Lagrangian and Hamiltonian Mechanics Notes

7/28/2019 Lagrangian and Hamiltonian Mechanics Notes

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Level 2 Classical Mechanics 2012

V. Eke (Dept. of Physics, Durham University)

Contents

1 Synopsis 1

2 Motivation 4

3 Generalised Description of Mechanical Systems 5

4 The Lagrangian 7

5 Variational Calculus 11

WORKSHOP 1: APPLICATIONS OF THE LAGRANGIAN APPROACH 14

6 Linear Oscillators 16

7 Driven Oscillators 18

8 Coupled Small Oscillations 20

9 Normal Modes 23

10 Central Forces 25

11 Gravitational Attraction 28

WORKSHOP 2: CENTRAL FORCES/R UTHERFORD SCATTERING 31

12 Noethers Theorem and Hamiltonian Mechanics 33

13 Canonical Transformations and Poisson Brackets 36

14 Rotating Reference Frames 40

15 Inertial Forces on Earth 43

WORKSHOP 3: CORIOLIS FORCE 46

16 Rotational Inertia, Angular Momentum and Kinetic Energy 48

17 The Parallel and Principal Axis Theorems 52

18 Rigid Body Dynamics and Stability 55

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A Second-Order Ordinary Linear Differential Equations with Constant Coefficients I

A.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I

A.2 Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I

A.2.1 Auxiliary Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I

A.2.2 General Solution (Roots not Repeated) . . . . . . . . . . . . . . . . . . I

A.3 Possible Complications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II

A.3.1 Repeated Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II

A.3.2 Additional Constant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . II

A.4 Representing Position in a Plane by a Single Complex Number . . . . . . . . II

B Taylor Series III

B.1 Taylors Theorem of the Mean . . . . . . . . . . . . . . . . . . . . . . . . . . . . III

B.2 Taylor Series in One Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . III

B.3 Taylor Series in Two Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . IV

B.4 Taylor Series in Three Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . IV

C Determinants, Eigenvalues and Eigenvectors VC.1 Symmetric Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V

C.2 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . V

C.3 The Conventional Eigenvalue Problem . . . . . . . . . . . . . . . . . . . . . . . V

C.4 Generalized Eigenvalue Problem . . . . . . . . . . . . . . . . . . . . . . . . . . VII

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1 Synopsis

Description of Mechanical Systems

SEE ALSO HAND AND FINCH, 1.11.5

Dynamical variables. Degrees of freedom. Constraints. Generalized coordinates.

The Lagrangian Approach


DAlemberts principle. Generalised equations of motion. The Lagrangian. Ignorable

coordinates.

Variational Calculus

SEE ALSO HAND AND FINCH, 2.12.8Euler equation. Hamiltons principle. Lagrange multipliers and constraints.

Linear Oscillators


Equilibrium. Lagrangian near to static equilibrium. Simple harmonic oscillator. Damped

oscillator and the quality factor, Q.

Driven Oscillators


Equation of motion. Dirac function. Impulsive driving force. Greens function, G. Gen-

eral driving force. Resonance.

Coupled Small Oscillations


Two Coupled Pendulums. Problem solving recipe. Matrix form of L.

Normal Modes


Normal coordinates. Mode orthogonality. Linear triatomic molecule.

Central Forces


Two interacting bodies. Translational symmetry and conservation of momentum. Ignor-able centre of mass motion. Rotational invariance and conservation of angular momen-

tum. Keplers 2nd law. Equivalent 1D problem.

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Gravitational Attraction

SEE ALSO HAND AND FINCH, 4.1, 4.54.6

Solving 1D systems by quadrature. Gravitational potential and the u = 1/r transformation.Keplers 1st and 3rd laws.

Noethers Theorem and Hamiltonian Mechanics


Invariance of L under continuous transformations. Noethers theorem. Legendre transfor-

mations. The Hamiltonian. Hamiltons equations of motion.

Canonical Transformations and Poisson Brackets


Canonical Transformations. Equivalence of Lagrangians. Generating functions. The trans-formed Hamiltonian. Poisson brackets.

Rotating Reference Frames


Accelerating reference frames. Rotated reference frames in 2D. Small rotations in 3D.

Velocity and acceleration. Inertial forces in a rotating frame.

Inertial Forces on EarthSEE ALSO HAND AND FINCH, 7.77.10

Procedure for determining a local coordinate system. Inertial forces. Centrifugal force.

Effective gravity on Earths surface. Coriolis Force. Euler force.

Rotational Inertia, Angular Momentum and Kinetic Energy


Displacement of a rigid body. Moment of inertia. Angular momentum and rotational

kinetic energy. The inertia tensor. Generalized angular momentum and rotational kineticenergy.

The Parallel and Principal Axis Theorems

SEE ALSO HAND AND FINCH, 8.2,8.7

Displaced axis theorem. Parallel axis theorem. Symmetry of the inertia tensor. Principal

axis transformation. Principal axis theorem.

Rigid Body Dynamics and StabilitySEE ALSO HAND AND FINCH, 8.48.5

Euler angles. Eulers equations of motion. Torque-free motion.

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Books

Hand & Finch, Analytical Mechanics, CUP. Appropriate level and style.

Thornton & Marion, Classical Dynamics of Particles and Systems, Thomson. Second half of

book pretty much covers this course.

Kibble & Berkshire, Classical Mechanics, Imperial College Press. Covers more than needed

for this course. Sometimes terse treatment. Lots of questions and answers.Brizard, An Introduction to Lagrangian Mechanics, World Scientific. Quite a mathemati-

cal approach. Covers almost all of the course, and not much else.

Goldstein, Classical Mechanics, Addison-Wesley. More formal treatment. Goes far be-

yond this course.

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2 Motivation

Newtons formulation

Formally, the system comprises N point particles. For particle i,

mi ri = Fi(ri, rj, ..., ri, rj, ..., t).

The scheme is conceptually simple if Fi is explicitly known, e.g. gravitation, Lorentz forces.

Microscopic systems are simple in this sense (provided that classical physics still applies

at the relevant scales). Macroscopic systems include forces due to constraints. All New-

ton says about such forces is that

Fi due to j = Fj due to i.

Features of this scheme1. Valid in an inertial (non-accelerating) frame for objects moving with speeds much

less than that of light and sizes much larger than those of atoms.

2. Conceptually simple, with direct solutions in many elementary cases.

3. Totally deterministic in principle, like any formulation of classical mechanics (but

might need to know initial conditions with infinite precision).

4. Constraints need to be handled explicitly.

5. Conservation of energy, momentum and angular momentum are derived, and thusnot fundamentalwithin this approach. For electromagnetism, a common approach

is to take Maxwells Equations and the Lorentz Force as fundamental. Then conser-

vation of energy follows from Poyntings theorem and a page of algebra. Surely

energy and momentum are more central than this?

6. Classical physics is only approximate and applies to the limiting case of large bodies

and low velocities. While force is central to the Newtonian approach, it doesnt

even appear in Schrdingers equation.

The aim of the present course is to explore the alternative formulations of classical me-chanics of Lagrange and Hamilton. These are important in the development of modern

physics (e.g. quantum mechanics, chaotic behaviour, ...). Each formulation is a new set

of first principles; provided they predict the correct behaviour, they need not be derived

from any other set of first principles.

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3 Generalised Description of Mechanical Systems


Dynamical variables

These are any set of variables that describe completely the configuration (positions of

parts) of a mechanical system. In 3D space, possible sets of dynamical variables include

(x,y,z), (r, ,z) and (r, , ).

They can change under the action of forces the equation of motion of a system

specifies the dynamical variables as functions of time.

Newtons laws imply that these functions are found by solving second order differen-

tial equations hence the initial positions/angles and velocities/angular velocities (or

equivalent information) must be known.

Degrees of Freedom (DoF)

The motion of a point mass r(t) is described by 3 independent dynamical variables, i.e.,the point mass has N = 3 DoF.

A system of M point masses has N = 3M DoF, but the existence of j independentconstraints reduces this number to N = 3M j DoF.

M=3j=3

M=3j=2

(flexible joint)

M=2j=1

N=5 DoF N=6 DoFN=7 DoF

Any rigid body containing more than 2 point masses connected by rigid rods has 6

DoF (or fewer, if there are additional constraints).

Types of constraintsIf constraints on a system of M point masses can be expressed in the form

f(r1, r2, . . . , rM, t) = 0,

then the constraints are called holonomic.

Rheonomic (running law) constraints have an explicit time dependence (e.g. bead

on a rotating wire). Scleronomic (rigid law) constraints do not, i.e., f(r1, r2, . . . , rM) = 0(e.g. pendulum).

Nonholonomic constraints involve either differential equations or inequalities (rather

than algebraic equations), e.g.: velocities (coin rolling without slipping on a 2D surface),

inequalities (particle sliding down the outside of a bowling ball), most velocity dependent

forces (friction).

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Generalised coordinates qk

The existence of j constraints means that the coordinates of e.g. M point masses are no

longer independent.

We require only as many coordinates qk as there are DoF (N = 3M j).If the constraints are holonomic then we can express the position of the ith part of a

system as

ri = ri(q1, . . . , qk, . . . , qN, t).

There may be an explicit time dependenceparticularly if the constraints are rheonomic.

