Lagrangian Mechanics and Special Relativity

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Lagrangian mechanics and special relativity

Harold Erbin Notes de cours de Magistère L3 donné par Mme Steer.

Ce texte est publié sous la licence libre Licence Art Libre : http://artlibre.org/licence/lal/

Version : April 27, 2011 Site : http://harold.e.free.fr/ Sommaire

I Introduction to Lagrangians in the context of non- relativistic systems2

1 Introduction: classical mechanics with Newtons equations against Lagrangian approach3

2 Calculus of variations: The Lagrangian, Euler-Lagrange equa- tions6

3 Calculus of variations: symmetry and conservation equations 11

4 The Lagrangian of classical mechanics 13

5 Smalls oscillations and normal modes 20

6 Constrained systems and Lagrange multipliers 26

7 From Lagrangians to Hamiltonians – The Hamiltonian formulation of classical mechanics 32

II Special relativity 35

8 Introduction 36

9 The postulates of special relativity and the Michelson–Morley experiment 38

10 Consequences of the constancy of c 42

11 Lorentz transformations 47

12 Minkowski spacetime, 4-vectors and Lorentz invariants 51

13 Diﬀerent 4-vectors and their invariant 56

14 Particle collisions 61

15 Electromagnetism 66

16 Relativistic lagrangian 71

ii III Exercices 73

17 TD 1 74

18 Partiel 2008 78

Appendix 83

A Quantum ﬁeld theory: an introduction 84

Index 89

List of Figures 91

Bibliography 92

Contents 93

iii Introduction

Organization

The lecture is composed ot two parts.

Introduction to Lagrangians in the context of non-relativistic systems In this chapter, we have v c where v is the velocity of the observer and c = 3 × 108 m s−1. Non-relativistic systems are described by classical mechanics and Newtons equations of motion.

All systems He re we have v ≤ c. Examples : photons, cosmics rays. . . The dynamic of relativistic systems is very different. All intuitions fails. It is the domain of all modern physics : – cosmology ; – string theory ; – physics of the early universe ; – quantum field theory. Lagrangian theory also applies for relativistic systems, but we will develop it mainly for non-relativistic systems in the first part.

Why revisit classical mechanics?

Becaus we will introduce very powerfull, diﬀerent method to determine the dynamics of a particle.

1 Part I

Introduction to Lagrangians in the context of non-relativistic systems

2 Chapter 1

Introduction: classical mechanics with Newtons equations against Lagrangian approach

1.1 Limitations of Newtons approach

Newtonians classical mechanics has (at least) three limitations : 1. It describes particles, which is inconvenient because often we want to study the dynamics of a number of particles, or the dynamics of extended objects (a spinning top, a bicycle. . . ), or the dynamics of ﬁelds. 2. Newtons equations are formulated in a special coordinate system — cartesian coordinates in an inertial frame.

Let xi (i = 1, 2, 3) be the coordinates of a particle of mass m in a inertial frame. Then, under the inﬂuence of a force with components fi(t), the dynamics of the particle is obtained by solving Newtons equations

mx¨i(t) = fi(t) (1.1)

3. You need to ﬁnd the force fi in order to solve Newtons equations: this is not always easy (think to the double pendulum — ﬁgure ??)! In principle, one can overcome these limitations: 1. For N particles you could write down N coupled newton’s equations and try to solve them.

2. One can pass from cartesian coordinates xi to other coordinate systems by using the "chair rule" of diﬀerentiation.

3. In principe it’s possible to find the fi, but it’s difficult. Conclusion: got rid of Newtons equations and approach classical mechanics differently: Lagrangian method which does not have the limitations. Lagrangian approach to classical mechanics is valid for any number of particles in any coordinate system, and you don’t need to know the fi.

3 Figure 1.1: Double pendulum.

1.2 Diﬀerence between Newtonian classical mechanics and Lagrangian classical mechanics

– Newtons equations are vector equations. Lagrangian approach puts great emphasis on scalar quantities. Equation for a particle with m and the position x(t): – x¨(t) and f are vectors (equation (1.1)) ; 1 – kinetic energy T = mx˙ 2 and potential energy V are scalar. 2 – In Newtonian mechanics, to determine the dynamics of a particle, one has to solve the second order equation (1.1). You need to specify x(ti) and ˙ ~x(ti). In Lagrangian mechanics the view point is totally diﬀerent: – We said that cartesian coordinates are not always the most useful : get rid of them! Introduce the conﬁguration space, which consist of the generalised coordinates qi of a system. Exemple 1.1. For one point (cartesian coordinates)

q1 = x, q2 = y, q3 = z

Exemple 1.2. For two points (spherical coordinates):

q1 = r, q2 = θ, q3 = ϕ

q4 = R, q5 = α, q6 = β

– We said that in Newtonian mechanics x(ti), x˙ (ti) and x(t), x˙ (t). In Lagrangian mechanics the approach is very diﬀerent. At time t and t+δt, the generalised coordinates take value qα(t) and qα(t+δt). As the

4 system we are describing evolves in time, it describes a trajectory/curve in configuration space. The trajectory in configuration space is defined uniquely by specifying either qα(ti) and q˙α(ti) (Newtonian approach), or qα(ti) and qα(tf ) (Lagrangian approach). The path followed is the one which minimizes a quantity (a functional) S[qα]. S[qα] is called the action and is directly related to the Lagrangian. Remarque : S[qα] is a functional of qα, to be distinguished from functions f(qα). The following questions are: – What is a functional? – How does one find the trajectory which minimize a functional? (the trajectory must satisfy Euler–Lagrange equations) – What is the definition of the action S[qα]? (We will see that it’s related to the Lagrangian, which itself depends on the scalar T and V ) – How two shows that we have the same solutions as that we would get from Newton’s equations?

5 Chapter 2

Calculus of variations: The Lagrangian, Euler-Lagrange equations

2.1 Functionals and the Euler–Lagrange equations 2.1.1 Examples 1. Consider 2-dimensional Euclidean space coordinates (x, y). Set two points A and B. The distance between A and B is an example of a functional: – functional: L[y]; – path : y(x). The path which minimizes L[y] is a straight line. 2. Two points A and B on the surface of the sphere. Functional L(θ] with path θ(ϕ). A geodesic minimizes L[θ]. 3. The brachistochrone. Consider two points A and B in the (x, z) plane, and a wire of shape z(x) linking A and B. Place bead of mass m at A, and under the inﬂuence of gravity, it goes to B (neglect all frictional forces). Which path z(x) minimizes the time it takes for the bead to go from A to B? – functional: T [z]; – path : z(x). 4. A place takes oﬀ from New-York bound for Paris. The on-board computer must choose the optimal path x(t) (sequence of latitudes, longitudes and altitudes) such that, given the wind direction, the total of fuel used is minimum. – functional: VF [x]; – path : x(t).

6 In all of the examples, one has to minimize F [qα] which depends on a path qα(ti). For the diﬀerent examples: 1. F = L, α = 1 2. F = L, α = 1 3. F = T , α = 1

4. F = VF , α = 3

From the examples, you can see that the functional F [qα], the path qα(τ) and the parameter τ, all change from one example to the next. We will introduce the calculus of variations for a functional F [x] with path x. It’s very easy to translate the results we will obtain from: – x(t) → qα(τ) – F (x) → F [qα]

2.2 Introduction to functional

Given a path x(t), the simplest functionals are integrals along the path of x(t) and its derivatives.

Exemple 2.1. – Z tf 2 F1[x(t)] = dt(x ) ti – Z tf F2[x(t)] = dt(x · x˙ ) ti Attention : A functional is a scalar.

f(·) F [·] f : R → R f : F → R x 7→ f(x) f 7→ F [f] n X ∂f Z δF [f] δf = δx δF = δf(x) dx ∂x i δf(x) i=1 i

Table 2.1: Diﬀerences between a function and a functional

The extremum for a functional is given by δF = 0 (2.1) δf

2.2.1 Examples of functional derivatives 1. Consider Z F [f] = f(x)dx

7 so Z δF = δf(x)dx

2. Consider Z 1 Z F [f] = a + b(x)f(x)dx + c(x, y)f(x)f(y)dxdy 2 so Z 1 Z δF = b(x)δf(x)dx + c(x, y)[δf(x)f(y) + δf(y)f(x)]dxdy 2 Z 1 Z = b(x)δf(x)dx + c(x, y)[f(y) + f(y)]δf(x)dxdy 2 Z 1 Z Z = b(x)δf(x)dx + δf(x) c(x, y)f(y)dy dx 2

and Z δF = b(x) + c(x, y)f(y)dy

2.3 Euler–Lagrange equations

We will derive the Euler–Lagrange equations for a function F [x] with path x(t). We deﬁne Z 2 F1[x] = |x| dt (2.2) Z F3[x] = xx˙ dt (2.3) Z 2 F3[x] = |x˙ | dt (2.4)

A general functional takes the form

Z tf F [x] = f(x, x˙ , t)dt (2.5) ti Remarques : 1. We suppose that f(x, x˙ , t) does not depend on higher time derivatives. 2. We will use the summation convention. How does we ﬁnd the path xˆ(t) which minimizes F [x] subject to the conditions x(ti) = xi x(tf ) = xf xˆ is the path we are looking for. Consider a small variation η(t) about that path x(t) = xˆ(t) + η(t) (2.6) where we suppose that η(ti) = 0 η(tf ) = 0

8 then

Z tf F [x] = f xˆ(t) + η(t), xˆ˙ (t) + η˙ (t), t dt ti Z tf ∂f ∂f = f xˆ(t), xˆ˙ (t), t + η + η˙ dt ti ∂x ∂x˙ Z tf ∂f ∂f = F [xˆ] + η + η˙ dt ti ∂x ∂x˙ Z tf ∂f Z tf d ∂f = F [xˆ] + ηdt − ηdt ti ∂x ti dt ∂x˙ because t Z tf ∂f ∂f f Z tf d ∂f η˙ dt = η − ηdt ∂x˙ ∂x˙ dt ∂x˙ ti ti ti | {z } =0 Z tf d ∂f = − ηdt ti dt ∂x˙ We ﬁnd Z tf ∂f d ∂f δF = − η dt (2.7) ti ∂x dt ∂x˙ where η(t) = δx(t) The functional has an extremum if δF = 0 for all inﬁnitesimal η = x. Assuming ηi are all independent, it follows that the Euler–Lagrange equation is

∂f d ∂f − = 0 (2.8) ∂x dt ∂x˙

Remarques : 1. The Euler–Lagrange equations are invariant under – f 7→ f 0 = f + c dΛ(q , τ) – f 7→ f 0 = f + α dτ 2. Euler–Lagrange equations hold in any coordinate system qα (for example, the qα could be a non interual coordinate system). Exemple 2.2 (Shortest distance on a plane). Which path y(x) is the shortest (in distance) between A and B?

B Z Z p L[y] = d` = dx2 + dy2 A s Z B dy 2 = dx 1 + A dx Z xb = f(x, y, dy/dx)dx xa

9 The Euler–Lagrange equation is ∂f d ∂f = ∂y dx ∂y0 where dy p y0 = f = 1 + y02 dx and ! d y0 0 = dx p1 + y02 ! y0 =⇒ = c p1 + y02 =⇒ y0 = c =⇒ y = cx + k

We need to ﬁx constants such that the straight line passes through A and B. Exemple 2.3 (Brachistochrone). Which curve y(x) minimizes the total time taken for the particles to go from A(0, 0) to B(x0, −y0) (y0 > 0) under the eﬀect of gravity? We have

Z B d` TA→B = A v(`) Z B pdx2 + dy2 = √ Energy conservation 2gy A √ Z −y0 1 + x02 = √ dy |{z} 0 2gy 0 dx x = dy Z −y0 = f(x, x0, y)dy 0 since f does not depend explicitly on x,

∂f d ∂f ∂f = 0 =⇒ = 0 =⇒ = cste ∂x dy ∂x0 ∂x0 where ∂f x0 1 = √ √ = c ∂x0 1 + x02 2gy

10 Chapter 3

Calculus of variations: symmetry and conservation equations

3.1 Conservation

In nearly all physical conservation phenomena, there exists quantities which are conserved during the evolution of the system. – How do these conserved quantities appear in the functional approach? Z tf – We will work with a function F [qα] = f(qα, q˙α, t). ti 1 Déﬁnition 3.1. A function C(qα(t), q˙α(t), t) is constant in time if, along the path qˆα(t) which is solution yo the Euler–Lagrange equations its total time derivative vanishes dC = 0 (3.1) dt 3.1.1 Time independance Suppose that f does not depend explicitly on t.

df d ∂f ∂f = q˙α + q¨α dt dt ∂q˙α q˙α d ∂f = q˙α dt ∂q˙α So we obtain ∂f h = q˙α − f (3.2) ∂q˙α is conserved if ∂f/∂t = 0 2.

1. It is the same thing as bo be a constant of motion, or a conserved quantity. 2. In french, the name is "identité de Beltrami".

11 Exemple 3.1. Minimal distance between two points in the plane. Z p L = dx2 + dy2 p = 1 + y02 dy

Here ∂f/∂x = 0, so ∂f h = y0 − f = c ∂y0 y0 p = y0 − 1 + y02 p1 + y02 1 = (y02 − (1 + y02)) p1 + y02 =⇒ y02 = cste

3.1.2 Coordinate independance

Suppose that f does not depend explicitly on qα ∂f = 0 ∂qα then, from the Euler–Lagrange equations d ∂f ∂f = = 0 dt ∂q˙α ∂qα

Thus ∂f/∂q˙α is a conserved quantity.

3.2 Noether’s theorem

Consider a 1-parameter family of transformations

qα(t) 7−→ Qα(s, t) s ∈ R where Qα(0, t) = qα(t). Exemple 3.2. qα are cartesian coordonates x 7−→ x + sn

This transformation is said to be a continuous symmetry of f if it leaves f invariant, or more explicitly

∂f (Qα(s, t), t) = 0 ∂s s=0 Théorème 3.1 (Noether’s theorem). For every continuous symmetry there exists a conserved quantity.

