Classical Mechanics (MP350)

Jon-Ivar Skullerud

with modiﬁcations by Brian Dolan

December 11, 2020 1 Contents

1 Introduction 5 1.1 Physics is where the action is ...... 5 1.2 Overview ...... 6

2 Lagrangian mechanics 8 2.1 From Newton II to the Lagrangian ...... 8 2.2 The principle of least action ...... 8 2.2.1 Hamilton’s principle ...... 11 2.3 The Euler–Lagrange equations ...... 12 2.4 Generalised coordinates ...... 15 2.4.1 The shortest path between two points (optional) ...... 20 2.4.2 Polar and spherical coordinates ...... 21 2.5 Lagrange multipliers [Optional] ...... 24 2.5.1 Constraints ...... 24 2.6 Canonical momenta and conservation laws ...... 29 2.6.1 Angular momentum ...... 30 2.7 Energy conservation: the hamiltonian ...... 32 2.7.1 When is H conserved? ...... 33 2.7.2 The Energy and H ...... 34 2.8 Lagrangian mechanics — summary sheet ...... 37

3 Hamiltonian dynamics 39 3.1 Hamilton’s equations of motion ...... 40 3.2 Cyclic coordinates and eﬀective potential ...... 42 3.3 Hamilton’s equations from a variational principle ...... 44 3.4 Phase space [Optional] ...... 45 3.5 Liouville’s theorem [Optional] ...... 48

2 3.6 Poisson brackets ...... 50 3.6.1 Properties of Poisson brackets ...... 50 3.6.2 Poisson brackets and conservation laws ...... 52 3.6.3 The Jacobi identity and Poisson’s theorem ...... 53 3.7 Noethers theorem ...... 54 3.8 Hamiltonian dynamics — summary sheet ...... 57

4 Central forces 59 4.1 One-body reduction, reduced mass ...... 59 4.2 Angular momentum and Kepler’s second law ...... 61 4.3 Effective potential and classification of orbits ...... 64 4.4 Integrating the energy equation ...... 64 4.5 The inverse square force, Kepler’s first law ...... 66 4.5.1 The shapes of the orbits ...... 68 4.6 More on conic sections ...... 69 4.6.1 Ellipse ...... 70 4.6.2 Parabola ...... 72 4.6.3 Hyperbola ...... 72 4.7 Kepler’s third law ...... 74 4.8 Kepler’s equations ...... 76 4.9 Runge-Lenz vector ...... 77 4.10 Central forces — summary sheet ...... 78

5 Rotational motion 80 5.1 How many degrees of freedom do we have? ...... 81 5.1.1 Relative motion as rotation ...... 82 5.2 Rotated coordinate systems and rotation matrices ...... 82 5.2.1 Active and passive transformations ...... 84 5.2.2 Elementary rotation matrices ...... 84 5.2.3 General properties of rotation matrices ...... 84 5.2.4 The rotation group [optional] ...... 87 5.3 Euler angles ...... 88 5.3.1 Rotation matrix for Euler angles ...... 89 5.3.2 Euler angles and angular velocity ...... 90 5.4 The inertia tensor ...... 91

3 5.4.1 Rotational kinetic energy ...... 91 5.4.2 What is a tensor? Scalars, vectors and tensors...... 96 5.4.3 Angular momentum and the inertia tensor ...... 98 5.5 Principal axes of inertia ...... 98 5.5.1 Rotations and the inertia tensor ...... 98 5.5.2 Comments ...... 101 5.6 Equations of motion ...... 102 5.6.1 The symmetric heavy top ...... 102 5.6.2 Euler’s equations for rigid bodies ...... 104 5.6.3 Stability of rigid-body rotations ...... 105

4 Chapter 1

Introduction

1.1 Physics is where the action is

In these lectures we shall develop a very powerful (and beautiful) way of formulating Newtonian mechanics. The basic idea is to derive Newton’s equations from a variational principle, meaning that for a given dynamical system we look for a function of the dynamical variables and velocities such that the time evolution of the system is obtained by minimising this function. The function is called the action for the system. This gives an extremely concise and elegant way of describing the dynamics: for systems with many degrees of freedom and/or many particles we do not need to write down a mess of complicated coupled differential equations to define the dynamics — we just write down a single function, the action. In principle we can write all the laws of physics on the back of a postage stamp. In order to solve the dynamics though we need to pick it apart, and that requires deriving dynamical equations from the action and solving them, which can still be quite complicated. But the simplicity and elegance of the variational formulation often points to a choice of variables that makes the solution easier. Moreover the action principle is the springboard to new physics. The methods introduced in this course can easily be extended to both special and general relativity and were instrumental in the development of quantum mechanics at the beginning of the 20th century. Indeed today the action is fundamental to our current understanding of the Standard Model of particle physics and relativistic quantum field theory, it is the principal tool used to study matter at the most fundamental level. We shall not sail into such exotic waters in this course though, we shall leave that to later modules. The focus here will remain on Newtonian mechanics, but there is a shift in emphasis, from Newtonian forces and acceleration to the more general and abstract formulations that were developed in the late 18th and the 19th century, associated with names like Euler, Lagrange, Hamilton and Jacobi. Therefore, this course is not more of the stuff you have already studied in modules like MP110, MP112 and MP205, but instead represents a completely new way of looking at mechanics, and one which forms the foundation of nearly all modern mathematical physics. The focus in this course is on methods and formulations rather than on answers or numbers. In part, this is because the key to solving complicated problems is very often

5 to formulate them properly and to select appropriate methods. However, there are other reasons for this shift in focus:

• Often, we are not that interested in numerical solutions, but more in the qualitative features of a system, and we can ﬁnd out a lot about this without doing any numerical calculations.

• We will see that wildly diﬀerent physical systems can look identical from a mathematical point of view, so solving one can immediately give us the solution to the other. Starting with numerical calculations can obscure this.

• Symmetries will play an extremely important role, and we will learn to identify and exploit symmetries to simplify and understand mechanical systems. Putting in numbers at the start will often hide the symmetries.

The Lagrange–Hamilton formalism and the symmetry principles which we will become acquainted with here, are used all throughout modern physics:

• quantum mechanics;

• statistical mechanics;

• condensed matter theory (quantum statistical mechanics)

• classical ﬁeld theory (electromagnetism, general relativity)

• particle physics (quantum ﬁeld theory and symmetry groups)

• chaos theory

• etc

1.2 Overview

The module will cover the following topics:

• The principle of least action (Hamilton’s principle), the lagrangian and the Euler– Lagrange equations.

• Generalised coordinates (how to formulate a mechanical problem in the most sensible way given symmetries and constraints).

• Canonical momenta and conservation laws; energy conservation.

• Hamilton’s equations of motion.

• Poisson brackets.

• Central force motion, angular momentum conservation.

• Planetary motion, Kepler’s laws.

6 • Rotations and rotation matrices.

• Inertia tensor, principal axes of inertia.

• Euler’s equations of (rotational) motion.

Learning outcomes

At the end of this course, you should be able to:

• formulate the basic principles of the Lagrange–Hamilton formalism;

• use these principles to derive equations of motion for dynamical systems;

• explain the relation between symmetries and conservation laws;

• apply conservation laws to analyse the motion of dynamical systems; and

• describe the mathematical properties of rotations and systems with rotational symmetry.

7 Chapter 2

Lagrangian mechanics

2.1 From Newton II to the Lagrangian

In the coming sections we will introduce both the notion of a Lagrangian as well as the principle of least action. This will be an equivalent, but much more powerful, formulation of Newtonian mechanics than what can be achieved starting from Newton’s second law. However, to introduce this new way of thinking, we will in this section give a short argument why the Lagrangian is a “natural” object to study. Consider now a single particle at position x in a potential V (x, t). The kinetic energy 1 2 of this particle is T = 2 mx˙ . The equation of motion for this particle is ∂ mx¨ = − V (x, t). (2.1) ∂x What we ultimately seek, is a way to generate this equation of motion from a simpler d ∂ object. Playing around with this equation we note that we can write mx¨ = dt ∂x˙ T . We may thus rewrite (2.1) as d ∂ ∂ T (x ˙) = − V (x, t). dt ∂x˙ ∂x Note that since T does not depend on x and V does not depend onx ˙ we can rewrite the equation further as d ∂ ∂ − (T − V ) = 0. (2.2) dt ∂x˙ ∂x This funny looking equation will be the starting point for this course. The difference d ∂ ∂ L = T − V we will call the Lagrangian, and the differential operator dt ∂x˙ − ∂x will be obtained from the principle of least action. We will find that (2.2) is more general than meets the eye. Especially, it will look the same irrespective of the coordinate system that we are working in. The same thing can not be said for Newton II, which becomes much more complicated when the coordinate system is not the Cartesian one.

2.2 The principle of least action

The starting point for the reformulation of classical mechanics is the principle of least action, which may be somewhat ﬂippantly paraphrased as “The world is lazy”, or in

8 the more ﬂowery words of Pierre Louis Maupertuis (1744), Nature is thrifty in all its actions:

The laws of movement and of rest deduced from this principle being precisely the same as those observed in nature, we can admire the application of it to all phenomena. The movement of animals, the vegetative growth of plants . . . are only its consequences; and the spectacle of the universe becomes so much the grander, so much more beautiful, the worthier of its Author, when one knows that a small number of laws, most wisely established, suﬃce for all movements.

This very general formulation does not in itself have any predictive power, but the idea that nature’s “thrift” could be used to derive laws of motion had already been successfully applied in optics for a long time:

Fermat’s principle The path taken between two points by a ray of light is the path that can be traversed in the least time.

This principle was first formulated by Ibn al-Haytham (aka Alhacen) in his Book of Optics from 1021, which formed one of the main foundations of geometric optics and the scientific method in general. He proved that it led to the law of reflection. It was restated by Pierre de Fermat in 1662, who also derived Snell’s law of refraction from this principle.

Example 2.1 Fermat’s principle of least time Consider a beam of light traveling across a planar interface from a point A in one medium (e.g. air) in which the speed of light is v1, to a point B in a diﬀerent medium (e.g. water) in which the speed of light is v2. What trajectory will minimise the time taken for the light to travel from A to B? The light will travel in a straight line in medium 1 and a straight line in medium 2, but we can vary the point O to try and minimise the time.

A x1

y 1 θ 1 v1 O v θ 2 2 y 2

x2 d B

Since A and B are ﬁxed y1 and y2 are ﬁxed and x1 + x2 = d, but we can vary x1 and x2 by moving the point O though only one of them is independent as x2 = d − x1.

9 The time taken for the light to travel from A to O, t1, is the length of AO, which is p 2 2 x1 + y1, divided by the speed of light in the medium 1,

p 2 2 x1 + y1 t1 = . v1 Similarly the time taken for light to travel from O to B is

p 2 2 p 2 2 x2 + y2 (d − x1) + y2 t2 = = . v2 v2 Hence the total time to travel from A to B is

p 2 2 p 2 2 x1 + y1 (d − x1) + y2 t = t1 + t2 = + . v1 v2

Now y1 and y2 are ﬁxed and we can minimise t(x1) by varying x1 and demanding that dt = 0. Now dx1

dt x (d − x ) x v p(d − x )2 + y2 − (d − x )v px2 + y2 = 1 − 1 = 1 2 1 2 1 1 1 1 . dx p 2 2 p 2 2 p 2 2p 2 2 1 v1 x1 + y1 v2 (d − x1) + y2 v1v2 x1 + y1 (d − x1) + y2

This vanishes, for ﬁnite x1, only when q q 2 2 2 2 x1v2 (d − x1) + y2 = (d − x1)v1 x1 + y1 (2.3) q q 2 2 2 2 ⇒ x1v2 x2 + y2 = x2v1 x1 + y1 x1 x2 ⇒ v2 = v2 p 2 2 p 2 2 x1 + y1 x2 + y2

⇒ sin θ1v2 = sin θ2v1,

or sin θ v n 1 = 1 = 2 , sin θ2 v2 n2

where n1 is the refractive index of medium 1 and n2 the refractive index of medium 2. This is Snell’s law for refraction (attributed to the Dutch astronomer Willebrord Snellius (1580-1626) but ﬁrst discovered in 984AD by the Persian scholar Ibn Sahl — it should be called Sahl’s law!). Snell’s law follows from the assumption that the light travels in a manner that minimises the time taken to go from A to B (we leave it as an exercise to show that this is a minimum and not a maximum, i.e. check that d2t 2 > 0 when x1 is given by (2.3). This way of viewing refraction, as minimising the dx1 travel time of a light beam, is known in optics as Fermat’s principle or the principle of least time. The law of reﬂection can be derived the same way.

10 2.2.1 Hamilton’s principle

In mechanics the proper mathematical formulation of Maupertuis’ principle is due to William Rowan Hamilton1, building on earlier work by Joseph Louis Lagrange. We will denote the kinetic and potential energy of a particle, or of a mechanical system in general, as

T = kinetic energy V = potential energy

dxi T usually depends on the velocities vi = dt ≡ x˙ i T = T (x ˙ i) but may also depend on position and explicitlyt on time T = T (x ˙ i, xi, t) V usually depends on the positions xi V = V (xi) but may also depend on velocities and explicitlyt on time V = V (x ˙ i, xi, t) (for example with time-varying external forces).

xi andx ˙ i here denote all the coordinates and their time derivatives. So for example we have

xi → x for a single particle in one dimension

xi → {x, y, z} for a particle in three dimensions

xi → {x1, y1, z1, x2, y2, z2, . . . , xN , yN , zN } for N particles in three dimensions

We now deﬁne the lagrangian L as the diﬀerence between kinetic and potential energy,

L(xi, x˙ i, t) = T − V. (2.4)

Note that L will be a function of the coordinates xi, the velocitiesx ˙ i, and the time t, although in many cases there is no explicit time dependence; ie, if we know the positions and velocities of all the particles we know the lagrangian.

A particular path is given by specifying the coordinates xi as a function of time, xi = xi(t). (Note that if xi(t) is known, its derivativex ˙ i(t) is also known.) For a given path, the action S is deﬁned as

t Z 2 S[x] ≡ L(x(t), x˙(t), t)dt . (2.5)

We are now in a position to formulate Hamilton’s principle of least action.

1On a General Method in Dynamics, Phil. Trans. Roy. Soc. (1834) 247; (1835) 95.

11 The principle of least action: The physical path a system will take between two points in a certain time interval is the one that gives the smallest action S.

Comments:

1. The potential energy V is deﬁned only for conservative forces, so the action as it is written here is deﬁned only for conservative forces. It is possible to generalise this to certain non-conservative forces and obtain the correct equations of motion (we will see examples of this later). However, all microscopic (fundamental) forces are conservative.

2. The action S is a “function of a function” since it depends on the function(s) xi(t). We call this a functional, and denote it by putting the function argument in square brackets, S = S[x].

2.3 The Euler–Lagrange equations

What does ‘the path that gives the smallest action’ actually x (t ) mean, and how can we ﬁnd it? To work this out, let us x’(t) 2 2 consider a path x(t) and another path x0(t) = x(t) + αh(t), where h(t) is some arbitrary smooth function of t, and α x(t) is a parameter that we will vary.

x t 1( 1) Since we are looking for the path the system will take between two speciﬁc points in a speciﬁc time interval, the endpoints of the two paths must be the same. We therefore have 0 0 x(t1) = x (t1) = x1 ; x(t2) = x (t2) = x2 ⇐⇒ h(t1) = h(t2) = 0 . (2.6) We can now write S[x + αh] = S(α), and treat it as a function of the parameter α. For dS a given h(t), the minimum of S will occur when dα = 0. This allows us to restate the principle of least action: For any smooth hi(t) with hi(t1) = hi(t2) = 0, the physical path xi(t) is such that

t2 d d Z S[x + αh] = L(x + αh , x˙ + αh˙ , t)dt = 0 . (2.7) dα dα i i i i t1

We often use the shorthands αh = δx and S[x + δx] − S[x] = δS = the variation of S, δS and call δx the functional derivative of S. The principle of least action is then often written as δS d δS = 0 or = 0 ⇐⇒ S[x + αh] = 0 for any h(t) . (2.8) δx dα

12 Let us now calculate the variation δS. For a single particle in one dimension, we have

d d Z t2 S[x + αh] = L(x + αh, x˙ + αh,˙ t)dt (2.9) dα dα t1 Z t2 ∂L ∂L = h + h˙ dt (2.10) t1 ∂x ∂x˙ Z t2 ∂L ∂L t=t2 Z t2 d ∂L = h dt + h − h dt (2.11) ∂x ∂x˙ dt ∂x˙ t1 t=t1 t1 Z t2 ∂L d ∂L = − h(t)dt . (2.12) t1 ∂x dt ∂x˙ In the ﬁrst step we used that L is a function of the three variables x, x,˙ t, but t does not depend on α. We can then use the chain rule for a function of two variables, d ∂f dx ∂f dy f(x, y) = + . dα ∂x dα ∂y dα In the second step we used integration by parts, Z Z ∂L uvdt˙ = uv − uvdt˙ with u = , v = h . ∂x

In the ﬁnal step the boundary term vanishes since h(t1) = h(t2) = 0. But h(t) is a completely arbitrary smooth function, and we must have δS = 0 for any h(t). This is only possible if the term within the brackets in (2.12) is 0 for all t, ie

d ∂L ∂L − = 0 The Euler–Lagrange equation (2.13) dt ∂x˙ ∂x

If we have N coordinates xi, the derivation proceeds following the same steps. Using the chain rule for a function of 2N variables, we ﬁnd d L(x + αh , x + αh , . . . , x + αh , x˙ + αh˙ , x˙ + αh˙ ,..., x˙ + αh˙ ) dα 1 1 2 2 N N 1 1 2 2 N N ∂L ∂L ∂L ∂L ˙ ∂L ˙ ∂L ˙ = h1 + h2 + ··· + hN + h1 + h2 + ··· + hN ∂x1 ∂x2 ∂xN ∂x˙ 1 ∂x˙ 2 ∂x˙ N N X ∂L ∂L = h + h˙ . (2.14) ∂x i ∂x˙ i i=1 i i Using integration by parts on the second term (for each i) gives us

N d X Z t2 ∂L d ∂L S[x + αh] = − h (t)dt = 0 . (2.15) dα ∂x dt ∂x˙ i i=1 t1 i i

Since all the hi are independent, arbitrary functions, the expression within the brackets must vanish for each i:

13 d ∂L ∂L − = 0 for all i = 1,...,N. (2.16) dt ∂x˙ i ∂xi

Example 2.2 Particle in a potential The kinetic energy of a single particle is 1 1 1 T = mv2 = m(v2 + v2 + v2) = m(x ˙ 2 +y ˙2 +z ˙2) . (2.17) 2 2 x y z 2 We take an arbitrary potential energy V = V (x, y, z, t). The Euler–Lagrange equations are d ∂L ∂L d ∂L ∂L d ∂L ∂L − = 0 ; − = 0 ; − = 0 . (2.18) dt ∂x˙ ∂x dt ∂y˙ ∂y dt ∂z˙ ∂z We ﬁnd ∂L ∂L ∂V d ∂L ∂L d(mx˙) ∂V ∂V = mx˙ ; = − ⇒ − = + = 0 ⇔ p˙ = − (2.19) ∂x˙ ∂x ∂x dt ∂x˙ ∂x dt ∂x x ∂x

where px = mx˙ is the x-component of the momentum. Likewise we get ∂V ∂V p˙ = − , p˙ = − or y ∂y z ∂z d~p = −∇V = F,~ Newton’s 2nd law! (2.20) dt This is the correct way to write Newton’s second law. The force equals the rate of change of momentum. This is correct even if m(t) is a function of time. Only when m is constant to we get the more familiar

m~a = m~r¨ = F.~

So the Euler–Lagrange equations are exactly equivalent to Newton’s laws.

So what is the point?

1. The equations are often easier: We get rid of complicated vectors and forces, and derive everything from scalars (energy). 2. It is easier to generalise to systems with constraints. 3. We can choose whichever coordinates we want. 4. The lagrangian formalism can be generalised to quantum mechanics (in the Feyn- man formulation: all paths are possible, but weighted by the action) and ﬁeld theory (with inﬁnitely many degrees of freedom).

We will look at points 2 and 3 next.

14 2.4 Generalised coordinates

It is often advantageous to change variables from the cartesian coordinates {xi, yi, zi} for each particle i = 1,...,N to some other variables {qj}, j = 1, . . . , n. These are called generalised coordinates. Consider for example a system of N particles. We need 3N independent coordinates to describe the system completely: we say that there are 3N degrees of freedom. Now, imagine that there is a constraint relating the 3N coordinates, for example:

1. Two particles are tied together with a rod of length l, so that

2 2 2 2 (x1 − x2) + (y1 − y2) + (z1 − z2) = l . (2.21)

2. The N particles are all moving on the surface of a sphere, ie

2 2 2 2 xi + yi + zi = R ∀i = 1,...,N. (2.22)

3. A ball in a squash court, 0 ≤ x ≤ L, 0 ≤ y ≤ L, z ≥ 0.

The ﬁrst two of these can be described by M equations of the form

fj(~x1, . . . , ~xN , t) = 0 , j = 1,...,M. (2.23) Such constraints are called holonomic (or integrable) constraints, and we will mostly focus on such constraints in the following (though a general procedure for dealing with non-holonomic constraints is described in §2.5). With such constraint equations, the coordinates xi, yi, zi are no longer independent. Instead we have

M relations =⇒ n = 3N − M real degrees of freedom.

By choosing n suitable generalised coordinates to describe these degrees of freedom, we achieve two things:

• We eliminate the forces of constraints which are required in the newtonian formulation. No net work is done by these forces, so they can safely be eliminated. • The Euler–Lagrange equations look exactly the same in the new coordinates, so the problem is no more diﬃcult (and probably easier) than the original one.

In the ﬁrst example above, the constraint (2.21) reduces the number of degrees of freedom from 6 to 5. The 5 coordinates can for example be chosen to be the centre of mass coordinates X,Y,Z for the two particles, and two angles θ, φ that describe the orientation of the rod.2 In the second example, each particle is described by 2 instead of 3 coordinates. These can be chosen to be the ‘latitude’ θ and ‘longitude’ φ of each particle (corresponding to spherical coordinates, see Sec. 2.4.2).

2In Chapter 5 we will look more at how these angles can be chosen.

15 Example 2.3 Simple pendulum Consider a simple pendulum with length `, mass m in a constant gravitational ﬁeld g (see Fig. 2.1). Here it is convenient to choose the angle θ as our coordinate. The x (horizontal) and z (vertical) coordinates A and their time derivatives can be written in terms of θ r A A ...... as ...... A ` θ A x = ` sin θ x˙ = `θ˙ cos θ , (2.24) A A z = −` cos θ z˙ = `θ˙ sin θ . (2.25) A A AA m The kinetic energy is | 1 1 T = m~v2 = m(x ˙ 2 +z ˙2) 2 2 Figure 2.1: A simple pen- 1 1 = m`2θ˙2(cos2 θ + sin2 θ) = m`2θ˙2 . (2.26) dulum 2 2 The potential energy is

V = mgz = −mg` cos θ . (2.27)

The lagrangian therefore becomes

1 L = T − V = m`2θ˙2 + mg` cos θ . (2.28) 2

The Euler–Lagrange equation is ∂L d ∂L d = =⇒ −mg` sin θ = m`2θ˙ (2.29) ∂θ dt ∂θ˙ dt g =⇒ θ¨ = − sin θ . (2.30) `

This is the equation of motion for the pendulum.

Once we have found the equation of motion for θ, and the solution to this equation, we can go back and calculate x and z as functions of time. However, in the example of the simple pendulum, we are not usually interested in this. We note that the mass m does not appear in the equation of motion. We could have seen this already by inspecting the lagrangian: the EL equations are unchanged if the lagrangian is multiplied by an overall constant α, L → αL. In this case, since the mass just enters as an overall factor in the lagrangian, the EL equation will not depend on the mass.

