

The potential energy of the upper mass is M g L(1− cosϑ) ≈ M g Lϑ2/2, and that of thelower mass is given as mg

[L(1−cosϑ)+ l (1−cosϕ)

], which, in the small angle approxima-

tion becomes

VLower(ϑ,ϕ) = 1

2mg

(Lϑ2 + lϕ2) . (6.22)

The total potential energy, written in terms of x and y , therefore reads:

V = (M +m)g

2Lx2 + mg

2l(y −x)2. (6.23)

6


The equations of motion can therefore be written as

(M 00 m

)(xy

)=

( −M+mL g − mg

lmg

lmgl −mg

l

)(xy

). (6.24)

This is the form given in (6.9). We shall return to this example in the next section.

6.3 NORMAL MODESThe equations of motion of the double pendulum would be easy to solve if both matrices Mand K would be diagonal. In that case, we would end up with a set of uncoupled harmonicoscillators. It is possible indeed to find a linear transformation of the generalised coordinatesq j such that these matrices are diagonal.

The key for finding the diagonalised representation resides in finding solutions to Eq. (6.16)of the form

δq j = A j e iωt (6.25)

where ω does not depend on j – all the degrees of freedom oscillate at the same frequency.Such a motion is called a normal mode. In the following we shall use q j instead of δq j : q j isthe generalised coordinate measured with respect to its equilibrium value.

We have q j =−ω2q j , so (6.9) reduces to

N∑k=1

M j kω2 Ak =

N∑k=1

K j k Ak . (6.26)

If the mass tensor M j k would be the identity, Eq. (6.26) would be an eigenvalue equation. Forgeneral mass tensors, the equation is a generalised eigenvalue equation. We can reduce thisequation to an ordinary eigenvalue equation by multiplying the left and right and side by theinverse M−1 of the mass matrix. We then have:

ω2 A j =N∑

k,l=1M−1

j k Kkl Al . (6.27)

In algebraic terms, the solutions to these equations are the eigenvectors A (with componentsA j ) and the corresponding eigenvalues ω2. In physical terms, the components A j are theamplitudes of the oscillatory motions of the generalised coordinates, and ω is the frequencyof the oscillation. In order for the normal modes to exist, the eigenvalues should be real. Thatthe eigenvalues are indeed real follows from the fact that both M and K are real, symmetricmatrices. It is a well-known result of linear algebra that the eigenvalues of the generalisedeigenvalue problem are then real.

Another question is whether the eigenvalues are positive or negative. Assuming that weare expanding the potential around a minimum, the matrix K can be shown to be positivedefinite. A positive definite matrix has only positive eigenvalues1. Moreover, the mass matrixcan be shown to be positive. Then its inverse M−1 is also positive. Multiplying two positivematrices yields a product matrix which is positive. Therefore M−1K is positive, and theω2 arepositive. Hence the frequencies of the oscillations are always real – we do not find expontialgrowth or decay. In physical terms one could say that perturbing the system from equilibriumalways pushes it back to this equilibrium – therefore the ‘spring force’ experienced by thecoordinates is always opposite to the perturbation, and therefore an oscillation arises, andnot a drift away from equilibrium, or some exponential decay. Such decay may however befound near a local maximum or near a saddle point of the potential.

1In fact, we shall occasionally allow for zero eigenvalues; in that case, the matrix is called positive semidefinite.

6.3. NORMAL MODES 65

6

6.3.1 NORMAL COORDINATES

We will now show that it is possible to linearly transform the generalised coordinates suchthat both M and K assume a diagonal form. First note that the matrix M is real and symmetric– therefore its eigenvalues are real. Furthermore they are positive, as we have just noted. So,M can by put in diagonal form by an orthogonal transformation O (as you should know fromyour linear algebra course):

OTMO = d, (6.28)

where the superscript ‘T’ denotes the transpose as usual, and d is a diagonal matrix with theeigenvalues d j of M on the diagonal. It is easy to see that we can bring d to unit form bymultiplying on the left and right by the matrix d−1/2, which is a diagnonal matrix with the

values 1/√

d j . The positivity guarantees that the 1/√

d j are real. So we can write

d−1/2OTMOd−1/2 = I (6.29)

(I is the unit matrix). Using the fact that

OOT = OTO = I, (6.30)

Starting now from the generalized eigenvalue equation

ω2MA = KA, (6.31)

where A is a vector and M and K are matrices, and multiply on both sides from the left withd−1/2OT we get

ω2d−1/2OTMA = d−1/2OTKA. (6.32)

Now we insert on the left and right hand side the unit matrix in the form

I = Od−1/2d1/2OT, (6.33)

to obtainω2d−1/2OTMOd−1/2d1/2OTA = d−1/2OTKOd−1/2d1/2OTA. (6.34)

Using (6.29), and definingK′ = STKS (6.35)

whereS = Od−1/2, (6.36)

we then haveω2d1/2OTA = Kd1/2OTA. (6.37)

If we now defineA′ = STA, (6.38)

the result can concisely be written in the form

ω2A′ = K′A′, (6.39)

in other words, we must diagonalise the matrix K′. Having done so, we have found a set ofvectors A′ which, after transforming them back to A:

A = (ST)−1

A′, (6.40)

give the eigenvectors of the original generalised eigenvalue problem (6.27). It is easy to showthat K′ is symmetric (use (AB)T = BTAT and the fact that K is symmetric):(

K′)T = (d−1/2OTKOd−1/2)T = d−1/2OTKOd−1/2. (6.41)

6


so the set of vectors A′n is, or can be chosen orthogonal. It may seem strange that we can

always find simultaneous eigenvectors of the transformed M′ (i.e. the unit matrix) and K′,while these two matrices need not commute. It should however be noticed that we do notdiagonalise as usual through a similarity transform (i.e. by an orthogonal transformation) butallow for a more general, linear transformation (a so-called congruence transform). It turnsout that the eigenvectors are not orthogonal as would be the case with a similarity transform(as is done in quantum mechanics).

We can now rewrite the Lagrangian as follows:

L = 1

2qTMq− 1

2qTKq =

1

2qT (

ST)−1STMSS−1q− 1

2qT (

ST)−1STKSS−1q =

1

2q′T ·q′− 1

2q′TK′q′, (6.42)

where in the first step, we have inserted unit matrices of the form SS−1 etcetera, and in thesecond step we have used q′ = S−1q.

