Special Relativity Christopher R. Prior Accelerator Science and Technology Centre Rutherford...

Special Relativity

Christopher R. Prior

Accelerator Science and Technology Centre

Rutherford Appleton Laboratory, U.K.

Fellow and Tutor in Mathematics

Trinity College, Oxford

2

Overview• The principle of special relativity• Lorentz transformation and consequences• Space-time• 4-vectors: position, velocity, momentum, invariants,

covariance.• Derivation of E=mc2

• Examples of the use of 4-vectors• Inter-relation between and , momentum and energy• An accelerator problem in relativity• Relativistic particle dynamics• Lagrangian and Hamiltonian Formulation• Radiation from an Accelerating Charge• Photons and wave 4-vector• Motion faster than speed of light

3

Reading• W. Rindler: Introduction to Special Relativity (OUP

1991)• D. Lawden: An Introduction to Tensor Calculus and

Relativity• N.M.J. Woodhouse: Special Relativity (Springer

2002)• A.P. French: Special Relativity, MIT Introductory

Physics Series (Nelson Thomes)• Misner, Thorne and Wheeler: Relativity• C. Prior: Special Relativity, CERN Accelerator

School (Zeegse)

4

Historical background• Groundwork of Special Relativity laid by Lorentz in

studies of electrodynamics, with crucial concepts contributed by Einstein to place the theory on a consistent footing.

• Maxwell’s equations (1863) attempted to explain electromagnetism and optics through wave theory

– light propagates with speed c = 3108 m/s in “ether” but

with different speeds in other frames

– the ether exists solely for the transport of e/m waves

– Maxwell’s equations not invariant under Galilean transformations

– To avoid setting e/m apart from classical mechanics, assume

• light has speed c only in frames where source is at rest• the ether has a small interaction with matter and is carried

along with astronomical objects

5

Contradicted by:

• Aberration of star light (small shift in apparent positions of distant stars)

• Fizeau’s 1859 experiments on velocity of light in liquids

• Michelson-Morley 1907 experiment to detect motion of the earth through ether

• Suggestion: perhaps material objects contract in the direction of their motion

21

2

2

0 1)(

c

vLvL

This was the last gasp of ether advocates and the germ This was the last gasp of ether advocates and the germ of Special Relativity led by Lorentz, Minkowski and of Special Relativity led by Lorentz, Minkowski and

Einstein.Einstein.

6

The Principle of Special Relativity

• A frame in which particles under no forces move with constant velocity is inertial.

• Consider relations between inertial frames where measuring apparatus (rulers, clocks) can be transferred from one to another: related frames.

• Assume:– Behaviour of apparatus transferred from F to F' is independent

of mode of transfer

– Apparatus transferred from F to F', then from F' to F'', agrees with apparatus transferred directly from F to F''.

• The Principle of Special Relativity states that all physical The Principle of Special Relativity states that all physical laws take equivalent forms in related inertial frames, so laws take equivalent forms in related inertial frames, so that we cannot distinguish between the framesthat we cannot distinguish between the frames.

7

Simultaneity• Two clocks A and B are synchronised if light rays

emitted at the same time from A and B meet at the mid-point of AB

• Frame F' moving with respect to F. Events simultaneous in F cannot be simultaneous in F'.

• Simultaneity is not absolute but frame dependent.

A BCFrame F

A’’ B’’C’’

A’

B’C’Frame F’

8

The Lorentz Transformation• Must be linear to agree with standard Galilean transformation

in low velocity limit• Preserves wave fronts of pulses of light,

• Solution is the Solution is the Lorentz transformationLorentz transformation from frame F from frame F ((t,x,y,z)t,x,y,z) to frame F to frame F''((tt'',x,x'',y,y'',z,z'')) moving with velocity moving with velocity vv along along the the xx-axis:-axis:

2

1

2

2

2

1where

'

'

'

'

c

v-

zz

yy

vtxx

c

vxtt

0xwhenever

0i.e.22222

22222

tczyQ

tczyxPct

9

1)-not and 1,, (e.g. sensecommon someApply

1 space ofIsotropy

. of tscoefficien Equate

''''

Then

'

'

'

'Set

222222222222

2222222222

k

)vk()vk(k

x,y,z,t

zyxtckzytxxtc

zyxtckzyxtc

kQP

zz

yy

txx

xtt

Outline of Derivation

10

2

21

c

xvtt

v

xvtvxx

General 3D form of Lorentz Transformation:

11

Consequences: length contraction

z

x

Frame F

v

Frame F’z’

x’

RodA B

Moving objects appear contracted in the direction of the Moving objects appear contracted in the direction of the motionmotion

Rod AB of length L' fixed in F' at x'A, x'B. What is its length measured in F?

