Unit - V Special Theory of Relatuivity

8/9/2019 Unit - V Special Theory of Relatuivity

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Mr. A.CHARIS ISRAEL. M.Sc., B.Ed., (Ph.D.)

Sr. Lecturer of PHYSICS

Mobile No: +91-9269680853



PHYSICS-I UNIT –V SPECIAL THEORY OF RELATIVITY 2009-2010

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PHYSICS - I

UNIT – V

SPECIAL THEORY OF RELATIVITY

Postulates of Special Theory of Relativity :

In 1905, Einstein propounded the special theory of relativity which makes a conceptual change in the

ideas of space and time. According to Newtonian mechanics, time and space are invariant for all observers,

while both are variable at high speeds for various observers in Special theory of relativity.

The two fundamental postulates of the Special theory of Relativity are:

(i) “All the laws of physics have the same form in all inertial frames of reference which are moving with constant velocity”. This postulate is just the principle of Relativity.

(ii) “The speed of light is constant in vacuum in every inertial frame of reference”. This postulate is an

experimental fact and asserts that the speed of light does not depend on the direction of propagation in

vacuum and the relative velocity of the source and the observer.

These two postulates of special theory of relativity look to be very simple, but they have revolutionized

the physics with far reaching consequences.

LORENTZ TRANSFORMATIONS

Suppose that S and S’ be two inertial frames of reference. S’ is moving along positive direction of x-axis

with velocity of ‘v’ relative to the frame S.

Now suppose that a source of light situated at the

origin O in the frame S emits a wavefront at t = 0.

(Assume that the origins O and O’ of the two coordinate

systems coincide at t = 0). When wavefront reaches the

point P, let the positions and times, measured by the

observers O and O’ be (x,y,z,t) and (x’ ,y’ ,z’ ,t ’ ) respectively.

If c is the velocity of light, then the equations

representing the wavefront in frame S and frame S’ are

x2+y

2+z

2=c

2t 2 (1)

x’ 2+y’ 2+z’ 2=c2t ’ 2 (2)Now the transformations relating x,y,z,t and x’ ,y’ ,z’ , t’

should be such that eqn(2) transforms to eqn(1).

Let us assume the transformations of x’ ,y’ ,z’ and t’ are as

x’ =α (x-vt), y’ =y, z’ =z and t ’ =α ’ (t+fx) (3)

where α, α’ , f are constants which are to be determined.

Now substitute x’ ,y’ ,z’ and t’ in eqn(2), we have

α 2(x-vt)

2+y

2+z

2=c

2α ’

2(t+fx)

2

2(x

2+v

2t 2-2xvt)+y

2+z

2=c

2 ’

2(t

2+f

2 x

2+2tfx)

2

x2

+ 2v

2t 2-2xvt

2+y

2+z

2=c

2 ’

2t 2+ c

2 ’

2 f

2 x

2+ 2c

2 ’

2tfx

x2(

2- ’

2c

2 f

2)+y

2+z

2-2xt(v

2+c

2 ’

2 f)=c

2t 2( ’

2-

2 2

2

v

c

α ) (4)

Comparing eqn(4) with eqn(1), we have

2- ’

2c

2 f

2=1 (5) vα

2+c

2 ’

2 f=0 (6) α ’

2-

2 2

2

v

c

α =1 (7)

eqn(6) ⟹ α ’ 2=1+

2 2

2

v

c

α (8)

eqn(5) ⟹ f =

2

2 2'

v

c

α

α

− (9)

S 'S

O'O

P

( ), , , x y z t

( )', ', ', ' x y z t

X

Y

Z

'Y

' X

' Z

v






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PHYSICS - I

substitute eqn(9) in eqn(5), we have

α 2

2 2 2 4

4 4

'