16 Solutions to the equations of motion?

Now we have found the equation of motion for the simple pendulum, and we may want to know the solutions to this equation, ie what the actual motion of the pendulum is for diﬀerent initial conditions. It is actually possible to integrate the equation (2.30) and write down a solution, but this involves elliptic integrals and lots of other complicated maths, and will not help us to understand the physical system. It will be more useful to ﬁnd numerical solutions, and in Computational Physics MP354 we will learn how this can be done. What we can do to understand the system better, is

• look at the general types of solutions we may have. We will do this when we discuss conservation of energy;

• consider limiting cases such as small oscillations. This is what we will do now.

If θ is small, we may approximate sin θ with the ﬁrst term in its power expansion (Taylor expansion), 1 1 sin θ = θ − θ3 + θ5 + · · · ≈ θ . (2.31) 3! 5! In that case (2.30) simpliﬁes to g θ¨ = − θ . (2.32) ` We recognise this as the equation for a simple harmonic oscillator,x ¨+ω2x = 0, with x → θ, ω2 → g/`. We therefore see that for small oscillations, the simple pendulum behaves p as a simple harmonic oscillator with angular frequency ωs = g/`, ie the frequency is inversely proportional to the square root of the length of the pendulum (and independent of the mass).

17 Example 2.4 Double Atwood machine Consider the double Atwood machine in Fig. 2.2. We assume that: ...... • the pulleys are light, so we can ...... ignore their kinetic energy; ...... x...... and x

• the ropes do not slip (or they slide `1 − x without friction). ? m1 ...... Here we have two independent degrees of ...... ? . .. .. freedom, which we can choose to be x and ...... s ...... y. In terms of these, the positions of the ...... `2 − y three blocks are y ? m x1 = −x , 3

x2 = −(`1 − x + y) , ? x3 = −(`1 − x + `2 − y) . m2 The kinetic and potential energy of the three blocks are 1 Figure 2.2: Double Atwood machine. T = m x˙ 2 1 2 1 1 d 1 T = m (` − x + y)2 = m (y ˙ − x˙)2 2 2 2 dt 1 2 2 1 T = m (x ˙ +y ˙)2 3 2 3 V1 = −m1gx

V2 = −m2g(`1 − x + y)

V3 = −m3g(`1 + `2 − x − y).

The lagrangian becomes

1 1 L = (m + m + m )x ˙ 2 + (m + m )y ˙2 + (m − m )x ˙y˙ 2 1 2 3 2 2 3 3 2 + (m1 − m2 − m3)gx + (m2 − m3)gy + m2g`1 + m3g(`1 + `2) .

Note that the last two terms are constants which do not play any role in the equations of motion. We get two equations of motion: d ∂L ∂L = (m + m + m )¨x + (m − m )¨y = = (m − m − m )g dt ∂x˙ 1 2 3 3 2 ∂x 1 2 3 d ∂L ∂L = (m + m )¨y + (m − m )¨x = = (m − m )g. dt ∂y˙ 2 3 3 2 ∂y 2 3

18 Example 2.5 Pendulum with rotating support ...... Consider a pendulum mounted on the edge of a disc ...... ...... ω C ... with radius a, rotating with constant angular velocity .. ... .. .. a .. .. . ...... sC . . .. ωt . . . . ω (see Fig 2.3). If the support point is in the hori- . . . C...... zontal position at t = 0, the angular position of the .. .. C ... s ...... C ...... support point at time t is φ = ωt, and the cartesian ...... C ...... coordinates of the bob at time t are ...... C C ...... C ` x = a cos ωt + ` sin θ θ C z = a sin ωt − ` cos θ C C C giving the velocities C C ˙ C m x˙ = −aω sin ωt + `θ cos θ C ˙ z˙ = aω cos ωt + `θ sin θ. | Figure 2.3: Pendulum with This gives us the lagrangian rotating support. 1 L = T − V = m(x ˙ 2 +z ˙2) − mgz 2 m = a2ω2 + `2θ˙2 + 2aω`θ˙ sin(θ − ωt) − mga sin ωt − ` cos θ. 2 This system has only one degree of freedom θ, but the lagrangian depends explicitly on time because of the rotation of the support point. The Euler–Lagrange equation is d ∂L d = m`2θ˙ + maω` sin(θ − ωt) = m`2θ¨ + maω`(θ˙ − ω) cos(θ − ωt) dt ∂θ˙ dt ∂L = = maω`θ˙ cos(θ − ωt) − mg` sin θ ∂θ aω2 g ⇒ `θ¨ − aω2 cos(θ − ωt) = −g sin θ =⇒ θ¨ = cos(θ − ωt) − sin θ ` ` Finding the equation of motion for this system becomes a bit complicated, but it is still far simpler than it would have been to compute the forces at each point and use Newton’s second law. We can check that our result is sensible by seeing what happens if there is no rotation, ie ω = 0. In this case the system reduces to the simple pendulum, and the equation of motion should be the same. We can immediately see that this is the case. It is worth noting that the potential energy contains a time-dependent term mga sin ωt, which one naively would think should contribute to the dynamics of the system — however, it plays no role since it does not contain the coordinate θ. There is also a constant term ma2ω2/2 in the kinetic energy which plays no role. NB: If you removed a sin ωt from the deﬁnition of z, you would for consistency also need to remove aω cos ωt fromz ˙, and this will change the dynamics.

19 2.4.1 The shortest path between two points (optional)

In deriving the Euler-Lagrange equations (2.16) we did not actually make use of the deﬁnition of L in terms of T and V : we could have used any functional evaluated along the path between the two points — for example the length of the path itself!

Example 2.6 The shortest path between two points

Consider a curve y = y(x) between two points (x1, y1) and (x2, y2). The length ds of an inﬁnitesimal segment (dx, dy) of this curve is given by Pythagoras: ds2 = dx2 + dy2 = dx2 + (y0(x)dx)2 = (1 + y0(x)2)dx2 (2.33) =⇒ ds = p1 + y0(x)2dx . (2.34) If, to make life simpler for ourselves, we assume that x is monotonically increasing along the curve, we ﬁnd that the total length of the curve is

Z x2 Z x2 q S = p1 + y0(x)2dx = L(y(x), y0(x), x)dx with L = 1 + y02 . (2.35) x1 x1 This looks like what we had before, but with t → x; x(t) → y(x);x ˙(t) → y0(x). The Euler–Lagrange equation becomes d ∂L ∂L − = 0 . (2.36) dx ∂y0 ∂y We see immediately that ∂L/∂y = 0. To ﬁnd ∂L/∂y0 we use the chain rule, ∂L dL ∂u √ = with u = 1 + y02 ; L = u ∂y0 du ∂y0 ∂L 1 y0 =⇒ = · 2y0 = . 0 p p ∂y 2 1 + y02 1 + y02

d ∂L To ﬁnd dx ∂y0 we use the product rule and the chain rule: ∂L 1 = vw with v = y0 , w = = u−1/2 (2.37) 0 p ∂y 1 + y02 d ∂L dv dw du dy0 dw du dy0 =⇒ = w + v = w + y0 dx ∂y0 dx du dx dx du dy0 dx 00 1 0 1 −3/2 0 00 = y p + y · − u · 2y · y 1 + y02 2 (2.38) 1 y02 y00 y02 = y00 − = 1 − p 02 3/2 p 02 1 + y02 (1 + y ) 1 + y02 1 + y y00 1 = . p 02 1 + y02 1 + y d ∂L So = 0 =⇒ y00(x) = 0 =⇒ y(x) = Ax + B. (2.39) dx ∂y0 This describes a straight line, so we have shown that the shortest path between two points is a straight line!

20 2.4.2 Polar and spherical coordinates

When we have rotational motion, or a system with rotational (or spherical) symmetry, it is very often most convenient to use polar coordinates (in 2 dimensions) or spherical coordinates (in 3 dimensions). The deﬁnition of these coordinates are given in Fig. 2.4. Since we will be using them often, we need to know what the kinetic energy of a particle is in terms of these coordinates.

z 6 y 6 ¡ ¡ ¡ r θ ¡ ...... ¡ ¡ ¡ ¨¨* ¡ ¨¨ r ¡ - . ... ¨ ... ¨.. θ y .. ¨ ...... ¨ ...... ¨ . - ...... x φ S S x ©

Figure 2.4: Plane polar coordinates (r, θ) (left) and spherical coordinates (r, θ, φ) (right).

Polar coordinates

The relation between cartesian and polar coordinates is given by

x = r cos θ =⇒ x˙ =r ˙ cos θ − rθ˙ sin θ (2.40) y = r sin θ =⇒ y˙ =r ˙ sin θ + rθ˙ cos θ (2.41)

This gives for the kinetic energy,

1 T = m(x ˙ 2 +y ˙2) 2 m = (r ˙2 cos2 θ + r2θ˙2 sin2 θ − 2rr˙θ˙ cos θ sin θ +r ˙2 sin2 θ + r2θ˙2 cos2 θ + 2rr˙θ˙ cos θ sin θ) 2 m = (r ˙2 + r2θ˙2) . (2.42) 2

21 Spherical coordinates

The relation between cartesian and spherical coordinates is given by

x = r sin θ cos φ =⇒ x˙ =r ˙ sin θ cos φ + rθ˙ cos θ cos φ − rφ˙ sin θ sin φ (2.43) y = r sin θ sin φ =⇒ y˙ =r ˙ sin θ sin φ + rθ˙ cos θ sin φ + rφ˙ sin θ cos φ (2.44) z = r cos θ =⇒ z˙ =r ˙ cos θ − rθ˙ sin θ (2.45)

Using this we ﬁnd that 1 1 T = m(x ˙ 2 +y ˙2 +z ˙2) = m(r ˙2 + r2θ˙2 + r2φ˙2 sin2 θ) . (2.46) 2 2 The complete derivation is left as an exercise.

22 Example 2.7 Coriolis force Strange things can happen in a rotating co-ordinate system. Let x and y be Cartesian co-ordinates in 2-dimensions and consider changing to a rotating co-ordinate system x˜ = x cos ωt + y sin ωt, y˜ = −x sin ωt + y cos ωt, where ω is a constant angular frequency. Conversely we can express (x, y) in terms of (˜x, y˜), x =x ˜ cos ωt − y˜sin ωt, y =x ˜ sin ωt +y ˜cos ωt, (2.47) and of course x2 + y2 =x ˜2 +y ˜2. From (2.47) x˙ = (cos ωt)x˜˙ − (sin ωt)y˜˙ − ω(˜x sin ωt +y ˜cos ωt) y˙ = (sin ωt)x˜˙ + (cos ωt)y˜˙ + ω(˜x cos ωt − y˜sin ωt) from which we get x˙ 2 +y ˙2 = x˜˙ 2 + y˜˙2 + ω2(˜x2 +y ˜2) + 2ω(˜xy˜˙ − y˜x˜˙). In the rotating co-ordinate system the Lagrangian for a free particle of mass m is m L = x˜˙ 2 + y˜˙2 + ω2(˜x2 +y ˜2) + 2ω(˜xy˜˙ − yx˜˙) . 2

Thex ˜(t) equation of motion follows from ∂L ∂L = m(x˜˙ − ωy˜), = m(ωy˜˙ + ω2x˜) ∂x˜˙ ∂x˜ from which d ∂L ∂L = dt ∂x˜˙ ∂x˜ ⇒ x˜¨ = ω2x˜ + 2ωy.˜˙ (2.48) Similarly they ˜(t) equation of motion is y˜¨ = ω2y˜ − 2ωx.˜˙ Deﬁne a vector in the z-direction ω = ωzˆ, then we can write the equation of motion in the rotating co-ordinate system as d2˜r d˜r = ω2˜r − 2 ω × . dt2 dt The ﬁrst term is the centrifugal force and the second is known as the Coriolis force. The Coriolis force is responsible for forcing winds moving into the centre of an area of low pressure to spiral rather than to move in straight radial lines and causes the beautiful spiral pattern of hurricane clouds.

23 2.5 Lagrange multipliers [Optional]

Using the constraint equations to reduce the number of coordinates is usually the most straightforward way of handling constraints. But it is not always practical:

• It may not be straightforward to solve the constraint equations.

• The constraint equations may involve velocities.

• The constraint equations may be expressed as diﬀerential rather than algebraic equations.

• We may want to know the forces of constraint (for example, to ﬁnd out when they become too large or too small to physically constrain the system).

It is useful to develop a technique for handling such situations.

2.5.1 Constraints

Suppose the generalised co-ordinates are not all independent but are constrained in some way. In particular we suppose that under an inﬁnitesimal variation

N X Ai(q, t)δqi + B(q, t)δt = 0 (2.49) i=1 where Ai(q, t) and B(q, t) are given functions and q represents the whole set of generalised co-ordinates, q1, . . . , qN . This constraint aﬀects the variational approach: when a path qi(t) is varied by qi(t) → qi(t) + αhi(t), with t ﬁxed, set δt = 0 and δqi(t) = αhi(t) in (2.49) and this enforces a constraint on the variation X Ai(q, t)hi = 0. (2.50) i

The constraint (2.50) can be incorporated into the variational approach by adding some multiple of it to the variational equations arising from the Lagrangian

X d δL X ∂L X d δL X ∂L X h = h → h = h + λ(t) A (q, t)h dt δq˙ i ∂q i dt δq˙ i ∂q i i i i i i i i i i i i (2.51) and considering the N equations

d δL ∂L = + λ(t)Ai(q, t). (2.52) dt δq˙i ∂qi λ can be eliminated by choosing one of the equations, say i = N, and solving for λ,

1 d δL ∂L λ = − . (2.53) AN dt δq˙N ∂qN

24 An example of the Coriolis force. A hurricane is caused by a small area of very low pressure at the centre making very strong winds blow toward the middle. Since the Earth is rotating there is a component 2π of angular velocity normal to the plane, ω = ω0 cos θ, where ω0 = T with T = 24 hours and θ the co-latitude (i.e. latitude−90◦). ω points upwards in the northern and downwards in the southern hemisphere and the resulting Coriolis force −2ω × ˜r˙ makes the wind bend to the right in the northern hemisphere and to the left in the southern hemisphere. Can you work out which hemisphere the above storm is in? Exactly on the equator the Coriolis force would vanish as ω would have no normal component there.

25 This can now be used in the other N − 1 equations in (2.52) to give N − 1 equations for the N functions qi(t) and the function λ(t) has been eliminated. One other equation comes from including (2.49) explicitly in the form

N X Ai(q, t)q ˙i + B(q, t) = 0. (2.54) i=1

A complete formulation of the problem is now given by

d δL ∂L = + λ(t)Ai(q, t), i = 1,...,N − 1; (2.55) dt δq˙i ∂qi N X Ai(q, t)q ˙i + B(q, t) = 0, (2.56) i=1

with λ given by (2.53). These are N equation for the N functions qi.

Physically what is happening here is that constraints must be implemented by forces, Fi (large forces so that the internal dynamics of the system can never overcome the force), and (2.52) is just a way of writing

d δL ∂L = + Fi, i = 1,...,N − 1, (2.57) dt δq˙i ∂qi and the constraining forces being applied externally are Fi = λAi. There may be more than one constraint, suppose there are M of them (M < N)

N X Aai(q, t)δqi + Ba(q, t)δt = 0 (2.58) i=1 with a = 1,...,M. Then path variations are constrained by M equations X Aai(q, t)hi = 0. (2.59) i

In that case introduce M functions λa(t) and the above procedure easily generalises.

In principle (2.56) can be solved to eliminate one of the unknown functions qi(t), for given Ai and B, leaving N − 1 equations to deal with, but in practice the procedure can be rather complicated. The situation is much simpler if the constraints are holonomic. For a single constraint (M = 1) there are N + 1 independent functions in (2.49) but, as described in §2.4, the constraint is holonomic if it arises from varying a single function, f(q, t) = C, with C a constant,

N X ∂f ∂f δf(q, t) = δq + δt = 0. (2.60) ∂q i ∂t i=1 i

26 We can incorporate this into the dynamics by introducing a new generalised co-ordinate λ(t) and adding a term to the Lagrangian, L = T (q, q˙) − V (q) → Lλ = T (q, q˙) − V (q) + λ f(q, t) − C . (2.61)

˙ Since there is no λ in the Lagrangian Lλ (there is no dynamics associated with λ) its equation of motion is particularly simple

d δL δL 0 = λ = λ = f − C. (2.62) dt δλ˙ δλ The equation of motion for λ is the constraint. The remaining equations are

d δL δL δf = + λ(t) . (2.63) dt δq˙i δqi δqi

The function λ(t) is called a Lagrange multiplier. With M holonomic constraints, fa(q, t) = Ca (2.64) with Ca constants, we need M Lagrange multipliers λa(t) and

M d ∂L ∂L X ∂fa = + λ (t) (2.65) dt ∂q˙ ∂q ∂q a i i a=1 i The Euler–Lagrange equations with Lagrange multipliers

We now have N + M unknown functions qi(t), λa(t), but we also have N + M equations: the N EL equations (2.65) and the M constraint equations (2.64). This will therefore completely determine the dynamics of the system once the initial conditions are given.

If we know the Lagrange multipliers, we can ﬁnd the (generalised) constraint forces Fi. These are given by X ∂fa F = λ . (2.66) i ∂q a a i

Example 2.8 A hoop rolling down an inclined plane without slipping

Consider a hoop of radius R and mass m rolling ...... down an inclined plane without slipping as shown in ...... @ .. . . . @ . . . .. θ .. Fig. 2.5. The condition of no slipping relates x to θ, .. .. H ... ¡ ... H ...... H ...... H ...... ¡ ...... under an inﬁnitesimal change in x there is a corre- HjH...... x HH sponding change in θ H HH H H...... H δx = Rδθ (2.67) .. . H . Φ H

Figure 2.5: A hoop rolling down an inclined plane 27 without slipping. or δx − Rδθ = δ(x − Rθ) = 0, this is a holonomic constraint. This forces the velocities to satisfy x˙ = Rθ.˙ (2.68) The kinetic energy is the sum of the translational and the rotational kinetic energies, 1 1 T = mx˙ 2 + mR2θ˙2. (2.69) 2 2 If the plane has length l and is inclined at an angle Φ to the horizontal then the centre of mass of the hoop is always a distance R cos Φ above the point of contact. The potential energy is mgh where h = R cos Φ + (l − x) sin Φ is the height of the centre of mass of the hoop above the foot of the plane. The R cos Φ can be ignored, it is just a constant and adding a constant to the potential energy changes nothing. So we take the potential energy to be

V (x) = mg(l − x) sin Φ. (2.70)

Including a Lagrange multiplier for the constraint the Lagrangian is 1 1 L = mx˙ 2 + mR2θ˙2 − mg(l − x) sin Φ + λ(x − rθ) (2.71) λ 2 2 giving the equations of motion

mx¨ = mg sin Φ + λ, mR2θ¨ = −λR.

From the second equation λ = −mRθ¨ and this can be used to eliminate λ from the ﬁrst

x¨ + Rθ¨ = g sin Φ.

Finally (2.68) tells us thatx ¨ = Rθ¨ and we only need solve simple equation 1 x¨ = g sin Φ (2.72) 2 to completely determine the motion. Assuming the hoop is initially at rest and starts rolling from the top of the plane the solution is 1 x(t) = (g sin Φ) t2. 4 The hoop arrives at the bottom of the plane after a time s l t = 2 . g sin Φ

28 with velocity v = plg sin Φ. This is actually rather a simple example because the constraint is linear in the generalised co-ordinates x and θ. We could just set Rθ˙ =x ˙ in the Lagrangian 1 1 L = mx˙ 2 + R2θ˙2 − V (x) = mx˙ 2 − V (x) 2 2 forget about λ and θ and just use the single co-ordinate x. The dynamics is exactly the same as for an unconstrained system with one degree of freedom, x, and twice the mass. For non-linear constraints however Lagrange multipliers are often the most eﬃcient way of solving the problem.

2.6 Canonical momenta and conservation laws

Assume the lagrangian L does not depend explicitly on the coordinate qi. Such coordinates are called cyclic. The Euler–Lagrange equations for the cyclic coordinate qi becomes d ∂L ∂L ∂L = = 0 =⇒ ≡ pi = constant . (2.73) dt ∂q˙i ∂qi ∂q˙i

We call the qunatity pi the canonical momentum conjugate to (or corresponding to) qi.

Why momentum?

Consider the ‘usual’ case where

1. we use cartesian coordinates qi = xi; 2. there are no constraints; and

3. the potential depends only on the coordinates, V = V (x).

In this case we have 1 X ∂L L = T − V = m x˙ 2 − V (x) =⇒ = mx˙ = p = ordinary momentum. 2 j ∂x˙ i i j i

So we have found the law of conservation of momentum pi if the potential V does not depend on the coordinate xi — ie, if the system is translationally invariant in the i- direction. Note that if V does not depend on xi this implies that there are no net forces in the i-direction. We may in a similar way demonstrate conservation of total momentum for a system of n particles if the potential energy does not depend on the centre of mass coordinate. But the concept of canonical momenta is much more general and powerful than this, and can be used to derive a whole host of other conservation laws. One of the most important is angular momentum, which we will look at next.

29 2.6.1 Angular momentum

Consider a one-particle rotationally symmetric 2-dimensional system, and let us use polar coordinates (r, θ) to describe the parrticle. Rotational symmetry then means that the potential energy V (r, θ) = V (r), independent of the angle θ. The lagrangian is then 1 L = T − V = m(r ˙2 + r2θ˙2) − V (r) . (2.74) 2

We see that θ is a cyclic coordinate, and the canonical momentum pθ is therefore conserved. What is this canonical momentum? We straightforwardly ﬁnd ∂L ∂ 1 2 ˙2 2 ˙ pθ = = mr θ = mr θ . (2.75) ∂θ˙ ∂θ˙ 2 ˙ But θ is the same as the angular velocity ω, and we know that the velocity vθ in the ˙ angular direction (perpendicular to the radius r) is vθ = rω = rθ, so pθ = r(mvθ). But this is exactly equal to the angular momentum of the particle,

Jz = (~r × ~p)z = mrvθ . (2.76)

So the canonical momentum conjugate to the angle θ is the angular momentum, which is conserved if the system is rotationally symmetric, ie the lagrangian does not depend on θ.

Angular momentum in spherical coordinates

In section 2.4.2 we found that the kinetic energy in spherical coordinates (see Fig. 2.4 is 1 T = m(r ˙2 + r2θ˙2 + r2 sin2 θφ˙2) . (2.46) 2 The angle φ corresponds to rotations about the z-axis: if a particle rotates about the z-axis, φ changes while r and θ are unchanged. If the potential energy does not depend on φ, we have rotational symmetry about the z-axis, and the canonical momentum pφ is conserved. From (2.46) we ﬁnd

2 2 ˙ ˙ pφ = mr sin θφ = r(mr sin θφ)(sin θ) . (2.77)

We now want to show that this is equal to the z-component of the angular momentum, Jz = (~r × ~p)z. We can put unit vectors (ˆr, θ,ˆ φˆ) in the direction of z increasing a coordinates at the point ~r and decompose 6 the velocity in its (r, θ, φ) components, ¡ rˆ φˆ££¡ ...... ¡£ ˆ ˆ ...φ...... ~v = vrrˆ + vθθ + vφφ (2.78) ...... ¡@ ¡ @ θ ¡ @R ...... ˆ The unit vectorr ˆ denotes the radial direction, ie the .....¡...... θ direction where r changes, while θ, φ are unchanged. ¡ Similarly, θˆ denotes the direction where θ changes ¡ Figure 2.6: Unit vectors in 30 spherical coordinates. while r, φ are unchanged, and φˆ denotes the direction where φ changes while r, θ are unchanged. The three vectors are orthogonal, and φˆ is also orthogonal toz ˆ, since motion in the φ-direction does not change z.

The velocity component vφ is the rotational velocity about the z-axis, which again is equal to the distance from the axis times the angular velocity about the axis. Since φ is the rotational angle about the z-axis, the angular velocity is dφ/dt = φ˙. The distance from the axis is r sin θ, so ˙ vφ = (r sin θ)φ . (2.79)

We can now work out the vector product ~r × m~v in the spherical coordinate system. Since ~r = rrˆ, we need the cross product ofr ˆ with each unit vector. Using the right-hand rule we ﬁnd rˆ × θˆ = φˆ , rˆ × φˆ = −θˆ , rˆ × rˆ = 0 . (2.80) Using these the angular momentum is

~ ˆ ˆ 2 ˙ ˆ ˙ ˆ J = ~r × (m~v) = m(rrˆ) × (vrrˆ + vθθ + vφφ) = mr (θφ − sin θφ θ) and the z-component of the angular momentum is therefore ˆ ˆ ˆ ˆ ˆ Jz = (~r × m~v)z = mr[ˆr × (vrrˆ + vθθ + vφφ)]z = mr[vθφ − vφθ]z = −mrvφθ · zˆ .