The generalised velocities are qk = dqk/dt.We regard qk and qk as independent variables (this is very important, and means for

example that qk/qk = qk/qk = 0).By the chain rule, one can derive the following identity, which can be a useful check:

vi = ri = k

riqk

qk + ri

t ri

qk= ri

qkcancelling the dots.

This last step comes from taking the partial derivative with respect to qk.

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4 The Lagrangian


Lagranges formulation of mechanics

We will assume:

1. Holonomic constraints

2. Constraining forces do no work

3. Applied forces are conservative, such that a scalar potential energy function exists.

These assumptions imply: no friction or viscous drag, reaction forces are normal to sur-

faces, strings are inextensible and always tight. Note that the potential function may

change with time.

DAlemberts Principle

Newtons 2nd law for particle i isp

i= Fi.

This implies that

i

(Fi pi) ri = 0,

where ri represents an arbitrary virtual displacement. A virtual displacement is a hypo-

thetical change of coordinates at one instant in time that is compatible with the con-

straints. This force includes both the constraint and applied forces, such that

Fi = F(c)i + F

(a)i .

Given that we have supposed the constraining forces do no work, this gives us DAlemberts

Principle:

i

(F(a)i pi) ri = 0. (1)

We have removed the constraint forces from this equation, but have yet to rewrite the

equation in terms of the generalised coordinates that are independentof one another.

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Generalised Equations of Motion (EoM)

We would like to rewrite DAlemberts Principle (i(F(a)i pi) ri = 0) in terms of the inde-

pendent, generalised coordinates, qk. Recall

vi =

d

dt (ri) = k

riqkqk +

rit . (2)

Taking the partial derivative of this with respect to qk yields

viqk

=riqk

. (3)

By analogy with Eq. (2), we can write

d

dt ri

qj = k2ri

qjqk

qk +2ri

qjt

. (4)

Taking the partial derivative of Eq. (2) with respect to qj gives

viqj

= k

2riqjqk

qk +2riqjt

. (5)

Comparing Eqs. (4) and (5) leads to

d

dt riq

j =viq

j

. (6)

The 1st term in Eq. (1), which represents the virtual work done by the applied force (now without

the (a), as we no longer need to worry about the constraint force) can be written as

W = i

Fi ri = i

Fi k

riqk

qk = k

i

Fi riqk

Generalised forces Fk = Wqk

qk = k

Fkqk.

The 2nd term in Eq. (1), can be written as

i

pi ri =

i

mivi ri = i

k

mi vi riqk

qk.

If we note that

vi riqk

=d

dt

vi

riqk

vi

d

dt

riqk

then we can write

i

pi ri =

i

k

mi ddt vi ri

qk vi d

dt ri

qk qk.

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Using Eqs. (3) and (6), we can write

i

pi ri =

i

k

d

dt

mivi

viqk

mivi

viqk

qk.

Changing the order of summation gives

i

pi ri =

k

d

dt

qk

i

1

2miv

2i

qk

i

1

2miv

2i

qk.

Therefore, DAlemberts Principle can be written as

k

Fk ddt

T

qk

+

T

qk

qk = 0,

where the kinetic energy, T = i12 miv

2i = T(q1, . . . , qN, q1, . . . , qN, t).

The qk are independent, so the generalised equations of motion are

Fk ddtT

qk

+

T

qk= 0. (7)

We are assuming that the applied forces are conservative, so Fi = iV(r1, . . . , ri, . . . , rM).Correspondingly,

Fk = i

iV riqk

= Vqk

(follows from the chain rule).

Substituting this into Eq. (7), and defining the Lagrangian by

L = TV,

gives the Euler-Lagrange equations:

d

dt

L

qk

L

qk= 0.

These are the fundamental equationsof Lagrangian mechanics. Note that L is a function

of all of the qk, qk and t.

Ignorable coordinates

If the time derivative of a coordinate appears in the Lagrangian, but the coordinate itself

does not, then this is an ignorable coordinate. As

L

qk= 0,

the Euler-Lagrange equation implies that

d

dt

L

qk

= 0.

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Thus, the canonically conjugate momentum to the coordinate, defined as

pk =L

qk,

is a constant of the motion.

Example: a free particle in 1D with displacement x, has a Lagrangian L =12 mx

2

. Thus,the coordinate x is ignorable, and p = Lx = mx is constant.

Having constants of motion can greatly simplify finding the equation of motion of a

system. The canonically conjugate momentum plays a central role in the Hamiltonian

approach to quantum mechanics.

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5 Variational Calculus


Calculus of Variations

What path y(x) leads to an extreme value of the functional

I[y] =x2

x1f(y,y, x)dx, (8)

where y = dy/dx and the trajectory goes through points (x1,y1) and (x2,y2)?

Consider a small variation such that the path becomes y(x, = 0) + (x), where (x1) =(x2) = 0. For I[y] to be stationary, we want (dI/d)=0 = 0. Note that y/ = and y

/ =.

dI

d=

x2x1

f

y

y

+

f

yy

dx

=x2

x1

f

y +

f

y

dx

Integrating the second term by parts givesx2x1

f

yd

dxdx =

f

y

x2x1

x2

x1

d

dx

f

y

dx.

The first term is zero because = 0 at the ends of the path. Therefore

dI

d= 0

x2x1

d

dx

f

y

f

y

(x)dx = 0 (x).

Hence we arrive at Eulers equation

d

dx

f

y

f

y= 0.

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Recall the Euler-Lagrange equations

d

dt

L

qk

L

qk= 0.

For the case where f does not explicitly depend upon x, ie f/x = 0, one can use

d f

dx=

f

x+

f

y

dy

dx+

f

ydy

dx

in conjunction with the Euler equation to show that

d

dx

fy f

y

= 0. (9)

This is a useful alternative form of the Euler equation.

Hamiltons Principle of Least Action

The Euler-Lagrange equations are the solution to the calculus of variations problem: for

what route q(t) between points q1

and q2

at t1 and t2 is the integral

S[q(t)] =t2

t1L(q(t), q(t), t)dt

a minimum? Hamiltons Principle states that a mechanical system moves in just such a

way as to minimise this integral, which is also known as the action functional, S[q(t)]. The

argument of a function is a number, and a number is returned the argument of afunctional is a function, and a number is returned.

Hamiltons Principle can be written as S = 0, where the represents a change withrespect to the path between the end points of the integral. The variational derivative of

L with respect to q is writtenL

q L

q d

dt

L

q

.

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Lagrange multipliers

If a mechanical system has an applied constraint and the number of DoF reduces, then this

is usually dealt with by eliminating a dynamical variable and finding the equation of motion

for a smaller set of generalised coordinates. An alternative method involves using Lagrange

multipliers.

Consider a pendulum of length l. The natural generalised coordinate to choose to describe themotion is the angle . Suppose you stubbornly and foolishly insist on using x and y, then you

also have the constraint x2 + y2 = l2. The variation of the action is

S =

L

xx +

L

yy

dt.

Hamiltons Principle implies that S = 0, but the non-independence ofx(t) and y(t) means thatwe cannot simply say that each variational derivative is separately zero. Our constraint can be

written as

G(x,y) = x2

+ y2

= c a constant.Now

G =G

xx +

G

yy = 0

is how the variations in x and y are connected. The variation of the action can then be written as

S =

L

x G

x

x +

L

y G

y

y

dt.

where is an arbitrary function of the independent variable (t in this case) called a Lagrange

multiplier. If we choose such that the coefficient ofx is set to zero, then S = 0 implies thatthe coefficient ofy must also vanish. It is as if we had independent variables x and y.

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Workshop 1: Applications of the Lagrangian Approach

1. Consider a point mass (of mass m) falling in the Earths gravitational field (take the acceler-

ation = g to be constant, i.e., independent of the distance of the point mass from the Earthssurface, and ignore air resistance).

(a) What is the kinetic energy Tof the point mass? What is the potential energy V? Hence,determine the Lagrangian L = TV.

(b) Use the Euler-Lagrange equation to determine the equation of motion for the vertical

motion (z direction) of the point mass.

(c) Solve the equation of motion, and express the solution in terms of the point masss

initial position z(0) and velocity z(0).

2. Consider a point mass (of mass m) attached to a spring (spring constant k), constrained

somehow to move only in the x direction, where x = 0 is taken to be the position of the

point mass when it is at rest. This is just a one-dimensional simple harmonic oscillator(undamped), with characteristic oscillation frequency =

k/m.

(a) What is the kinetic energy Tof the point mass? What is the potential energy V? Hence,

determine the Lagrangian L = TV.(b) Use the Euler-Lagrange equation to determine the equation of motion of the point

mass.

(c) The equation of motion is a linear, homogeneous, second-order differential equation.

Solve this using the trial solution x et.

(d) Find the general solution of the equation of motion, in terms of the point masss initialposition x(0) and velocity x(0).

(e) Two independent real constants A and B can equally well be represented in terms of

two other real constants A and , through A = A sin , and B = A cos . Use this,along with the trigonometric identity sin( + ) = sin cos + cos sin , to showyour solution can also be expressed in the form

x(t) = A sin(t + ),

and express A and in terms of the initial position and velocity.