'

c v

c

α α

α

− =1 ⇒ α 2

2 4

2 2'

v

c

α

α

− =1 ⇒ α 2

2 2

2 21

'

v

c

α

α

−

= 1 ⇒

2 2

2 2 2

11

'

v

c

α

α α

− = (10)

substitute eqn(8) in eqn(10)2 2

22 22

2

11

1

v

vc

c

α

α α

− =

+

⇒2 2

22 2 22

2

11

v

c vc

c

α

α α

− =

+

⇒2 2 2 2 2

2 2 2 2

1c v v

c v

α α

α α

+ −=

+

⇒2 2 2

2

2

c v

c

α α

+= ⇒

2 22

21

v

c

α α = + ⇒

22

21 1

v

cα

− =

⇒ 2

2

2

1

1v

c

α =

−

(11)

Substitute eqn(11) in eqn(8)

α ’2=1+

2

22

2

1

1

v

cv

c

−

⇒

2 2

2 22

2

2

1

'

1

v v

c c

v

c

α

− +

=

−

⇒ 2

2

2

1'

1v

c

α =

−

(12)

substituting eqn(11) and eqn(12) in eqn(9), we get

f = 2

vc

− (13)

Now substitute eqn’s 11, 12 & 13 in x’ and t ’ of eqn(3) we get the transformations as

2

2 2

2 2

' , ' , ' '

1 1

vxt

x vt c x y y z z and t v v

c c

−−

= = = =

− −

(14)

The above eqn(14) represents the Lorentz Transformations for space and time coordinates.

Inverse Lorentz Transformations:

In the above derivation, we assumed that frame S’ is moving in positive X-direction with velocity v relative to the frame S. But if we observe S from S’ , S appears to move with ‘-v’ velocity relative to S’ along

negative X-direction, then the

transformations are

These are known as Inverse Lorentz transformations.

CONSEQUENCES OF LORENTZ TRANSFORMATIONS

(1) Relativity of Space (OR) Length Contraction (OR) Lorentz-Fitzerald Contraction:

Consider two systems S and S’ among which S’ is

moving with velocity v relative to S along positive direction

of X-axis. Let a rod of length ‘lo’ be placed in system S’ . Let

the x-coordinates of the two ends of this rod in frame S’ be

1 2' ' x and x . Then the length of the rod in frame S’ is

1 2' '

ol x x= −

(1)

Let an observer O be located in system S who

measures the length of the moving rod. Let x1 and x2 be the

S 'S

O

'O X

Y

Z

'Y

' X

' Z

v

1 2' ' x x

1 x2 x

0l

2

2 2

2 2

''

' ', ' , '

1 1

vxt


c c

++

= = = =

− −






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PHYSICS - I

coordinates along x-axis of the two ends of the rod measured simultaneously at an instant t .

Thus length of the rod measured by an observer in frame S is 2 1l x x= −

(2)

Now from Lorentz transformations, 1 21 2

2 2

2 2

' '

1 1

x vt x vt x and x

v v

c c

− −= =

− −

(3)

Substituting the values of 1 2' ' x and x from eqn(3) in eqn(1), we get

( )( )2 12 1

2 2 2 2

2 2 2 2

(2)

1 1 1 1

o o o

x x x vt x vt ll l l from eqn

v v v v

c c c c

−− −

= − ⇒ = ⇒ =

− − − −

2

21

o

vl l

c∴ = −

(4)

The above eqn(4) shows that l < lo. Thus the length ‘l’ of the rod measured by the observer O from

frame S is smaller than the proper length ‘lo’ measured by an observer O’ from the frame S’ .Such a contraction of length in the direction of motion relative to an observer is called Lorentz-

Fitzgerald Contraction.

(2) Time Dilation:

Let a frame S’ be moving along X-axis with velocity v relative to S. Let a clock be placed in frame S which is at rest. Let the distance of clock from O is x1. Suppose that this clock gives a first signal at time t 1 in S and

t 1’ is the time measured by the observer in S’ simultaneouslycorresponding to the time t 1.