We now need to work out the scalar product θˆ · zˆ. Looking at the ﬁgure on the right, we see that since θ is the angle of ~r with the z ¡rˆ 6θ ¡ ˆ ...... z-axis, and θ is orthogonal to ~r (but still in the z − r plane), the ...... ¡. projection of θˆ onto the z-axis is θˆ · zˆ = − sin θ. ¡ ¡ . . H .. H... θ Therefore we ﬁnd that the z-component of the angular momentum H ˆ HHjθ is -sin θ ˙ Jz = (~r × m~v)z = −mrvφ(− sin θ) = r · (mr sin θφ) · sin θ = pφ . (2.81)

So the canonical momentum pφ is indeed the angular momentum about the z-axis, and it is conserved if we have rotational symmetry about the z-axis. If we have full spherical symmetry, this means we have rotational symmetry about all 3 3 axes, so by the same argument as above Jx and Jy must also be conserved. Therefore, for a spherically symmetric system, the angular momentum vector L~ = ~r×~p is conserved. Na¨ıvely one would think that if we have full rotational symmetry, the angle θ should also be irrelevant, and the canonical momentum pθ should also be conserved. However, this is not the case: although the potential energy does not depend on θ, the kinetic 1 2 2 ˙2 energy does, through the term 2 mr sin θφ . This θ-dependence is an artefact of how 3This is not obvious in the chosen co-ordinate system. It is left as an exercise to show that the x and y components of angular momentum are

2 ˙ ˙ Jx = −mr (θ sin φ + φ cos θ sin θ cos φ) 2 ˙ ˙ Jy = mr (θ cos φ − φ cos θ sin θ sin φ) and that these are constant if V (r) is independent of θ and φ. Our choice of spherical co-ordinates singles out z and makes the symmetry around the z-axis obvious, but it hides a similar symmetry about the x and y-axes. This is an important lesson — sometimes a Lagrangian can have a hidden symmetry that is not obvious in the chosen co-ordinate system.

31 we have chosen the coordinate system, but it is an unavoidable artefact: no matter how we choose our spherical coordinate angles, these coordinates must break the full spherical symmetry somehow. We realise the full symmetry by noting that we could have chosen the coordinates differently, eg we could have chosen θ to be the angle with the x-axis and φ to correspond to rotations about the x-axis — which would have led us to find that Lx is conserved. Similarly, if we choose θ to be the angle with the y-axis we will find that Ly is conserved.

2.7 Energy conservation: the hamiltonian

We know that when we have conservative forces, the potential energy depends only on positions, and not on time, and the total energy is conserved. We have derived conservation of linear and angular momentum in lagrangian mechanics, so we may ask ourselves if we can also derive energy conservation within the same framework? The answer to this is that not only can we do this, but the energy conservation theorem we arrive at is more general than the one we already know! To see how this works, let us take the (total) time derivative of the lagrangian L = L(qi(t), q˙i(t), t). Using the chain rule and the Euler–Lagrange equations we get

dL X ∂L X ∂L ∂L = q˙ + q¨ + dt ∂q i ∂q˙ i ∂t i i i i X d ∂L X ∂L dq˙i ∂L = q˙i + + (2.82) dt ∂q˙ ∂q˙ dt ∂t i i i i d X ∂L ∂L = q˙ + dt ∂q˙ i ∂t i i d X ∂L ∂L dH ∂L ⇐⇒ q˙ − L + ≡ + = 0 , (2.83) dt ∂q˙ i ∂t dt ∂t i i where we have deﬁned

X ∂L X H = q˙ − L = p q˙ − L = the hamiltonian (2.84) ∂q˙ i i i i i i

So we ﬁnd that if the lagrangian does not depend explicitly on time, then the hamiltonian or energy function H is conserved. To see how this relates to energy conservation as we know it from before, consider a system of particles in cartesian coordinates, described by the lagrangian

1 X L = T − V = m q˙2 − V (q) . 2 i i i

32 The hamiltonian for this system is

X ∂L X h1 X 2 i H = q˙i − L = (miq˙i)q ˙i − miq˙i − V (q) ∂q˙i 2 i i i (2.85) 1 X = m q˙2 + V (q) = T + V. 2 i i i So we ﬁnd that the hamiltonian is equal to the total energy, so conservation of the hamiltonian is the same as energy conservation in this particular (most common) case.

2.7.1 When is H conserved?

We found that H is conserved if L does not depend explicitly on time, ie L(q, q,˙ t) = L(q, q˙). We would like to understand in what circumstances an explicit time dependence could appear in the lagrangian. One possibility would be that the potential energy depends explicitly on time in the ﬁrst place. But there are also other possibilites. The kinetic energy, written in terms of the original cartesian (or, for that sake, ordinary polar or spherical coordinates) does not have any explicit time dependence. But time dependence could still appear in either the kinetic or the potential energy when we write it in terms of generalised coordinates. To see how this can happen, let us recall why we introduced generalised coordinates in the ﬁrst place:

1. Constraints: There are fewer actual degrees of freedom in the system because of constraints. We use generalised coordinates to denote the real (relevant) degrees of freedom. An example of this would be the pendulum, where the original x and z coordinates are reduced to the single coordinate θ.

2. Symmetries: There are symmetries in the system which mean that using non- cartesian coordinates may give a simpler description. An example of this would be using polar coordinates for a system with rotational symmetry.

Explicit time-dependence can appear in both those types of cases, leaving us with three possibilites for how explicit time-dependence could appear in the lagrangian:

1. The potential energy is explicity time-dependent, V = V (x, t). Physically, this means that there are external or non-conservative forces, so the energy of the system is not conserved.

2. The constraints are time-dependent. An example of that would be Example 5, the pendulum with rotating support. In such cases, external forces are usually required to maintain the constraint, so the energy of the system is not conserved.

3. We have chosen to use time-dependent transformations xi = fi(q, t) between the old coordinates x and the new coordinates q because this may simplify the description of the system. In this case, the hamiltonian may not be conserved even if the total energy is conserved.

33 2.7.2 The Energy and H

We showed the hamiltonian H is equal to the total energy E = T + V when

1 X L = T − V = m q˙2 − V (q) . 2 i i i More generally, it is the case when

1. V is independent of the velocitiecq ˙i, V = V (q, t), and 2. T is a homogeneous quadratic function of q˙, X T = aij(q, t)q ˙iq˙j,. ij

Proof

Take X L = T − V = aij(q, t)q ˙iq˙j − V (q, t) (2.86) ij

We note that we can always arrange it so that aij = aji, sinceq ˙iq˙j =q ˙jq˙i. The canonical momenta are ∂L X X X p = = a q˙ + a q˙ = 2 a q˙ . (2.87) k ∂q˙ ik i kj j kj j k i j j The two terms appear because we get a contribution both from the k = j term and from the k = i term in the sum. The hamiltonian is then X X X H = piq˙i − L = 2 aijq˙iq˙j − aijq˙iq˙j + V (q, t) = T + V = E, (2.88) i ij ij which completes the proof. We have proven that, if the potential V is independent of the velocities and the kinetic energy is a homogeneous quadratic function of the velocities, then H = E = T + V .

Example 2.9 Spring mounted on moving platform Consider a body with mass m sitting at the end of a horizontal spring with spring constant k, with the other end attached to a ﬁxed point on a platform moving with a constant velocity v. Since one end of the spring is ﬁxed to the moving platform, the equilibrium point x0 of the body on the spring is also moving with velocity v. If we say that x0 = 0 when t = 0, we therefore have x0 = vt.

The potential energy of the body is given by the displacement x−x0 from equilibrium, 1 2 1 2 V = 2 k(x−x0) = 2 k(x−vt) . The kinetic energy is the usual one, so the lagrangian is 1 1 L = T − V = mx˙ 2 − k(x − vt)2 . (2.89) 2 2

34 The canonical momentum is px = mx˙, which gives us the hamiltonian 1 1 H = p x˙ − L = mx˙ 2 + k(x − vt)2 = T + V = E. (2.90) x x 2 2 Since the lagrangian depends explicitly on time, the hamiltonian (and the total energy) is not conserved. We can understand this by noting that the motor driving the platform will have to do work to maintain a constant velocity; in the absence of this the platform will undergo oscillations along with the body attached to the spring. We can now introduce a new coordinate

q = x − vt =⇒ q˙ =x ˙ − v (2.91) 1 1 1 1 mv2 =⇒ L(q, q,˙ t) = m(q ˙ + v)2 − kq2 = mq˙2 + mvq˙ − kq2 + . (2.92) 2 2 2 2 2

The canonical momentum is pq = m(q ˙ + v), and the hamiltonian is

1 1 mv2 1 1 mv2 H = pq˙−L = mq˙2+mvq˙− mq˙2−mvq˙+ kq2− = mq˙2+ kq2− . (2.93) q 2 2 2 2 2 2 When written in terms of q, the lagrangian does not depend explicitly on time, and therefore the hamiltonian (2.93) is conserved! However, it is not equal to the total energy, rather

Hq = Hx + pqq˙ − pxx˙ = E + px(q ˙ − x˙) = E − mvx.˙

So, by changing coordinates, we have here traded a non-conserved hamiltonian, for a conserved hamiltonian that is not equal to E.

Example 2.10 Electrodynamics One case where the distinction between ordinary and canonical momentum is important is electrodynamics. A particle with charge Q moving with velocity ~v in an electric field E~ and a magnetic field B~ is F~ = Q(E~ + ~v × B~ ), (2.94) which is called the Lorentz force law. Using Maxwell’s laws, we can introduce the electrostatic and vector (‘magnetic’) potentials φ, A~: ∇ · B~ = 0 ⇐⇒ B~ = ∇ × A,~ (2.95) ∂B~ ∂A~ ∇ × E~ = − ⇐⇒ E~ = −∇φ − , (2.96) ∂t ∂t where φ is the electric potential, so the potential energy of a charged particle Q is ~ V = Qφ, and Ai(x, t) is called the vector potential for the magnetic field B. The Lorentz force law (2.94) can be derived from the Lagrangian 3 3 1 X X L = m x˙ 2 − Qφ + Q A x˙ . (2.97) 2 i i i i=1 i=1

35 We have ! ∂L d ∂L ∂Ai X ∂Ai = mx˙ + QA ⇒ = mx¨ + Q + x˙ ∂x˙ i i dt ∂x˙ i ∂t j ∂x i i j j

∂L ∂φ X ∂Aj = −Q + Q x˙ ∂x ∂x j ∂x i i j i giving equations of motion ∂φ ∂Ai X ∂Aj ∂Ai mx¨ = −Q − Q + Q x˙ − i ∂x ∂t j ∂x ∂x i j i j ~ = Q(E + ~v × B)i , which is the Lorentz force law (2.94), since

~ ! ∂A X ∂Aj ∂Ai (~v × B~ ) = ~v × (∇ × A) = ~v. − (~v.∇)A = x˙ − . i i ∂x i j ∂x ∂x i j i j

The canonical momentum is ∂L pi = = mx˙ i + QAi . (2.98) ∂x˙ i This is not the ordinary momentum, a distinction which becomes quite important in quantum mechanics, where it is the canonical momentum that enters into the commutation relations that are used to quantise the system. Note that in general

∂φ X ∂Aj mx¨ + QA˙ = −Q + Q x˙ i i ∂x j ∂x i j i so momentum is not conserved. The hamiltonian of the particle is

3 3 3 3 X X 1 X X H = p x˙ − L = (mx˙ + QA )x ˙ − m x˙ 2 + Qφ − Q A x˙ i i i i i 2 i i i i=1 i=1 i=1 i=1 (2.99) 3 1 X = m x˙ 2 + Qφ . 2 i i=1 We see that the vector (magnetic) potential does not contribute to the energy. Phys- ically this is because no net work is done by the magnetic ﬁeld,

~v.F~ = QE.~

36 2.8 Lagrangian mechanics — summary sheet

1. Lagrangian L = T − V = kinetic energy − potential energy. L = L(q, q,˙ t) is a function of the coordinates qi, their time derivatives q˙i and time t. 2. Generalised coordinates For a system of N particles, we may instead of cartesian coordinates ~ri = (xi, yi, zi), i = 1 ...N, use any set of coordinates

qj = fj(~r1, . . . , ~rN ) , j = 1 ...M. M is the number of degrees of freedom of the system. For an unconstrained system M = 3N, but if there are constraints then M < 3N. 3. Principle of least action Nature “chooses” the path q(t) that minimises the action

Z t1 S = L(q(t), q˙(t), t)dt t0 with q(t0) = q0, q(t1) = q1 kept ﬁxed, or S[q(t)] − S[q(t) + αh(t)] δS = lim = 0 α→0 α

for arbitrary h(t) with h(t0) = h(t1) = 0. This leads to 4. Euler–Lagrange equations d ∂L ∂L − = 0 dt ∂q˙i ∂qi 5. Canonical momentum ∂L pi = ∂q˙i (a) Linear momentum 1 P 2 If qi = xi and L = 2 i mix˙ i − V (x), then pi = mx˙ i . (b) Angular momentum If qi is a rotational angle φ about some axis, then pi is the angular momentum [L~ = ~r × (m~v)] about that axis. 6. Conservation laws From the Euler–Lagrange equations we see that if L does not depend explicitly on the coordinate qi then dp i = 0 ⇐⇒ p is conserved. dt i 37 7. Hamiltonian X H = piq˙i − L i If there are no time-dependent constraints or velocity-dependent forces (or potentials) then H = T + V = total energy. dH ∂L = − , dt ∂t so if the lagrangian L does not explicitly depend on time, then the hamiltonian H is conserved.

38 Chapter 3

Hamiltonian dynamics

The main idea in hamiltonian dynamics is that instead of using only the coordinates qi(t) and their derivatives to describe the system, we think of the coordinates qi and the momenta pi as independent variables.

This may seem like an odd idea, since once we know the coordinates qi(t) as a function of time, we also know their time derivativesq ˙i(t), and through that the momenta pi(t). So how can q and p be considered independent? One way of making sense of this is to note that knowing the position of a body at a particular time does not in itself tell us anything about its velocity (or momentum), or vice-versa. It is only if we know the position at several diﬀerent times that we will be able to work out its velocity. And the full relation between q(t) andq ˙(t), or between q(t) and p(t), can only be known if we know the coordinate q(t) at all times t — but this amounts to having solved the problem of the motion of the body! So in this sense, q and p (or, indeed, q andq ˙) can be considered independent variables. Secondly, the relation between coordinates and canonical momenta is not a simple one like the relation between a coordinate and its derivative, but is related to the dynamics of the system as encoded in the lagrangian (or, as we shall see, the hamiltonian). Therefore, it makes sense to consider p as independent of q in a way thatq ˙ cannot be. Thirdly, as you will see in quantum mechanics, the two must be treated as independent quantities there. From the quantum mechanical commutation relations between coordinates and canonical momenta, [qi, pk] = ihδ¯ jk, one can derive Heisenberg’s uncertainty relation, ∆qi∆pi ≥ h¯, which holds for all (q, p) pairs. Therefore, in quantum mechanics, it is impossible to know the coordinate and momentum of a particle at the same time. Also, statistical mechanics, which forms the basis of the modern treatment of thermal physics, is formulated in the phase space where coordinates and momenta are considered as independent variables. So our ﬁrst step in arriving at the hamiltonian formulation of mechanics will be to use the relation between momenta pi and velocitiesq ˙i to eliminate q˙, and write the hamiltonian as a function of coordinates and momenta,

H = H(qi, pi, t) as opposed to L = L(qi, q˙i, t) . (3.1)

This will be our starting point in this chapter.

39 3.1 Hamilton’s equations of motion

The Euler–Lagrange equations are equations of motion written in terms of the lagrangian. We now want to find similar equations in terms of the hamiltonian. To achieve this, we first note that the equations of motion are differential equations, so we can try to differentiate the hamiltonian H. Starting from the definition of H in terms of the lagrangian, N X H = piq˙i − L, (3.2) i=1 we note that the left hand side of this equation is now a function of q, p and t, while the right hand side is a function of q, q,˙ t and p, since p enters into the first term.

We now vary a solution qi,q ˙i (hence also pi) on both sides of (3.2) by a small amount. Allowing for possible explicit time dependence we should also vary t and the resulting inﬁnitesimal variation of H is

N N X X ∂L ∂L ∂L δH = δp q˙ + p δq˙ − δq + δq˙ − δt . (3.3) i i i i ∂q i ∂q˙ i ∂t i=1 i=1 i i Using the deﬁnition of the canonical momentum and the Euler–Lagrange equation,

∂L ∂L d ∂L dpi = pi and = = =p ˙i , (3.4) ∂q˙i ∂qi dt ∂q˙i dt we ﬁnd

N X ∂L δH = q˙ δp + p δq˙ − p˙ δq − p δq˙ − δt i i i i i i i i ∂t i=1 N X ∂L (3.5) = q˙ δp − p˙ δq − δt i i i i ∂t i=1 .

But H is a function of qi, pi, t, so using the chain rule we also have

N X ∂H ∂H ∂H δH = δdq + δdp + δt . (3.6) ∂q i ∂p i ∂t i=1 i i The two expressions (3.5) and (3.6) must be the same so comparing the two expressions we see, for any solution of the equations of motion

∂H ∂H ∂H ∂L q˙i = ;p ˙i = − ; = − . (3.7) ∂pi ∂qi ∂t ∂t The boxed equations are called Hamilton’s equations of motion or the canonical equations.

40 Using Hamilton’s equations of motion, it is very straightforward to show that the hamiltonian is conserved if it does not depend explicitly on time:

N N dH X ∂H ∂H ∂H X ∂H ∂H = q˙ + p˙ + = p˙ q˙ +q ˙ p˙ + = . (3.8) dt ∂q i ∂p i ∂t i i i i ∂t ∂t i=1 i i i=1

Example 3.1 Consider a particle constrained to move on the cylindrical surface x2 + y2 = R2, subject to a central force F~ = −k~r. Using cylinder coordinates (θ, z) with x = R cos θ, y = R sin θ we ﬁnd 1 1 V = kr2 = k(R2 + z2) , (3.9) 2 2 1 1 T = m(x ˙ 2 +y ˙2 +z ˙2) = m(R2θ˙2 +z ˙2) (3.10) 2 2 1 1 =⇒ L = m(R2θ˙2 +z ˙2) − k(R2 + z2) . (3.11) 2 2 The canonical momenta are

∂L 2 ˙ ∂L pθ = = mR θ , pz = = mz˙ . (3.12) ∂θ˙ ∂z˙ ˙ We can use this to ﬁnd θ, z˙ in terms of pθ, pz: p p θ˙ = θ , z˙ = z . (3.13) mR2 m The hamiltonian is p p 1 h p 2 p 2i 1 H = p z˙ − p θ˙ − L = p z + p θ − m R2 θ + z + k(R2 + z2) z θ z m θ mR2 2 mR2 m 2 p2 p2 1 1 = z + θ + kz2 + kR2 = H(z, p , p ) . (3.14) 2m 2mR2 2 2 z θ It is equal to the total energy since the potential energy does not depend on velocities and the kinetic energy has the usual form. It is conserved since there is no explicit time-dependence in L (or H). Hamilton’s equations of motion for this system are

∂H ˙ ∂H pθ p˙θ = − = 0 , θ = = 2 (3.15) ∂θ ∂pθ mR ∂H ∂H pz p˙z = − = −kz , z˙ = = (3.16) ∂z ∂pz m

We can use (3.15), (3.16) to arrive at p˙ k p˙ = mR2θ˙ = constant, z¨ = z = − z . (3.17) θ m m These are the Euler–Lagrange equations for the system, so we have shown that Hamilton’s equations are exactly equivalent to the Euler–Lagrange equations, as they should be.

41 In the example above we see that θ does not appear in the expression for H: it is a cyclic coordinate. This implies that the canonical momentum pθ is conserved, but in the hamiltonian framework it actually simpliﬁes the system even further: in the remaining equations we can simply treat pθ as any other constant, so the whole motion in the θ-coordinate decouples from the remaining equations: instead of 3 variables z, z,˙ θ˙ we now just have 2: z, z˙. This decoupling is a generic feature which can simplify the analysis of the system con- siderably, as we shall see below.

3.2 Cyclic coordinates and eﬀective potential

To see how this works out, let us look at a slightly more complex system, that of the spherical pendulum. This is a pendulum that can swing freely in all directions, not just in a plane. Using spherical coordinates (θ, φ), where θ is the angle with the vertical axis, we ﬁnd that the lagrangian of this system is 1 L = m`2(θ˙2 + φ˙2 sin2 θ) + mg` cos θ . (3.18) 2 The canonical momenta are

∂L 2 ˙ pθ = = m` θ (3.19) ∂θ˙ ∂L 2 2 ˙ pφ = = m` sin θφ (3.20) ∂φ˙

From (3.19), (3.20) we ﬁnd p p θ˙ = θ , φ˙ = φ . (3.21) m`2 m`2 sin2 θ Since there is nothing funny going on in this system, the hamiltonian is equal to the total energy, p2 p2 H = T + V = θ + φ − mg` cos θ . (3.22) 2m`2 2m`2 sin2 θ Hamilton’s equations of motion are then ∂H p ∂H p2 cos θ θ˙ = = θ , p˙ = − = φ − mg` sin θ , (3.23) 2 θ 2 3 ∂pθ m` ∂θ m` sin θ ∂H p ∂H φ˙ = = φ , p˙ = − = 0 . (3.24) 2 2 φ ∂pφ m` sin θ ∂φ

Since pφ is constant, the two last terms in (3.22) depend only on θ. They can be taken to define an effective potential, p2 V (θ) = φ − mg` cos θ (3.25) eff 2m`2 sin2 θ with dV p˙ = − eff θ dθ

42 (we use an ordinary derivative with respect to θ here, rather than a partial derivative, because it is understood that everything else is constant). Let us now look at a system with a single degree of freedom θ, with kinetic energy 2 2 pθ/(2m` ) and potential energy Veﬀ(θ). The hamiltonian of this system is exactly the same as (3.22), and therefore Hamilton’s equations of motion for θ, pθ are exactly the same as (3.23). 3

2 gl 1 /m eff V

k = 0 0 k = 0.02 k = 0.1 k = 0.5 k = 1.0 k = 1.5

-1 0 0.5 1 1.5 2 2.5 3 θ

Figure 3.1: The eﬀective potential for the spherical pendulum, for diﬀerent values of 2 2 3 k = pφ/2m g` .

We can use the effective potential to find out what types of motion are possible in the θ-direction. Figure 3.1 shows Veff as a function of θ for several values of pφ. We can see that for all pφ 6= 0 the effective potential goes to infinity at θ = 0 and θ = π. This means that only bounded motion exists for pφ 6= 0. The minimum of the potential corresponds to circular motion at a fixed angle θ, given by p2 mg` sin4 θ = φ cos θ . (3.26) m`2

In contrast, for pφ = 0 (the solid line), both bounded motion (through θ = 0) and unbounded motion are possible, for −mg` < E < mg` and E ≥ mg` respectively. In this case, the spherical pendulum reduces to the plane (simple) pendulum. What we have seen in this example is quite typical of what happens if one or more of the coordinates are cyclic. In general, if the hamiltonian can be written as

2 2 H(q1, q2, p1, p2) = f(q1)p1 + g(q1)p2 + V (q1) , (3.27) where f and g can be any function of the coordinate q1, then we immediately see that the second coordinate q2 is cyclic, and the momentum p2 is therefore conserved, ie it is

43 a constant. We can then deﬁne an eﬀective potential

2 Veﬀ(q1) = g(q1)p2 + V (q1) , (3.28) and the hamiltonian will be equivalent to that of a 1-dimensional system with kinetic energy T1 given by

2 T1 = f(q1)p1 =⇒ H(q1, p1) = T1 + Veﬀ(q1) . (3.29)

It is clear that the equations of motion of q1 and p1 we obtain from (3.29) are the same as what we obtain from (3.27) since the two hamiltonians are exactly the same: all we have done is group the terms in a diﬀerent way,

If are only interested in the motion in the q1 coordinate we can stop as soon as we have solved the equations of motion resulting from (3.29). If we also want to determine q2(t), we use its equation of motion,

Z t ∂H 0 0 q˙2 = = 2p2g(q1) =⇒ q2(t) = 2p2 g(q1(t ))dt . (3.30) ∂p2 q2(t0)

Once we have found q1(t) it is in principle straightforward to perform this integral to obtain q2(t).

