relativity for accelerator physicists - cern · §examples: phase of a wave, rate of radiation of...

Relativity for Accelerator Physicists

Christopher R. Prior

Accelerator Science and Technology Centre

Rutherford Appleton Laboratory, U.K.

Fellow and Tutor in Mathematics

Trinity College, Oxford

CERN School on Small Accelerators

2

Overview

q The principle of special relativityq Lorentz transformation and its consequencesq 4-vectors: position, velocity, momentum,

invariants. Einstein’s equation E=mc2

q Examples of the use of 4-vectors

q Inter-relation between β and γ, momentum and energy

q Electromagnetism and Relativity


3

Readingq W. Rindler: Introduction to Special Relativity

(OUP 1991)

q D. Lawden: An Introduction to Tensor Calculus and Relativity

q N.M.J. Woodhouse: Special Relativity (Springer 2002)

q A.P. French: Special Relativity, MIT Introductory Physics Series (Nelson Thomes)


4

Historical backgroundq Groundwork by Lorentz in studies of

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

q Maxwell’s equations (1863) attempted to explain electromagnetism and optics through wave theory§ light propagates with speed c = 3×108 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§ And the ether has a small interaction with matter and is carried along with

astronomical objects


5

Nonsense! Contradicted by:q Aberration of star light (small shift in apparent

positions of distant stars)q Fizeau’s 1859 experiments on velocity of light in

liquidsq Michelson-Morley 1907 experiment to detect motion

of the earth through etherq Suggestion: perhaps material objects contract in the

direction of their motion 21

2

2

0 1)( ��

��

�−=

c

vLvL

��

��


6

The Principle of Special Relativityq A frame in which particles under no forces move with

constant velocity is “inertial”q Consider relations between inertial frames where

measuring apparatus (rulers, clocks) can be transferred from one to another.

q Behaviour of apparatus transferred from F to F' is independent of mode of transfer

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

qq 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

Simultaneityq 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

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

q Simultaneity is notabsolute but frame dependent.

� ��

� ��

� ��


8

The Lorentz Transformationq Must be linear to agree with

standard Galilean transformation in low velocity limit

q Preserves wave fronts of pulses of light,

qq Solution is the Solution is the LorentzLorentztransformationtransformation from frame from frame F (F (t,x,y,zt,x,y,z)) to frame to frame FF''((tt'',x,x'',y,y'',z,z'')) moving with moving with velocity velocity vv along the along the xx--axis:axis:

( )2

1

2

2

2

1where

'

'

'

'−

��

��

�=

��

�

��

==

−=

��

��

� −=

c

v-

zz

yy

vtxx

c

vxtt

γγ

γ

0xwhenever

0i.e.22222

22222

=′−′+′+′≡=−++≡

tczyQ

tczyxP


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

ςε

ςεδγβα

ςε

δγβα

��

Outline of Derivation


10

Consequences: length contraction

� ��

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

>=−=′−′=′−=′−=′

γγγγ

�

!

"��"

�

"��"#�#

!#

� �$ %


11

Consequences: time dilatationq Clock in frame F at point with coordinates (x,y,z)

at different times tA andtBqq In frame In frame FF'' moving with speed moving with speed v, v, LorentzLorentz

transformation givestransformation gives

qq So:So:

��

��

� −=′��

��

� −=′22 c

vxtt

c

vxtt BBAA γγ

( ) ttttttt ABAB ∆>∆=−=′−′=′∆ γγ

� �� &�� &��


12

Schematic Representation of the LorentzTransformation

Frame F′

Frame F

t

x

t′

x′

L

L′

Length contraction L<L′

t

x

t′

x′

∆t′∆t

Time dilatation: ∆t<∆t′Rod 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′


13

v = 0.8c

v = 0.9c

v = 0.99c

v = 0.9999c


14

Example: Rocket in Tunnel

q All clocks synchronised.

q Observers X and Y at exit and entrance of tunnel say the rocket is moving, has contracted and has length

q But the tunnel is moving relative to the ends A and B of the rocket and obesrvers here say the rocket is 100 m in length but the tunnel has contracted to 50 m

m504

311001100

100 212

1

2

2

=��

��

� −×=��

��

�−×=

c

v

γ

cv2

3=Tunnel 100m long

Rocket length 100m


15

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

Questions

q If X’s clock reads zero as the A exits tunnel, what does Y’s clock read when the B goes in?

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

q Where is the B when his clock reads 0?

ns2003

10050 −≈−=−cv

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

ns2003

10050 +≈+=+cv

B’s clock reads 0 when A’s clock reads 0, which is as A exits the tunnel. To A and B, tunnel is 50m, so B is 50m from the entrance in the rocket’s frame, or 100m in tunnel frame.