Then from Lorentz transformation equations, we have

11 2

12

2

'

1

vxt

ct v

c

−

=

−

(1)

Suppose that the same clock gives its second signal at

time t 2 measured by an observer in S and t 2’ measured by an observer in S’ , then1

2 2

22

2

'

1

vxt

ct v

c

−

=

−

(2)

Thus the time duration between the two clock signals measured from S is t o= t 2 – t 1 (3)

Similarly, the time duration measured by the observer from S’ is t = t 2’ – t 1’ (4)

Substituting the values of t 1’ and t 2’ from eqn’s 1&2 in eqn(4), we get

( )1 1 1 1

2 1 2 12 2 2 22 1

2 2 2 2

2 2 2 2

1 1 1 1

vx vx vx vxt t t t t t c c c ct t t

v v v v

c c c c

− − − − +−

= − ⇒ = ⇒ =

− − − −

(5)

Hence from eqn’s(3)&(5), we have 0

2

2

1

t t

v

c

=

−

(6)

From eqn(6), it is clear that the time duration t measured by the moving observer will be dilated or lengthened

by a factor2

2

1

1v

c−

where v is the velocity of the moving observer.

S 'S

O

'O X

Y

Z

'Y

' X

' Z

v

1 x

clock






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PHYSICS - I

(3) Relativistic addition of velocities:

Let the coordinates of a particle in frame S be (x,y,z,t) and in frame S’ is (x’,y’,z’,t’). Now components

of velocity of the particle in two frames can be written as

u x = dx/dt, u y = dy/dt and u z = dz/dt in frame S (1)

and u x’ = dx’ /dt ’ , u y’ = dy’ /dt ’ and u z’ = dz’ /dt ’ in frame S’ (2)

According to the Inverse Lorentz transformations,

2

2 2

2 2

''

' ', ' , '

1 1

vxt


c c

++

= = = =

− −

(3)

The differentials of eqn(3) can be written as

2

2 2

2 2

''

' ', ' , '

1 1

vdxdt

dx vdt cdx dy dy dz dz and dt v v

c c

++

= = = =

− −

(4)

Substitute eqn(4) in eqn(1), then

( )

( )( )

( )( )2 2

2 22 2

2

'' ' 1 ' ' '

' ' '' 1 ' 1

'

x

dx vdx vdt v c dx vdt dx dt u

vdt vdx vdx dxdt v c dt c c

dt c

++ − +

= = = =

+ − + +

( )

2

'

1 '

x

x

x

u vu

vu

c

+

∴ =

+

(5)

Also,

( )2 2 2 2

22

'' '1 1

' '' 1'

y

dydy dy dt u v c v c

vdt vdx dxdt c dt c

= = − = −

+ +

2 2

2

'1

1 '

y

y

x

uu v c

vu

c

= −

+

(6)

Similarly,2 2

2

'1

1 '

z z

x

uu v c

vu

c

= −

+

(7)

This is the relativistic law of addition of velocities.

In eqn(3), we have used inverse Lorentz transformation equations and arrived at eqn’s(5),(6)&(7).

If we use Lorentz transformations, we can prove that

( ) 2 2 2 2

2 2 2

' , ' 1 and ' 1

1 1 1

y x z x y z

x x x

uu v uu u v c u v c

v v vu u u

c c c

−

∴ = = − = −

− − −

(8)

Special case: If the particle is photon and is moving with a velocity c in the frame S’ and also S’ is moving with

velocity of light c relative to S along positive x-axis, then from eqn(5) we have i.e., u x’= c, v = c then

( )2

2

2

21

x x

c c cu u c

c

c

+

= = ⇒ =

+

From the above result it is clear that, the speed of the photon in frame S is also equal to c. Therefore,

velocity of light is the same for all inertial frames.






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PHYSICS - I

(4) Relativity of Mass (Variation of mass with velocity ):According to the Newtonian mechanics, the mass of a body

does not change with velocity i.e., it is independent of velocity. But

according to Einstein, the mass of the body in motion is different

from the mass of the body at rest i.e., mass is a function of velocity

of the body.

