RELATIVITY
(1) Some Important Terms
(i) Particle : A particle is a small amount of matter, having no linear dimension, but only a position at a
point. It has mass as well charge.
(ii) Observer : A person who locates, records, measures and interprets an event is called an observer.
(iii)Event : It implies anything that occurs suddenly or instantaneously at a point in space. It involves a
position and a time of occurrence.
(iv) Frame of Reference : A system of coordinate axes which defines the position of a particle or specifies
the location of an event is called a frame of reference. E.g. Cartesian coordinate system which is
specified by x, y and z coordinates.
(2) Frame of references are of two types:(i) Inertial Frame of Reference : Let us consider a system S of coordinates (x, y, z) with respect to which
a body is in uniform motion. The velocity of the body is given by
dx
dt
dz
 ux ,  u y ,  uz
dt
dt
dt
where ux, uy and uz are the velocity components along x, y and z directions respectively. Also, since
the body is in uniform motion,
d 2 x dv x
d 2 y dv y
d 2 z dvz


a

0
,


a

0
,

 az  0
x
y
dt 2
dt
dt 2
dt
dt 2
dt
where ax, ay, az denote the acceleration along x, y and z directions.
An inertial frame of reference may be defined as a frame of reference in which Newton’s Laws of
mechanics are valid. Such a frame of reference is also referred to as Newtonian or Galilean frame of
reference. All frames of reference, moving with a constant velocity with respect to an inertial frame
are also inertial frames of reference. The inertial frames are non-accelerating frames.
(ii) Non-inertial Frame of Reference : A frame of reference in which Newton’s laws of motion are not
valid is referred to as a non-inertial frame of reference.
(3) GALILEAN RELATIVITY
Let us consider a frame of reference S which is inertial. There is another frame of reference S’ with axes x’, y’,
z’ parallel to the axes x, y, z of the frame S. Let the frame S’ is moving with a uniform velocity v along the xaxis. Let at t = 0 the two frames coincided. Let at any time t, the x-coordinate in S exceeds that in S’ by vt.
Therefore, the observed coordinates in the two frames are given as:x'  x  vt 

y'  y 
z '  z 
(1)
Also, in Newtonian mechanics, t’ = t.
The above four equations are known as Galilean transformations equations.
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The inverse transformation equations are given by :
x  x' vt 
y  y ' 

z  z' 
t  t ' 
(2)
Differentiating equation (1), we obtain velocity transformation equations as:-
u x'  u x  v 

u 'y  u y 
u z'  u z 
(3)
Thus, velocity is not invariant in Galilean transformation.
Differentiating equation (3) we get:-
a x'  a x 

a 'y  a y 
a z'  a z 
(4)
Thus, the acceleration is invariant in Galilean transformation.
According to Newton’s second law F = ma we get:-
Fx  max'  max'  Fx' 

Fy'  Fy

'

Fz  Fz

(5)
Thus, the Newton’s law is valid in both S and S’ i.e., both S and S’ are inertial frames.
Therefore, the inertial frames are also known as Galilean frames.
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(4) LUMINIFEROUS ETHER HYPOTHESIS
 Previously, it was assumed that all the waves known to mankind were mechanical waves which require
a material medium for their propagation. Also, a medium called ether filled the entire universe, due to
which light waves could travel to us from the distant stars.
 It was assumed that all planets, stars and heavenly bodies move through ether which offers no resistance
to their motion. Even the empty spaces within atoms and molecules was assumed to be filled with ether.
All the efforts to measure the weight of the ether failed and so it was supposed to the weightless.
 The formulation of Maxwell’s equations in the year 1862, showed that electromagnetic waves
propagate through the free space with the speed of light c.
 Hertz later generated and detected electromagnetic waves in the laboratory. It was then confirmed that
light is an electromagnetic wave.
 In order to establish the existence of ether, Michelson and Morley conducted an experiment. The
negative result of this experiment ruled out the existence of ether.
(5) MICHELSON-MORLEY EXPERIMENT
Michelson devised an apparatus to detect the motion of the earth relative to ether at rest. It was expected that
light would propagate with different speeds in different directions as viewed from the earth. Therefore, the time
taken by light to travel equal distance in different directions would be different. The objective of the experiment
was to find this time difference from which the relative velocity between ether and the earth could be estimated.
Let a beam of light from a monochromatic source S falls upon a half silvered glass plate P1 inclined at 45o to
the direction of propagation of light. It is partly reflected and partly transmitted. The reflected ray travels
towards mirror M2, gets reflected at B and retraces its path and finally enters the telescope T. Another part of
ray gets refracted and transmitted through P1. The ray gets reflected by mirror M1, retraces its path and finally
meets the other ray in the telescope. Depending upon the path (or phase) difference between the two rays, bright
or dark interference fringes are formed.
The time taken by light in the to and fro journey to the mirror M1 is :
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l
l
l c  v   l1 c  v 
2l c
t1  1  1  1
 2 1 2 
2
2
cv cv
c v
c v
l
21
c
v2
1 2
c
(1)
Similarly, the time taken by light in the to and fro journey to the mirror M2 is given as:-
x
l 2  x2
x2
v2
 2
 2