Must measure positions of ends in F at the same time, so events in F are (t,xA) and (t,xB). From Lorentz:

LLxxxxL

vtxxvtxx

ABAB

BBAA

12

Consequences: time dilation

• Clock in frame F at point with coordinates (x,y,z) at different times tA and tB

• In frame In frame FF'' moving with speed moving with speed v, v, Lorentz Lorentz transformation givestransformation gives

• So:So:

22 c

vxtt

c

vxtt BBAA

ttttttt ABAB

Moving clocks appear to run slowMoving clocks appear to run slow

13

Schematic Representation of the Lorentz Transformation

Frame F

Frame F

t

x

t

x

L

L

Length contraction L<L

t

x

t

x

tt

Time dilatation: t<tRod at rest in F. Measurement in F at fixed time t, along a line parallel to x-axis

Clock at rest in F. Time difference in F from line parallel to x-axis

Frame F

Frame F

2c

vxttvtxx

14

v = 0.8c

v = 0.9c

v = 0.99c

v = 0.9999c

15

Example: High Speed Train

• Observers A and B at exit and entrance of tunnel say the train is moving, has contracted and has length

• But the tunnel is moving relative to the driver and guard on the train and they say the train is 100 m in length but the tunnel has contracted to 50 m

m504

311001100

100 212

1

2

2

c

v

All clocks synchronised.

A’s clock and driver’s clock read 0 as front of train emerges from tunnel.

16

Moving train length 50m, so driver has still 50m to travel before his clock reads 0. Hence clock reading is

Question 1A’s clock reads zero as the driver exits tunnel. What does B’s clock read when the guard goes in?

ns2003

10050

cv

17

Question 2

What does the guard’s clock read as he goes in?

To the guard, tunnel is only 50m long, so driver is 50m past the exit as guard goes in. Hence clock reading is

ns2003

10050

cv

18

Question 3

Where is the guard

when his clock

reads 0?

Guard’s clock reads 0 when driver’s clock reads 0, which is as driver exits the tunnel. To guard and driver, tunnel is 50m, so guard is 50m from the entrance in the train’s frame, or 100m in tunnel frame.

So the guard is 100m from the entrance to the tunnel when his clock reads 0.

19

F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard.

Question 1A’s clock reads zero as the driver exits tunnel. What does B’s clock read when the guard goes in?

2

2

c

vxttvtxx

c

xvtttvxx

vvt

xx

B

GB

501100

100,100

20

Question 2

What does the guard’s clock read as he goes in?


2

2

c

vxttvtxx

c

xvtttvxx

vvt

xx

G

G

501100

100,100

21

Question 3

Where is the guard

when his clock

reads 0?


2

2

c

vxttvtxx

c

xvtttvxx

mxx

tx GG

200

0,100

Or 100m from the entrance to the tunnel

22

Question 4Where was the driver

when his clock reads

the same as the guard’s

when he enters the

tunnel?


2

2

c

vxttvtxx

c

xvtttvxx

mtvxv

tx

D

DD

100

50,0

Or 100m beyond the exit to the tunnel

23

Example: Cosmic Rays

-mesons are created in the upper atmosphere, 90km from earth. Their half life is =2 s, so they can travel at most 2 10-6c=600m before decaying. So how do more than 50% reach the earth’s surface?

• Mesons see distance contracted by , so

• Earthlings say mesons’ clocks run slow so their half-life is and

• Both give 150,,150km90

cv

cc

v

km90

v

km90v

24

Space-time• An invariant is a quantity that has the same

value in all inertial frames.

• Lorentz transformation is based on invariance of

• 4D space with coordinates (t,x,y,z) is called space-time and the point

is called an event.