16

Example: π-mesons

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

q Mesons see distance contracted by γ, so

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

q Both give

150,,150km90 ≈≈== γτ

γcv

cc

v

km90��

��

�≈γ

τv

( ) km90≈γτv


17

Invariantsq An invariant is a quantity that has the same value in all inertial frames.

§ Examples: phase of a wave, rate of radiation of moving charged particle

q Lorentz transformation is based on invariance of

q Write this in terms of the 4-position vector as

q Fundamental invariant (preservation of speed of light):

q Write ∆τ=∆t/γ, τ is the proper timeq When v=0, τ = t, so τ is the time in the rest-frame.

2

22

222

22

2222222222

1

1

��

��

� ∆=��

��

�−∆=

��

��

�

∆∆+∆+∆−∆=∆−∆−∆−∆

γt

cc

vtc

tc

zyxtczyxtc

yxyxyyxx�� ⋅−=Υ⋅Χ=Υ=Χ 0000 productinvariant thedefine),,(),,(If

xxctzyxtc�� ⋅−=++− 222222 )()(

( )xct�

,=Χ Χ⋅Χ


18

4-VectorsThe Lorentz transformation can be written in matrix form

( )

zz

yy

vtxx

c

vxtt

=′=′

−=′

��

��

� −=′

γ

γ2

��

�

�

��

�

�

��

�

�

��

�

�

−

−

=

��

�

�

��

�

�

′′′′

z

y

x

ct

c

vc

v

z

y

x

tc

1000

0100

00

00

γγ

γγ

�

Position 4-vector ( )xct�

,=Χ

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

(analogous to the 3-vector of classical mechanics)


19

4-Vectors in Special Relativity Mechanics

q Velocity:

q Note invariant

q Momentum

),(),( vcxctdt

d

dt

d

d

dV

�� γγγτ

==Χ=Χ=

2222 )( cvcVV =−=⋅ �γ

),(),(00 pmcvcmVmP�� === γ

momentum-3 theis

mass icrelativist is

0

0

vmvmp

mm�� ==

=γγ


20

Example of Transformation: Addition of Velocities

q A particle moves with velocity in frame F, so has 4-velocity

q Add velocity by transforming to frame F′ to get new velocity .

q Lorentz transformation gives

( )ucV u

�

,γ=),,( zyx uuuu =�

( )uxt�� γγ ↔↔ ,

)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

γγγγ

γγ

γγγγ

γγγγ

w�


21

Einstein’s relation

q Momentum invariant

q Differentiate

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

q But

q So

220

20 )( cmVVmPP =⋅=⋅

��

��

�=��

��

�==== fdt

dmc

dt

pd

dt

dmcpmc

dt

d

dt

dP

d

dPF

�

�

�

,,),( γγγγτ

00 =⋅�=⋅ FVd

dPV

τ

00 =⋅�=⋅ττ d

dPV

d

dPP

( ) 02 =⋅− fvmcdt

d �

�

�� '��

��

E=mcE=mc22 is total energyis total energy

fv�

� ⋅

( )1constant 20

2 −=+= γcmmcT


22

Basic quantities used in Accelerator calculations

( )1energy Kinetic

Momentum

cVelocity

cv velocityRelative

20

20

0

−==

==

=

=

γ

βγ

β

β

cm)c(m-mT

cmmvp

v

( )

( )22

222

2

222

2

12

2

1

2

2

111

11

γβγγβγ

βγ

−==�−==�

−=��

��

�−= −

−

c

v

c

v

c

v


23

Velocity as a Function of Energy

( )

βγγβ

γ

γ

cmp

cmT

cmT

o=

−=

+=

−=

2

20

20

11

1

1

( ) 20

20

2

221

2

2

2

11so

2

111,c

vsmallFor

vmcmT

c

v

c

v

≈−=

+≈��

��

�−=

−

γ

γ


24

Relationships between small variations in parameters ∆E, ∆T, ∆p, ∆β, ∆γ

( )( )

( )

)2(1

11

)1(

1

3

22

22

ββγγ

βγ

γβγβγγβγβγ

γβγ

∆=∆�

−=

∆=∆�

∆=∆�

−= ( )

T

T

E

E

cm

cm

p

p

∆+

=

∆=

∆=∆=

∆=∆=∆

1

11

)(

2

22

0

0

γγ

ββγ

βγγ

β

βγβγ

βγβγ

(exercise)


25


26

4-Momentum Conservation

q Equivalent expression for 4-momentum

q Invariant

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

( )pcEpmcvcmP��

,),(),(0 === γ

22

222

0 pc

EPPcm

�−=⋅= 22202

2pcm

cE �+=

constant and

constant

iparticles,i particles,

i particles,

��

�

�

=

ii

i

pE

P

�


27

Example of use of invariants

q 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


28

� ��(

)"�!��!��*

� ��+

), ��!��*

��

��

� −= pc

cmEP

�

,2

011

( )0,02 cmP =

��

��

� ′= pc

EP

�

,2

21 �

�

��

� ′−= pc

EP

�

,2

22

( )212:Invariant PPP +⋅

221

20

2210

2

02

0

EEcm

pc

E

c

Ep

c

Ecm

=�

×′+×=×−×


29

Electromagnetism and Relativityq Maxwell’s equations are relativistically invariant

q Lorentz force law:

