Special theory of relativity:Michelson-Morley experiment,Galilean and Lorentz Tranformation,Length contraction,Time Dilation,Relativity of mass,mass-energy equivalence

1. Sarveshwar S.Kasarla
Assistant Professor of Physics
Institute of Science, Nagpur
(An Autonomous Institute)
phy.kasarla@iscnagpur.ac.in, 9890452763, 8329874268
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2. Fow
 Michelson-Morley Experiment
 Special theory of Relativity
 Frame of Reference
 Galilean Transformation
 Postulates of Special theory of Relativity
 Lorentz Transformation.
 Length Contraction
 Time dilation
 Velocity compounding
 Relativity of mass
 Mass energy equivalence
Relativity
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B.Sc. Sem.-VI Physics-I Unit-I

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Michelson Morley
experiment(1887)
B.Sc. Sem.-VI Physics-I Unit-I

4. Contents:
 Introduction
 Luminiferous ether
 Experiment
 Result
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5. The Michelson-Morley experiment was
a scientific experiment to find the presence and properties of
a substance called A ether. The experiment was done by Albert
Michelson and Edward Morley in 1887.
Since waves in water need something to move in
(water) and sound waves do as well (air), it was believed that
light also needed something to move in. Scientists in the 18th
century named this substance "a ether”.
They believed that a ether was all around us and
that it also filled the vacuum of space. Michelson and Morley
created this experiment to try and prove the theory that a
ether existed. They did this with a device called an
interferometer.
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6. What is the Luminiferous ether ?
The Earth travels very
quickly (100,000 km ph)
around the Sun. If a ether
exists, the Earth moving
through it would cause a
"wind" .
In the same way, a ether
should seem like a moving
substance to things on
Earth.
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7. Requirements :
Light Source
Glass plate
Mirror
Telescope
Compensating glass plate
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Semisilvered glass
plate
Mirror 1
Mirror 2
Compensating glass plate is not
shown
A
O B

9. ct
ct
o
O’ B B’
A
vt
p
Reflected ray
Transmitted ray
G G’
A’
E’
telescope
E
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A A’
O
B B’

10. Total time taken by the reflected wave (OA’+ A’O):
Total time taken by the transmitted wave(OB+BO)
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𝑡2 =
𝑙
𝑐 + 𝑣
+
𝑙
𝑐 − 𝑣
=
2𝑙
𝑐(1 −
𝑣2
𝑐2)
𝑡1 =
𝑙
𝑐2 − 𝑣2
+
𝑙
𝑐2 − 𝑣2
=
2𝑙
𝑐 1 −
𝑣2
𝑐2
∆𝑡 = 𝑡1 − 𝑡2 =
2𝑙
𝑐 1 −
𝑣2
𝑐2
−
2𝑙
𝑐 1 −
𝑣2
𝑐2
=
2𝑙
𝑐
1
1 −
𝑣2
𝑐2
−
1
1 −
𝑣2
𝑐2
∆𝑡 =
𝑙𝑣2
𝑐3
Time difference is given by:

11. The whole interferometer is rotated through 90º. Hence
the time difference △t’ between the reflected and
transmitted waves is given by :
∆𝑡′
= −
𝑙𝑣2
𝑐3
Hence, on rotation through 90º the time difference
change by:
∆𝑡 − ∆𝑡′
=
𝑙𝑣2
𝑐3 − (−
𝑙𝑣2
𝑐3 ) =
2𝑙𝑣2
𝑐3
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12. Effective path difference=velocity of light (△t-△t’)
Effective path difference = 𝐜
2𝑙𝑣2
𝑐3
=
2𝑙𝑣2
𝑐2
Fringe shift ,△ N =
effective path difference
wavelength
=
2𝑙𝑣2
𝜆𝑐2
l = 11 m
c = 3 x 108 m/s
ƛ = 5.5 x 10-7 m
v = 3 x 104 m/s
△N=0.4
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13. Michelson Morley tried to observe the fringe shift by
continuously rotating the interferometer during day and
night at different place on the earth. But ,could not observe
any fringe shift. Hence, as there is no fringe shift the result
of the experiment is negative.
The experiment result showed that the speed of light is
same for all observers and for propagation of light in free
space no medium is necessary.
Result:
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15. Special theory of relativity
Einstein has written two separate theory's of relativity
 Special theory of relativity (1905)
 General theory of relativity (1915)
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16.  A system of coordinate axes which define the position
of a particle in two or three dimensional space is
called a frame of reference.
 In a Cartesian system of coordinates , the position of
a particle is specified by its three coordinates x, y , z
along the three perpendicular axes.
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Frame of Reference
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Frame of Reference
phy.kasarla@iscnagpur.ac.in
Inertial frames of reference: The frame of reference in a state of rest or in
a uniform motion with respect to one another.
These are neither accelerating nor rotating with respect to the distant
galaxies(a=0)
Non-Inertial Frames: The frame of reference which is in Rotational motion with
constant angular velocity, acceleration in curved paths(a≠0)
B.Sc. Sem.-VI Physics-I Unit-I

18. Galilean Transformation
<- - - - - - - - X - - - - - - - - - - - ->
Y
Z
-
-
-
--
-
Inertial
Frame
S
Inertial
Frame
S’
V
Y’
V t X’
O O’
Z’
X’
P(x, y, z)
or
(x’ ,y’ ,z’)
X
Let the O’ of the frame S’ move relative to the origin O of the Frame S With
Constant velocity “V” along the Common X-axis . If the two origin coincide at time
t=0 , then the separation (I.e. OO’) after time t is equal to vt.
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Consider two internal frame of references S and S’ as shown in the
figure .
Let the X-axis of two frames lie Along the Same line
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B.Sc. Sem.-VI Physics-I Unit-I

19. <- - - - - - - X- - - - - - - - - - - - - ->
Y
Z
-
-
-
Inertial
Frame
S
Inertial
Frame
S’
V
Y’
V t X’
O O’
Z’
X’
(x, y, z)
or
(x’ ,y’ ,z’)
X
OO’ = vt
Now the point “P” may be Described by the Coordinates (x,y,z) in the
inertial frame S or by the coordinates (x’,y’,z’) in the inertial frames S’
Hence , from the figure these coordinates are related by:
x’= x- vt
y’ = y
z’ = z and t’=t ………(1)
These equations are called as the Galilean Coordinate Transformation
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20. Galilean Transformation
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The Galilean velocity and acceleration transformation
equations by differentiating these equations with respect to
time. (u = dx/dt and a= du/dt)
for the Galilean transformation, the increment of time used in
differentiating to calculate the particle velocity is the same in
both frames, dt=dt′ .
Differentiation yields
ux = u′x+v , uy = u′y , uz = u′z
and
ax = a′x , ay = a′y , az = a′z .
Newton’s laws are invarient (consistent) in all inertial frames
B.Sc. Sem.-VI Physics-I Unit-I

21. Galilean Transformation
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The failure of the Galilean Transformation
The position of a light-wave front speeding from the origin at time
zero should satisfy equation in the frame (x,y,z,t) and equation in the
frame (x’,y’,z’,t′)
x2 +y2 +z2=c2t2 ---i
x’2 +y’2 +z’2 =c2t’2 ---ii
Formula (ii) does not transform into formula (i) using the Galilean
transformations (1).
Maxwell's equations of electromagnetism are not invariant under the
Galilean Transformation
B.Sc. Sem.-VI Physics-I Unit-I