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ADDITIONAL EXERCISE

3. Consider the situation of a small block (SB) of mass m sliding on a frictionless inclined plane

(IP) of height h, which itself has a mass M and rests on a flat surface, without any friction

between this surface and the inclined plane. Consider a set of cartesian axes such that the

Y axis points up along the side of the initial location of the inclined plane, and the X axis

points along the underside of the inclined plane (we ignore the Z direction). Take (XSB, YSB)and (XIP, YIP) to be the locations of the centre of mass of the SB, and the right-angled cornerof the IP in this coordinate system, respectively.

(a) Because the motion of the SB is constrained to remain on the upper surface of the IP,the value for the dynamical variable YSB can be determined exactly if the values for the

dynamical variables XSB, XIP are known. Derive an equation (a constraint equation)

describing this relationship.

(b) Consider an alternative dynamical variable d, the distance of the SB centre of mass from

the top of the IP. Determine expressions for XSB

, YSB

in terms ofXIP

, d, h and . Taking

the time-derivative, determine equivalent expressions for XSB, YSB.

(c) Trivially, the IP is constrained not to move in the Y direction, and so YIP and YIP can

be neglected. Determine an expression for the total kinetic energy T of the SB and IP

system in terms ofXIP and d. Determine an expression for the total potential energy Vin terms ofXIP, h and d.

(d) Write down the Lagrangian L using your expressions for T and V. From the La-

grangian, derive Lagranges equations of motion for d(t) and XIP(t). Solve the equa-tions of motion, writing the solutions in terms of the initial positions and velocities

d(0), XIP(0), d(0), XIP(0).

(e) Assuming that the SB is initially at rest at the top of the IP, which is also at rest, derive

an expression for the distance the IP has moved once the SB has reached the bottom of

the IP, and hence (assuming = /4 radians) determine how large m must be relativeto M for this distance to be = h/2.

X

Y M

m

frictionless surfaces

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6 Linear Oscillators


Equilibrium

A mechanical system that remains at rest is in equilibrium. This occurs at points in config-

uration space where all generalised forces Fk vanish.For a conservative system, this corresponds to configurations where the potential en-

ergy V(q1, . . . , qN) is stationary, i.e., its first derivatives with respect to the qk vanish.

Lagrangian Near Static Equilibrium (1 DoF)

Carry out a 2nd order Taylor series expansion of some L(q, q) around q = qS [where qS is astationary point, i.e., V/q

|q=qS= 0] and q = qS = 0. Choose qS = 0, then

L Lapprox =L(qS, qS) + q Lq

qS,qS

+ qL

q

qS,qS

+1

2

q2

2L

q2

qS,qS

+ 2qq2L

qq

qS,qS

+ q22L

q2

qS,qS

= A + Bq + Cq + Dq2 + Eqq + Fq2.

Substitute the resulting Lapprox into the Euler-Lagrange equation (n.b., B = 0 by the definition ofequilibrium)

d

dt(

Lapproxq

C + Eq + 2Fq)

Lapproxq

(2Dq + Eq) = 0.

This yields q (D/F)q = 0, which is identical to the EoM for a particle of mass m derived from

L =m

2

q2 +

D

Fq2

. (10)

Simple Harmonic Oscillator (SHO)

The Lagrangian for a simple harmonic oscillator, such as a mass m attached to a spring

with spring constant k, can be written as

L =1

2mq2 1

2kq2.

If D/F < 0, then Eq. (10) is the Lagrangian for a simple harmonic oscillator, with k =m(D/F).

Close to equilibrium one therefore observes simple harmonic motion with character-

istic frequency = D/F. Hence, the motion is described as stable (if D/F > 0, it isunstable).

q + 2q = 0 is a linear, homogeneousdifferential equation, and can be solved to give

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q(t) = q(0) cos(t) +q(0)

sin(t).

Damping Force

A frictional force Fd, acts to suppress motion, i.e., it must act in opposition to motion, andvanish when the motion ceases.

The simplest such force is Fd = q = (m/Q)q (expressed in convenient form for adamped SHO), where Q is the dimensionless quality factor of the oscillator.

Damped Simple Harmonic Oscillator

We insert the SHO Lagrangian L = (m/2)(q2 2q2) into the Euler-Lagrange equation andincorporate the (non-conservative) damping force separately

ddt

Lq

Lq

= Fd.

This corresponds to taking Eq. (7), grouping the conservative forces (in the form of a

potential energy function V) with T to make L, and keeping the non-conservative forces

separate.

The result is a linear, homogeneousdifferential equation, q + (/Q)q + 2q = 0.This is solved by setting q = et and determining the allowed values of from the

resulting auxiliary equation. The solutions differ qualitatively depending on the value of Q.

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7 Driven Oscillators


Oscillator Driven by an External Force

We consider an undamped oscillator, driven by a time-dependent external force F(t)with no spatial dependence.

Whatever supplies the driving force is not considered to be part of the dynamical

system. In particular we do not consider any effect of forces the oscillator exerts on the

source of the external force.

The corresponding Lagrangian is given by

L =mq2

2T

m2q2

2 F(t)q

V

,

which yields a linear, inhomogeneous differential equation:

q + 2q = F(t)/m. (11)

Dirac -Functions

It is instructive to start by considering an impulsive (instantaneously applied and infinitely

intense) force. For this we make use of -functions. Recall that the effect of (t t) isdefined (necessarily under an integral) as

f(t) =

(t t)f(t)dt.

(t t) is the zero-width limit of a function/distribution of t, sharply peaked at t = t,with unit area:

(t t)dt = 1 (dimensionless). Consequently, if t has the dimension of

time, then (t t) must therefore have the dimension of inverse time, i.e., frequency.

Response to a q-Independent Impulsive Force

Consider an impulsive force F(t) = K(t t). Integrating the EoM for the driven oscillatoraround t = t yields

t+t

(q + 2q)dt =t+

tK

m(t t)dt,

q(t + ) q(t ) + 2t+

tq(t)dt

0 as 0, unless q(t)

=K

m.

Hence, there is an instantaneous change in the velocity: q(t+

) = q(t

) + K/m, wheret = lim0 t (this is the definition of an impulsive force).

Note that q(t+) = q(t), i.e., there is no instantaneous change in position when theimpulsive force is applied.

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Time-Evolution Sequence of a SHO with an Impulsive Force

1. t < t system evolves as a SHO, to position q(t), velocity q(t).

2. t = t instantaneous jump in velocity: q(t) = q(t), q(t) = q(t) + K/m.

3. t > t system evolves as a SHO:

q(t) = q(t) cos([t t]) +q(t) + K/m

sin([t t]). (12)

(Causal) Greens Function G

Consider G(t t) = q(t)m/K [equal to G(0) when the impulsive force is applied]. FromEq. (11), the EoM of G is given by G(t t) + 2G(t t) = (t t).

We consider the oscillator to be at rest prior to application of the impulsive force.

Hence, by comparison with Eq. (12),

G(t t) =0, t t 0,G(t t) = 1

sin([t t]), t t 0. (13)

General Driving Force F(t)

By definition of the -function, the equation of motion for the driven oscillator [Eq. (11)]

can be written as

q + 2q = (1/m)

dtF(t)(t t).

We can then eliminate (t t) by substituting in the EoM of the Greens function G:

q + 2q = (1/m)

dtF(t)[G(t t) + 2G(t t)].

Comparing terms, and substituting in the solution to G(t t) [Eq. (13)] it follows that

q(t) =1

m t

F(t)G(t

t)dt =

1

m t

F(t) sin([t

t])dt.

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8 Coupled Small Oscillations


Lagrangian formulation particularly useful for small oscillations of a set of coupled os-

cillators about equilibrium positions. Important in acoustics, molecular spectroscopy, me-

chanical systems and coupled electrical circuits.

We consider small oscillations around a stable equilibrium point of a mechanical sys-

tem. If this motion does not depart too far from a stable equilibrium point, then the system

resembles a set of coupled oscillators.

Two coupled pendulums

Consider a system with two pendulums of length l, connected half way down by a

spring with constant k, and unstretched length equal to the horizontal separation of the

pivot points. There are 2 DoF - use 1 and 2 as generalised coordinates. If the rigid

pendulums have all of their mass, m, contained in the bob, and the spring is massless,

then the kinetic energy is

T =ml2

2

21 +

22

.

Without coupling, and assuming small oscillations,

V =mgl

2(21 +

22).

The spring provides an extra potential energy,

Vcoupling =k

2

l

2

2(2 1)2.

Hence

L =

ml2

2 (

2

1 +

2

2) 1

2 mgl(21 + 22) + k l22

(2 1)2 .

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The 12 term makes life awkward, so transform to centre of mass c = (1 + 2)/2 andrelativer = 2 1 coordinates. Defining 0 =

g/l, and the coupling constant = kl4mg ,

L = ml2

2c 202c

+1

4

2r 20(1 + 2)2r

.

The Euler-Lagrange equations yield two separated harmonic oscillator EoMs:

C + 20C = 0

and

r + 20(1 + 2)r = 0.

In a system of N coupled oscillators there always exist N generalised coordinates that

separate, yielding N independent, homogeneous, linear differential equations.

Those generalised coordinates undergoing simple harmonic motion are called normal

coordinates.

A more general recipe

1. Find an equilibrium configuration, i.e., values of the generalised coordinates where

all generalised forces Fk = V/qk are = 0.2. Taylor expand the Lagrangian L to second order in the generalised coordinates and

velocities, around values for the generalised coordinates qk given by the equilibrium

configuration, with the values of the generalised velocities qk set = 0.