20000

1in vacuum

densitiescurrent and charge

free-source00

cEDHB

jt

DHD

t

BEB

===

=∂∂−×∇=⋅∇

=∂∂+×∇=⋅∇

µεεµ

ρ��

�

�

��

�

��

( )ondistributi chargefor

charge particle singlefor

BjEf

qBvEqf��

�

�

�

��

∧+=∧+=

ρ


30

Lorentz force lawq Relativistic equation of motion

§4-vector form:

§3-vector component:

§Lorentz force derives naturally from relativistic 4-vector

(4×4 matrix) formulation of Maxwell’s equations and is

not an additional hypothesis.

( ) ( )BvEqfvmdt

d �

�

��

� ∧+==γ0

��

��

�=��

��

� ⋅�=

dt

pd

dt

dE

cf

c

fv

d

dPF

�

�

�

�

,1

, γγτ


31

Motion of charged particles in constant electromagnetic fields

q Constant E-field gives uniform acceleration in straight line

q Solution of

q Energy gain

q Constant magnetic field gives uniform spiral about B with constant energy (and ).

cmqEtm

qE

cm

qEt

qE

cmx

02

0

2

0

20

for2

1

11

<<≈

�

�

��

�

�−��

�

��

�+=

( ) ( )BvEqfvmdt

d �

�

��

� ∧+==γ0

( ) Em

qv

dt

d

0

=�γ

qEx=

Bvm

q

dt

vd �

�

�

∧=γ0 constantx

constant//

=

=

⊥�

�

v


32

Relativistic Transformations of E and B

q In a frame F, a particle moves with velocity , under fields and , and Lorentz force is

§ In particle’s rest frame F', and force is

§ Assume measurements give same charge and force, so

q Point charge q at rest in F generates fields:

§ See a current in F′, giving a field (Biot-Savart)

§ Suggests

( )BvEqf�

�

��

∧+=

v�

E�

B�

Eqf ′′=′��

BvEEqq�

�

�

∧+=′′= and

0,4 3

0

== Br

rqE

�

�

�

πε

Evcr

rvqB

�

�

��

�

∧−=∧−=23

0 1

4πµ

Evc

BB�

�

��

∧−=′2

1

Rough

idea

0=v�

( )

////2

////

,

,

BBc

EvBB

EEBvEE

��

�

�

�

��

�

�

=′��

��

� ∧−=′

=′∧+=′

⊥⊥

⊥⊥

γ

γ

Exact:


33

Electromagnetic Field of a Single Particleq Charged particle moving along x-axis of Frame F

q In F′, P is at

q And fields are only electrostatic (B=0), given by

Pp

pPpP tc

vxtttvbrxbvtx γγ =��

�

��

�−=+==−=

2222 ','' so ),,0,'('

��

Origins coincide at t=t′=0

Observer P

�

b

charge q!

"��" � "��"#�#

!#

333 '',0',

'

'''

''

r

qbEE

r

qvtEx

r

qE zyxP ==−=�= �

�


34

q Transform to laboratory frame F:

q As v c, 1, and magnetic induction cBy -Ez

q At non-relativistic energies, γ 1, and restores the Biot-Savart law:

( )

( ) 23

2222

23

2222

'

0

'

tvb

bqEE

E

tvb

vtqEE

zz

y

xx

γ

γγ

γ

γ

+==

=

+−==

0

'2

==

−=−=

zx

zzy

BB

Ec

Ec

vB

βγ

3r

rvqB

��

� ∧∝


35

Electromagnetic Field of a Beam of Particles

q Coasting beam, momentum p±∆p

q In effective rest frame, see only an electrostatic field, E′⊥, and B′⊥=0

q Transform to laboratory frame:

q So particle in beam is affected by a space charge force proportional to

2////

////

',0'

',0'

c

EvBBB

EEEE

⊥⊥

⊥⊥

×===

===�

�

γ

γ

( ) ⊥⊥

⊥⊥ ��

��

�−=��

�

��

� ××+=×+ '1'

'2

2

2E

c

v

c

EvvEBvE

�

�

�

�

�

�

�

γγ

Electrostatic repulsion between particles

Magnetostatic attraction between thin current wires

minus


36

University of Maryland Electron Ring

A small ring used to explore aspects of beam dynamics, including effects of strong space-charge forces in the beam.


37

Effect of Space Charge on an Intense BeamInjected beam in Proposed Fermilab Booster

With space charge Without space charge


38

relativity for accelerator physicists - cern · §examples: phase of a wave, rate of radiation of...

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