3.3 Hamilton’s equations from a variational principle

We arrived at Hamilton’s equations from the lagrangian and using the Euler–Lagrange equations, but it is also possible to derive them directly from a variational principle, as we shall now see. To this end, we can rewrite the action S = R Ldt as Z S = pq˙ − H(q, p, t)dt . (3.31)

The equations of motion will now be derived by requiring δS = 0, where now p and q can be varied independently. We proceed analogously to how we derived the Euler–Lagrange equations in Section 2.3, using the shorthand δS instead of dS/dα: Z δS = δ pq˙ − H(q, p, t)dt

Z t2 h X X ∂H ∂H i = δp q˙ + p δq˙ − δp + δq dt i i i i ∂p i ∂q i t1 i i i i t (3.32) X Z 2 h d ∂H ∂H i = q˙ δp + (p δq ) − p˙ δq − δp − δq dt i i dt i i i i ∂p i ∂q i i t1 i i Z t2 X t2 X h ∂H ∂H i = piδqi + q˙i − δpi − p˙i + δqi dt = 0 . t ∂p ∂q i 1 i t1 i i

The ﬁrst term on the last line is zero because the endpoints qi(t1), qi(t1) are ﬁxed, so δqi(t1) = δqi(t2) = 0. In the second term, the integral must be zero for any arbitrary

44 variations δpi, δqi, which can both be chosen indenpendently for all degrees of freedom i. The only way this can be the case is if both terms inside the ordinary brackets are zero at all times, and for all i. This gives us ∂H ∂H q˙i = , p˙i = − , (3.33) ∂pi ∂qi which is Hamilton’s equations of motion.

3.4 Phase space [Optional]

We can consider the values of {p, q} = (p1, . . . , pN , q1, . . . , qN ) as coordinates of a point in a 2N-dimensional space. This space is called phase space. The evolution of the system in time can then be thought of as a trajectory in this space. The state of a system is given by its location in phase space, ie by the values of all the generalised coordinates and momenta — in other words, every point in phase space describes a unique state. There are two important aspects of this:

1. The trajectory of a system in phase space provides a complete description of the motion of the system.

2. The complete trajectory is uniquely determined by the initial coordinates {pi(t0), qi(t0)}. Mathematically, this is because Hamilton’s equations of motion constitute a set of ﬁrst order ODEs for the phase space coordinates as a function of time, and the solution of a ﬁrst order ODE is uniquely determined by the initial values. Physically, it is because the position and momentum together completely determine the motion of any particle or system.

An important corrollary of this is that trajectories in phase space never cross. Identical states will have identical futures (and identical pasts). The description of a system in terms of states in phase space is a cornerstone of statistical mechanics. It is worth looking at how this changes in quantum mechanics. Here, Heisenberg’s uncertainty relation ∆q∆p ≥ h/¯ 2 implies that a system cannot be identiﬁed with a unique point in classical phase space. This means that identical states can have diﬀerent futures. The appropriate description in quantum mechanics is in terms of vectors (points) in a Hilbert space of quantum states rather than in classical phase space, and it is possible to make connections between the two — but this goes way beyond the topic of this module.

45 Example 3.2 Particle in constant gravitational ﬁeld

p The hamiltonian is here given by

p2 H = + mgz . (3.34) 2m z Since H is conserved the trajectories in phase space are given by those curves in the (p, z) plane where H is constant. We can describe these curves in two (equivalent) ways:

Figure 3.2: Phase space trajecto- 1. In terms of initial coordinates p , z : ries for a particle in a constant 1 1 gravitational ﬁeld. p2 p2 + mgz = 1 + mgz (3.35) 2m 2m 1 p2 − p2 ⇐⇒ 1 = mg(z − z) (3.36) 2m 1 1 ⇐⇒ z = z + (p2 − p2) . (3.37) 1 2m2g 1 This describes a parabola. The resulting trajectories are shown in ﬁgure 3.2

2. In terms of the total energy E:

2 2 2 p z p p + mgz = E =⇒ mgz = E − =⇒ = 1 − 2 (3.38) 2m z0 p0

with

E √ z = ; p = 2mE = mp2gz . (3.39) 0 mg 0 0

46 Example 3.3 Plane pendulum The hamiltonian is

2 pθ p c) H = θ + mgl(1 − cos θ) = E, (3.40) 2m`2

b) where we have chosen the potential energy to a) be zero at the stable equilibrium point of the π θ pendulum. Reorganising this we ﬁnd

p2 θ = E − mg`(1 − cos θ) (3.41) 2m`2 p 2 1 θ =⇒ θ = 1 − sin2 , (3.42) Figure 3.3: Phase space trajecto- p0 Q 2 ries for a plane pendulum. Note the bifurcation point at pθ = with 0, θ = π. See text for discussion √ E of the trajectories a), b), c). p = ` 2mE , Q = . (3.43) 0 2mg` Depending on the value of Q, we have three classes of trajectories:

θ a) 0 < Q < 1. this gives ellipse-like trajectories: if we write q = sin 2 , then (3.42) can be rewritten as 2 2 pθ q p + = 1 , with q0 = Q. (3.44) p0 q0

The motion is bounded in both pθ and θ.

b) Q = 1. Here the trajectories touch at pθ = 0, θ = π — but strictly speaking they do not actually cross A system moving along the trajectory b) in Fig. 3.3 will take an inﬁnite amount of time to reach this point, and a system ﬁnding itself at this point will be in an unstable equilibrium. If a system does end up at this point and is subject to a small perturbation, it can continue either down along the solid line, or up along the dotted line. This is called a bifurcation point. Bifurcation points are important in chaos theory.

c) Q > 1. Here the motion is unbounded in θ but bounded in pθ, with

1 p 2 1 − ≤ θ ≤ 1 . (3.45) Q p0

Example 3.4 Particle on cylinder Take a particle moving on the surface of a cylinder with radius R, subject to a harmonic force F~ = −k~r. In cylinder coordinates the potential energy is V =

47 1 2 1 2 2 2 kr = 2 k(z +R ) and the hamiltonian is (ignoring the constant zero-point energy) p2 p2 1 H = θ + z + kz2 . (3.46) 2mR2 2m 2

Since θ is cyclic, pθ is constant and we can ignore it. The resulting phase space is three-dimensional: (pz, z, θ). We have

p2 1 z + kz2 = const , (3.47) 2m 2

giving an ellipse in the (pz, z) plane, or a uniform elliptic spiral in (pz, z, θ)-space.

3.5 Liouville’s theorem [Optional]

In statistical mechanics we do not know the detailed (microscopic) state of the system, ie the position and momentum of every single particle (for example, every molecule in a gas). Instead, we consider ensembles of possible states, ie collections of states that are consistent with what we know about the system. Every point in phase space corresponds to a possible state, and each has its own unique trajectory. We are interested in finding out what happens to nearby trajectories as they evolve in time (see figure on the right). At time t1, these points are located in a particular region of phase space. At some later time t2, where will they be relative to each other, and what kind of region of phase space will the occupy? We assume that a region ∆V in phase space contains N (representative) states. We can then define the density of states as

N ρ = with ∆V = ∆p ∆p ··· ∆p ∆q ∆q ··· ∆q . (3.48) ∆V 1 2 N 1 2 N

Let us now consider a small volume dpdq and look at the ﬂow of states in and out of this volume. This is illustrated in ﬁg. 3.4. In a small time dt, a point near (p, q) will move dq a distancepdt ˙ to the right, and a distanceqdt ˙ up.

The inflow dNl across the left boundary and dNb across the bottom boundary are given by (p,q) dp dN = ρ(p, q)pdqdt ˙ , (3.49) Figure 3.4: Flow of phase l space trajectories through dNb = ρ(p, q)qdpdt ˙ . (3.50) a small volume dpdq. Both ρ, p˙ andq ˙ vary with p and q, so the outflow across the right and top boundaries will be slightly different. They are

48 given by h ∂ρ ∂p˙ i dN = ρ(p + dp, q)p ˙(p + dp, q)dqdt = ρ(p, q)p ˙ + p˙ + ρ dp dqdt , (3.51) r ∂p ∂p h ∂ρ ∂q˙ i dN = ρ(p, q + dq)q ˙(p, q + dq)dqdt = ρ(p, q)q ˙ + q˙ + ρ dq dpdt . (3.52) t ∂q ∂q The total net inﬂow is then ∂ρ ∂p˙ ∂ρ ∂q˙ dN = dN + dN − dN − dN = − p˙ + ρ dpdqdt − q˙ + ρ dqdpdt , (3.53) l b r t ∂p ∂p ∂q ∂q and the change in density is

∂ρ dN dN ∂ρ ∂p˙ ∂ρ ∂q˙ = = = − p˙ + ρ + q˙ + ρ . (3.54) ∂t dV dt dpdqdt ∂p ∂p ∂q ∂q

But according to Hamilton’s equations of motion we have

∂H ∂H ∂p˙ ∂q˙ ∂2H ∂2H p˙ = − q˙ = =⇒ + = − + = 0 , (3.55) ∂q ∂p ∂p ∂q ∂p∂q ∂q∂p so we end up with ∂ρ ∂ρ ∂ρ dρ + p˙ + q˙ ≡ = 0 . (3.56) ∂t ∂p ∂q dt

If we have N coordinates p1, . . . , pN , q1, . . . , qN , we do the same for each pk, qk. The left boundary is now a hyper-surface with area dV/dpk, and similarly for the other boundaries. Taking the sum over all k, we ﬁnd

Liouville’s theorem

N dρ ∂ρ X ∂ρ ∂ρ = + p˙k + q˙k = 0 (3.57) dt ∂t ∂pk ∂qk k=1

Liouville’s theorem can be expressed in words as

As the (representative) states of the system evolve in time, their density in phase space remains constant.

Another way of expressing it is that the volume in phase space occupied by N trajectories remains the same throughout the evolution of the system. This result is a cornerstone of statistical mechanics, as well as of chaos theory.

49 3.6 Poisson brackets

Consider some function f(q1, . . . qN , p1, . . . , pN , t), ie an arbitrary function of coordinates, momenta, and time. Let us now ﬁnd the total time derivative (ie rate of change) of this function. Using the chain rule we have

N N df ∂f X ∂f ∂f ∂f X ∂f ∂H ∂f ∂H = + q˙k + p˙k = + − , (3.58) dt ∂t ∂qk ∂pk ∂t ∂qk ∂pk ∂pk ∂qk k=1 k=1 where in the last step we have made use of Hamilton’s equations of motion. We can write this in shorthand form as

df ∂f = + {f, H} . (3.59) dt ∂t {f, H} is called the Poisson bracket of H and f.

For two general functions f and g of coordinates and momenta, we deﬁne the Poisson brackets as

X ∂f ∂g ∂f ∂g {f, g} ≡ − (3.60) ∂qk ∂pk ∂pk ∂qk k

3.6.1 Properties of Poisson brackets

From the deﬁnition (3.60) we can immediately see that the Poisson bracket is antisymmetric, {f, g} = −{g, f}. We can also prove the following relations for arbitrary functions f, g, h and constants a, b, c:

{f, f} = 0 , (3.61) {f, c} = 0 , (3.62) {af + bg, h} = a{f, h} + b{g, h} (3.63) {fg, h} = f{g, h} + g{f, h} (3.64) ∂f {f, pk} = , (3.65) ∂qk ∂f {f, qk} = − . (3.66) ∂pk

For example, the proof of (3.62) is

X ∂f ∂c ∂f ∂c X ∂f ∂f {f, c} = − = · 0 − · 0 = 0 , (3.67) ∂qk ∂pk ∂pk ∂qk ∂qk ∂pk k k

50 since c is a constant and therefore does not depend on either p or q. To prove (3.66) we note that all coordinates and momenta are independent variables and therefore depend only on themselves, ∂p ∂q ∂p ∂q ∂p ∂q k = k = 0 ∀i, k ; k = k = 1 ∀k ; k = k = 0 if i 6= k. (3.68) ∂qi ∂pi ∂pk ∂qk ∂pi ∂qi We therefore have X ∂f ∂pk ∂f ∂pk ∂f ∂f {f, p } = − = · 1 = . (3.69) k ∂q ∂p ∂p ∂q ∂q ∂q i i i i i k k

Proving the other relations will be left as an exercise. From (3.66), (3.65), we immediately ﬁnd the fundamental Poisson brackets, which are the Poisson brackets of coordinates and momenta. They are

{qj, qk} = {pj, pk} = 0 ∀j, k (3.70) ( 1 , j = k {qj, pk} = δjk = (3.71) 0 , j 6= k

You may note the similarity between these expressions and the commutation relations between the position and momentum operators in quantum mechanics,

[p ˆi, xˆj] = −ihδ¯ ij . (3.72)

This analogy was noted early on, and is often used to formulate the quantum mechanical version of a classical mechanics system. In general, one may ﬁnd the Poisson brackets of any two quantities f and g, and this will give the commutation relation between the quantum mechanical operators f,ˆ gˆ, i {f, g} −→ − [f,ˆ gˆ] . (3.73) h¯ In particular, applying this to (3.59) yields Heisenberg’s equation of motion,

dfˆ i ∂fˆ = [f,ˆ Hˆ ] + , (3.74) dt h¯ ∂t which is valid for all quantum mechanical operators.1

1In the Heisenberg picture, where the operators depend on time. This is in contrast to the Schr¨odingerpicture, where the states depend on time and the operators are in general independent of time.

51 3.6.2 Poisson brackets and conservation laws

From (3.59) we can see that if f does not depend explicitly on time, ie f = f(p, q), then df/dt = {H, f}. Speciﬁcally, for any motion integral (conserved quantity which is a function of coordinates and momenta) I, we have

{I,H} = 0 . (3.75)

We can use this to demonstrate that a quantity is conserved, even if it is not the canonical momentum of a cyclic coordinate. The example below will illustrate this.

Example 3.5 Angular momentum Consider a particle with mass m moving in a central force (spherically symmetric) potential V (r). The hamiltonian for this system is, in spherical coordinates (r, θ, φ),

p2 p2 p2 H = T + V = r + θ + φ + V (r) . (3.76) 2m 2mr2 2mr2 sin2 θ Since the azimuthal angle φ does not appear in H (ie, it is a cyclic coordinate), we can immediately see that the canonical momentum pφ is conserved: ∂H p˙ = {p ,H} = − = 0 . (3.77) φ φ ∂φ In a spherically symmetric system all angles should be irrelevant, so one would na¨ıvely expect that pθ would also be conserved. However, because of the way we singled out the z-axis when deﬁning our coordinates, θ is not cyclic, and we get

∂H p2 cos θ p˙ = − = φ 6= 0 . (3.78) θ ∂θ mr2 sin3 θ Let us now consider instead the quantity

p2 J 2 ≡ p2 + φ . (3.79) θ sin2 θ We can use Poisson brackets to ﬁnd the rate of change of J 2:

dJ 2 = {J 2,H} dt ∂J 2 ∂H ∂J 2 ∂H ∂J 2 ∂H ∂J 2 ∂H ∂J 2 ∂H ∂J 2 ∂H = − + − + − ∂r ∂pr ∂pr ∂r ∂θ ∂pθ ∂pθ ∂θ ∂φ ∂pφ ∂pφ ∂φ (3.80) 2p2 cos θ p p2 cos θ = 0 − 0 + − φ · θ − 2p · − φ + 0 − 0 sin3 θ mr2 θ mr2 sin3 θ = 0 .

Therefore, J 2 is conserved.

52 What is the physical meaning of J 2?

We can write the angular momentum J~ in spherical coordinates as ~ ˆ ˆ ˙ˆ ˙ ˆ J = ~r × ~p = m~r × (vrrˆ + vθθ + vφφ) = mrrˆ × (r ˙rˆ + rθθ + r sin θφφ) . (3.81)

The vector products ofr ˆ with the unit vectors are

rˆ × rˆ = 0 ;r ˆ × θˆ = φˆ ;r ˆ × φˆ = −θˆ . (3.82)

Therefore, p J~ = mr2θ˙ φˆ − mr2 sin θφ˙ θˆ = p φˆ − φ θˆ (3.83) θ sin θ p 2 J~2 = p2 + φ = J 2 , (3.84) θ sin θ

since the two vectors θ,ˆ φˆ are orthogonal. We see that J 2 is just the square of the 2 2 2 2 ~ ~ total angular momentum, J = Jx + Jy + Jz = J.J.

We have already shown (2.81) that pφ = Jz is the z-component of the angular momentum. It may also be shown, using Poisson brackets, that

Jx = −pθ sin φ − pφ cot θ cos φ (3.85)

2 is conserved, this is left as an exercise. In fact all four of Jx, Jy, Jz and J are conserved, but they are not all independent. Only two of them are independent 2 conserved quantities and it is often convenient to use J and Jz. For a rotationally symmetric system we can always choose the axes so that the conserved vector J~ points in the z-direction and Jx = Jy = 0.

3.6.3 The Jacobi identity and Poisson’s theorem

One important relation that Poisson brackets obey is the Jacobi identity, {f, {g, h}} + {g, {h, f}} + {h, {f, g}} = 0 , (3.86) for any three arbitrary functions f, g, h. The proof of this is straightforward but extremely tedious, so we will not show it here. We will however note that the same identity holds for other antisymmetric products, including commutators and vector products: [A, [B,C]] + [B, [C,A]] + [C, [A, B]] = 0 , (3.87) A~ × (B~ × C~ ) + B~ × (C~ × A~) + C~ × (A~ × B~ ) = 0 . (3.88) The proof for commutators is very straightforward and relatively short: [A, [B,C]] + [B, [C,A]] + [C, [A, B]] = [A, BC − CB] + [B,CA − AC] + [C, AB − BA] = ABC − ACB − (BCA − CBA) + BCA − BAC − (CAB − ACB) (3.89) + CAB − CBA − (ABC − BAC) = 0 .

53 We can use this identity to prove that the Poisson brackets {F,G} of two conserved quantities F and G is also conserved: d ∂ ∂ {F,G} = {F,G} + {{F,G},H} = {F,G} − {H, {F,G}} (3.90) dt ∂t ∂t ∂F ∂G = { ,G} + {F, } + {F, {G, H}} + {G, {H,F }} (3.91) ∂t ∂t ∂F ∂G = { + {F,H},G} + {F, + {G, H}} = 0 . (3.92) ∂t ∂t In (3.90) we have used (3.59); in (3.91) we have used the product rule for ∂/∂t and used the Jacobi identity to rewrite {H, {F,G}}. Finally, in (3.92) we have used the general properties of Poisson brackets, {f, g} = −{g, f} and {f, g + h} = {f, g} + {f, h}. This gives us

Poisson’s theorem: dF dG d = = 0 =⇒ {F,G} = 0 . (3.93) dt dt dt

We can use this to construct further conserved quantities from ones we already know, although this is not quite as useful as it may at ﬁrst seem: in many cases the Poisson brackets of two conserved quantities will simply vanish.

Example 3.6

Given the hamiltonian (3.76) and Jx deﬁned in (3.85), we can, with a minimum of calculation, construct a further conserved quantity Jy. We already know that pφ is conserved, and using the Poisson brackets we have found that Jx is also conserved. Therefore the Poisson brackets of the two, ∂J J = {p ,J } = − x = p cos φ − p cot θ sin φ , (3.94) y φ x ∂φ θ φ is also conserved.

3.7 Noethers theorem

We have earlier seen that if a variable qi is cyclic then there is a conserved quantity ∂L pi = related to it as ∂q˙i ∂L p˙i = = 0. ∂qi But it goes deeper than that. As soon as the Lagrangian as a symmetry, then there will be a conserved quantity related to that quantity. This result is called Noethers theorem.

54 Assume you can change your variables q → q(α) andq ˙ → q˙(α) by a parameter α and still keep the Lagrangian invariant ∂ L (q (α) , q˙ (α) , t) = 0. (3.95) ∂α i i

This could be simple things like q → q + α or something more complicated, like q1 → q1 cos α − q2 sin α, q2 → q1 sin α + q2 cos α. Note here that α is independent of time and also that we do not change the time coordinate — if we do the result bellow will need to be generalized. We have ∂ X ∂L ∂qi ∂L ∂q˙i L (q (α) , q˙ (α) , t) = + . ∂α i i ∂q ∂α ∂q˙ ∂α i i i Assuming that α is independent of time and derivatives with respect to α and t commute, ∂q˙i ∂ ∂qi ∂α = ∂t ∂α this can be written as ∂ X ∂L ∂qi d ∂L ∂qi d ∂L ∂qi L (q (α) , q˙ (α) , t) = − + ∂α i i ∂q ∂α dt ∂q˙ ∂α dt ∂q˙ ∂α i i i i X ∂L d ∂L ∂qi d X ∂qi = − + p ∂q dt ∂q˙ ∂α dt i ∂α i i i i = 0.

This vanishes because we assume that varying α is a symmetry of the Lagrangian (3.95), we have not assumed that qi is a solution of the equations of motion. If qi is a solution then we can deduce that ! d X ∂qi(α) p = 0. dt i ∂α i We have shown that

X ∂qi(α) C = p (3.96) i ∂α i

is a conserved quantity.

From this immediately follows something we already knew: If qi is cyclic then pi is conserved. This follows since if ∂L = 0 then the Lagrangian is also invariant under ∂qi ∂qi(α) under qi → qi(α) = qi + α. We therefore have ∂α = 1 and so dq (α) C = p i = p i i dα i which shows that pi is conserved when qi is cyclic.

55 Example 3.7 Assume a spherically symmetric potential V (r) with Lagrangian

m L = x˙ 2 +y ˙2 +z ˙2 − V (r) 2 We now perform a rotation in the x, y-plane. We then have

x (α) = x cos α − y sin α y (α) = x sin α + y cos α.

Note that x2 + x2, and so r2 = x2 + y2 + z2, is invariant under this rotation, and therefore also the Lagrangian. We then have ∂x = −x sin α − y cos α ∂α = −y (α) ∂y = x cos α − y sin α ∂α = x (α) ∂z = 0. ∂α The conserved quantity is then ∂x ∂y ∂z C = p + p + p xy x ∂α y ∂α z ∂α = −pxy (α) + pyx (α) + 0

= Jz, which is the angular momentum. In a similar way we may show that

Cyz = Jx

Czx = Jy

We thus conclude that invariance under rotations are intimately connected with angular momentum conservation.

56 3.8 Hamiltonian dynamics — summary sheet

1. Hamiltonian formalism We consider coordinates qi and momenta pi as independent variables, and write the hamiltonian as a function of coordinates and momenta, H(q, p, t). 2. Hamilton’s equations of motion ∂H ∂H q˙i = , p˙i = − . ∂pi ∂qi 3. Modiﬁed principle of least action Hamilton’s equation can be derived from the condition Z t2 X δ piq˙i − H(q, p, t) dt = 0 , t1 i

where now qi(t) and pi(t) can both be varied independently, but are kept ﬁxed at the endpoints t1, t2. 4. Poisson brackets For a system with N degrees of freedom,

N X ∂f ∂g ∂f ∂g {f, g} ≡ − . ∂q ∂p ∂p ∂q i=1 i i i i 5. Equation of motion for a quantity f(q, p, t) df ∂f = {f, H} + . dt ∂t 6. Integrals of motion For a motion integral or conserved quantity F , dF/dt = 0, so ∂F {F,H} + = 0 . ∂t If F does not depend explicitly on time, this reduces to {F,H} = 0 .

