by varun menon · 2018. 1. 26. · [email protected], [email protected], united kingdom...

ISSN (Print) 2313-4410 & ISSN (Online) 2313-4402Volume 22, 2018

© ASRJETS THESIS PUBLICATION http://asrjetsjournal.org/

By Varun Menon

The Lagrangian and Hamiltonian Formalisms of Mechanics

http://asrjetsjournal.org/

ASRJETS research papers are currently indexed by:

© ASRJETS THESIS PUBLICATION http://asrjetsjournal.org/

http://asrjetsjournal.org/

Copyright © 2018 by Varun Menon

All rights reserved. No part of this thesis may be produced or transmitted in any form or by any means without written

permission of the author.

ISSN (Print) 2313-4410 & ISSN (Online) 2313-4402

The ASRJETS is published and hosted by the Global Society of Scientific Research and Researchers (GSSRR).


Members of the Editorial Board EDITOR IN CHIEF

Dr. Mohammad Othman, Global Society of Scientific Research and Researchers. Tel: 00962788780593, [email protected], Jordan

EXECUTIVE MANAGING EDITOR Dr. Mohammad Othman, Global Society of Scientific Research and Researchers. Tel: 00962788780593, [email protected], Jordan

COORDINATING EDITOR Dr. Daniela Roxana Andron, Lucian Blaga University of Sibiu, Departamentul Drept Privat și Științele Educației, Universitatea Lucian Blaga din Sibiu, http://dppd.ulbsibiu.ro, http://drept.ulbsibiu.ro/centrul-ccap, tel. +40-269-211 602, fax +40-269-235 [email protected], Romania

SECTION EDITORS Prof. Dr. Hye-Kyung Pang, Business Administrations Department, Hallym University, #1, Ockcheon-dong, Chooncheon-si, Kangwon-do, [email protected], [email protected]/ (office)82-33-248-3312/ (cell) 82-10-5311-6131, Korea, Republic of

Prof. Dr. Felina Panas Espique, Professor, School of Teacher Education, Saint Louis University, Baguio City. Dean of the School of Teacher Education. Bonifacio St., Baguio City, 2600, (074) 447-0664 (landline), Mobile Number 099-897-63236, [email protected], Philippines

Prof. Dr. Indrani Pramod Kelkar, Department of Mathematics,Chief Mentor, Acharya Institute of Technology, Soldevanahalli, Hesaraghatta Main Road, Banavara Post, Bangalore- 560 107. Cell : 9164685067,[email protected], India

LAYOUT EDITORS & COPYEDITORS Dr. Christina De Simone, Faculty of Education, University of Ottawa.333 Chapel Street, Suite 305, Ottawa, Ontario, K1N 8Y8, tel. 613-562-5800,4112, [email protected], Canada

EDITORIAL BOARD MEMBERS AND REVIEWERS Associate Professor Zaira Wahab, Iqra University, Pakistan

Asso. Prof. Girish Chandra Mishra, O.P.Jindal Institute of Technology,Raigarh(C.G.), [email protected], Tel:

09827408085., India

Dr. Ankur Rastogi, O.P.Jindal Institute of Technology, Raigarh(C.G.),Mob.: 91 97559 27688, [email protected],

[email protected], [email protected], India

Dr. Kabita Kumari Satapathy, O.P.Jindal Institute of Technology,Raigarh, C.G., phone: 9770695159,

[email protected], India

Dr. Salman Shahza, Institute of Clinical Psychology, University of Karachi, Pakistan

Dr. Daniel Hailegiorgis Haile, Wollo University, Dessie, Ethiopia

Dr. Lowida L. Alcalde, Northwestern Mindanao State College of Science and Technology, Philippines

Dr. Fazal Shirazi, MD Anderson Cancer Center, Texas,Phone: 7133064951, [email protected], United States

Asso. Prof. Mario Amar Fetalver, College of Arts & Sciences, Romblon State University, Philippines

Dr. Esther Patrick Archibong, Dept. of Sociology, University of Calabar, [email protected], Nigeria

Prof. Dr. Samir El-Sherif, College of Agriculture, Cairo University, Egypt

Dr. Chandan Kumar Sarkar, IUBAT- International University of Business Agriculture and Technology, Telephone:

01717314333, 01726077766, E- Mail : [email protected], Bangladesh

Dr. Narayan Ramappa Birasal, KLE Society’s Gudleppa Hallikeri College Haveri (Permanently affiliated to Karnatak

University Dharwad, Reaccredited by NAAC),[email protected], 94491 22732, India

Dr. Rabindra Kayastha, Kathmandu University, Nepal

mailto:[email protected]


http://dppd.ulbsibiu.ro/

http://drept.ulbsibiu.ro/centrul-ccap




mailto:9164685067%[email protected]










mailto:NAAC)%[email protected]

Dr. Rasmeh Ali AlHuneiti, Brunel university, Telephone: (+974) 443070364, Mobile: (+97466190008) Email:

[email protected], [email protected], United Kingdom

Dr. Florian Marcel Nuta, Faculty of Economics/Danubius University of Galati,[email protected], 040722875933,

Romania

Dr. Arfan Yousaf, Department of Clinical Sciences, Faculty of Veterinary and Animal Sciences, PMAS-Arid Agriculture

University Rawalpindi .Pakistan ,[email protected], +92-333-6504830, Pakistan

Dr. Daniela Roxana Andron, Lucian Blaga University of Sibiu, Departamentul Drept Privat și Științele Educației,

Universitatea Lucian Blaga din Sibiu, http://dppd.ulbsibiu.ro, http://drept.ulbsibiu.ro/centrul-ccap, tel. +40-269-211 602,

fax +40-269-235 [email protected], Romania.

Asso. Prof. Zakaria Ahmed, PRIMEASIA UNIVERSITY, MICROBIOLOGY DEPARTMENT, HBR TOWER, 9 BANANI,

DHAKA 1213, Bangladesh

Prof. Dr. Hamed Hmeda Kassem, Department of Zoology , Faculty of Science, Benghazi University, phone : +

218925476608, [email protected], Libya

Dr. Hulela Keba, Botswana College of Agriculture Faculty of Agriculture Private Bag 0027. [email protected],

[email protected], Botswana

Dr. Aleksandra K. Nowicka, MD Anderson Cancer Center, Houston, Texas, [email protected], Phone: +1 713-

792- 7514, United States

Prof. Dr. Mahmoud Younis Mohammed Taha, Department of Dental Basic Sciences, College of Dentistry, Mosul

University, Iraq

Prof. Dr. Akbar Masood, Dean, Faculty of Biological Sciences Head, Department of Biochemistry, the University of

Kashmir,[email protected], +91-9906966281, +91-194-2415697, India

Prof. Dr. Ramel D. Tomaquin, Dean, CAS Surigao Del Sur State University Tandag City Surigao Del Sur, Philippines

Prof. Dr. Iryna Morozova, Romance-Germanic Faculty, Odesa Mechnikov University, [email protected]

(professional), [email protected] (personal) tel. +380506572043, +380487254143, Ukraine

Prof. Dr. Ayhan Esi, Adiyaman University, Science and Art Faculty, Department of Mathematics, Turkey

Prof. Dr. Felina Panas Espique, Professor, School of Teacher Education, Saint Louis University, Baguio City. Dean of the

School of Teacher Education. Bonifacio St., Baguio City, 2600, (074) 447-0664 (landline), Mobile Number 099-897-

63236, [email protected], Philippines

Prof. Dr. Hye-Kyung Pang, Business Administrations Department, Hallym University, #1, Ockcheon-dong, Chooncheon-

si, Kangwon-do, [email protected], [email protected]/ (office)82-33-248-3312/ (cell) 82-10-5311-6131, Korea,

Republic of

Prof. Dr. Aghareed M. Tayeb, Chemical Engineering Dept., Faculty of Engineering, Minia University Minia, Egypt

Prof. Dr. Maria Luisa A. Valdez, Dean of Colleges and Head of Graduate School, Batangas State University ARASOF

Nasugbu Campus, Nasugbu, Batangas, Philippines

Dr. Tahira Naz, Centre of excellence in Marine biology, karachi University pakistan., Pakistan

Prof. Dr. Indrani Pramod Kelkar, Department of Mathematics,Chief Mentor, Acharya Institute of Technology,

Soldevanahalli, Hesaraghatta Main Road, Banavara Post, Bangalore- 560 107. Cell :

9164685067,[email protected], India

Dr. Agustin Nuñez Arceña, School Administrator/Dean of Colleges/Principal - University of the Visayas – Danao City

Campus, Philippines

Dr. Joseph Djeugap Fovo, University of Dschang, Faculty of Agronomy and Agricultural Science (FASA) Department of

Plant Protection, Plant Pathology Laboratory, [email protected], Cameroon

Asso. Prof. Fateh Mebarek-Oudina, Département des Sciences de la Matière, Faculté des Sciences, Université 20 Août

1955- Skikda (Skikda University, Algeria), Route El-Hadaeik, B.P. 26, Skikda 21000, Algeria

Asso. Prof. Evangelos P. Hinis, National Technical University of Athens - NTUA, [email protected],Tel. +30

2107722911, Greece

Prof. Dr. Manu Rathee, Head of Department Department of Prosthodontics and Crown & Bridge Post Graduate Institute

of Dental Sciences Pt. B. D. Sharma University of Health Sciences, +91-9416141376 Rohtak, Haryana, India

Dr. Christina De Simone, Faculty of Education, University of Ottawa.333 Chapel Street, Suite 305, Ottawa, Ontario, K1N

8Y8, tel. 613-562-5800,4112, [email protected], Canada.




mailto:Galati%[email protected]

mailto:%[email protected]

http://dppd.ulbsibiu.ro/

http://drept.ulbsibiu.ro/centrul-ccap






mailto:Kashmir%[email protected]





mailto:9164685067%[email protected]




Varun Menon

Abstract

This paper serves to demonstrate and summarize my research into Analytical Mechanics. Advanced mechanics

is an area of physics that Physics students are certainly never exposed to due to the mathematical complexity of

the subject. However, concepts in analytical mechanics are extended to cutting edge areas of modern physics,

particularly quantum physics, statistical physics and relativistic physics. The Lagrangian and Hamiltonian

reformulations of classical mechanics are rich in beautiful, elegant mathematics and present a far more general

and abstract way of understanding observable phenomena than purely Newtonian approaches. Moreover,

these concepts are extended and abstracted to other areas of theoretical physics, so an understanding of these

formulations is invaluable to a physics student.

Computing physical problems using Lagrangians and Hamiltonians are remarkably easy and reduce the computational

complexity of solving for the dynamics of large complex physical systems. The difficult part is following and

understanding the logic behind the development of these esoteric concepts. The mathematical proofs are not

arduous by any means, yet the challenge lies in understanding how and why some of the greatest physicists and

mathematicians of antiquity even thought of these things. This was by far the biggest enigma I faced in my

research as there are little to no resources that explain the motivation and more importantly, the thought process

behind analytical mechanics. A lot of proofs in textbooks, though logically sound, will work backwards from the

premise instead of working towards the premise which leaves one wondering how some concepts were conceived

of at all. Thus, this paper will largely focus on logic and intuition with minimal computation and problem

solving as there are already a plethora of detailed resources available that provide myriad rigorous examples and

practice problems. Furthermore, I will also investigate the Lagrangian and Hamiltonian formalisms through a

geometric interpretation.

Key Words

Lagrangian, Hamiltonian, Mechanics, Physics, Quantum, Analytical, Calculus, Variations, Legendre, Lie,

Algebra, Poisson, Euler, Coordinates, Phase, Transformations

Contents

1 Introduction 3

2 Calculus of Variations 11

2.1 Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.1.1 The Brachistochrone problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.1.2 Generic variational problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.2 The Euler-Lagrange equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.2.1 Solving the Euclidean Geodesic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.2.2 Functionals with more than one dependent variable . . . . . . . . . . . . . . . . . . . . . 19

3 Space and constrained motion 20

3.1 Degrees of freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.2 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.3 Generalized coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.3.1 Holonomic constraints and systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

4 Lagrange’s equations of motion 24

4.1 Virtual work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4.2 D’alembert’s principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.3 Deriving a law of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

5 Hamilton’s Principle 31

6 More on the Lagrangian Formalism 35

6.1 Cyclic coordinates and conservation laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

6.2 Using the Lagrange equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

7 Hamiltonian mechanics 38

7.1 Time variance of the Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

7.2 Conservation of Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

7.3 The Legendre Transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

7.4 Hamilton’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

7.4.1 Hamiltonian analysis of Atwood’s machine . . . . . . . . . . . . . . . . . . . . . . . . . . 44

7.5 Ignorable coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

7.6 Phase Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

7.7 Geometrical interpretation of Hamiltonian Dynamics . . . . . . . . . . . . . . . . . . . . . . . . 50

7.8 Canonical transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

7.9 Poisson Brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

7.10 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

2

Chapter 1

Introduction

What is Lagrangian and Hamiltonian mechanics about?