12 Chapter 4

The Lagrangian of classical mechanics

Recall: the Lagrangian formulation of classical mechanics introduces: – conﬁguration space with generalized coordinates qα (supposed independent); – the path which the system follows in time as it evolves dynamically is that which minimizes the functional S[qα] called the action; – hence the path followed is that which satisﬁes the Euler–Lagrange equations (2.8). Question: Which Lagrangian gives us back Newtons laws? Answer: There is no general rule to write down L, valid for relativistic, non- relativistic, etc., systems. But for the majority of non-relativistic systems, a Lagrangian of the form

L = T − V (4.1) where T is the kinetic energy and V the potential energy, gives the correct equations of motion. Exemple 4.1. Let’s check that this Lagrangian "works" in the simplest case: a particle of mass m, moving in a conservative potential V , and which we describe using cartesian coordinates in an inertial frame. For this particle 1 L = mx˙ 2 − V (x) 2 then d ∂L ∂L = dt ∂x˙ ∂x d (mx˙ ) = −∇V dt mx¨ = −∇V

Remarques :

13 1. The Lagrangian depends on physical (scalar) quantities T and V , which are completely independent of the choice of the generalized coordinates. 2. The Lagrangian is not deﬁned uniquely

L0 = L + const dΛ(q , t) L0 = L + α dt give the same equations of motions. 3. Conserved quantities: if L is explicitly time independent, then ∂L h =q ˙α − L ∂q˙α is constant. h corresponds to the total energy: h = T + V , if

(a) V is independent of q˙α;

(b) T is quadratic in q˙α, in other words 1 T = m g (q )q ˙ q˙ 2 αβ α α β Exemple 4.2. i. Particle of mass m in cartesian coordinates 1 1 T = mx˙ 2 = m(x˙ 2 + x˙ 2 + x˙ 2) 2 2 1 2 3

with gij = δij. ii. Particle of mass m in spherical polar coordinates (r, θ, ϕ) 1 T = m(r ˙2 + r2θ˙2 + r2ϕ˙ 2 sin2 θ) 2

2 2 2 with g11 = 1, g22 = r and g33 = r sin θ.

4.1 Exemples 4.1.1 Ball slides A ball slides (without friction) on a wire of shape y = Ax2 (A > 0) under the eﬀect of gravity. 1. Write down the Lagrangian. 2. Calculate the Euler–Lagrange equation. 3. Are there any conserved quantities? Lets choose x as generalized coordinate.

L = T − V = h(x, x,˙ t)

14 1 1 T = mx˙ 2 + my˙2 where y = Ax2 2 2 1 = m(x ˙ 2 + 4A2x2x˙ 2) 2 1 = x˙ 2(1 + 4A2x2) 2 V = mgy = mgAx2 then 1 L = mx˙ 2(1 + 4A2x2) − mgAx2 2

Conserved quantity: ∂tL = 0 and T is quadratic in x˙, V is independent of x˙, so the total energy is conserved (by Beltrami) 1 h = mx˙ 2(1 + 4A2x2) + mgAx2 2 Equations of motion: let’s calculate the Euler–Lagrange equations d ∂L ∂L = dt ∂x˙ ∂x x¨(1 + 4A2x2) = −4xA ˙ 2x − 2gAx

You can obtain the same equation with dh/dt = 0. In order to solve for x(t), it is much simple to use the conserved quantity h: in that case, you "only" need to solve a ﬁrst order diﬀerential equation.

4.1.2 Simple pendulum with variable height Simple pendulum in which the point of support of the pendulum changes height with time. 1. Write down Lagrangian. 2. Determine equations of motion and any conserved quantity. Let’s choose θ as generalized coordinate. 1 1 T = mx˙ 2 + y˙2 2 2 1 1 = m`2(cos θ)2θ˙2 + m h˙ 2 + 2h`˙ (sin θ)θ˙ + 2h`(sin θ)2θ˙2 2 2 1 1 = m`θ˙2 + m h˙ 2 + 2h`˙ (sin θ)θ˙ 2 2 and V = mgy = mg h(t) − ` cos θ then 1 L = m `2θ˙2 + h˙ 2 + 2h`˙ (sin θ)θ˙ − mg h(t) − ` cos θ 2

15 Equations of motion d ∂L ∂L = dt ∂θ˙ ∂θ d (m`2θ˙ + mh`˙ sin θ) = mh`˙ θ˙ cos θ − mg` sin θ dt `2θ¨ + h`¨ sin θ + h`˙ θ˙ cos θ = h`˙ θ˙ cos θ − g` sin θ m`θ¨ = −m(g + h¨) sin θ

It’s equivalent to a standard simple pendulum in a gravitational ﬁeld g + h¨.

4.1.3 Free particle in non-inertial frame Free particle of mass m in three spatial dimensions. Its lagrangian in an inertial frame 1 L = m(x2 + y2 + z2) 2 1 = mr˙ 2 2 What is the dynamics of the particle in a coordinate system which is rotating relative to this one? It will be subjected to so-called "ﬁctitious forces" which are forces which result from the acceleration of the coordinate system itself (rather than any physical force which acts on the particle). Lagrangian formulation provides a particulary straightforward way of deter- mining equations of motion in the news coordinate system (one simply has to write down the Lagrangian in new coordinate system and then write down the Euler–Lagrange equations). Let r0 = (x0, y0, z0) be the new coordinates, which share the same origin as the old coordinates, but they rotate with angular velocity ω = (0, 0, θ˙(t)) where θ is a given function of t. z0 = z x0 = x cos θ + y sin θ y0 = −x sin θ + y cos θ We need to invert to ﬁnd (x, y, z) in terms of (x0, y0, z0) z0 = z x = x0 cos θ − y0 sin θ y = x0 sin θ + y0 cos θ We need x˙, y˙ and z˙. z˙ =z ˙0 x˙ = cos θ(x ˙ 0 − y0θ˙) − sin θ(y ˙0 + x0θ˙) y˙ = cos θ(y ˙0 + x0θ˙) − sin θ(x ˙ 0 − y0θ˙)

16 substituting into 1 L = m(x2 + y2 + z2) 2 1 = m z˙0 + (x ˙ 0 − y0θ˙)2 + (y ˙0 + x0θ˙)2) 2 1 = m x02 + y02 + z02 + 2θ˙(x0y˙0 − y0x˙ 0) + θ˙2(x02 + y02) 2 The Euler–Lagrange equations in the (x0, y0, z0) coordinate system are d ∂L ∂L = dt ∂r˙ 0 ∂r0 In terms of r 1 L = m r˙ 02 + 2ω · (r0 × r˙ 0) + (ω × r0)2 2 then d 2r˙ 0 + 2(ω × r0) = 2(r˙ × ω) − 2ω × (ω × r0) dt r¨0 = −ω˙ × r0 − 2(ω × r˙ 0) − ω × (ω × r0) | {z } | {z } | {z } Euler Coriolis Centrifugal

4.1.4 Spherical pendulum Spherical pendulum. A pendulum of mass m on a rod of length ` of negligible mass. The pendulum can move in 3D. Choose as generalized coordinates θ ∈ [0, π[ and ϕ ∈ [0, 2π[. 1 T = m(`2θ˙2 + `2ϕ˙ 2 sin2 θ) 2 V = −mg` cos θ 1 L = T − V = m(`2θ˙2 + `2ϕ˙ 2 sin2 θ) + mg` cos θ 2

Conserved quantities: ∂tL = 0 and T is quadratic in ϕ and θ, hence the total energy is 1 H = T + V = m(`2θ˙2 + `2ϕ˙ 2 sin2 θ) − mg` cos θ 2 is conserved. Equations of motion – For ϕ d ∂L ∂L = = 0 dt ∂ϕ˙ ∂ϕ ∂L =⇒ = cste ∂ϕ˙ 2 =ϕm` ˙ sin θ = Jz – For θ d ∂L ∂L = dt ∂θ˙ ∂θ −∂V =⇒ mθ¨ = eff ∂θ

17 4.2 Consequences of Noether’s theorem

Recall: for each transformation qα(t) → Qα(s, t) which leaves L invariant to linear order s (i.e continuous symmetry of L), there is a conserved quantity

∂L ∂Qα = cste ∂q˙α ∂s s Consider N particles interaction between each other via a potential V

N X 1 L = mr˙ 2 − V (4.2) 2 i i=1 where V only depends on the relative coordinates of the particles V ({ri − rj})

4.2.1 Momemtum conservation Proposition 4.1. Homogeneity of space: the system is invariant under spatial translations. The Lagrangian (4.2) has translational symmetry

L(ri, r˙ i) = L(ri + sn, r˙ i) From Noether’s theorem, the conserved quantity is n X ∂L · n = cste ∂r˙ i=1 i

⇐⇒ P tot · n = cste (4.3)

If system is invariant for any n, then the total momentum P tot is conserved. If system is invariant only for a speciﬁc n, for example n = ez, then the P tot|z is conserved.

4.2.2 Angular momemtum conservation Proposition 4.2. Isotropy of space: the system is invariant under spatial rotations. The Lagrangian (4.2) is invariant under rotations about any axis n. All the xi are rotated by the same angle δθ = s, then 0 xi −→ xi = xi + δθ(n × xi) (4.4) so L(xi, x˙ i, t) = L(xi + δθ(n × xi), x˙ i + δθ(n × x˙ i), t) (4.5) The corresponding conserved quantity is 0 X ∂L ∂xi cste = ∂x˙ ∂θ i i θ=0 X = mix˙ i(n × xi) i cste = nJ (4.6)

18 and X X J = x˙ i × mixi = xi × p˙ i (4.7) i i If system is invariant under rotations about any vector n, then the total angular momentum J tot is conserved.

19 Chapter 5

Smalls oscillations and normal modes

“Physics is that subject of human experience which can be reduced to coupled harmonic oscillations.” — Peskin – Here we will dealed with uncoupled harmonic oscillators. – The Lagrangian formulation of classical mechanics gives us a relatively straightforward way in which to write down the dynamical equations of motion for a system. But, actually solving those equations is generally very diﬃcult: one has to use some kind of simplication or approximation schemes. – Here we study (exactly) the dynamics of a system only about its equilibrium positions. – If the system is disturbed slightly about a stable equilibrium position, it will oscillate with small amplitude oscillations. – In the Lagrangian formulation, we can study the dynamics of these oscillations and the problem reduced to studying N uncoupled harmonic oscillators, each ringing with a diﬀerent frequency ωα. – These uncoupled oscillating modes are called normal modes.

5.1 The equations

We work with generalized coordinates qα. Potential is V = V (qα) (so V is explicitly independent of t, qα). Kinetic energy is of the form T = 1/2 mαβ(qα)q ˙αq˙β. From the Euler–Lagrange equations, the system is in equilibrium when

∂V = 0 (5.1) ∂q α qα,0

If initially the generalized coordinates take values qα,0 and if q˙α = 0, then the system stays in equilibrium indeﬁnitely. What happens when the system is displaced from an equilibrium position? – Stable equilibrium: small displacement → small bounded oscillations ( eiωt, with ω ∈ R).

20 – Unstable equilibrium: small displacement → unbounded oscillations ( eωt, with ω ∈ R+). – Partially equilibrium: All will be classiﬁed according to the signs of

∂2V (5.2) ∂qαqβ Aim: departures from equilibrium are small: Taylor expand the Lagrangian about an equilibrium position, heeping only lowest order terms

qα(t) = qα,0 + ηα(t) (5.3) where qα 1, and qα,0 is the equilibrium position, and is solution of (5.1).

V (qα) = V (qα,0 + ηα) 2 ∂V 1 ∂ V = V (qα,0) + ηα + ηαηβ ∂qα 2 ∂q q qα,0 α β qα,0 | {z } =0

Set V (qα,0) to zero by choosing the zero of potential energy at the equilibrium position. Rewritten in matrix form 1 V = ηtV η (5.4) 2 where V is a matrix with components Vαβ (5.2), and   η1  .  η =  .  ηn

Kinetic energy is 1 T = m (q )q ˙ q˙ 2 αβ α α β where q˙α(t) =q ˙α,0 +η ˙α(t) =η ˙α(t). So 1 T = m (q )η ˙ η˙ 2 αβ α α β 1 = m (q )η ˙ η˙ + O(η2 ) 2 αβ α,0 α β α 1 = η˙tm η˙ (5.5) 2 where m is the matrix with components mαβ(qα,0). m is a symmetric real positive matrix. V is a real symmetric matrix. We have 1 L = (η ˙tm η˙ − ηtV η) (5.6) 2

21 The Euler–Lagrange equation (2.8) gives

mη¨ = −V η (5.7)

It’s N coupled linear second order diﬀerential equations with constant coeﬃ- cients (in components of mαβη¨β = −Vαβηβ). Solutions of (5.7): we try a solution of the form

η = A eiωt (5.8)

Notice that ω is α independent: each ηα is assumed to oscillate with the same frequency ω. So substitude (5.8) in (5.7). We have η¨ = −ω2η, so

ω2mA = V A (5.9)

A "generalized" eigen-value equation, where ω2 ≡ λ is the eigen-value and A is (not quite) the eigen-vector. Solution only if det(ω2m − V ) = 0 (5.10) It is an N th order diﬀerential equation in ω2, with generally N diﬀerent solutions for ω2. Is ω2 > 0 (oscillatory solutions) or ω2 < 0 (expontential growth)? Consider (5.9), and multiply on the left by A†

ω2(A†m A) = (A†V A) A†V A ω2 = A†m A

† † since V and m are hermitian, both A V A and A m A are real, so ω2 ∈ R. How about the sign of ω2? A†V A can be positive or negative, depending the sign of (5.2)(∀α, β, Vαβ > 0). The equilibrium is: – stable, ω2 > 0; – partially stable if some of Vαβ < 0; 2 – unstable if all Vαβ < 0, and ω < 0. The general solution of (5.7) is

n ! X iωnt η = < cn(An e ) (5.11) n=1 5.2 Exemples 5.2.1 Normal modes of a linear triatomic molecule Three atoms (of mass m, M and m) are assumed to be aligned along the same axis, and we only consider motion along that axis. Also assume nearest neighbour interactions 1 L = (mx˙ 2 + Mx˙ 2 + mx˙ 2) − V (x − x ) − V (x − x ) (5.12) 2 1 2 3 2 1 1 3 2

22 Let xi,0 be the equilibrium positions of the atoms with

x3,0 − x2,0 = x2,0 − x1,0 = r0

Denote xi(t) = xi,0 + ηi(t) departures from equilibrium positions. Expand the Lagrangian to second order in ηi 2 1 ∂ V 2 V (x2 − x1) = (η2 − η1) 2 ∂r2 r0 | {z } ≡k where r = x2 − x1 = r0 + η2 − η1, and k > 0. Equations of motion are

mη¨1 = −k(η1 − η2) Mη¨2 = −k (η2 − η1) + (η2 − η3)

mη¨3 = −k(η3 − η2)

iωt Substituting ηi = Ai e         m 0 0 A1 1 −1 0 A1 2 ω  0 M 0  A2 = k −1 2 −1 A2 0 0 m A3 0 −1 1 A3 Solution only if ω2m − k 1 0  det  1 ω2M − 2k 1  = 0 0 1 ω2m − k