v
c
l2  x 2 c 2
x2
v2
or, 1  2

1

l2  x 2
x2
or,
l22
v2

1

l22  x 2
c
or,
l22  x 2 
l22
1
(2)
v2
c2
l22  x 2

The time taken by the light ray to cover this distance =
c
l2
c 1
v2
c2
Hence, the time for the to and fro motion of the second ray :
t2 
2l2
(3)
v2
c 1 2
c
Interference fringes can be produced due to difference between t 1 and t2. If mirrors M1 and M2 are exactly
perpendicular to each other, the interference fringes will be concentric circles. Michelson-Morley kept the angle
between the mirrors slightly less than 90o so that the fringes were parallel lines as are obtained from a wedge.
From equations (1) and (2):



2
l1

t  t1  t2 
 l2 
2 
2

v
v
c 1  2  c 1 2

c 
c

(4)
By rotating the interferometer through 90o we get:_
t1' 
2l1
c 1
v2
c2
and t2' 
2l2
 v2 
c1  2 
 c 
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



2
l2 
'
'

t '  t1  t2 
l1 
v2 
v2 
c 1 2 
1 2 
c 
c 
(5)
The time difference is given by:



2l1  l2   1

t ' t 

1


v2 
v2

c 1 2  1 2
c 
c

Using Binomial expension and leaving the higher order terms we get:-
t 't 
2
1
l1  l2  v 2
c
c
(6)
v2
The corresponding path difference ct ' t   l1  l2  2 .
c
Hence, the shift in the number of fringes is : n 
l1 = l2 = l, then n  
If
2l v 2
 c2
l  l  v 2
c t ' t 
 1 2 2