The straightforward answer is that both are essentially reformulations of Newtonian mechanics. Although

Newtonian mechanics provides the foundational framework for all of physics, it’s structure is not ideal for a

number of different problems in physics. In particular, when dealing with mechanical systems of a large number

of interacting components with several internal and constraint forces, using Newton’s second law to solve for

the dynamics of the system becomes exponentially harder the more number of interactions you have. Moreover,

using Newton’s second law in non-cartesian coordinate systems becomes complicated and messy. This forces us

to either use cartesian coordinates to model systems in which they are not ideal, or to use contrived version of

Newton’s second law in different coordinate systems. The best way to illustrate this is with a simple example[17]

let us consider the motion of a simple pendulum-a system that we are thoroughly used to examining- but we

are going to do it the painful way with cartesian coordinates.

Figure 1.1: Simple pendulum in cartesian coordinates

Let us say that the pendulum in figure 1.1 has string length l with a Tension T and a ball of mass m. We will

assume that the string is massless and that the only external force is gravity.

Recall that in 2D cartesian coordinates, Newton’s second law takes the simple form

F = mr

⇓

Fx = mx

Fy = my

3

Lagrangian and Hamiltonian mechanics Varun Menon

Where r is a the position vector of the pendulum and is a function of x and y.1 Fx is the x component of the

net force acting on the pendulum and Fy is the corresponding y component. The origin is chosen where the

pendulum originates with the downward direction as positive y and the rightward direction as positive x. Let

us call the unit vectors x and y respectively. Our problem entails resolving for forces in the x and y directions

and hopefully we will end up with some sort of discernible differential equation for the accelerations in each

direction.

Resolving in x:

mx = −T(xl

)x

Here, the x component of the tension T - which we will denote as Tx - is the only force in the x direction. To

calculate the x component, all we do is scale the tension T by a factor of(xl

)to get the proportion of T that

acts in the x direction.2 Note that the negative sign is to account for Tx acting to the left while we defined

positive x as to the right.

Likewise, resolving in y, we get:

my =(mg − T

(yl

))y

Here too, we have scaled T by a factor of(yl

)to get Ty and the same logic applies as in the x case. The

negative sign accounts for Ty acting upwards while positive y was defined to be downwards. We can rearrange

our two equations as3:

T =−mxlx

T =(mg −my)l

y

Thus we can eliminate T 4 and cancel the m and l terms on both sides to give a single equation:

xy = xy + gx (1.1)

This is a relation between x, y, and their time derivatives. In order to simplify this any further we need another

equation. By now we should realize that x and y are not strictly independent of each other. The motion of the

pendulum is constrained to move in a certain way. Constraints will be dealt with in detail in a later chapter

but the principle we need to understand now is that the string and the tension in the string restrict the possible

motion of the pendulum. Analytically, we see that the mass can only move in the arc of a circle with radius l

centred at the origin. Thus we should immediately be able to write down a relation between x and y for the

possible positions of the mass as the equation of a circle5 :

x2 + y2 = l2

1Note that r is the time derivative of r and r is the second time derivative of r. The dot notation will be used as such throughout

this paper to represent the time derivatives of variables.2One way to think about this is to consider using the angle θ of the displacement of the pendulum to resolve the force. We

would have T sin θ as Tx and T cos θ as Ty . Since sin θ = xl, T sin θ = T

(xl

)and so on for Ty .

3I will discard the unit vectors for the rest of the problem since they do not take part in the calculation in cartesian coordinates.4We are interested in the motion of the pendulum rather than the tension in this problem.5Alternatively we could use the Pythagorean theorem on a right angle triangle with sides x and y and hypotenuse l to get

x2 + y2 = l2

4


Let’s see if we can use this constraint equation to simplify equation 1.1. Remember, we are trying to find

equations for x and y in terms of x, y and their first derivatives to solve for the dynamics of the system. Thus

if we want to eliminate either x or y with the constraint equation, it is a reasonable step to differentiate the

constraint equation twice with respect to time and see if the result is of any use to simplify our expression.

We get:

2xx+ 2yy = 0

↓

xx+ yy = 0

for the first derivative with respect to time.

Differentiating once more using the product rule, we get:

xx+ xx+ yy + yy = 0

↓

x2 + y2 + xx+ yy = 0

Now we can rearrange the double differentiated constraint equation for x or y and plug it into equation 1.1.

Thus, solving for y in the constraint equation and substituting into 1.1 we get (after a bit of rearranging and

simple algebraic manipulation that I will not go through here), an equation for x6:

x =−gxy − xx2 − xy2

(x2 + y2)

Remember that our constraint requires that x2 + y2 = l2. Thus we can simplify the denominator and we arrive

at the result:

x =−gxy − xx2 − xy2

l2(1.2)

Following the exact same procedure to eliminate x to arrive at an equation for y, we get:

y =gx2 − yx2 − yy2

l2(1.3)

We now have two differential equations for the accelerations in both independent directions. However, this

was pretty much an exercise in futility. For starters, both equations 1.2 and 1.3 have no obvious physical

interpretation from just looking at them. You cannot deduce anything relevant about the system unlike the

usual pendulum equation with l and θ. Moreover, it turns out these equations cannot be solved analytically in

terms of elementary functions. There is no concrete way to represent the solution of this equation unlike our

typical sin and cos solutions for a harmonic oscillator. Even worse, it turns out that we cannot even solve these

equations numerically to an approximation![2] Hopefully, it is clear by now that Cartesian coordinates were a

terrible choice for this problem even if we got to use a neat form of Newton’s second law. Next, we will use the

same pendulum problem to discuss Newton’s second law in polar coordinates. We will see that although the

solution takes a nice form at the end for this particular problem, Newton’s law itself does not take the simple

aesthetic form of Fx = mx and Fy = my in polar coordinates.

6Because we eliminated y using the constraint equation.

5


Figure 1.2: Simple pendulum in polar coordinates

Let’s begin the analysis of Newton’s second law in polar coordinates. This is the same pendulum with tension

T , length l and mass m.7 Recall that:

r =√x2 + y2

θ = arctan(yx

)This is just the regular transformation from cartesian to polar coordinates. We need two new unit vectors for

our coordinates r and θ which we will call r and θ. Just as in Cartesian coordinates where x is the unit vector

in the direction of increasing x with constant y and y is the unit vector in the direction of increasing y with

constant x, we can say that r is the unit vector in the direction of increasing r at constant θ and θ is the unit

vector in the direction of increasing θ at a constant r. We can see that in figure 1.2 that r and θ are orthognal8

to each other and therefore should span the coordinate space we are interested in. Therefore we can resolve the

net force F on the pendulum into components:

F = Fr r + Fθ θ

Where Fr is the radial component of the force and Fθ is the tangential component of the force. However, as

we will see, Newton’s laws will not take the neat form of Fr = mr and Fθ = mθ.9

7Not shown in the diagram to prevent clutter8Perpendicular9Here, r is the radial coordinate, not the position vector of the pendulum mass.

6


Figure 1.3: Changing position vectors

(a) Position vectors in Cartesian coordinates (b) Position vectors in polar coordinates

In the figures 1.3a and 1.3b we see the fundamental difference between a cartesian coordinate system and polar

coordinate systems.10 In cartesian coordinates, the unit vectors do not change regardless of what the position

vector is. However, when we look at the case for polar coordinates, it’s apparent that both r and θ vary with

the position vector. We should therefore be able to intuitively predict that Newton’s second law will involve

some derivatives of the unit vectors since they are no longer constant11. Note that we from now on we will

use the nomenclature r for the position vector while we will use a standard r for the radial coordinate.12 Let

us now derive Newton’s second law for polar coordinates. The position vector of any particle or individual

component in polar coordinates takes the simple form:

r = rr (1.4)

This is apparent as the position of the particle can be described just by the radial component r and the radial

unit vector r since the radial component always points in the same direction as the position vector. Of course,

r depends on θ and the tangential component θ may be involved in the motion of the particle, nonetheless r

and r are sufficient to describe the position. A simple proof follows from looking at the transformation between

cartesian and polar coordinates: x = r cos θ and y = r sin θ. In cartesian coordinates the position vector is

given by r = xx+ yy. It is also easy to see that r = r cos θx+ r sin θy. Since we are considering a unit vector,

the radial magnitude r is just equal to one and so we are left with r = cos θx + sin θy. Therefore, trivially

putting all three equations together, we get r = rr.

We’ve made an educated guess that the derivatives of r and θ will be involved, so it is worthwhile to solve for

these first and use these results in our derivation.

Consider a change in the unit vector r i.e r2 − r1, which we will call ∆r. We

will assume that the time interval ∆t over which this happens is small. Since

∆r is perpendicular to r1 and r2, it is in the same direction as θ. Now, to an

approximation, ∆r ≈ r∆θ = ∆θ13 if we think of ∆r as the arc of a sector of

a circle instead of a straight line (it doesn’t matter whether it’s r1 or r2, both

have magnitude 1). This ’approximation’ becomes increasingly accurate as ∆r

becomes smaller. Dividing both sides by ∆t we get:

∆r

∆t≈ ∆θ

∆t10As well as the vast majority of other coordinate systems.11Ofcourse they are of constant length of 1 unit, but their direction will change with the position vector, and thus with time

since the position is a function of time in this dynamical system.12By convention the letter r is used for position vectors. For the sake of completeness with the rest of the paper, we will stick

with this convention despite the possible confusion in the case of polar coordinates.13since r = 1

7


Now if we take the lim∆t→0 of the above expression, all the ∆ quantities become infinitesimal quantities since

both r and θ are also functions of time. The approximated ’arc’ of the circle becomes exactly equal to ∆r (Now

dr) and so we are left with the exact differential expression:

dr

dt=

dθ

dtθ = θθ (1.5)

where the θ appears to signify direction.

Likewise, for θ: ∆θ = θ2 − θ1. From the diagram we can see that ∆θ ≈ θ∆θ.

Since the magnitude of θ is 1, it follows that ∆θ = ∆θ. Dividing both sides by

∆t, we get:

∆θ

∆t≈ ∆θ

∆t

Using the same logic as above: Taking the lim∆t→0:

dθ

dt=

dθ

dt= θ

∆θ is perpendicular to θ and is therefore parallel to r. Notice however that ∆θ points in the opposite direction

as r (refer to figure 1.3b) and so it points in the −r direction. Therefore, we get the exact differential expression:

dθ

dt= −dθ

dtr = −θr (1.6)

Equations 1.5 and 1.6 give us the time derivatives of the unit vectors r and θ. We can now return to the

problem of finding the equations of motion for the pendulum using these results. We have:

r = rr

differentiating the above with respect to time:

r = rdr

dt+ rr

↓

v = r = rθθ + rr

Here v is the velocity (v = r). This should look very familiar, as the tangential velocity in angular motion (in

the θ direction) is given by vθ = rω wherein ω = θ. We also have a radial component of velocity vr (in the r

direction) although this will not be involved in our pendulum motion as we will see since the radius is of fixed

length l and does not change. Differentiating once more, we have:

a = r =d

dtr =

d

dt(rθθ + rr)

Using the product rule:

⇓

a = rr + rdr

dt+ (rθ + rθ)θ + rθ

dθ

dr

Substitutingdθ

dt= −θr and

dr

dt= θθ and rearranging the terms:

↓

8


a =(r − rθ2

)r +

(rθ + 2rθ

)θ (1.7)

We now have Newton’s second F = ma in its components as:

Fr = m(r − rθ2

)Fθ = m

(rθ + 2rθ

) (1.8)

(1.9)

This awkward form of Newton’s second law in polar coordinates is nowhere near as aesthetically pleasing as with

Cartesian coordinates. With other stranger choices of coordinate systems, this is only aggravated further and

becomes increasingly difficult to derive a succinct form of Newton’s second law. For the sake of completeness,

let’s solve our pendulum problem. In this case, Fr and Fθ become far simpler as r is a fixed length. Therefore,

all derivatives of r go to zero. Thus, for the case where r does not change, Newton’s law becomes:

Fr = −mrθ2 = mrω2

Fθ = mrθ = mrα

Where ω and α are the familiar angular velocity and angular acceleration respectively. Fr is simply the

centripetal force while Fθ is the tangential force or torque. Referring to figure 1.2, we can resolve Fθ = −mg sin θ,

therefore: mrθ = −mg sin θ. If we rearrange this equation we get the all too familiar differential equation for

the motion of a harmonic oscillator:

θ =g

rsin θ

Without digressing into the details of simple harmonic motion, if we restrict the motion to small angles and

approximate sin θ = θ and make the substitution ω2 =g

r, the solution to the differential equation through

inspired guessing turns out as:

θ(t) = A cos(ωt+ φ) (1.10)

So thus far we’ve used the example of a pendulum to illustrate some of the issues with using Cartesian

coordinates and using different coordinate systems with the traditional version of Newton’s second law. Often

in physics it is worthwhile to use somewhat arbitrary choices of coordinates. For example, if we were interested

in the motion of a bead along a twisted wire and we knew the equation of the function that described the

shape of the wire, it would make sense to simply use the length along the wire that the bead has travelled

as a coordinate instead of using traditional coordinate systems. We call these sort of coordinates generalized

coordinates and using generalized coordinates reduces the complexity of problems exponentially when we have

large systems of interacting objects and forces. We will look at generalized coordinates in detail in the following

chapters.