• Fundamental invariant (preservation of speed of light): 2

22

222

22

22222222222 11

t

cc

vtc

tc

zyxtczyxtcs

2222222 )()( xctzyxtc

),(,,, xtzyxt

dt is called proper time, time in instantaneous rest frame,

an invariant. s=c is called the separation between two events

t

x

Absolute future

Absolute past

Conditional present

25

4-VectorsThe Lorentz transformation can be written in matrix form as

zz

yy

vtxx

c

vxtt

2

z

y

x

ct

c

vc

v

z

y

x

tc

L 1000

0100

00

00

Position 4-vector xct

,

An object made up of 4 elements which transforms like X is called a 4-vector

(analogous to the 3-vector of classical mechanics) LXX

26

Invariants

XXgXX

z

y

x

ct

zyxctzyxtc T

1000

0100

0010

0001

,,,22222

),(),,( 00 bbBaaA

babababababagAAAA T 0033221100

Basic invariant

Inner product of two 4-vectors

Invariance:

AAgAAAgLLALAgLAAA TTTT

27

4-Vectors in S.R. Mechanics

• Velocity:

• Note invariant

• Momentum

),(),( vcxctdt

d

dt

d

d

dV

222

22222

1)( c

cv

vcvcVV

),(),(00 pmcvcmVmP

momentum-3 theis

mass icrelativist is

0

0

vmvmp

mm

28

Example of Transformation:

Addition of Velocities• A particle moves with velocity in

frame F, so has 4-velocity

• Add velocity by transforming to frame F to get new velocity .

• Lorentz transformation gives

ucV u

,

),,( zyx uuuu

uxt uu

,

)0,0,(vv

22

2

2

11

1

c

vuu

w

c

vu

uw

c

vuvu

w

uw

uw

vuw

c

uv

xv

zz

xv

yy

x

xx

zuzw

yuyw

uxuvxw

xuuvw

w

2

2

c

vxttvtxx

c

xvtttvxx

29

4-ForceFrom Newton’s 2nd Law expect 4-Force given by

fdt

dmc

dt

pd

dt

dmcpmc

dt

ddt

dP

d

dPF

,

,),(

vdt

dm

dt

pdf

0equation force3 :Note

30

• Momentum invariant

• Differentiate

Einstein’s Relation

0.

0,,

000

2

fvmcdt

d

fdt

dmcvc

FVd

dPV

d

dPP

220

20 )( cmVVmPP

Therefore kinetic energy

1constant 20

2 cmmcT

= rate force does work = rate of change of kinetic energyfv

.

E=mcE=mc2 2 is total energyis total energy

31

Basic Quantities used in Accelerator Calculations

22

222

2

222

2

12

2

1

2

2

111

11

c

v

c

v

c

v

1

c

20

20

0

cm)c(m-mT

cmmvp

vc

vRelative velocity

Velocity

Momentum

Kinetic energy

32

Velocity v. Energy

cmp

cm

T

cmT

o

2

20

20

11

1

1

20

20

2

221

2

2

2

11so

2

111,For

vmcmT

c

v

c

vcv

33

Energy/Momentum Invariant

220

20 cmVVmPP

0

00

20

2220

22222

2

2ETT

EEEE

EEcmEcppc

E

Example: ISIS 800 MeV protons (E0=938 MeV)

=> pc=1.463 GeV

84.0

85.11

56.1

22

02

0

20

E

pc

cm

cm

34

Relationships between small variations in parameters

E, T, p, ,

)2(1

11

)1(

1

3

22

22

T

T

E

E

cm

cm

p

p

1

11

)(

2

22

0

0

(exercise)

Note: valid to first order only

36

4-Momentum Conservation• Equivalent expression

for 4-momentum

• Invariant

• Classical momentum conservation laws conservation of 4-momentum. Total 3-momentum and total energy are conserved.

pcEpmcvcmP

,),(),(0

22

222

0 pc

EPPcm

222

02

2pcm

cE

constant and

constant

iparticles,i particles,

i particles,

ii

i

pE

P

37

A body of mass M disintegrates while at rest into two parts of rest masses M1 and M2. Show that the energies of the parts are given by

M

MMMcE

M

MMMcE

2,

2

22

21

22

2

22

21

22

1

38

SolutionBefore:

)0,(

McP

After:

pc

EP

,11

pc

EP

,22

Conservation of 4-momentum:

M

MMME

cMcMMEcM

PPPPPPPP

PPPPPP

PPPPPP

2

2

2

22

21

2

1

222

2211

22

22111

2211

2121

39

Example of use of invariants

• Two particles have equal rest mass m0.

– Frame 1: one particle at rest, total energy is E1.

– Frame 2: centre of mass frame where velocities are equal and opposite, total energy is E2.