7. Jacobi identity {f, {g, h}} + {g, {h, f}} + {h, {f, g}} = 0 for any functions f, g, h of coordinates and momenta.

57 8. Poisson theorem If F and G are motion integrals, then their Poisson bracket {F,G} is also a motion integral: dF dG d = = 0 =⇒ {F,G} = 0 . dt dt dt 9. Noethers theorem If the Lagrangian is invariant under a tranformation qj → qj(α), q˙j → q˙j(α), then X dqi C = p i dα i is a conserved quantity.

58 Chapter 4

Central forces

If the force between two bodies is directed along the line connecting the (centres of mass of) the two bodies, this force is called a central force. Since most of the fundamental forces we know about, including the gravitational, electrostatic and certain nuclear forces, are of this kind, it is clear that studying central force motion is extremely imortant in physics. Moreover, the motion of a system consisting of only two bodies interacting via a central force is one of the few problems in classical mechanics that can be solved completely (once you add a third body it becomes, in general, unsolveable). Examples of such systems are the motion of planets and comets around a star, or satellites around a planet, or binary stars; and classical scattering of atoms or subatomic particles. The full description of atoms and subatomic particles requires quantum mehcanics, but even here the classical analysis of central forces can yield a great deal of insight. In the two-body problem we start with a description in terms of 6 coordinates, namely the three (cartesian) coordinates of each of the two bodies. We shall see that it is possible to reduce this to just 2, and for some purposes only 1 eﬀective degree of freedom. This reduction will happen in 3 steps:

1. We can treat the relative motion as a 1-body problem.

2. The relative motion is 2-dimensional (planar).

3. We can use angular momentum conservation to treat the radial motion as 1- dimensional motion in an eﬀective potential.

4.1 One-body reduction, reduced mass

We start with a system of two particles, with coordinates ~r1 and ~r2. We need six coordinates to describe this system, and this is provided by the three components of ~r1 and the three components of ~r2. However, since we know that the potential energy only depends on the combination r = |~r| = |~r1 − ~r2|, we may want to describe it instead in terms of the three components of the relative coordinate ~r = ~r1 − ~r2 and a second

59 vector, which we can take to be the centre-of-mass vector m ~r + m ~r R~ = 1 1 2 2 . (4.1) m1 + m2

~ First we need to express ~r1 and ~r2 in terms of the new coordinates R, ~r. We have

~r = ~r1 − ~r2 ⇐⇒ ~r1 = ~r + ~r2 (4.2) ~ m1 ~r1 + m2 ~r2 m1(~r + ~r2) + m2 ~r2 m1 R = = = ~r2 + ~r, (4.3) m1 + m2 m1 + m2 m1 + m2 which gives

~ m1 ~r2 = R − ~r (4.4) m1 + m2 ~ m2 ~r1 = ~r + ~r2 = R + ~r (4.5) m1 + m2

We now plug (4.4), (4.5) into the expression for the kinetic energy, 1 1 T = m ~r˙2 + m ~r˙2 2 1 1 2 2 2 2 2 1 ~˙ m2 ˙ 1 ~˙ m1 ˙ = m1 R + ~r + m2 R − ~r 2 m1 + m2 2 m1 + m2 2 1 ~˙ 2 2m2 ~˙ ˙ m2 ˙2 = m1 R + R · ~r + 2 ~r 2 m1 + m2 (m1 + m2) (4.6) 2 1 ~˙ 2 2m1 ~˙ ˙ m1 ˙2 + m2 R − R · ~r + 2 ~r 2 m1 + m2 (m1 + m2) 2 2 1 ~˙ 2 1 m1m2 m1m2 ˙2 = (m1 + m2)R + 2 + 2 ~r 2 2 (m1 + m2) (m1 + m2) 1 ˙ 1 = MR~ 2 + µ~r˙2 . 2 2 The total lagrangian is therefore 1 ˙ 1 L = T − V = MR~ 2 + µ~r˙2 − V (r) (4.7) 2 2 where m m µ = 1 2 . m1 + m2 We see that the lagrangian splits into two separate parts: one describing the motion of the centre of mass, and one describing the relative motion. We can therefore analyse the relative motion without any reference to the overall motion of the centre of mass. More- over, R~ is cyclic, so its canonical momentum is conserved. The canonical momentum conjugate to R~ is ∂L ˙ P = = MR˙ =⇒ P~ = MR.~ (4.8) i ˙ i ∂Ri This is just the total momentum of the system: ˙ ˙ ~ ~˙ m1 ~r1 + m2 ~r2 ˙ ˙ P = MR = (m1 + m2) = m1 ~r1 + m2 ~r2 . (4.9) m1 + m2

60 Therefore, the absolute motion is merely linear motion with constant total momentum. From now on, we will ignore the absolute motion completely, and focus only on the relative motion — ie, we will drop the ﬁrst term in (4.7). The lagrangian then becomes 1 L = µ~r˙2 − V (r) . (4.10) 2 This looks exactly like the lagrangian for a single particle with position ~r in a potential V (r), but with mass m m µ = 1 2 = the reduced mass (4.11) m1 + m2 We have therefore reduced two-body motion to the equivalent motion of a single body with mass µ. It is worthwhile looking more closely at the reduced mass and its relation to the masses m1, m2. We can assume without any loss of generality that m1 ≥ m2 (since we could just swap the labels if it were the other way around). Then we have m 1 m 1 1 ≥ , 2 ≤ (4.12) m1 + m2 2 m1 + m2 2 m m m m =⇒ 1 ≥ 1 2 ≥ 2 . (4.13) 2 m1 + m2 2 So we see that the reduced mass has a value that lies between half the larger mass and half the smaller mass. Two special cases of particular interest are where the two masses are equal, and where one mass is much larger than the other. The ﬁrst includes scattering of identical particles (for example two α-particles) as well as some binary stars. The second includes the motion of a planet or comet around the sun, or satellites around a planet.

In the ﬁrst case, m1 = m2 = m, we get that the reduced mass is µ = m/2, ie the reduced mass is half the mass of each body.

In the second case, m2 m1, we can rewrite the reduced mass as 1 m m 2 µ = m = m 1 − 2 + 2 + ... . (4.14) 2 m2 2 1 + m1 m1 m1

If m2 is small enough compared to m1 (for example, for the earth–sun system we have −6 M⊕/M = 3 · 10 ) we can just set µ = m2, ie the reduced mass equals the smaller of the two masses.

4.2 Angular momentum and Kepler’s second law

Our system is now equivalent to a single particle with mass µ moving in a spherically symmetric potential V (r). Since we have spherical symmetry, the angular momentum J~ = ~r × ~p is conserved, both in magnitude and in direction. We will use this to simplify the problem further. We have seen that, when the potential V (r) depends only on r and is independent of the angular variables θ and φ, then the total angular momentum (3.84) and the z-component

61 of angular momentum (2.81) are conserved. We can choose the z-axis of our coordinate system to be pointing in the direction of J~, ie J~ = `zˆ, and in this case both ~r and ~p must be in the xy-plane. We can get a more physical understanding of this by noting that as long as ~p = m~r˙ remains in the xy-plane, ~r will not move out of this plane, while as long as ~r remains in that plane there is no force that will move ~p our of the plane, since the central force always points towards the centre (or away from it, in the case of a repulsive force), ie along the vector ~r. We have therefore reduced the motion to planar (2-dimensional) motion, and we can use polar coordinates (r, θ) to describe this motion (where θ is the angle with some arbitrarily chosen direction in the plane of motion). The lagrangian for the system, in those coordinates, becomes 1 1 L = µr˙2 + µr2θ˙2 − V (r) . (4.15) 2 2 Since L does not depend on the angle θ (this is the remaining rotational symmetry), the angular momentum ` = pθ is conserved, ` = µr2θ˙ = constant . (4.16) This is exactly equivalent to Kepler’s second law for planetary motion, which gives a nice geometrical interpretation of angular momentum conservation. Consider the area dA swept out by the radius vector in a small (inﬁnitesimal) time dt. The angle swept out in that time is dθ = θdt˙ , and the length of the arc swept out is ds = rdθ = rθdt˙ (see ﬁg. 4.1). If dθ is small we can approximate the area by a triangle with length r and height ds, ie 1 1 dA 1 ` dA = rds = r · rθdt˙ ⇐⇒ = r2θ˙ = = constant . (4.17) 2 2 dt 2 2µ

r(t+dt) rdθ dθ r(t)

Figure 4.1: The area swept out by a radius vector.

Kepler’s second law reads, in words,

A line joining a planet and the Sun sweeps out equal areas during equal intervals of time.

62 Kepler came to this conclusion through painstaking observation of planetary motions, and the apparatus of newtonian and lagrangian mechanics, which we are using here, was developed long after his time. But from this derivation we can see that this law is valid not just for planetary motion, but for all central force motion, whatever the force is, and is equivalent to conservation of angular momentum.

Example 4.1 The planet Mercury orbits the Sun in 87.97 days, in an elliptic orbit with semimajor axis a = 57.91 · 106km and semiminor axis b = 56.67 · 106km.

1. What is the areal velocity of Mercury?

2. Use the relation between areal velocity and angular momentum to ﬁnd the speed of Mercury

(a) at its perihelion (closest to the sun), 46.00 · 106km from the sun; (b) at its aphelion (furthest from the sun), 69.82 · 106km from the sun.

Answer:

1. Since the areal velocity is constant, it is just equal to

dA A πab π · 57.91 · 56.67 · 1012km2 = = = = 1.3565 · 109 km2/s , (4.18) dt T 87.97d 87.97 · 86400s where we have also used that the area of an ellipse is A = πab.

2. From the relation between areal velocity and angular momentum `, we have

dA ` |~r × µ~v| 1 = = = |~r × ~v| . (4.19) dt 2µ 2µ 2 At perihelion and aphelion, the radial velocity is zero, so ~v is orthogonal to the radius vector ~r. At these points we therefore have

dA 1 2dA/dt = rv =⇒ v = . (4.20) dt 2 r At perihelion:

2 · 1.3565 · 109km2/s v = = 58.98km/s. (4.21) 46.00 · 106km

At aphelion:

2 · 1.3565 · 109km2/s v = = 38.86km/s. (4.22) 69.82 · 106km

63 4.3 Eﬀective potential and classiﬁcation of orbits

Now that we have shown that angular momentum is conserved, we can use this to simplify the problem further. We have already seen how a 2-dimensional system can be treated as one-dimensional motion in an eﬀective potential when one coordinate is cyclic. In our case, we have ` p = ` = µr2θ˙ ⇐⇒ θ˙ = , (4.23) θ µr2 so we can write the hamiltonian or total energy as

p2 p2 H = r + θ + V (r) , (4.24) 2µ 2µr2 or 1 `2 1 E = µr˙2 + + V (r) = µr˙2 + V (r) . (4.25) 2 2µr2 2 eff 2 2 The term Vc(r) = ` /2µr is sometimes called the centrifugal potential, and can be −3 thought of as giving rise to a (fictitious) “centrifugal force” Fc ∝ r . By investigating the shape of the effective potential Veff we can find out what kinds of motion are possible in the radial direction. Some examples of effective potentials are shown in figure 4.2.

In general, if ` 6= 0, the centrifugal potential Vc(r) → +∞ as r → 0, providing a barrier against the bodies getting too close. This term will dominate at short distances unless V (r) is strongly attractive, meaning that V (r) → −∞ fast enough. “Fast enough” here means that 1 V (r) ∼ − , n > 2 , (4.26) rn since if n < 2 we will always ﬁnd that ar−n < br−2 for any a, b if r is small enough.

4.4 Integrating the energy equation

We can also use (4.25) to completely solve the motion in r, by using energy conservation and rewriting it to get 1 µr˙2 = E − V (r) (4.27) 2 eff s dr r 2 2 `2 ⇐⇒ r˙ = = ± E − V (r) = ± E − V (r) − (4.28) dt µ eff µ 2µr2 Z dr Z =⇒ = dt = t − t0 . (4.29) q 2 µ E − Veff(r)

Note that this does not tell us anything directly about the shape of the orbit, since for that we still need to know θ(t), but once we have found r(t) it is straightforward to obtain θ from (4.23). In particular, if the motion is bounded, we can use this to ﬁnd the period T of the radial motion, ie the time it takes to complete one full oscillation in

64 V(r) = k/r

2 V(r)=kr

V(r) = -k/r V(r) = -k/r³

Figure 4.2: The effective potential (4.25) for different types of potential V (r). The dotted red curves denote the centrifugal potential, and the dotted black curves V (r). The thick blue curve is the effective potential Veff(r). Top left: a quadratic (harmonic oscillator) potential. The motion in r is bounded for all values of ` > 0. Top right: a repulsive, inverse-square force law, V (r) = k/r. In this case, only unbounded motion is possible.Bottom left: an attractive, inverse-square force law. The different solid curves correspond to different values of `. Here the motion is always bounded if E < 0, and unbounded if E > 0. Bottom right: an attractive multipole force with V (r) = −k/r3. Here we can have bounded motion through the origin or unbounded motion.

the radial direction, from rmin to rmax and back again. This is given by

rmax Z pµ/2 dr T = 2 p , (4.30) E − Veff(r) rmin where the factor 2 accounts for the “return journey” from rmax to rmin. However, if we want to obtain the shape of the orbit, and are not particularly interested in the motion in time, we can use a similar trick to obtain r as a function of θ or vice-versa. Using (4.23) and (4.28) together, we can write dθ dθ dt ` ±pµ/2 = = (4.31) 2 p dr dt dr µr E − Veff(r) ` Z dr Z =⇒ √ = dθ = θ − θ0 . (4.32) 2p 2µ r E − Veff(r) This gives us θ(r), which we can then invert to find r(θ), which defines the shape of the orbit.

65 Again, if the orbit is bounded, we can ﬁnd the angular period ∆θ, which is the angle swept out in the course of one full radial oscillation. It is given by

rmax r 2 Z dr ∆θ = ` . (4.33) 2p µ r E − Veﬀ(r) rmin

If ∆θ = 2πm/n, where n and m are integers, then the system will return to the same place after n radial oscillations, having completed m revolutions of an angle of 2π. Such an orbit is closed. A remarkable result is that closed (non-circular) orbits are extremely rare: Bertrand’s theorem (1873) states that the only potentials that give rise to such orbits are the harmonic oscillator V (r) = kr2 and the inverse-square force V (r) = −k/r (see Goldstein, pp. 89–92 for an explanation).

4.5 The inverse square force, Kepler’s ﬁrst law

From now on we will concentrate on the attractive inverse-square force law, ie

k `2 k V (r) = − =⇒ V (r) = − . (4.34) r eﬀ 2µr2 r

This describes the gravitational force between two bodies, with k = Gm1m2 (where m1, m2 are the masses of the two bodies). It could also describe the electrostatic at- traction between two opposite charges Q1, −Q2, in which case we would have k = Q1Q2/4πε0.

Looking at Veff(r), we find that bounded motion is possible if Emin ≤ E < 0, where Emin is the minimum value of Veff (which we will derive in a moment). For E ≥ 0 the radial motion is unbounded, although for any ` 6= 0 there is a minimum distance rmin. If E = Emin we have a stable circular orbit. Let us now find the minimum and maximum distances for a particular energy and angular momentum. They are given by

`2 k V (r) = − = E (4.35) eﬀ 2µr2 r `2 ⇐⇒ Er2 + kr − = 0 (4.36) 2µ q 2 2E`2 s −k ± k + µ k 2E`2 ⇐⇒ r = r = = − 1 ± 1 + = a(1 ± e) , min,max 2E 2E µk2 (4.37) with s k 2E`2 a = − , e = 1 + . (4.38) 2E µk2

Inspecting (4.37) we see that

66 2 2 • If E < Emin = −µk /2` , the expression inside the square root becomes negative, and there is therefore no solution.

2 • If E = Emin the square root (e) is zero, and we have rmin = rmax = a = ` /µk. 2 This corresponds to a stable circular orbit with r = r0 = ` /µk. You can verify 0 that this also corresponds to the minimum of Veﬀ, where Veﬀ(ro) = 0.

• If Emin < E < 0 there are two solutions, rmin = a(1 − e), rmax = a(1 + e), and the radial motion is bounded between these two distances.

2 • If E = 0, a → ∞, but (4.35) still has a single solution at r = rmin = ` /2µk. There is no maximum value for r, so the motion becomes unbounded.

• If E > 0 we have e > 1 and a < 0. Since the distance r must be positive, only the minus sign in (4.37) gives a physically acceptable solution, rmin = a(1 − e) = (e−1)k/2E. There is again no maximum value for r, and the motion is unbounded.

Example 4.2

The asteroid Pallas orbits the sun in an orbit with perihelion distance rmin = 3.19 · 1011m and e = 0.231. The speed of Pallas relative to the sun at perihelion is 4 v = 2.26 · 10 m/s. Find the aphelion distance rmax of Pallas and its speed at that point.

Answer: From (4.37) we ﬁnd that the aphelion distance rmax is 1 + e 1.231 r = r = · 3.19 · 1011m = 5.11 · 1011m . (4.39) max 1 − e min 0.769 To ﬁnd the speed at aphelion, we need to use either conservation of angular momentum or conservation of energy. Using angular momentum is easier. At perihelion and aphelion we have ~v ⊥ ~r and therefore ` = mvr where m is the mass of the asteroid. The aphelion and perihelion speeds va, vp are therefore related by

` = mvprmin = mvarmax (4.40)

rmin 3.19 4 4 =⇒ va = vp = · 2.26 · 10 m/s = 1.41 · 10 m/s . (4.41) rmax 5.11

We can now use the methods of Section 4.4 to ﬁnd an equation for the orbit. For V (r) = −k/r (4.32) becomes

Z `dr Z dr θ(r) = ± r = ± q . (4.42) 2 2 2Eµ 2µk 1 2 k ` r 2 + 2 − 2 r 2µ E + r − 2µr2 ` ` r r

We now make the substitution 1 dr u = =⇒ du = − , (4.43) r r2

67 and also introduce the parameter α = `2/µk. This gives us Z du Z dv θ(r) = ± q = ± q , (4.44) 2Eµ u 2 e2 2 `2 + 2 α − u α2 − v where in the second step we have made the further substitution v = u − 1/α and introduced the parameter e from (4.38). We can look up this integral in a table, or solve it using additional clever substitutions. We can introduce the angle φ given by r e e e2 e p e v = cos φ =⇒ dv = − sin φ dφ , − v2 = 1 − cos2 φ = ± sin φ , α α α2 α α (4.45) and therefore Z dv Z θ = q = − dφ = −φ , (4.46) e2 2 α2 − v where we have chosen the integration constant to be zero. We can do this because the rotational symmetry of the problem means that we can choose θ = 0 to be any direction in the plane. Working our way back, we then ﬁnd e 1 1 α v = cos θ = − ⇐⇒ = e cos θ + 1 (4.47) α r α r s α `2 2E`2 ⇐⇒ r = with α = , e = 1 + . (4.48) 1 + e cos θ µk µk2

This is the equation for a conic section, where e is the eccentricity, and governs the shape of the orbit (while α governs the size).

• If e = 0, r is constant and the orbit is a circle, as can be easily seen from (4.48).

• If 0 < e < 1, the orbit is closed, with 1/(1 + e) ≤ r ≤ 1/(1 − e). In this case, the orbit is an ellipse.

• If e = 1, r → ∞ as θ → π, so the orbit is open (just). In this case the orbit is a parabola.

• If e > 1, the orbit is again open (unbounded), but there is a limit to the possible angles, cos θ < 1/e. In this case the orbit is a hyperbola.

4.5.1 The shapes of the orbits

Starting from equation (4.48) we may now write the shape of the orbit in Cartesian x coordinates. We remember that cos θ = r and insert that into (4.48), yielding α r = ex 1 + r This equation can be rewritten as r = α−ex. By squaring the equation and substituting r2 = x2 + y2 we get y2 = α2 − 2αex + (e2 − 1)x2.

68 This is the equation for a conic section, which has a solutions ellipses, parabolas and hyperbolas. We note that the sign in front of x2 depends on whether e > 1 or e < 1. Completing the square over x then reads αe y2 = α2 + e2 − 1 x2 − 2x (4.49) e2 − 1 αe α2e2 α2e2 = α2 + e2 − 1 x2 − 2x + − e2 − 1 (e2 − 1)2 (e2 − 1)2 α2e2 αe 2 = α2 − + e2 − 1 x − e2 − 1 (e2 − 1) α2 αe 2 = − + e2 − 1 x − e2 − 1 e2 − 1 which can be re-arranged to give (1 − e2)2 αe 2 (1 − e2) x − + y2 = 1. (4.50) α2 e2 − 1 α2

2 α2 2 α2 Now if we identify a = (1−e2)2 and deﬁne b = |1−e2| we can reformulate this as

(x + ea)2 y2 2E`2 ± = 1 with e2 − 1 = (4.51) a2 b2 µk2 where the sign depends on whether e > 1 or e < 1, i.e. on the sign of E, the plus sign for E < 0 and the minus sign for E > 0. We can now see that when:

• E < 0 ⇒ 1 − e2 > 0. This equation describes an ellipse with semi-major axis a and semi-minor axis b. See Figure 4.3. • E = 0 ⇒ e = 1. Then it is best to go back to (4.49) from which we see that y2 = α2 − 2αx, which is a parabolic curve. • E > 0 ⇒ e > 1, we use the minus sign in (4.51) and the above equation reads (x + ea)2 y2 − = 1. a2 b2 This equation describes a hyperbolic curve with the closet passage to the origin b being ea and the asymptotic lines being y = a x. See Figure 4.4.

4.6 More on conic sections

Conic sections are all solutions of equations of the type Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 . They are called conic sections because this type of equation appears when you intersect a cone, described by the equation x2 +y2 = kz2, with a plane, described by the equation αx + βy + γz = δ. The type of curve described by these equations depends on the parameters A, B, C:

69 • If B2 − 4AC < 0, we get an ellipse. After a change of variables, this curve can be written on the form x2 y2 + = 1 . (4.52) a2 b2 • If B2 − 4AC = 0, we get a parabola. The equation can then be written, after a change of variables, as y = ax2 . (4.53)

• If B2 − 4AC > 0, we get a hyperbola. We can then make a change of variables to write the resulting equation as

x2 y2 − = 1 or xy = a2 . (4.54) a2 b2

That this definition is equivalent to (4.48) is by no means obvious. In the following we will look in some more detail at the different definitions of conic sections, and establish that they are indeed equivalent. In the process, we will also find some important relations between the various parameters of the conic sections, in particular the ellipse. There is yet another general definition of a conic section, namely the curve defined by r = ep, where p is the distance to a straight line, called the directrix. It is straightforward to prove the equivalence of this and (4.48). If we place a straight line parallel to the y-axis at x = α/e, then the distance p from a point on the curve (4.48) to this line is

α α α α cos θ p = − x = − r cos θ = − e e e 1 + e cos θ α(1 + e cos θ) − eα cos θ α r = = = . (4.55) e(1 + e cos θ) e(1 + e cos θ) e

4.6.1 Ellipse

An ellipse, pictured in Figure 4.3, can be defined by first identifying two points, called the foci (denoted by crosses in Fig. 4.3). If `1 and `2 are the distances of a point from each of the two foci, then an ellipse is any set of points where the sum of `1 and `2 is constant, `1 + `2 = 2a . (4.56) This definition has an optical interpretation: if you place a light source at one focus (and a screen between to the two foci), then light that is reflected off any point of the ellipse will gather at the second focus — which is indeed why it is called a focus. This definition can also be used to draw an ellipse, using two pins and a piece of string. You pin the ends of a string of length 2a at each focus, pull the string tight with the tip of your pen, and move the pen around while keeping the string tight. This will trace out an ellipse. We will now show that (4.56) is equivalent to (4.52), if we take the origin to be at the centre of the ellipse (so x + ea → x in (4.51)). The two foci are then located at (−c, 0) and (c, 0), and the distances `1, `2 from a point (x, y) to each focus are

2 2 2 2 2 2 `1 = (x + c) + y , `2 = (x − c) + y . (4.57)

70 y

l1 r l2 c=ea θ a O x b

Figure 4.3: An ellipse with major semiaxis a, minor semiaxis b and foci at a distance c = ea from the centre of the ellipse. The distance l1 + l2 = 2(rmin + c) = 2a is the same for all points on the ellipse.