9


The objective of this section was to get one to start questioning the practicality of using F = ma for problems

in dynamics. As an overly simplified example, If we wanted to analyze the motion of say, molecules of gas in an

ideal deterministic classical world, we would have to use F = ma on each individual particle of gas and account

for all the forces between them and the forces of the particles hitting the walls of the container and we would

end up with a hideous mathematical quagmire.

By now, it should seem apparent that there is a need for an alternative formulation of Newtonian mechanics.

Lagrangian and Hamiltonian mechanics -named after the French-Italian mathematician Joseph Louis Lagrange

and the Irish mathematician Sir William Rowan Hamilton- is exactly that. It is merely a reformulation pioneered

by the aforementioned gentlemen, albeit a far more abstract one. No ‘new physics’ is introduced in either

formulation of mechanics. The same laws of physics hold true and the same classical principles such as work

and energy can be derived from these formulations too. They merely provide a different, more holistic structure

for analyzing dynamical systems. That may sound like an extremely vague meaningless statement now, but

hopefully through the course of this paper it will become appreciable. For now, let us move on with knowing

that Newtonian mechanics is somewhat naive in a physical sense and that there must be a more pragmatic way

of analyzing complex dynamical systems.

10

Chapter 2

Calculus of Variations

2.1 Functionals

Elementary calculus is mostly concerned with functions and how they change infinitesimally. A function is

simply a rule that relates one independent variable to one (or more) dependant variables[7]. The mathematical

foundation of analytical mechanics and the ’new formulations’ whose existence was conjectured in the last

chapter is based on a type of calculus called the calculus of variations. The calculus of variations is also

concerned with infinitesimal changes, rates of change and finding stationary points, but of a mathematical

device called a functional. Over the course of this paper, functionals will appear very often and they generally

do in all of physics. The entirety of Lagrangian and Hamiltonian mechanics is built upon variational principles

and so a discourse on the calculus of variations is an essential mathematical digression.

A functional, if overly simplified, is just a function of a function. It is something that does not depend on just

one value of a variable but on an entire function in question and all the possible values it takes. A regular

function y(x) can be described mathematically as a map of R ⇒ R since both the domain and range of the

function are real numbers. Similarly, a functional can also be described by a map. If C is the set of all

continuous, differentiable functions, then a functional J[y] is a map of C ⇒ R that associates a real number

to a chosen function y. Thus, a functional has a domain that is a function and a range that is a real number,

unlike regular functions whose domain and range are both real numbers. A function is often described as a

machine that takes in a number and ’spits out’ another number. Likewise, a functional is a machine that

takes in an entire function and spits out a number. The calculus of variations is largely concerned with the

optimization, i.e finding maximum, minimum and stationary points of functionals. This is best described by

a very simple, almost cliched example that is seen in virtually all courses and textbooks that introduce the

calculus of variations. This is for good reason, as it is an intuitive example and one whose solution we already

know.

What is the shortest distance between two points?

This turns out to be a far more interesting question than expected. Of course, we know the answer is a straight

line. It’s obvious, but lets actually prove it.

11


Figure 2.1

Objects that are the shortest distance between two points on various surfaces are called Geodesics[9] and we

are in particular looking for the Euclidean geodesic since we are interested in the shortest distance between

two points in regular Euclidean space. Figure 2.1 shows a generic curve y(x) that connects the two points 1

and 2 with coordinates (x1, y1) and (x2, y2) respectively. The goal is to find the particular function y(x) that

minimizes the distance between the two points. Consider an infinitesimal distance ds along the curve. Then

the total length s of the curve is given by:

s =

∫ 2

1

ds

From figure 2.1 it is apparent than ds can be written in terms of component infinitesimal displacements in the

x and y directions: dx and dy. Thus from the Pythagorean theorem, ds =√dx2 + dy2 which leads to:

s =

∫ x2

x1

√dx2 + dy2

=

∫ x2

x1

√(1 +

(dy2

dx2

))dx2

=

∫ x2

x1

√1 +

(dy

dx

)2

dx

=

∫ x2

x1

√1 + (yx)2 dx (2.1)

Recall that yx stands for the first derivative of y with respect to x.

We now have an expression for the length of the curve. The task at hand is to pick a y(x) that will minimize

this expression. This integral expression is a functional, since the value of the integral depends on what function

y(x) we choose to plug into it. The real number output (distance in this case) depends on all the values y(x)

takes on the interval in question and not on any particular value of x. The minimization of this functional

is a problem that the calculus of variations deals with and we will solve this particular example in the next

few sections. First we will look at another famous problem in mathematics lest readers should think that the

calculus of variations is restricted to trivialities such as figuring out that a straight line is the shortest distance

between two points.

12


2.1.1 The Brachistochrone problem

The Brachistochrone problem was first considered by Galileo, but was made famous by Johann Bernoulli.

Bernoulli’s statement of the problem was as follows:

”Given two points A and B in a vertical plane, what is the curve traced out by a particle acted on only by

gravity, which starts at A and reaches B in the shortest time.”

A more intuitive way to think of the problem is to consider a bead of mass m subject only to the force of

gravity that can slide freely along a wire between two points A and B with coordinates (x1, y1) and (x2, y2)

respectively. The problem entails finding that function (the shape of the wire) that minimizes the time taken

for the bead to reach the other point when released from the first point.

The total time t taken by the bead would be:

t =

∫ t2

t1

dt

where dt is an infinitesimal amount of time. If v is the instantaneous velocity

of the bead at a given moment and ds is an infinitesimal displacement along

the wire, then dt =ds

v. Thus:

t =

∫ b

a

1

vds

=

∫ x2

x1

√1 + yx2

vdx

using the same expansion of ds as in equation 2.1. From the conservation of energy, we know that 12mv

2 = mg∆y

and therefore v =√

2g∆y. Thus we have an expression for the time taken:

t =

∫ x2

x1

√1 + yx2

√2g∆y

dx (2.2)

where ∆y is the change in the vertical height y upto a given point as y changes i.e: y − y1

This is yet another problem that uses the calculus of variations to find a solution. Once again, we have a

functional that is an integral expression that depends on both y and yx in this case, instead of just depending

on yx as the geodesic problem did. Although I will not go through solving the Brachistochrone problem since

it involves slightly cumbersome algebra and is an unnecessary digression, the solution has some interesting

properties. For a long time it was thought that the fastest path would be the arc of a circle. However, the

solution to this problem is a class of curves called Cycloids. This is the shape that a point on the rim of a

wheel would trace out when rolling forward without any slippage (somewhat similar to my illustration above).

A peculiar property of this cycloid trajectory is that no matter where on the cycloid you release the bead from,

it will reach the other point in the same time period. Hence it is also called a tautochrone -loosely translating

to ”same time”. We’ve seen two examples of variational problems to illustrate the concepts of functionals and

their optimization. Let us now develop the mathematical framework to solve such problems. We will use the

methods we develop to solve for the Euclidean geodesic towards the end of the chapter.

13


2.1.2 Generic variational problem

The so called variational problem is just a generalization of problems of the kind we have looked at. It involves

the minimization of the functional:

J [y] =

∫ x2

x1

F (y, y′, x) dx

Here F is just a function of y and y′ (like we saw in the Brachistochrone problem) and we have allowed for an

explicit dependence on x as well. An implicit dependence on x would be when F only depended on x through y

since y is a function of x but does not directly depend on x. An explicit dependence on the other hand is where

F directly depends on x. This dichotomy will become quite important when we consider implicit and explicit

time dependence in physics problems. When we call F a function we really mean that F is an expression that

is some combination of y, y′ and x. The integral is with respect to x because ultimately, the function y depends

on x and so the entire integral will depend only on x. The reason we consider functionals that depend only upto

the first derivative is that in physics, the dynamics of a system are inherently second order. Newton’s second

law is a second order linear differential equation, and thus the time evolution of a system can be determined

given two initial conditions: the position and velocity. Functionals of higher order derivatives are not involved

in the framework of analytical mechanics, atleast at the level of this paper. As of now, our interest is merely

in finding the y(x) that minimizes the above integral J [y].1

2.2 The Euler-Lagrange equation

Theorem 1. If the particular function y(x) minimizes the integral J [y] =

∫ x2

x1

F (y, y′, x) dx over the closed

interval [x1, x2] , then y(x) must satisfy the Euler-Lagrange differential equation[8]

∂F

∂y− d

dx

∂F

∂yx= 0 (2.3)

This is one of the most profound mathematical results in mechanics and so deserves to be formally stated as

a theorem first. Over the course of this section I will develop the idea of the Euler-Lagrange equation from a

purely mathematical standpoint, and we will see how it is relevant to mechanics and how it provides a more

general scheme of analysis than Newton’s mechanics in later sections.

Before, we get into proving this result and going through how it was conceived, it is worthwhile to just think

about what this differential equation is. The Euler-Lagrange equation is a non-linear second order differential

equation and in essence, is a necessary condition to extremize2 an integral functional of the above form. Non

linear means that higher powers of variables may be involved3 (such as y2 or y3) and second order means that

the highest order derivative involved is the second derivative of the function. Much like we have dydx = 0 as

a necessary condition for finding extrema of a typical function y(x) in single variable calculus, the necessary

condition to extremize the integral above is that the input function y(x) must satisfy the Euler-Lagrange

differential equation. Thus, this equation is essentially the solution of our generic variational problem in the

sense that the integral will be extremized if and only if y(x) satisfies this equation. Now let us prove it.

1It is just convention to represent functionals by a capital J and I genuinely have no idea why.2An extremum is either a minimum, maximum or saddle point.3this isn’t strictly true but will suffice for our purposes since the classification of the differential equation isn’t really important

from a physical persepctive.

14


Proof. In ordinary differential calculus, we find out at which x coordinates the derivative of a function f(x) is

zero to find stationary points of the function. This means that the variation in f(x) is 0 when x is varied at

stationary points[13]. A variation is just a very small change. Thus, an infinitesimal change dx would cause no

change in f(x) at a stationary point. How would we find where the derivative is zero for a functional? What

would the derivative of a functional even mean? Consider:

J [y] =

∫ x2

x1

F (y, yx, x) dx

When we ask what the derivative of a functional is, we are really asking, how does the value of J change as the

entire function y changes. Remember, a functional’s dependant variable is not really a variable, it is a function

itself. J’s output changes as we change what function y we put in, not as we change particular values that y

takes. Thus, we need a way to cause a variation in what function y(x) we choose so that a particular functional

will extremize the functional. Let us call the correct input function that extremizes the functional y(x).4

Figure 2.2: Variation of y(x) using another function and a variable parameter.

If this is true, then the functional evaluated for y(x) is less than the functional evaluated for any other curve

that we will call Y (x) that is around y(x). Now we can create any function Y (x) by using the function η(x).5

All we do is add η(x) to y(x) and using a parameter6 ε that we multiply η(x) by, we can attain any function

Y (x) by varying ε such that Y (x) = y(x) + εη(x). This is shown in figure 2.2 wherein we are able to create a

whole class of functions Y (x) by adding another function to our candidate function y(x). Since we are interested

in the class of curves that start and end at the same points (because this is the interval of integration), it is

necessary that η(x1) = η(x2) = 0. This way, at the end points of the interval, η(x) vanishes and we are left with

the initial y(x). Note that we can choose absolutely any function η that we like, provided that this condition

is met and that η is continuous and differentiable over the interval.

4We are first assuming that some y(x) is the solution making the integral stationary and then working backwards to find a

necessary condition for extremization.5This the Greek letter ”eta”6Think of this as a third variable that we use to cause some change. Here we use the greek letter ’epsilon’

15


Given the class of functions7 Y = y + εη it is obvious that Yx = yx + εηx. This will be useful soon.

Let us consider the functional J [Y ] which is equivalent to J [y + εη]. Notice that this functional is now purely

dependent on ε since we can change the input function just by changing ε and thus we can change the value of

the functional itself. Therefore, let us make a substitution

φ(ε) = J [Y ]

This is really just a substitution for the sake of conciseness in the steps to come. It is essentially a restatement

that J [Y ] can be written as a function of just ε, and we call this function φ. We have developed 3 pieces of key

information:

• Y = y + εη

• Yx = yx + εηx

• φ(ε) = J [Y ] =

∫ x2

x1

F (Y, Yx, x) dx

We can now differentiate the functional J [Y ] with respect to ε. this will tell us how the functional J varies as we

vary the input function Y (x), since Y (x) is varied by changing ε. We can do this because φ is a function of ε and

thus we can arrive at notion of stationary points by setting this derivative equal to zero as with conventional

calculus. This is analogous to taking the directional derivative of the functional J in the η direction. In essence,

we ask how the functional changes if we change the input function in the direction of another function η. As a

matter of fact, if you are curious, what we are really doing is using something called the Frechet derivative and

the Gateaux derivative which are generalizations of the directional derivative at higher levels of abstraction. It

is not necessary to know any of this as a formal rigorous derivation of the functional derivative is far beyond

the scope of this paper. What is important is that we understand why the functional is being differentiated

with respect to ε. We are particularly interested in finding the condition for y(x) which we assumed to be the

function that minimizes the functional J . Since Y = y + εη, Y = y when ε = 0 and so we will evaluate the

derivative of the functional with respect to ε at ε = 0 (we will do this towards the end of the proof). Thus the

necessary condition for an extrema is φ′(0) = 0. By differentiating the third item with respect to ε we have:

φ′(ε) =d

dεJ [Y ]

=d

dε

∫ x2

x1

F (Y, Yx, x) dx

=

∫ x2

x1

∂

∂εF (Y, Yx, x) dx

In the last step, we have brought the derivative into the integral and becomes a partial derivative of the inner

function. This is a trivial step and I would ask you to take my word that this swap can be done. This is the

Leibniz Integral Rule[13], more commonly known as ”differentiating under the integral”. It is just a corollary

of the fundamental theorem of calculus.