Solutions are – ω2 = 0 and 1 A = 1 1 because     1 −1 0 A1 −1 2 −1 A2 = 0 0 −1 1 A3 – ω2 = k/m and  1  A =  0  −1 – M + 2m ω2 = k Mm and  1  A = −2m/M 1

23 The general vibrational motion of the molecule is a sum of these three normal modes " 1  1  r ! k η =< (C + Dt) 1 + E 0 exp i t     m 1 −1  1  r !# M + 2m + F −2m/M exp i k t   Mm 1

5.2.2 The double pendulum Two massless rods of length `. The extremity of one is ﬁxed at (0, 0), and has a mass m1, at the other end. The extremity of the other is ﬁxed at m1, and has a mass m2 at the other end. 1. Write down the Lagrangian. 2. Find the equilibrium positions. 3. Find the normal modes about the stable equilibrium. Kinetic energy: – mass 1:

x1 = ` sin θ1

y1 = −` cos θ1 1 1 T = m (x ˙ 2 +y ˙2) + m (x ˙ 2 +y ˙2) 2 1 1 1 2 2 2 2 – mass 2:

x2 = `(sin θ1 + sin θ2)

y2 = −`(cos θ1 + cos θ2) 1 1 T = m `2θ˙2 + `2m θ˙2 + θ˙2 + 2θ˙ θ cos(θ − θ ) 2 1 1 2 2 1 2 1 2 1 2 Potential energy:

V = −m1g` cos θ1 − m2g`(cos θ1 + cos θ2)

Equilibirum positions when ∂V ∂V = 0 = 0 ∂θ1 ∂θ2 so ∂V = (m1 + m2) sin θ1 = 0 ∂θ1 ⇐⇒ θ1 = 0 or θ1 = π and ∂V = m2g` sin θ2 = 0 ⇐⇒ θ2 = 0 or θ2 = π ∂θ2

24 There are four equilibrium positions   (0, 0) stable  (0, π) partially stable (θ1, θ2) =  (π, 0) partially stable  (π, π) unstable

At the stable position, we have θ1 = θ2 = 0. Denote the perturbations about the stable equilibrium by η1(t) and η2(t), and expand the Lagrangian to the second order in the ηi. 1 1 L = m `2η˙2 + m `2(η ˙2 +η ˙2 + 2η ˙ η˙ ) 2 1 1 2 2 1 2 1 2 m g` m g` − 1 η2 + 2 (η2 + η2) 2 1 2 1 2

The Euler–Lagrange equations for ηi are in matrix form

M m η¨ M 0 η `2 1 = g` 1 m m η¨2 0 m η2 where M = m1 + m2 and m = m2. iωt Set ηi = Ai e , solve det(ω2m − V ) = 0 to ﬁnd g M r m ω2 = 1 ± ` m M with eigen-vectors √ m A = √ ± M

25 Chapter 6

Constrained systems and Lagrange multipliers

So far, we have always assumed that the generalized coordinates qα were independent. We will now consider situations in which the qα are not longer independent. In this case, the Euler–Lagrange equations (2.8) are not valid. Recall the example of the particle moving on the wire of shape y = Ax2. We used x as the single generalized coordinates, and then used the Euler–Lagrange to determine the dynamics of the particle. If you decide to use x and y as generalized coordinates, these would no longer be independent as y = Ax2, and then you can not use the Euler–Lagrange equations. When deriving the Euler–Lagrange equations, we minimized the action and arrived at the equation

Z tf ∂L d ∂L δS = − δqαdt = 0 ti ∂qα dt ∂q˙α For example, if α = 1, 2, then

Z tf ∂L d ∂L ∂L d ∂L δS = − δq1 + − δq2 dt = 0 ti ∂q1 dt ∂q˙1 ∂q2 dt ∂q˙2

Only if δq1 and δq2 are independent can we conclude from that ∂L d ∂L ∂L d ∂L = = ∂q1 dt ∂q˙1 ∂q2 dt ∂q˙2

If δq1 and δq2 are related then you can not obtain the Euler–Lagrange equations. There are diﬀerent types of constraints: – holonomic constraints: These are constraints of the form

hi(q1, . . . , qN ) = 0 i ∈ 1, n (6.1) J K where n is the number of holonomic constraints, so there are N − n independent coordinates.

26 Exemple 6.1. Read on the wire of shape y = Ax2. If one solves this problem using both x and y as generalized coordinates, then these are related by the holonomic constraints

h(x, y) = y − Ax2 = 0

Exemple 6.2. Cylinder of radius a rolling down a shape. Here y and φ are obvious generalized coordinates, but they are linked by the holonomic constraints

h = y − aφ = 0

Notice N X ∂hi δh = δq = 0 i ∂q α α=1 α – non-holonomic constraints: these come in diﬀerent forms – diﬀerential constraints between the generalized coordinates

N X aiαdqα + bidt = 0 (6.2) α=1 which can not be integrated. – inequalities. Exemple 6.3. A ball on a curcular table of radius a. You would need to impose x2 + y2 ≤ a2.

6.1 Lagrange multipliers

Suppose we make a proﬁt of G(x, y, z) euros when we sell a cereal bar with x ,y, z grams of additives. But european regulations impose that x, y, z must satisfy h(x, y, z) = 0 where h is a given function. The aim is to manufacture a cereal bar which – the (x, y, z) lie on the 2-dimensional surface h = 0. – on this surface you want to maximize your proﬁt. Consider small variations of the inputs x, y, z. Then G varies by ∂G ∂G ∂G δG = δx + δy + δz = 0 ∂x ∂y ∂z but since h(x, y, z) = 0, we also have

∂h ∂h ∂h δh = δx + δy + δz = 0 ∂x ∂y ∂z

27 Hence λh = 0 also, where λ(x, y, z) (Lagrange multiplier) is an arbitraty function. Then we have

δG − λδh = 0 ∂G ∂h ∂G ∂h ∂G ∂h − λ δx + − λ δy + − λ δz = 0 ∂x ∂x ∂y ∂y ∂z ∂z We only have two independent variables (x and y) since h = 0 impose the relationship. Let use choose the arbitraty function λ such that ∂G ∂h − λ = 0 (6.3) ∂z ∂z and since x and y are independent, equations which must be satisﬁed are ∂G ∂h − λ δx = 0 ∂x ∂x ∂G ∂h − λ δy = 0 ∂y ∂y To solve the problem, you need to solve these equations plus the constraint equation h = 0. Let’s return to the action S which we want to minimize subject to the constraint h = 0. Since both δS = 0 and δh = 0, consider

Z tf δS − dt(λδh) = 0 (6.4) ti where λ is an arbitraty function of (qα, t).

N Z tf X d ∂L ∂L ∂h dt δq − − λ = 0 (6.5) α dt ∂q˙ ∂q ∂q ti α=1 α α α There are N − 1 independent coordinates in the problem, which we take to be (q1, . . . , qN−1). Let’s choose λ such that for α = N d ∂L ∂L ∂h − = λ dt ∂q˙N ∂qN ∂qN Then (6.5) becomes

N−1 Z tf X d ∂L ∂L ∂h dt δq − − λ = 0 α dt ∂q˙ ∂q ∂q ti α=1 α α α

But the (q1, . . . , qN−1) are independent generalized coordinates so d ∂L ∂L ∂h − = λ α ∈ 1,N − 1 (6.6) dt ∂q˙α ∂qα ∂qα J K plus the equation h = 0. If you have (λ1, . . . , λn) constraints, one simply replaces

n ∂h X ∂h λ −→ λ (6.7) ∂q i ∂q α i=1 α

28 One can show that the Lagrange multipliers have a physical meaning. They must be related to the external forces which are imposing the constraint between the generalized coordinates. Indeed

constraint ∂h Fα = λ (6.8) ∂qα Exemple 6.4 (Cylinder rolling down an inclined plane). Initially the cynlinder is at rest at y = 0. Find the Lagrangian, the Lagrange multipliers, forces on the cylinder and solve the problem. Use as generalized coordinates y and φ which are linked through

h = y − aφ

The Lagrangian is 1 1 L = my˙2 + Iφ˙2 + mgy sin α 2 2 The Euler–Lagrange equations with the constraints are d ∂L ∂L ∂h − = λ dt ∂y˙ ∂y ∂y =⇒ my¨ − mg sin α = λ and d ∂L ∂L ∂h − = λ dt ∂φ˙ ∂φ ∂φ =⇒ Iφ¨ = −aλ and constraint equation y = aφ. With I = ma2/2 1 y¨ ma2 = −aλ 2 a 1 =⇒ my¨ = −λ 2 Eliminate λ in the ﬁrst equation 2 y¨ = g sin α 3 Integrate and use initial conditions

1 y = gt2 sin α 3 hence from the third equation

y 1 φ = = gt2 sin α a 3a and −1 λ = gm sin α 3

29 m

θ a

Figure 6.1: Particle on an hemisphere.

Finally the constraint forces on the cylinder (making it roll rather than slip) are −1 F constraint = gm sin α y 2 a F constraint = gm sin α φ 3 Exemple 6.5. A particle of mass m starts at rest on the top of a smooth ﬁxed hemisphere of radius a (ﬁgure ??). Find the constraint force, and determine when the particle leaves the hemisphere. Choose generalized coordinates (r, θ). There are not independent when the particle is on the hemisphere, since then r = a. It’s an holonomic constraint : h = r − a = 0. The Lagrangian in terms of (r, θ) is 1 L = m(r ˙2 + r2θ˙2) − mgr cos θ 2 The Euler–Lagrange equations with constraints are d ∂L ∂L ∂h − = λ dt ∂r˙ ∂r ∂r d ∂L ∂L ∂h − = λ dt ∂θ˙ ∂θ ∂θ so we need to solve  mr¨ − mrθ˙2 − mg cos θ = λ  d (mr2θ˙) − mgr sin θ = 0  dt  h = 0 ⇐⇒ r = a From the third we deduce r˙ = 0, so

maθ˙2 − mg cos θ = λ g θ¨ − sin θ = 0 a Aim is to ﬁnd the constraint force and hence we need to ﬁnd λ. To do so multiply the previous equation by θ˙ and integrate

θ˙2 g + cos θ = K 2 a

30 But initially the particle is at rest, at θ = 0 (θ˙ = 0) so K = g/a. Hence

θ˙2 g g + cos θ = 2 a a −λ aθ˙2 − g cos θ = m Eleminate θ˙2 between these equations:

λ = mg(3 cos θ − 2)

and constraint force is Fα = λ∂qα h. The particle falls oﬀ the hemisphere when there is no force to hold on it, so when λ = 0: 2 cos θ = 3

31 Chapter 7

From Lagrangians to Hamiltonians – The Hamiltonian formulation of classical mechanics

Lagrangian approach: – Lagrangian L(qα, q˙α, t) – N independent coordinates qα → N-dimensional configuration space – Euler–Lagrange equations : N second order equations – To solve these equations, one need 2N initial/boundary conditions – Equations invariant under qα → Q(q1, . . . , qN , t) Hamiltonian approach: – Hamiltonian H(qα, pα, t) – 2N dimensional phase space spanned by (qα, pα) – 2N first order equations – 2N boundary conditions – Equations are invariant under a much larger set of "canonical transformation" qα → Q(q1, . . . , qN , t) and pα → P (p1, . . . , pN , t) The Hamiltonian formulation is not particularly better than the Lagrangian are for solving the problems we have studied so far. But the Hamiltonian formulation provides links to many other differents branches of physics : – H and L are linked by a Legendre transformation ←→ thermodynamics (energy E and enthalpy H). – Phase space dynamics ←→ chaos, QM. . . – Poisson brackets ←→ commutators in quantum mechanics (groups, Lie algebras. . . ). The Hamiltonian approach is very powerful but not adapted to the study of relativistic systems (particle physics, classical field theory, general relativity. . . ). Here you need to use the Lagrangian which can be written in a "covariant" way.

32 As a mathematical problem, the transformation from L to H is a change of variables from L(qα, q˙α, t) to H = H(qα, pα, t), where ∂L pα ≡ (7.1) ∂q˙α called the generalized momenta. Exemple 7.1. Consider a function f(x, y) with diﬀerential ∂f ∂f df = dx + dy ≡ udx + ydy ∂x ∂y Suppose we want to change the basis of our description from the variables (x, y) to (u, y). In terms of (u, y) diﬀerentials are expressed in terms of du and dy. Let g = f − ux (Legendre transformation). Then

dg = df − udx − xdu = ydy − xdu so that g = g(u, y). Going from L to H is an identical procedure:

H(pα, qα, t) = pαq˙α − L What are the Hamiltons equations? Calculate the diﬀerential of H

dH = pαdq ˙α +q ˙αdpα − dL ∂L ∂L =q ˙αdpα + dqα + dt ∂qα ∂t ∂H ∂H ∂H = dpα + dqα + ∂pα ∂qα ∂t and ∂H ∂L = − ∂qα ∂qα d ∂L = − = −p˙α dt ∂q˙α hence ∂H = −p˙α (7.2a) ∂qα ∂H =q ˙α (7.2b) ∂pα Remembering Beltrami (3.2): dH ∂L = − (7.3) dt ∂t H is constant if L is explicitly time independent, and corresponds to the total energy in certain cases.