3. For N DoF, the Taylor expansion must in principal be in 2N variables (all generalisedcoordinates and velocities). In practice, expansions in fewer variables may be ade-

quate, e.g. if some terms are already in quadratic form.

Matrix form of L

If ak are the equilibrium values of an initially used set of generalised coordinates qk, then

we can define a new set of generalised coordinates qk = qk ak.

These qk will have equilibrium values = 0, and L can then take the form

L = qTq T

qTqV

+c (a constant),

where q is a column vector, and its transpose qT = (q1, . . . , qN) is a row vector.The matrices and are symmetric (i.e., equal to their transposes, so that e.g. kj =

jk), with matrix elements given by

jk =1

2

2T

qjqk

qj,qk=0

, jk =1

2

2V

qjqk

qj,qk=0

.

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Equations of Motion

Note that partial differentiation of the Lagrangian L with respect to the generalised coor-

dinates and velocities yields

L

qj= 2

N

k=1

jkq

k= 2(q)

j,

L

qj=

2N

k=1

jkq

k=

2(q)j

,

and that the Euler-Lagrange equations therefore yield a linear system of differential equa-

tions: q + q = 0.

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9 Normal Modes


The EoM for a system of coupled oscillators can be written compactly in matrix nota-

tion as q + q = 0.

Working Towards a Solution

Inserting a trial solution q = beit into the EoM yields

( 2)b = 0. (14)

We also know (due to a theorem from linear algebra) that, unless b = 0, this can only befulfilled if | 2 + | = 0.

(This strictly only represents an eigenvalue problem if is a multiple of the identity

matrix. If different masses are associated with the different generalised velocities, thena variable transformation i =

miqi rectifies this. In practice, we can treat the solutions

to Eq. (14) as eigenvectorswith their associated eigenvalues.)

Calculating the determinant will yield an Nth order polynomial in 2. Hence, thereare Ngeneralised values 2j (possible values of

2), each with its corresponding vector bjthat will solve Eq. (14).

Returning to the example of the coupled pendulums, before we used our intuition to

convert from (1, 2) to (c, r), we had

L =ml2

2(2

1

+ 2

2

)

1

2 mgl(21 + 22) + kl

22

(21)

2=

mgl

2

1

20(21 +

22) ((1 + )(21 + 22) (12 + 21))

,

where 0 =

g/l and = kl4mg . We can thus infer that

2=mgl

2

0

0

,

where = 2

20, and

=mgl

2

1 + 1 +

The values of for which the system oscillates at a single frequency are thus2 + = (1 + )2 2 = 0,i.e. (1) = 1 and (2) = 1 + 2, which correspond to (1) = 0 and (2) = 0

1 + 2.

To find the mode vectors associated with these modes of oscillation, substitute (j) back

into Eq. (14). This gives

b(j)2b(j)1

= 1 +

,

from which, b(1)1 = b(1)2 and b(2)1 = b(2)2.

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Normal Coordinates

It is common to normalise the mode vectors b(j), i.e., to have bT(j) = (b(j)1, b(j)2, . . . , b(j)N)

such that

Nk=1 b

2(j)k

= 1. Using the vector elements, we may define a new set of sepa-

rated generalised normal coordinates rj via

rj =N

k=1

b(j)kqk.

Note that, for the coupled pendulum example, r1 (1 + 2) and r2 (2 1).If 2j > 0, the equilibrium configuration is stable, and the normal coordinate rj evolves

as rj(t) = rj(0) cos(jt) + (rj(0)/j) sin(jt). Such motion is called a normal mode, whereall parts of the system oscillate with the same frequency. Motion in general can be de-

scribed by a superposition of normal modes.

If 2j < 0, the equilibrium is unstable, and if 2j = 0, the motion is also not oscillatory,

usually corresponding to e.g. rotation or translation of the centre of mass.

A normal mode is the motion of a normal coordinate.

Mode orthogonality

As we have effectively been finding eigenvectors of a real symmetric matrix, they will be

orthogonal, ie

bTi bj = 0 if i = j.1. If is are all different, then bis are unique.

2. If some is are the same, then there is a choice of bis, but can always find or-

thogonal bis (if bi and bj are eigenvectors with the same , then so is any linear

combination bi + bj).

The bis are the natural axes in N-dimensional space. With respect to them, the trans-formed potential energy, , is diagonal (i.e. contains no cross terms). In these coordi-nates, the system undergoes simple harmonic oscillations.

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10 Central Forces


Two Interacting Bodies

Consider two bodies (with position vectors r1 and r2) interacting via a central force, de-

fined as a force depending only on the distance r between the two bodies, directed

along the line between them.

The potential energy V12 from which such a force is derived is only dependent on r,

i.e., V12 = V12(r1, r2) = V12(|r1 r2|) = V(r) for such a force.If we consider the general case of 2 point masses interacting via a central force, with

additional external forces, then the Lagrangian is given by

L =1

2(m1r

21 + m2r

22) V1(r1) V2(r2) external potentials

V12(|r1 r2|). interaction potentialIn the absence of external potentials (V1 = V2 = 0), this Lagrangian is translationally

invariant: L(r1, r2, r1, r2) = L(r1 + a, r2 + a, r1, r2).

Translational Symmetry and Conservation of Momentum

Consider an infinitesimal translation a, and Taylor expand the Lagrangian L to first order in :

L(r1 + a, r2 + a, r1, r2) = L(r1, r2, r1, r2) + a (1 + 2)L + O(2).

We neglect the O(2) term. Since a is arbitrary, ifL is to be translationally invariant, then (1 +2)L must = 0, i.e.,

L

x1+

L

x2= 0,

L

y1+

L

y2= 0,

L

z1+

L

z2= 0.

From the Euler-Lagrange equations for x1 and x2 (and equivalently for y and z):

d

dt

L

x1

+

d

dt

L

x2

=

L

x1+

L

x2= 0.

L/x1 = p1x defines the canonically conjugate momentum p1x to x1 (here this is equal to themechanical momentum mx1, but this need not in general be so). Hence, P = p1 +

p2

= 0, andso the total momentum P is conserved.

Ignorable Centre of Mass (CoM) Motion and Conservation of Momentum

Conservation of the total momentum P implies that the CoM [coordinate R = (m1r1 +m2r2)/(m1 + m2)] moves like a free particle.

Rewrite L in terms of R and the relative coordinate r = r1

r2:

L =1

2MR

2+

1

2r2 V12(r),

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where M = m1 + m2 (total mass) and = m1m2/(m1 + m2) (the reduced mass).R does not appear in the Lagrangian L. Such coordinates are called ignorable. For

any ignorable generalised coordinate q, it follows from the Euler-Lagrange equation that

d(L/q)/dt pq = 0, and hence that P is conserved in this case.

Rotational Invariance and Conservation of Angular Momentum

If we transform to the centre of mass frame (a frame moving with the CoM), then R =R = 0, and L = (1/2)r2 V12(r). Note that V12(r) is dependent on distance only, notdirection, and hence that L is rotationally invariant.

In terms of spherical coordinates (x = r cos sin , y = r sin sin , z = r cos ), we de-duce that r2 = x2 + y2 + z2 = r2 + r22 + r22 sin2 . Hence,

L =1

2[r2 + r22 + r22 sin2 ]V12(r). (15)

is ignorable, as it does not appear in L. Therefore

p L

= r2 sin2 = constant Jzcan show to be the z component of J = r r

Without loss of generality, we can choose the coordinates such that the total angular

momentum, J, lies along the z axis. Therefore J is always constant.

Motion Takes Place in a Plane

As J = r r, both p r and r must be perpendicular to J.Hence, p and r must be in a plane perpendicular to J, and remain in that plane if J is

to remain constant.

Choose J to lie along the z direction, so that the angle of elevation = /2, = 0,

and J = |J| = Jz = r2.

Equal Areas are Swept out in Equal Times (Keplers Second Law)

d

r

r+dr

dr{height h

Consider motion in the = /2 plane. If the position vector changes incrementallyfrom r to r + dr, with the angle changing by d, then the area swept out, dA, is given bythe area of the triangle formed by r, dr, and r + dr.

By straightforward trigonometry dA = (1/2)|r + dr||r| sin(d). Keeping only first-orderterms, dA = (1/2)r2d.

It follows thatdA

dt=

1

2r2 =

J

2. (16)

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J is conserved, so the rate at which area is swept out is constant.

Keplers second law is therefore a consequence of angular momentum conservation,

itself a consequence of considering a central force, and has nothing specifically to do

with gravity as such.

Equivalent 1D Problem

We can remove and entirely from the Lagrangian.

Following Eq. (15), we can write E = T+ V = (1/2)[r2 + r22 + r22 sin2()] + V12(r),which, if we make use of = 0, = /2, = J/r2, simplifies to

E =1

2r2 +

J2

2r2+ V12(r).

