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EPPT M2
INTRODUCTION TO RELATIVITY
EPPT M2
INTRODUCTION TO RELATIVITY
K Young, Physics Department, CUHKThe Chinese University of Hong Kong
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CHAPTER 6
VELOCITY, MOMENTUM and
ENERGY
CHAPTER 6
VELOCITY, MOMENTUM and
ENERGY
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Displacement
Velocity
Momentum,Conservation
ForceNewton's second law
x
xv
t
mvp
pF
t
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ObjectivesObjectives
Momentum Collisions
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MomentumMomentum
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MomentumMomentum
Newtonian momentum is wrong Should transform as 4-vector Form of p and E
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Four-velocityFour-velocity
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Coordinates = ( time, space )
Displacement = change of postion
ctx
x
c tx
x
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ExampleExample
P travels to a star 5 ly away At a speed 0.5c
5 10
10
5
x t
c tx
x
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Four velocity
1u x
Displacement per unit proper time
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c tx
x
/1
/
c tu x
x
/
/ /
c tu
x t t
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/
/ /
c tu
x t t
1t
u c
x v t c t
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u c
tu c
xu c
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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
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ExampleExample
Particle is travelling at 0.6c
0.60
1
2 21 1.25
1.25tu 0.75xu
1.25 0.6 0.75
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Case of low velocitiesCase of low velocities
tu c c
xu c c v
is just ordinary velocityxu t xu u carries no informationtu
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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
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Three spatial componentsThree spatial components
u c
x
y
z
u c
/x xv c etc.
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Four-MomentumFour-Momentum
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Momentum = mass velocity Now is more convenientu
p mu����������������������������
p m u ����������������������������
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Explicit expressionExplicit expression
t xp mc p cm tu c xu c
2 21 /v c
mc
2 21 /v c
mv
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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
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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
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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
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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
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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
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2
2 21 /t
mcE p c
v c
2
0E mc
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Provided m 0, takes E = to reach v = c Therefore can never attain v = c
v
E
v = c
E0 = m c2
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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?
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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
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Application to collisionsApplication to collisions
"Classical" collisions / Elastic collisions
dcba dcba
dcba EEEE
)()(
)()(
22
22
ddcc
bbaa
KcmKcm
KcmKcm
dcba KKKK dbca mmmm
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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
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AnalogyAnalogy
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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
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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
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Particle Mass (MeV/ c2)
electron 0.5110
muon 105.7
proton 938.3
neutron 939.6
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Conservation of four -momentum
Conservation of four -momentum
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The four-momentumThe four-momentum
/ , p E c p��������������
Recall
Contains
energy + momentum
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Conservation lawConservation law
For an isolated system, the total 4 – momentum is conserved.
p��������������
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CollisionsCollisions
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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
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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)
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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
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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
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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
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Energy in the CM frameEnergy in the CM frame
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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
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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
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Principle of RelativityPrinciple of Relativity
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ConservationConservation
dcba dcba
a b c dp p p p
0a b c dP p p p p
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Linear transformation Principle of Relativity
0a b c dP p p p p
0P ��������������
0P L P ����������������������������
0P ��������������
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ObjectivesObjectives
Momentum Collisions
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AcknowledgmentAcknowledgment
I thank Miss HY Shik and Mr HT Fung for design