7I will omit the (x) since it is obvious that Y , y and η are all functions of x.

16


We can apply the chain rule to the integrand which leaves:∫ x2

x1

(∂F

∂Y· ∂Y∂ε

+∂F

∂Yx· ∂Yx∂ε

+∂F

∂x· ∂x∂ε

)dx

Since Y = y + εη and Yx = yx + εηx, it follows that∂Y

∂ε= η and

∂Yx∂ε

= ηx. Also note that∂x

∂ε= 0 since x is

independent of ε. Thus the third term disappears and the integral can be rewritten as:

∫ x2

x1

(∂F

∂Yη +

∂F

∂Yxηx

)dx

=

∫ x2

x1

∂F

∂Yη dx+

∫ x2

x1

∂F

∂Yxηx dx (2.4)

The second integral in 2.4 can be integrated by parts. Recall that integration by parts states that:∫u

dv

dxdx = uv −

∫v

du

dxdx

Setting u =∂F

∂Yxand

dv

dx= ηx, then:∫ x2

x1

∂F

∂Yxηx dx =

[∂F

∂Yxη

]x2

x1

−∫ x2

x1

d

dx

∂F

∂Yxη dx

The first term on the right hand side goes to 0 since η(x1) = η(x2) = 0 (recall that we stated this was a

necessary condition for the choice of η(x)) Thus, equation 2.4 simpifies to:∫ x2

x1

(∂F

∂Y− d

dx

∂F

∂Yx

)η dx

and therefore:

φ′(ε) =

∫ x2

x1

(∂F

∂Y− d

dx

∂F

∂Yx

)η dx

The required condition for extremization was that φ′(0) = 0. We saw that if ε = 0 then, Y = y and Yx = yx.

Therefore, we have:

φ′(0) =

∫ x2

x1

(∂F

∂y− d

dx

∂F

∂yx

)η dx = 0 (2.5)

17


Now since η(x) could be any function provided that η(x1) = η(x2) = 0, it does not necessarily have to be 0

elsewhere in the interval. Thus, the only way the integral can equal 0 is if the other part of the integrand is 0.8

Therefore, we have the Euler Lagrange equation:

∂F

∂y− d

dx

∂F

∂yx= 0 (2.6)

QED

Note that although we proved the specific case for minimizing the integral, the same proof can be applied to

maximizing or extremizing the integral above. The usage of the term minimum is customary because most

of the time, it is the minimum that is of interest to physicists. This is really a misnomer and in general the

Euler-Lagrange equation is a necessary condition for a function y(x) to make the integral J[y] stationary, not

just a minimum.

2.2.1 Solving the Euclidean Geodesic

Lets us now go back to our problem of solving for the shortest distance between two points and gain some

insight into how the Euler-Lagrange equation (which i will abbreviate as EL from now on) is used. The task

was to minimize9 the integral:

s =

∫ x2

x1

√1 + (yx)2 dx

We can use the EL equation:∂F

∂y− d

dx

∂F

∂yx= 0

with F =√

1 + (yx)2.

Thus, if y(x) is the curve that minimizes the distance between two points, it follows that:

∂

∂y

(√1 + (yx)2

)− d

dx

∂

∂yx

(√1 + (yx)2

)= 0

Just from computing the respective partial derivatives, we have:

∂

∂y

(√1 + (yx)2

)= 0

(since F = F (yx) and does not depend on y itself).

and

∂

∂yx

(√1 + (yx)2

)=

yx√1 + (yx)2

8This is called the first fundamental lemma of the calculus of variations and will not be formally proved here. A lemma stands

for some intermediate step in a larger proof that is assumed to be true and then proven such that the larger proof works. The

lemma states that if

∫ x2

x1

α(x)η(x) dx = 0 given that η(x1) = η(x2) = 0, then α(x) is necessarily 0 over the entire interval [x1, x2].

Several proofs of this trivial result are easily accessible and are beyond the scope of this paper.9Here we are genuinely interested in only the minimum and not any extrema

18


Therefore:

− d

dx

yx√1 + (yx)2

= 0

→ d

dx

yx√1 + (yx)2

= 0

The only way the derivative of something is 0 is if that something is a constant. Thus:

yx√1 + (yx)2

= C → (yx)2

1 + (yx)2= C2

where C is some constant. Rearranging this a bit, it isn’t hard to see that:

yx = m

where m is some other constant.

This is a simple differential equation and if we integrate this, we get:

y = mx+ b (2.7)

where b is an other constant (the y-intercept of the line). Thus, we have finally proved that a straight line is

the minimum distance between two points!

2.2.2 Functionals with more than one dependent variable

In Physics problems, we will see that we generally have several dependent variables and not just one y(x)[17].

Usually, we will have only one independent variable: the time t. This greatly simplifies considerations of

severable variables, nevertheless we will see what the solutions to functionals of more than one dependent

variable look like in this quick aside.

Say we have the integral

J [y, z] =

∫ x2

x1

F (y(x), z(x), x) dx

Here, y and z are the dependent variables and x is the independent variable. In such a case, the solution to

extremizing the integral will be two EL equations:

∂F

∂y− d

dx

∂F

∂yx

∂F

∂z− d

dx

∂F

∂zx

In general, if we have a functional with one independent variable and N dependent variables, then the solution

to extremizing the integral will be N Euler-Lagrange equations[7]. The proof for this result is exactly the same

as with a single variable. We would just consider different variations εη for each dependant variable and follow

the same process as in section 2.2 to derive an EL equation for each variable.

19

Chapter 3

Space and constrained motion

3.1 Degrees of freedom

The mechanical degrees of freedom of a system are essentially the number of ways you can independently specify

the configuration of the system[18]. The configuration of a system is how we would describe the system’s position

and arrangement in space. This is best explained through examples.

(a) 1 degree of freedom (b) 2 Degrees of freedom

(c) 3 Degrees of freedom

Here, the red object - which we will assume to be a dimensionless particle - is shown to have various degrees

of freedom depending on whether it can only move along a line, surface or all of 3 dimensional space. Thus

far, degrees of freedom and dimensionality of space seem like the same thing. Now instead of a point particle,

consider a sphere, cube or any other rigid body in 3 dimensional space. Such a body would have 6 degrees

of freedom. 3 translational independent directions of motion and also 3 independent rotational angles, that

would completely determine how the body was arranged in space. In general, any rigid body in free space (an

object where the distance between any two points remains constant) has 6 independent degrees of freedom as

described above. If instead we considered a system of two point particles, then the system has 6 degrees of

freedom, since each particle has 3. Likewise, a system of two rigid bodies would have 12 degrees of freedom: 6

for each. In general, if we have a system of N particles, then the system has 3N degrees of freedom since each

particle has 3 independent degrees of freedom.

20


The degrees of freedom of a system can now be viewed as the dimensionality of the configuration space of the

system. We imagine this to be some N dimensional vector space where each dimension is a component that

describes the positions of the elements of a system. Thus, for a rigid body we would have a 6 dimensional

configuration space and for a system of N particles we would have a 3N dimensional configuration space. This

is simply an abstraction for the sake of visualization and can be ignored if it is confusing.

3.2 Constraints

Suppose we had a system of two particles in 3 dimensions. Normally, the system would have 6 degrees of freedom

as we saw earlier. Now suppose the two particles were connected to each other by some sort of massless rod.

How they are connected is not particularly important here, what matters is that the distance between them is

fixed.

The particles are effectively constrained to move in a certain

way. The system has one less degree of freedom now since both

particles must be at a fixed distance from each other. The

tension in the rod would be the constraint force. This force

is what keeps the motion of the object constrained in the first

place.

Another simple example would be a block moving on a surface. In this case, the block is constrained to move on

the surface and would have only two degrees of freedom. The force of constraint would be the familiar normal

force. This force restricts the block from leaving the surface. In essence, a constraint is simply a restriction on

the motion of an object.

Lastly, consider a disc rolling along a straight line without slipping. If there was slippage, the disc would have

two degrees of freedom: one translational along the line and one rotational about the disc’s axis. However, the

force of friction constrains the disc to move in a certain way. yes, the disc still rotates, but now the rotation

is predetermined and dependent on the translation of the disc. It can no longer freely rotate since the friction

makes it roll without slipping. Thus, the friction constrains a system that would normally have two degrees of

freedom to a system with just one degree of freedom. Usually, we are interested in the motion of a body, not in

the constraint forces. So one of the main motives of Lagrangian mechanics is to find a way to eliminate these

forces of constraint as they simply add to the computational complexity of the problem.

3.3 Generalized coordinates

Consider the pendulum we used as an example in chapter 1. Here, although the pendulum was technically

moving in 2 dimensions, the pendulum actually had only one degree of freedom: motion along the arc of a circle

of radius l. Here we see the difference between degrees of freedom and dimensionality of space. The pendulum

could not move freely in two dimensions because of the constraint. It could only move freely in one way (the

arc of the circle). We saw that the motion of the pendulum could be described entirely by one single parameter:

the angle θ. This is an example of a generalized coordinate. Generalized coordinates are a set of parameters

that describe the configuration of a system with reference to some other coordinate system[16].

21


Thus, we can describe the position of each of N components of a system using n generalized coordinates:

ri = r(q1, q2, q3, . . . , qn)

where i goes from 1 to N and the letter q is used to signify a generalized coordinate. The converse is also

true, that is, we must be able to express the generalized coordinates as a function of position. We use the

minimum number of generalized coordinates possible to greatly simplify problems in mechanics. For example, by

parameterizing the motion of the pendulum with θ, we reduced a two dimensional problem to a one dimensional

problem and the relation between the position and generalized coordinate θ was given by:

r = (l sin θ, l cos θ)

Thus, there must always be a two way relation between the generalized coordinates and the position of the

system (defined in any other coordinate system) and a generalized coordinate is simply a ’convenient’ choice

of coordinates[1]. These coordinates do not have to be vector based position coordinates like the cartesian

coordinates we are used to. It can be an angle, a scalar distance, an amplitude or anything that helps define

the configuration of a system in an independent way, and can be transformed to and from the original cartesian

/ position vector based coordinates.

We call the generalized coordinates natural, if the relation between the position vector (in cartesian coordinates)

and the generalized coordinates does not explicitly depend on time (such as the ones we discussed above with

the pendulum). A set of generalized coordinates is called forced if there is an explicit time dependence between

the position vector and the generalized coordinates, i,e:

r = r(q1, q2, . . . , qn, t)

[17]

3.3.1 Holonomic constraints and systems

A constraint is called holonomic when the constraint can be expressed as an equation that is a function of

the coordinates of the system[9]. Thus a constraint C is expressed as:

C = f(x1, x2....xn)

The pendulum we examined is subject to a holonomic constraint, specifically:

l2 = x2 + y2

22


Thus, a holonomic constraint is also relation between the coordinates of the system and consequently implies

that the coordinates are not strictly independent of each other. Typically, the constraint force in a holonomic

constraint will be perpendicular to the motion of the particle, for example: the normal reaction force. Therefore,

for such systems, the constraint forces do no work[7]. We can use holonomic constraints to make some

coordinates redundant and arrive at the minimum number of generalized coordinates. In general if we have a

system of N particles subject to M constraints, it follows that the system has 3N-M degrees of freedom and

consequently 3N-M independent generalized coordinates that describe the configuration of the system. This

is made obvious in the pendulum problem. The pendulum moves in two dimensions but had one constraint

equation. Therefore, the pendulum had 2-1=1 degree of freedom and one generalized coordinate. A system is

called a holonomic system if the degrees of freedom of the system and the number of generalized coordinates

of the system are the same and so each generalized coordinate describes one dimension of the configuration

space[7]. In this sense, we use a set of generalized coordinates as independent parameters when the usual

cartesian coordinates are no longer independent because of a holonomic constraint imposed on the system.

