Relativistic Mechanics
Relativistic Mass and
Momentum
Relativistic Mechanics

In classical mechanics, momentum is
always conserved.
Relativistic Mechanics


In classical mechanics, momentum is
always conserved.
However this is not true under a
Lorentz transform.
Relativistic Mechanics

If momentum is conserved for an
observer in a reference frame S
Relativistic Mechanics

If momentum is conserved for an
observer in a reference frame S , it is
not conserved for another observer in
another inertial frame S .
Relativistic Mass
V=0
v
v
1
2
Before
After
Stationary Frame
Relativistic Mass
V′
v′x
1
2
Before
After
Moving Frame of mass 1
Relativistic Mass

In the stationary frame,
Pi = mv – mv = 0 (assume the masses are equal

However in the moving frame, P final >
P initial.
Relativistic Momentum

For momentum to be conserved the
expression for the momentum must be
rewritten as follows:
P
m0 v
1 v
2
c
2
(Relativistic Momentum)
Relativistic Mass

Where m0 is the rest mass.
Relativistic Mass


The expression is produced by
assuming that mass is not absolute.
Instead the mass of an object varies
with velocity i.e.mv .
Relativistic Mass

The expression for the relativistic mass
is
m
m0
1 v c
2
2
(Relativistic mass)
Relativistic Mass

From the expression for the
momentum, the relativistic force is
defined as,
F 
dP
dt
Stationary Frame
V=0
v
v
1
2
Before
Pi = mv – mv = 0 (assume the masses are equal)
Pf = (m1+m2)x0 = 0
Pi = Pf . Therefore Momentum is conserved.
After
Moving Frame of mass1
V′
v′x
1
2
Before

 u v
Pi  m1   x1
 1  vux1

c2

After



 u v
  m2   x 2

 1  vux 2


c2








Moving Frame of mass1
V′
v′x
1
2
Before



 u v 
 u v
x
1
  m2   x 2
Pi  m1  
 1  vux1 
 1  vux 2



c2 
c2


After






 vv 
 vv 
  m
  m

vv
v

(

v
)

1 
1






c2 
 c2 


Moving Frame of mass1
V′
v′x
1
2
Before

 u v
Pi  m1   x1
 1  vux1

c2

 2m v

1 v2 c2



 u v
  m2   x 2

 1  vux 2


c2


After






 vv 
 vv 
  m
  m

vv
v

(

v
)

1 
1






c2 
 c2 


Moving Frame of mass1
V′
v′x
1
2
Before

 V v
Pf  2m   x
 1  vVx

c2







After
Moving Frame of mass1
V′
v′x
1
2
Before

 V v
Pf  2m   x
 1  vVx

c2





 0v 
  2m  


1 0  v 



c2 


After
Moving Frame of mass1
V′
v′x
1
2
Before

 V v
Pf  2m   x
 1  vVx

c2





 0v 
  2m  
  2mv
0

v

1




c2 


After
Moving Frame of mass1

Hence momentum is not conserved.
Moving Frame of mass1

Now consider what happens when the
modification to the mass is introduced.
Moving Frame of mass1

Recall the equations for the final and initial
momentum.
 2v
 2m v
where v2 
Pi  mv2 
2
2
1 v2 c2
1 v c
Pf  (m1  m2 )V   M V   M v where V   v
Moving Frame of mass1

Consider the initial momentum
 2m v
Pi 
1 v2 c2
where m really corresponds to the mass m2
NB:m1  m0 Since it is at rest in the moving frame
Moving Frame of mass1

Consider the initial momentum
 2m v
Pi 
1 v2 c2
m
m0
v
1
c
2
2
2
(Relativistic mass)
Moving Frame of mass1
 2v
where v2 
m
2
2
1

v
c
2
v2
1 2
c
m0
Moving Frame of mass1
 2v
where v2 
m
2
2
1

v
c
2
v2
1 2
c
m0
m 
2
1   2v 

1  2  
2
2 
c 1 v c 
m0
Moving Frame of mass1
 2v
where v2 
m
2
2
1

v
c
2
v2
1 2
c
m0
m0
m 

2
2
2
(
2
v
)
c
1   2v 
1

1  2  
2
2 2
2
2 
(
1

v
c
)
c 1 v c 
m0
Moving Frame of mass1
 2v
where v2 
m
2
2
1

v
c
2
v2
1 2
c
m0
m0
m 

2
2
2
(
2
v
)
c
1   2v 
1

1  2  
2
2 2
2
2 
(
1

v
c
)
c 1 v c 
m0
simplify
m
m0
(1  v 2 c 2 ) 2
(1  v 2 c 2 ) 2
Moving Frame of mass1
 2v
where v2 
m
2
2
1