“The mass increases with increased velocity” according to

the relation

0

2

21

m

m v

c

=

−

where mo is the rest mass of the body, ‘c’

is the velocity of light and ‘v’ is the velocity of the body.

Let us prove the above statement.

Consider two coordinate systems S and S’ among which S’ is moving with a velocity v. Now, let us

consider two bodies which undergo collision in system S’ and it is observed from the system S. Suppose that

the two bodies have masses m1 and m2 and are moving with velocities u’ and –u’ along positive X-axis in the

system S’ . During this collision, we suppose that two bodies coalesce into one body after collision. So,

according to conservation of mass, the mass of the coalesced bodies after collision is equal to (m1+m2). As the

two bodies were moving with the same velocity in opposite directions, hence after collision they are at rest in

system S’ .

Now, let us see how this collision experiment appears to an observer in system S. Using the law of addition of velocities, the velocities u1 and u2 in system S are given by

1 2

2 2

' '

' '1 1

u v u vu and u

u v u v

c c

+ − += =

+ −

(1)

where u1 and u2 are the velocities of the bodies m1 and m2 respectively in system S.

Applying the principle of conservation of momentum, we have

m1 u1 + m2 u2 = (m1+m2)v (2)

substituting the values of u1 and u2 from eqn(1) in eqn(2), we get ( )1 2 1 2

2 2

' '

' '1 1

u v u vm m m m v

u v u v

c c

+ − ++ = +

+ −

1 1 2 2 1 2

2 2 2 2

' ' ' '

' ' ' '1 1 1 1

u v u v u v u vm m v m v m m v m v

u v u v u v u v

c c c c

+ − + + − +

− = − ⇒ − = − + − + −

2 2

2 2 21

1 2

22 2 2

' ' '' ' 1

' ' '

1 1 1

u v u v u vu v v v u v

mc c cm mu v u v u vm

c c c

+ − − − + − +

= ⇒ = + − −

(3)

From eqn(1),

2 2 22 2

2 22 1

1 2 22 2

2 22 2

' ' '1

' 1 '

1 1 1' ' ' '1 1 1 1

u v u v u v

uu v u v c c c

u u v u vc c u v u vc c c c

+ + + − + +

= ⇒ − = − = − = + ++ +

2 2 2 2 2

2 22 2 2 21

2 22

2 2

'' ' ' '1 11 2 2

1' '

1 1

u vu v u v u v u v

c cu c c c c c

c u v u v

c c

− −+ + − − −

⇒ − = =

+ +

S 'S

O 'O

P

X

Y

Z

'Y

' X

' Z

v

1m2m

'u 'u−






Page 7 of 9

PHYSICS - I

122 2 2 2

2 2 2 2 2

2 22 2

1 1

2 2

' '1 1 1 1

' '1 1

1 1

u v u v

c c c cu v u v

c cu u

c c

− − − −

+ = ⇒ + =

− −

(4)

Similarly,

122 2

2 2

2 2

2

2

'1 1

'1

1

u v

c cu v

c u

c

− −

− =

−

(5)

Substitute eqn’s(4)&(5) in eqn(3), we get

12 2

2

2

1

12 2

21

2

1

1

u

cm

m u

c

−

=

−

(6)

Let us consider that the body of mass m2 be moving with zero velocity in system S before collision i.e.,

u2=0, then

1

22

1

2

1

1

m

m u

c

=

−

(7)

For commonly used notation, m1=m, m2=m0 and u1=v

0

2 20

2 2

1

1 1

mmm

m v v

c c

∴ = ⇒ =

− −

(8)

where 0m is called as the rest mass and m is the effective mass.

The above eqn(8) is the relativistic formula for the variation of mass with velocity.

Case(i): When v << c,

2

2

v

c will be negligible in comparison to unity, therefore m=m0.i.e., at velocities which are much smaller than the velocity of light, the mass of the moving object is the same

as its rest mass.