Likewise, a non-holonomic constraint is one that cannot be expressed as a function of just the position

coordinates of a system and may involve constraints on the velocities too. An example of a non-holonomic

constraint is a gas inside a container. The gas molecules are constrained to move only within the walls of the

vessel, but can move anywhere inside. Thus the constraint would be some sort of inequality rather than a

specific function that defined how the molecules can move. A non-holonomic system is one which would require

more generalized coordinates to determine its configuration than the degrees of freedom of the system. Most of

the ’normal’ systems you can come up with in mechanics involve holonomic constraints and holonomic systems

so we will only focus on such relatively simple systems and this will not be a limiting factor in understanding

the subject.

23

Chapter 4

Lagrange’s equations of motion

4.1 Virtual work

In physics, we know that the work done on an object is given by W =

∫ r2

r1

F · dr where F is the net force on

the object and dr is an infinitesimal displacement. it is important to realize that dr is a real displacement that

occurs in some time dt. A virtual displacement, which we call δr is a an imagined displacement to the system.

It does not actually occur in some time period t and is something we just consider as a tool of analysis. A

virtual displacement must be an imagined displacement in accordance with the constraints on the system. For

example, if we were considering the virtual displacement of a pendulum, it must be some small change in the

pendulum’s position along the arc of the circle that the pendulum is constrained to move on. Any random

imagined changed in position will not be an appropriate virtual displacement.

Now consider a system of n particles. If the system is in dynamical equilibrium, we know that the net force

must vanish:n∑i=1

F = 0

Fi is the force on each individual particle and from now on, the sum will just be denoted as∑i

Fi = 0 since

it is understood that we are summing over all n particles and all sums will be depicted as such1, wherein it is

assumed that the sum is over the relevant set of quantities.

It follows that if we consider some virtual displacement δr to the system, then it is obvious that:

∑i

Fi · δri = 0

Where δri is the virtual displacement to each ith particle. The force on each particle can be decomposed into

the applied force F a and the forces of constraint F c :

Fi = F ai + F ci

Therefore, we can rewrite the above equation as:

∑i

F ai · δri +∑i

F ci · δr = 0

1Think of this as the ”sum over all i” or ”sum over all j” for some other sum.

24


If we are dealing with holonomic constraints, then the force of constraint will always be perpendicular to the

motion or displacement of the particle (like the normal force from a surface or the tension in the string of

the pendulum). This does not include constraint forces such as friction that are parallel or antiparallel to the

displacement. In such cases, the second term in the above equation vanishes since the dot product of two

perpendicular vectors is always 0. Thus we are left with:

∑i

F ai · δri = 0 (4.1)

This is called the principal of virtual work[7]. It states that the total virtual work done on a system must be

equal to zero. this seemingly obvious result has the huge advantage that we have completely eliminated the

need to solve for constraint forces in a static equilibrium problem. By changing the problem from a vector

statics problem where we need to solve for all the forces in the system, to a work based problem using imagined

displacements, we have made the constraint forces irrelevant (because they do no work). Note however, this does

NOT mean that the work done in each individual component (x,y, or z) of the displacement δr is zero, neither

does it mean that the work F ai · δr done on any particular particle i is zero. This is because the components

of r and the positions of each particle are not independent of each either because they are still connected by

constraints. Therefore the principle of virtual work in this form does not imply that each applied force F ai = 0.

It only implies that the SUM of the work done by the applied forces is zero. In-order for the former to be true,

we would have to write the principal of virtual work in terms of independent generalized coordinates. We can

transform the virtual displacements δri to some virtual displacements in terms of m generalized coordinates.

Then the position vector of any particle i can be written as:

ri = ri(q1, q2, . . . , qm, t)

Then, it follows from the chain rule that:

δri =∂ri∂q1

δq1 +∂ri∂q2

δq2 . . .+∂ri∂qm

δqm +∂ri∂t

=∑j

∂ri∂qj

δqj +∂ri∂t

=∑j

∂ri∂qj

δqj

The term∂ri∂t

is discarded because the displacement is virtual, i.e does not happen in a real time frame.

Therefore the derivative of ri with respect to t must be 0.

Since the component forces and the component generalized displacements are independent of each other, the

total virtual work can be zero only if every component is also zero. Therefore, the principal of virtual work in

generalized coordinates can be written as:

∑i,j

Fi ·∂ri∂qj

δqj = 0 (4.2)

It is understood that Fi is only the applied force and the superscript a is excluded. The sum is over all relevant

i and j i.e from i = 1 to i = n and j = 1 to j = m.

We call∑i

Fi ·∂ri∂qj

the generalized force and denote it with Qj . Since all Qj are independent, it follows that

each generalized force must be equal to zero if the virtual work done by the sum of the generalized forces is

zero2.

2Since the generalized forces are not related by constraints like the regular forces were.

25


4.2 D’alembert’s principle

The principle of virtual work is useful in analyzing static systems, but we now need a way to extend such an

idea to dynamics. Let’s see what we can do with Newton’s second law for a dynamic system of n particles:

Fi = pi

Where Fi is the total force on the ith particle and pi is the momentum.

We will use a similar analysis, as with the principle of virtual work, to make constraint forces irrelevant.

Fi = F ai + F ci

Therefore:

Fai + F ci = pi

We can now eliminate constraint forces by considering the work done when the particles are virtually displaced:

∑i

(F ai + F ci ) · δri =∑i

pi · δri

↓∑i

F ai · δri +∑i

F ci · δri =∑i

pi · δri

Since the constraint forces do no work:∑i

F ci · δri = 0

Therefore: ∑i

Fi · δri =∑i

p · δri

⇓∑i

(Fi − pi) · δri = 0 (4.3)

Here the superscript a has been omitted without ambiguity since it will now be assumed that F refers to only

the applied force. Equation 4.3 is called D’alembert’s principle[7] and is the dynamical analogue of the principal

of virtual work. It essentially turns a dynamics problem to a pseudo-statics problem by introducing a fictitious

force3 p that is of the same magnitude as the rate of change of momentum and cancels out F . Although

it looks like a simple rearrangement of Newton’s second law, it is important to distinguish that the second

term is not a real force and is a an ’imaginary’ term we introduce to make it a statics problem. By doing so,

we can transform the problem to generalized coordinates just as with the virtual work case and thus set the

independent components equal to 0. We would then have a law of motion in each generalized coordinate.

3Sometime called an inertial force, these are forces that manifest due to the nature of the system and not real forces as we know

them. In this case, a fictitious force is introduced to reduce a dynamical problem to a statics problem.

26


4.3 Deriving a law of motion

Let us try deriving a law of motion from D’alembert’s principle:

∑i

(Fi − pi) · δri = 0

Consider a transformation between the position vector r and some set of m generalized coordinates q (just as

we did with virtual work):

ri = ri(q1, q2 . . . , qm)

As discussed earlier:

δri =∑j

∂ri∂qj

δqj (4.4)

We can now write D’alembert”s principle in two parts as:

∑i,j

(Fi ·

∂ri∂qj

)δqj −

∑i,j

(pi ·

∂ri∂qj

)δqj = 0 (4.5)

The first term cannot be simplified any further and we will just rewrite it in terms of the generalized force as∑j

Qjδqj .

Let us try and simplify the second term further:

∑ij

p · ∂ri∂qj

δqj =∑ij

mri ·∂ri∂qj

δqj

This is just the definition of momentum4. Let us now use the product rule in reverse. Consider the time

derivative of the expression∑ij

mri ·∂ri∂qj

(we will drop the δqj terms throughout the derivation until the end

for lack of confusion) :

∑ij

d

dt

(mri ·

∂ri∂qj

)=∑ij

mri ·∂ri∂qj

+∑ij

mri ·d

dt

(∂ri∂qj

)

The above is just the regular product rule. The term we are interested in solving for is boxed, Thus we can

re-arrange as follows:

∑ij

mri ·∂ri∂qj

=∑ij

d

dt

(mri ·

∂ri∂qj

)−∑ij

mri ·d

dt

(∂ri∂qj

)=∑ij

[d

dt

(mri ·

∂ri∂qj

)−mri ·

d

dt

(∂ri∂qj

)](4.6)

4Note that we assume that all particles i have the same mass m. This is to avoid another index i for the masses for the sake of

simplicity.

27


Two more identities can be used to simplify this further. The first identity comes from simplifying the last

term in equation 4.6 as:d

dt

(∂ri∂qj

)=∂ri∂qj

(1)

We have just switched the order of differentiation since the derivatives are commutative on linear expressions.

The second identity we need will come from the transformation between r and q:

ri = ri(q1, q2 . . . , qm)

Differentiating with respect to time:

ri =∂ri∂q1· q1 +

∂ri∂q2· q2 · · ·+

∂ri∂qm

· qm

↓

ri =∑j

∂ri∂qj

qj

Differentiating the above with respect to qj , the second identity5 is:

∂ri∂qj

=∂ri∂qj

(2)

The two identities can now be plugged back into equation 4.6 which results in:

∑ij

mri ·∂ri∂qj

=∑ij

[d

dt

(mri ·

∂ri∂qj

)−mri ·

∂ri∂qj

]

We can now ’reverse engineer’ the two terms by considering their antiderivatives. This reduces to:

∑ij

mri ·∂ri∂qj

=∑ij

[d

dt

[∂

∂qj

(∑i

1

2mr2

i

)]− ∂

∂qj

(∑i

1

2mr2

i

)]

One can see that this is entirely equal to the previous step by simple evaluating the partial derivatives involved.

We know that∑i

1

2mr2

i is the kinetic energy of the system of particles, which we will denote with T . This is

just a strange convention that has stuck for some reason. It would be interesting to know why physicists use

half the letters they use.

Using the two terms from equation 4.5 in their simplified form, D’alembert”s principle is now entirely equivalent

to:

∑j

[d

dt

(∂T

∂qj

)− ∂T

∂qj−Qj

]δqj = 0 (4.7)

5Note that i have removed the sigma on the term for conciseness. The remaining sum at the end will be assumed to sum over

ALL is and js.

28


Where Qj =∑i

Fi ·∂ri∂qj

.

If all forces Fi are conservative forces, then each force must be derivable from a potential energy function

Vi(r1, r2 . . . , rn). Once again, the choice of the letter V for potential energy is entirely arbitrary. In a

one dimensional case we would say the the force is the negative derivative of the potential. Likewise, the

multivariable extension of the same is that the force is the negative gradient of the potential, since we are

taking the partial derivative in each direction r.

Fi = −∇V (r1, r2 . . . rn, t)

Therefore:

Qj = −∑i

∇iV ·∂ri∂qj

Upon closer inspection, this is equivalent to the partial derivative of the potential with respect to the generalized

coordinate qj :

Qj = −∂V∂qj

Equation 4.7 can now be written as:

∑j

[d

dt

(∂T

∂qj

)− ∂T

∂qj+∂V

∂qJ

]δqj = 0

↓∑j

[d

dt

(∂T

∂qj

)−(∂T

∂qj− ∂V

∂qj

)]δqj = 0

↓∑j

[d

dt

(∂T

∂qj

)−(∂(T − V )

∂qj

)]δqj = 0

29


Since the potential V depends only upon the generalized6 coordinates qj and not the generalized velocities7 qj .

Thus, a V can be subtracted from the first term since∂T

∂qj=∂(T − V )

∂qj.

This is because∂V

∂qj= 0. This allows us to rewrite D’alembert’s principle in a symmetrical way:

∑j

[d

dt

(∂(T − V )

∂qj

)− ∂(T − V )

∂qj

]δqj = 0

If we define a function L(q, q, t) where L = T − V , then:

∑j

[d

dt

(∂L

∂qj

)− ∂L

∂qj

]δqj = 0 (4.8)

This function L is called the Lagrangian of the system, and is defined as L = T−V for most ordinary mechanical

systems[16]. note however, that this is not the only Lagrangian that satisfies D’alembert’s principle. It is just

the most obvious, useful one. Also, this definition only holds true for holonomic systems acted on by potential

derivable conservative forces that depend only on the objects position[17]. Part of our original goal was to

make the components of D’alembert’s principle independent so that each would all be equal to zero. This is

now possible, since everything in the inside the sum in equation 4.8 is now independent of other components

of the sum. Therefore, we can discard the sum and we have:

d

dt

(∂L

∂q

)− ∂L

∂q= 0 (4.9)

For each q that goes from 1 to m (the total number of generalized coordinates). This is called Lagrange’s

equation: a law of motion. It is entirely equivalent to Newton’s second law. The motion of a given system

is defined by m Lagrange equations which are second order non linear ordinary differential equations. Looks

familiar?

6Although we looked at a potential that depends on regular coordinates, this automatically implies that the potential V also

depends on the generalized coordinates since we can freely transform between position vectors and generalized coordinates.7This is not always true, for example, the force on a charged particle in a magnetic field is F = Bqv. Here the potential does

depend on velocities. In such cases, we wouldn’t be able to make the following substitution.