33 Exemple 7.2. A particule of mass m in a central potential V (r). Work in spherical polar coordinates (r, θ, φ). 1 L = m(r ˙2 + r2 sin2 θφ2 + r2θ˙2) − V 2 2 2 2 ˙ pr = mr˙ pφ = mr φ sin θ pθ = mr θ

Replace in

H = pαq˙α − L

= prq˙r + pφq˙φ + pθq˙θ − L ! 1 p2 p2 = p2 + θ + φ + V 2m r r2 r2 sin2 θ

7.1 Poisson brackets

df ∂f ∂f ∂f = p˙ + q˙ + dt ∂p ∂q ∂t ∂f ∂H ∂f ∂H ∂f = − + + ∂p ∂q ∂q ∂p ∂t ∂f = [f, H] + PB ∂t where the Poisson bracket of two function f(q, p, t) and g(q, p, t) is

∂f ∂g ∂f ∂g [f, g] = − (7.4) ∂q ∂p ∂p ∂q You can check that

[f, g] = −[g, f] [f + g, h] = [f, h] + [g, h] h i h i h i [f, g], h + [g, h], f + [h, f], g = 0 Jacobi identity [p, q] = 1 [p, p] = [q, q] = 0

The link between classical mechanics and quantum mechanics is made by: – replacing allfunctions by operators:

f −→ fˆ H −→ Hˆ

– −i [f, g] −→ [f,ˆ gˆ] commutators ~ dfˆ ∂fˆ = i [f,ˆ Hˆ ] + dt ~ dt

34 Part II

Special relativity

35 Chapter 8

Introduction

The fundamental ingredients of special relativity are – the speed of light c, which is a constant of nature ; – nothing (no signal, electromagnetic wave, information. . . ) car propagate faster than c. These are encoded into the two postulates on which special relativity is based (Einstein, 1905). Postulat 8.1 (Principle of relativity). The laws of physics are the same in inertial frames. In other words, there is no preferred rest frame. Postulat 8.2 (Universality of the speed of light). The speed of light is the same in all inertial frames. What are the consequences? 1. Our intuitions on relative speeds is wrong. Exemple 8.1. Suppose we are sitting in a rocket which has a constant velocity v, relative to the earth’s frame (which, for the purposes of the example, is taken to be an inertial frame). Suppose that a light signal is emitted from the back of the rocket, outside. According to an observer on earth, the light signal propagates with speed c. According to our classical mechanics intuition, the observer in the rocket should measure the speed of light signal to be c0 = c − v. Conversly: if you measure c0 from within the rocket, et c outside, then you should be able to determine v. But all experiments of this kind have always given c0 = c, hence the origin of the second postulate of special relativity. This result is in complete agreement with Maxwell’s equations in vacuum. Special relativity resolves the contraction, which reigned in physics at the time of Maxwell, Lorentz, Minkowski, Einstein. 2. Galilean transformations must be wrong. Consider two reference frames R and R0, where R0 moves at constant speed v relative to R along their common x-axis. Galilean transformations

36 are

x0 = x − vt y0 = y z0 = z t0 = t

They are wrong because they give c0 = c − v which is incompatible with experiment. They must be replaced by the Lorentz transformations

x0 = γ(v)(x − vt) y0 = y z0 = z vx t0 = γ(v) t − c2 where γ(v) is the Lorentz factor 1 γ(v) = p1 − v2/c2

Note that coordinates y and z which are perpendicular to the velocity are unchanged. In non-relativistic limit, v/c 1, the Lorentz transformations reduce to the Galilean transformations. 3. Loss of simultaneity: if two events happen at the same time in one reference frame, they occur at diﬀerent times in other reference frame. 4. Lengths are not constant: the lengths measures depend on the frame you are in (it’s not a scalar). 5. Newton’s law must be replaced by d F = (γ(v)mv) dt 6. Space and time are no longer separated: one must work in 4-dimensional space-time. 7. The famous equation E = mc2 or more exactly E = pp2c2 + m2c4. Remarque : In special relativity, space-time is non-dynamical. In general relativity, space-time is dynamical.

37 Chapter 9

The postulates of special relativity and the Michelson–Morley experiment

The first postulate 8.1 can be reworded in different ways: – all laws of physics are the same in inertial frames; – no experiment can measure the absolute velocity of an observer: the result of all experiments made by an observer is independent of his speed relative to other observers not involved in the experience; – there is no special/privileaged frame or observer from the point of view of physics: the vacuum is not absolute. Définition 9.1 (Frame). A frame is an ensemble of observers who are at rest relative to eachother, and each of wchich has a clock. All the clocks are syn- chonised.

Déﬁnition 9.2 (Inertial frame). An inertial frame is one in which, in the ab- sence of forces, a particle moves with constant velocity v. There is an inﬁnite number of inertial frames. In non-relativistic physics, the principle of relativity is encoded in the invariance of physical laws and equations under Galilean transformations. Given a frame R(t, x, y, z) and a frame R0(t0, x0, y0, z0) moving with constant veolocity u along teir common x-axis, then the Galilean transformations are

x0 = x − ut y0 = y z0 = z t0 = t =⇒ v0 = v − u a0 = a

38 Maxwells equations in SI 1 units, and in the vacuum are

∇ E = 0 (9.1a) ∇ B = 0 (9.1b) −∂B ∇× E = (9.1c) ∂t ∂E ∇× B = µ ε (9.1d) 0 0 ∂t These can be combined, giving

∂2 − c2 ∇2 E = 0 (9.2a) ∂t2 ∂2 − c2 ∇2 B = 0 (9.2b) ∂t2 with 1 c = √ (9.3) µ0ε0 the speed of light in vacuum. Maxwells equations (9.1) are written in an inertial frame and hence c is the speed of light in that frame. Contrary to Newtons equations, the equations change under a Galilean transformations, one obtains for the wave equations (9.2)

∂ ∂ ∂ ∂ + (c − u) − (c + u) E0 = 0 ∂t0 ∂x0 ∂t0 ∂x0

Equations consisting of waves propagating with velocity c0 = c ± u. This equation is not (9.2), so Maxwells equations are not invariant under Galilean transformations. As we discussed, experiments were designed to measure "our" speed relative to a prefered inertial frame in which light waves propagate with veolocity c0, but one found c0 = c. What to do? 1. Perhaps Maxwell’s equations are wrong, and need to be modified in order to make Galilean invariant? Lots of trials, but all such modification lead to new electrical phenomena which are ruled experimentally. 2. Perhaps Maxwell’s equations are only valid in a priveledged "ether frame"? And hence in a different inertial frame, light travels with c0 = c ± u. 3. Perhaps the Galilean transformations are wrong and Maxwell’s equations are true in all inertial frames? In this case, one should replace the Galilean transformations by other transformations, which will then no longer leave Newtons laws invariant. 1. Standard international.

39 9.1 The Michelson–Morley interferometer

Aim was to measure c0 and hence ﬁnd our velocity relative to the "ether" frame. At the time, the largest velocity accessible was that of the earth around the sun (30 km/s). – If the experiment is at rest in the ether frame, the times are equal, and hence the beams arrive on the screen in phase. – Now suppose that the experiment moves with speed u relative to the ether frame. Let’s analyse it in the ether frame, where the speed of light is always c. 0 0 Let t1 be the time for light to go from B to C . In this time, light travels a 0 distance L1 + ut1 (emitted at B and arrives at C because C has moved). But we are working in the ether frame where light travels at speed c. Hence L + ut1 = ct1 and L t = (9.4) 1 c − u 0 0 Let t2 be the time taken for light to go from C back to B (d = L − ut2) L t = (9.5) 2 c + u Hence

∆tk = t1 + t2 L L = + c − u c + u 2Lc = c2 − u2 2L/c = (9.6) 1 − v2/c2

0 0 Let t3 be the time taken for light to go from B to D p 2 2 L + (ut3) = ct3 2 2 2 2 t3(c − u ) = L so L/c t3 = (9.7) p1 − u2/c2

Let t4 be the time to go back. It is also t3 (symmetry).

∆t⊥ = t3 + t4 2L/c = (9.8) p1 − u2/c2

On a ∆t⊥ 6= ∆tk. Hence the two beams arrive at the screen with phase diﬀerent

∆φ = ω(∆tk − ∆t⊥) L u2 u4 = 2π + θ (9.9) λ c2 c4

40 which is suﬃciency big to be measured (8u ≈ 30 km/s: but no eﬀect has ever been measured.

41 Chapter 10

Consequences of the constancy of c

About 18 years after the Michelson–Morley experiment, Einstein postulates the second principle of relativity 8.2. Corollaires: Galilean transformations have to be replaced by a new set of transformations that leave Maxwell equations invariant.

10.1 Space-time diagram

A space-time diagram is the (t, x) plane (we suppress y and z because we can not draw in four dimensions). From now, we set c = 1: t and x are both measured in meter. Définition 10.1 (Event). An event is an event which happens at a given value of (t, x, y, z). Hence it’s a point on a space-time diagram. Définition 10.2 (World-line). A world-line of a particle is a curve x(t), which gives the position of the particle as a function of time. We have dt 1 = dx v(t) Since no information can travel faster than light dt v(t) ≤ 1 =⇒ ≥ 1 dx The slope of the world line of a photon is ±1. p 2 2 Careful: this is a space-time diagram: xe + te is not the "distance" between the origin and the event. You can not use Pythagoras theorem, etc., on these space-time diagram. Since v ≤ 1, the concepts of future and past are modified with respect to Newtonian physics. In Galilean physics, space-time is the union of the future (everything that occures after a giving event) and the past (before that event). It is no longer valid in special relativity. The space-time diagram is divided in three areas.

42 Now we want to include on the space-time diagram of O a second observer O0 moving with constant velocity u along their common axis x. Suppose that at t = x = 0, t0 = x0 = 0. We now need to construct the x0 axis (set of all events which occurs at t0 = 0). Let’s consider the space-time diagram of O0. All events on the x0 axis have the following property: a light ray is emitted at x0 = 0 and t0 = −a, and arrives on the x0 axis at t0 = 0 and x0 = a. If it’s reﬂected, it returns to the t0 axis at t0 = 0. Let’s try to draw this on the space-time diagram of O. Remarque : On a space-time diagram, x0 and x axis do not coincide. And t0 and x0 are not perpendicular. The angle θ is given by x tan θ = = u (10.1) t

10.2 Simultaneity

Consider two events A and B.

(tA, xA) = (0, xA)(tB, xB) = (0, 0)

These events are simultaneous for the observer O because they both occur at t = 0. 0 0 0 0 For observer O , tB = 0 = xB, and tA < 0. The two events are not simulta- 0 0 neous since tA < tB. In special relativity, one therefore talks of the "relativity of simultaneity", i.e simultaneity is observer-dependent.

10.3 Invariant interval

Consider E and P on the world line of a photon. Both observer agree that

(∆s)2 = (∆t)2 − (∆x)2 = 0 (∆s0)2 = (∆t0)2 − (∆x0)2 = 0 because of the speed of light. So if ∆s = 0 in one frame then ∆s0 = 0 in the other. Now consider any two arbitrary events B and Q: tB = xB = 0 in O, and 0 0 0 0 0 (tQ, xQ) in O, tB = xB = 0, and (tQ, xQ) in O . We deﬁne the invariant interval by

(∆s)2 = (∆t)2 − (∆x)2 − (∆y)2 − (∆z)2 (10.2) (∆s0)2 = (∆t0)2 − (∆x0)2 − (∆y0)2 − (∆z0)2

One can show that ∆s = ∆s0 (10.3) i.e. ∆s is an invariant on which all observers agree.

43 Proof. – First note that if Q and O are on a photon worldline then ∆s = ∆s0 = 0. – More generally, we use as basic input that space and time are assumed homogeneous 1 and that space is isotropic 2.

Careful: ∆s2 is not necessarly positive (think with ∆t2 = 0). The consequence of this is that the transformations between (x0, t0) and (x, t) must be linear: x0 = αx + βt t0 = γx + δt Hence

(∆s0)2 = (∆t0)2 − (∆x0)2 − (∆y0)2 − (∆z0)2 02 02 = tQ − xQ = Q(t, x, y, z) where 3 X i j i 2 Q(t, x, y, z) = (Aijx x + Bix t + ct ) i=1 since the transformation is linear, and the coeﬃcient Aij, Bi and C are all functions of kuk = u by isotropy, and (x1, x2, x3) = (x, y, z). 2 2 2 2 We know that Q = 0 on a photon world line, i.e. when t` = x +y +z (pho- 2 2 ton at c, t = x in one dimension). Hence Q(t`, x, y, z) = Q(t`, −x, −y, −z) = 0 meaning that Bi = 0 and

X i j 2 Aijx x + Ct` = 0

Hence:

0 2 2 2 (∆s ) = −Ct` + Ct = −C(x2 + y2 + z2) + Ct2 = C(u)∆s2

We want to show that c = 1. We have observers: – O. – O0 moving with velocity u relative to O. – O00 moving with velocity v relative to O and velocity w relative to O0. then

(∆s00)2 = C(v)(∆s)2 = C(w)(∆s0)2 = C(w)C(u)(∆s)2 =⇒ C(v) = C(u)C(w) with solution C = 1. Hence (∆s0)2 = (∆s)2, all observers agree on the value of ∆s. 1. Invariant under translation: there are no prefered position in space or time. 2. Invariant under rotations.

44 10.3.1 Interval types The interval is said to be – time-like if ∆s2 > 0; – space-like if ∆s2 < 0; – light-like if ∆s2 = 0. If ∆s2 ≥ 0, the two events are causally connected since

∆s2 ≥ 0 =⇒ ∆t2 − ∆x2 ≥ 0 ∆x2 =⇒ 1 − ≥ 0 ∆t =⇒ v2 ≤ 1

There exists a reference frame moving with constant velocity v such that, in this frame, the two events occur at the same spatial position. If ∆s2 < 0, the two events are causally disconnected: v > 1, so it is impos- sible. Remarque : Some authors prefer to use the convention

(∆s)2 = −(∆t)2 + (∆x)2 + (∆y)2 + (∆z)2

10.4 Calibration of (t0, x0) axis

How can we callibrate the (t, x0) axes on the space-time diagram of O? Where is t0 = 1 mark? To do this, we use the invariant interval, taking one event to be the origin (t = x = t0 = x0 = 0) and the other being a general event (t, x) which has coordinates (t0, x0) in O0.

∆s2 = t2 − x2 = t02 − x02

∆t = t − 0 Which set of events have space-like invariant interval given by ∆s2 = −a2? p t = ± x2 − a2

Which set of events have time-like invariant interval given by ∆s2 = b2? p t = ± x2 + b2

If we choose b = 1, then, from the invariance of the interval, t02 − x02 = 1. But by deﬁnition, on the t0 axis, x0 = 0. Hence the intersection of the t0 axis with curve t2 − x2 = 1 corresponds to t0 = ±1.

10.5 Proper time and time dilatation

For a time-like interval, the proper time ∆τ is deﬁned by

∆s2 = c2∆τ 2 (10.4)

45 What is the physical meaning of τ? Consider the causally connected events O and A(tA, xA). There exists a frame moving woth velocity v in which two diﬀerent events occur at the same position x0 = 0 but at diﬀerent time. Using the invariant interval

∆s2 = ∆t2 − ∆x2 = ∆t02 − ∆x02 | {z } =0

So, with c = 1, τ is nothing other that the time t0. From above p ∆t0 = ∆t2 − ∆x2 (10.5) p = ∆t 1 − v2 (10.6)

Since v < 1, ∆t0 < ∆t. The time that you measure between two events is not the same depending of the observer you are.