Let us now writeq′ =∑

nCnA′

n , (6.43)

then, using the fact that the A′n can be taken orthonormal, we see that the Lagrangian takes

the form

L = 1

2

∑n

(C∗

nCn −C∗nω

2nCn

). (6.44)

This represents a system consisting of a set of uncoupled harmonic oscillators. The new coor-dinates Cn are called normal coordinates. Note that the vectors An = SA′

n satisfy the equations(6.26). These vectors need not be orthonormal, but the vectors A′

n are.As a first example of normal modes, we consider the system consisting of two masses and

three springs. For simplicity, we take the masses and the spring constants equal m and κ

respectively). We then obtain the following equation:∣∣∣∣ mω2 −2κ −κ−κ mω2 −2κ

∣∣∣∣ . (6.45)

This yields a quadratic equation in ω2 with solutions

ω2 = κ

mand ω2 = 3

κ

m. (6.46)

The eigenvectors are (1−1

)and

(11

)(6.47)

respectively. The first mode corresponds to the two masses moving parallel and feeling onlythe springs with which they are attched to the walls. The middle spring does not influencethe motion and the frequency is given by a single mass and single spring (ω2 = κ/m). In thesecond mode, the masses move in opposite directions. If we introduce the coordinates

u = x1 +x2p2

(6.48a)

v = x1 −x2p2

, (6.48b)

we see that the Lagrangian takes the form

L = m

2

(u2 + v2)− κ

2v2 −3

κ

2u2, (6.49)

6.4. VIBRATIONAL ANALYSIS IN MOLECULES 67

6

so that we have indeed written the Lagrangian as a sum of two Lagrangians of single harmonicoscillators: this is the normal coordinate representation discussed above.

As a second example, we consider finding the normal modes for the coupled pendulums.Note that this problem is relatively simple as a result of the fact that the mass matrix is diag-onal and therefore trivial to invert. After multiplying both sides of Eq. (6.24) by M−1, we havea standard diagonalisation problem for the matrix:( M+m

ML g + mgMl −mg

Ml− g

lgl

). (6.50)

The eigenvalues are the solutions of the so-called secular equation which has the from∣∣∣∣ M+mML g + mg

Ml −ω2 −mgMl

− gl

gl −ω2

∣∣∣∣= 0. (6.51)

This reduces to the following quadratic equation in ω2:

ω4 − M +m

M

( g

L+ g

l

)ω2 + M +m

M

g 2

Ll= 0. (6.52)

This equation has two solutions for ω2.We will examine some special cases. If M À m, then, provided that l is not too close to L,

the two roots with corresponding eigenvectors (Ax , Ay ) are given by

ω≈√

g

l;

Ax

Ay≈ m

M

L

l −L(6.53)

and

ω≈√

g

L;

Ax

Ay≈ L− l

L. (6.54)

The first solution describes an almost stationary motion of the upper pendulum with thelower one oscillating at its natural frequency. In the second case, the motion of the upper andlower are of the same order of magnitude with the natural frequency of the upper pendulum.

If M ¿ m, the solutions are

ω2 ≈ g

L+ l

Ax

Ay= L

L+ l(6.55)

and

ω2 ≈ m

M

( g

L+ g

l

),

Ax

Ay≈−m

M

L+ l

L. (6.56)

The first case describes a motion in which the two rods are aligned so that we have essentiallya single pendulum of length l +L and mass m. The second case corresponds to a very highfrequency of the upper mass with an almost stationary lower mass.

6.4 VIBRATIONAL ANALYSIS IN MOLECULESThe way in which atoms are bound together in molecules is described by quantum mechan-ics. There is a long standing tradition in the quantum mechanical calculation of stationarystates of molecules. In the last fifteen years or so it has become possible to perform dynamicalcomputations of molecules to very good accuracy using fully quantum mechanical calcula-tions. These calculations are quite demanding on computer resources and they do not alwaysgive a very good insight into the dynamics of interest. Therefore, a semi-classical approach isoften adopted in order to calculate vibration spectra for example.

First, the total energy of the molecule is calculated as a function of the nuclear positionsRi , i = 1, . . . , N for an N -atomic molecule. There is however a problem in doing this. Suppose

6


stretch

torsion

bend

FIGURE 6.2: Interactions in a molecule.

1

2

3

m

µ µ

FIGURE 6.3: Triatomic molecule.

.

we want to calculate this energy for 10 values of all the coordinates of a 10-atom molecule.As there are 30 coordinates, we need to perform 1030 stationary quantum calculations, whichwould require the age of the universe. Therefore the potential is parametrised in a sensibleway, which we now describe. All the chemically bonded atoms are described by harmonicor goniometric interactions. The degrees of freedom chosen for this parametrisation are thebond length, bond angle and dihedral angle. The forces associated with these degrees of free-dom are called stretch, bend, and torsion respectively. These degrees of freedom are shown infigure 6.2 The form of the potentials associated with bond stretching is given as

VStretch = κ

2(l − l0)2 (6.57)

where l is the bond length and l0 is the equilibrium bond length. The spring constant κdetermines how difficult it is to stretch the bond. The bending potential is given in terms ofthe bond angle ϕ:

VBend = α

2(ϕ−ϕ0)2 (6.58)

A similar expression exists for the torsional energy.The constants κ and α can be determined from stationary quantum mechanical calcu-

lations. Assuming that these parameters are known, we shall now use the given form of thepotential to calculate the vibration spectrum of a triatomic, linear molecule, such as CS2 orCO2 (see figure 6.3). We neglect bending here, so only bond stretching is taken into account.

