# CHAPTER 6 RELATIVISTIC KINEMATICS - Department of Physics ...

Mechanics

Nov 14, 2013 (4 years and 8 months ago)

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EPPT M2

INTRODUCTION TO
RELATIVITY

K Young
, Physics Department, CUHK

The Chinese University of Hong Kong

CHAPTER 6

VELOCITY,
MOMENTUM and
ENERGY

Displacement

Velocity

Momentum,

Conservation

Force

Newton's second law

x
x
v
t

mv
p

p
F
t

Objectives

Momentum

Collisions

Momentum

Momentum

Newtonian momentum is wrong

Should transform as 4
-
vector

Form of
p

and
E

Four
-
velocity

Coordinates

= ( time, space )

Displacement = change of postion

ct
x
x
 
 

 
 
 
c t
x
x

 
 
 
 
 

 
Example

P
travels to a star 5 ly away

At a speed
0.5c

5 10
10
5
x t
c t
x
x
   

   
 
  
   
 

   

Four velocity

1
u x

   
 
   

Displacement per unit
proper

time

c t
x
x

 
 
 
 
 

 
/
1
/
c t
u x
x

 
 
   
  
 
   
 

 

/
//
c t
u
x
t t

 
 
 
 

 
 
 
 

/
//
c t
u
x
t t

 
 
 
 

 
 
 
 
1
t

  
u c


 
 

 
 
 
x v t c t

    
u c


 
 

 
 
 
t
u c

x
u c


Example

Particle is travelling at 300 m s
-
1

8
300
3 10

6
10 1

 

1 1
2 12
2 2
1 1 10 1.00
 
 

    
t
u c c

 
-1
300 m s
x
u c c v
 
   
Example

Particle is travelling at 0.6
c

0.60

1
2
2
1 1.25
 

  
1.25
t
u

 
0.75
x
u

 
1.25 0.6 0.75

  
Case of low velocities

t
u c c

 
x
u c c v
 
  

is just ordinary velocity

x
u

t x
u u


carries no information

t
u
Time component carries no
extra information

True in general

2 2
2 2
t x
u u c c
 
  

2 2 2 2
1
c c
 
  
Three spatial components

u c


 
 

 
 
 
x
y
z
u c




 
 
 
 

 
 
 
 
 
/
x x
v c

etc.

Four
-
Momentum

Momentum = mass velocity

Now is more convenient

u
p mu

p m u
   

   
Explicit expression

t x
p mc p
c
m
 

 
t
u c

x
u c


2 2
1/
v c
mc

2 2
1/
v c
m
v

2 2
1/
t
mc
c
p
v

2 2
1/
x
m
v
c
p
v

If , = ordinary expression

2 2
v c

x
p

Recover Newtonian physics

If

0
v

0 , 0
x t
p p mc
  

as

,
t x
p p

/1
v c

Spatial component

v
=
c

p = m

v

v

p
x

Do not call this effective
mass
M
!

2 2

1/
x
m
p v
v c

Time component

2 2
1/
t
mc
p
v c

2 2
/1
v c

Consider (non- relativistic)
2 2 1 2
(1/)
t
p mc v c

 
Assuming mass does not change

2 2 1 2
(1/)
t
p mc v c

 

2
2 2 2 2
1 1 2//
mc v c v c
     

2 2
1 2
t
pc mc mv
  

2
const 1 2
mv
  
2
1 1
2
mc mv
c
 
  
 
 
Apart from additive constant, which does
not matter

/
x
E c
p
p
 

 
 

p
E
 
Conservation of
cons. of cons. of p

2
1 2
t
pc mv
  
const
t
p c E

2
2 2
1/
t
mc
E p c
v c
 

2
0
E mc

Provided
m

0,

takes
E

=

to reach
v

=
c

Therefore can never attain
v

=
c

v

E

v
=
c

E
0

= m c
2

T
here 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?

Kinetic 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 collisions

"Classical" collisions / Elastic collisions

d
c
b
a

d
c
b
a
E
E
E
E

)
(
)
(
)
(
)
(
2
2
2
2
d
d
c
c
b
b
a
a
K
c
m
K
c
m
K
c
m
K
c
m

d
c
b
a
K
K
K
K

d
b
c
a
m
m
m
m

Nuclei / Elementary particles

( ) ( ) 0
a b c d
m m m m m
     
a b c d
Q
K K K K
   
2
Q
mc
 
Mass is "converted" to energy

Analogy

Relation between
E

and
p

Newtonian

2
2
1
2
2
p mv
E mv
p
E
m

2
2 2
2 2 2
2
2
2
4
2
2 2
1/
/
1/
/
mv
p
v c
mc
E c
v c
c c
c c
E p m
E p m

 

 

Relativistic

System of units

E

:

eV

MeV

GeV

pc

:

eV

MeV

GeV

p

:

eV/
c

MeV/
c

GeV/
c

mc
2

:

eV

MeV

GeV

m

:

eV/
c
2

MeV/
c
2

GeV/
c
2

2
2
2 2
E pc mc
 
Particle

Mass
(MeV/
c
2
)

electron

0.5110

muon

105.7

proton

938.3

neutron

939.6

Conservation of

four
-
momentum

The four
-
momentum

/,
p E c

p

Recall

Contains

energy + momentum

Conservation law

For an
isolated

system, the
total 4

momentum is
conserved
.

p
Collisions

Example

1

1/3

2

0

2

v

1

u

2
2 2
1 2 1 2
1 3
0
1 1
1 1 3
:
u v
p
u v
    
 

2
2 2
1 2 1 2
:

1 1 1
1
1 1
1 1 3
E
u v
    
 

1
c

Better to analyze in terms of
p

p, E

directly measured and quoted

v

= 0.999… inconvenient

formulas apply to massless particles
(photons)

d
c
b
a

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( )
2
d
E M M
P
m M M
p
E

 
2 2
2 2
2
c
p
m M
P
m M EM

 

p
p
p
p
p
p

4
0
P
M M M
E M
P
E

Mass
Momentum
Energy
known

Example
Production of
p

at threshold

7 7 0.94 GeV 6.58 GeV
E M
   
E M E

 
2
2 2
2
( )
EM M
E E

 
2 2
2
2 2
( ) )
2
(4
EM M
P M P M
 

e
e
Z

Example

P

= 150 GeV

M

= 90 GeV

q

Q

Q

Z

e
+

e

2 2 2 2
2 2
P M Q m Q
   
2 cos
P Q
q

2 2
cos
P
P M
q

2 61.82
q

Energy in the CM frame

I
n 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

Example

M

E

Both of mass M

E
*

in CM = ?

M
E
E
P
P
t
t

M
M
E
M
E
P
E
P
t
t
2
1
2
1
2

)
(
t
t
P
E
E

)
(
2
M
E
M
E

E

Fixed target experiments are inefficient

Colliding beams much better

Principle of Relativity

Conservation

d
c
b
a

a b c d
p p p p
  
0
a b c d
P p p p p
    
Linear transformation

Principle of Relativity

0
a b c d
P p p p p
    
0
P
 

 

0
P L P
   

 
   
0
P
 

 
Objectives

Momentum

Collisions

Acknowledgment

I thank Miss HY Shik and Mr HT Fung for
design