Exemple 10.1 (Cosmic rays). More radioactivity was measured on earth that could actually be explained by known sources of radioactivity. Hess (1912) went up in a balloon to altitudes higher than 6000 m and discovered that radioactivity increases with altitude. He deduced that there were radioactivity sources outside the earth: the cosmic rays. Cosmic rays consist not only of rays (X- and γ-rays), but mostly charged particles (protons. . . ). Their energy are between 108 eV (v ≈ 0.4c — assuming protons) and 1021 eV (v ≈ 0.999c). When a cosmic ray interacts with the particles in the upper atmosphere, − − they produce muons (µ). . . Muons decay trhough µ → e + νe + νµ with −5 τµ = 2.2 × 10 s in the frame in which the muon is at rest. Without special relativity: the muon travels a distance d = v×τµ = 659 m < 12 km, but muons are detected on earth in large quantities. This is because of the time dilatation. On the earth frame, the muon decay is a time tµ = γτµ and v × tµ 12 km. So the distance they actually travel aroud 15 km.

46 Chapter 11

Lorentz transformations

Let’s derive them: to do so, we again – assume homogeneity and isotropy of space-time; – recall that c = cste only aﬀects the axis parallel to the direction of motion. As usual, take two frames O and O0 with constant velocity v along their common axis x. We can already say that y0 = y and z0 = z. Homogeneity of space and time implies that the transformations are linear:

0 0 t = α1t + α2xx = α3x + α4t (11.1)

(αi depend on |v| by isotropy). Recall that the t0 axis has equation t/x = 1/x. Hence

0 = α3tv + α4t

0 = α3v + α4

The x0 axis has equation t/x = v. Hence

0 = α1v + α2

Hence

0 t = α1(t − vx) 0 x = α3(x − vt)

Now let’s use the invariant interval, which tells us that

t2 − x2 = t02 − x02 2 2 2 = α1(t − vx) − α3(x − vt) 2 2 2 2 2 2 2 2 2 2 = (α1 − α3v )t + (α1v − α3)x − 2vxt(α1 − α3) so α1 = α3 and 2 2 1 α (1 − v ) = 1 =⇒ α1 = √ ≡ γ 1 1 − v2

47 The Lorentz transformations are t0 = γ(t − vx) x0 = γ(x − vt) (11.2) y0 = y z0 = z

The inverse transformations are t = γ(t0 + vx0) x = γ(x0 + vt0) (11.3) y = y0 z = z0 But for the other observer, the frame moves with velocity −v. Caution: never use the Lorentz transformations to relate two diﬀerent events. In matrix form we have t0   γ −vγ 0 0 t x0 −vγ γ 0 0 x   =     (11.4) y0  0 0 1 0 y z0 0 0 0 1 z It’s useful to write down the Lorentz transformations for the frame O0 moving relating to O in some arbitrary direction v. To do so, decompose the spatial position vector r = (x, y, z) of the event into components parallel and perpendicular to v. Let ev k v a unit vector so that

rk = (r · ev)ev = (r · v v) | {z } =kvk2

r⊥ = r − rk Under a Lorentz boost in the v direction: 0 r⊥ = r⊥ 0 rk = γ(rk − vt) so that 0 0 0 r = r⊥ + rk

= r⊥ + γ(rk − vt)

= (r⊥ + rk) + (γ − 1)rk − γvt r · v = r + (γ − 1) v − γvt v2 In matrix form   γ −γvx −γvy −γvz  0    t (γ−1)v2 . . t 0  x .. ..  x −γvx 1 + v2  x   =     (11.5) y0  ......  y   −γvy . . .    z0   z ...... −γvz . . .

48 11.1 Time dilatation with Lorentz transformations

Consider two events A and B in frame O, which take place at diﬀerent tA and tB but at the same position xA = xB = x0. So the time between the events 0 o, frame O os δt = tB − tA. Now go to a moving frame O . 0 0 0 ∆t = tB − tA

= γ(tB − vx0) − γ(tA − vx0)

= γ(tB − tA) = γ∆t

11.2 Length contraction

Another remarquable consequence of Lorentz transformations regards the lengths of a given object, say a ruler, measured in different inertial frames. Let’s fix the ruler to be at rest in frame O, and parallel to the x axis. The observer will be in frame O0 (with velocity v). – If v = 0, so that O0 = O and the observer is at rest relative to the ruler. Length is defined by ∆x = xR − xL = xR. In other words is defined to be the difference in positions of the end points of the ruler at the same time. – If v 6= 0, the length is again defined by a photo taken by the observer in O0. 0 0 0 0 0 L = xR(t ) − xL(t ) 0 0 = xR(0) − xL(0) What is the relationship between L0 and L? 0 0 0 – Event B : xB = L, tB =?, xB = L , tB = 0. 0 0 – Event O : xO = tO = xO = tO = 0. 0 0 – Event A : xA = L, tA = 0, xA =?, tA =?. To find the relationship, we have three methods: 1. Inverse Lorentz transformations gives 0 xB = γxB L = γL0 i.e. the moving observer measures the ruler to have a length L0 smaller than L. 0 2. The x axis has equation t/x = v and hence tB = vxA = vL. Using the Lorentz transformations applied to B we find the relationship. 3. Using the invariant interval between O and B. Remarque : If the ruler is moving and the observer is stationary, does he see a ruler with a longer length? We find actually the same relation.

11.3 Doppler eﬀect

Time dilatation plays an important role in many relativistic phenomena, including the Doppler eﬀect. For example, the universe is expanding, with

49 the velocity v of galaxies a distance at a distance r away from us with v ∝ r (Hubble’s law, 1929). This measurement is based on the Doppler eﬀect. Consider a source emitting light with wavelength λ (and frequency ν). If the source moves relative to the observer, then the observer measures light with wavelength λ0 (frequency ν0). In non-relativistic case, then t0 = t (no time dilatation. . . ), and according to the observer the wave fronts are separated by λ + v∆t = c∆t, so that λ ∆t = (11.6) c − v from which v ν0 = ν 1 − (11.7) c λ λ0 = (11.8) 1 − v/c

In the relativistic case t0 6= t. The Doppler effect formule is modified because of time dilatation and in fact ∆t must be replace with γ∆t. The relativistic Doppler effect is v ν0 = γν 1 − (11.9) c s 1 + v/c λ0 = λ (11.10) 1 − v/c

11.4 Composition of velocities

Recall that from the Galilean transformations, the velocities of a particle viewed in two diﬀerent frames was w0 = w−v. This can not be righr since it does not leave the speed of light invariant. We must use the Lorentz transformations. Consider two frames O and O0 with O0 moving along the x axis with velocity v. In frame O a particle moves with velocity w. What is its velocity w0 in frame O0? dx0 w0 = dt0 γ(dx − vdt) = γ(dt − vdx) (dx/dt − v) = 1 − vdx/dt w − v = 1 − vw If w = 1 we ﬁnd w0 = 1.

50 Chapter 12

Minkowski spacetime, 4-vectors and Lorentz invariants

We have studied the components of (t, x, y, z). We have shown how it transforms under Lorentz transformations. We have also constructed the invariant interval ∆s2. Many other quantities transform in a similar way under Lorentz boost. We will see, for example, that (E, px, py, pz) – where E is the energy of the particle and p its momemtum — transform in the same way:

0 E = γ(E − vpx) 0 px = γ(px − vE) and the invariant is m2 = E2 − p2 In the following, we will construct general 4-vectors, that is vectors in space- time which transform in the same way as (t, x, y, z) under a Lorentz transformations.

12.1 4-vector

4-vectors will be denoted by A. We will introduce a four dimension (Minkowski) space-time and deﬁne a scalar product between 4-vectors (A · B). Remarque : All of this will apply to general relativity. We will start with something much simpler than 4-dimension space-time: the plane in 2D. With this example we can understand the diﬀerence between 1 x and x1, and also introduce the concept of metric. Let B = (e1, e2) (ke1k = ke2k = 1) be a base. There are two ways in which to express a vector a in this basis:

1 2 a = a e1 + a e2

a · e1 = a1 a · e2 = a2

51 1 1. If B is orthonormal, a = a1, since

1 1 a1 = a · e1 = (a e1 + a2e2)e1 = a

2. If B is non-orthonormal, we have

1 2 a1 = a · e1 = a + a e1 · e2

A 4-vector A in space-time with similarly be described either by: 1. its contravariant components Aµ;

2. its covariant components Aµ. µ In general A 6= Aµ. Back in the 2D Euclidean plane, consider the scalar product between two vectors a and b: 1 2 1 2 a · b = (a e1 + a e2)b = a b1 + a b2 From now we will use a modiﬁed summation convention (Einstein summation convention): we only sum over two repeated indices if one of them is "upstairs" and the other is "downstairs". Hence

1 2 i a · b = a b1 + a b2 = a bi (12.1)

The same will apply with 4-vectors. i i Reconsider the scalar product a · b. Since a = a ei and b = b ei, then i j a · b = a b (ei · ej ). Using this expression, we deﬁne the metric

gij = ei · ej (12.2) and i j a · b = a b gij (12.3) and we deduce that j bi = gijb (12.4) The metric is the object which enables one to raise and lower indices. Rappel: Using the example of the 2D Euclidean plane, we discussed the i diﬀerence between ai and a . i – One goes between a and ai using using the metric gij

j ai = gija

– Modiﬁed summation convention (used all the time here after): sum only on repated indices when one is "downstairs" and the other "upstairs". Hence

2 j X j k ai = gija = gija = gika j=1

– Scalar product between two vectors a and b is given by

i j i a · b = a b gij = a bi

52 – The distance between x and x + dx is

2 2 i j d` = dx = gijdx dx (12.5)

– By deﬁnition the metric is gij = ei · ej so

gij = gji (12.6)

If ei are orthogonal and normalised then 1 0 0 1 The distance d`2 = dx2 + dy2 is invariant under rotations, translations and reﬂections. For example, a rotation might be written as

i i j x = R jx We wan also write x0 cos θ − sin θ x = y0 sin θ cos θ y i If ei are non-orthogonal, gij change and a 6= ai. The metric tells us about the intrinsec geometry of a surface. Back to special relativity. – Here we deal with vectors A in 4-dimensional space-time. – A can be represented either by its contravariant or covariant components. – Greek indices take values 0 to 3. That is they run over the 4-space-time directions. – Latin indices take values 1 to 3. That is they run over the 3 spatial directions. The time component of a vector is the µ = 0 component. – The metric of special relativity (Minkowski metric) is conventionnally denoted by ηµν rather than gµν (which is reserved to general relativity). 1 0 0 0  0 −1 0 0    (12.7) 0 0 −1 0  0 0 0 −1

– ηµν is constant in special relativity does not change from a frame to another. – As before we raise and lower indices using the metric

ν Aµ = ηµν A (12.8) 0 i For example A0 = A but Ai = −A . µν – The inverse metric η is the inverse matrix of ηµν . −1 ij −1 M with component Mij and M with components M . We have MM = kj j µν I and MikM = δi . So η = ηµν . µν The components (numerical value) η are the same as those of ηµν , but in order to use the summation convention we must diﬀerentiate between the two µα µα ν η Aµ = η ηµν A | {zα } = δ µ

53 µα α ν α α µα η = δ ν A = A so A = η Aµ. We can go the inverse matrix to go from covariant to contravariant indices:

ν α αβ Aµ = ηµν A ⇐⇒ A = η Aβ

α αβ Rules for indices: An expression such as A = η Aβ contains two types of indices: – α indices which are free indices (not summed over). – β indices which are dummy indices (which are summed over). The ﬁnal expression Aα does not depend on β since we sum over β. The free indices must appear in the same position (that is upstairs or downstairs) on each side of the equation. Scalar product, as before

µ ν µ A · B = ηµν A B = A Bµ

i Hence B = −Bi and

A2 = (A0)2 − (A1)2 − (A2)2 − (A3)2 It reminds the invariant ∆s and it is not positive deﬁnite. As before, in the context of special relativity (where we discuss events, invari- 0 1 2 3 ants. . . ), we deﬁne (x , x , x , x ) = (t, x, y, z) and (x0, x1, x2, x3) = (t, −x, −y, −z), hence x2 = t2 − x2 | {z } invariant interval Lorentz transformations:

µ 0 µ ν (x ) = Λ ν x (12.9)

Since the interval is invariant under Lorentz transformations, x02 = x2 or

0µ 0ν α β ηµν x x = ηαβx x but there is a problem for repeated indices so

µ α ν β α β ηµν Λ αx Λ βx = ηαβx x hence µ ν ηµν Λ αΛ β = ηαβ (12.10) This identity defines the Poincaré group. Définition 12.1 (4-vector). A vector A is said to be a 4-vector if its contravari- µ µ 0 µ ν µ 0 µ ν ant components A transform as (A ) = Λ ν A when (x ) = Λ ν x . Hence, by construction, A2 is invariant under Lorentz transformations. A·B is also invariant under Lorentz transformations. (A + B)2 is also invariant. Just as the invariant interval x2, a 4-vector A is classified according to the sign of A2: – if A2 = 0 then A is light-like; – if A2 > 0 then A is time-like; – if A2 < 0 then A is space-like.

54 4-vectors can have some very strange properties: for example, consider two light-like 4-vectors A and B. Suppose that B k A, i.e. B = λA (λ ∈ R). But B is also perpendicular to A:

A · B = A · λA = λA2 = 0

Consider again two general 4-vectors A and B, then AB is also a 4-vector, since µ µ 0 0µ 0µ µ ν µ ν µ (A + B ) = A + B = Λ ν A + Λ ν B = Λ ν (A + qB) Hence (A + B)2 is Lorentz-invariant. But (A + B)2 = A2 +2A B + B2 | {z } |{z} |{z} invariant invariant invariant hence A B is invariant. More generally, a quantity is Lorentz-invariant (by construction) if it contains no free indices. Exemple 12.1. µ ν 1. A B = A B ηµν contains no free indices and is Lorentz-invariant. µ 2. A Bν contains the two free indices µ and ν (which are not summed over). This is not a Lorentz-invariant. µ 3. C Bν Cµ is not Lorentz-invariant.

12.2 Some 4-vectors

Some 4-vectors contravariant are: – The position xµ. dxµ – The velocity uµ = 1. ds d2xµ – The acceleration aµ = . ds2 – The momemtum pµ = muµ. µ Other 4-vectors such as ∂µ = ∂/∂x are covariant. A 4-vector A with covariant components Aµ transforms as

0 ν Aµ = Λµ Aν under a Lorentz transformation. ν What is Λµ ? α Apply the rules starting from Λ β. 1. Lower the index α: α Λµβ = ηαµΛ β 2. Raise the index β: βν ν Λµβη = Λµ One can also use the transformation properties of Aα:

0 0ν ν γ ν γε Aµ = ηµν A = ηµν Λ γ A = ηµν Λ γ η Aε

| {z ε } =Λµ

We will see later that ∂/∂µ indeed transforms as a covariant-vector.