If the initial configuration is linear, the motion takes place along a straight line, which wetake as our X -axis. The coordinates of the three atoms are x1, x2 and x3. The kinetic energy

6.4. VIBRATIONAL ANALYSIS IN MOLECULES 69

6

can therefore be written down immediately:

T = µ

2

(x2

1 + x23

)+ m

2x2

2 . (6.59)

The potential energy is given by

V = κ

2(x2 −x1 − l )2 + κ

2(x3 −x2 − l )2 . (6.60)

Here, l is the equilibrium bond length. The centre of mass of the system will move uniformlyas there are no external forces acting, and we take this centre as the origin. The equilibriumcoordinates are then x1 =−l , x2 = 0 and x3 = l . The deviations from these values are

δx1 = x1 + l ; δx2 = x2 and δx3 = x3 − l . (6.61)

In this representation, we have

T = µ

2

(δx2

1 +δx23

)+ m

2δx2

2 . (6.62)

andV = κ

2(δx2 −δx1)2 + κ

2(δx3 −δx2)2 . (6.63)

We can find the matrices K and M directly from these expressions:

M = µ 0 0

0 m 00 µ

, (6.64)

and

K = κ 1 −1 0

−1 2 −10 −1 1

. (6.65)

The normal modes can now be found by solving (6.26) with these matrices. The eigenvectorscan be found by solving the secular equation:∣∣∣∣∣∣

κ−µω2 −κ 0−κ 2κ−mω2 −κ0 −κ κ−µω2

∣∣∣∣∣∣= 0. (6.66)

This leads to:(κ−µω2)ω2(µmω2 −κm −2κµ) = 0, (6.67)

from which we find:

ω1 = 0; ω2 =√κ

µ; ω3 =

√κ

(1

µ+ 2

m

). (6.68)

The corresponding eigenvectors can be found after some algebra:

A1 = 1

11

; A2 = 1

0−1

; A3 = 1

−2µ/m1

. (6.69)

The first of these, corresponding to ω1 = 0, is a mode in which the atoms all slide in the samedirection with the same speed. This is a manifestation of the translational symmetry of theproblem, which has been recovered by our procedure. The second one represents a modein which the middle atom stands still and the two outer atoms vibrate oppositely. Obviously,

6


1

2

3

m

µ µ

Mode 1

Mode 2

Mode 3

FIGURE 6.4: The three modes of the triatomic molecule.

the frequency of this mode is ω2 =pκ/m, corresponding to the two springs. Finally, the last

mode is one in which the two outer atoms move in one direction, and the central atom in theopposite direction. The motion can be understood by replacing the two outer masses by asingle one with mass 2µ at their midpoint, coupled by a spring with spring constant 2κ to thecentral mass. The reduced mass of this system (1/(2µ)+1/m)−1 then occurs in the expressionfor the resonance frequency. The three modes are depicted in figure 6.4.

The vibrations of the CO2 molecule are very important: CO2 is one of the notorious‘greenhouse gases’ which are held responsible for part of the global warming. This is becauseinfrared radiation emanating from the earth to cool down its surface, are partly absorbed bythe vibrational modes of the greenhouse gases in the atmosphere. After some time, a vibra-tionally excited molecule relaxes by emitting radiation, part of which reaches the earth again.In addition to the modes we have studied above, there are bending modes and rotations be-ing excited in the molecule.

The vibrational spectrum of a molecule can be measured by studying the absorption ofinfrared radiation or Raman scattering. Obviously, the frequencies depend on the value κ ofthe spring constant, which is difficult to find. But the ratio between the two frequencies turnsout independent of κ: we find

ω3

ω2=

√1+2

µ

m≈ 1.9 (6.70)

where we have used the masses for the carbon (12 amu) and oxygen (16 amu). The frequen-cies have been measured at 2349 cm−1 and 1388 cm−1 which have a ratio of about 1.7. The dif-ference with our result is explained by the interaction between the two oxygen atoms whichwe have neglected in our simple analysis.

6.5 THE CHAIN OF PARTICLESIn the previous section we have analysed a triatomic molecule. Now we shall analyse a largersystem: a chain of N particles. We assume that all particles have the same mass, and thatthey are connected by a string with tension τ. The particles are assumed to move only in thevertical (y) direction, and the x-components of adjacent particles differ by a separation d .The first and last spring are connected to points at y = 0. The chain is depicted in figure 6.5.The chain is a model for a continuous string, which is obtained by letting N →∞ and d → 0while keeping the string length N d fixed.

Let us consider particle number k. The springs connecting this particle to its neighboursare stretched, and this may result in a net force acting on particle k. The spring between

6.5. THE CHAIN OF PARTICLES 71

6

k

k+

k−1

1yk

d

FIGURE 6.5: The harmonic chain of particles.

particle k and k +1 has a length

l =√

d 2 + (yk+1 − yk )2 ≈ d + (yk+1 − yk )2

2d(6.71)

where a first order Taylor expansion is used to obtain the second expression. The potentialenergy for this link is equal to the tension τ times the extension of the string, and thereforewe find for the total potential energy:

V = τ

2d

N∑k=0

(yk+1 − yk )2 (6.72)

withy0 = yN+1 = 0. (6.73)

The kinetic energy is given by

T =N∑

k=1

m

2y2

k . (6.74)

We now find the matrices Mkl and Kkl as:

Mkl = mδkl (6.75)

where δkl is the Dirac delta function, in other words, Mkl is m times the unit matrix. For Kkl

we find:

K =

2 τ

d − τd 0 0 0 · · · 0

− τd 2 τ

d − τd 0 0 · · · 0

0 − τd 2 τ

d − τd 0 · · · 0

......

......

......

. . .

. (6.76)

Note that the indices of this matrix run from 1 to N : the end points are not degrees of freedom.The normal mode equation (6.26) can be solved analytically for arbitrary N by substitutingfor the eigenvector Ak = γexp(iαk), where γ is some constant. This trial solution does notsatisfy the boundary equations (6.73), but we do not bother about this for the moment. Thenfor 2 ≤ k ≤ N −1 we find

mω2e i kα = τ

d

(−e iα(k−1) +2e i kα−e iα(k+1)

)(6.77)

Dividing left and right hand side by exp(i kα), we find

mω2 = 2τ

d(1−cosα). (6.78)

For each α, there is also a solution for −α for the same ω. This can be used to construct asolution

Ak = γ(e i kα−e−i kα) = 2iγsinkα. (6.79)

6


This solution always vanishes at k = 0 and it vanishes also at k = N+1 when (N+1)α= nπ, forinteger n. So the conclusion is that for each n = 0, . . . , N , we have a solution which vanishesat the two ends of the string. For values of n higher than N , or lower than 0, the solutionsobtained are identical to the solutions with 0 ≤ n ≤ N . For each solution, all particles moveup and down with the same frequency, given by (6.78). The wavelength is given by kd suchthat kα= 2π, so λ= 2πd/α, and the wavevector q =α/d .