v
c
2
v2
1 2
c
m0
m0
m 

2
2
2
(
2
v
)
c
1   2v 
1

1  2  
2
2 2
2
2 
(
1

v
c
)
c 1 v c 
m0
simplify
m
(1  v 2 c 2 )
 m0 
2
2
2
2 2
(
1

v
c
)
(1  v c )
(1  v 2 c 2 ) 2
m0
Moving Frame of mass1

Therefore the initial momentum
 2m v
Pi 
1 v2 c2
Moving Frame of mass1

Therefore the initial momentum
m0 (1  v 2 c 2 )
 2m v
 2v
Pi 


2
2
2
2
1 v c
1 v c
(1  v 2 c 2 )
Moving Frame of mass1

Therefore the initial momentum
m0 (1  v 2 c 2 )
 2m v
 2v
Pi 


2
2
2
2
1 v c
1 v c
(1  v 2 c 2 )
 2m0v
Pi 
1 v2 c2
Moving Frame of mass1

Now consider the final momentum
Pf  (m1  m2 )V   M V   M v
M 
M0
2m0

2
2
v
v
1 2 1 2
c
c
 2m0 v
 Pf 
1 v2 c2
Relativistic Momentum

Hence momentum is conserved.
Relativistic Momentum

Consider the difference in the
relativistic momentum and classical
momentum.
Relativistic Momentum
Relativistic Energy
Relativistic Energy

To derive an expression for the
relativistic energy we start with the
work-energy theorem
Relativistic Energy

To derive an expression for the
relativistic energy we start with the
work-energy theorem
W  K 
 Fdx
dP
W  
dx
dt
(Work done)
Relativistic Energy

Using the chain rule repeatedly this
can rewritten as
dP
W 
vdv
dv
Relativistic Energy

Using the chain rule repeatedly this
can rewritten as
dP
W 
vdv
dv

m0 v
d 

vdv

2
2
dv


1

v
c
0

v
Relativistic Energy

Using the chain rule repeatedly this
can rewritten as
dP
W 
vdv
dv
(Work done


m
v
d
0

vdv accelerating an object

2
2
dv


from rest some
1

v
c
0

v
velocity)
Relativistic Energy
v

0

m0 v
d 

 vdv
dv  1  v 2 c 2 


v
W 
m0 v
 1  v
0
2
c2

3
dv
2
Relativistic Energy
v

0

m0 v
d 

 vdv
dv  1  v 2 c 2 


v
W 
m0 v
 1  v
0
2
c2

3
2
dv  m0 
1  v
v
2
c
2

3
dv
2
Relativistic Energy
v

0

m0 v
d 

 vdv
dv  1  v 2 c 2 


v
W 
m0 v
 1  v
0
2
c2

3
2
dv  m0 
Integrating by substitution:
1  v
v
2
c
2

3
dv
2
Relativistic Energy
v

0

m0 v
d 

 vdv
dv  1  v 2 c 2 


v
W 
m0 v
 1  v
0
2
c2

3
2
dv  m0 
Integrating by substitution:
Let u  v 2  du  2vdv
du
 dv 
2v
1  v
v
2
c
2

3
dv
2
Relativistic Energy
v

0

m0 v
d 

 vdv
dv  1  v 2 c 2 


v
W 
m0 v
 1  v
0
2
c2

3
2
dv  m0 
Integrating by substitution:
Let u  v 2  du  2vdv
du
 dv 
2v
1  v
v
2
c
2

3
dv
2
Relativistic Energy
W  m0
v
 1  u c 
2
3
2
du
2v
Relativistic Energy
W  m0
v
 1  u c 
2
3
2
du m0

2
2v
du
 1  u
c
2

3
2
Relativistic Energy
W  m0
v
 1  u c 
2

m0  1  u c 2


1
2 
2c 2


3
du m0

2
2v
2
u

1

2



0
du
 1  u
c
2

3
2
Relativistic Energy
W  m0
v
 1  u c 
2

m0  1  u c 2


1
2 
2c 2


3
du m0

2
2v
2
du
 1  u
c
2

u

1

2

1
2
2
  m0 c 1  u c  2

0
u
0
3
2
Relativistic Energy

W  m0 c 1  u c
2
2

1
2
 m0 c 2
Relativistic Energy

W  m0 c 1  u c
2
Therefore W 
2

1
2
 m0 c 2
m0 c 2
1  v
2
c
2

1
2
 m0 c 2
Relativistic Energy

W  m0 c 1  u c
2
Therefore W 
2

1
2
 m0 c 2
m0 c 2
1  v
2
c
2

1
2
However this equal to  K
 m0 c 2
Relativistic Energy

W  m0 c 1  u c
2
Therefore W 
2

1
2
 m0 c 2
m0 c 2
1  v
2
c
2

1
2
 m0 c 2
However this equal to  K
therefore the Relativistic Kinetic energy is
K
m0 c 2
1  v
2
c
2

