EPPT M2

INTRODUCTION TO RELATIVITY

EPPT M2

INTRODUCTION TO RELATIVITY

K Young, Physics Department, CUHKThe Chinese University of Hong Kong

CHAPTER 6

VELOCITY, MOMENTUM and

ENERGY

CHAPTER 6

VELOCITY, MOMENTUM and

ENERGY

Displacement

Velocity

Momentum,Conservation

ForceNewton's second law

x

xv

t

mvp

pF

t

ObjectivesObjectives

Momentum Collisions

MomentumMomentum

MomentumMomentum

Newtonian momentum is wrong Should transform as 4-vector Form of p and E

Four-velocityFour-velocity

Coordinates = ( time, space )

Displacement = change of postion

ctx

x

c tx

x

ExampleExample

P travels to a star 5 ly away At a speed 0.5c

5 10

10

5

x t

c tx

x

Four velocity

1u x

Displacement per unit proper time

c tx

x

/1

/

c tu x

x

/

/ /

c tu

x t t

/

/ /

c tu

x t t

1t

u c

x v t c t

u c

tu c

xu c

ExampleExample

Particle is travelling at 300 m s -1

8

300

3 10

610 1

1 1

2 122 21 1 10 1.00

tu c c -1300 m sxu c c v

ExampleExample

Particle is travelling at 0.6c

0.60

1

2 21 1.25

1.25tu 0.75xu

1.25 0.6 0.75

Case of low velocitiesCase of low velocities

tu c c

xu c c v

is just ordinary velocityxu t xu u carries no informationtu

Time component carries no extra informationTime component carries no extra information

True in general

2 22 2t xu u c c

2 2 2 21c c

Three spatial componentsThree spatial components

u c

x

y

z

u c

/x xv c etc.

Four-MomentumFour-Momentum

Momentum = mass velocity Now is more convenientu

p mu����������������������������

p m u ����������������������������

Explicit expressionExplicit expression

t xp mc p cm tu c xu c

2 21 /v c

mc

2 21 /v c

mv

2 21 /t

mc

cp

v

2 21 /x

mv

cp

v

If , = ordinary expression2 2v c xp

Recover Newtonian physics

If 0v 0 , 0x tp p mc

as , t xp p / 1v c

Spatial componentSpatial component

v = c

p = m v

v

px

Do not call this effective mass M!

2 2

1 /x

mp v

v c

Time componentTime component

2 21 /t

mcp

v c

2 21 /t

mcp

v c

2 2 / 1 v c Consider (non - relativistic)

2 2 1 2(1 / )tp mc v c

Assuming mass does not change

2 2 1 2(1 / )tp mc v c

22 2 2 21 1 2 / /mc v c v c

2 21 2tp c mc mv

2const 1 2 mv

21 1

2mc mv

c

Apart from additive constant, which does not matter

/

x

E cp

p

p

E

Conservation of

cons. of cons. of p

21 2tp c mv const

tp c E

2

2 21 /t

mcE p c

v c

2

0E mc

Provided m 0, takes E = to reach v = c Therefore can never attain v = c

v

E

v = c

E0 = m c2

There was a young fellow named Bright

Who travelled much faster than light.

He set off one day, in a relative way

And come back the previous night!

Faster than light?Faster than light?

Kinetic energyKinetic energy

2

2

E mc

mc K

2 2

2 2

1( 1) 1

1 /K mc mc

v c

2 2

1

1 /v c

Application to collisionsApplication to collisions

"Classical" collisions / Elastic collisions

dcba dcba

dcba EEEE

)()(

)()(

22

22

ddcc

bbaa

KcmKcm

KcmKcm

dcba KKKK dbca mmmm

Nuclei / Elementary particles

( ) ( ) 0a b c dm m m m m

a b c dQK K K K 2Q mc

Mass is "converted" to energy

AnalogyAnalogy

Relation between E and pRelation between E and pNewtonianNewtonian

2

2

1

22

p mv

E mv

pE

m

2 2 2

2 2 2

2

2

2

4

2

2 2

1 /

/1 /

/

mvp

v cmc

E cv c

c c

c c

E p m

E p m

RelativisticRelativistic

System of unitsSystem of units

E : eV MeV GeV

pc : eV MeV GeV

p : eV/c MeV/c GeV/c

mc2 : eV MeV GeV

m : eV/c2 MeV/c2 GeV/c2

222 2E pc mc

Particle Mass (MeV/ c2)

electron 0.5110

muon 105.7

proton 938.3

neutron 939.6

Conservation of four -momentum

Conservation of four -momentum

The four-momentumThe four-momentum

/ , p E c p��������������

Recall

Contains

energy + momentum

Conservation lawConservation law

For an isolated system, the total 4 – momentum is conserved.

p��������������

CollisionsCollisions

Example Example

11/3

20

2v

1u

2 2 21 2 1 2

1 3 0

1 11 1 3:

u vp

u v

2 2 21 2 1 2:

1 1 1 1

1 11 1 3E

u v

1c

Better to analyze in terms of pBetter to analyze in terms of p

p, E directly measured and quoted v = 0.999… inconvenient formulas apply to massless particles (pho

tons)

dcba dcba

0

c d

c d

m M m M

P

E M E

p

E

p

Mass

Momentum

Energy known

c d

c d

P

E E

p p

M E

2 2

2( )

2d

E M MP

m M Mp

E

2 2

2 2 2cpm M

Pm M EM

pppppp pppppp

4

0P

M M M

E M

P

E

Mass

Momentum

Energy known

Example Production of p at thresholdExample Production of p at threshold

7 7 0.94 GeV 6.58 GeVE M

E M E 22 22 ( )EM ME E

2 2 2 2 2( ) )2 (4EM MP M P M

eeZ eeZ

Example Example

P = 150 GeVM = 90 GeV

Q

Q

Z

e+

e

2 2 2 22 2P M Q m Q

2 cosP Q

2 2cos

P

P M

2 61.82

Energy in the CM frameEnergy in the CM frame

In a collision, much of the energy of the

projectile is used to carry the whole system

forward; only a small fraction is used to

produce new particles

ExampleExampleME Both of mass M

E* in CM = ?

MEEPP tt

M

ME

ME

P

E

P

t

t

21

212

)( tt PEE

)(2 MEME E

Fixed target experiments are inefficientColliding beams much better

Principle of RelativityPrinciple of Relativity

ConservationConservation

dcba dcba

a b c dp p p p

0a b c dP p p p p

Linear transformation Principle of Relativity

0a b c dP p p p p

0P ��������������

0P L P ����������������������������

0P ��������������

ObjectivesObjectives

Momentum Collisions

AcknowledgmentAcknowledgment

I thank Miss HY Shik and Mr HT Fung for design