 c
or, n 
2l v 2
 c2
Substituting the orbital velocity of the earth v = 3 x 104 m/s, so that
wavelength  = 6000 Å = 6 x 10–7 m:-
(7)
v2
= 10–8, l = 1.2 m and for the light of
c2
n
 0.04
n
Although, the instrument was capable of measuring such a small shift, no fringe shift was actually observed.
Michelson improved the accuracy of measurement and repeated the experiment by substituting l1 + l2 = 22 m, 
= 5500 Å. The fringe shift worked out to be 0.4
Again no fringe shift was observed.
To avoid the effect of vibrations on the instrument and the stress produced by rotating the arm, they fixed the
instrument to a stone slab and floated it in mercury. In subsequent years, they repeated the experiment many
times with more accurate instruments, but still no fringe shift could be detected.
Conclusions of Michelson-Morley experiment :
(i)
The negative result of this experiment proved that the space or medium in which light propagated is
not moving relative to the earth.
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(ii)
Ether has no observable properties and there is nothing like absolute space or a fixed fundamental
frame of reference with respect to which absolute motion of the bodies can be determined. In other
words, absolute motion is meaningless.
(iii)
The velocity of light is same in all directions and is independent of the relative motion of the source
or the observer or both.
(6) EINSTEN’S SPECIAL THEORY OF RELATIVITY
Two postulates of Einstein’s special theory of relativity:(i) First postulate:-All laws of physics remain invariant in all inertial or Galilean frames. In other words, the
fundamental laws of physics are same in all inertial frames of reference. All inertial frames are equivalent and
no scientific experiment will be able to find a difference between them.
(ii) Second postulate:- The velocity of light in free space is constant and independent of the direction of
propagation as well as the relative velocity between the source of light and the observer.
Lorentz Transformations.
Any physical event can be specified by space time coordinates (x, y, z, t). Lorentz transformations corelate
these coordinates in two inertial frames. Let there are two observers O and O’ in two frames of reference S and
S’ respectivly such that frame S’ is moving with uniform velocity v along the x-axis. Let initially the two
coordinate frames coincident. Let a light signal be emitted at this instant of time. For both the observers, signal
develops into a spherical wavefront with its point of origin as centre. For observer O, the equation of spherical
wavefront is:x2 + y2 + z2 – c2t2 = 0
(1)
Similarly, for the observer O’ in the frame S’ the equation of spherical wavefront is:x’2 + y’2 + z’2 – c2t’2 = 0
(2)
From (1) and (2):x’2 + y’2 + z’2 – c2t’2 =  (x2 + y2 + z2 – c2t2)
(3)
where  is any constant.
Also, y = y’ and z = z’
(4)
For  = 1, we have:- x’2 – c2t’2 = x2 – c2t2
The linear and simplest transformation between x and x’, for v << c, is given as:x’ = (x – vt)
(5)
where constant  is independent of x and t.
For frame S moving with relative velocity – v along x-axis:Using (5) in (6) we get:-
x = (x’ + vt’)
(6)
x = [(x – vt) + vt’]
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 x
1 
or t '   t  1  2 
 v   
(7)
 x
1 
x  c t    x  vt   c  t  1  2 
 v   
2
Using (5) & (7) in ((3) :-
2 2
2
2
2
2 2
2
 x
1 
or, x  c t   x  2vxt  v t  c  t  1  2   0
 v   
2
2 2
2

2
2 2

2 2
2
 2 2 xt 
1  x2 
1  
or, x  c t   x  2vxt  v t  c  t 
1    1     0
v  2  v 2  2  

2
2 2
2

2
2 2

2 2
In order to satisfy this identity, the coefficient of various powers of x and t should vanish separately.
Here equating coefficient of xt to zero, we get:-
22v 
2c 22 
1
1  2   0
v   
or, (v2 – c2) 2 + c2 = 0
or,  
1
(8)
v2
1 2
c
Putting the value of  in equation (5), we get
x' 
x  vt
v2
1 2
c
(9)
From equation (7), we have :
 x
1 
t '   t  1  2   
 v   
 x  c 2  v 2 
 xv 2 
1



t

1



t  2 

c 2 
v 2  vc 
v2  v 
1 2
1 2
c
c
1
vx
c2
or, t ' 
v2
1 2
c
t
(10)
Combining equations (9), 10.16 and (10), we have
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x' 
x  vt 

v2 
1 2
c 
y'  y 
z'  z 



vx 
c2 
t'

2
1 v 2 
c 
t
(11)
These are called Lorentz transformation equations.
Inverse Lorentz Transformations :
The transformation equations relating coordinates in the frame S’ (x’, y’, z’, t’) to that in the frame S (x, y, z, t)
are given by:x ' vt ' 

v2
1 2 
c 

y  y'


z  z'

vx'
t ' 2 
c 
t
v2 
1 2 
c 
x
(12)
If v << c, i.e., Lorentz transformations reduce to Galilean transformations.
x’ = x – vt
y’ = y
z’ = z
t’ = t
APPLICATIONS OF LORENTZ TRANSFORMATION
(i) Length Contraction
Let a rod be placed along the x-axis of the frame S with coordinates of its end points being (x1, 0, 0) and (x2, 0,
0). So that its length is given by:- l0 = x2 – x1
For an observer on the frame S’, moving with uniform velocity v, the length of the rod will be measured as l =
x’2 – x’1, where x’1 and x’2 are the end points (x’1, 0, 0) and (x’2, 0, 0) of the rod in the frame S’.
Using Lorentz transformations, we have
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x1 
x '1 vt '
and x2 
2
1
v
c2
x'2  vt '
1
v2
c2
x'2  x'1
 x2  x1 
v2
1 2
c
l
or, l0 
l
or, l  l0 1 
v2
c2
v2
c2
So, the observer S’ finds the rod AB in the frame S to be contracted. This is called Lorentz contraction.
Reciprocity of Length Contraction : Let a rod be placed at rest in the frame S’ with space coordinates (x’1,0,
0) and (x’2, 0, 0), so that length of the rod measured by the observer O’ in the frame S’ is l0 = x’2 – x’1
For an observer O in frame S measures the length of the rod as:-
x'1 
x1  vt
2
1
v
c2
and x'2 
x2  vt
1
v2
c2