It is possible to formulate the Lagrangian directly in a continuum form, and derive thewave equation from this. Note that in the continuum limit, α and d small, we obtain from(6.78) for the frequency:

ω2 = τα2

md= τ d

mq2. (6.80)

Comparison with the well known dispersion equation ω= cq , we learn that the sound speedc is given as

pτd/m. Defining the density ρ = m/d , we have

c =√τ

ρ. (6.81)

6.6 SUMMARYIn this chapter, we have considered normal modes: vibrational excitations of a system closeto equilibrium where all the degrees of freedom vibrate at the same frequency. Our analysisstarted by Taylor-expanding the potential leading to

V = V + 1

2δqTKδq, (6.82)

where δq = q− q is the vector of displacements with respect to the equilibrium position (forwhich ∂V /∂q j = 0 for all j ). In this expression, K is the matrix containing the second deriva-tives of the potential:

K j k = ∂2V

∂q j∂qk. (6.83)

We have also written the kinetic energy as a quadratic expression in the velocities:

T = 1

2δqTMδq, (6.84)

where M is the mass matrix:

M j k = ∂2M

∂q j∂qk. (6.85)

Note that δq j = q j .The Euler Lagrange equations can be formulated concisely using the matrices M and K:

N∑k=1

M j kδqk =N∑

k=1K j kδqk . (6.86)

Requiring that all degrees of freedom δq j vibrate at the same frequency ω yields the matrixequation:

ω2MA = KA, (6.87)

where the vector A should satisfy this generalised eigenvalue problem defined by the matricesM and K. This equation can sometimes be solved by educated guessing of the eigenvectors.More generally, the frequencies are found as the zeroes of the polynomial∣∣ω2M−K

∣∣= 0. (6.88)

For each frequency found, putting it into the generalised eigenvalue equations allows us tofind the corresponding eigenvectors.

It turns out that the Lagrangian can be written as a sum of Lagrangians, each of whichdescribes an independent harmonic oscillator.

APROBLEMS AND EXERCISES CLASSICAL

MECHANICS

73

A

74 A. PROBLEMS AND EXERCISES CLASSICAL MECHANICS

* = Problems which test you elementary knowledge, you should be able to do these correctly!** = Problems on the level of the exam*** = Challenging (difficult) problems

Answers can be found at the end.

NEWTON’S LAWS AND CONSERVATION LAWS* 1. A chain of length l is lying on a frictionless table and then released. A length y0 of

the chain hangs freely over the edge, so that the chain is pulled downward. Give thelength of the chain as a function of time until the moment where the chain is fallingcompletely freely.

* 2. (a) What is the moment of inertia with respect to an exis through the center of a (non-massive) wheel with radius r and mass m?

(b) The same question for an axis through the rim of the wheel.

(c) Give the acceleration with which such a wheel rolls down a slope which makes anangle α with the horizon.

(d) What should be the minimal friction coefficient to prevent the wheel from sliding?

* 3. A comet approaches the solar system with velocity v and would, if the sun would notattract the comet, pass the sun at a distance d . What is in reality the shortest distancebetween comet and sun. (Hint: use conservation of angular momentum and energy).

CONVENIENT COORDINATES

* 4. Two rigid bodies, attached with a small ball bearing, can move freely through space.How many degrees of freedom does this system possess? Explain your answer.If one of the two bodies is attached to a second ball bearing which is itself fixed, howmany degrees of freedom are then left?

** 5. A uniform rod of mass M and length l is resting with one end against a wall and startssliding. There is no friction with the soil and the wall. How many degrees of freedomdoes this system have if you consider it to be two-dimensional? Express the kineticenergy in terms of these degrees of freedom.

** 6. A bead of mass m slides along a smooth conic spiral. Use cylinder coordinates in thisproblem. Assuming that ρ = az and φ = −bz, with a and b constants, show, usingd’Alembert’s principle, that the equation of motion reads:

..z (a2 +1+a2b2z2)+a2b2zz2 =−g . (A.1)

LAGRANGE-HAMILTON FORMALISM

** 7. (a) Construct the Lagrangian for a mass m attached to a pendulum of fixed length land moving in a plane.

(b) Construct the Lagrangian for two masses m1 and m2 attached to both ends of arope hanging over a pulley. The mass m1 is connected to one end of the ropethrough a spring with spring constant k. The radius of the pulley is R and itsmoment of inertia is I .

75

A(c) A beam of length L and mass M is connected on one end through a pivot on theceiling such that it can move in a plane. At the other end, a mass m is attached tothe beam. The size of this mass is small enough for it to be considered as a pointmass. The moment of inertia I of the beam with respect to an axis through its endis ML2/3. How many degrees of freedom does this system have? Choose suitablecoordinates and give the Lagrangian.

(d) A trolly on four wheels can move frictionless on a table in the x-direction. Themass of the trolley (including its wheels) is M and the wheels each have a radius aand moment of inertia I . On top of the trolley a second mass m is placed, which isattached to the trolly through a spring with spring constant γ. This second mass isalso allowed to move in the x-direction only. Give the Lagrangian for this system.

** 8. A mass m is attached to a pendulum with a fixed length l which is suspended at a pointp. The point p itself can move freely along a horizontal axis, but it is attached to twosprings along the x-direction and both with spring constant k.

(a) Choose suitable generalised coodinates.

(b) Construct the Lagrangian for this system and derive the equations of motion fromit.

*** (c) Assume that the mass is displaced a little from its equilibrium state and then re-leased. Neglecting the mass of the springs, show that the pendulum starts oscil-lating with period

T = 2π

√mg +2kl

2kg. (A.2)

Hints: (1) Use the small-angle approximation, (2) neglect all terms in the equationof motion of second order in the generalised coordinates and (3) assume that thependulum and the springs oscillate with the same frequency ω (this implies thatthe acceleration of the angle θ of the pendulum woth the vertical axis is ω2θ, andsimilar for the motion of the springs).

** 9. Assume that the block B , to which is attached a spring, is driven in the vertical directionaccording to s = A sin(ωt ). Figure (a) shows the position of the spring if there is no ex-ternal force (so also no gravitational force) acting on B and m is in equilibrium. Figure(b) shows an arbitrary position of the system. The spring has length l in equilibrium.Show that the height of the mass m with respect to its equilibrium position is given by

q =C sin(p

k/m t +δ)+ Ak

m(k/m −ω2)sinωt .