1. Where ds is the invariant interval.

55 Chapter 13

Diﬀerent 4-vectors and their invariant

13.1 Velocity

By deﬁnition dxµ uµ = (13.1) dτ where τ is the proper-time deﬁned by ds2 = c2dτ 2. uµ are indeed the contravariant components of a 4-vector (u) since τ is Lorentz invariant: hence under a Lorentz transformation,

0µ µ ν u = Λ ν u

Notice that dxµ/dt can not be the contravariant components of any 4-vector since under a Lorentz transformation

µ 0 µ ν γ dx Λ ν dx µ dx = 0 α 6= Λ γ dt Λ α dx dt So dx0 dt u0 = = c dτ dτ dxi dt dxi ui = = dτ dτ dt What is the invariant

2 µ ν u = u u ηµν = (u0)2 − (u)2 dt 2 dx2 = c2 − dτ dt dt 2 c2 = (c2 − v2) dτ

56 and dt = γ dτ Hence u = (γc, γv) (13.2) then c2 = u2. Thus u is a time like 4-vector.

13.1.1 Composition of velocities Let v and v0 be the velocities of particle in the original frame and the frame moving with velocity β. Since (u0)0 = γ(β)(u0 − βu1) and (u1)0 = γ(u1 − βu0) and uµ = γ(v)(c, v) and u0µ = γ(v0)(c, v0), the equation is

0 γ(v ) = γ(β) γ(v) − βγ(v)vx

13.2 Acceleration

The 4-vector acceleration is deﬁned by

du d2x A = = (13.3) dτ dτ 2 Its components are Aµ = γ4v · a, γ4(v · a)v + γ2a (13.4)

Since u2 = c2 we have du 2u = 0 dτ so u · A = 0 (13.5) Thus A is perpendicular to u. Since u is time-like vector, it follows from (13.4) that A · u is a space-like vector.

13.3 Derivative ∂

Consider ∂ ∂ = , ∇ (13.6) ∂xµ ∂t Are there components of a 4-vector? If so, is it "naturally" covariant or contravariant? ∂ ∂xα ∂ = (13.7) ∂x0µ ∂x0µ ∂xα 0β β α α α 0µ µ ε where since x = Λ αx , it follows that x = Λµ x (recall that Λε Λ ω = µ δ ω), thus ∂xα = Λ α (13.8) ∂x0µ µ

57 and ∂ ∂ = Λ α (13.9) ∂x0µ µ ∂x0α This is the transform rule for the covariant components of a 4-vector. Hence we deﬁne ∂ ∂ = (13.10) µ ∂xµ to be the covariant of the 4-vector ∂. Why is this useful? Exemple 13.1. ∂2 is Lorentz invariant and

2 2 2 ∂ = ∂0 − ∇

13.4 Wave k

The wave 4-vector is deﬁned by

kµ = (ω, k) (13.11) where ω is the angular frequency and x the wave vector (with k2 = ω2). Indeed, the phase of an electromagnetic wave

ϕ = ωt − k · x = x · k (13.12) is a Lorentz scalar. k is a light-like 4-vector since k2 = 0.

13.4.1 Application Consider a source emitting light with angular frequency ω at an angle θ to the x-axis in a frame O. According to an observer in frame O0 moving with velocity v along the x-axis, the angular frequency is ω0 and the source is at angle θ0 with the x0-axis. The diﬀerent components of kµ in the two frames are

kµ = (ω, −ω cos θ, −ω sin θ) k0µ = (ω0, −ω0 cos θ0, −ω0 sin θ0) since |k| = ω. But kµ are the components of a 4-vector then

ω0 = γ(ω − vk) = γω(1 + cos θ)

(relativistic Doppler eﬀect).

58 13.5 Momentum

The momentum 4-vector is deﬁned by

P = mu (13.13) where m must be a Lorentz invariant. Hence m is an invariant characteristic of a particle: its mass. Remarque : Note that P is not deﬁned for massless particles propagating at the speed of light . For photons, we use k. The invariant P 2 = m2u = m2c2 (13.14) and the components of P are

P µ = (γmc, γmv) (13.15)

What is P 0? Consider the non-relativistic limit, then 1 v2 P 0 ≈ mc 1 + + ··· 2 c2  

1  2 1 2  ≈ mc + mv + ···  c  2  | {z } kinetic energy This motivates the deﬁnition, valid for all v of E P 0 = (13.16) c where E is the total energy of the particle. The invariant is E2 = p2c2 + m2c4 (13.17) because E2 P 2 = m2c2 = − p2 c2 If the particle is at rest, p = 0 and E = mc2, the rest mass of a particle. For a massless particle E = |p|.

13.6 Force

We must write down a generalisation of Newton law, namely an equation of the form "f = ma", but which transform correctly under Lorentz transformation. Hence it must be in terms of 4-vectors

F = mA (13.18) and hence F · u = 0 (13.19) The obvious question is: what is the link between the components F µ and the usual 3-force f?

59 Recall (13.4). Does there exist a frame in which A reduces to a = dv/dt? If so, in that frame F = f. The answer is yes: when v = 0 so that γ = 1 and µ µ AIRF = (0, a) (IRF means instantaneous rest frame). In this frame FIRF = µ (0, f) and uIRF = (1, 0). Since F µ is a 4-vector, we can now ﬁnd its components in any frame using the inverse Lorentz transformations:

F µ = γ(v · f, f) (13.20) and we have dP µ F µ = (13.21) dτ v · f is the work done by the force.

60 Chapter 14

Particle collisions

In classical mechanics, particle collisions are straight forward to study if the particle are free (i.e. no interactions between them):

p1 + p2 + · · · −→ q1 + q2 + ···

We use the conservations of energy and momemtum: X X Ei = Ef ini fin X X pi = pf ini fin

In special relativity, the unique equation is X X P i = P f (14.1) ini fin since P µ = (E, p). Furthermore, since each particle also has a rest-mass energy (mc2), regarding energy conservation, some of that mass can be converted into energy and vice versa. Hence the initial number of particles is not necessarly equal to the ﬁnal number of particles:

p1 + ··· + pr −→ q1 + ··· + qs

Generally r 6= s and r s X X P i = P f i=1 f=1 A collision is said to be elastic if the initial and ﬁnal particles are the same:

a + b −→ a + b e− + γ −→ e− + γ (Compton eﬀect)

61 else, it’s said to be inelastic:

e+ + e− −→ γ (annihilation of particles) π0 −→ γ + γ (uncharged pion) n −→ p + e− + ν (β-decay/radioactivity) p + p −→ p + p + p + p (proton-proton scattering)

The analysis of such collisions can be simpliﬁed by recalling that: – For each collision you can construct many invariants, for example p , (p + 1 1 p )2... 2 – One can easily pass from one reference frame to another using Lorentz transformations. Indeed in certain reference frames, calculations can be much simpler. Typically three frames come into these types of calculations: – The laboratory frame: the one in which the experiment is done and hence the one in which you must present your results! – Center of mass frame: in which the total spatial momentum vanishes:

r s X X pi = pf i=1 f=1

– Frame attached to one of the particles.

14.1 Examples 14.1.1 An elastic collusion: the Compton eﬀect

e− + γ −→ e− + γ One use also the name "Compton scattering". Take as laboratory frame the frame in which the initial electron is at rest. The photon and the electron are scattered with an angle θ and ϕ. Find the wavelength λ if the scattered photon in terms of the wavelength λ of the initial photon, the electron mass m and the photon diﬀusion angle θ.

Initial Final ˆµ 2 2 – Pe− = (E, p) with E = m + µ 2 – Pe− = (m, 0) |p| µ – P = (|pγ | , pγ ) ˆµ 0 0 γ – Pγ = (|pγ | , pγ ) – |p | = ω = h/λ 0 0 γ ~ – |pγ | = h/λ

Conservation of 4-momentum: ˆ ˆ P e− + P γ = P e− + P γ

It’s not a good idea to square the expression, because we would have ϕ.

62 ˆ ˆ P e− + P γ − P γ = P e− ˆ 2 ˆ2 2 (P e− + P γ − P γ ) = P e− = m ˆ =⇒ P e− + P γ − P γ = m |{z} |{z} =0 =0 ˆ ˆ 2 2 =⇒ 2(P e− · P γ − P e− · P γ − P γ · P γ ) + m = m ˆ ˆ =⇒ P e− · P γ = P e− · P γ + P γ · P γ h h h h =⇒ m = m + − p · p 0 λ λ0 λ0 λ γ γ 0 0 where pγ · pγ = |pγ | |pγ | cos θ, and since |pγ | = h/λ h λ0 − λ = (1 − cos θ) (14.2) m

14.1.2 Elastic proton scattering

p + p −→ p + p – Suppose that in the experiment (laboratory frame) which studies this collision, one of the initial protons is at rest. Let E be the energy of moving proton. – What is the energy E0 of each of the initial protons in the center of mass frame? (see table 14.1)

Laboratory Center of mass µ 0µ 0 Proton 1 p1 = (m, 0) p1 = (E , p) µ 0µ 0 Proton 2 p2 = (E, q) p2 = (E , −p)

Table 14.1: Impulsion 4-vector in the laboratory and center of mass frames

We have 2 0 0 2 (P 1 + P 2) = (P 1 + P 2) 2 2 0 2 P 1 + P 2 + 2P 1 · P 2 = (2E ) m2 + m2 + 2(Em) = 4E02 m2 + Em = 2E02 so r1 E0 = m(E + m) (14.3) 2

14.1.3 Desintegration

a −→ b + c a, b, and c are some particles with masses ma, mb and mc.

63 1. Given mb and mc, can this desintegration occur for all values of ma? 2. In the center of mass frame, what is the 3-momentum of b and c as a function of ma, mb and mc? To answer to the ﬁrst question, we must work in the center of mass frame, since particles b and c have minimum energy when their 3-momentum vanishes (pb = pc = 0). But this is the deﬁnition since pb + pc = 0! In the center of mass frame:

0µ Pa = (ma, 0) q 0µ 0b 2 2 Pb = (E , p) = ( mb + p , p) 0µ 0c p 2 2 Pc = (E , −p) = ( mc + p , −p)

Conservation of P : from the µ = 0 component of

0µ 0µ 0µ Pa = Pb + Pc q 2 2 p 2 2 ma = mb + p + mc + p

ma ≥ mb + mc since p2 > 0. What is |p|?

0b 2 2 2 (E ) = mb + p 0c 2 2 2 (E ) = mc + p 0b 2 2 0c 2 2 =⇒ (E ) − mb = (E ) − mc and we also have 0 0 ma = Eb + Ec 0 0 so substitute Eb = ma − Ec and this gives

0 2 2 0 2 2 (ma − Ec) − mb = (Ec) − mc 2 2 2 2 0 2 2 ∗ ma − mb + mc =⇒ ma − 2Ecma − mb = −mc =⇒ Ec = 2ma

0 2 2 2 Finally since (Ec) = mc + p :

2 2 2 2 2 0 2 2 ma − mb + mc 2 |p| = (Ec) − mc = − mc 2ma

14.1.4 Inelastic collision of protons

p + p −→ p + p + p + p In the laboratory frame, one of the protons is at rest. What is the minimum of energy Emin that the other proton must have in order for process the reaction to occur?

64 It’s easiest to calculate the minimum energy in the center of mass frame, and then to transform to the laboratory frame. The conservation of energy in center of mass frame is 0 2Emin = 4m

Now need to transform to the laboratory frame in order to determine Emin. Again use the invariant:

2 0 0 2 (P 1 + P 2) = (P 1 + P 2) 2 2 02 02 0 0 P 1 + P 2 + 2P 1 · P 2 = P 1 + P 2 + 2P 1 · P 2 2 2 2 2 0 2 2 m + m + 2mEmin = m + m + 2 (Emin) + q 0 2 2 mEmin = (Emin) + q thus

0 2 0 2 2 mEmin = (Emin) + (Emin) − m = 4m2 + (4m2 − m2) = 7m2

0 2 2 2 so Emin = 7m. But (Emin) = m + q .

65 Chapter 15

Electromagnetism

Our starting point was Maxwells equations: c is a Lorentz invariant and the wave equations are invariant under Lorentz transformations. Now we want to try to rewrite Maxwells equations themselves in the language of 4-vectors. We are used to solve Maxwells equations in simple situation (eg: a charge Q at rest). But what if the charge is moving? But to go from a stationary to a moving charge we only need to do a Lorent transformation: if we know E and B transform under a Lorentz transformation, we could easily ﬁnd E0 and B0. Can one construct some 4-vectors from E and B? The answer is no: but they can be uniﬁed into a 4-tensor Fµν , which we will construct.