1
2
 m0 c 2
Relativistic Energy

The velocity independent term ( m0 c 2 ) is
the rest energy – the energy an object
contains when it is at rest.
Relativistic Energy
(Correspondence Principle)

At low speeds v   c we can write the
kinetic energy as

K  m0 c 1  v
2
2
c

1
2  2

1
Relativistic Energy
(Correspondence Principle)

At low speeds v   c we can write the
kinetic energy as

K  m0 c 1  v
2
2
c

1
2  2

1
Using a Taylor expansion we get,
Relativistic Energy
(Correspondence Principle)

At low speeds v   c we can write the
kinetic energy as

K  m0 c 1  v
2
2
c

1
2  2

1
Using a Taylor expansion we get,

K  m0 c 1   v
2
1
2
2

c  v c
2
3
8
2

2 2

5
16
v
2
c

2 3
 
 ...  1
Relativistic Energy
(Correspondence Principle)

Taking the first 2 terms of the
expansion
Relativistic Energy
(Correspondence Principle)

Taking the first 2 terms of the
expansion


K  m0 c 2 1  12 v 2 c 2  1
Relativistic Energy
(Correspondence Principle)

Taking the first 2 terms of the
expansion


K  m0 c 2 1  12 v 2 c 2  1
So that
K  m0 v
1
2
2
(classical expression for the Kinetic energy)
Relativistic Energy
Note that
even an
infinite
amount of
energy is not
enough to
achieve c.
Relativistic Energy

The expression for the relativistic
kinetic energy is often written as
m0 c 2
1 v c
2
2
 K  m0 c
Relativistic Energy

The expression for the relativistic
kinetic energy is often written as
m0 c 2
1 v c
2
2
 K  m0 c
or equivalent as mc2  K  m0 c 2
Relativistic Energy

The expression for the relativistic
kinetic energy is often written as
m0 c 2
1 v c
2
2
 K  m0 c
or equivalent as mc2  K  m0 c 2
where E  m c2 is the total relativistic energy
Relativistic Energy

So that E  K  m0 c
2
Relativistic Energy


So that E  K  m0 c
2
If the object also has potential energy
2
2
mc

K

V

m
c
it can be shown that
0
Relativistic Energy

The relationship E  m c2 is just Einstein’s
mass-energy equivalence equation, which
shows that mass is a form of energy.
Relativistic Energy

An important fact of this relationship is
that the relativistic mass is a direct
measure of the total energy content of
an object.
Relativistic Energy


An important fact of this relationship is
that the relativistic mass is a direct
measure of the total energy content of
an object.
It shows that a small mass
corresponds to an enormous amount
of energy.
Relativistic Energy


An important fact of this relationship is
that the relativistic mass is a direct
measure of the total energy content of
an object.
It shows that a small mass
corresponds to an enormous amount
of energy. This is the foundation of
nuclear physics.
Relativistic Energy

Anything that increases the energy in
an object will increase its relativistic
mass.
Relativistic Energy

In certain cases the momentum or
energy is known instead of the speed.
The relationship involving the
momentum is derived as follows:
Relativistic Energy

In certain cases the momentum or
energy is known instead of the speed.
Relativistic Energy
2 4
m
2
2 4
0c
Note that E  m c 
1 v2 c2
Relativistic Energy
2 4
m
2
2 4
0c
Note that E  m c 
1 v2 c2


 m2 c 4 1  v 2 c 2  m02 c 4
Relativistic Energy
2 4
m
2
2 4
0c
Note that E  m c 
1 v2 c2


 m2 c 4 1  v 2 c 2  m02 c 4
 m2 c 4  E 2  m2 c 2 v 2  m02 c 4
Relativistic Energy
2 4
m
2
2 4
0c
Note that E  m c 
1 v2 c2


 m2 c 4 1  v 2 c 2  m02 c 4
 m2 c 4  E 2  m2 c 2 v 2  m02 c 4
Since
P  mv
Relativistic Energy
2 4
m
2
2 4
0c
Note that E  m c 
1 v2 c2


 m2 c 4 1  v 2 c 2  m02 c 4
 m2 c 4  E 2  m2 c 2 v 2  m02 c 4
Since
P  mv
E 2  P 2 c 2  m02 c 4
Relativistic Energy

When the object is at rest, p  0 so that
E  m c2
Relativistic Energy

When the object is at rest, p  0 so that
E  m c2

For particles having zero mass (photons)
we see that E  pc the expected relationship
relating energy and momentum for photons
and neutrinos which travel at the speed of
light.