30

Chapter 5

Hamilton’s Principle

Thus far, it has been proved that the motion of a particle is defined by the differential equation:

d

dt

(∂L

∂qj

)− ∂L

∂qj= 0

where L is a function that is called the Lagrangian and is defined as L(q, q, t) = T − V for all the generalized

coordinates q and generalized velocities q. However, it has also been proven that the equation:

d

dx

(∂F

∂yx

)− ∂F

∂y= 0

is the necessary condition for extremizing an integral of the form

∫ x2

x1

F (y, yx, x) dx. Therefore, with a simple

replacement of F with L, y with q and x with t, it follows that Lagrange’s equation of motion implies that the

integral:

∫ t2

t1

L(q, q, t) dt

will be a minimum1 over the motion of the system and there will be such an integral that is minimized for

each generalized coordinate and velocity of the system. This integral of T −V over a time interval is called the

action of the system. Note that it is not the total energy of the system, and is instead the difference between

the kinetic and potential energy of the system. Thus, the principle of least action states that the action of a

system will be a minimum over the path of motion or trajectory in the configuration space that the system

follows. Thus, the principle defines how a mechanical system evolves with time. The Lagrangian L need not

always be equal to T − V and it is quite possible that other Lagrangian functions may define the motion of a

system. The principle of least action however, is known to work and is derivable as has been shown.

It might seem peculiar that this ’physically meaningless’ integral should be a minimum over the motion of a

system. Feynman best explains the physical interpretation of the principle of least action in one of his lectures

at Caltech.

1or more accurately, stationary

31


Figure 5.1: True path of a projectile in one dimension

Feynman’s analysis of the principle of least action aims to make it physically intuitive by describing it as

nature’s balancing act between kinetic and potential energy. If a ball was thrown up, then its trajectory when

plotted against time would look like a parabola[6]. The ball would keep rising until it is stationary and then

fall back down.

Why is this trajectory the ’true path’ of the ball. Why cant the ball follow some arbitrary path? We now

know that the path the ball will follow is the one which minimizes the action integral. Any other path between

the two points must have a greater value of action than the true path. In some sense, this is axiomatic, and

is as much of a ’law’ as Newton’s second law. Trying to reduce it further as to why this is true is almost

futile[6]. We can however analyse what this minimization of action would look like. Nature balances kinetic

and potential energy by trying to get the most potential energy out of the least kinetic energy. If a ball is

thrown, it will go as high as possible to maximize its potential energy, but if it is thrown too hard, then the

kinetic energy increases. Thus, the ball follows such a trajectory that gets as high as possible with the least

initial jolt of kinetic energy, therefore minimizing the difference between T and V. This strange behaviour of

mechanical systems is yet another case where nature tries to make things as efficient as possible. The principle

of least action is an aggregate law, that considers the motion of a system over an entire time frame. However,

this aggregate law also implies an infinitesimal law- one that is true for each instance of time t. If the action is

minimized over the entire time period, it follows that the action must be a minimum during each differential

amount of time dt. If the action were greater at one small time interval and at a minimum during the rest of

the motion, then the overall action would not be a minimum since it is possible to reduce the action further by

reducing the action over that particular small interval. Thus the principle of least action also holds true at a

differential level[6].

The principle of least action is more formally know as Hamilton’s principle. It is formally stated as:

Theorem 2. The actual path followed by a system of particles in configuration space in the time interval t1 to

t2 is such that the integral: ∫ t2

t1

L dt

is stationary over the path.

32


Thus, calling it the principle of least action is a misnomer and is really the principle of stationary action. Most

textbooks will introduce this first as a law, and then derive Lagrange’s equations from this using variational

principles, assuming a true physical path, much like in the Calculus of Variations chapter. Then it will be

shown that the Euler Lagrange equation is equivalent to Newton’s second law for L = T − V as follows in a

one dimensional case:

L = T − V =1

2mx2 −mgx

↓

∂L

∂x= −∂V

∂x= Fx

∂L

∂x=∂T

∂x= mx = p

Therefore :

∂L

∂x− d

dr

(∂L

∂x

)= 0 ⇒ Fx − px = 0

⇓

Fx = px

Here the principle of least action has been assumed as an axiom first thus implying the Euler Lagrange equations

hold. Then it is shown that the Euler Lagrange equation works with the particular choice of L = T − V and

is equivalent to F = p in each dimension. Although such a derivation is logically sound, it is important to

understand that Lagrange’s equations were originally derived from D’alembert’s principle and Newton’s second

law. Only then did Hamilton see that Lagrange’s equations implied a variational principle and then formulated

Hamilton’s principle and postulated that the action integral must be stationary.

Hamilton’s principle is often stated first as the foundational principle because it has applications in several

other areas of physics, particularly quantum mechanics, and isn’t restricted to classical systems. However, this

paper aims to explain the logic behind Hamilton’s principle. The variational form 2 was Hamilton’s contribution

to the framework of Lagrangian mechanics; however, Lagrange had independently developed his equations of

motion prior to this.

2Hamilton’s principle does not have anything to do explicitly with Hamiltonian mechanics. Hamilton’s principle is the defining

principle behind Lagrangian mechanics. Confusing, but this is because Hamilton was the one who saw the variational principle in

Lagrange’s equations of motion. Hamilton’s reformulation of mechanics is a another paradigm altogether.

33


Hamilton’s principle however, provides a far more general view of mechanics than Newton’s law does and acts

as a convenient scaffolding that circumvents the messy algebra that Newtonian mechanics would bring about

in complex systems. For one, it is impartial to the choice of coordinates. The value of the Lagrangian remains

the same regardless of what generalized coordinates we use. This was one of the goals of analytical mechanics,

most of which uses Hamilton’s principle as its foundational axiom. Secondly, as we saw in the derivation of

Lagrange’s equations from D’alembert’s principle, the need to account for constraint forces goes away when

appropriate generalized coordinates are chosen. The redundancy of constraint forces can be proved equivalently

from Hamilton’s principle, although I will not do so here. The Lagrangian formulation of mechanics is an

abstraction of Newtonian mechanics. Newton’s second law is twisted and concealed within the action integral

and the EL equation, but still pulls all the strings from deep within. With this new formalism, the need to

explicitly rely on computation with forces goes away. The time evolution of any holonomic system can be

determined through a series of relatively simple steps:

1. Determine a set of appropriate generalized coordinates.

2. Find the kinetic and potential energies - T and V - in terms of the generalized coordinates.

3. Plug the difference into the Euler-Lagrange equation.

4. Solve for an equation of motion for the system.

Elegant and relatively effortless. Although this method will always give a differential equation as an answer for

each coordinate, some systems may still not have solutions to these equations. this does not mean they cannot

be solved, but it does mean that the solutions cannot be expressed in terms of standard functions. Thus, most

solutions will use numerical methods and mathematical tricks to extract useful information from the equations,

both of which are beyond the scope of this paper.

34

Chapter 6

More on the Lagrangian Formalism

6.1 Cyclic coordinates and conservation laws

Lagrange’s equation is analogous to Newton’s law in the sense that Newton’s law relates force and momentum

while Lagrange’s equation relates generalized forces and generalized momenta.

Thus∂L

∂qis the generalized force and

∂L

∂qis the generalized momentum, then the equation of motion

∂L

∂q=

d

dt

(∂L

∂q

)

states that the generalized force equal the rate of change of the generalized momentum. Thus in Cartesian

coordinates, each generalized force would just be the regular force in the x,y and z directions. This is because∂L

∂q=∂(T − V )

∂q=∂V

∂x= Fx and so on for y and z. Likewise the generalized momentum would just be the

regular Cartesian momentum since∂L

∂q=

∂T

∂q=

∂(

12mx

2)

∂x= mx = px. This proves that when we choose

Cartesian coordinates, the generalized forces and generalized momenta are the same as the regular forces and

momenta. However, if we chose an angle as a generalized coordinate, then the generalized force would be a

torque and the generalized momentum would be an angular momentum.

If the Lagrangian does not depend on or change with q, then it follows that∂L

∂q= 0. If this is the case, then

the Lagrange equation implies that:

d

dt

(∂L

∂q

)= 0

If the derivative of something is 0 then that something must be a constant. In other words, it doesn’t change and

is conserved. If the Lagrangian is independent of a particular coordinate q, then we say that the coordinate is

cyclic[8]. Thus, if a coordinate is cyclic then its conjugate momentum1 is conserved[8]. In the Cartesian case,

if a potential does not depend on one coordinate, say x, then Fx is 0 and therefore px is conserved. Likewise,

for angular polar coordinates, a cyclic angle would imply that the torque is 0 and that the angular momentum

is conserved. Thus, cyclic coordinates will greatly simplify problems in Lagrangian dynamics. This sort of

conservation law that arises from a redundant coordinate is called a symmetry. For example, if a cartesian

system’s Lagrangian is independent of the coordinates x,y and z then the system is translationally symmetric

1Associated momentum with respect to the particular coordinate

35


which implies that the conjugate momentum - the regular linear momentum in this case - is conserved. The

generalized version of the inference that symmetries imply conservation laws, is called Noether’s theorem.

6.2 Using the Lagrange equation

We will look at an example[17]2 to demonstrate the Lagrangian method.

Consider a block of mass m on a wedge of mass M. The wedge lies on top of a surface and the block is free to

slide down the wedge and the wedge is free to slide across the surface, both without friction. The wedge is at

an angle α to the surface and the length of the slope of the wedge is l. Find the acceleration of the block down

the wedge and the time taken for the block to reach the bottom of the wedge.

The system has two degrees of freedom, one for the wedge and one for the block. It is a holonomic system

with both bodies constrained by the respective normal forces. Thus, our most obvious choice of generalized

coordinates would be the distance from the origin to the wedge: q1 and the distance from the top vertex of

the wedge to the block q2. The next step is to write the Lagrangian for the system. The Kinetic energy of the

wedge is1

2Mq2

1 . The velocity of the block will have components from the velocity of the block relative to the

wedge AND the velocity of the wedge itself. Resolving horizontally (x) and vertically (y) for the velocity of the

block v:

v = (q2 cosα+ q1, q2 sinα)

Therefore, the kinetic energy of the block is:

1

2m[(q2 cosα+ q1)2 + (q2 sinα)2]

↓1

2m(q2

2 cos2 α+ 2q1q2 cosα+ q21 + q2

2 sin2 α)

↓

Since sin2 α+ cos2 α = 1

↓1

2m(q2

2 + q21 + 2q1q2 cosα)

The total Kinetic energy of the wedge-block system is then:

T =1

2(M +m)q2

1 +1

2m(q2

2 + 2q1q2 cosα)

The potential energy of the wedge is 0 and the potential energy of the block is just mgh and since the height

is −q2 sinα, we have::3

V = −mgq2 sinα

2Example 7.5, Classical Mechanics, John R. Taylor, 2005 edition.3The minus sign is involved because q1 is positive acting down the slope while height is defined as positive going vertically up.

36


The Lagrangian of the system in terms of the generalized coordinates is then:

L = T − V =1

2(M +m)q2

1 +1

2m(q2

2 + 2q1q2 cosα) +mgq2 sinα (6.1)

The EL equation for q1 is:

∂L

∂q1=

d

dt

(∂L

∂q1

)

Since L is independent of q1, the conjugate momentum is conserved and is some constant c. This is just

conservation of momentum in the x direction since q1 is the same as x. Computing this gives:

Mq1 +m(q1 + q2 cosα) = c

Differentiating both sides with respect to time:

Mq1 +mq1 +mq2 cosα = 0

↓

q1 = −mq2 cosα

M +m(1)

Likewise, the EL equation for q2 is:

∂L

∂q2=

d

dt

(∂L

∂q2

)↓

mg sinα =d

dtm(q2 + q1 cosα)

↓

mg sinα = m(q2 + q1 cosα)

↓

q2 = g sinα− q1 cosα (2)

Equation 2 can be eliminated by substituting in equation 1 for q1:

q2 = g sinα+mq2 cos2 α

M +m

↓

q2 −mq2 cos2 α

M +m= g sinα

↓

q2 =g sinα

1− m cos2 αM+m

(6.2)

Since q2 is the acceleration of the block down the slope, we know from elementary kinematics that the time it

takes to reach the bottom - thus travelling a distance l - is given by:

t =

√2l

q2

37

Chapter 7

Hamiltonian mechanics

7.1 Time variance of the Lagrangian

Thus far, it has been shown that translational invariance of the Lagrangian in a coordinate qi results in the

conservation of its conjugate momentum∂L

∂q= pi. Now consider how the Lagrangian L(q1, . . . qn, q1, . . . qn, t)

varies with time. From the chain rule:

dL

dt=∑i

∂L

∂qiqi +

∑i

∂L

∂qiqi +

∂L

∂t(7.1)

The first term on the right can be simplified from the El equation:

∂L

∂qi=

d

dt

(∂L

∂qi

)=

d

dtpi = pi

The second term on the right of equation 7.1 is just the generalized momentum pi The time derivative of the

Lagrangian is then:dL

dt=∑i

(piqi + piqi) +∂L

∂t

Using the product rule in reverse for the first term on the right:

dL

dt=∑i

d

dt(piqi) +

∂L

∂t

When the Lagrangian is time symmetric, there is no explicit dependence on t, and the last term on the right

cancels out, leaving:

dL

dt=∑i

d

dt(piqi)

↓∑i

d

dt(piqi)−

dL

dt= 0

↓

d

dt

(∑i

(piqi)− L

)= 0

Since the time derivative of the expression is 0, the quantity inside the brackets, which we will call E is

conserved. Thus it follows that:

38


Theorem 3. If the Lagrangian L is explicitly time invariant, the quantity

E =∑i

(piqi)− L

is conserved.