15.1 Tensors

A number of physical quantities are neither scalars (Lorentz invariants), nor the components of 4-vectors, but the components of 4-tensors. A tensor generally has n ≥ 2 indices which can be covariant or contravariant. µν µ An example is η , ηµν , η ν . A general tensor is typically denoted by the letter T (for tensor). Exemple 15.1. µναβ µ δ For n = 4 : T , T γ ε... Indices on a tensor are raised and lower using the same rules as a 4-vector, but with the additional rule that order of indice on the tensor is left unchanged. Exemple 15.2. α αβ Since u = η uβ α αβ Tγ = ηβγ Under a Lorentz transformation, an n = 2 transforms in the same way as two 4-vectors with the same index structures. So since α β α β γ δ A B −→ Λ γ Λ δ(A B ) then αβ 0αβ α β γδ T −→ T = Λ γ Λ δT (15.1) α 0α α β γ T β −→ T β = Λ γ Λδ T δ (15.2)

66 15.2 Electromagnetic 4-tensor

E and B will be the components of "naturally" contravariant 4-tensor with two indices F µν , the electromagnetic 4-tensor. To construct F µν , we need to go back to Maxwell’s equations ∇ E = ρ (15.3a) ∂B ∇× E = − (15.3b) ∂t ∇ B = 0 (15.3c) ∂E ∇× B = j + (15.3d) ∂t From (15.3a) and (15.3d) we ﬁrst derive the continuity equation ∂ρ ∂E = ∇ = ∇(∇× B − j) = − ∇ j ∂t ∂t ∂ρ =⇒ + ∇ j = 0 (15.4) ∂t Time in all inertial frames, i.e if must be possible to write this equation in explicitly Lorentz invariant form: µ ∂µj = 0 (15.5) where jµ is the current 4-vector with components jµ = (ρ, j) (15.6) Now consider the two Maxwell equations (15.3c) and (15.3b), and as usual introduce the potentials φ and A. B = ∇× A (15.7a) ∂A E = − ∇ φ − (15.7b) ∂t However it’s important to note that given E and B, there is no unique way of constructing φ and A: the transformations A0 = A − ∇ f (15.8a) ∂f φ0 = φ + (15.8b) ∂t leave E and B invariant. It’s useful to restrict the possible choices for φ and A by imposing some extra-conditions: a particularly useful choice is the so-called Lorenz gauge: ∂φ + ∇ A = 0 (15.9) ∂t In terms of φ and A, Maxwell equations with sources (15.3a) and (15.3d) become (using (15.9) and (15.7)) 2 2 (∂t − ∇ )φ = ρ µ 0 (∂µ∂ )φ = j

67 and

2 2 (∂t − ∇ )A = j µ (∂µ∂ )A = j

0 µ Since (j , j) are the components of a 4-vector, and since ∂µ∂ is a Lorentz invariant, (φ, A) are the components of a 4-vector

Aµ = (φ, A) (15.10) and the previous equation become

µ ν ν (∂µ∂ )A = j (15.11)

From (15.7), the derivatives of Aµ give the electric and magnetic ﬁelds, hence deﬁne F µν = ∂µAν − ∂ν Aµ (15.12) and   0 −Ex −Ey −Ez µν Ex 0 −Bz By  F =   (15.13) Ey Bz 0 −Bx Ez −By Bx 0 For example

F 01 = ∂0A1 − ∂1A0 = ∂0A1 − ∂1φ ∂A1 ∂φ = + ∂t ∂x Note that F µν = −F νµ, i.e F µν is an antisymmetric tensor. In terms of F µν , Maxwells equations (15.3a) and (15.3d) can be written as

µν ν ∂µF = j (15.14) and Maxwells equations without sources (15.3c) and (15.3b) are

∂λFµν + ∂µFνλ + ∂ν Fλµ = 0 (15.15)

Finally one can construct from Fµν a number of diﬀerent Lorentz invariant. For example µν 2 2 Fµν F ∝ E − B (15.16) which is, in fact, the Lagrangian of electromagnetism.

15.3 Transformations of E and B

By deﬁnition µν µ ν αβ F = Λ αΛ βF

68 so for example take µ = 0 and ν = 1.

0 Λ 0 = γ 0 Λ 1 = −vγ 1 Λ 0 = −vγ 1 Λ 1 = γ so

001 0 0 1 0β 0 1 1β F = −Ex = Λ 0Λ βF + Λ 1Λ βF 0 1 01 0 1 10 = Λ 0Λ 1F + Λ 1Λ 0F 2 2 = γ (−Ex) + (−vγ) Ex 2 2 = −γ Ex(1 − v )

= −Ex so 0 Ex = Ex Taking µ = 1 and ν = 3

013 0 1 3 αβ F = By = Λ αΛ βF 1 3 0β 1 3 3β = Λ 0Λ βF + Λ 1Λ βF 1 3 03 1 3 13 = Λ 0Λ 3F + Λ 1Λ 3F

= (−vγ)(−Ez) + γBy so 0 By = γ(By + vEz) Doing the same for all values of µ and ν gives

0 E⊥ = γ(E⊥ − v × B⊥) (15.17a) 0 Ek = Ek (15.17b) 0 B⊥ = γ(B⊥ − v × E⊥) (15.17c) 0 Bk = Bk (15.17d) (15.17e) i.e

0 Ex = Ex 0 Ey = γ(Ey − vBz) 0 Ez = γ(Ez − vBy)

15.4 An example: moving charge

In frame O, a charge q has constant velocity v along the x-axis. At t = 0, the charge is at (x, y) = (0, 0). Calculate the E and B ﬁelds generated by the charge, measured by an observer at rest in O, at position (0, −b).

69 Introduce a frame O0 in which the charge is at rest at (x0, y0) = (0, 0). In this frame

B0 = 0 q E0 = r0 |r0|3 where r0 = (x0, y0). At the position of the observer

y0 = −b x0 = γ(x − vt) = −γvt so q q E0 = = × (−γvt, −b) (x02 + y02)3/2 (γ2v2t2 + b2)3/2 hence −γvtq E0 = x (γ2v2t2 + b2)3/2 and −bq E0 = y (γ2v2t2 + b2)3/2

70 Chapter 16

Relativistic lagrangian

Clearly L = T − V can not work for relativistic systems, since we know that for relativistic systems Newton equation is wrong and should be replaced by d (γmx˙ ) = 0 dt – How do we solve this problem? – For relativistic systems we still use an action principle, but some things must change in order to be consistent with the two postulates of special relativity. 1. Time t has been treated as a special coordinates, distinct from (x, y, z). This is contrary to special relativity in which we have seen that space and time must be placed on an equal footing. So we must replace t in the definition of S by a parameter which is Lorentz invariant, and which traces the position of the particle in configuration space. A good candidate is τ, the proper time of the particle. 2. Physics must be the same in all inertial frame (so the Euler–Lagrange equations) and so L must be a Lorentz invariant, and hence also S. Back to the free massive particle: let’s choose it’s proper time τ to label the position of the particle in configuration space. Z S = Ldτ

L must have dimension of energy: the simplest Lorentz invariant we could write down for a particle od mass m is

L = −mc2 (16.1)

Let’s see what it gives! Z S = − mc2dτ

Z mc2 = dt γ

71 and mc2 L = − (16.2) γ Notice that in non-relativistic limit Z 1 S = dt −mc2 + mx2 + ··· (16.3) 2

The Euler–Lagrange equations are d ∂L ∂L = dt ∂x˙ ∂x d (γmx˙ ) = 0 dt

72 Part III

Exercices

73 Chapter 17

TD 1

17.1 Rappels

Exercice 17.1. Soit f = f(z(x), y(x), x). Exprimer la dérivée totale df/dx en fonction des diﬀérentes dérivées partielles de f. Soit f = f(z(x), y(x), x). Alors

f(x + δx) = f(z(x + δx), y(x + δx), x + δx) dz dy = f z(x) + δx, y(x) + δx, x + δx dx dx ∂f dz ∂f dy ∂f = f(z(x), y(x), x) + δx + δx + δx ∂z dx ∂y dx ∂x d’où df ∂f dz ∂f dy ∂f = + + dx ∂z dx ∂y dx ∂x

∂f ∂f ∂f ∂f df Exercice 17.2. Calculer , , , et (où x˙ = dx/dt, x = x(t) et ∂x ∂y ∂x˙ ∂t dt y = y(t)) pour 1. f = x2 2. f = y ln x avec x(t) = t3 3. f = sh(x ˙ 4 + x) + t avec x(t) = ln t 4. f = t2xy3 avec x(t) = t2 et y(t) = 1/t 1. Soit f = x2, alors ∂f = 2x ∂x ∂f ∂f ∂f = = = 0 ∂y ∂x˙ ∂t df = 2xx˙ dt

74 2. Soit f = y ln x avec x(t) = t3, alors

∂f y = ∂x x ∂f = ln x ∂y ∂f ∂f = = 0 ∂x˙ ∂t df 3yt2 = +y ˙ ln x dt x

3. Soit f = sh(x ˙ 4 + x) + t avec x(t) = ln t, alors

∂f = ch(x ˙ 4 + x) ∂x ∂f = 0 ∂y ∂f = 4x ˙ 3 ch(x ˙ 4 + x) ∂x˙ ∂f = 1 ∂t df = (4¨xx˙ 3 +x ˙) ch(x ˙ 4 + x) dt

4. Soit f = t2xy3 avec x(t) = t2 et y(t) = 1/t, alors

∂f = t2y3 ∂x ∂f = 3xt2y2 ∂y ∂f = 0 ∂x˙ ∂f = 2xty3 ∂t df = 2t2y3x˙ − 3xy2y˙ + 2xty3 dt

Exercice 17.3. Soit f = f(z(t), z˙(t), t). Réﬂéchir à la diﬀérence entre ∂tf et dtf. Exercice 17.4. Passage des coordonnées cartésiennes (vecteurs unitaires i et j) aux coordonnées polaires (vecteurs unitaires er(t) et eθ(t)). Soit r(t) = x(t)i + y(t)j la position d’une particule au temps t en coordonnées cartésiennes. En coordonnées polaires nous écrivons

r(t) = r(t)er(t)

Exprimer r˙ (t) et r¨(t) dans les deux jeux de coordonnées.

75 17.2 Introduction aux fonctionnelles

Exercice 17.5. Sur le plan euclidien, la distance parcourue L entre deux points donnés O et A est un exemple de fonctionnelle : cette distance, L[y], dépend du chemin y(x) qu’on emprunte entre les points. Soient O(0, 0) et A(x1, y1), et soit y(x) une courbe qui va de O à A. Utiliser votre intuition pour dire quelle courbe va minimiser L[y].

1. Pour y(x) = (y1/x1)x, quelle est la valeur de cette distance minimale Lmin ? 2. Écrire la forme de L[y] pour une courbe générale y(x). 1. Puisque OA est une droite, on a q 2 2 Lmin = x1 + y1

2. On a Z L = d` avec p d` = dx2 + dy2 s dy 2 = dx2 + dx2 dx

d’où

s 2 Z x1 dy L[y] = 1 + dx 0 dx

Exercice 17.6. Cet exercice est une introduction au fameux problème de la courbe brachistochrone. Considérons dans le plan vertical deux points A et B. On lâche une masse, sans vitesse initiale, en A. La masse glisse sans frottements sur un guide sous l’effet de la pesanteur et arrive en B. Le problème entier du brachistochrone consiste à trouver la forme de la courbe y(x) du guide telle que le temps néces- saire à la masse pour aller de A en B soit minimum. Ici nous allons faire un exemple plus simple : on calcule ce temps pour une série de courbes différentes. Soient A(0, 0) et B(−L, L). On considère quatre courbes différentes :

1. y1(x) consiste en l’axe des y pour 0 > y ≥ −L suivi par l’axe des x pour 0 < x < L.

2. y2(x) consiste en l’axe des y pour 0 > y ≥ −L/2 suivi par la droite y = −1/2(x + L) pour 0 < x ≤ L.

3. y3(x) est la droite y2 = −2x pour 0 < y ≤ L/2 suivi par l’axe des x pour L/2 < x ≤ L.

4. y4(x) = −x pour 0 ≤ x ≤ L. 1. Dessiner les quatre courbes. 2. Utiliser la conservation d’énergie totale pour exprimer la vitesse v de la particule en une position (x, y) quelconque en fonction de g et y.

76 3. Écrire la fonctionelle T [x], le temps nécessaire à la masser pour aller de Z yf A en B, en forme intégrale (soit T [x] = f(x, dx/dy, y)dy, soit T [y] = yi Z yf f(x, dy/dx, y)dx). yi 4. Calculer T pour les quatre chemins donnés ci-dessus. On suppose que l’énergie est toujours conservée et qu’il n’y a pas de chocs. 5. Quel chemin a le plus petit T ?

17.3 Rayons courbes et mirages

77 Chapter 18

Partiel 2008

18.1 Oscillateur simple

1. The Lagrangian is 1 1 L = m(r ˙2 + r2θ˙2) + K(r − `)2 + mgr cos θ 2 2 2. The equations are

mr¨ = mrθ˙2 + mg cos θ − K(r − `) r2θ¨ + 2rr˙θ˙ = −gr sin θ

3. Since ∂tL = 0, the corresponding conserved quantity is ∂L ∂L H = r˙ + θ˙ − L ∂r˙ ∂θ˙ = T + V

It is the total energy. 4. THe Lagrangian is now 1 1 L = m(r ˙2 + r2θ˙2) + K(r − `)2 2 2 A rotation by a ﬁxed angle α takes

θ −→ θ + α θ˙ −→ θ˙ L −→ L

The angular momentum mr2θ˙ is conserved. 5. The equation for r is J 2 K r¨ = − (r − `) m2r3 m Initial conditions are ˙ r(0) = r0 r˙(0) = 0 θ(0) = 0

78 Since J = mr2θ˙ is time independent, J = 0. Let z = r − `, then K z¨ = − z m r ! r ! K K =⇒ z = A cos t + B sin t m m r ! r ! K K =⇒ r = ` + A cos t + B sin t m m

We have B = 0 and A = r0 − `, so r ! K r = ` + (r − `) cos t 0 m

18.2 Oscillateurs couplés

1. We have 1 T = m(x ˙ 2 +x ˙ 2) 2 1 2 1 1 1 V = K(x − `)2 + λ(x − x − `)2 + K(x − 2`)2 2 1 2 2 1 2 2 so 1 1 1 1 L = m(x ˙ 2 +x ˙ 2) − K(x − `)2 + λ(x − x − `)2 + K(x − 2`)2 2 1 2 2 1 2 2 1 2 2 √ √ 2. Let 2X = x1 + x2 and 2Y = x1 − x2, then X + Y X − Y x1 = √ x2 = √ 2 2 and

2 2 ˙ 2 ˙ 2 x˙ 1 +x ˙ 2 = X + Y √ 2 2 2 1 2 2 2 (x1 − `) = (x1 − 2x1` + ` ) = (X + Y + 2XY ) − 2(X + Y )` + ` √ 2 √ 2 2 2 2 (x2 − x1 − `) = (− 2Y − `) = 2Y + 2 2`Y + ` ) √ 1 (x − 2`)2 = x2 + 4`2 − 4`x = 4`2 − 2 2`(X − Y ) + (X2 + Y 2 − 2XY ) 2 2 2 2 so 1 1 3 L = m(X˙ 2 + Y˙ 2) − Y 2(K + 2λ) + √ KX 2 2 2 1 √ √ 5 λ − Y `( 2K + 2 2λ) − `2 K + 2 2 2

79 3. Let X = X + a` and Y = Y + b`, then

2 αX2 = αX + α2Xa` + αa2`2 γX` = γX` + γa`2

To remove terms from the Lagrangian which are linear in X, one must choose 2αa` + γ` = 0 −γ =⇒ a = 2α and similarly −δ b = 2β So 1 2 2 2 2 L = m(X˙ + Y˙ ) + αX + βY 2 4. Euler–Lagrange for X and Y are 2α X¨ + ω2X = 0 ω2 = 1 1 m 2β Y¨ + ω2Y = 0 ω2 = 2 2 m 18.3 Gagner un slalom