A


q

m

m

B

B

kl = constant

s

(a) (b)

** 10. A uniform rod AB with mass M and length l is suspended to a frictionless hinge A. Theend B is attached to a spring as shown in the figure. For θ = 0, the spring is stretched.Neglect the mass of the spring and show that V for this system is given by:

V =−1

2M g l cosθ+ 1

2k[(s2 + l 2 −2sl cosθ)1/2 − l0]2.

Assume θ to be small and show that the oscillation frequency is given by:

ω=√

3g

2l+ 3ks(l − s + l0)

Ml (l − s).

l

θ

M

k

B

A

s

** 11. Consider the following:

77

A

m

m

l

r

k 1

2

ϑ

The mass m1 can move freely on the table; the mass m2 can only move along the verticaldirection.

(a) Show that the Lagrangian has the form:

L = 1

2m1

(r 2 + r 2ϑ2)+ 1

2m2 l 2 +m2g l − 1

2k(l + r −b)2,

where b is the length of the string plus the length of the spring in equilibrium.

(b) Show that the quantity Jz = m1r 2ϑ is constant in time. What does this quantityrepresent?

*** (c) Solve the equation of motion for Jz = 0.

** 12. Deze opgave is bedoeld voor de geïnteresseerden – niet verplicht!

Beschouw een functie f (pk , qk ) die afhangt van alle impulsen pk en coordinaten qk .We kijken naar de tijdsafhankelijkheid van een dergelijke functie.

De tijdsafhankelijkheid van f wordt gegeven als

d f

d t= f , H .

Werk de uitdrukking f , H in het rechterlid uit. De algemene definitie van f , g is:

f , g =∑j

[∂ f

∂q j

∂g

∂p j− ∂ f

∂p j

∂g

∂q j

].

Neem nu f = p j , respectievelijk f = q j en laat zien dat je zo de Hamilton vergelijkingenterugvindt.

* 13. From the definition of the Lagrangian in terms of the Hamiltonian, L = pq−H(p, q) viadL derive the Hamilton equations:

q = ∂H

∂p

p =−∂H

∂q

af.Hint: use q and q as independent coordinates.

* 14. See problem ** 11

A


(a) Find the momenta corresponding to the gegeneralised coordinates used in thatproblem.

(b) Formulate the Hamiltonian for this problem.

** (c) Consider the same problem, but now without the spring. When the motion starts,the mass m2 is at rest and m1 is at a distance r0; r is then 0. The value of Jz isthen J0. J0 and r0 are therefore assumed to be known. Calculate r at the momentwhere m2 has descended over a distance d .Hint: Use the conservation of energy.

** 15. A particle experiences a conservative force Fc in addition to a frictional force of theform:

Fw =−γr .

Is it possible to derive this force from a (generalised) potential W (r, r) as is done in thelecture notes for the electromagnetic force? If the answer is yes, give W ((r, r), otherwise,explain why this is not possible.

What would be the best way to analyse a problem with constraints and friction?

*** 16. A coin with radius a is dropped on a stone table and starts to spin. After some time, thecoin will come to a rest. The precise mechanism leading to this damping are still subjectto debate. Without damping, the motion will persist forever; the total energy will thenremain constant. Considering the axis d perpendicular through the centre of the coin,we assume that the angle θ between this axis and the vertical remains constant andthat the angular velocity φ with which this axis precesses around the vertical, is alsoconstant in time. We us the notation of subsection 2.9.2.

(a) If the coin is rolling on its edge, the contact point with the table describes a circlewith radius a cosθ. If the coin does not slip, we can obtain a relation between theprecision speed φ and the angular spin velocity ψ around d. Find this relation.

(b) Give an expression for the angular velocity vectorωωω of the coin.

(c) Because of the symmetry of the coin, it is convenient to use instead of the threeaxes z, e, d, the perpendicular axes f, e, d. Give an expression for ωωω in the f, e, dsystem.

(d) Write down the Lagrangian for the rolling coin.

(e) Show that ϕ is given by:

ϕ= 2

√g

a sinθ.

(f) Show that the component ofωωω along the vertical axis z is given by

ωz = 2

√g sin3θ

a.

** 17. Two particles with mass m1 and m2 are connected by a string and a spring as in thefigure. The particles can move frictionless through the horizontal tubes. The tubeshave a moment of inertia I around the vertical axis. Use θ, r1 and r2 as coordinates andassume that no torque is acting on the vertical axis. Show that in that case:

(I +m1r 21 +m2r 2

2 )θ = Pθ = constant, (A.3)

m1..r 1 −m1r1θ

2 =−k(r1 + r2 − c), (A.4)

m2..r 2 −m2r2θ

2 =−k(r1 + r2 − c). (A.5)

79

Ar1

r2

m1

m2

θ

k

TWO-BODY PROBLEM

** 18. In the two-body problem, we use polar coordinates r and φ.

(a) What are the expressions for the angular momentum r×p and for the velocity r inpolar coordinates?

(b) The equations of motion for a particle moving in the x y plane under the influenceof a central force fr are:

mr −mr ϕ2 = fr ;

pϕ = mr 2ϕ= constant.

We recognise in pϕ the angular momentum.

Used

d t= dϕ

d t

d

dϕ= pϕ

mr 2

d

dϕ

and the substitution u = 1/r to show that the equation of motion for u has thefollowing form:

d 2u

dϕ2 +u =− m fr

p2ϕu2

.

(c) Assume that the path is a conic section with r = A1−εcosϕ , where ε is the eccentricity

and A a constant which depends on the system and the starting conditions of themotion. Show that fr =−P 2

ϕ/(m Ar 2).Note: whether the actual path is an ellipse, parabola or hyperbola, depends onthe starting conditions; in all three cases, fr is proportional to 1/r 2.

(d) Show that for an orthogonal hyperbola with x y =C , it holds that fr = P 2ϕr

4mC 2 .

(e) Show that for a cardioid r = a(1+cosϕ), we must have fr = −3p2ϕa

mr 4 and sketch theorbit of m for this case.

** 19. A rocket orbits around the earth in an ellipsoidal trajectory with eccentricity ε andsemi-major axis a. By letting the rocket engine work for some short time, the rocketis given a change in velocity ∆v which may be considered to be instantaneous. Themass mr of the rocket may be considered to remain constant.

