Lecture 1. Basic concepts (book: p.1-19; 28-34)

Reference body. ‘Observe’ = deduce. Postulates. Standard configuration.

Spacetime diagrams; axes and simultaneity

Lorentz transformation:

t′ = γ(t− vx/c2)x′ = γ(−vt+ x)y′ = yz′ = z

,

ct′

x′

y′

z′

=

γ −γβ 0 0

−γβ γ 0 00 0 1 00 0 0 1

ctxyz

Minkowski metric and definition of Λ:

g =

−1 0 0 00 1 0 00 0 1 00 0 0 1

,

ΛTgΛ = g

⇒ A′TgB′ = ATgB ≡ A · B Invariant

Postulate 1, “Principle of Relativity”:

The motions of bodies included in a given space are the same

among themselves, whether that space is at rest or moves uniformly

forward in a straight line.

Postulate 2, “Light speed postulate”:

There is a finite maximum speed for signals.

Alternative statements:

Postulate 1:

The laws of physics take the same mathematical form in all inertial

frames of reference.

Postulate 2:

There is an inertial reference frame in which the speed of light in

vacuum is independent of the motion of the source.

x

t

x

t phot

on

t and t′: worldlines for two inertially moving particles (or observers)

x and x′: lines of simultaneity for those observers

(b)(a)

Lecture 2. 4-vectors; Proper time; Method of invariants; transformationof velocity (book p.14; 399-400; 22-25; 39-44)

Proper time Familiarity with γ :dt

dτ= γ γ = (1− β2)−1/2,

dγ

dv= γ3v/c2,

d

dv(γv) = γ3

symbol definition components name(s) invariantX X (ct, r) 4-displacement, interval −c2τ 2

U dX/dτ (γc, γv) 4-velocity −c2

P m0U (E/c,p) energy-momentum, 4-momentum −m20c

2

A dU/dτ γ(γc, γv + γa) 4-acceleration a20

Timelike U · U < 0 e.g. 4-velocityspacelike A · A > 0 e.g. 4-accelerationnull P · P = 0 e.g. energy-momentum of light pulse

Method of invariants = “Try using an invariant if you can, and pick an easyreference frame.”

4-acceleration is orthogonal to 4-velocity: U · A = 0.

Transformation of velocity:

u′∥ =

u∥ − v

1− u · v/c2, u′

⊥ =u⊥

γv (1− u · v/c2). ︸︷︷︸ ︷︸︸︷ −→ particle jets

Method of invariants

“If a question is of such a nature that its answer will always be

the same, no matter in which inertial frame one starts, it must

be possible to formulate the answer entirely with the

help of those invariants which one can build with the

available 4-vectors.

One then finds the answer in a particular inertial frame

which one can choose freely and in such a way that the answer is

there obvious or most easy.

One looks then how the invariants appear in this particular system,

expresses the answer to the problem by these same in-

variants, and one has found at the same time the general answer.”

(statement by Hagedorn, but he was not the first to think of it)

1. Calculate U · A for an arbitrarily moving particle.

2. Suppose an observer moving with velocity u in the x-direction relative

to the lab observes a particle of rest mass m0 moving with velocity v in

the (1, 1, 0)-direction relative to the lab. What is the energy of the particle

relative to the observer?

-U

t P

u’S’

S

v

u’ < v u’ > v

v << c

v ~ c

Where is the spaceship when the lightfrom the glowing hoop enters the windowat normal incidence relative to the rocket?

v

a

Doppler effect

A source moving at velocity v in the laboratory emits waves of frequencyω0 in its own rest frame. What is the frequency observed in the laboratory?

v

1k

ω0⋆

θ

Lecture 3. Doppler effect; Headlight effect; rapidity; Introducing force

1. Doppler effect

4-wave vector K ≡ (ω/c, k), ei(k·r−ωt) = eiK·X

Doppler effect: K · U ⇒ ω

ω0=

1

γ(1− (v/vp) cos θ).

2. Headlight effect or ‘aberration’:

Doppler effect: K · U ⇒ ω

ω0=

1

γ(1− (v/vp) cos θ).

Headlight effect: K = Λ−1K0 ⇒ cos θ =cos θ0 + v/c

1 + (v/c) cos θ0⇒ dΩ

dΩ0=

(ω0

ω

)2

.

−→ brightness (power per unit solid angle) transforms as (ω/ω0)4 for isotropic

source

3. Rapidity: tanh(ρ) ≡ v

c, ⇒ cosh(ρ) = γ, sinh(ρ) = βγ, exp(ρ) =

(1 + β

1− β

)1/2

.

Λ =

cosh ρ − sinh ρ 0 0

− sinh ρ cosh ρ 0 00 0 1 00 0 0 1

, Colinear rapidities add.

−1

cos v/c

v

−2 0 2 4 6 8

−2

−1

0

1

2

3

4

5

6

c

−2 0 2 4 6 8

−2

−1

0

1

2

3

4

5

6

c

v

Lecture 4. Force; simple dynamical problems

1. Force

F ≡ dP

dτ, U · F = γ2

(−dE

dt+ u · f

)= −c2

dm0

dτ.

‘Pure’ force:

U · F = 0 ⇒ m0 = const,dE

dt= f · u

2. Transformation of force: use ΛF and γ, or (d/dt′)(ΛP) :

f ′∥ =f∥ − (v/c2)dE/dt

1− u ·v/c2, f ′⊥ =

f⊥γv(1− u ·v/c2)

⇒

f ′∥ = f∥f ′⊥ = f⊥/γ

for u = 0

3. Equation of motion in any given reference frame:

f =dp

dt=

d

dt(γmv) = γma+m

dγ

dtv

⇒

(i) acceleration is not necessarily parallel to force!(ii) f = ma only valid at v = 0(iii) for ‘pure’ force (m=constant): f∥ = γ3ma∥, f⊥ = γma⊥.