1. The total energy E of the skier is a constant. 1 T = m(x ˙ 2 +y ˙2) 2 V = −mgx sin α 1 =⇒ E = m(x ˙ 2 +y ˙2) − mgx sin α 2 and E = 0 with the initial conditions. 2. We have d` dt = v and (d`)2 (dt)2 = v2 dx2 + dy2 = x2 + y2 dx2 + dy2 = 2gx sin α 3. x p Z 1 Z n 1 + y02 T = dt = √ √ dx 2g sin α 0 x | {z } = f

80 4. The optimal trajectory must satisfy the Euler–Lagrange equation d ∂f ∂f = = 0 dt ∂y0 ∂y

so ∂y0 = cste. 5. Then y0 = c px(1 + y02) 6. We have dy dy dt y0 = = dx dt dx and r y˙ 1 − c2x x˙ = =y ˙ y0 c2x But x2 + y2 = 2gx sin α. Hence substituting this in the previous equation

y˙2 = 2gc2 sin α x˙ 2

81 Appendix

82 1

83 Appendix A

Quantum ﬁeld theory: an introduction

A.1 Relativistic particle kinematics

In special relativity one have space-time 4-vector xµ = (ct, x) and a metric µ ηµν . One can construct a new 4-vector xν = ηµν x = (ct, −x) where I 0˜ ηµν = (A.1) 0˜ −I3 We determine the particle kinematics with dxµ E pµ = m = , p (A.2) dτ c and one can construct quantities independent of the frame in which we are moving:

µ 2 2 2 xµx = c t − x (A.3) E2 p pµ = − p2 = m2c2 µ c2 (A.4) E2 = m2c4 + p2c2 so r p2 p2 E = mc2 1 + ∼ mc2 + + ··· (A.5) m2c2 2m A.2 Non relativistic quantum mechanics

One have a wave function ψ(x, t). |ψ(x, t)|2 is a probability density. The Schrödinger equation’s for a particle of mass m is p2 Hψˆ = ψ (A.6) 2m ∂ − 2 i ψ(x, t) = ~ ∇2 ψ(x, t) (A.7) ~∂t 2m

84 A possible way to make particle relativistic is r p2 E = mc2 1 + (A.8) m2c2 and r ! ∂ 2 ∇2 i ψ(x, t) = mc2 1 − ~ ψ(x, t) (A.9) ~∂t m2c2 ∂2 2 ∇2 − 2 ψ(x, t) = m2c4 1 − ~ (A.10) ~ ∂t2 m2c2 1 ∂2 m2c2 − ∇2 ψ + ψ = 0 (A.11) c2 ∂t2 ~2 We use the notation 1 ∂2 = − ∇2 = ∂ ∂µ (A.12) c2 ∂t2 µ where ∂ µ ∂ ∂µ = µ ∂ = (A.13) ∂x ∂xµ If we put a particle in a box, we have

∆x∆p ∼ ~ (A.14) so p −→ p + (p + p) (A.15)

A.3 Simple harmonic oscillator

We work yet in natural units with

c = ~ = 1 (A.16) and the units change to

[momentum] = [energy] [time] = [space] [space-time] = [energy-momentum] so 2 ( + m )ψ = 0 (A.17) Let be m = 1 and the hamiltonian become pˆ2 1 Hˆ = + ωxˆ2 (A.18) 2 2 and we look for energy eigenstates

Hˆ |Ei = E |Ei (A.19)

85 We have hx|Hˆ |Ei = E h x| Ei (A.20) | {z } =ψE (x,t) The commutator between pˆ and xˆ is deﬁned by

[ˆx, pˆ] = i (A.21)

We define ωxˆ + ipˆ aˆ = √ (A.22) 2ω ωxˆ − ipˆ aˆ† = √ (A.23) 2ω and so 1 1 Hˆ = ω(â†aˆ +a âˆ†) = ωaˆ†aˆ + ω (A.24) 2 2 The commutator between aˆ† and aˆ is 1 a,ˆ aˆ† = [ωxˆ + ip,ˆ ωxˆ − ipˆ] 2ω 1 = ([ωx,ˆ −ipˆ] + [ip,ˆ ωxˆ]) = 1 2ω We denote the ground state of the simple harmonic oscillator by |0i and 1 H |0i = ω |0i aˆ |0i = 0 (A.25) 2 and since h i H,ˆ aˆ† = ωaˆ† (A.26) we find Hˆ (â† |0i) =a ˆ†Hˆ |0i +ω(â† |0i) (A.27) | {z } 1 † = 2 ω(â |0i) Let |1i =a ˆ† |0i so 1 Hˆ |1i = ω + ω |1i (A.28) 2

A.4 Relativistic quantum mechanical particles

Particles with only mass m: – bosons; – non-interacting are put in a square box of length L. We have eik·x with 2π k = x (A.29) ω and eik1L = eik2L = eik3L = 1 (A.30) The ground state (no particles — vacuum) is denoted by |0i.

86 We deﬁne the operator aˆ†(k), which creates a particle of momemtum k and hence p 2 2 ωk = m + k (A.31) so aˆ†(k) |0i = |ki (A.32) and

† † aˆ (k1)ˆa (k2) |0i = |k1 · k2i † † † † =⇒ aˆ (k1), aˆ (k2) = 0 (A.33) aˆ (k2)ˆa (k1) |0i = |k2 · k1i

1 since |k1 · k2i = |k2 · k1i because it is bosons : we don’t know which is which. aˆ(k) removes a particle of momemtum k from the box, so

aˆ |0i = 0a ˆ(k) |ki = |0i (A.34) hence

aˆ(k), aˆ(k0) = 0 † 0 aˆ(k), aˆ (k ) = δkk0

We want an Hamiltonian Hˆ0

Hˆ0 |0i = 0

p 2 2 Hˆ0 |ki = ω + k |ki | {z } =ωk We can write ˆ H0 |k1, k2i = (ωk1 + ωk2 ) |k1, k2i (A.35) so X † Hˆ0 = ωkaˆ (k)ˆa(k) (A.36) k And

X † † Hˆ0 |pi = ωkaˆ (k)a ˆ(k)ˆa (p) |0i | {z } =δkp X † = ωkδkpaˆ (k) |0i

= ωpaˆ(p) |0i

= ωp |ki

We deduce Hˆ from Hˆ0: X 1 Hˆ = ω aˆ†(k)ˆa(k) +a ˆ(k)ˆa†(k) 2 k Λ X 1 = H + ω ∼ Λ4 0 2 k k

1. which are indistinguishible.

87 Remarque : The operators all depend of time. The Heisenberg’s equations are h i aˆ(k, t), Hˆ0 = iaˆ(k, t) (A.37) and

aˆ(k, t) =a ˆ(k) e−iωkt aˆ†(k, t) =a ˆ(k) eiωkt so ω ϕˆ(k, t) + iπˆ(k, t) aˆ(k, t) = k √ 2ω ω ϕˆ(k, t) − iπˆ(k, t) aˆ†(k, t) = k √ 2ω and we have X 1 ϕˆ(k, t) = √ ϕˆ(k, t) ek·x k V X 1 = √ aˆ(k) eik·x−iωkt +a ˆ†(k) e−ik·x+iωkt 2ωkV X 1 = √ aˆ(k) eik·x +a ˆ†(k) e−ik·x 2ωkV where V is the volume of the box. So

k = |ω, ki (A.38) x = |t, xi (A.39)

Remarque : We have

ϕˆ†(k, t) =ϕ ˆ(x, t) ϕˆ†(k, t) =ϕ ˆ(−k, t) non charge

And hence, with (A.17)

2 ±ik·x 2 2 2 ±ik·x ( + m ) e = ± ωk − (k + m ) e = 0 2 =⇒ ( + m )ϕ ˆ(t, x) = 0 We can also write the Lagrangian: 1 1 L = (∂ ϕ)(∂µϕ) − m2ϕ2 (A.40) 2 µ 2 and the action is Z S = L dxdt (A.41)

88 Index

4-vector, 54 Event, 42 light-like, 54 space-like, 54 Frame, 38 time-like, 54 center of mass, 62 inertial, 38 Action,5 Function constant in time, 11 Beltrami identity, 11 Functional,7,8 Brachistochrone,6, 10 extremum,7

Causality, 45 Independance Conservation coordinate, 12 angular momemtum, 19 time, 11 momemtum, 18 Interval, 43 Constraint light-like, 45 holonomic, 26 space-like, 45 non-holonomic, 27 time-like, 45 Contravariant Invariant components, 52 interval, see Interval Covariant components, 52 Lagrangian electromagnetism, 68 Eﬀect non-relativistic, 13 Doppler (-), 50 Length non-relativistic, 50 contraction, 49 Energy, 33 Lorentz factor, 37 Equation Lorentz-invariant, 55 Euler–Lagrange,9 Hamilton, 33 Metric, 52 matrix form, 21 Momemtum Maxwell conservation, 61 in vacuum, 39 generalized, 33 Maxwell (-), 67 Newton (-),3 Noether theorem, 12, 18 wave, 39 Normal mode, 20 Equilibrium condition, 20 Poincaré group, 54 partially (-), 21 Poisson bracket, 34 stable (-), 20 Potential, 67 unstable (-), 21 Principle of relativity, 36

89 Quantity conserved, 18

Scalar product, 52 Space homogeneity, 18 isotropy, 18 Space-time diagram, 42 Minkowski (-), 51 Speed of light, 39 universality, 36 Summation convention, 52

Tensor, 66 electromagnetic, 67 Lorentz (-), 54 Time dilatation, 46, 49 proper (-), 45 Transformation Galilean (-), 36 Legendre, 33 Lorentz (-), 37, 48, 54

Velocity 4-vector, 56 composition, 50, 57

World line, 42

90 List of Figures

1.1 Double pendulum...... 4

6.1 Particle on an hemisphere...... 30

91 Bibliography

[1] Jean-Louis Basdevant and Christoph Kopper. Principes variationnels et mécanique analytique, 2008. http://catalogue.polytechnique.fr/site.php?id=88&fileid=488. [2] Richard Feynman, Robert Leighton, and Matthew Sands. Mécanique 1. Les cours de physique de Feynman. Dunod, 1999. [3] Jean Hladik and Michel Chrysos. Introduction à la relativité restreinte — Cours et exercices corrigés. Dunod, 2001. [4] David W. Hogg. Special Relativity, December 1997. http://cosmo.nyu.edu/hogg/sr/. [5] L. Marleau. Mécanique classique I. Université Laval, Québec, Canada, 2006. http://feynman.phy.ulaval.ca/marleau/notesdecours.htm.

92 Contents

I Introduction to Lagrangians in the context of non- relativistic systems2

1 Introduction: classical mechanics with Newtons equations against Lagrangian approach3 1.1 Limitations of Newtons approach...... 3 1.2 Diﬀerence between Newtonian classical mechanics and Lagrangian classical mechanics...... 4

2 Calculus of variations: The Lagrangian, Euler-Lagrange equa- tions6 2.1 Functionals and the Euler–Lagrange equations...... 6 2.1.1 Examples...... 6 2.2 Introduction to functional...... 7 2.2.1 Examples of functional derivatives...... 7 2.3 Euler–Lagrange equations...... 8

3 Calculus of variations: symmetry and conservation equations 11 3.1 Conservation...... 11 3.1.1 Time independance...... 11 3.1.2 Coordinate independance...... 12 3.2 Noether’s theorem...... 12

4 The Lagrangian of classical mechanics 13 4.1 Exemples...... 14 4.1.1 Ball slides...... 14 4.1.2 Simple pendulum with variable height...... 15 4.1.3 Free particle in non-inertial frame...... 16 4.1.4 Spherical pendulum...... 17 4.2 Consequences of Noether’s theorem...... 18 4.2.1 Momemtum conservation...... 18 4.2.2 Angular momemtum conservation...... 18

5 Smalls oscillations and normal modes 20 5.1 The equations...... 20 5.2 Exemples...... 22 5.2.1 Normal modes of a linear triatomic molecule...... 22

93 5.2.2 The double pendulum...... 24

6 Constrained systems and Lagrange multipliers 26 6.1 Lagrange multipliers...... 27

7 From Lagrangians to Hamiltonians – The Hamiltonian formulation of classical mechanics 32 7.1 Poisson brackets...... 34

II Special relativity 35

8 Introduction 36

9 The postulates of special relativity and the Michelson–Morley experiment 38 9.1 The Michelson–Morley interferometer...... 40

10 Consequences of the constancy of c 42 10.1 Space-time diagram...... 42 10.2 Simultaneity...... 43 10.3 Invariant interval...... 43 10.3.1 Interval types...... 45 10.4 Calibration of the (t’,x’) axis...... 45 10.5 Proper time and time dilatation...... 45

11 Lorentz transformations 47 11.1 Time dilatation with Lorentz transformations...... 49 11.2 Length contraction...... 49 11.3 Doppler eﬀect...... 49 11.4 Composition of velocities...... 50

12 Minkowski spacetime, 4-vectors and Lorentz invariants 51 12.1 4-vector...... 51 12.2 Some 4-vectors...... 55

13 Diﬀerent 4-vectors and their invariant 56 13.1 Velocity...... 56 13.1.1 Composition of velocities...... 57 13.2 Acceleration...... 57 13.3 Derivative 4-vector...... 57 13.4 Wave k...... 58 13.4.1 Application...... 58 13.5 Momentum...... 59 13.6 Force...... 59

14 Particle collisions 61 14.1 Examples...... 62 14.1.1 An elastic collusion: the Compton eﬀect...... 62 14.1.2 Elastic proton scattering...... 63 14.1.3 Desintegration...... 63

94 14.1.4 Inelastic collision of protons...... 64

15 Electromagnetism 66 15.1 Tensors...... 66 15.2 Electromagnetic 4-tensor...... 67 15.3 Transformations of E and B...... 68 15.4 An example: moving charge...... 69

16 Relativistic lagrangian 71

III Exercices 73

17 TD 1 74 17.1 Rappels...... 74 17.2 Introduction aux fonctionnelles...... 76 17.3 Rayons courbes et mirages...... 77

18 Partiel 2008 78 18.1 Oscillateur simple...... 78 18.2 Oscillateurs couplés...... 79 18.3 Gagner un slalom...... 80

Appendix 83

A Quantum ﬁeld theory: an introduction 84 A.1 Relativistic particle kinematics...... 84 A.2 Non relativistic quantum mechanics...... 84 A.3 Simple harmonic oscillator...... 85 A.4 Relativistic quantum mechanical particles...... 86

Index 89

List of Figures 91

Bibliography 92

Contents 93