Transcript Mass energy equivalence

Mass energy equivalence
RELATION BETWEEN MOMENTUM AND
ENERGY
 In quantum mechanics , we considered that kinetic
energy could be increased only increasing by its
velocity
 But now dealing with relativistic mechanics we take
mass variation into account
Relationship between mass and energy
 If force F acting on a particle ,produces a
displacement dx then the work done by the force =
Fdx the work done must be equal to the gain in the
kinetic energy dE of the particle

dE= Fdx
 But force being rate of change of linear momentum
of the particle , is given by

d
dv
dm
F  mv   m  v
dt
dt
dt
dv
dm
 dE  m dx  v
dx
dt
dt
 dx

2
 dE  m vdv v dm   v 
 dt

m0
2
m0
2 2
2
2 2
2 2
m
m 
 m c  m v  m0 c
2
2
1 v c
1  v2 c2
differentiating
 c 2 2m dm m 2 2vdv  v 2 2m dm  0
c 2 dm  m vdv v 2 dm  0
 dE  c 2 dm
m
E  c 2  dm  c 2 m  m0   E  m c2   m0c 2
m0
putting
m0
m
1  v2 c2
v  c, we
for
1  v
E
2
c

2 1 2
m0c 2
1  v
2
c2 
 m0c 2
have
 1  v 2 2c 2  3 v 4 8c 4  ....
4
1
3
v
 E  m0c 2 1  v 2 2c 2  3 v 4 8c 4  ....  1  m0v 2 
2
8 c2
Examples for proving equivalence
between energy and mass
NUCLEAR FISSION
NUCLEAR FUSION
NUCLEAR REACTION PROCESSES
PHENOMENON OF PAIR PRODUCTION
Nuclear fission
Example 1
 What is the annual loss in the mass of the sun , if the
earth receives heat energy approximately 2
cal/cm2/min , The earth sun distance is about
150x106 km
 Solution
 Rate of energy radiated
 2  4.2  107
erg / cm2 / min
Total energy radiated per min ute
 4  150 10
  2  4.2  10
11 2
7
energy radiated per year
 4  150 10
  2  4.2  10  5.3  10
11 2
7
5
E
annual loss in m ass is 2
c

4  150 10
  2  4.2  10  5.3  10
11 2
7
5
9  1020
 1.4  1014 tons per year  1.4  1014 tons per year
Example 2
 A nucleus of mass m emits a gamma ray of
frequency 0 .Show that the loss of internal energy
h
by the nucleus is not h0 but is h0 1  02 

2m c 
 Solution
 The momentum of gamma ray photon is p 
h 0
c
According to the law of conversation of momentum ,
the nucleus having mass m will recoil with the
momentum h in the back ground direction .
c
Therefore, the loss of energy recoiling is
0
1 2 m 2v 2
p 2 h0 
E  mv 


where photon energy h0
2
2
2m
2m 2m c
 h0 2 
h0 

  h0 1 
the total loss  h0  

2 
2
m
c
2
m




2
Example 3
 A certain accelerator produces a beam of neutral K-
mesons or kaons mkc2=498 MeV . Consider a kaon
that decays in flight into two pions (mc2140 MeV)
 Show that the kinetic energy of each pion in the
special case in which the pions travel parallel or anti
parallel to the direction of the kaon beam or 543
MeV and 0.6 Mev
Solution
 The initial relativistic total energy
 Ek= K+ mkc2 =325 MeV + 498 MeV =823 MeV
 Total initial momentum
Pk c  E  mk c
2
k

2 2
8232  4982

 655MeV
 Total energy for final system consisting of two pions
is
E  E1  E2 
i
 p1c
2
 m c

2 2

 p2c 
2
 m c

2 2
 823 MeV
 Applying conservation of momentum , the final
momentum of the two pions system along the beam
direction is P1 + P2 and setting this equal to the initial
momentum Pk , one obtains
 P1c +P2c =Pkc = 655 MeV
ii
 We have now two equations in the two unknown P1
and P2 , solving we find P1c= 668 MeV or -13 MeV iii

K
Pc2  m0c 2 2  m0c 2
K1 
6682  1402  140  543
MeV
K2 
 132  1402  140  0.6
MeV
Relation between momentum and energy
 Relativistic momentum of a particle moving with a
velocity v is given by P=mv
 Where
m0
m
(1)
(2)
 M0 being the rest mass of the particle , from
relativity we have E= mc2
(3)
 From 1 and 2 we have
2 4
2 2
2
m0 c
c m0 v
2
2 2
2 4
2 2 2
2 4
E c P m c  c m v 


m
c
2
2
2
2
0
1 v c 1 v c

1  v 2 c2
or E 2  c 2 P 2  m 02c 4
( 4)
Particles with zero rest mass
 Photon and Graviton are the familiar examples of
particles with zero rest mass . a particle with zero
rest mass always moves with the speed of light in
vacuum . According to 4 , if m0 =0 , we have E= Pc

E 
2
p  m v  v 2  as E  m c 
c


E
Pc
pv 2 v 2
c
c
or v  c
 i.e., a particle with zero mass ( rest mass ) always
moves with the speed of light in vacuum . The velcity
of the particle observed in some other inertial frame
S` is U   U  v 2
1  Uv c
 Where v is the velocity of the frame S` with respect to
the frame S in which the velocity of the particle is U,
cv

U

c
hence U=c we have
2
1  cv c
 Clearly the particle has the same speed c and zero
rest mass for all observers in inertial frames.