Veff(r)

(17)

The effective potential Veff(r) combines the interaction potential with the so-called angu-lar momentum barrier (responsible for the centrifugal force).The total energy E is the same as that corresponding to an effective Lagrangian Leff =

(1/2)r2 Veff, from which the Euler-Lagrange equation for r yields

r =J2

r3 V12

r. (18)

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11 Gravitational Attraction

SEE ALSO HAND AND FINCH, 4.1, 4.54.6

Gravitational Interaction Potential

The potential energy between two gravitating bodies of masses m1 and m2 is V12(r) =k/r, where k = Gm1m2 and G is the gravitational constant. The EoM is then

r =J2

r3 k

r2

We would like to solve this equation to determine the orbits of the bodies.

The u = 1/r Transformation

It is helpful to change the dependent variable from r to u = 1/r. From J = r2(d/dt), we candetermine that dt = [/(Ju2)]d and hence that

d

dt=

Ju2

d

d dr

dt= 1

u2du

dt= J

du

d.

Expressing the energy in the CoM frame [Eq. (17)] in terms ofu and , gives

E =J2

2

du

d

2+ u2

ku. (19)

Differentiating with respect to (note that E is constant, i.e., dE/d = 0),

J2

2

2

du

d

d2u

d2+ 2u

du

d

kdu

d= 0,

from whichd2u

d2+ u =

k

J2. (20)

This can be solved to give (1/r) u = (k/J2) + B cos (B is an integration constant, and wehave set a second integration constant, equivalent to an arbitrary phase added to , to 0).

Elliptical Orbits (Keplers First Law)

Define p J2/kand pB. Then putting the solution for u into Eq. (19) yields

E =k2

2J2(2 1) = k

2p(2 1).

If 0 < 1, the energy E is negative, implying a bound state.The solution to Eq. (20) can be written as

p = r + r cos.

Expressed in cartesian coordinates, this becomes p =

x2 + y2 + x, or equivalently

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(1 2)x2 + 2px + y2 p2 = 0. (21)

Restricting ourselves to 0 < 1, Eq. (21) describes an ellipse (Keplers first law):1

a2 x +

p

1 2

2

+y2

b2

= 1, (22)

where the semimajor axis a = p/(1 2) and the semiminor axis b = p/1 2 (withthese quantities defined, the energy takes the simple form E = k/2a).

is the eccentricity of the elliptical orbit. = 0 for a circular orbit, = 1 describes aparabolic trajectory and > 1 for a hyperbola.

Keplers Third Law

From Keplers second law dA/dt = J/(2) [Eq. (16)], the period of an elliptical orbit

follows from the total area A = ab of that orbit , i.e., = A2/J. A little algebra yields

= 2

1

G(m1 + m2)a3/2

Keplers third law states that 2 a3 with the same proportionality constant for all the

planets. As the mass of the sun dominates the masses of the individual planets, this is

approximately true.

Bertrands Theorem

For force laws of the form

F cr,only = 2 or = 1 give rise to closed, non-circular orbits. (Note that a circular orbit is possiblefor any .) A closed orbit is one for which the radius at a fixed azimuthal angle, , is described

by a function r(). = 2 corresponds to the familiar inverse square law, where the orbit is anellipse and the central force originates from one focus. = 1 is Hookes law, which gives rise toelliptical orbits that are centred on the source of the force.

The precession of the perihelion of Mercury was one hint that the inverse square law of gravity

was not the whole story. The General Theory of Relativity accurately predicts the size of the

effect.

Solving 1D Systems by Quadrature

Consider a 1D system such as a point mass in an external potential. The Lagrangian

L = (1/2)mq2 V(q), and it has energy E = (1/2)mq2 + V(q). This implies

q dqdt

=

2[EV(q)]m

dt = dq

m

2[EV(q)]

.

The time t taken to move from q = 0 to q(t) is then

t =

m

2

q0

dq

1

E V(q) . (23)

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The elapsed time as a function of the coordinate, t(q), must then be inverted if wewish to obtain q(t) (this may have to be carried out numerically).

A problem solved as an integral [i.e., put into integral form, as in Eq. (23)] is said to

be solved by quadrature.

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Workshop 2: Central Forces/Rutherford Scattering

1. A particle of mass m moves in a central force field with potential energy given by V(r). Inspherical polar coordinates, the Lagrangian of the particle can be written as

L =1

2

m(r2 + r22 + r2 sin2 2)

V(r).

(a) Using the fact that the azimuthal angle, , is an ignorable coordinate, determine the as-sociated constant of the motion, J. What does this constant represent? Express the total

energy of the particle as a function of the single variable, r, in a way that incorporates

the effective potential energy

Veff = V(r) +J2

2mr2

and explain why no dependence is required.

(b) Assuming that V(r) = Ar

n+1

/(n + 1) where A>

0, for each of the cases n + 1 0, sketch the radial variation of the effective potential.In each case, mark the radius, r0, where the potential is stationary with respect to ra-

dius and a circular orbit exists, and state whether or not the orbit is stable or unstable.

Calculate an expression for the radius r0.

(c) By performing a Taylor series expansion of the effective potential about the point r = r0,show that, for small perturbations away from a stable circular orbit, the radius of the

particle performs simple harmonic motion with an angular frequency

=

n + 3J

mr

2

0

.

(d) By comparing this angular frequency with that for a circular orbit at radius r0, deter-

mine which values of n give rise to closed orbits. Describe the orbits for the cases

n = 2 and n = 1.ADDITIONAL EXERCISE

2. Consider two positively charged point masses of (identical) mass m and charge c. Use the

Coulomb potential VC = [c2/(40)](1/r) to model how they interact, and consider them

to be in free space (no external potentials).

(a) Determine the Lagrangian describing the motion of the relative coordinate r = r1 r2.Hence, determine an expression for the total energy E in the CoM frame as a function

ofr = |r| and the angular momentum J.(b) Define an effective 1D Lagrangian Leff which yields the same value of E, but with

T = (1/4)mr2 (where r = |r|). With the Euler-Lagrange equation, derive from Leffthe equation of motion for r.

(c) Express the energy in the centre of mass frame in terms ofu 1/r and the azimuthalangle , and solve the differential equation (of u, differentiated with respect to ) re-sulting from the expression for dE/d, to find a relationship between r and .

(d) Express this solution in terms of cartesian coordinates x and y rather than polar coor-dinates r and , and hence show that all trajectories must take the form of hyperbolae,

of the general form (x xc)2/a2 y2/b2 = 1, in the centre of mass frame.

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(e) Use the identity cosh2 sinh2 = 1 to parametrize x and y in terms of, and hencedetermine an equivalently parametrized form of r. Use this to solve the effective 1D

problem by quadrature and determine an equivalent to Keplers equation for this sys-

tem.

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12 Noethers Theorem and Hamiltonian Mechanics


Invariance of L under Continuous Transformations

Consider a set of continuous transformations, i.e. depending on a continuously variable pa-rameter, s (e.g. translations along or rotations about the z axis are single-variable families of

transformations).

When s = 0, this should yield the identity transformation, i.e., no change.Ifqk(t) is a solution of the original EoM for a Lagrangian L, then Qk(s, t) will be a solution for theEuler-Lagrange equations evaluated within a transformed set of coordinates, where Qk(0, t) =qk(t).Given that the Lagrangian L is invariant (independent ofs) under this transformation,

L L(

Q1(

s, t)

, . . . , QN(

s, t)

,

Q1(

s, t)

, . . . ,

QN(

s, t)

, t)

=L(q1, . . . , qN, q1, . . . , qN, t).

As the Lagrangian is independent ofs, dL/ds = 0, and dropping the time variation for simplic-ity,

dL

ds=

N

k=1

L

Qk=

d

dt L

Qk

dQkds

+L

Qk

dQkds

=d

dt dQkds

=

d

dt

N

k=1

L

Qk

dQkds

= 0.

This implies the existence of a constant of the motion

I(q1, . . . , qN, q1, . . . , qN) N

k=1

pk(dQk/ds)|s=0,

which is evaluated at s = 0 for convenience.For multiple parameters s1, s2, . . ., the procedure can be repeated for each to find associated

constants I1, I2, . . . .

This procedure effectively describes Noethers theorem.

Noethers Theorem

If the Lagrangian is invariant under a continuous symmetry transformation, then there

are conserved quantities associated with that symmetry, one for each parameter of the

transformation. These can be found by differentiating each coordinate with respect to the

parameters of the transformation in the immediate neighbourhood of the identity trans-

formation, multiplying by the conjugate momentum, and summing over the degrees of

freedom.

Take the example of a point mass, m, moving in free space with L = 12 mq2. For trans-

formed coordinates Q = q + s, this gives

dQ

ds

s=0

= 1 I(q) = mq.

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Legendre Transformations

Consider a function A(x,y) of two variables. Introduce a third variable z and define B(x,y,z) yz A(x,y). By application of the chain rule,

dB = zdy + ydz

A

x y dx A

y x dy = zA

y x vanishes, as we define z = z(x,y) A

y

x

dy + ydz

A

x y dx. (24)

For this choice of z, B has no explicit y dependence. Thus y can be written in terms of x and zgiving

B = B(x,y(x,z),z) = B(x,z) = yz A(x,y)This is a Legendre transformation (cf thermodynamics). Furthermore, it follows from Eq. (24)

that

Bz

x= y, Bx

z

= Ax y .Note that the same Legendre transformation takes A(x,y) to B(x,z) as takes B(x,z) back toA(x,y). x is called a passive variable, and y an active variable.