The ellipse is given by

p 2 2 p 2 2 `1 + `2 = (x + c) + y + (x − c) + y = 2a (4.58) ⇐⇒ p(x + c)2 + y2 = 2a − p(x − c)2 + y2 (4.59) ⇐⇒ (x + c)2 + y2 = 4a2 + (x − c)2 + y2 − 4ap(x − c)2 + y2 (4.60) ⇐⇒ 2cx = 4a2 − 2cx − 4ap(x − c)2 + y2 (4.61) c ⇐⇒ p(x − c)2 + y2 = a − x (4.62) a c2 ⇐⇒ x2 − 2cx + c2 + y2 = a2 − 2cx + x2 (4.63) a2 a2 − c2 ⇐⇒ y2 + x2 = a2 − c2 (4.64) a2 x2 y2 ⇐⇒ + = 1 with b2 = a2 − c2 . (4.65) a2 b2

The two main (“long” and “short”) diameters of the ellipse, with lengths 2a and 2b, are called the major axis and the minor axis, respectively. The parameters a and b, being half of the major and minor axes, are called the major semiaxis and minor semiaxis. To show that this is also equivalent to (4.48), we will have to put the origin at one of

71 the foci. Let us choose the right one to be speciﬁc. Instead of (4.52) we now have (x + c)2 y2 + = 1 (4.66) a2 b2 ⇐⇒ a2b2 = b2x2 + 2b2cx + b2c2 + a2y2 = (b2 − a2)x2 + a2(x2 + y2) + 2b2cx + b2c2 (4.67) = −c2x2 + a2r2 + 2b2cx + b2c2 ⇐⇒ a2r2 = b2(a2 − c2) − 2b2cx + c2x2 = b4 − 2b2cx + c2x2 = (b2 − cx)2 (4.68) b2 − cx b2 − cr cos θ ⇐⇒ r = = (4.69) a a c b2 ⇐⇒ r 1 + cos θ = (4.70) a a α c b2 ⇐⇒ r = with e = , α = = a(1 − e2) (4.71) 1 + e cos θ a a

So we see that (4.48) indeed corresponds to an ellipse, with the origin at one of the foci, if e < 1. This corresponds to Kepler’s ﬁrst law:

The planets move in elliptical orbits, with the sun at one focus.

4.6.2 Parabola

If one of the foci is taken away to inﬁnity, the resulting curve becomes a parabola. The optical interpretation of this is that parallel rays (corresponding to rays coming in from a source at inﬁnity) are focused in a single point. This property is widely used in telescopes and satellite dishes, which all tend to have a parabolic shape. It will be left as an exercise to prove that (4.48), with e = 1, is equivalent to the usual y = Ax2.

4.6.3 Hyperbola

If the eccentricity e > 1, the resulting curve is a hyperbola. On inspection of (4.48) you −1 will see that r → ∞ as θ → ± cos (1/e) = ±θmax. The curve will therefore approach, but never touch, the two straight lines deﬁned by θ = ±θmax. One may also construct the mirror image of this curve, α α r = = . (4.72) 1 − e cos θ 1 + e cos(π − θ) The two mirror images, called the two branches of the hyperbola, can be shown to be π ◦ equivalent to the expressions (4.54). In the second case, we see that θmax = 4 or 45 .

Example 4.3 A satellite in orbit around the earth has speed v = 7400m/s at its apogee, 630km above the surface of the earth. What is its distance from the surface of the earth at perigee, and what is its speed at that point?

72 y

O x

2 2 Figure 4.4: A hyperbola with equation (x+ea) − y = 1.The curve stays within (but a2 b2 √ b 2 approaches asymptotically) the cone y = ± a (x + ea) = ± e − 1 (x + ea)

Answer: The energy of a satellite with mass m, orbiting the earth at a distance r and travelling with a speed v is 1 mM E = mv2 − G ⊕ . (4.73) 2 r The angular momentum is

~ ` = |J| = m|~v × ~r| = mvr sin θvr . (4.74)

At perigee or apogee the velocity (and momentum) is perpendicular to the radial direction, so ` = mvr at these points of the orbit. Since the mass of the satellite is much smaller than the mass of the earth, the reduced mass µ can be replaced by m:

mM mM µ = ⊕ ≈ ⊕ = m . (4.75) m + M⊕ M⊕

The constant k in the inverse-square force is in this case k = GmM⊕. With this knowledge we can work out the parameters α (or major semiaxis a) and eccentricity e for the orbit. There are several ways doing this. The simplest is probably using the relation between the energy and semimajor axis, which holds for an elliptical orbit: k GmM⊕ GM⊕r a = − = − 2 = 2 . (4.76) 2E mv − 2GmM⊕/r 2GM⊕ − rv

73 In our problem, the speed of the satellite at apogee is v = 7400m/s. The distance 6 from the centre of the earth is r = 630 km + R⊕ = 7.000 · 10 km. This gives

6.674 · 10−11m3/kg s2 · 5.9736 · 1024kg · 7.000 · 106m a = 2 · 6.674 · 10−11m3/kg s2 · 5.9736 · 1024kg − 7.000 · 106m · (7400m/s)2 = 6740km . (4.77)

The sum of the apogee distance r+ and the perigee distance r− is twice the major semiaxis,

r+ + r− = 2a ⇐⇒ r− = 2a − r+ = 2 · 6740km − 7000km = 6481km . (4.78)

Therefore, the height of the satellite above earth at perigee is r− − R⊕ =111km. The speed may be found from the angular momentum,

r− 7000 ` = mv+r+ = mv−r− =⇒ v+ = v− = 7400m/s = 7994m/s . (4.79) r+ 6481

4.7 Kepler’s third law

Let us look again at the elliptic orbit. We can use the relations we have found above, together with Kepler’s second law, to derive an expression for the period, the tine T it takes to complete one full orbit. From Kepler’s second law we have dA ` Z T dA ` = =⇒ A = dt = T. (4.80) dt 2µ 0 dt 2µ The area of an ellipse is A = πab, so we will need to ﬁnd a suitable expression for the minor semiaxis b. We know from above that b2 = a2(1 − e2). Using the expression for the eccentricity e, we get s s 2 2 √ −2E` ` ` 1 b = a 1 − e2 = a = a = √ a 2 . (4.81) µk2 µka µk Putting this together, we have 2µ 2πµ 2πµ ` rµ T = A = ab = √ a3/2 = 2π a3/2 , (4.82) ` ` ` µk k or

4π2µ T 2 = a3 . (4.83) k

For planets orbiting the sun, the ratio µ/k = 1/G(m + M ) ≈ 1/(GM ) to a very high approximation, so the proportionality constant will be the same for all planets! This

74 is Kepler’s third law: The square of the orbital period varies like the cube of the major axis.

Example 4.4 A comet is observed travelling at a speed of 64.0 km/s at is closest approach to the sun, 64.5 million km from the sun. Will this comet ever be seen again, and if so, when? What would the answer be if the closest distance to the sun was 65.0 million km? Answer: The solution of this problem follows the same lines as that of Example 3, with the mass of the sun, M , replacing the mass of the earth, M⊕. The comet will be seen again if the orbit is closed, which happens if the total energy E < 0. Alternatively, it is possible to calculate the eccentricity e and determine whether e < 1 (closed orbit) or e > 1 (open orbit).

We may ﬁrst note that Newton’s constant G and the mass of the sun M always occur in the combination GM so we may calculate this product once and for all,

GM = 6.674 · 10−11m3/kg s2 · 1.9881 · 1030kg = 1.3275 · 1020m3/s2 (4.84) = 1.3275 · 1011km3/s2 .

If the speed of the comet at perihelion is v = 64.0 km/s and the distance is r = 64.5 · 106 km, we ﬁnd that

E v2 GM 64.02 1.3275 · 1011km3/s2 = − = km2/s2 − = −10.14km2/s2 . (4.85) m 2 r 2 64.5 · 106km So E < 0 and the orbit is closed (an ellipse). The comet will therefore be seen again. To find out when, we use Kepler’s third law, 2π T = √ a3/2 . (4.86) GM We must first find the semimajor axis a, which is given by

GM 1.3275 · 1011km3/s2 a = − = = 6.5152 · 109km . (4.87) 2E/m 2 · 10.14km2/s2

Inserting this into (4.86) gives 2π T = (6.5152 · 109km)3/2 = 9.069 · 109s = 104963d = 287yr . p1.3275 · 1011km3/s2 (4.88) The comet will be seen again in 287 years. Making the perihelion distance just a bit larger, 65.0 million km, we ﬁnd that

E 642 1.3275 · 1011 = − km2/s2 = 5.69km2/s2 . (4.89) m 2 65 · 106 This comet now moves in a hyperbolic orbit, and will never be seen again.

75 4.8 Kepler’s equations

It is possible to integrate the angle equation, dθ ` `(1 + e cos θ)2 dθ µk2 = = ⇐⇒ = dt . (4.90) dt µr2 µ(`2/µk)2 (1 + e cos θ)2 `2 However, the integral that we obtain from this is very ugly and must be expressed in terms of special functions (elliptic integrals) that cannot be straightforwardly inverted to obtain θ as a function of the time t. Instead, we can go back to the energy equation, and the integral (4.29) we obtained from this,

r rµ Z dr t − t0 = 2 q k `2 E + − 2 rmin r 2µr r r µ Z rdr = q (4.91) −2E 2 k k k2 2`2E −r − r − 2 + 2 (1 + 2 ) rmin E 4E 4E µk r r µ Z rdr = . −2E pa2e2 − (r − a)2 rmin We can now make the substitution

r − a = −ae cos ψ (4.92) =⇒ pa2e2 − (r − a)2 = ae sin ψ , (4.93) rdr = (r − a + a)dr = (−ae cos ψ + a)(ae sin ψdψ) (4.94)

Putting this into (4.91) we get s r µ Z rµ k t = (−ae cos ψ + a)dψ = a(ψ − e sin ψ) + C 2|E| k 2|E| (4.95) rµ = a3/2(ψ − e sin ψ) + C. k To determine the integration constant C, we take t = 0 at perihelion. At this point we have

r0 − a = a(1 − e) − a = −ae = −ae cos ψ0 =⇒ ψ0 = 0 ,C = 0 . (4.96)

This gives us r = a(1 − e cos ψ) rµ Kepler’s equations (4.97) t = a3/2(ψ − e sin ψ) k The parameter ψ is called the eccentric anomaly. This name dates back to medieval, ptolemean astronomy where all the heavenly bodies were assumed to move in perfect circles. To rescue this assumption the planets were assumed to sit on circles which

76 were themselves orbiting the earth (or the sun in the Copernicus picture). This epicycle motion motion was called the ‘anomaly’. The angle θ is called the true anomaly. We have already seen that θ = ψ = 0 at the perihelion, and we can also see that θ = ψ = π at aphelion. If e = 0 (circular motion) we have ψ = θ always. Solving Kepler’s equations is not straightforward. In fact, Kepler himself said:

I am sufficiently satisfied that it cannot be solved a priori, on account of the different nature of the arc and the sine. But if I am mistaken, and any one shall point out the way to me, he will be in my eyes the great Apollonius.

4.9 Runge-Lenz vector

We have seen that angular momentum is conserved for central forces, f = −f(r)ˆr with dV (r) potential V (r) so f(r) = − dr , dL d 1 = (r × p) = r˙ × p + r × p˙ = p × p + r × f = 0. dt dt m k For the particular case of an inverse square force, with f = − r3 r, there is another conserved vector, called the Runge-Lenz vector, A = (p × L) − kmˆr.

To see that this is conserved ﬁrst consider the time derivative of the unit vector ˆr. Since the magnitude does not change the derivative can only be a rotation dˆr 1 = ω × ˆr = (L × r). dt mr3 Thus dA k k k = p˙ × L − (L × r) = − (r × L) − (L × r) = 0. dt r3 r3 r3 To understand the physical signiﬁcance of A we calculate its direction and magnitude. Since it is a constant vector we can calculate it at any point in the orbit and, for a bound orbit, it is convenient to do this at the point of closest approach: r = r−a(1 − e) where l = mr−v and L, p and r are all mutually perpendicular. Then p × L = m2v2r so l2 l2 A = rm2v2 − km ˆr = km − 1 ˆr = km − 1 ˆr. kmr− kma(1 − e) k Since a = − 2E this is

2El2 2 2 1 −2El − mk2 − (1 − e) (−e + e) A = − 1 ˆr = ˆr = ˆr = e ˆr. km k2m(1 − e) 1 − e 1 − e We have evaluated this at the point of closest approach, were r is parallel to the major axis of the ellipse, but it is a constant hence A is always parallel to the major axis of A the ellipse and km is the eccentricity e. Thus the constant vector A reproduces facts that we already knew: the closed orbits have constant eccentricity e and keep a constant orientation in space.

77 4.10 Central forces — summary sheet

1. Relative motion, reduced mass We can write the kinetic energy of a two-body system as 1 ˙ 1 T = MR~ 2 + µ~r˙2 , with 2 2 m ~r + m ~r R~ = 1 1 2 2 = centre of mass, m1 + m2 ~r = ~r1 − ~r2 = relative coordinate, m m µ = 1 2 = reduced mass m1 + m2 ˙ For an isolated system, V (~r1, ~r2) = V (~r) =⇒ P~ = MR~ =constant and the CM motion is trivial. 2. Planar motion For V = V (r), all motion is in the plane spanned by ~r and ~p, orthogonal to the constant angular momentum L~ = ~r × ~p. 3. Angular momentum µ L = (r ˙2 + r2θ˙2) − V (r) 2 Since L does not depend on θ,

∂L 2 ˙ pθ = = µr θ = constant = ` ∂θ˙ 4. Kepler’s second law Area swept out in time dt: 1 dA 1 dθ ` dA = r(rdθ) =⇒ = r2 = = constant 2 dt 2 dt 2µ Bodies in a central force field move with constant areal velocity. 5. Effective potential ` 1 `2 θ˙ = =⇒ µr2θ˙2 = µr2 2 2µr2 1 `2 1 H = E = T + V = µr˙2 + + V (r) = µr˙2 + V (r) 2 2µr2 2 eff

78 6. Exploiting energy conservation E = constant r 2 rµ Z r dr0 r˙ = E − Veff(r) =⇒ t(r) = p 0 µ 2 r0 E − Veff(r ) dr dθ dθ ` dθ ` pµ/2 = = =⇒ = 2 2 p 0 dt dr dt µr dr µr E − Veff(r )

7. Kepler problem: orbit equation Integrating dθ/dr with V (r) = −k/r gives s α `2 2E`2 r(θ) = , with α = , e = 1 + . 1 + e cos θ µk µk2

This is the equation for a conic section:

e = 0 circle 0 < e < 1 ellipse e = 1 parabola e > 1 hyperbola

8. Kepler’s ﬁrst law The planets move in elliptic orbits, with the sun at one focus. k GmM Major semiaxis: a = α(1 − e2) = = −2E −2E p rα ` Minor semiaxis: b = a 1 − e2 = a = a1/2 √ a kµ

9. Kepler’s third law dA ` A πab rµ = = = =⇒ T = 2π a3/2 dt 2µ T T k

For planets, µ/k ≈ 1/(GM ) = constant. The square of the orbital period varies like the cube of the major axis. 10. The Laplace–Runge–Lenz vector mk~r A~ = ~p × L~ − , r p2 cos θ A = p p sin θ + θ − mk cos θ , A = {p ,A } x r θ r y θ x

79 Chapter 5

Rotational motion

A rigid body is deﬁned as a body (or collection of particles) where all mass points stay at the same relative distances at all times. This can be a continuous body, or a collection of discrete particles: the same equations of motion hold for both cases. A rigid body will move as a single entity, but it can change its orientation, and this motion can be highly nontrivial as we shall see. The notion of a rigid body is an idealisation, since no real completely rigid bodies exist in the real world. Firstly, real bodies consist of atoms and molecules which always undergo vibrations (and electrons, as quantum particles, are never at rest). These complications can be ignored for macroscopic bodies. Secondly, it is always possible to deform an actual body, and this can happen even in the absence of external forces if for example the internal forces keeping the constituents apart are not strong enough to balance the attractive forces — or vice-versa! Still, the idealised description is reasonable for many macroscopic solids, and provide a good description of for example tops, gyroscopes, bicycle wheels, falling sandwiches and tumbling cats.

Learning outcomes

At the end of this section, you should be able to

• identify appropriate degrees of freedom and coordinates for a rigid body;

• describe rotations using rotation matrices, and explain the general properties of rotation matrices;

• deﬁne the inertia tensor and explain the relation between the inertia tensor, rotational kinetic energy and angular momentum;

• calculate the inertia tensor for simple objects;

• explain what is meant by principal axes of inertia and how they may be found;

• use the equations of motion for rotating bodies (Euler equations) to analyse the motion of rotating systems.

80 5.1 How many degrees of freedom do we have?

Let us imagine a rigid body consisting of N discrete particles. Altogether this gives us 3N coordinates. The requirement that the body is rigid, ie all the internal distances are ﬁxed, imposes constraints on these coordinates. We denote the distance between particles i and j by rij. We can work out how many constraints and hence degrees of freedom we have:

N = 2 Here we have a single constraint r12 = r, and hence we have a total of 3 · 2 − 1 = 5 degrees of freedom.

N = 3 We now have 3 internal distances to be ﬁxed: r12, r13 and r23, giving 3 constraints and 3N − 3 = 6 degrees of freedom.

N = 4 Here there are 6 internal distances to be fixed, so we have 3 · 4 − 6 = 6 degrees of freedom. Strictly speaking, there are 2 different configurations which both satisfy these 6 constraints, corresponding to rigid bodies that are mirror images of each other. This ambiguity does not correspond to any physical degree of freedom, however.

N = 5 Now it appears there would be 10 constraints, corresponding to the 10 pairs of particles we have. However, if you construct a 5-particle body from a 4-particle one, you will see that once the ﬁfth particle has been positioned relative to 3 of the others, the position relative to the fourth one is also given. (You may try this for yourself!) We therefore only have 9 constraints, and 3 · 5 − 9 = 6 degrees of freedom.

N ≥ 6 As was the case for N = 5, we will need to specify 3 relative distances to position the 6th particle relative to the other 5, 3 more for the 7th particle, etc. This cancels out the 3 coordinates that each new particle comes with, so we end up with 6 degrees of freedom in every case.

In summary, we ﬁnd that a rigid body has 6 degrees of freedom, except for N = 2, which is a special case, and has only 5 degrees of freedom. A more careful analysis will reveal that there are only 5 degrees of freedom whenever the rigid body is 1-dimensional (all the constituent particles are located on a single line), and 6 otherwise. Three of these degrees of freedom can be taken to represent the position of the body, and it is natural to use the centre of mass coordinates of the body as corresponding generalised coordinates. The three remaining degrees of freedom represent the orientation of the body, and it is natural to choose three angles as coordinates. We will come back to how these angles may be chosen later on. Dynamically, the 3+3 degrees of freedom correspond to two diﬀerent kinds of motion: the linear motion of the centre of mass, and the rotation of the body about its centre of mass. We can now see why there are 3 such degrees of freedom: they correspond to rotations (changes in orientation) about 3 axes going through the centre of mass. In the case of a 1-dimensional body, there are only 2 rotational degrees of freedom, since rotation about the line the body is located on does not correspond to any real motion.

81 This discussion can be summarised in Chasles’ theorem, which states

Any motion of a rigid body is the sum of a translation and a rotation.

From now on we will focus on how we can describe and study the rotational motion and degrees of freedom.

5.1.1 Relative motion as rotation

Suppose a rigid body is rotating about some axis. Choose some particular point in the body on the axis of rotation — we can choose this point to be the origin of our coordinate system. For any other point in the body the position ~r changes if the point is not on the axis of rotation but the distance r of the point from the origin does not change, because the body is assumed to be rigid, only the direction from the origin can change. In a short time the displacement of the point d~r will be at right angles to the position, d~r ⊥ ~r. This defines an infinitesimal rotation angle dφ about the axis passing through the origin. Since all the points in the body remain at a fixed distance relative to the origin, they all rotate about an axis through the origin, and since the body retains its shape, they all rotate about the same axis. This can be summarised in Euler’s theoremx:

Any movement of a rigid body with one point ﬁxed is a rotation about some axis.

If we say that the vector dφ~ points along the axis of rotation, and this vector forms and angle θ with the position vector ~r, we have |d~r| = |~r| sin θ|dφ~| or d~r = dφ~ × ~r. (5.1)

The velocity of the point ~r relative to the origin is then d~r dφ~ ~v = = × ~r = ~ω × ~r . (5.2) dt dt If in addition, the whole body moves with some linear velocity V~ , the total velocity of the point P is ~ ~ ~v = V + ~vrel = V + ~ω × ~r (5.3) You have in the past studied the case where the axis of rotation, ie the direction of ~ω is ﬁxed. But in the general case, ~ω = ~ω(t) can change both magnitude and direction, and we need to describe this general situation.

5.2 Rotated coordinate systems and rotation matrices

Before we go on to discuss the kinematics and dynamics of rotational motion, let us look at the coordinates we can use to describe the rigid body. It is clearly convenient to

82 dφ= d φ n

θ dr O r dφ

Figure 5.1: A clockwise rotation through an inﬁnitesimal angle dφ about an axis pointing in the direction of a unit vector ~n is represented by a vector dφ~ = dφ~n whose length is dφ and direction is ~n.

describe the shape of the body, or the relative positions of the constituent parts of the body, using a coordinate system that is sitting in the body and moving with it. It will also turn out to be convenient to describe the rotational motion of the body in such a system, since it is natural to describe this as the body rotating about its own axes. We call such a coordinate system the body coordinate system. But to describe the motion of the body in space we need a coordinate system fixed in space, not moving with the body. We therefore need to know the relation between the body coordinate system and this fixed coordinate system. Ignoring any linear displacements, these coordinate systems will be rotated relative to each other. We therefore need to know how coordinates change when the coordinate system is rotated. Let us call the original coordinates of a point (eg in the fixed coordinate system) ~r = (x, y, z) and the coordinates in the rotated (eg the body) coordinate system r~0 = (x0, y0, z0). We call the angles between the axes in the original and the rotated 0 coordinate systems θij, i = x, y, z: for example, θxz is the angle between the x -axis and the z-axis. We can then write the position vector (or indeed any vector) as ~r = xxˆ + yyˆ + zzˆ = x0xˆ0 + y0yˆ0 + z0zˆ0 . (5.4) Since xˆ0, yˆ0, zˆ0 are orthogonal, we can find the rotated coordinate x0 by 0 0 0 0 0 x = ~r · xˆ = xxˆ · xˆ + yyˆ · xˆ + zzˆ · xˆ = x cos θxx + y cos θxy + z cos θxz . (5.5) Similarly we find 0 0 y = x cos θyx + y cos θyy + z cos θyz , z = x cos θzx + y cos θzy + z cos θzz , (5.6) This can be written as a matrix equation,  0     x cos θxx cos θxy cos θxz x 0 0 y  = cos θyx cos θyy cos θyz y ⇐⇒ ~r = R~r (5.7) 0 z cos θzx cos θzy cos θzz z The 3 × 3 matrix R is called the rotation matrix.

83 5.2.1 Active and passive transformations

There are two equivalent ways of thinking about rotations:

1. You rotate the coordinate system, as described above. This is called a passive rotation, since it is not doing anything to the world, only to the mathematical description of it. The rotation matrix then gives the relation between the old and the new coordinates.

2. You leave the coordinate system in place, but rotate your points (for example the body you want to describe) in the opposite direction. This is called an active rotation, since you are now doing something to the world. The rotation matrix then gives the positions of the body after it has been rotated, given the position before.

The two points of view are mathematically equivalent: the relation between old and new coordinates are exactly the same. The two pictures: active and passive, and the equivalence between them, can be extended to other transformations, such as translations, where it is the same whether you move an object from a position x to a position x + a or you shift the coordinate system by −a.