Case(ii): When v is comparable to c, then2

21

v

c

−

will be less than unity or m>m0

i.e., at velocities which are comparable to velocity of light, the mass of the moving object appears to be

greater than at rest.

Case(iii): When v=c or v>c, then2

2

v

cwill be equal to or greater than unity, so m = ∞ or imaginary.

i.e., at velocities which are equal to or greater than c, the mass of the object becomes infinite or imaginary,

which is a nonsense concept.






Page 8 of 9

PHYSICS - I

EINSTEIN’S MASS-ENERGY RELATION

( Equivalence of mass and energy)

According to classical mechanics, the force is the rate of change of momentum,

i.e., ( )d

F mvdt

= (1)

But according to theory of relativity, the mass as well as velocity are variable, thus

dv dmF m v

dt dt = + (2)

When a particle is displaced through a distance dx by the application of a force F , then the increase in

kinetic energy dk is given by dk = F dx (3)

Substituting the value of F from eqn(2) in eqn(3), we get

2

dv dm dx dx dxdk m dx v dx dk m dv v dm dk mvdv v dm v

dt dt dt dt dt

= + ⇒ = + ⇒ = + =

(4)

But, the variation of mass with velocity is given by

0

2

21

mm

v

c

=

−

(5)

Squaring eqn(5) on both sides, we get2 2

2 2 2 2 2 2 2002 2

m c

m m c m v m cc v

= ⇒ − =

−

(6)

Differentiating eqn(6), we get

( )2 2 2 2 2 2 22 2 2 0 2 0 0mc dm mdmv vm dv m c dm v dm vmdv c dm v dm vmdv− − = ⇒ − − = ⇒ − − =

2 2c dm v dm mvdv= + (7)

Comparing eqn’s(4)&(7), we have 2dk c dm= (8)

Now consider that the body is at rest initially and by the application of force it acquires a velocity v.

Hence the mass of the body increases from m0 to m. Therefore the total kinetic energy acquired by the body is

given by

0

2 2

0( )

m

m

dk c dm K c m m= ⇒ = −∫ ∫ (9)

Eqn(9) gives the increase in kinetic energy due to the increase in mass.

Case(i): The total energy of a moving particle is the sum of its kinetic energy and the energy at rest. Thus total

energy, E = K + m0 c2

(10)

Substituting the value of K from eqn(9) into eqn(10), we have

E = c2(m-m0) + m0 c

2= c

2m - c

2m0 + m0 c

2

2 E mc∴ = (11)

The above eqn(11) gives the universal equivalence between mass and energy and is called as “Einstein’s

Mass-Energy relation”.Case(ii): Relativistic Kinetic Energy:

Substituting eqn(5) in eqn(9), we get

12

2 200 0 2

2

2

1 1

1

m vK c m K m c

cv

c

−

= − ⇒ = − −

−

(12)

The above eqn(12) represents the Relativistic Kinetic Energy of the body.






Page 9 of 9

PHYSICS - I

ENERGY – MOMENTUM RELATION

Consider a body of mass m is moving with velocity v and having a rest mass m0. Then the total

relativistic energy E is given by2

2 0

2

21

m c E mc

v

c

= =

−

(1)

And the relativistic momentum p is given by0

2

21

m v p mv

v

c

= =

−

(2)

Multiplying eqen(2) with c and squaring on both sides, we get2 2 2

2 2 0

2

21

m c v p c

v

c

=

−

(3)

Squaring eqn(1), we have2 4

2 02

21

m c E v

c

=

−

(4)

Subtracting eqn(3) from (4), we have

( )2 2 2 22 4 2 2 202 2 2 2 40 0

02 2 2

2 2 2

1 1 1

m c c vm c m c v E p c m c

v v v

c c c

−

− = − = =

− − −

2 2 2 2 4

0 E p c m c⇒ = + (5)

The above eqn(5) is the relativistic relation between total energy E and momentum p of a body.

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Unit - V Special Theory of Relatuivity