4. Uniform B field: just like Newtonian result, but with m replaced by γm.Hence ω = qB/γm and p = qBr.

5. Motion parallel to E field: hyperbolic motion, x2 − c2t2 = (c2/a0)2 and U ∝ U.

Constant proper acceleration.

a t

v a t+

v

f tp f t+

p

f

Force and acceleration are usually not parallel.

+

−f

fv

+

−θ

f

f

S S

Trouton-Noble experiment

v

+

−

+

−SS

Explanation of Trouton-Noble experiment

Lecture 5. Rigidity and acceleration; The conservation of energy andmomentum

1. Rigidity; the Great Train Disaster

2. Lewis and Tolman argument: p = mα(v)v ⇒ p = γmv.

3. Impact of simultaneity on “Ptot = P1 + P2 + P3 + . . .Pn”.

4. “Zero component lemma”: if one component of a 4-vector is zero in all referenceframes, then the whole 4-vector is zero.

Hence momentum conservation ⇔ energy conservation.

Main postulates,momentum conservation

⇒ E0 = mc2, equivalence of rest mass and rest energy.

x

t

y

A rod undergoing acceleration: spacetime diagram.

plane ofsimultaneity

yy

worldline

xx

tt

y

A rod undergoing acceleration: spacetime diagram.

−1 0 1−1

−0.9

−0.8

−0.7

−0.6

−0.5

−0.4

−0.3

−0.2

−0.1

0

x

S−2 −1 0 1 2

−1

−0.9

−0.8

−0.7

−0.6

−0.5

−0.4

−0.3

−0.2

−0.1

0

−2

−1

0

1

2

x’

y’

S’

A rod undergoing acceleration: observations in frames S and S′.

Situation in rest frame of the bridge. The train is contracted by γ = 3.

The cable cannot support the train. Disaster!

Situation in rest frame of the train. How can the disaster happen?

Explanation:

1. The cable is weaker by factor γ = 3.

2. Downward force on train is larger by factor γ = 3.

3. Downwards acceleration appears as bending in this frame.

Lecture 5. Rigidity and acceleration; The conservation of energy andmomentum

1. Rigidity; the Great Train Disaster

2. Lewis and Tolman argument: p = mα(v)v ⇒ p = γmv.

3. Impact of simultaneity on “Ptot = P1 + P2 + P3 + . . .Pn”.

4. “Zero component lemma”: if one component of a 4-vector is zero in all referenceframes, then the whole 4-vector is zero.

Hence momentum conservation ⇔ energy conservation.

Main postulates,momentum conservation

⇒ E0 = mc2, equivalence of rest mass and rest energy.

x

x

Lecture 6. Collisions.Methods:

P · P = −m2c2 ⇒ E2 − p2c2 = m2c4 (1)

p = γmv, E = γmc2 ⇒ v =pc2

E(2)

“Isolate and square.” (3)

1. Decay at rest. e.g. find the energy of one of the products:

E1 =M 2 +m2

1 −m22

2Mc2, Ephoton = ∆Erest − Erecoil.

2. In-flight decay. e.g. Find the rest mass of the original particle:

M 2 = m21 +m2

2 +2

c4(E1E2 − p1 · p2c

2)

⇒ it suffices to measure m1,m2, p1, p2, θ.

3. Particle formation. e.g. threshold energy (stationary target):

Eth =(∑

imi)2 −m2 −M2

2Mc2.

4. Elastic collision. e.g. find angles in the lab frame

5. Compton effect (A. H. Compton, Physical Review 21, 483 (1923)):

λ′ − λ =h

mc(1− cos θ).

[other: 3-body decay, inverse Compton effect, etc.]

Lecture 7. Composite body; 4-gradient; flow

1. The idea of a composite body with a net energy and momentum, hence rest massand velocity. p =

∑i pi, E =

∑iEi, P ≡ (E/c,p), m ≡

√−P2/c2, v = pc2/E.

2. Concept of a Lorentz-invariant scalar field.e.g. B · E, E2 − c2B2, but NOT charge density or potential energy.

3. 4-gradient operator:

≡(−1

c

∂

∂t,

∂

∂x,

∂

∂y,

∂

∂z

)If ϕ is a Lorentz-invariant scalar quantity, then: ′ϕ = Λϕi.e. ϕ is a 4-vector.

4. 4-divergence

· F =1

c

∂F 0

∂t+∇ · f , 2 = · = − 1

c2∂2

∂t2+∇2.

5. Wave phase ϕ is Lorentz-invariant.

⇒ K ≡ ϕ is a 4-vector

2ϕ = 0 is the wave equation.

6. Flow and conservation:

4-current density J ≡ ρ0U = (ρc, j)

continuity equation · J = 0.

B C D

A

All frames agree on events when A is opposite B,C,D.

(6. Thomas precession:) net excess angle for a circle ∆θ = 2π(γ − 1),

dθ

dt= (γ − 1)

a

v, More generally: ωT =

a ∧ v

c2γ2

1 + γ

L0θθ

L0

γ

L0sinθ

cosθ

a) b)

B

x

y

S

A

u

v

x’

y’

A

B

S’

A

x’’

y’’

S’’

B