7.2 Conservation of Energy

We will now analyse how the quantity E and the energy of the system are related. Recall that a set of generalized

coordinates is natural if the relation between the coordinates and the underlying Cartesian coordinates is

independent of time[17]. That is:

r = r(q1 . . . qn)

Most holonomic systems will have this property, unless the system is being acted upon by an external force or

torque that is driving the motion, in which case the generalized coordinates will vary with time. For systems

characterized by natural generalized coordinates, it turns out that the quantity E is just the total energy of

the system. Let us look at an example.

Consider a particle moving in the x-y plane, connected to the origin by a spring of force constant k. In Cartesian

coordinates, T = 12m(x2 + y2) and V = 1

2k(x2 + y2) and L = T − V = 12m(x2 + y2)− 1

2k(x2 + y2)

Since p =∂L

∂xi:

E =∑

pq − L =∂L

∂xx+

∂L

∂yy − 1

2m(x2 + y2) +

1

2k(x2 + y2)

= mx2 +my2 − 1

2mx2 − 1

2my2 +

1

2k(x2 + y2)

=1

2m(x2 + y2) +

1

2k(x2 + y2)

= T + V

Now in Polar coordinates, a choice of generalized coordinates that we know is natural, T = 12m(r2 + r2θ2) and

V = 12kr

2 and L = T − V = 12m(r2 + r2θ2)− 1

2kr2

39


Therefore:

E =∑

pq − L =∂L

∂rr +

∂L

∂θθ − 1

2m(r2 + r2θ2) +

1

2kr2

= mr2 +mr2θ2 − 1

2mr2 − 1

2mr2θ2 +

1

2kr2

=1

2m(r2 + r2θ2) +

1

2kr2

= T + V

We have shown that E is the total energy for this system with a choice of natural coordinates. Now it must be

proved that this is true for all natural systems. Let the position vectors of the system be:

ra = ra(q1 . . . qn)

From the chain rule, it follows that:

ra =

n∑i

∂ra∂qi

qi

The Kinetic energy T = 12m∑a r2

a then simplifies to:

T =1

2m∑a

∑i

(∂ra∂qi· ∂raqi

)q2i =

1

2

∑i

Fi q2i (7.2)

In general, for any natural system will take the form T = 12Fiq

2i where F is a function of the generalized

coordinates such that

F (q1 . . . qn) = m∑a

∂ra∂qi· ∂ra∂qi

The kinetic energy will always be a homogeneous quadratic function of q with some additional term F that

is a function of only q, resulting from the interdependence of the Cartesian positions to the generalized

coordinates[7].

Likewise, the generalized momenta pi =∂L

∂qi. Assuming that the potential V is conservative and is independent

of generalized velocity q, then∂L

∂qi=∂T

∂qi

Therefore:

pi =∂T

∂qi=

∂

∂qi

(1

2Fi q

2i

)=∑i

Fi qi (7.3)

Substituting the results from equations 7.2 and 7.3 into E =∑i

piqi − L :

E =∑i

(∑i

Fi qi

)qi −

1

2

∑i

Fi q2i + V

=∑i

Fi q2i −

1

2

∑i

Fi q2i + V

= 2T − T + V

= T + V (7.4)

40


We have used the fact that as long as the coordinate transformation is natural, the kinetic energy T will be

quadratic in q which gives 2T when differentiated to find generalized momenta. When L is subtracted from 2T ,

we get E = T + V . However, if the coordinates had an explicit time dependence, an additional term involving

the partial time derivative of q would arise and the above result would no longer be true.

41


7.3 The Legendre Transform

The quantity E =∑i

piq − L has special properties. This operation on L, p and q is actually called the

Legendre Transform. From Lagrange’s equation[1]

p =∂L

∂q

p =∂L

∂q

How can we make these two equations more symmetrical? Notice that in the first equation we have a time

derivative in denominator of the partial derivative, while in the second equation, the time derivative is on

the left hand side with no partial derivative encompassing it. Being able to switch the q and the p in the

first equation would result in a more symmetrical set of equations[10]. This sort of symmetry has a number

of benefits, namely that the variables p and q would be almost interchangeable[19]. However, such a switch

cannot be made out of nowhere. The Legendre Transform has the effect of making such a switch, in effect

substituting q with p. Thus, being able to write our equation of motion as some function of q, p and t would

give us ’independent’ variables that are truly independent unlike the Lagrangian’s arguments where q is the

time derivative of q. However, E in its current form is still a function of q so we must first eliminate q and

rewrite it in terms of our new independent variables q and p.

E =∑i

piqi − L

However,

pi =∂L

∂qi

Thus it should be possible to solve this equation for qi = qi(qi, pi).

The Legendre transform on the Lagrangian L then takes the form:

H =∑i

piqi(q, p)− L(qi, qi(q, p), t) (7.5)

Where E is rewritten as H and is called the Hamiltonian of the system when it is expressed in terms of q

and p which are called canonical coordinates.

7.4 Hamilton’s equations

Now let us see if we can extract any information on the dynamics of the system from the Hamiltonian. To do so,

consider how the Hamiltonian varies with q and p respectively[17]. For now, we will assume only 1 dimension

with a single q and p. Differentiating equation 7.5 with respect to q:

∂H

∂q= p

∂q

∂q− ∂L

∂q− ∂L

∂q· ∂q∂q

42


However, since∂L

∂q= p , the first and third terms on the right hand side are identical, thus simplifying to:

∂H

∂q= −∂L

∂q

From Lagrange’s equation,∂L

∂q=

d

dt

(∂L

∂q

)= p

Substituting this in and bringing the minus sign to the other side, it follows that:

p = −∂H∂q

(7.6)

Likewise, consider how the Hamiltonian varies with p by differentiating equation 7.5:

∂H

∂p= q + p

∂q

∂q− ∂L

∂q· ∂q∂p

Once more, recognizing that ∂L∂q = p , the last two terms on the write cancel each other out. Therefore:

q =∂H

∂p(7.7)

Equations 7.6 and 7.7 together are called Hamilton’s equations of motion and are the defining equations of

Hamiltonian mechanics. In a multi dimensional system of n generalized coordinates and conjugate momenta:

pi = −∂H∂qi

qi =∂H

∂pi

Where i goes from 1 to n.

The goal of a more symmetric set of equations has been achieved, at the cost of L(q, q, t) being replaced by a

new function H(q, p, t) and the nuance of the minus sign[19]. One may ask why we went through the trouble

of replacing q with p. The variables q and p are now strictly independent. But isn’t p = mq ? How does

the addition of a scale factor m change the nature of the variable? While it is true in cartesian coordinates

that the generalized momentum is just the the generalized velocity scaled by a factor of m, this is not true in

general. The generalized momenta is defined as p =∂L

∂q. In other choices of generalized coordinates, p does

not follow the same rules that linear momentum follows. For example, in rotational motion, the generalized

momentum is NOT equal to mω. In fact, it is equal to Iω, where I is the moment of inertia, which is clearly

distinct from linear momentum. Also note that since we proved that E is conserved for natural systems in the

previous section, it follows that the Hamiltonian is a constant of the motion for natural systems[17].

43


The process for using the Hamiltonian to solve for the dynamics of the system is slightly more cumbersome

than the Lagrangian formalism, but is still straightforward nonetheless. In summation, it involves[2]

• Find the Lagrangian of the system

• Use the Legendre transform to find the Hamiltonian

• Write the Hamiltonian in terms of the generalized momenta pi by solving pi =∂L

∂qiand eliminating q

• Solve Hamilton’s equations for the time evolution of the system

However, we proved that the quantity E - and therefore H - is equivalent to the total energy of the system for

holonomic systems with natural coordinates. In such situations, we can skip most of the steps and write down

H = T + V .

7.4.1 Hamiltonian analysis of Atwood’s machine

This simple example will illustrate how the Hamiltonian formalism is used in mechanics. Consider the typical

Atwood’s machine with masses m and M where m < M . It is assumed that the string of length l is massless

and that the pulley is frictionless. This implies the following constraint : q1 +q2 +C = l where C is the constant

length of the string that is wrapped around the top of the pulley (2πr). Therefore:

q2 = l − C − q1

l − C is still a constant term and so can be rewritten as C again1:

q2 = −q1 + C

Differentiating the above:

q2 = −q1

Thus the kinetic energy T can be expressed as:

T =1

2mq2

1 +1

2Mq2

2

=1

2mq2

1 +1

2M (− q1)2

=1

2mq2

1 +1

2Mq2

1

=1

2(m+M)q2

1

1This is not the same constant, but is still a constant term nonetheless.

44


Likewise, the potential energy V is given by:

U = −mgq2 −Mgq1

= −(mg(−q1 + C))−Mgq1

= mgq1 − C −Mgq1

The constant C can be dropped since it will be eliminated when differentiating the Lagrangian or Hamiltonian

= mgq1 −Mgq1

= (m−M)gq1

The Lagrangian of the system is:

L = T − V =1

2(m+M)q2

1 − (m−M)gq1

The generalized momentum is then given by:

p1 =∂L

∂q1= (m+M)q1

Solving for q1:

q1 =p1

(m+M)

We can now apply the Legendre transform to the system:

H = p1q1 − L

= p1q1 −1

2(m+M)q2

1 − (m+M)gq1

= p1 ·p1

(m+M)− 1

2(m+M) · p2

1

(m+M)2+ (m−M)gq1

=p2

1

(m+M)− 1

2

p21

(m+M)+ (m−M)gq1

=1

2

p21

(m+M)+ (m−M)gq1

Alternatively, we could’ve used H = T +V (since this is a holonomic, natural system) to reach the same result.

Now we can apply Hamilton’s equations.

p = −∂H∂q1

=∂

∂q1

(1

2

p21

(m+M)+ (m−M)gq1

)= (m−M)g (1)

q1 =∂H

∂p1

=∂

∂p1

(1

2

p21

(m+M)+ (m−M)gq1

)=

p1

(m+M)(2)

45


It follows from equation 2 that p1 = q1(m + M) and thus p1 = q1(m + M). This can be used to eliminate p1

from equation 1 which leaves:

q1 =m−Mm+M

g

which is the familiar equation for the Atwood’s machine from Newtonian mechanics.

46

Appendix: Essence of the Hamiltonian:

Abstract visualizations

Hamilton’s equations are entirely equivalent to Lagrange’s equation and consequently, Newton’s second law.

At first glance, using Hamilton’s method seems like it adds unnecessary complications to Lagrange’s method.

Lagrange’s method already eliminates the need to account for constraints, allows interchangeability of generalized

coordinates and considers the evolution of the system as a whole. Hamilton’s equations are a coupled system

of 2n first order differential equations in contrast to the Lagrangian formalism that entails n second order

differential equations[19]. Although the reduction in order does not simplify computation, it has profound

implications on the underlying geometry and mathematical structure of Hamiltonian mechanics. This geometry

is what makes the Hamiltonian formalism so useful in other domains such as statistical and quantum mechanics

which are interested in the average behaviour of a system over time. Over this chapter I will try and showcase

the elegance and utility of the Hamiltonian method without a deep digression into differential geometry.

7.5 Ignorable coordinates

In the Lagrangian method, a cyclic coordinate implied that its corresponding generalized momentum was

conserved. The Hamiltonian approach extends this benefit by effectively reducing the dimensionality of the

problem. For example, a system with two degrees of freedom may have a Lagrangian L(q1, q1, q2), i.e L is

independent of q2. Although the generalized momentum is constant, it is still possible for the Lagrangian

to depend on q since it is the generalized momentum p that is conserved and generalized momenta are not

necessarily the same as linear momenta. However, in the Hamiltonian formulation, every position coordinate is

paired with its conjugate momentum. The symmetry of Hamilton’s equations imply that if H is independent of

a coordinate q, both that coordinate and its conjugate momentum can be ignored[15]. To prove this, consider

once more a system of two degrees of freedom with coordinates and momenta (q1, q2, p1, p2). Now suppose

H is independent of q2. From Hamilton’s equation, p2 = −∂H∂q2

it follows that p2 is conserved. Therefore,

the pair q2, p2 becomes ignorable and the system is effectively one dimension lower than it was. Thus, an

n dimensional system becomes an n − 1 dimensional system for every independent coordinate qi, from an

analytical perspective.

7.6 Phase Space

We have already explore the configuration space of a system as the n dimensional vector space that represents the

position and orientation of a system. In the Lagrangian formulation, another level of dimensionality was added

in terms of the generalized velocities of the system. This 2n dimensional space that includes n dimensions for

the degrees of freedom and n dimensions for the generalized velocities is called the state space. The Lagrangian

determines a path through the state space that shows how the system evolves and changes its state space

47


coordinates (q1 . . . qn, q1 . . . qn) with time. That is, given a set of initial conditions (q, q), the Lagrange equation

defines a unique path in state space.