The Hamiltonian

A Legendre transformation of a (time-independent, 1 DoF) Lagrangian L(q, q), where theactive variable is the velocity q, the passive variable is the coordinate q, and the third

variable is the canonically conjugate momentum p, yields the Hamiltonian H:

H(q, p) pq L(q, q), where p Lq

constant q

. (25)

Differentiating this equation and using the Euler-Lagrange equation, we find Hamiltons

equations.

H

p

q

= q,H

q

p

= Lq

q

= ddt

L

q

q

= p.

Note that, while the Lagrangian formulation of mechanics produces N second order

ordinary differential equations (in theqk

), the Hamiltonian treatment yields2N

first order

ordinary differential equations (in qk and pk).

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Hamiltons Principle

From Hamiltons principle, variations in the action S =

Ldt are given by S = S[q + q] S[q] =

Ldt = 0. Using the relationship between L and Hgiven by Eq. (25),

L=

qp+

pq

H

q q+

H

p p H

= qp +d

dt(pq) pq H

qq H

pp =

q H

p

p

p +

H

q

q +

d

dt(pq).

Note that d(pq)/dt vanishes on integration, as q = 0 at the endpoints of the physical path,and that Hamiltons equations imply that the q, p terms also vanish. Hence,

Ldt = 0 and

independent infinitesimal variations in q, p from the physical path in coordinate-momentum

space (called phase space as opposed to the configuration space of Lagrangian mechanics) do not

change the action, S.

Hamiltons Equations of Motion

For N DoF, H Nk=1 pkqk L (where L may now be explicitly time-dependent). Thereforeinclude a time differential in

dH =N

k=1

pkdqk + qkdpk Lqkdqk

L

qkdqk

L

tdt

=

N

k=1(qkdpk pkdqk)L

t dt.

We can also consider time as a passive variable in the Legendre transformation, which

yields

H

t

q1,...,p1,...

= Lt

q1,...,q1,...

dHdt

=H

t= L

t.

If there is no explicit time dependence in L, H is a constant of the motion. If T is also a

quadratic form in the qk, H is the total energy E = T+ V.Finally, Hamiltons equations of motion in complete generality are:

qk =Hpk

, pk = Hqk,dHdt

= Lt

.

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13 Canonical Transformations and Poisson Brackets


Canonical Transformations

The Lagrangian formulation of mechanics makes coordinate transformations (i.e. Q =Q(q, t)) fairly straightforward. These types of change to new generalised coordinates andtheir associated generalised velocities are commonly called point transformations.

Coordinate transformations take place in configuration space. In phase space one

can consider new coordinates Q and momenta P that are functions of the original coor-

dinates q and their momenta p (and possibly the time t):

Q = Q(q, p, t), P = P(q, p, t).

These are commonly called contact transformations.

A transformation is by definition canonical, if it preserves the structure of Hamiltons

equations for all dynamical systems. Hence, for a specific dynamical system, a new

Hamiltonian H that is a function of Q and P is produced, and Hamiltons equations forH, P, and Q must describe the correct transformed dynamics.

Equivalence of Lagrangians

Consider two Lagrangians L and L such that L(q, q, t) = L(q, q, t) dF(q, t)/dt (F is indepen-dent ofq). Substituting L into the Euler-Lagrange equation yields

d

dt

L

q

L

q=

d

dt

q

dF

dt

q

dF

dt

=d

dt

q

qF

q+

F

t

q

qF

q+

F

t

=d

dt

F

q

q2F

q2+

2F

qt

= 0,

i.e., the Euler-Lagrange equation derived from L. Hence, the descriptions of a physical system

provided by L and L are equivalent.We now consider instead a starting Lagrangian L(q, q) and a transformed Lagrangian L(Q, Q)related by a similar F that is a function of both q and Q, and perhaps t, such that

L(Q, Q, t) = L(q, q, t) dF(q, Q, t)dt

(26)

( = 1 is associated with a change of units or scale transformation. We will always consider = 1)We next integrate Eq. (26) with respect to t (and with = 1):

t2

t1Ldt =

t2

t1Ldt + F(q(t1), Q(t1), t1) F(q(t2), Q(t2), t2) (27)

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Recalling Hamiltons principle, if we take variations of Eq. (27) [assuming that arbitrary vari-

ations q(t) imply arbitrary variations Q(t)] the variation oft2

t1Ldt vanishes. If we further

assume that F = 0 at the end points, then we can say that the variation oft2

t1Ldt vanishes,

and it follows that the two descriptions are equivalent.

The Generating Function

F(q, Q, t) is called a generating function. From the chain rule, the time derivative for thegenerating function is dF/dt = (F/q)q + (F/Q)Q + F/t

By construction, q does not appear explicitly in L, and so, from Eq. (26)

0 =L

q=

L

q

q

F

qq +

F

QQ +

F

t

=

L

q F

q p = F

q. (28)

Similarly

LQ = FQ P = FQ . (29)

These are known as the implicit transformation equations.

With the implicit transformation equations we have two equations and two unknowns,

P(p, q) and Q(p, q). To find an explicit form for the transformation, solve Eq. (28) to expressQ = Q(q, p, t), and then insert this relation into Eq. (29) to get P = P(p, q, t).

Extension to N> 1 DoF is in principle straightforward, although in practice the task of

determining 2N unknowns from 2N equations may be formidable.

For example, consider the generating function F = qQ, then

p = Fq

= Q, P = FQ

= q.

This particular generating function essentially interchanges the coordinates and momenta.

The Transformed Hamiltonian

To find the new Hamiltonian H(Q, P), return to the definition of H in terms of a Legendretransformation [Eq. (25)]. Hence,

H(Q, P, t) PQ L = FQ

Q L + Fq

q + FQ

Q + Ft

= pq L + Ft

=H(q, p, t) +F(q, Q, t)

t

(30)

where we have made use of both Eq. (29) and Eq. (28). Finally, p and q appearing in the

RHS of Eq. (30) must be expressed in terms of P and Q.

If there is no explicit time dependence to F, then the new Hamiltonian H is simply theoriginal Hamiltonian H, rewritten by inserting the inverse of the transformation equations

expressing P and Q in terms of p and q.

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Four Forms of Generating Function

The generating function need not be written in terms of q and Q. Any combination of

one original and one new dynamical variable may be employed, in conjunction with

appropriately modified implicit transformation equations:

F1(q, Q, t), p =F1q

, P = F1Q

;

F2(q, P, t), p =F2q

, Q =F2P

;

F3(p, Q, t), q = F3p

, P = F3Q

;

F4(p, P, t), q = F4p

, Q =F4P

.

(31)

The different forms of generating function may be related via Legendre transforma-

tions, or the implicit transformation equations can simply be determined independently

in a similar manner to that used for F = F1 above.

Poisson Brackets

For N DoF, a Poisson bracket for two arbitrary functions F and G is defined as

{F, G} N

k=1 Fqk Gpk Fpk Gqk

For the case where N = 1,

{F, G} Fq

G

p F

p

G

q.

This is just the Jacobian defined by

J

F, G

q, p

(F, G)

(q, p)

Fq

Gq

Fp

Gp

Clearly, {q, p} = 1. If we consider a Poission bracket ofq and p defined with respect to the

canonically transformed dynamical variables Q and P, then [using Eq. (31) and the chain rule]

q

Q

p

P q

P

p

Q=

q

Q

2F2Pq

+2F4Pp

p

Q=

Q

q

q

Q+

Q

p

p

Q=

Q

Q= 1,

i.e., the value of the Poisson bracket for q and p is independent of the representation. Conversely,{Q, P} = 1 is a necessary and sufficient condition for a transformation to be canonical.There is an equivalent chain rule for Jacobians whereby

(F, G)

(Q, P) =

(F, G)

(q, p)

(q, p)

(Q, P) ,

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soF

Q

G

P F

P

G

Q=

F

q

G

p F

p

G

q.

We therefore do not need to specify a particular set of coordinates and conjugate momenta when

discussing Poisson brackets.

Poisson Bracket Formulation of Hamiltons Equations

Hamiltons equations of motion can be reformulated in terms of Poisson brackets:

q = {q, H}, p = {p, H}. For an arbitrary function F(q, p, t), F = {F, H}+ Ft

.

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14 Rotating Reference Frames


Accelerating Reference Frames

One can only specify the position rS of a point mass with respect to a chosen reference

frame S.

One can equally choose another reference frame B to specify the position rB of the

same point mass, displaced by R from S. The positions of the point mass relative to the

two frames are then related by rB = rS R.If S is an inertial frame where Newtons laws hold (fixed in Space), but R = 0, then B

is not an inertial frame (attached to an accelerating Body). Hence, in frame S, mrS = F,but in frame B,

mrB

= m(rS

R) = Ftrue force

mR.

fictitious force

Rotated Reference Frames in 2D

(a)

xS

yS

xByB

r

(b)

xB

yB

r

(c)

x

y

rS

rB

(a) In frame S, the position vector r has positive x and y components. A new fixed

reference frame B is defined with respect to an original reference frame S by an

anticlockwiserotation through angle .

(b) In frame B, r has a positive x component and a negativey component.

(c) An observer who changes their observation frame from S to B finds that r appears

to have rotated clockwisethrough an angle , from rS to rB.