1
x2  vt  x1  vt  x2  x1


or, l0  x'2  x'1 

2
2
2 
v
v2
v
v 

1

1

1 2  1 2 
c2
c2
c
c 

or, l  l0 1 
v2
c2
Thus, rod appears to be contracted. If the rod moves with the velocity of light, its length reduces to zero.
(ii) Time Dilation Relativity of Time : Let a clock is placed at a point in the frame S and another at a point in
the frame S’. According to Lorentz transformation :
v
x
2
c
t
v2
1 2
c
t '
Let a light signal is emitted at a point x’ at time t’1 and another at the same location at time t’2 in the frame S’.
The corresponding times recorded by the clock in the frame S is given by :
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v
v
t '2  2 x'
t '1  2 x'
c
c
t1 
and t2 
2
v
v2
1 2
1 2
c
c
vx'
vx '
t '1  2
2
c 
c
 t  t 2  t1 
2
v
v2
1 2
1 2
c
c
t '2 
or, t 
t '2 t '1
1
Since

v2
c2
1
t '
1
v2
c2
> 1,  t’ >  t.
v2
1 2
c
(iii) Transformation of Velocities
Let the velocity in the frame S be u (ux, uy, uz) and that in S’ be u’ (u’v, u’y, u’z). Then :
ux 
dx
dx'
and u 'x 
dt
dt '
From Lorentz transformations equation:-
dx' 
dx  vdt
v2
1 2
c
v
dx
c2
v2
1 2
c
dt 
, dy '  dy , dz '  dz and dt ' 
Dividing dx’, dy’, dz’ by dt’, we get :
dx' dx  vdt

dt ' dt  v dt
c2
dy 1 
v2
c2
dy '

dt ' dt  v dx
c2
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v2
dz 1  2 dx
dz'
c

v
dt '
dt  2 dx
c
Dividing the numerator and denominator of the RHS by dt, we get:





v2 
uy 1  2
c 
u' y 

v
1  2 ux 
c

v2 
uz 1  2 
c 
u'z 
v
1  2 u x 
c

u v
u'x  x
v
1  2 ux
c
The inverse transformations are :


u ' x v
ux 

v
1  2 u'x 
c

2 
v
u'y 1  2 
c 
uy 

v
1  2 u 'x 
c

2 
v
u 'z 1  2 
c 
uz 
v
1  2 u 'x 
c

For the non-relativistic approximation
v
<< 1,
c
ux = u’x + v
uy = u’y
uz = u’z
which are the Galilean transformation.
(iv) Relativity of Simultaneity
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Let two events occur simultaneously at the instant t at x = x1 and x = x2 in the frames S. The times of
occurrences in the frame S’ at the two locations x1 and x2 respectively are :
vx1
vx
t  22
2
c and t ' 
c
t '1 
2
2
v
v2
1 2
1 2
c
c
t
v
x  x2 
2  1
 t '1  t '2  c
v2
1 2
c
As, x  x2, therefore, t’1  t’2.
(v) Relativity of Mass
Let us consider two identical and perfectly elastic particles of masses m’1 and m’2 in the frame S’, moving with
velocities + u’ and – u’ parallel to the x’ – axis, undergo a head-on collision. After collision, the particles
rebounds with velocities – u’ and + u’ respectively.
Let us observe from the frame S, which is moving with velocity – v relative to S’ along the x-axis. Let the
particles of masses m1 and m2 have velocities u1 and u2, before the collision with respect to the frame S. At the
instant of collision applying the law of conservation of momentum, we get:m1u1 + m2u2 = M v
Using the relation, u x 
Particle 1 : u1 
Particle 2 : u2 
where m1 + m2 = M
(1)
u 'x  v
for
vu 'x
1 2
c
u ' v
vu '
1 2
c
 u ' v
vu '
1 2
c
(2)
(3)
Substituting (2) & (3) in (1) we get:



 u ' v 
  u ' v 
  m2 
  m1  m2 v
m1
 1  uv 
 1  vu ' 




c2 
c2 


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



 u ' v


u ' v 

or m1
 v   m2  v 
 1  uv '