In the Hamiltonian formulation, the Lagrangian is replaced with the Hamiltonian operator. It defines the

time evolution of a system over the coordinates (q1 . . . qn, p . . . pn) in a 2n dimensional phase space[19]. This

change to the momentum domain has profound geometrical implications. First, let’s visualize the phase space.

Since I am constrained to 2 dimensional graphics on paper, consider a system with one degree of freedom, say,

a spring-block oscillator with force constant k. Its phase space would be two dimensional with a dimension

for a generalized coordinate x that represents the distance from the equilibrium point of the oscillation and a

conjugate momentum p.

The potential energy of the system V is1

2kx2. Since the angular frequency ω of a spring block oscillator is

given by√

km it follows that V = 1

2mω2x2. Consequently, the total energy of the system is some constant E

since energy is conserved.

Thus the relation between p and x is H = T +V =p2

2m+

1

2mω2x2 = E which can be re-written as

p2

a+x2

b= E

where a = 2m and b =2

mω2. This is the equation of an ellipse and can be used to plot a phase space diagram.

Figure 7.1: Phase portrait of a one dimensional harmonic oscillator

Consider the case where the block is released from the amplitude A. The displacement of the block is positive

but reduces to 0 due to the restoring force of the spring, eventually reaching maximum kinetic energy, and thus

maximum momentum at x = 0. However, since the momentum directs the block away from A, it is negative

and so it is really a point of minimum momentum (but still a maximum in absolute terms).

48


The second stage is when the block crosses the point of equilibrium and the displacement and momentum are

both negative until it reaches −A. In the third stage, the block accelerates back towards equilibrium and so

displacement is negative while momentum is positive until the system reaches maximum positive momentum

at x = 0 yet again. Finally, both displacement and momentum are positive as the block returns to A. This is

one oscillation. If the arrows were drawn anticlockwise, it is easy to see that the portrait would not represent

typical behaviour of harmonic motion. Note that higher total energy E would correspond to larger concentric

ellipses while smaller values of E correspond to smaller concentric ellipses. Also note that no two phase

space trajectories can intersect since each trajectory is unique to a particular set of initial conditions. If

two trajectories intersected, it implies that the system could evolve in two different ways from the point of

intersection, which violates the law of motion.

We can define the phase space vector z as a vector with 2n components z(q1 . . . qn, p1 . . . pn) that defines a

unique point in phase space[3]. Think of this as the position vector of a point in 2n dimensional phase space.

Then, the Hamiltonian governs the time evolution of the phase space vector. Since qi =∂H

∂piand pi = −∂H

∂qithe Hamiltonian tells us how each phase space coordinate changes with time. Thus, at any point in time t, the

Hamiltonian defines the phase space vector at t+∆t. For this reason, the Hamiltonian is called the infinitesimal

generator of time translations in phase space. The rate of change of the phase space vector is then Hamilton’s

equations applied to each coordinate[19]

z = (qi . . . qn, pi . . . pn) =

(∂H

∂pi. . .

∂H

∂pn,−∂H

∂qi· · · − ∂H

∂qn

)(7.9)

∂H

∂pi= fi(q1 . . . qn, p1 . . . pn) and −∂H

∂qi= gi(q1 . . . qn, p1 . . . pn) where f and g are some functions of the 2n

generalized coordinates. Now we can define a vector P who’s components are the 2n functions such that

P = (f1 . . . fn, g1 . . . gn) Since f and g are functions of q and p and z is also a function of q and p, we can write:

z = P (z)

This implies that the rate of change of the phase space vector is a function of itself[17]. This type of differential

equation is well understood and easy to solve, making dynamical analysis is state space far easier. The significant

benefit of using the Hamiltonian formulation is that the 2n functions f ad g are all determined by a single scalar

function H. Particularly, it almost looks like z = ∇H but not quite, since the p′s and q′s must be interchanged

for the expression to look like the Gradient, in addition to getting rid of the minus sign.

49


7.7 Geometrical interpretation of Hamiltonian Dynamics

This analysis will be restricted to 1 dimension. It is easy to extend the concept to any number of phase space

coordinates, but with obvious geometric limitations to visualizations.

Figure 7.2: Hamiltonian Vector Field

Hamilton’s equations define a Hamiltonian vector field on the phase space[3]. A vector field is a map that

associates a vector to every coordinate (q, p). In this case we define the vector field−→S (q, p), where every point

has an associated vector (−→Sq,−→Sp) where Sq = q =

∂H

∂pand Sp = p = −∂H

∂q

The phase space trajectory is then an integral curve of the vector field or a field line, much like the solutions to

a differential equation are represented as an integral curve of a slope field. The vectors are tangent to the phase

space trajectory at every point (in red), which shows how the phase space coordinates will evolve with time[3].

Thus, given a state (q(t), p(t)), the future state of the system after a time dt is given by q(t+dt) = q(t)+Sq ·dtand p(t+ dt) = p(t) + Sp · dt

On the other hand, the Hamiltonian itself H(q, p) is a scalar field, associating a value to every coordinate.

Consider the lines of equal values of H or the contour lines of the scalar field.

Figure 7.3: Contour lines of the Hamiltonian

On each purple curve, the value of the Hamiltonian is constant (k0, k1, k2).

50


Now, suppose we took the Gradient of this scalar field. ∇H would form a vector field with components(∂H

∂q,∂H

∂p

)that points in the direction of optimally increasing2 H. If this is true, all the gradient vectors

would be perpendicular to the contour lines of H. Since the contour lines show constant values of H, no gradient

vector can have a component that is tangent to it, else the vectors would not point in the direction of greatest

increase. Consequently, each gradient vector points towards a contour line of a higher constant value of H.

Figure 7.4: ∇H vector field

Remarkably, with a simple rotation of 90◦ clockwise, the gradient vector field and the Hamiltonian vector field

are exactly the same[19]!

Figure 7.5: z = S(q, p) = rot 90◦∇H(q, p)

It isn’t too hard to imagine that this rotation could be accomplished with multiplication by some transformation

matrix J. Without going into the specifics of what J would look like, it is now possible to write:

z(q, p) = J∇H(q, p)

This is a remarkable result, since the entire evolution of all components of the phase space vector has been

condensed into a gradient field. Nevertheless, it is not quite the gradient, and this form of the gradient

supplemented with the J matrix is called a symplectic gradient [19]. It is a result of the beautiful symplectic

geometry of the phase space, a subject far beyond the scope of this paper that delves into differential geometry.

2Review Chapter 1

51


7.8 Canonical transformations

Just like the Lagrangian formalism, the Hamiltonian formalism is invariant to coordinate transformations. That

is, we can replace a set of generalized coordinates q1 . . . qn with a new set of generalized coordinates (T1 . . . Tn)

as long as every Ti is a definitive function of the original generalized coordinates:

Ti = Ti(q1 . . . qn)

for all i.

The same holds true for the Hamiltonian formalism, which takes it a step further by allowing the same sort of

transformation from a set of conjugate momenta (p1 . . . pn) to another (L1 . . . Ln) such that:

Li = Li(p1 . . . pn)

This is possible since p is an independent variable just as q is in the Hamiltonian formalism.

However, what truly makes the Hamiltonian formalism powerful is that it is invariant under transformations

that depend on both p and q simultaneously as if they were interchangeable:

Ti = Ti(q1 . . . qn, p1 . . . pn) Li = Li(q1 . . . qn, p1 . . . pn)

Explaining why this is true with rigorous mathematics is beyond the scope of this paper, but intuitively,

the symplectic geometry of the phase space renders q and p symmetric in some way, allowing a degree of

interchangeability between them. The Hamiltonian is then invariant under these canonical transformations

of both variables[1]. This result is extremely useful in statistical mechanics, and such canonical transformations

transform the phase space vector itself, while preserving information about the time evolution of the system.

7.9 Poisson Brackets

What would a constant of motion of a system look like in general? Suppose we have some function of phase

space variables F (q, p, t). F is a constant of motion ifdF

dt= 0. Expanding the left hand side using the chain

rule:

dF (q, p, t)

dt=

n∑i=0

{∂F

∂qiqi +

∂F

∂pipi

}+∂F

∂t

=

n∑i

{∂F

∂qi· ∂H∂pi− ∂F

∂pi· ∂H∂qi

}+∂F

∂t

Thus, F is a constant of the motion if

n∑i

{∂F

∂qi· ∂H∂pi− ∂F

∂pi· ∂H∂qi

}+∂F

∂t= 0

The identity in the curly braces can be abstracted in general to any two phase space functions A and B. We

define the poisson bracket of A(q, p) with B(q, p) as:

{A,B} =

n∑i

{∂A

∂qi· ∂B∂pi− ∂A

∂pi· ∂B∂qi

}

52


It follows that a phase space function is a constant of the motion if its Poisson bracket with the Hamiltonian

is zero (assuming time independence). It is said that F commutes with H. Likewise, any two phase space

functions commute with each other if their Poisson bracket is identically zero[19]. The Poisson bracket can also

be used to prove the earlier result that the Hamiltonian is a constant of the motion if it is independent of time.

If we take {H,H} it is easy to see that this is equal to zero.

The Poisson bracket is a profound identity because it is responsible for the natural transition between classical

and quantum mechanics. It is essentially a bilinear operator on two phase space functions and with a minimal

exposure to introductory Quantum Mechanics, this will look familiar. It is said that Dirac had an epiphany in

which he saw the connection between the Poisson bracket and Quantum Mechanics during an evening walk: the

Poisson bracket between two phase space functions plays the same role as the commutator between two matrix

operators in quantum mechanics. The bracket has a mathematical structure called a Lie Algebra [19] - a

set of identities and properties that make the Poisson Bracket extremely useful in several areas of Physics and

although, the specifics of how it is used are too complicated, it is an integral and beautiful result of Hamiltonian

mechanics that can atleast be appreciated in abstraction.

7.10 Summary

Why is the Hamiltonian formulation useful (or more useful than the Lagrangian formalism)?:

• The Hamiltonian is a constant of the motion (conserved) for time independent holonomic systems.

• The Hamiltonian is simply the total energy of the system T+V for (time independent) natural coordinate

transformations.

• A pair of coordinates (q,p) are completely ignorable if the Hamiltonian is symmetric with respect to q

• The phase space is endowed with a symplectic structure that allows the Hamiltonian and the time evolution

of the system to be written as a symplectic Gradient

• The Hamiltonian is invariant under canonical transformations of q and p due to an underlying symmetry

and interchangeability between them.

• The Poisson Bracket and its Lie Algebra have a deep mathematical structure that extends to several other

areas of modern Physics. A phase space function is a constant of the motion if the Poisson bracket (and

the partial time derivative if there is explicit time dependence) is equal to zero.

∗∗∗∗∗∗∗∗∗∗∗∗

53

References

[1] Vladimir Igorevich Arnold. Mathematical Methods of Classical Mechanics. Springer, 1989.

[2] Mary L. Boas. Mathematical Methods in the Physical Sciences. Wiley India Pvt. Limited, 2006.

[3] Gabriel Carcassi. Understanding the hamiltonian. YouTube Video- Gabriel Carcassi, June 2013.

[4] Lamar University Math Department. Gradient. Article on Website- Paul’s online math notes.

[5] Lamar University Math Department. Partial derivative. Article on Website- Paul’s online math notes.

[6] Richard P. Feynman. Ch 19: The principle of least action. In The Feynman Lectures on Physics Vol II.

[7] Herbert Goldstein et al. Classical Mechanics. Pearson, 2013.

[8] L.D Landau and E.M Lifshitz. Mechanics. Butterworth-Heinemann, 1976.

[9] Simon J.A. Malham. An introduction to lagrangian and hamiltonian mechanics. Heriot Watt School of

Mathematics and Computer Science.

[10] David Morin. Introduction to Classical Mechanics. Cambridge University Press, 2008.

[11] PhysicsHelps. Euler-lagrange equation. YouTube Video-PhysicsHelps, April 2013.

[12] PhysicsHelps. The principle of least action. YouTube-PhysicsHelps, April 2013.

[13] Nikolaj S. Piskunov. Differential And Integral Calculus. Peace Publishers, Moscow, 1974.

[14] ShareLaTeX. All latex reference materials. Website. https://www.sharelatex.com.

[15] Leonard Susskind. Legendre transform. Website: http://www.lecture-notes.co.uk/susskind/classical-mechanics,.

[16] Leonard Susskind. Lecture 3: Theoretical foundations of modern physics. YouTube Video - Stanford,

October 2011.

[17] John R. Taylor. Classical Mechanics. University Science Books, 2005.

[18] J. Kim Vandiver. Introduction to lagrange with examples. Video: MIT OpenCourseWare. MIT 2.003SC

Engineering Dynamics, Fall 2011.

[19] V.Balakrishnan. Lecture series on classical physics: Hamiltonian dynamics. YouTube Video- nptelhrd-IIT

Madras, May 2009.

54

by varun menon · 2018. 1. 26. · [email protected], [email protected], united kingdom...

Documents