Problem: Relate E1 to E2

40

Total energy E1

(Fixed target experiment)

Total energy E2

(Colliding beams expt)

p

c

cmEP

,

201

1 0,02 cmP

p

c

EP

,

22

1

p

c

EP

,

22

2

212:Invariant PPP

221

20

2210

2

02

0

EEcm

pc

E

c

Ep

c

Ecm

41

Collider Problem

• In an accelerator, a proton p1 with rest mass m0 collides with an anti-proton p2 (with the same rest mass), producing two particles W1 and W2 with equal mass M0=100m0

– Expt 1: p1 and p2 have equal and opposite velocities in the

lab frame. Find the minimum energy of p2 in order for W1

and W2 to be produced.

– Expt 2: in the rest frame of p1, find the minimum energy

E' of p2 in order for W1 and W2 to be produced.

42

Note: same m0, same p mean same E.220

22

2cmp

cE

p1

W1

p2

W2

Total 3-momentum is zero before collision and so is zero after impact

4-momenta before collision:

pcEPpc

EP ,, 21

4-momenta after collision:

qcEPqc

EP ,, 21

Energy conservation E=E > rest energy = M0c2 = 100 m0c2

43

p2

p1

W1

W2

Before collision:

pcEPcmP

,0, 201

Total energy is

201 cmEE

Use previous result 2m0c2 E1=E22 to relate E1 to total energy E2 in

C.O.M frame

2

02

04

220

220

20

221

20

000,201102

200)2(2

2

cmcmE

cmEcmEcm

EEcm

44

4-Acceleration

vc

ava

c

avA

233 ,

vcdt

d

d

dVA

,Acceleration = “rate of change of velocity”

22322

11

1

c

av

c

vv

dt

d

c

vv

Use

2||,,0 aAAaA

Instantaneous rest-frame

45

Radiation from an accelerating charged particle

• Rate of radiation, R, known to be invariant and proportional to in instantaneous rest frame.

• But in instantaneous rest-frame• Deduce

• Rearranged:

2a

2aAA

22

26 1

ac

avAAR

2

226

3

2

3

2

c

vaa

c

eR

Relativistic Larmor Formula

46

Motion under constant acceleration; world lines

• Introduce rapidity defined by

• Then

• And

• So constant acceleration satisfies

sinh,cosh),( cvcV

cosh1

1tanh

2

c

v

d

dc

d

dVA cosh,sinh

c

a

c

a

d

d

d

dcAAaa

,

2222

47

World line of particle is hyperbolic

c

a

a

ct

d

dt

c

a

c

a

a

cxx

d

dx

dt

dx

c

acc

sinh

coshcosh

1cosh

sinhsinh

2

0

Particle Paths

2

422

22

0

22 1sinhcosh

a

ctc

a

cxx

parabola,

2

11

2

11 2

02

222

0

c

a

a

ct

axc

a

a

cxx

48

Relativistic Lagrangian and Hamiltonian Formulation

etc/1

2/1220 x

V

cv

x

dt

dm

dt

pdf

x

V

x

L

cv

xm

x

Letc

x

L

x

L

dt

d

,/1

2/122

0

2222 zyxv

3-force eqn of motion under potential V:

Standard Lagrangian formalism:

Since , deduce

Vcm

Vc

vcmL

20

21

2

22

0 1

Relativistic Lagrangian

49

,

1

20

202

0

222

21

2

2

0

VE

VcmVcm

vm

Lzyx

cv

mL

x

LxH

Hamiltonian

total energy

Hamilton’s equations of motion

420

222 cmcpE

Since

VcmpcH 21

220

2

etc,,x

Hp

p

Hx x

x

50

Photons and Wave 4-Vectors

Monochromatic plane wave:

Phase is number of wave crests

passing an observer, an invariant.

)sin( xkt

xkt

21

k

cxctxkt

,,

2frequencyangular ;2

vector,wave λ

πkk

Position 4-vector, X Wave 4-vector, K

51

Relativistic Doppler Shift

ck

n n

cK

,1

For light rays, phase velocity is

So where is a unit vector

v

F F'

'

Lorentz transform (t↔ω/c2, x↔ωn/c):

cv

c

v

cos

sintan

cos1

Note: transverse Doppler effect even when θ=½

52

Motion faster than light1. Two rods sliding over each

other. Speed of intersection point is v/sinα, which can be made greater than c.

2. Explosion of planetary nebula. Observer sees bright spot spreading out. Light from P arrives t=dα2/2c later.

c

x

c

x

c

dt

22

2

xc

P

Od

Special Relativity Christopher R. Prior Accelerator Science and Technology Centre Rutherford...

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