A


(a) At which point of the orbit and in which direction one should do this in order tolet the rocket escape from the gravity field of the earth?

(b) Express |∆v | of the rocket in terms of a, ε, G and M (with M À mr ).Does |∆v |min increase or decrease with increasing ε when a is kept constant?

VARIATIEREKENING

* 20. De hyperbolische functies cosh(x) en sinh(x) worden gegeven door

cosh(x) ≡ (ex +e−x )/2

sinh(x) ≡ (ex −e−x )/2

(a) Laat zien dat

cosh2(x)− sinh2(x) = 1.

(b) Laat zien dat cosh(x) een oplossing is van de vergelijking

y′′ =

√1+ y ′2.

** 21. We proberen de vorm van een ketting die symmetrisch aan twee uiteinden is opge-hangen, zie de figuur, af te leiden met behulp van een krachtenbalans. In de ket-ting heerst een spankracht. De ketting heeft een massadichtheid ρ (massa per lengte-eenheid). Omdat het gewicht van de ketting niet verwaarloosd kan worden is dezespankracht niet constant. Wel kunnen we zeggen dat de spankracht altijd gericht islangs de raaklijn aan de ketting. We definiëren de x-coordinaat zo dat x = 0 het middenvan de ketting vormt, en dat de ophangpunten liggen op x = L en x = −L. De vormvan de ketting wordt aangeduid met y(x). Beschouw het segment van de ketting tussenx = 0 en x = a ¿ L. Op dit segment werken drie krachten: de spankracht ~Fs in de ket-ting op x = 0, de spankracht ~T in x = a en de zwaartekracht ~Fzw.

−L L

0xa

(a) Laat zien dat de krachtenbalans voor dit segment leidt tot

ρ g∫ a

0d x

√1+ y ′2 = Fs y

′(a) (A.6)

81

A(b) Differentieer het linker- en rechterlid naar a en laat zien dat de resulterende tweedeorde differentiaalvergelijking de kettinglijn A cosh(x/A)+B als oplossing heeft.Wat wordt Fs?Hint: d

d a

∫ a0 d x f (x) = f (a).

(c) Geef een uitdrukking voor de lengte van de ketting.

** 22. In deze opgave gebruiken we bolcoördinaten en variatierekening om de geodeet (decurve die de kortste afstand tussen twee punten geeft) op een bol te vinden. Pilotennoemen dit de “great circle route”.

(a) Leid eerst m.b.v. variatierekening de differentiaalvergelijking voor deze kortsteroute af.

*** (b) Gebruik nu de substitutie w = cotθ, dus sin2θ = 11+w 2 in deze differentiaalvergeli-

jking en leid af dat de

oplossing gegeven wordt door θ = arccot[p

A sin(φ−φ0)], met A enφ0 constanten.

** 23. Bereken het pad van een lichtstraal in een medium waarin de index van refractie gegevenwordt door n(x, y) =p

y .

** 24. Beschouw de kegel x2 + y2 = z2. Gebruik cylindercoordinaten, en beschouw een padvan de vorm (z,φ(z)).

(a) Laat zien dat de lengte d s van een klein lijnsegment op de kegel gegeven wordtdoor

d s =√

2d z2 + z2dφ2.

(b) Schrijf de Euler-Lagrange vergelijking op en laat zien dat deze leidt tot de differ-entiaal vergelijking:

dφ

d z=

p2C

zp

z2 −C 2.

(c) Los deze vergelijking op door gebruik te maken van∫d x

C

xp

x2 −C 2= arccos

C

x.

SYSTEMEN NABIJ EVENWICHT

** 25. Beschouw een potentiaal van een systeem met gegeneraliseerde coordinaten θ1 en θ2.De potentiaal van dit systeem is gegeven door

V =−m1g l cosθ1 −m2g l cosθ2 + 1

2κl 2(sinθ1 − sinθ2)2,

waarbij g de valversnelling is enκ een veerconstante; m1 en m2 zijn massa’s. Wat zijn dewaarden van de gegeneraliseerde coördinaten in de evenwichtsstand? Vind de matrixK (zie dictaat).

Met welk fysisch systeem correspondeert deze potentiaal?

A


** 26. Beschouw een systeem bestaande uit twee bollen met massa’s m1 en m2. De bovenstebol (m1) is met een veer met veerkonstante κ1 aan het plafond bevestigd en boven-dien is zij met een veer (veerconstante κ2) aan m2 bevestigd. De onderste massa m2

is op haar beurt met een veer (κ3) aan de vloer bevestigd. We beschouwen uitsluitendverticale bewegingen van de bollen.

Vind de normal modes van het systeem (houd er rekening mee dat de veren niet in hunruststand zijn als het systeem in evenwicht is).

** 27. Beschouw onderstaand systeem:

κ

M, I

mm

2

1

R

l

Vind de normal modes.

** 28. Beschouw twee slingers met gelijke lengte l . Aan beide slingers hangt een massa m.De massa’s zijn met elkaar verbonden door een veer met veerconstante κ (zie de figuurhieronder).

m m

lθ θ1 2

l/2

(a) Geef de Lagrangiaan voor de slingers zonder veer.

(b) Idem, voor de gekoppelde slingers. Leid ook de bewegingsvergelijkingen af.

(c) Stel de matrix voor de kinetische en potentiële energie op en vind de normalmodes. Beschijf de beweging behorend bij elk van de gevonden modes.

83

A** 29. Beschouw twee massa’s m die met wrijvingloos draaibare, onbuigzame en massalozestaven van gelijke lengte l zijn opgehangen aan het plafond. De slingers zijn gekopp-peld door een veer met veerconstante κ en met een massa M , die bevestigd is aan demiddens van de beide staven:

m m

lθ θ1 2

l/2

(a) Geef de Lagrangiaan voor de slingers zonder veer. Geef de Lagrangiaan ook voorkleine uitwijkingen ten opzichte van de evenwichtstoestand.

(b) Druk de positie van het zwaartepunt van de veer uit in termen van θ1, θ2 en l , allesvoor kleine uitwijkingen. Geef de kinetische en potentiele energie van het gehelesysteem in de limiet voor kleine uitwijkingen.

(c) Geef de matrices voor de kinetische en potentiële energie en vind (m.b.v. Maple)de normal modes.