 1  vu ' 




c2
c2 



vu'
1 2
m1
c

m2 1  vu '
c2
(4)
Using (2) and (3) :-


2
2
2
2
2
2
2
u ' v 2c  u1  u2  2 c  u1 c  u2

c2
u1  u2

(5)
Using (5) in (4) we get:u22
m1
c2

m2
u12
1 2
c
1
(6)
If u2 = 0, then:-
m1 
m2
(7)
u2
1  12
c
Using m1 = m, m2 = m0 and u1 = v. we get:-
m
m0
(8)
v2
1 2
c
(vi) Relativistic Momentum and Force
Momentum of a particle is given by:-
p  mv 
mo v
1
According to Newtons Law of Motion :-
For DTU by:- Dr. Sameer Jain (9811153371)
v2
c2




dp d  m0v 
m0
dv
F

 
3

2
2
2 dt
dt dt
 1  v 2  1  v 
c   c 2 

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For a partical of zero mass:
p
m0 v
1
v2
c2
(vii) EINSTEIN’S MASS-ENERGY RELATION
According to work-energy theorem, the kinetic energy of a moving body is equal to the amount of work done
The work done by the force for a small displacement ds is given as:u
W   F .ds
0
Kinetic energy of the body Ek is :
v
Ek  W   F .ds
0
v
v
  F.
0
ds
dt  F .vdt
dt
0
v
dp
.vdt
dt
0
dp 

 F 

dt 


v
 v
0
d
mvdt
dt
 p  mv
v
  v.d mv 
0
m0
As we know m 
v2
1 2
c
.




v
  v 2  12 
 m0

Ek   v.d 
v   m0  vd v1  2  
2
  c  
v
0
0
 1  2 


c 

v
Therefore,
3
 v 2  12
 v 2  2  2v  
1
 m0  v 1  2  dv  v. 1  2    2 dv 
 c 
2 c   c  
0


v
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14
v
 v2 
 m0  v1  2 
c 
0 
v
 m0 
0
3
2
 v2 v2 
1  c 2  c 2  dv


dv
3
 v2  2
1  2 
 c 
Substituting 1 
v2
2vdv
 y , so that  2  dy , we get
2
c
c
 c2 
1 dy
2
Ek  m0   3
y 2
1
v2
1 2
c
 1 
m0 c  y 2 



2 1 
 2
1
 v  0, y  1 

2
 v  v, y  1  v 

c 2 
v2
c2
2




2
1
  m0c  m0c 2
m0c 2 

1


v2
v2
1

1



c2
c2


or Ek = (m – m0)c2
This is a relativistic expression for kinetic energy.
The total energy of the body is the sum of the rest energy and the relativistic kinetic energy of the body, i.e. E =
rest energy + relativistic kinetic energy
= m0c2 + (m – m0)c2
= mc2
Non-relativistic case : The relativistic kinetic energy :
Ek = (m – m0)c2




 m0


 m0 c 2
2
 1  v 2

c


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1
2


 v2 

 1  2   1 m0c 2
 c 



In non-relativistic cases, v <<c, Using bionomial expension and leaving higher order terms:
 1 v2 3 v4

Ek  1 
 4  ... – 1 m0c 2
2
8c
 2c

1 v2
1
Ek  . 2  m0c 2  m0v 2
2 c
2
(vii) MOMENTUM-ENERGY RELATION
As we know:-
p  mv 
m0v
1
v2
c2
The total energy of this particle is E  mc 2 
m0c 2
1
v2
c2
 E2 
m02c 4
v2
1 2
c
Putting the above values in the expression p2c2 + m02 c4, we get
p 2 c 2  m02 c 4 

m02v 2c 2
m02 v 2c 2  m02c 4  m02 v 2c 2
2 4

m
c

0
v2
v2
1 2
1 2
c
c
m02c 4
 E2
v2
1 2
c
 E 2  p 2c 2  m02c 4
or, E 
p 2c 2  m02c 4
This is Einstein’s relation between momentum and energy,
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