5.2.2 Elementary rotation matrices

For a rotation about the z-axis, it is clear that the z-coordinates are unchanged. If we rotate an angle θ, then we have that θxx = θyy = θ (the angles between the old and new x- and y-axes respectively). The angle θxy between the old y-axis and the new x-axis is ◦ ◦ 90 − θ, while the angle θyx between the old x-axis and the new y-axis is 90 + θ. The rotation matrix is therefore  cos θ cos(90◦ − θ) 0  cos θ sin θ 0 ◦ Rz(θ) = cos(90 + θ) cos θ 0 = − sin θ cos θ 0 (5.8) 0 0 1 0 0 1

Similarly we ﬁnd for rotations about the x- and y-axes,

1 0 0  cos θ 0 − sin θ Rx(θ) = 0 cos θ sin θ ,Ry(θ) =  0 1 0  . (5.9) 0 − sin θ cos θ sin θ 0 cos θ

5.2.3 General properties of rotation matrices

1. Any combination of two successive rotations about the same point (albeit about diﬀerent axes through that point) is a rotation about that point. This is ‘obvious’, since the relative distance of each point from the ﬁxed point (origin) remains unchanged throughout. We can therefore describe the combined rotation by a rotation matrix which is the product of the two rotation matrices.

84 If we call the coordinates after the first rotation ~r0 and after the second rotation ~r00, and the two rotation matrices R (first rotation) and R0 (second rotation), we have ~r00 = R0~r0 = R0(R~r) = (R0R)~r = R00~r with R00 = R0R. (5.10) Note that the first rotation matrix is on the right and the second rotation matrix is on the left.

Example 5.1 Find the rotation matrix corresponding to a 30◦ rotation about the x-axis followed by a 45◦ rotation about the z-axis. Solution: A rotation of an angle 30◦ about the x-axis is, according to (5.9),     1 0 0 1√ 0 0 R (30◦) = 0 cos 30◦ sin 30◦ = 0 1 3 1 . (5.11) x    2 √2  0 − sin 30◦ cos 30◦ 1 1 0 − 2 2 3 The matrix for a 45◦ rotation about the z-axis is √ √  cos 45◦ sin 45◦ 0  1 2 1 2 0 2 √ 2 √ ◦ − sin 45◦ cos 45◦ 0 1 1 Rz(45 ) =   = − 2 2 2 2 0 . (5.12) 0 0 1 0 0 1

The combined rotation is √ √  1 2 1 2 0 1 0 0  2 √ 2 √ √ R = R R = 1 1 0 1 3 1 z x − 2 2 2 2 0  2 √2  0 0 1 0 − 1 1 3 2 2 √ √ √  1 2 1 6 1 2 2 √ 4 √ 4 √ = − 1 2 1 6 1 2 (5.13)  2 4 4 √  1 1 0 − 2 2 3

2. Rotations (about diﬀerent axes) are not commutative: the order in which you do them matters. We know that matrix multiplication is not commutative: AB 6= BA if A and B are matrices. You can also show for yourself that if you for example rotate a book 90◦ about the x-axis followed by 90◦ about the y-axis, you get something diﬀerent from doing the two in reverse order.

Example 5.2 The rotation matrix for a 45◦ rotation about the z-axis followed by a 30◦

85 rotation about the x-axis is √ √ 1 0 0   1 2 1 2 0 √ 2 √ 2 √ R0 = R R = 0 1 3 1 1 1 x z  2 √2  − 2 2 2 2 0 0 − 1 1 3 0 0 1 2 2 √ √  1 2 1 2 0  2 √ 2 √ = − 1 6 1 6 1 6= R R (5.14)  4√ 4 √ √2  z x 1 1 1 4 2 − 4 2 2 3

3. All rotation matrices are orthogonal, RT R = 11. Proof: We know that the distance of any point from the origin is the same before and after the rotation. Therefore we have that x0 02 0 0 0 0 0T 0 T T T 2 T ~r = x y z y  = r r = (Rr) (Rr) = r R Rr = ~r = r r (5.15) z0 ⇐⇒ RT R = 11 . (5.16)

We can check explicitly that all the matrices in the examples above are orthogonal. We can also show that if R and R0 are orthogonal, then their product RR0 is also orthogonal:

(RR0)T (RR0) = (R0T RT )(RR0) = R0T (RT R)R0 = R0T R0 = 11 . (5.17)

4. For every rotation R there is an inverse rotation given by RT which brings us back to our starting point. It is physically obvious that you can always get back to your starting point by reversing all your rotations in reverse order. It follows from point 3. that the inverse rotation matrix is given by RT .

5. (a) Any (proper) rotation can be expressed as a combination of elementary rotations about coordinate axes. (b) No combination of such rotations can produce a reflection ~r → −~r. (c) All proper rotations have det R = 1. Improper rotations (involving an odd number of reflections) have det R = −1. The proof of (a) is complicated, but we will use this fact later on when we will define coordinates corresponding to the the rotational degrees of freedom. Statement (c) follows from the properties of determinants. It is straightforward to show that the elementary rotation matrices all have det R = 1. We also have

det(RR0) = (det R)(det R0) (5.18)

so any combination of elementary rotations must have determinant 1. Moreover, any orthogonal matrix R must have det R = ±1:

det 11= 1 = det(RT R) = det RT det R = (det R)2 . (5.19)

86 Finally, the reﬂection matrix, which takes (x, y, z) → (−x, −y, −z) is

−1 0 0  I =  0 −1 0  , det I = −1 (5.20) 0 0 −1

(the notation is meant to convey I for invert in the origin). It is clearly impossible to construct this from a product of matrices with determinant 1. Physically, this means that it is impossible (in 3 dimensions) to rotate a body into its mirror image.

5.2.4 The rotation group [optional]

A group is deﬁned as a set of elements, together with a composition (multiplication) operation, with the following properties:

1. There exists an identity element (called 1 or e) which is a member of the group.

2. The combination a · b of two elements a and b is also a member of the group.

3. For every member a of the group there is an inverse a−1 which is also a member of the group, such that a · a−1 = a−1 · a = 1.

We see that rotations form a group according to this definition: properties 1 and 4 of Sec. 5.2.3 correspond to properties 2 and 3 above. The identiy element is the operation of doing nothing, corresponding to the identity matrix. We can use the rotation matrices to define this group, and this gives the rotation group its name: O(3), or “the group of real orthogonal 3×3 matrices. The set of all proper rotations also form a (smaller) group, called SO(3), or “the group of all special [determinant 1] real orthogonal 3×3 matrices”. However, this is merely a particular representation of the general operation that we call rotations. We could equally well represent the group elements by actual rotations in space, or by 3 angles (with a suitably defined multiplication operation), or in many other ways. All these different representations would share the same multiplication table, which is what ultimately defines the group. This theory of different representations of the same group is mathematically extremely powerful, and the group-theory properties of rotations are extremely important in modern physics. For example, particles and bodies in general may be classified according to how they behave when rotated, and this turns out to be a fundamental classification. It is possible to express any rotation matrix formally as

0 0 0 0 0 −1  0 1 0 L~ ·T~ R = e , with T1 = 0 0 1 ,T2 = 0 0 0  ,T3 = −1 0 0 . 0 −1 0 1 0 0 0 0 0 (5.21) This expression is very useful theoretically, but useless if you want to ﬁnd an explicit form for R. It does however give a direct connection between rotations and angular momentum. The T matrices obey the commutation relations

[T1,T2] = −T3 , [T2,T3] = −T1 , [T3,T1] = −T2 . (5.22)

87 These are essentially the commutation relations of the angular momentum operators in quantum mechanics, and the vector L~ is proportional to the angular momentum of the particle in question. But it turns out that there is another group with the same multiplication table, namely the group of 2 × 2 complex unitary matrices with determinant 1, called SU(2). The operations of this group actually describe the rotations of fermions, while bosons such as photons are described by the usual 3 × 3 rotation matrices. The SU(2) matrices can be written in a similar form to (5.21),

R = eiS~·~σ , (5.23) where 0 1 0 −i 1 0 σ = , σ = , σ = , (5.24) 1 1 0 2 −i 0 3 0 −1 are the three Pauli matrices, which obey the same commutation relations as the T - matrices in (5.21) (up to a factor i). The vector S~ is a new form of angular momentum called spin, which corresponds to an ‘internal rotation’ degree of freedom. A curious result is that we must rotate a fermion by 4π (2 full rotations) to get back to where we started!

5.3 Euler angles

We now go on to discuss which coordinates we can use to describe the orientation of a rigid body, or alternatively, which three parameters can be used to uniquely obtain a rotated coordinate system from an original one. There are several possibilities:

• The most natural would be to use rotation angles about the x-, y- and z-axes. This leads to the three Tait–Bryan angles, which are widely used for aircraft. However, since rotations do not commute, unless supplemented with a prescription for the order of the three rotations, these angles are not unique, except for small rotations. • The next most natural parameter would be to ﬁnd the axis and angle of rotation, ie the vector φ~ above. This is called the Cayley–Klein or Euler parameters. These are mathematically very nice, and can be related to the vector L~ given above, but are not very practical. • The third, most widely used parametrisation in mechanics is in terms of the three Euler angles. Here, a general rotation is constructed from 3 successive elementary rotations:

1. The body is rotated an angle φ ∈ [0, 2πi about the z-axis. The x- and y-axes move to x0 and y0, while the z = z0 axis is unchanged. 2. The body is rotated an angle θ ∈ [0, π] about the new x0-axis. The z0- and y0-axes move to z00 = z0 and y00 = y0, but the x00 = x0 axis is unchanged. 3. The body is rotated an angle ψ ∈ [0, 2πi about the new z00-axis. The x00 and y00-axes move to x000 and y000, while the z000 = z00-axis is unchanged.

Note that there is no rotation about any y-axis here; instead there are two rotations about (diﬀerent) z-axes. These three rotations are summarised in ﬁgure 5.2. It can be proved that any rotation can be expressed in this way.

88 Figure 5.2: The Euer angles

5.3.1 Rotation matrix for Euler angles

We can now construct the general rotation matrix explitly as a function of the Euler angles. Calling the vectors in the intermediate coordinate systems ~ρ, ρ~0 respectively, the 0 first rotation gives us ~ρ = Rz(φ)~r. The second rotation gives us ρ~ = Rx(θ)~ρ, and the 0 0 final rotation r~ = Rz(ψ)ρ~ . Putting all this together, we find

R = Rz(ψ)Rx(θ)Rz(φ)  cos ψ sin ψ 0 1 0 0   cos φ sin φ 0 = − sin ψ cos ψ 0 0 cos θ sin θ − sin φ cos φ 0 0 0 1 0 − sin θ cos θ 0 0 1  cos ψ cos φ − sin ψ cos θ sin φ cos ψ sin φ + sin ψ cos θ cos φ sin ψ sin θ = − sin ψ cos φ − cos ψ cos θ sin φ − sin ψ sin φ + cos ψ cos θ cos φ cos ψ sin θ . sin θ sin φ − sin θ cos φ cos θ (5.25)

Example 5.3 Find the Euler angles corresponding to a rotation of an angle ϑ about the y-axis. The rotation matrix is cos ϑ 0 − sin ϑ Ry(ϑ) =  0 1 0  . (5.26) sin ϑ 0 cos ϑ

Comparing the bottom right (zz) element of the matrices, we immediately see that ϑ = θ. Comparing the bottom left elements we then must have sin φ = 1 ⇒ φ = π/2. While comparing the top right elements requires sin ψ = −1, so ψ = 3π/2. You may then conﬁrm that all the other elements come out as desired. You may also check for yourself that you may indeed achieve a rotation about the y axis by the following combination of rotations:

1. rotate it by 90◦ about the z-axis;

2. rotate it about the x-axis (by your desired angle);

89 3. rotate it back by 90◦ about the z-axis.

5.3.2 Euler angles and angular velocity

The total angular velocity can be constructed as the sum of angular velocities that result from the changes in each of the three Euler angles. Note that simply adding these three vectors together is ok, since these correspond to inﬁnitesimal changes in orientation, and for such changes the order does not matter.  1 δφ 0 2 Rz(δφ) = 1 + −δφ 1 0 + O(δφ ) 0 0 1 1 0 0  2 Rx0 (δθ) = 1 + 0 1 δθ + O(δθ ) 0 −δθ 1  1 δψ 0 2 Rz00 (δψ) = 1 + −δψ 0 0 + O(δψ ) 0 0 1

Hence  1 δφ 0 1 0 0   1 δψ 0 Rz00 (δψ)Rx0 (δθ)Rz(δφ) = 1+−δφ 1 0+0 1 δθ+−δψ 0 0+··· . 0 0 1 0 −δθ 1 0 0 1 (5.27) To lowest order it does not matter what order we combine the rotations, and this is all that matters for derivatives. For angular velocities we can therefore write

~ω = φ~˙ + θ~˙ + ψ~˙ . (5.28) The magnitude of each of these three vectors is the respective angular velocity: φ,˙ θ,˙ ψ˙. But in which directions are they pointing? We will in the end want to express the angular velocity in the body coordinate system, since it will turn out that the equations of motion are best expressed in these coordinates. This requires performing a further ﬁnite rotation through Euler angles (φ, θ, ψ) on the angular velocity ~ω:

φ~˙ This vector points along the original z-axis, which is unchanged by the first rotation. The second rotation is about the intermediate x0-axis, and changes the vector (0, 0, φ˙) → (0, φ˙ sin θ, φ˙ cos θ). After the final rotation of an angle ψ about the final z00-axis, we get

φ~˙ = (φ˙ sin θ sin ψ, φ˙ sin θ cos ψ, φ˙ cos θ) . (5.29)

θ~˙ This vector points along the intermediate x0-axis, so after the second rotation we have θ~˙ = (θ,˙ 0, 0). After the ﬁnal rotation about the ﬁnal z00-axis it becomes

θ~˙ = (θ˙ cos ψ, −θ˙ sin ψ, 0) . (5.30)

90 ψ~˙ This vector points along the body’s ﬁnal z00-axis, so ψ~˙ = (0, 0, ψ˙).

Adding up these, we get

φ˙ sin θ sin ψ + θ˙ cos ψ ~ ~ ~ ~ω = φ˙ + θ˙ + ψ˙ = φ˙ sin θ cos ψ − θ˙ sin ψ (5.31) φ˙ cos θ + ψ˙ We will use this to derive the equations of motion. But ﬁrst we need to determine the kinetic energy.

5.4 The inertia tensor

5.4.1 Rotational kinetic energy

Let us now work out the total kinetic energy of a rigid body. From (5.3) we get that 1 X 1 X h i T = m (V + ~ω × r )2 = m V2 + 2V · (~ω × r ) + (~ω × r )2 2 α α 2 α α α α α ! (5.32) 1 X 1 X = MV2 + V · ~ω × m r + m (~ω × r )2 2 α α 2 α α α α P where M = α mα is the total mass. If one point P in the body is fixed, ie the motion is pure rotation, then the one can chose the origin of the body coorinate system at P . Then V = 0 so the first two terms vanish, and the third is the rotational energy Trot, where rα is the distance from particle number α to the point P . More generally suppose the body is rotating about an axis and at the same time a point P on an axis of rotation, but fixed withing the body, is moving with linear velocity V~ . Let R be the position of the centre of mass relative to the point P and rα the position of a particle α in the body relative to P . Then we write

rα = R + erα (5.33) where erα is the position of the particle α relative to the centre of mass. The centre of mass is deﬁned by X mαerα = 0. (5.34) α In terms of erα (5.32) can be simpliﬁed using (5.34) and written ! 1 X 1 X 2 T = MV2 + V · ~ω × m (R + r ) + m ~ω × (R + r ) 2 α eα 2 α eα α α 1 1 1 X = MV2 + MV · (~ω × R) + M(~ω × R)2 + m ~ω × r )2 (5.35) 2 2 2 α eα α 1 1 X = MV2 + m ~ω × r )2, 2 CM 2 α eα α

91 where VCM = V + (~ω × R) is the velocity of the centre of mass. We have derived the simple result that

1 2 1 X 2 T = TCM + Terot = MV + mα(~ω × rα) , (5.36) 2 CM 2 e α where erα is the distance of particle α in the body from the centre of mass. We now use that the angular velocity ~ω is the same for the whole body. We furthermore use the identity

~ ~ 2 2 2 2 2 2 2 2 2 ~ ~ 2 (A × B) = A B sin θAB = A B (1 − cos θAB) = A B − (A · B) . (5.37)

With this the rotational kinetic energy in (5.36) becomes

1 X 2 2 2 Terot = mα r ω − (rα · ω) (5.38) 2 eα e α 1 X X X = m r2 ω2 − (x ω )(x ω ) (5.39) 2 α eα i eα,i i eα,j j α i ij 1 X X X = m x2 δ − x x ω ω (5.40) 2 α eα,k ij eα,ieα,j i j ij α k 1 X 1 = Ieijωiωj = ~ω · eI · ~ω. (5.41) 2 2 ij

We end up with

1 X Terot = Ieijωiωj , (5.42) 2 ij Z X 2 2 3 Ieij = mα(reαδij − xeα,ixeα,j → ρ(xe)(xe δij − xeixej)d x (5.43) α

The parameters Ieij are called the moments of inertia and the 3 × 3 matrix eI = {Ieij} is called the moment of inertia tensor or often just the inertia tensor of the body. It characterises the way the body rotates about any particular axis and how it responds to torques.. We have defined the inertia about the centre of mass (CM). If we had chosen a different point P we would have got a different inertia tensor — it depends on the choice of P (figure 5.3). More generally the inertial tensor is defined as

X 2 Iij = mα(rαδij − xα,ixα,j (5.44) α

92 ω

~ r mα α CM

R r α

Figure 5.3: A rigid body rotating about an axis through point P ﬁxed in the body. The centre of mass (CM) is displaced by R from P . with rα = R + erα. Then X 2 Iij = mα(rαδij − rα,irα,j) (5.45) α X 2 = mα |R + erα| δij − (Xi + xeα,i)(Xj + xeα,j α X 2 2 = mα (R + reα)δij − (XiXj + xeα,ixeα,j α 2 X 2 = M(R δij − XiXj) + mα(reαδij − xeα,ixeα,j) α 2 ⇒ Iij = M(R δij − XiXj) + Ieij. (5.46) where Xi are the Cartesian co-ordinates of the centre of mass. Iij are the components of the inertia tensor about a general point P , relative to which the centre of mass is at R. With the moment of inertia tensor in hand we can calculate the moment of inertia associated with rotation of any axis. Let n = ~ω/ω be a unit vector along the axis of rotation, then the moment of inertia about n is deﬁned to be

3 X I(n) = Iijninj (5.47) i,j=1 and the rotational kinetic energy is 1 T = I(n)ω2. (5.48) Rot 2

93 n

ω =ω n

CM d θ

Figure 5.4: Geometry of the Parallel Axes Theorem.

If ~ω is directed purely along one of the coordinate axes, for example ~ω = (0, 0, ω), we see that Trot reduces to the expression we have encountered before, 1 X 1 T = I ω ω = I ω2 , (5.49) rot 2 ij i j 2 zz ij where Izz is the moment of inertia about the z-axis. But the expression (5.42) is much more general, and will hold in any (orthogonal) coordinate system, regardless of which direction the rotation vector is pointing.

The inertia tensor Iij depends on the point about which it is calculated but does not depend on any particular axis. The moment of inertia I(n) depends on the point about which it is calculated and on a choice of axis. We can see how I(n) depends on the point from (5.46). Referring to ﬁgure 5.4

I(n) = MR2(1 − sin2 θ) + Ie(n) = Md2 + Ie(n). (5.50)

This result is known as the Parallel Axes Theorem, the diﬀerence between I(n) and Ie(n) depends only on the distance between the two axes and not on their direction. We can now calculate the inertia tensor once and for all. Once we know I in one coordinate system, we shall see that we can relatively easily ﬁnd it in any other coordinate system.

94 Example 5.4 Find the inertia tensor of the dumbbell pictured in ﬁg. 5.5, and ﬁnd the kinetic energy if it rotates with y 6 m1 angular velocity ω ¡ ¡ z ¡ 1. about the y-axis, ¡ r1 ...... θ...... ¡ ¡ 2. about the z-axis, ¡ - ¡ x 3. about its own axis. ¡ r2 ¡ ¡ Answer: ¡ m2 x Particle 1 is located at ~r1 = (r1 sin θ, r1 cos θ, 0). Particle 2 is located at ~r2 = (−r2 sin θ, −r2 cos θ, 0). Figure 5.5: A dumbbell diag prod diag P 2 We divide the computation of I as I = I − I where Ii,j = δi,j α mαrα and prod P Ii,j = α mαxα,ixα,j. The two terms are computed as

diag 2 2 I = (m1r1 + m2r2)1  2 2  m1r1 + m2r2 0 0 2 2 =  0 m1r1 + m2r2 0  2 2 0 0 m1r1 + m2r2 and

 2 2 2   2 2 2  r1 sin θ r1 sin θ cos θ 0 r2 sin θ r2 sin θ cos θ 0 prod 2 2 2 2 2 2 I = m1 r1 sin θ cos θ r1 cos θ 0 + m2 r2 sin θ cos θ r2 cos θ 0 0 0 0 0 0 0  sin2 θ sin θ cos θ 0 2 2 2 = (m1r1 + m2r2) sin θ cos θ cos θ 0 0 0 0

Subtracting them we get the full intertial tensor I as

 cos2 θ  z }| {  2   1 − sin θ − sin θ cos θ 0 2 2  sin2 θ  I = (m1r1 + m2r2)    z }| 2 {  − sin θ cos θ 1 − cos θ 0 0 0 1

1. For rotation about the y-axis, ~ω = (0, ω, 0), so 1 1 T = I ω2 = (m r2 + m r2)ω2 sin2 θ rot 2 yy 2 1 1 2 2

2. For rotation about the z-axis, ~ω = (0, 0, ω), so 1 1 T = I ω2 = (m r2 + m r2)ω2 . rot 2 zz 2 1 1 2 2

95 3. For rotation about the body axis, ~ω = (ω sin θ, ω cos θ, 0), so 1 1 1 1 T = I ω2 + I ω ω + I ω ω + I ω2 rot 2 xx x 2 xy x y 2 yx y x 2 yy y 1 = (m r2 + m r2)ω2[cos2 θ sin2 θ − 2(sin θ cos θ)(sin θ cos θ) + sin2 θ cos2 θ] 2 1 1 2 2 = 0 .

It should not be a surprise that we get Trot = 0 here, since nothing is actually moving in this case!

Example 5.5 Find the inertia tensor for a homogeneous cube with mass M and length L with the origin at one corner and edges along coordinate axes. Answer: The density of the cube is ρ = M/L3. We ﬁnd

L L L L L Z Z Z M M Z Z I = (y2 + z2)dxdydz = (y2 + z2)dydz xx L3 L2 0 0 0 0 0 (5.51) L M Z L3 M L4 L3 2 = + z2L dz = + L = ML2 , L2 3 L2 3 3 3 0 L L L Z Z Z M I = (−xy)dxdydz xy L3 0 0 0 (5.52) L M Z 1 M 1 1 = − L2ydy = − L2 = − ML2 . L2 2 2 2 4 0

Since the x, y and z axes are completely symmetric, it is clear that Ixx = Iyy = Izz and Ixy = Ixz = Iyz, and the inertia tensor is

 2 2 1 2 1 2  2 1 1  3 ML − 4 ML − 4 ML 3 − 4 − 4 1 2 2 2 1 2 2 1 2 1 I = − 4 ML 3 ML − 4 ML  = ML − 4 3 − 4  . (5.53) 1 2 1 2 2 2 1 1 2 − 4 ML − 4 ML 3 ML − 4 − 4 3

5.4.2 What is a tensor? Scalars, vectors and tensors.

In practice, you can think of a tensor as a kind of matrix, where the rows and columns correspond to directions in space. You may then treat the expression for Trot as a vector–matrix–vector multiplication.

96 In principle, a tensor is a generalisation of a vector, ie a physical (or mathematical) quantity with more than one direction. Tensors (and vectors) are defined by their transformation properties, and in particular how they change when they, or the coordinate system, are rotated. We shall see that in a rotated coordinate system, the inertia tensor is given by 0 X Iij = RikRjlIkl . (5.54) kl This is the defining property of a tensor (to be precise, a rank-2 tensor, ie a tensor with 2 indices). This mirrors the definition of scalars and vectors, which are defined as follows:

• A scalar is a quantity that does not change when you rotate your coordinate system. Examples of this is the length of a vector, or the kinetic energy. • A vector is a quantity v that transforms in the same way as the position vector, ie

0 X vi = Rijvj . (5.55) j

There are however some vector and scalar type quantities that transform diﬀerently under reﬂections:

• An ordinary vector will change its sign when seen in a mirror, but for example the angular momentum vector, L~ = ~r × ~p, will keep the same sign (since both ~r and ~p change sign). Such vectors are called pseudovectors or axial vectors. • Finally, ordinary scalars will be unchanged under reﬂections, but some quantities change sign. These are called pseudoscalars. For example, ~v ·L~ (the scalar product of the velocity and angular momentum) would be a pseudoscalar.