** 30. Een dunne, homogene staaf van lengte l en massa m rust op twee veren met veercon-stante κ die verticaal kunnen trillen.

y y1 2

Bij de volgende vragen gaan we ervan uit dat de veren dicht in de buurt van hun even-wichtsstand zijn.

(a) Wat is de positie van het massamiddelpunt van de staaf, uitgedrukt in termen vande hoogten y1 en y2 van de veren? Idem voor de hoek θ die de staaf maakt met dehorizontale as.

(b) Schrijf de Lagrangiaan op. Gebruik: het traagheidsmoment van de staaf is 112 ml 2.

(c) Vind de normal modes van de staaf. Leg uit met welke bewegingen deze normalmodes corresponderen.

** 31. Beschouw het onderstaande systeem, bestaande uit drie karretjes, waarvan de boven-ste twee verbonden met een veer. Het linker onder karretje heeft massa m1 en hetrechter m3. Het karretje dat op het linker karretje staat heeft massa m2. Verder hebbende twee wielen van m2 elk een traagheidsmoment I en wordt het traagheidsmomentvan de overige wielen verwaarloosd.

A


x

x

l

x

1

2

3

κ

1

2 3I I

Laat zien dat

T = 1

2

[(m1 +m2) x2

1 +(m2 +2I /r 2) q2

2 +m3x23 −2m2x1q2

];

en

V = 1

2κ

[x2

1 +q22 +x2

3 −2x1x3 +2x3q2 −2x1q2]

,

waarbij q2 = const.−x2 en waarbij r de straal van de wielen van m2 is.

Laat zien dat er twee normal modes zijn met frequentie nul. Beschrijf de bijbehorendebeweging. Vind de frequenties van de derde normal mode.

BANTWOORDEN

85

B

86 B. ANTWOORDEN

1. De massa van de ketting is m. De zwaartekracht die werkt op de los hangende lengte yis g m(l/y). Hieruit volgt voor de versnelling my = g m(y/l ) ⇒ y = g

y . Een oplossing van

deze differtiaalvergelijking is y(t ) = Aeαt +Be−αt met α=√

gl . Uit de startvoorwaarde

˙y(0) = 0 rekenen we uit dat A = B en uit y(0) = y0 dat A = B = y0/2. De oplossing isy(t ) = y0

2

(eαt +e−αt

).

2. (a) ma2

(b) 2ma2

(c) 12 g sinα

(d) 12 tanα

3.√

d 2 +d 20 −d0, met d0 =GM/v2, M de massa van de zon

4. (a) 9

(b) 6

5. 16 Ml 2θ2

6. -

7. (a) 12 ml 2θ2 +mg l cosθ

(b) 12 m1(r1 + u)2 + 1

2 m2r 21 +m1g (r1 +u)+m2g (l − r1)+ 1

2 I (r1/R)2 − 12 ku2

(c) 12 ( M

3 +m)L2α2 + g (m + M2 )L sinα, met α de hoek tussen balk en plafond

(d) L = 12 (M + 4I

a2 )x2 + 12 m(x + q)2 − 1

2γq2

8. (a) 2 vrijheidsgraden. Kies u (de uitwijking van de veren uit de evenwichtsstand langsde horizontale as) en θ (de hoek van de slinger met de verticaal).

(b) L = 12 m(u2 +2l cosθuθ+ l 2θ2)+mg l cosθ−ku2.

De bewegingsvergelijkingen worden:cosθu + l θ+ g sinθ = 0m(u − l θ2 sinθ+ l θcosθ) =−2ku

(c) -

9. -

10. -

11. (a) -

(b) -

(c)

l (t ) = 1

m1 +m2

(m1 A cos(

√k

µt )+m1B sin(

√k

µt )+ m1

k(µg +kb)+ 1

2m2g t 2 −C t −D

)

r (t ) = 1

m1 +m2

(m2 A cos(

√k

µt )+m2B sin(

√k

µt )+ m2

k(µg +kb)− 1

2m2g t 2 +C t +D

)

met µ≡ m1m2/(m1 +m2), en A, B , C en D constanten.

12. -

13. -

14. (a) pr = m1r , pθ = m1r 2θ, pl = m2 l

87

B

(b) H = p2r

2m1+ p2

θ

2m1r 2 + p2l

2m2−m2g l + 1

2 k(l + r −b)2

(c) 12 (m1 +m2)r 2 =− J 2

0 d2m1

(2r0−d)(r0−d)2r 2

0+m2g d

15. Nee - probeer maar een potentiaal te vinden die voldoet. Gebruik wetten van Newton.

16. (a) ψ=−ϕ(b) −cosθϕ

(c) ω= ϕ(z−cosθd)

(d) ω=−ϕsinθf

(e) L = 12 I1ϕ

2 sin2θ−M g a sinθ

(f) -

(g) -

17. -

18. (a) Impulsmoment mr 2φ z. Snelheid x = r cosφ− r φsinφ, y = r sinφ+ r φcosφ.

(b) -

(c) -

(d) -

(e) -

19. Beschouw de totale energie voor en na de snelheidsverandering.

(a) Ena−Evoor = 12 m

(2~v · (∆~v)+ (∆~v)2

), dus moeten∆~v en~v in dezelfde richting gericht

zijn. Het verschil is maximaal in het perihelion.

(b) ∆vmi n =√

2GM(1−ε)a

(1−

√1+ε

2

), dus neemt ∆vmi n bij toenemende ε en constante a

af.

20. -

21. (a) -

(b) Fs = ρg A

(c) lengte = 2A sinh(L/A)

22. (a) θ2 = sin4 θC 2 − sin2θ, met θ de polaire hoek, d.w.z. de hoek met de z-as en C een

constante.

(b) -

23. Differentiaalvergelijking y′ =

√y

A2 −1 met A een constante, en oplossing y = x2

4A2 + A2.

24. (a) Gebruik dat r 2 = z2.

(b) -

(c) φ(z) =p2arccos(C /z).

25. In evenwicht geldt: θ1 = θ2 = 0.

K =(

m1g l +κl 2 −κl 2

−κl 2 m2g l +κl 2

)Fysisch systeem: twee slingers met elk een massa eraan die verbonden zijn door eenveer.