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Rotations in 3D

O

N

P

Q

n

N

P

Q

rS n

V

Figure 1: Left: Overall and Right: top views of a general clockwise rotation of a vector OP = rSthrough an angle about an axis defined by the unit vector n, to become OQ = rB. This is what

we would observe if the reference frame we were in rotated in an anticlockwise direction.

We wish to determine OQ = ON+ NV+ VQ. Firstly,

ON = (n rS)n (component ofrS in n direction).

Noting that OP = ON + NP NP = rS (n rS)n, and that NQ has the same magnitude asNP, it follows that

NV = NP|NP| |NQ| cos() = [rS (n rS)n] cos()

We note that rS n has direction perpendicular to rS = OP and n (as does VQ), and magnitude|rS n| = |NQ|. Hence,

VQ =(rS n)|rS n|

|NQ| sin() = (rS n) sin().

Gathering the terms calculated above, we have

OQrB =

ON (n rS)n +

NV [rS (n rS)n] cos() +

VQ (rS n) sin(),

which is generally rearranged slightly to obtain the rotation formula:

rB = rS cos() + (n rS)n[1 cos()] + (rS n) sin(). (32)

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Infinitesimal Rotations

Consider a point fixed in frame S at rS = r. If frame B is infinitesimally rotated w.r.t. S byd, then rB = r + dr ( dr = rB rS). Taking the small-angle limit (i.e. = d), the rotationformula [Eq. (32)] transforms to r + dr = r + (r n)d.

Hence, (noting that for any two vectors U

V =

V

U) the velocity of the point in

reference frame B relative to that in frame S is given by:dr

dt

in B

=

nd

dt

r =

angular velocity

r.

Note that even if r is constant in reference frame B, it is time dependent if viewed from

reference frame S.

Velocity and Acceleration

If the vector r does have a time-dependence in reference frame S (e.g. represents theposition of a moving particle), then this must also be accounted for:

dr

dt

in B

velocity in B

=

dr

dt

in S

velocity in S

r. velocity due to motion of B relative to S

This equation is completely general for any vector A, so

dA

dt in B=

dA

dt in S A.

We can use the operator d

dt

in B

=

d

dt

in S

in conjunction with vB = vS r to determine the acceleration in the non-inertial frameB.

dvBdt

in B

=

d

dt

in S

vB

= d

dtin S (vS r) vB= aS r vS vB= aS r (vB + r) vB= aS 2 vB ( r) r

Inertial Forces in a Rotating Frame

If considering motion of a point mass, then multiplying

dvBdt

in B

by m and noting that

F = m(dvS/dt)S, we deduce that

mr = F 2m r coriolis force

m ( r) centrifugal force

m r. Euler force

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15 Inertial Forces on Earth


Procedure for Determining a Local Coordinate Systems

x

z

y

z

y

Figure 2: Left: Overall view of a local coordinate system in the region of a point on the surface of

the spherical earth. The positive x direction is east, the positive y direction is north, the positive z

direction is out from the centre of the earth. Right: Cross-sectional view, showing the local y and z

axes only (the x axis points perpendicularly into the page). is the latitude, and the colatitude

(in spherical coordinates, the angle of elevation).

1. Model the Earth as a perfect, solid sphere, and require an appropriate coordinate

system to describe motion near a point on its surface. Start with a cartesian coordi-

nate system with the z axis running through the poles, and the x and y axes emerging

through the equator [in this frame = (0,0,)].

2. Rotate this coordinate system around its x axis so that the z axis emerges at a chosen

point on the Earths surface. In these new coordinates, this looks like a rotation of about the x axis, thus the angular velocity in the local frame is

=

1 0 00 cos sin

0 sin cos

00

=

0 sin

cos

3. Finally, we displace the origin of the coordinate system by R to the point where thelocal z axis emerges from the Earths surface (where R = |R| is the radius of theEarth).

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Inertial Forces in the Local Coordinate System

Note that the derivation of centrifugal, coriolis, and Euler force terms assumes r to be

defined with respect to a point on the axis of rotation.

The final displacement of the local coordinate system origin means the associated

position vector r does not begin on the axis of rotation.One must therefore substitute r = r + R into the inertial force terms, where r is the

position vector associated with the local coordinate system.

Centrifugal Force m ( r)The direction of the centrifugal force can be determined quite generally: r is perpen-dicular to , r, and ( r) is perpendicular to both and r.

Hence, the centrifugal force m ( r) points directly away from the axis of rota-tion, meaning it can counteract gravity.

Effective Gravity on Earths Surface

(a) (b) (c)

(a) Direction and magnitude of the force due to earths gravitation on the surface of a

spherical earth model.

(b) Direction of the centrifugal force on the surface of a spherical earth model (magni-

tude relative to the gravitational force greatly exaggerated for effect).

(c) Effect of adding the gravitational and centrifugal forces together. The net force for

a static object does not point exactly to the centre of the earth except at the poles

and the equator.

Coriolis Force

2m

r

In the local coordinate system defined for the surface of the Earth, the coriolis force is

given by

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2m r = 2m( z sin y cos , x cos ,x sin )Note that it is proportional to velocities, and therefore only comes into play when there is

motion.

One often considers phenomena when motion is effectively constrained to be in

plane, when we ignore the z dimension. In this case the coriolis force x component 2m y cos , and the coriolis force y component 2mx cos .

Loose Ends

The Euler force is not zero on Earth, but is commonly taken to be negligible.

In a rotating reference frame that is also accelerating translationally with acceleration

R, the general EOM is given by

mr = F

2m

r

m

(

r)

m

r

mR.

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Workshop 3: Coriolis Force

1. A stone is dropped from a helicopter hovering a height h over a point on the equator. h is

small in comparison with the radius of the Earth, R, so that the acceleration due to gravity

can be taken as a constant, g. Ignore air resistance.

(a) The operator relating time derivatives of a vector viewed in an inertial frame, S, to thosemeasured in a frame B, rotating with angular velocity (for example that associated

with the rotation of the Earth), isd

dt

in B

=

d

dt

in S

.

Use this to show that the force measured in the rotating frame satisfies

mr = F m ( r) 2m r m r,where r represents the position measured in frame B, and F is the force in the inertialframe.

In which directions do the three inertial forces point for the falling stone?

The rate at which the Earths rotation is decreasing is very small, so you should now

assume = 0.

(b) Ignoring inertial forces, find expressions for the speed and height of the stone as a

function of time, and the time taken before the stone hits the ground, t0. What condition

must be satisfied for the inertial forces to be unimportant in determining t0?

(c) Using the Coriolis force and the approximation that it always acts in the same direction,

how far, in terms of t0, does the stone land from the point beneath the helicopter, and

in which direction?

(d) Solve the same problem by using the conservation of angular momentum, and compare

your answer with that from (c).

ADDITIONAL EXERCISES

2. A Foucault pendulum is a heavy weight suspended by a cable from a bearing that does not

restrict the motion (unlike e.g. a typical clock pendulum, the motion of which is constrained

to a plane), thus making it sensitive to the earths angular velocity .

If the oscillation amplitude is small compared to the length of the pendulum l (small angle

approximation), then we can neglect motion in the vertical (z) direction, and approximate

the relevant forces in the (x,y) plane tangential to the earths surface by F = m20(x +y),where 20 = g/l, at the earths surface.

(a) Using

mr = F m ( r) 2m r m r,derive expressions for x and y for the Foucault pendulum in the appropriate local co-

ordinate system ( pointing east, pointing north, kpointing out from the centre of the

earth).

(b) In what follows we will neglect the centrifugal force terms. This is acceptable if0 is

significantly larger than ||. Why?

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(c) Determine the general solution of the motion of the pendulum bob in the plane tan-gential to the earths surface (to the assumed level of approximation) in terms of the

initial position and velocity. The mathematics will be easier if you consider the time

evolution of the complex number (t) = x(t) + iy(t). Also, you may like to define = cos . Make sure that you consistently drop all terms of the same size or smallerthan the neglected centrifugal force terms (to do this you will need the approximate

identity 1 + 1 + /2 for small ).(d) The pendulum is allowed to start oscillating after being held at an amplitude A due

east from the origin. What is the position of the pendulum bob after each complete os-

cillation? Determine how many such oscillations must elapse before the pendulum bob(approximately) returns to its original position. How many hours does this correspond

to:

i. At the north pole;

ii. At the equator;

iii. In Durham ( latitude 5447 north)?3. What angular velocity would be required for your effective weight at the equator to be half

that at the north pole? (Take the earths radius to be 6378 km). How many minutes would a

full rotation of the earth take?

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16 Rotational Inertia, Angular Momentum and Kinetic Energy


DoF of Rigid Bodies

A rigid body is a system of mass points (or a continuum) constrained so that distances

between all points are constant (e.g. connected by massless rigid rods). A system of 2

connected point masses has 5 DoF, and a system of 3 connected point masses has 6 DoF.

Connecting a further point mass (e.g. by massless r igid rods) to 3 original point

masses completely specifies its position relative to the rigid body, and contributes 3 coor-

dinates 3 constraints = 0 additional DoF. Any rigid body more complex than two con-nected point masses has 6 DoF.

These 6 DoF correspond to a translation (3 DoF) and a rotation (2 DoF to define a

rotation axis and 1 DoF for the rotation).

