Download Relativistic Momentum - UCF College of Sciences

Relativistic Momentum
In classical mechanics, the momentum of a
particle is defined
as a product of its mass and its

velocity, m v .
In an isolated system of particles, with no net
force acting on the system, the total momentum of
the system remains the same. However , we can see
from a simple though experiment that the quantity
Σmivi is not conserved in isolated system.
As was shown before, the Galilean transformations leads to
same acceleration in different inertial frames:
ax’ = ax
and forces are also the same in both frames.
However, according to Lorentz transformations, acceleration is
not the same in two such reference frames. If a particle has
acceleration a, and velocity ux in frame S, it’s acceleration in S’,
obtained by computing dux’/dt’ is:
a 
'
x
ax
 vux 
 1  2 
c 

3
3
We consider two observers: observer A in reference frame S and
observer B in frame S’, which is moving to the right in the x
direction with speed v with respect to the frame S. Each has a ball
of the mass m.
One observer throws his ball up with a speed u0 relative to him and
the other throws his ball down with a speed u0 relative to him, so
that each ball travel a distance L, makes an elastic collision with the
other ball, and returns.
Classically, each ball has vertical momentum of magnitude mu0.
Since the vertical components of the momenta are equal and
opposite, the total vertical momentum are zero before the
collision. The collision merely reverses the momentum of each
ball, so the total vertical momentum is zero after the collision.
Relativistically, however, the vertical components of the velocities of
the two balls as seen by the observers are not equal and opposite.
Thus, when they are reversed by the collision, classical momentum is
not conserved. As seen by A in frame S, the velocity of his ball is
uAy=+u0. Since the velocity of ball B in frame S’ is u’Bx=0, uBy’=-u0, the
y component of the velocity of ball B in frame S is uBy=-u0/γ. So, the Σ
miui is conserved only in the approximation that u<<c.
We will define the relativistic momentum p of a particle to have the
following properties:
1. In collisions , p is conserved.
2.As u/c approaches 0, p approaches mu.
We will show that quantity
mu
p  mu 
u2
1 2
c
is conserved in the elastic collision and we take this equation for the
definition of the relativistic momentum of a particle.
Conservation of the Relativistic Momentum
p
mu
u2
1 2
c
One interpretation of this equation is that the mass of an object
increases with the speed. Then the quantity
mrel 
m
u2
1 2
c
is called the relativistic mass. The mass of a particle when it is at
rest in some reference frame is then called its rest mass .
mrel= γmrest
Conservation of the Relativistic Momentum
The speed of ball A in S is u0, so the y component of its relativistic
momentum is:
p Ay 
mu0
u02
1 2
c
The speed of B in S is more complicated. Its x component is v and
its y component is –u0/γ.
2 

v 
2
2
2
2

u B  u Bx  u By  v   u 0 1  2


c


2 2
u0 v
2
2
2
u B  v  u0  2
c
2
Using this result to compute √1-(u2B/c2), we obtain
2
B
2
2
2
0
2
2
u
v u
2 v
1
 1  2   u0 4 
c
c
c
c
 v 2  u02 
 1  2 1  2 
 c  c 
and
2
B
2
2
2
0
2
u
u
u
v
1
1
 1 2  1

1
c
c
c

c
2
0
2
Conservation of the Relativistic Momentum
The y component of the relativistic momentum of ball
B as seen in S is therefore
 mu0 
 

muBy
 

pBy 

u B2
u02
1
1 2
1 2
c

c
pBy 
 mu0
u02
1 2
c
Conservation of the Relativistic Momentum
Since pBy=-pAy’ the y component of the total
momentum of the two balls is zero. If the speed of
each ball is reversed by the collision, the total
momentum will remain zero and momentum will be
conserved.
p  mu
with

1
2
u
1 2
c
Relativistic momentum as given by equation
p
mu
u2
1 2
c
versus u/c, where u is speed of the object relative to an
observer. The magnitude of momentum p is plotted in units
of mc. The dashed line shows the classical momentum mu.
Relativistic Energy
In classic mechanic: The work done by the net force
acting on a particle equals the change in the kinetic
energy of the particle.
In relativistic mechanic: The net force acting on a
particle to accelerate it from rest to some final velocity
is equal to the rate of change of the relativistic
momentum. The work done by the net force can then
be calculated and set equal to the change of kinetic
energy.
Relativistic Energy
As in classical mechanic, we will define the kinetic energy
as the work done by the net force in accelerating a particle
from rest to some final velocity uf.




uf
uf
uf
uf
 mu 
dp

K  Fnet ds 
ds  udp  ud 
2 
dt

u
0
0
0
0
 1  2 
c 




where we have used u=ds/dt.

d
mu
1
u2
c
2
 
 
 
 m d 
 
 

 






1   2u 
 
m u  




2
2  c













 2 
1
u 
 m 2 
c 

u2


1



c2







 m





1
2

1  u

c2





3
1
1
u2
c2



u 




1
2

u
1 

c2





3




3












2


u
1


c2








2
 du  m
1  u

2

c








1
du  

u2
1

c2








1
  du 

2


u
1



2 
c











1

  du 

3

2 


u
1 



2




c 







3
2
du
Relativistic Energy
Substituting the last result in the equation for kinetic energy we
obtain:




uf
uf
2
 mu 

u
du  m1  2
K  ud 

2 
c

u

0
0
 1  2 
c 







1
2
2
2
 mc 
 1  mc  mc


u2
 1  2

c








3
2
udu 
Relativistic Energy
K
mc2
u2
1 2
c
 mc
2
The second part of this expression, mc2, is independent of
the speed and called the rest energy Eo of the particle.
E0=mc2
The total relativistic energy:
E  K  mc 
2
mc2
1
u
c
2
2
Relativistic Energy
The work done by unbalanced force increases the
energy from the rest energy to the final energy
mc2
u2
1 2
c
where
mrel 
is the relativistic mass.
 mrel c 2
m
2
u
1 2
c
Relativistic Energy
We can obtain a useful expression for the velocity
of a particle by multiplying equation for relativistic
momentum by c2:
p
pc 
2
mu
2
u
1 2
c
2
mc u
2
u
1 2
c
 Eu
u pc

c E
Energies in atomic and nuclear physics are usually
expressed in units of electron volts (eV) or megaelectron volts (MeV):
1eV = 1.602 x 10-19J
A convenient unit for the masses of atomic
particles is eV/c2 or MeV/c2, which is the rest
energy of the particle divided by c2.
The expression for kinetic energy
K
mc
2
2
 mc
2
u
1 2
c
does not look much like the classical expression ½(mu2).
However when u<<c, we can approximate 1/√1-(u2 /c2)
using the binomial expansion:
(1+x)n =1 + nx + n(n-1)x2/2 + …….≈ 1 + nx
Then
 u2 
 1  2 
2
u
 c 
1 2
c
1

1
2
1 u2
 1
2 c2
and when v<<c




2

 1
1
1u

2
2
2

K  mc 
 1  mc 1 

1

mu
2
 2
2
2
c


 1  u

2
c


The equation for relativistic momentum:
p
mu
u2
1 2
c
and equation for the total energy:
E  K  mc2 
mc2
1
u2
c
2
can be combined to eliminate the speed u:
E2 = p2c2 +(mc2)2
E2 = p2c2 +(mc2)2
We received the relation for total energy,
momentum, and rest energy.
If the relativistic energy of a particle is much grater
than its rest energy mc2, the second term on the
right side of the last equation can be neglected,
giving the useful approximation:
E ≈ pc (for E>>mc2)