Other examples of tensors

The stress tensor. The picture on the right depicts a fluid flowing - with a velocity ~u in the x-direction. The fluid flows faster at larger - y 6 - y, so that ∂yux 6= 0. In a viscous fluid this will create stress (shear) - forces in the y-direction. This can be expressed through the stress - - tensor σxy. The diagonal components of this tensor represent the x pressure of the fluid, eg σxx is the pressure in the x-direction. The same picture also governs stress forces in solid materials, if spatially varying forces are applied.

The (outer) product of two vectors. If we deﬁne ~ T = ~a ⊗ b ⇐⇒ Tij = aibj , (5.56) 0 then we can easily see that this satisﬁes the rotation transformation property Tij = P ~ kl RikRjlTkl. The vector product ~c = ~a × b can be obtained as the antisymmetric part of this tensor,

cx = Tyz − Tzy , cy = Txz − Tzx , cz = Txy − Tyx . (5.57)

97 The electromagnetic field. In 4-dimensional spacetime, the electric field E~ and ~ magnetic field B form the (antisymmetric) field tensor Fµν = −Fνµ,   0 Ex Ey Ey −Ex 0 Bz −By {Fµν} =   . (5.58) −Ex −Bz 0 Bx  −Ez By −Bx 0

(Note that in 4-dimensional space-time, ‘rotations’ include Lorentz boosts.)

5.4.3 Angular momentum and the inertia tensor

We want to ﬁnd the angular momentum L~ of a rigid body about some point O. This point can be

• the centre of mass (for a body tumbling freely in space, for example), or

• some point ﬁxed in space (for example a spinning top).

The momentum of some particle α is ~pα = mα~vα = mα~ω ×~rα. Using the vector identity

A~ × (B~ × A~) = A2B~ − (A~ · B~ )A~ (5.59) we can write the total angular momentum as ~ X X X 2 L = ~rα × ~pα = mα ~rα × (~ω × ~rα) = mα rα~ω − (~rα · ~ω)~rα (5.60) α α α X 2 X X X 2 Li = mα(rαωi − xα,jωjxα,i) = mα (rαδij − xα,jxα,i)ωj (5.61) α j α j X h X 2 i X = mα(rαδij − −xα,jxα,i) ωj = Iijωj . (5.62) j α j

So we ﬁnd

X ~ Li = Iijωj or L = I · ~ω (5.63) j

5.5 Principal axes of inertia

5.5.1 Rotations and the inertia tensor

The rotational kinetic energy depends on the point about which it is calculated and on the direction of the axis about which the body is rotating, but it should not depend on our choice of x, y and z axis in space. How does the inertia tensor change if we rotate

98 our coordinate system? To answer this question, we can use the fact that the kinetic energy must be the same regardless of how we choose our coordinates (it is a scalar):

1 X 1 X T = ω I ω = ω0I0 ω0 . (5.64) rot 2 i ij j 2 i ij j ij ij

But ~ω is a vector, and changes in the same way as the position vector when rotated1,

0 X ωi = Rikωk . (5.65) k Substituting this into (5.64), we get

1 X X X 1 X X 1 X T = R ω I0 R ω = ω R I0 R ω = ω I ω . rot 2 ik k ij jl l 2 k ik ij jl l 2 k kl l ij k l kl ij kl (5.66) P 0 T 0 −1 0 So we ﬁnd that Ikl = ij RikIijRjl or I = R I R = R I R. Life would be a lot simpler if, in some coordinate system, the inertia were diagonal,   I1 0 0 I =  0 I2 0  (Iij = Iiδij) . (5.67) 0 0 I3

We would then have 1 X L = I ω ; T = I ω2 . (5.68) i i i rot 2 i i i In particular, if the body rotates about one of the axes of such a coordinate system, we ~ 1 2 have L = I~ω, Trot = 2 Iω . The good news is that it is always possible to ﬁnd such a set of (body) coordinate axes. These axes are called principal axes of inertia, and the corresponding I1,I2,I3 are the principal moments of inertia. The question then is how we ﬁnd these axes. There are two methods:

• Find a rotation matrix R such at R IRT = I0 is diagonal. The coordinate axes in the rotated system are then the principal axes.

• Find vectors (directions) ~ω such that L~ = I · ~ω = I~ω. These vectors form the principal axes of inertia, and the numbers I are the principal moments.

In fact, both these methods are mathematically identical. Let us ﬁrst look at method 2. If ~ω points along a principal axis of inertia, we have X Li = Iijωj = Iωi , (5.69) j

1Strictly speaking, this is not quite the case: under a reﬂection, ~r → −~r, but ~ω is unchanged. We call a vector which behaves this way a pseudovector or axial-vector. The angular momentum is another example of such a pseudovector.

99 where I is the corresponding moment of inertia, or

L1 = Iω1 = I11ω1 + I12ω2 + I13ω3

L2 = Iω2 = I21ω1 + I22ω2 + I23ω3 (5.70)

L3 = Iω3 = I31ω1 + I32ω2 + I33ω3

This is an eigenvalue problem. The condition for a nontrivial solution is det(I−I11)= 0. This gives us a (third-degree) equation for I, called the characteristic equation. The solutions are the principal moments of inertia (or eigenvalues of I). Once we have found one of the solutions (eigenvalues), we can ﬁnd the corresponding principal axis by substituting the values of I into the equation for L~ = I~ω = I · ~ω. This gives us the direction of ~ω, or equivalently, the ratios ω1 : ω2 : ω3. The vectors ~ω are the eigenvectors of I.

Example 5.6 Show that the homogeneous cube with the origin at one corner has a principal 1 2 moment of inertia I = 6 ML , and ﬁnd the corresponding principal axis of inertia. Answer: We found that the inertia tensor was

 2 1 1  3 − 4 − 4 2 1 2 1 I = ML − 4 3 − 4  . (5.71) 1 1 2 − 4 − 4 3

1 2 I is a principal moment of inertia if det(I − I11)= 0. Setting I = 6 ML , we ﬁnd

2 2 1 2 1 2 1 2 1 2 1 2 3 ML − I − 4 ML − 4 ML 2 ML − 4 ML − 4 ML 1 2 2 2 1 2 1 2 1 2 1 2 − 4 ML 3 ML − I − 4 ML = − 4 ML 2 ML − 4 ML 1 2 1 2 2 2 1 2 1 2 1 2 − 4 ML − 4 ML 3 ML − I − 4 ML − 4 ML 2 ML

2 −1 −1 1 23 1 23 = ML −1 2 −1 = ML 2(2 · 2 − 1) + 1(−1 · 2 − 1) − 1(1 − 2 · (−1) 4 4 −1 −1 2 = 0 . (5.72)

The corresponding axis is given by 1 2 1 1 Iω = I ω + I ω + I ω =⇒ ω = ω − ω − ω (5.73) 1 11 1 12 2 13 3 6 1 3 1 4 2 4 3 1 1 2 1 Iω = I ω + I ω + I ω =⇒ ω = − ω + ω − ω (5.74) 2 21 1 22 2 23 3 6 2 4 1 3 2 4 3 1 1 1 2 Iω = I ω + I ω + I ω =⇒ ω = − ω − ω + ω (5.75) 3 31 1 32 2 33 3 6 3 4 1 4 2 3 3 Only two of these equations are independent, since (5.75) is equivalent to

100 (5.73)+(5.74). We thus have 1 1 1 ω − ω − ω = 0 (5.73) 2 1 4 2 4 3 1 1 1 − ω + ω − ω = 0 (5.74) 4 1 2 2 4 3 3 3 (5.73) − (5.74) : ω − ω = 0 ⇐⇒ ω = ω (5.76) 4 1 4 2 2 1 1 1 1 1 1 (5.73) =⇒ ω − ω − ω = ω − ω = 0 ⇐⇒ ω = ω (5.77) 2 1 4 1 4 3 4 1 4 3 3 1

We thus have ω1 = ω2 = ω3, or the principal axis is along the diagonal (1,1,1). We might have guessed this from the symmetry of the cube, and rotated our system so that the new x-axis is along the diagonal. This can be obtained by a 45◦ rotation about the x-axis, followed by a rotation of cos−1 p1/3 about the z0-axis. The matrix for this rotation is q q  1 2   √ √ √  3 − 3 0 1 0 0 2 2 2   1 1 1 q 2 1 0 √ √ √ R =  √ 0  2 2  = −2√ 1√ 1  . (5.78)  3 3  1 1 6 0 − √ √ 0 − 3 3 0 0 1 2 2

The rotated inertia tensor is √ √ √ √  2 2 2  2 − 1 − 1   2 −2 0  2 3 4 4 √ √ 0 T ML 1 2 1 I = RIR = −2 1 1  − 4 3 − 4   2 1 − 3 6 √ √ 1 1 2 √ √ 0 − 3 3 − 4 − 4 3 2 1 3 2 0 0  ML2 = 0 11 0 . (5.79) 12   0 0 11

This also gives us the other two principal moments of inertia, which we ﬁnd to be equal.

5.5.2 Comments

1. Finding the principal moments and axes of inertia (diagonalising the inertia tensor) by hand can be a very tedious process. Sometimes it can be simpliﬁed by symmetry considerations (see below), but in most cases it is better left to computers.

2. For the cube with the origin at one corner, we found that two of the principal 11 2 moments of inertia were equal, I2 = I3 = 12 ML . This means that the two corresponding principal axes can be any orthogonal pair of axes perpendicular to the diagonal — ie, the moment of inertia is the same about any axis orthogonal to the diagonal.

3. For a body where all three principal moments of inertia are equal, all directions or axes are equivalent. We can see this by noting that in this case the inertia tensor in

101 the coordinate system deﬁned by the principal axes is proportional to the identity (matrix), I = I11,which commutes with all rotation matrices, so I0 = IR11RT = I11 for all R ∈ O(3).

4. Any body which is rotationally symmetric about some axis has one principal axis of inertia along that axis. The two other axes can be arbitrarily chosen in the plane perpendicular to the ﬁrst principal axis, ie I2 = I3.

Deﬁnitions

• A body with I1 = I2 = I3 is called a spherical top.

• A body with I1 = I2 6= I3 is called a symmetric top. • A body where all principal moments are diﬀerent is called an asymmetric top.

• A body with I1 = 0,I2 = I3 is called a rotor.

For example, the cube with the origin at the centre is a spherical top; the cube with the origin at one corner is a symmetric top, and the dumbbell (or a diatomic molecule) is a rotor.

5.6 Equations of motion

Having found the principal axes of inertia, we can now use them to deﬁne the body coordinate system. This will greatly simplify the equations of motion. The kinetic energy is given by 1 X T = I ω2 . (5.80) 2 i i i

Using (5.31) we can write this as 1 1 1 T = I (φ˙ sin θ sin ψ+θ˙ cos ψ)2 + I (φ˙ sin θ cos ψ−θ˙ sin ψ)2 + I (φ˙ cos θ+ψ˙)2 . (5.81) 2 1 2 2 2 3 The equations of motion for generic values of the moments of inertia and a generic potential energy will become very complicated. We will instead focus on two specific cases: a symmetric top (I1 = I2) with one point fixed and under the influence of constant gravity, and force-free motion of an asymmetric top (all I1,I2,I3 are different).

5.6.1 The symmetric heavy top

For a symmetric top, we can take I1 = I2, and (5.81) becomes 1 1 T = I (θ˙2 + φ˙2 sin2 θ) + I (φ˙ cos θ + ψ˙)2 . (5.82) 2 1 2 3 We now take the top to be rotating about a fixed point at the bottom of the symmetry axis, under the influence of a constant gravitational field g. We take θ to be the angle

102 the symmetry axis forms with the vertical. In that case, the potential energy of the top is V = Mgh cos θ, where M is the mass of the top, and h is the distance from the base of the top to the centre of mass along the symmetry axis. The lagrangian is therefore 1 1 L = I (θ˙2 + φ˙2 sin2 θ) + I (φ˙ cos θ + ψ˙)2 − Mgh cos θ . (5.83) 2 1 2 3

The meaning of the three Euler angles in this case is:

• θ denotes the angle the axis of the top makes with the vertical. • φ denotes the orientation of the (tilted) axis relative to the ﬁxed reference coordinate system. • ψ denotes the orientation of the top relative to its own axis.

We will ﬁnd that in general, θ will oscillate between a minimum and a maximum, while both φ and ψ will be monotonically increasing (or decreasing). The motion in ψ reﬂects the top spinning around its own axis. The motion in φ corresponds to the orientation of the axis precessing around the vertical axis. Finally, the motion in θ corresponds to periodic “wobbles” in the tilt of the top, called nutation. We may now derive the Euler–Lagrange equations for the top and use them to study this motion in detail. We will not do this here, but instead perform a qualitative analysis of the possible motion. First we determine the canonical momenta,

∂L ˙ 2 ∂ω3 ˙ 2 ˙ ˙ pφ = = I1φ sin θ + I3ω3 = I1φ sin θ + I3 cos θ(φ cos θ + ψ) , (5.84) ∂φ˙ ∂φ˙ ∂L ˙ pθ = = I1θ (5.85) ∂θ˙ ∂L ∂ω3 ˙ ˙ pψ = = I3ω3 = I3(φ cos θ + ψ) . (5.86) ∂ψ˙ ∂ψ˙ Since L does not depend explicitly on either φ or ψ, ∂L/∂φ = ∂L/∂ψ = 0, and we see straightaway that the canonical momenta pφ, pψ are conserved. In particular, pψ = I3ω3, so the top spins with a constant angular velocity ω3 about its own axis. However, the precession rate, governed by the constant pφ is more complicated. We now proceed to derive the hamiltonian of the system. From (5.86) we see that ˙ ˙ ω3 = φ cos θ + ψ = pψ/I3. We can use this to rewrite (5.84), ˙ 2 ˙ 2 pφ = I1φ sin θ + I3 cos θω3 = I1φ sin θ + pψ cos θ (5.87) ˙ 2 ˙ pφ − pψ cos θ ⇐⇒ pφ − pψ cos θ = I1φ sin θ ⇐⇒ φ = 2 . (5.88) I1 sin θ Since the kinetic energy is quadratic in the generalised velocities, and the potential energy does not depend on velocities, the hamiltonian is equal to the total energy, H = T + V 2 2 2 1 pφ − pψ cos θ 2 pθ 1 pψ = I1 2 sin θ + + I3 + Mgh cos θ 2 I1 sin θ I1 2 I3 (5.89) 2 2 2 pθ pψ (pφ − pψ cos θ) = + + 2 + Mgh cos θ . 2I1 2I3 2I1 sin θ

103 Since pφ and pψ are both constant, there is effectively only one degree of freedom θ, and the second and third term in H can be combined with the final term V = Mgh cos θ to form an effective potential Veff(θ). The first term is the usual kinetic energy term.

The first term in Veff is just a constant, so it has no effect on the dynamics of the system, other than increasing the minimum energy. Unless pφ and pψ are both precisely zero 2 or pφ = ±pψ, the second term is nonzero and diverges to +∞ as θ → 0 and θ → π. Therefore, the motion in θ is bounded for all but very special values of pφ, pψ, with θmax in general being smaller the larger the value of pψ. If θmax < π/2 this ensures the top does not fall over.

5.6.2 Euler’s equations for rigid bodies

Consider now force-free motion of a rigid body. In this case, we have3

X 2 L = T = Iiωi . (5.90) i The Euler–Lagrange equations are given by d ∂L ∂L d ∂L ∂L d ∂L ∂L = ; = ; = . (5.91) dt ∂φ˙ ∂φ dt ∂θ˙ ∂θ dt ∂ψ˙ ∂ψ

We will however only consider the third of those, and then derive two additional equations of motion from symmetry considerations. We can write the Euler–Lagrange equation for ψ as

3 d ∂L d d ∂L X ∂ωi = I (φ˙ cos θ + ψ˙) = (I ω ) = = I ω . (5.92) dt ˙ dt 3 dt 3 3 ∂ψ i i ∂ψ ∂ψ i=1 We ﬁrst note that ∂ω 1 = φ˙ sin θ cos ψ − θ˙ sin ψ = ω , (5.93) ∂ψ 2 ∂ω 2 = −φ˙ sin θ sin ψ − θ˙ cos ψ = −ω . (5.94) ∂ψ 1

Since ω3 does not depend on ψ the last expression in (5.92) becomes ∂L ∂ω ∂ω = I ω 1 + I ω 2 = I ω ω + I ω (−ω ) = (I − I )ω ω , (5.95) ∂ψ 1 1 ∂ψ 2 2 ∂ψ 1 1 2 3 2 1 1 2 1 2 so (5.92) becomes dω I 3 = (I − I )ω ω . (5.96) 3 dt 1 2 1 2 However, the labels 1, 2 and 3 for the three axes is arbitrary, and we may just as well rename them, as long as we ensure the coordinate system remains right-handed (corresponding to cyclic permutations. This gives us the three equations

2 If pφ = pψ the second term in Veﬀ diverges as θ → π but goes to 0 as θ → 0, and conversely if pφ = −pψ. This, however, requires ﬁnely balanced initial conditions. 3Note that the translational energy can be ignored since R~ is cyclic.

104 dω I 1 = (I − I )ω ω , (5.97) 1 dt 2 3 2 3 dω I 2 = (I − I )ω ω , (5.98) 2 dt 3 1 3 1 dω I 3 = (I − I )ω ω . (5.99) 3 dt 1 2 1 2 These are Euler’s equations for force-free motion.

5.6.3 Stability of rigid-body rotations

Let us now look at what happens when we set a body rotating about one of the three principal axes. For example, if it rotates purely about the first axis, we have ω1 6= 0, ω2 = ω3 = 0. In practice, it is not possible to have exactly zero rotation about the other two axes, so what we have is that ω2 and ω3 are both much smaller (in magnitude) than ω1. We now want to find out how ω1, ω2 and ω3 each evolve with time. If ω2 and ω3 remain small and are either damped to zero or fluctuate around zero, the rotation about the first axis is said to be stable. On the other hand, if the magnitude of ω2 and/or ω3 grows with time, they will eventually become as large as ω1 and the body is no longer rotating about its original axis. In that case, the rotation is unstable.

Without any loss of generality, we can take I1 > I2 > I3 since we are allowed to label our axes as we wish. We then have three diﬀerent cases to deal with:

1. If ω2 ∼ ω3 ω1 the three equations become dω I 1 = (I − I )ω ω ≈ 0 =⇒ ω = constant , (5.100) 1 dt 2 1 2 3 1 dω2 I3 − I1 I2 = (I3 − I1)ω3ω1 =⇒ ω˙ 2 = ω1 ω3 , (5.101) dt I2 dω3 I1 − I2 I3 = (I1 − I2)ω1ω2 =⇒ ω˙ 3 = ω1 ω2 , (5.102) dt I3 Diﬀerentiating (5.101) and using (5.102) we get

I3 − I1 (I3 − I1)(I1 − I2) 2 ω¨2 = ω1ω˙ 3 = ω1ω2 (5.103) I2 I2I3

2 2 (I3 − I1)(I2 − I1) 2 ⇐⇒ ω¨2 + Ω1ω2 = 0 with Ω1 = ω1 > 0 . (5.104) I2I3 This has the solution

ω2(t) = A cos Ω1t + B sin Ω1t = C cos(Ω1t + δ) (5.105) 0 0 0 0 ω3(t) ∝ ω˙ 2(t) = A cos Ω1t + B sin Ω1t = C cos(Ω1t + δ ) . (5.106)

ω2 and ω3 oscillate about equilibrium values ω2 = ω3 = 0. The tip of ~ω describes 2 2 ω2 ω3 and ellipse, C2 + C02 = 1 and the body wobbles about the direction of the principle associated with the larges moment of inertia I1.

105 2. If ω1 ∼ ω2 ω3, by the same procedure as in the ﬁrst case, we obtain

2 2 (I3 − I2)(I3 − I1) 2 ω3 = const , ω1 ∝ ω˙ 3 , ω¨2 + Ω3ω2 = 0 , Ω3 = ω3 > 0 , I1I2 which again has solutions

0 0 ω1(t) = A cos(Ω3t + δ) , ω2(t) = A cos(Ω3t + δ ) . (5.107)

3. If ω1 ∼ ω3 ω2, then, using the same procedure again, we now ﬁnd the equations

ω2 = constant , (5.108)

I1 ω3 = ω˙ 1 , (5.109) (I2 − I3)ω2

(I3 − I2)(I2 − I1) 2 2 ω¨1 = ω2ω1 = Ω2ω1 . (5.110) I1I3 The general solution to (5.110) is

Ω2t −Ω2t ω1(t) = Ae + Be , (5.111)

so ω1 (and ω3) will increase exponentially with time, at least until the approxima- tions ω1 ω2 and ω3 ω2 are no longer valid.

The upshot of this is that rotations about the “long” and “short” axes are stable, while those about the intermediate axis are unstable. You can verify this for yourself by tossing a rectangular block (for example a a book held together by an elastic band) up in the air, rotating it about each of its three axes.

If two of the principal moments of inertia are equal, I1 = I2 say, then ω3 is strictly constant and

I1 − I3 I3 − I1 ω˙ 1 = ω2ω3, ω˙ 2 = 2 ω1ω3 I1 I1 2 2 (I1 − I3) 2 (I1 − I3) 2 ⇒ ω¨1 = − ω3ω1, ω¨2 = − 2 ω3ω2, I1 I1

|I1−I3| so ω1 and ω2 oscillate with the same frequency Ω3 = ω3 and there is no instability. I1

106 5.7 Rigid body motion — summary sheet

1. Linear and angular velocity The total velocity of a particle rotating with angular velocity ~ω about a point moving with linear velocity V~ is ~v = V~ + ~ω × ~r. 2. Rotation matrices The relationship between coordinates before and after a rotation is given by the rotation matrix A: x0 x r~0 = y0 = A~r = A y z0 z

Rotation A1 followed by A2 gives the matrix B = A2A1 6= A1A2. All rotations are orthogonal: AT A = 11. Elementary rotations about x-, y- and z-axis: 1 0 0  cos θ 0 − sin θ  cos θ sin θ 0 Ax = 0 cos θ sin θ Ay =  0 1 0  Az = − sin θ cos θ 0 0 − sin θ cos θ sin θ 0 cos θ 0 0 1 3. Scalars, vectors and tensors When you rotate a coordinate system, • a scalar is a quantity that does not change; 0 P • a vector v transforms as vi = j Aijvj; 0 P • A tensor T transforms as Tij = kl AikAjlTkl . 4. Euler angles All rotations can be represented as a series of 3 rotations: (a) An angle φ about the (old) z-axis, (b) An angle θ about the (intermediate) x-axis, (c) An angle ψ about the (new) z-axis. 5. Inertia tensor The kinetic energy of a rotating rigid body is 1 X T = I ω ω with rot 2 ij i j i,j X 2 Iij = mα(rαδij − xαixαj) =inertia tensor α

107 In a rotated coordinate system the inertia tensor is

0 X 0 T Iij = AikAjlIkl or I = AIA kl

6. Angular momentum X Li = Iijωj j

7. Principal axes of inertia We can always rotate our coordinate system so that I is diagonal,   I1 0 0 I =  0 I2 0  0 0 I3

I1,I2,I3 are the principal moments of inertia. The corresponding coordinate axes are the principal axes of inertia. These deﬁne the natural body coordinate system. 8. Euler’s equations for a rigid body (force-free motion)

I1ω˙ 1 − (I2 − I3)ω2ω3 = 0

I2ω˙ 2 − (I3 − I1)ω3ω1 = 0

I3ω˙ 3 − (I1 − I2)ω1ω2 = 0 These are given in the body coordinate system deﬁned by the principal axes of inertia.

108