B

88 B. ANTWOORDEN

26. Zie voorbeeld in Hoofdstuk 6.2.1 van het diktaat. Oplossen van de eigenwaardenvergeli-jking (met κ2 en κ3 verwisseld) geeft de normal modes

ω2 = B ±p

B 2 −4AC

2A,

met A ≡ m1m2, B ≡ m1(κ2 +κ3)+m2(κ1 +κ2) en C ≡ (κ1 +κ2)(κ2 +κ3)−κ22.

(Zie voor een speciaal geval (alle κ’s hetzelfde en alle massa’s hetzelfde) ook (6.33) inhet diktaat).

27. Kies gegeneraliseerde coordinaten y (de hoogte van het wiel) en θ (de rotatiehoek). Wenemen aan dat het touw niet slipt. De positie van m1 en m2 varieert dan met Rθ (voorm1) en -Rθ (voor m2). De Lagrangiaan wordt nu gegeven door

L =1

2m1(y +Rθ)2 + 1

2m2(y −Rθ)2 + 1

2M y2 + 1

2I θ2+

1

2κy2 + termen lineair in y (doen niet mee voor κ).

De normal modes vind je uit de eigenwaardenvergelijking∣∣∣∣ (m1 +m2 +M)ω2 −κ (m1 −m2)Rω2

(m1 −m2)Rω2 ((m1 +m2)R2 + I )ω2

∣∣∣∣= 0.

Met A ≡ (m1 +m2 +M), B ≡ (m1 −m2)R en C ≡ (m1 +m2)R2 + I .geeft dit:

ω2 = Cκ

AC −B 2

28. (a) Lagrangiaan zonder veer:

L = 1

2ml 2(θ2

1 + θ22)+mg l (cosθ1 +cosθ2).

(b) Lagrangiaan met veer:

L =ml 2

2

(Θ2

1 + Θ22

)+mg l (cosΘ1 +cosΘ2)

− κ

2

√(cosΘ1 −cosΘ2)2 l 2

4+

(l

2sinΘ1 − l

2sinΘ2 −L0

)2

−L0

2

De bewegingsvergelijkingen worden (tot op eerste orde, dus kwadratische ter-men in de snelheden verwaarlozend, en in de kleine-hoek benadering (cosθ ∼ 1,sinθ ∼ θ)):

ml 2Θ1 +mg lΘ1 + κl 2

4(Θ1 −Θ2) = 0

ml 2Θ2 +mg lΘ2 − κl 2

4(Θ1 −Θ2) = 0

(c)

K =(

mg l + κl 2

4 −κl 2

4

−κl 2

4 mg l + κl 2

4

).

M =(

ml 2 00 ml 2

).

89

B

Oplossen van de eigenwaardenvergelijking leidt dan tot normal modes met fre-quentie

ω2 = g /l

ω2 = g

l+ κ

2m

De eerste beschrijft de frequentie van de gekoppelde slingers wanneer zij met geli-jke fase bewegen (de slingers merken niets van de koppeling), en de tweede defrequentie van de beweging wanneer de slingers in tegenfase bewegen.

29. (a) Lagrangiaan zonder veer:

L = 1

2m

[θ2

1 + θ22

]+mg l [cosθ1 +cosθ2] ≈ 1

2m

[θ2

1 + θ22

]+mg l

[2− 1

2θ2

1 −1

2θ2

2

](b) Het zwaartepunt van de veer is gegeven door:

xm = 1

2(x1 +x2) = a

2+ l

4[sinθ1 + sinθ2] ≈ a

2+ l

4[θ1 +θ2] (B.1)

ym = 1

2(y1 + y2) = l

4[cosθ1 +cosθ2] ≈ l

2(B.2)

a is de afstand tussen de ophangpunten. De kinetische energie van het gehelesysteemis gegeven door

T = 1

2ml 2[θ2

1 + θ22] + 1

2M [x2

m + y2m] ≈ 1

2ml 2[θ2

1 + θ22] + l 2

32M

[θ1 + θ2

]2.

De potentiele energie is gegeven door:

V =−mg l cosθ1 −mg l cosθ2 −M gl

4(cosθ1 +cosθ2)+Vveer ,

met

Vveer = 1

2κ

√l 2

4(cosθ1 −cosθ2)2 +

(l

2(sinθ1 − sinθ2)+a

)2

−L

2

(B.3)

Dus, we vinden voor de potentiele energie in de limiet voor kleine uitwijkingen(constante termen weggelaten):

V = mg l1

2

[θ2

1 +θ22

] + 1

8M g l

[θ2

1 +θ22

] + κ

4l 2

[1

2θ2

1 −θ1θ2 + 1

2θ2

2

](c) De matrices voor de kinetische en potentiele energie zijn

M =(

ml 2 + Ml 2

16Ml 2

16Ml 2

16 ml 2 + Ml 2

16

), K =

(g l

(m + M

4

)+ κl 2

4 −κl 2

4

−κl 2

4 g l(m + M

4

)+ κl 2

4

)

en dus zijn de normal modes (gebruik Maple)

ω2 = g

l

(m + M4 )

(m + M8 )

, ω2 = g(m + M

4

)+ κl2

ml.

B

90 B. ANTWOORDEN

30. (a) Voor kleine veranderingen ten opzichte van de evenwichtstoestand, is het zwaartepuntgegeven door:

xM = l

2, yM = y1 + y2

2

en dus is de hoekΘ:sinΘ= y2 − y1

l.

(b) De kinetische energie is gegeven door:

T = 1

2M yM + 1

24Ml 2Θ2 .

De potentiele energie is:

V = M g ym + κ

2

((y1 − y0)2 + (y2 − y0)2)

Druk y1 en y2 uit in ym en sinΘ. Maak de kleine hoek benadering dan L = T −Vmet

T = 1

2my2

m + 1

2I Θ2

V = mg ym + 1

2κ(ym − 1

2lΘ− y0)2 + 1

2κ(ym + 1

2lΘ− y0)2

(c) De matrices voor de kinetische en potentiele energie zijn

M =(m 00 I

), K =

(2κ 00 1

2κl 2

)Het oplossen van de eigenwaarden vergelijking geeft de normal modes:

ω21 =

6κ

m, ω2

2 =2κ

m.

ω1 correspondeert met de rotatie- en ω2 met de verticale vibratie beweging.

31. -

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Classical Mechanics