Moment of Inertia I about an Axis of Rotation

Consider N mass points, rotating with angular velocity anticlockwise about an axisof rotation (assumed to pass through the origin), where the direction of the axis of rotation

is defined by the unit vector n (i.e., the angular velocity vector = n).Each mass point has a moment of inertia that is the square of its distance from the axis

of rotation multiplied by its mass m. For N mass points, I is the sum of such terms. Hence,

I =N

k=1

mk|(rk n)n rk|2 =

k=1

mk[r2k (rk n)2].

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Angular Momentum J and Rotational Kinetic Energy

The angular momentum (about the origin) of a system of mass points is given by J = Nk=1 rkp

k= Nk=1 mkrk rk, (if the point about which we calculate J is R = 0, then we replace rk, rk by

rk, rk, where r

k = rk R).

If rigid body constraints are imposed, any velocity rk is due to rotation only. rk is the velocity of the kth mass point due to its anticlockwise rotation.Using the vector identity a (b c) = (a c)b (a b)c and recalling = n,

J =N

k=1

mk[rk ( rk)] =N

k=1

mk[r2k (rk )rk]

J n = N

k=1

mk[r2k (rk n)2] = I.

Similarly, the kinetic energy reduces to T = 12

N

k=1m

kr2

k=

2

2

N

k=1m

k|n

rk|

2. Noting that

|n rk| = |(rk n)n rk|, the rotational kinetic energy takes a very simple form:

T =2

2

N

k=1

mk|(rk n)n rk|2 =I2

2.

Definition of the Inertia Tensor I

From J = Nk=1 mk[r2k

(rk

)rk], we infer that

Jx =N

k=1

mk(r2k x2k)x mkxkyky mkxkzkz,

Jy =N

k=1

mk(r2ky2k)y mkykxkx mkykzkz,

Jz =N

k=1

mk(r2kz2k)z mkzkxkx mkzkyky,

and use the linear transformation relating the angular velocity vector to the angularmomentum vector J to define the inertia tensor I:

JxJy

Jz

J

=

Ixx Ixy IxzIyx Iyy Iyz

Izx Izy Izz

I

xy

z

with I =N

k

mk(r2k rk,rk,), where =

1 if = 0 if

=

For continuous rigid bodies I =

volumedxdydz(x,y,z)(r2 rr), where is a mass

density.

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Generalised Description of the Rotational Kinetic Energy

Using the vector identity a (b c) = b (c a) = c (a b), the rotational kinetic energy canbe phrased more generally in terms of the moment of inertia tensor as

T =

1

2

N

k=1 mkr2

k =

1

2

N

k=1 mkrk ( rk) =1

2 N

k=1 mk(rk rk) . J = I

Hence T = (1/2) ( I) = (2/2)nTIn, where nTIn is the moment of inertia I about the axisdefined by n.

Centre of Mass (CoM)

The centre of mass position and velocity for a system of N mass points (mass of kth point

= mk) are defined by (total mass M = Nk=1 mk):

RC =

Nk=1 mkrk

M, RC =

Nk=1 mkrk

M.

It immediately follows that the total momentum P = Nk=1 mkrk = MRC is equivalent to themomentum of a single particle of mass M = Nk=1 mk moving with the CoM.

Separating the CoM Angular Momentum and Kinetic Energy

The total angular momentum about some point (situated at the origin of the coordinate system)is J = Nk=1mk(rk rk).Let rk = r

k + RC. Hence,

J =N

k=1

mk(rk + RC) (rk + RC)

=N

k=1

mk(rk rk + RC rk + rk RC + RC RC)

=

N

k=1 mk(rk rk) + MRC

Nk=1 mkr

kM +

Nk=1 mkr

kM

CoM velocity and position with respect to CoM, i.e., = 0

MRC + MRC RC.

J = Nk=1 mk(rk rk) + MRC RC can be expressed as the angular momentum about the CoM

plus the angular momentum of the CoM about the origin.

Similarly, the CoM kinetic energy, and that for motion with respect to the CoM can be separated:

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T =1

2

N

k=1

mkr2k =

1

2

N

k=1

mkrk rk =1

2

N

k=1

mk(rk + RC) (rk + RC)

=1

2

N

k=1

mkr2k + RC

N

k=1

mkrk

= 0

+1

2MR2C.

In general, if the external forces acting on a rigid body act through the CoM (e.g. gravity), then

the CoM motion can be considered separately.

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17 The Parallel and Principal Axis Theorems

SEE ALSO HAND AND FINCH, 8.2,8.7

Displaced Axis Theorem

In a similar fashion to the angular momentum and rotational kinetic energy, one can insert the

CoM position RC into the elements of the Inertia tensor I . With rk = rk + RC,

I =N

k=1

mk(r2k rk,rk,) =

N

k=1

mk[|rk + RC|2 (rk, + RC,)(rk, + RC,)]

=N

k=1

mk[(r2k + 2r

k RC + R2C) (rk,rk, + rk,RC, + RC,rk, + RC,RC,)]

=

N

k=1 mk(r2

k rk,rk,) + M(R2

C RC,RC,)

+ 2RC N

k=1

mkrk

=0

RC,N

k=1

mkrk,

=0

RC,N

k=1

mkrk,

=0

We have used the fact that the CoM position with respect to the CoM is = 0.

Hence, the inertia tensor of a rigid body, defined with respect to rotations about the

origin can be related to the inertia tensor of the same rigid body, defined with respect to

rotations about its CoM, through

I = ICoM + M A,

where A can be represented by a matrix, the elements of which are determined by the

elements of the CoM position vector: A = R2C RC,RC,. This is the displaced axis

theorem.

Parallel Axis Theorem

Consider a moment of inertia ICOM about an axis of rotation passing through the CoM.

This is equivalent to considering a diagonal element of some representation of the inertia

tensor.

From the displaced axis theorem, the moment of inertia I about a parallel axis is

I = ICoM + Md2,

where d is the distance of parallel axis from the axis of rotation passing through the CoM.

This is the parallel axis theorem.

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Symmetric Nature of the Inertia Tensor

The inertia tensor I is defined for a particular rigid body, and for rotations of that body

about a particular point (frequently the centre of mass).

In any representation (e.g. for some set of cartesian axes, conveniently chosen with

the origin at the point of rotation) the inertia tensor is represented by a symmetric matrix,

i.e., I = I. We say I is a symmetric tensor.The matrix elements depend on the chosen representation (the tensor is defined rep-

resentation free). We wish to know if there is a diagonal matrix representation of the

inertia tensor.

Orthogonal Matrices and Orthogonal Diagonalization

The inverse P1 of a matrix P is defined such that P1P = PP1 = 1, the identity matrix. Amatrix P for which an inverse P1 exists is called invertible.

A real invertible square matrix P such that its inverse P1

= PT

, its transpose, is said to be anorthogonal matrix. (So-called because both its rows and columns form orthonormal sets).

A general 2 2 orthogonal matrix has a single free parameter , and describes a 2-dimensionalrotation when acting on a 2-dimensional column vector:

P =

cos sin sin cos

.

The inverse matrix describes a rotation back in the opposite direction.

Similarly, in 3 dimensions, one can consider separate rotations about each axis:

Px =

1 0 00 cos sin 0 sin cos

, Py = cos 0 sin0 1 0

sin 0 cos

,

Pz =

cos sin 0sin cos 0

0 0 1

.

(PzPyPx)TPzPyPx = PTx PT

y PT

z PzPyPx = 1, and the product PzPyPx is a way of formulating a gen-eral 3 3 orthogonal matrix, describing a general rotation in 3-dimensions.A square matrix

A is called orthogonally diagonalizable if there is an invertible matrix

P such thatP1 AP = PT AP = D is diagonal; P orthogonally diagonalizes A. If A is an N N matrix, then

the following are equivalent. (a) A is orthogonally diagonalizable; (b) A has an orthonormal

set of Neigenvectors; (c) A is symmetric.Consider the 3 3 case. Let us say that a 3 3 matrix A has 3 orthonormal eigenvectors p

1=

(p11, p21, p31), p2 = (p12, p22, p32), p3 = (p13, p23, p33) with corresponding eigenvalues 1, 2,3.

Let P be the matrix whose columns are these eigenvectors:

P =

p11 p12 p13

p21 p22 p23p31 p32 p33

This implies that

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AP =

1p11 2p12 3p131p21 2p22 3p23

1p31 2p32 3p33

=

p11 p12 p13p21 p22 p23

p31 p32 p33

P

1 0 00 2 0

0 0 3

D

.

Since the columns of

P are orthonormal,

P is an orthogonal matrix. Hence,

A

P =

P

D P1 AP = PT AP = D, i.e., A is orthogonally diagonalizable.To help see that A must be symmetric, i.e., that AT = A, we use D = P1 AP = PT AP PDP1 = PDPT = A. Hence,

AT = (PDPT)T = PDTPT = PDPT = A.

Principal Axis Theorem

I

is a symmetric tensor, and is always represented by a3 3

symmetric matrix. Therefore,

I has three eigenvalues (which may be repeated). These are the principal moments of

inertia, called I1, I2, I3.

I must also have three orthonormal eigenvectors (in the case of repeated eigenval-

ues, there is some freedom in choosing what form the corresponding eigenvectors should

take). The directions the eigenvectors point in determine the principal axes, labelled 1,

2, 3.

It is always possibl

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Lagrangian and Hamiltonian Mechanics Notes