22. (Transforming position and time)
Relativity
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23. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
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24. Postulates of Special Theory of Relativity
 Einstein's Special theory of relativity deals with the problems in which one frame of
reference moves with a constant linear velocity relative to another frame of reference
. The fundamental postulates of special theory of relativity are as follows:
 The laws of physics are the same in all inertial frames of reference.
 The speed of light in free space has the same value in all internal frames
of reference .
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25. LORENTZ TRANSFORMATION
Let S and S’ be two rectangular inertial frame of reference with there axes O’X’ collinear.
Let v=uniform velocity of the frame S’ relative to frame S in positive X-direction.
Suppose the two origins O and O’ coincide at time, t=t’=0 and at that instant a pulse of light is initiated from
the source, Which is at the origin O of the frame S.
X’
Y
S S’ V
Y’
V t
O O’
Z’
X
Since the speed of light is constant and is independent of the motion of the frame
of reference , we have
for the observer at O in frame S : x = ct  t=x/c ----(1)
For the observer at O’ in the frame S’ : x’=ct’ t’=x’/c ----(2)
Z
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26. Since, there is one relative “V” the transformation equation must
satisfy the following conditions:
1. The coordinates of O’ in the frame S is vt.
2. The Coordinates of O in frame S’ is – vt.
The Simplest equation which satisfy these conditions are :
X’= k (X-vt) ……... (3) and
X= k (X’+vt’) .......... (4)
the factor k must be same in both the frames .
as the relative motion of S and S’ is confined to only X-direction :
Y’ = Y and Z’= Z
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27. substituting the value of t and t’ from eq.(1) and (2) in eq.(3) and (4) we
get :
……….(5)
……….(7)
Multiplying eq. (5) and (6) we get :
……….(6)
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28. substituting the value of k in eq.(3) we get : ….(8)
Now, from eq.(2) we get :
…...(9)
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29. Hence , Lorentz Transformation
Equation for space and time
coordinates are :
y’ = y ; z’ = z and
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Also , Lorentz Inverse
Transformation Equation for
space and time coordinates are :
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B.Sc. Sem.-VI Physics-I Unit-I

30. Lorentz Transformation
Example:
 As measured by S, a flash of light goes off at x=100km, y =
10 km, z = 1 km at t = 0.5 ms . What are the coordinates of
x′, y′, z′ and t′ of this event as determined by an observer S′
moving relative to S at -0.8c along the common x-x′ axis?
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( x′ = 367 km, y′=10km, z′=1 km and t′=1.28 ms)

31. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
Length Contraction:
L0=length observed by observer in S‘ frame
L=length observed by observer in S frame
𝐿0 = 𝐴𝐵 = 𝑋2′ − 𝑋1′
𝐿 = 𝑋2 − 𝑋1
𝑋1′ =
𝑋1−𝑣𝑡
1−
𝑣2
𝑐2
𝑋2′ =
𝑋2 − 𝑣𝑡
1 −
𝑣2
𝑐2
𝑋2′ − 𝑋1′ =
𝑋2−𝑋1
1−
𝑣2
𝑐2
𝐿0 =
𝐿
1−
𝑣2
𝑐2
 𝐿 = 𝐿0 1 −
𝑣2
𝑐2
𝐿 < 𝐿0
Y
Z
Inertial
Frame
S
Inertial
Frame
S’
V
Y’
V t
O
O’
Z’
X’
A(x1’, y’, z’) B (x2’ ,y’ ,z’)
X
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32. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
Time Dilation:
𝑡′
= t2′ − t1
′
= 𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑏𝑦 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑟 𝑖𝑛 𝑆′
𝑓𝑟𝑎𝑚𝑒
𝑡 = t2 − t1 = 𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑏𝑦 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑟 𝑖𝑛 𝑆 𝑓𝑟𝑎𝑚𝑒
t1
′
=
𝑡1−
𝑣𝑋
𝑐2
1−
𝑣2
𝑐2
t2
′
=
𝑡2−
𝑣𝑋
𝑐2
1−
𝑣2
𝑐2
t2′ − t1
′
=
𝑡2−
𝑣𝑋
𝑐2
1−
𝑣2
𝑐2
−
𝑡1−
𝑣𝑋
𝑐2
1−
𝑣2
𝑐2
=
𝑡2−𝑡1
1−
𝑣2
𝑐2
𝑡′ =
𝑡
1−
𝑣2
𝑐2
; 𝑡′
> 𝑡
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33. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
Addition of Velocities:
𝑋 = 𝑘 𝑥′
+ 𝑣𝑡′
 𝑑𝑥 = 𝑘 𝑑𝑥′
+ 𝑣𝑑𝑡′
t = k 𝑡′ +
𝑣𝑥′
𝑐2  𝑑t = k 𝑑𝑡′ +
𝑣𝑑𝑋′
𝑐2
𝑑𝑥
𝑑𝑡
=
𝑑𝑥′+𝑣𝑑𝑡′
𝑑𝑡′+
𝑣𝑑𝑥′
𝑐2
𝑑𝑥
𝑑𝑡
=
𝑑𝑥
𝑑𝑡′
′
+𝑣
1+
𝑣𝑑𝑥′
𝑐2𝑑𝑡′
 𝑢𝑥 =
𝑢𝑥
′+𝑣
1+
𝑣
𝑐2𝑢𝑥′
𝑦 = 𝑦′  𝑑𝑦 = 𝑑𝑦′
𝑑𝑦
𝑑𝑡
=
𝑑𝑦′
𝑘(𝑑𝑡′+
𝑣𝑑𝑋′
𝑐2 )
𝑑𝑦
𝑑𝑡
=
𝑑𝑦′
𝑑𝑡′
𝑘(1+
𝑣𝑑𝑥′
𝑐2𝑑𝑡′)
𝑢𝑦 =
𝑢′𝑦
𝑘(1+
𝑣
𝑐2 𝑢′𝑥)
=
𝑢′𝑦 1−
𝑣2
𝑐2
(1+
𝑣
𝑐2 𝑢′𝑥)
𝑢𝑧 =
𝑢′𝑧
𝑘(1+
𝑣
𝑐2 𝑢′𝑥)
=
𝑢′𝑧 1−
𝑣2
𝑐2
(1+
𝑣
𝑐2 𝑢′𝑥)
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34. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
𝑢𝑥 =
𝑢′
𝑥+𝑣
1+
𝑣
𝑐2𝑢′𝑥
𝑢𝑦 =
𝑢′𝑦 1−
𝑣2
𝑐2
(1+
𝑣
𝑐2 𝑢′𝑥)
𝑢𝑧 =
𝑢′𝑧 1−
𝑣2
𝑐2
(1+
𝑣
𝑐2 𝑢′𝑥)
𝑢𝑥′ =
𝑢𝑥−𝑣
1−
𝑣
𝑐2𝑢𝑥′
𝑢𝑦′ =
𝑢𝑦 1−
𝑣2
𝑐2
(1−
𝑣
𝑐2 𝑢𝑥)
𝑢𝑧′ =
𝑢𝑧 1−
𝑣2
𝑐2
(1−
𝑣
𝑐2 𝑢𝑥)
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35. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
Relativity of mass:
Y
Z
Inertial
Frame
S
Inertial
Frame
S’
V
Y’
V t
O O’
Z’
X’
u’ -u’ before collision
After collision
X
𝒎𝟏, 𝒎𝟐 𝒃𝒆 𝒕𝒉𝒆 𝒎𝒂𝒔𝒔𝒆𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆𝒔
𝒕𝒐 𝒃𝒆 𝒄𝒐𝒍𝒍𝒊𝒅𝒆𝒅.
𝒖′
𝒂𝒏𝒅 𝒖′
𝒃𝒆 𝒕𝒉𝒆 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒊𝒆𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆𝒔
𝒎𝟏 𝒂𝒏𝒅 𝒎𝟐 𝒓𝒆𝒔𝒑. 𝒊𝒏 𝒇𝒓𝒂𝒎𝒆 𝑺′
𝑢1 =
𝑢′+𝑣
1+
𝑣
𝑐2𝑢′
𝑢2 =
−𝑢′+𝑣
1−
𝑣
𝑐2𝑢′
𝑚1𝑢1 + 𝑚2𝑢2 = (𝑚1+𝑚2)𝑣
𝑚1
𝑢′ + 𝑣
1 +
𝑣
𝑐2 𝑢′
+ 𝑚2
−𝑢′ + 𝑣
1 −
𝑣
𝑐2 𝑢′
= (𝑚1+𝑚2)𝑣
𝑚1
𝑢′
+ 𝑣
1 +
𝑣
𝑐2 𝑢′
− 𝑚1𝑣 = 𝑚2𝑣 − 𝑚2
−𝑢′
+ 𝑣
1 −
𝑣
𝑐2 𝑢′
𝑚1(
𝑢′ + 𝑣
1 +
𝑣
𝑐2 𝑢′
− 𝑣) = 𝑚2(𝑣 −
−𝑢′ + 𝑣
1 −
𝑣
𝑐2 𝑢′
)
P
H
Y
S
I
C
S

36. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
𝑚1(
𝑢′ −
𝑣2
𝑐2 𝑢′
1 +
𝑣
𝑐2 𝑢′
) = 𝑚2(
−
𝑣2
𝑐2 𝑢′ + 𝑢′
1 −
𝑣
𝑐2 𝑢′
)
Relativity of mass:
𝑚1(
𝑢′ −
𝑣2
𝑐2 𝑢′
1 +
𝑣
𝑐2 𝑢′
) = 𝑚2(
𝑢′ −
𝑣2
𝑐2 𝑢′
1 −
𝑣
𝑐2 𝑢′
)
𝑚1
1+
𝑣
𝑐2𝑢′
=
𝑚2
1−
𝑣
𝑐2𝑢′
𝑚1
𝑚2
=
1+
𝑣
𝑐2𝑢′
1−
𝑣
𝑐2𝑢′
𝑢1 =
𝑢′+𝑣
1+
𝑣
𝑐2𝑢 ′
𝑢2 =
−𝑢′+𝑣
1−
𝑣
𝑐2𝑢 ′
𝑝𝑢𝑡 𝑥 =
𝑣
𝑐2 𝑢′
𝑢1 =
𝑢′+𝑣
1+𝑥
𝑢2 =
−𝑢′+𝑣
1−𝑥
P
H
Y
S
I
C
S

37. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
1 −
𝑢1
2
𝑐2 = 1 −
1
𝑐2 (
𝑢′+𝑣
1+𝑥
)2
= 1 −
1
𝑐2 (
𝑢′2
+𝑣2+2𝑢′𝑣
(1+𝑥)2 )
= 1 − (
𝑢′2
𝑐2 +
𝑣2
𝑐2 +
2𝑢′𝑣
𝑐2
(1+𝑥)2 )
1 −
𝑢1
2
𝑐2 =
1+2𝑥+𝑥2−
𝑢′2
𝑐2 −
𝑣2
𝑐2 −2𝑥
(1+𝑥)2
1 −
𝑢1
2
𝑐2 =
1+𝑥2−
𝑢′2
𝑐2 −
𝑣2
𝑐2
(1+𝑥)2
1 −
𝑢2
2
𝑐2 = 1 −
1
𝑐2 (
−𝑢′+𝑣
1−𝑥
)2
=
(1+𝑥)2−
𝑢′2
𝑐2 −
𝑣2
𝑐2 −
2𝑢′𝑣
𝑐2
(1+𝑥)2
P
H
Y
S
I
C
S

38. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
1 −
𝑢2
2
𝑐2 = 1 −
1
𝑐2 (
−𝑢′+𝑣
1−𝑥
)2
= 1 −
1
𝑐2 (
𝑢′2
+𝑣2−2𝑢′𝑣
(1−𝑥)2 )
= 1 − (
𝑢′2
𝑐2 +
𝑣2
𝑐2 −
2𝑢′𝑣
𝑐2
(1−𝑥)2 )
=
(1−𝑥)2−
𝑢′2
𝑐2 −
𝑣2
𝑐2 +
2𝑢′𝑣
𝑐2
(1−𝑥)2
=
1−2𝑥+𝑥2−
𝑢′2
𝑐2 −
𝑣2
𝑐2 +2𝑥
(1−𝑥)2
1 −
𝑢2
2
𝑐2 =
1+𝑥2−
𝑢′2
𝑐2 −
𝑣2
𝑐2
(1−𝑥)2
1 −
𝑢1
2
𝑐2
=
1 + 𝑥2 −
𝑢′2
𝑐2 −
𝑣2
𝑐2
(1 + 𝑥)2
1 −
𝑢1
2
𝑐2
1 −
𝑢2
2
𝑐2
=
(1 − 𝑥)2
(1 + 𝑥)2
P
H
Y
S
I
C
S

39. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
1−
𝑢1
2
𝑐2
1−
𝑢2
2
𝑐2
=
1−𝑥
1+𝑥
1+𝑥
1−𝑥
=
1−
𝑢2
2
𝑐2
1−
𝑢1
2
𝑐2
𝑚1
𝑚2
=
1+
𝑣
𝑐2𝑢′
1−
𝑣
𝑐2𝑢′
=
1+𝑥
1−𝑥
=
1−
𝑢2
2
𝑐2
1−
𝑢1
2
𝑐2
𝑚1 1 −
𝑢1
2
𝑐2 = 𝑚2 1 −
𝑢2
2
𝑐2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑖𝑓 𝑚2 = 𝑚0 = 𝑟𝑒𝑠𝑡 𝑚𝑎𝑠𝑠, 𝑢2 = 0
𝑎𝑛𝑑 𝑚1 = 𝑚, 𝑢1 = 𝑣 , 𝑡ℎ𝑒𝑛
𝑚 1 −
𝑣2
𝑐2
= 𝑚0
 𝑚 =
𝑚0
1 −
𝑣2
𝑐2
P
H
Y
S
I
C
S

40. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
𝐹 =
𝑑(𝑚𝑣)
𝑑𝑡
𝐹 = 𝑚
𝑑𝑣
𝑑𝑡
+ 𝑣
𝑑𝑚
𝑑𝑡
𝑑𝐸 = 𝐹. 𝑑𝑥
𝑑𝐸 = (𝑚
𝑑𝑣
𝑑𝑡
+ 𝑣
𝑑𝑚
𝑑𝑡
)𝑑𝑥
𝑑𝐸 = 𝑚𝑣𝑑𝑣 + 𝑣2𝑑𝑚 …(i)
Mass-Energy equivalence: 𝑚 =
𝑚0
1−
𝑣2
𝑐2
 𝑚2
=
𝑚0
2
1−
𝑣2
𝑐2
𝑚2
=
𝑚0
2𝑐2
𝑐2−𝑣2
𝑚2(𝑐2 − 𝑣2) = 𝑚0
2𝑐2
𝑚2𝑐2 − 𝑚2𝑣2 = 𝑚0
2𝑐2
2𝑚𝑑𝑚𝑐2 − 2𝑚𝑑𝑚𝑣2 − 2𝑣𝑑𝑣𝑚2 = 0
𝑑𝑚𝑐2 − 𝑑𝑚𝑣2 − 𝑣𝑑𝑣𝑚 = 0
(𝑐2
−𝑣2
)𝑑𝑚 − 𝑚𝑣𝑑𝑣 = 0
𝑐2
𝑑𝑚 = 𝑣2
𝑑𝑚 + 𝑚𝑣𝑑𝑣 ……(ii )
P
H
Y
S
I
C
S

41. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
𝑐2
𝑑𝑚 = 𝑑𝐸
𝐸0 = 𝑑𝐸 = 𝑐2𝑑𝑚
𝑚
𝑚0
𝐸0 = 𝑚𝑐2
− 𝑚0𝑐2
𝑚𝑐2 = 𝐸0 + 𝑚0𝑐2 = 𝐸
𝐸 = 𝑚𝑐2
Mass-Energy equivalence
P
H
Y
S
I
C
S
P
H
Y
S
I
C
S

42. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
Total Energy and momentum:
𝑚 =
𝑚0
1−
𝑣2
𝑐2
𝑚2 =
𝑚0
2
1−
𝑣2
𝑐2
𝑚2 =
𝑚0
2𝑐2
𝑐2−𝑣2
Multiplying by c 2 on both sides
𝑚2
𝑐4
− 𝑚2
𝑣2
𝑐2
= 𝑚0
2
𝑐4
𝑚2(𝑐2 − 𝑣2) = 𝑚0
2𝑐2
𝑚2
𝑐2
− 𝑚2
𝑣2
= 𝑚0
2
𝑐2
As E= mc2, p= mv
𝐸2 − 𝑝2𝑐2 = 𝑚0
2𝑐4
𝐸2 = 𝑝2𝑐2 + 𝑚0
2
𝑐4
𝐸 = 𝑝2𝑐2 + 𝑚0
2
𝑐4
P
H
Y
S
I
C
S

43. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
As measured by S, a flash of light goes off at x=100km, y=10km,
z=1km at t=0.5ms. What are the coordinates of x′,y′,z′ and t′ of this
event as determined by an observer S′ moving relative to S at -0.8c
along the common x-x′ axis?
20.13 km
10 km
1 km
1.27 ms
P
H
Y
S
I
C
S

44. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
𝐿 = 𝐿0 1 −
𝑣2
𝑐2
Length Contraction:
Time Dilation:
𝑡′ =
𝑡
1−
𝑣2
𝑐2
P
H
Y
S
I
C
S

45. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
2.How fast must a space craft travel relative to the earth so that one day
on the earth corresponds with 2 days on the space craft
1.A rod of length 1 m is moving with velocity 0.8 C relative to a laboratory.
What will be the length as measured by the observer of the laboratory
3.The mean life of p meson is 2 × 10–8 s. Calculate mean life of a meson
moving with a velocity of 0.8 C.
5.With what velocity should a rocket move so that every year spent on it
corresponds to 4 years on earth ?
4.The length of a rocket ship is 100 m on the earth, when it is moving
with velocity v ,its length observed is 99 m. Calculate its velocity,
Assignment A:
P
H
Y
S
I
C
S

46. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
𝑚 =
𝑚0
1 −
𝑣2
𝑐2
𝐸 = 𝑚𝑐2
Mass-Energy equivalence
Relativity of mass
𝑢𝑥 =
𝑢′
𝑥+𝑣
1+
𝑣
𝑐2𝑢′𝑥
𝑢𝑦 =
𝑢′𝑦 1−
𝑣2
𝑐2
(1+
𝑣
𝑐2 𝑢′𝑥)
𝑢𝑧 =
𝑢′𝑧 1−
𝑣2
𝑐2
(1+
𝑣
𝑐2 𝑢′𝑥)
𝑢′𝑥 =
𝑢𝑥−𝑣
1−
𝑣
𝑐2𝑢𝑥′
𝑢′𝑦 =
𝑢𝑦 1−
𝑣2
𝑐2
(1−
𝑣
𝑐2 𝑢𝑥)
𝑢′𝑧 =
𝑢𝑧 1−
𝑣2
𝑐2
(1−
𝑣
𝑐2 𝑢𝑥)
Addition of Velocities:
P
H
Y
S
I
C
S

47. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
9.A particle of mass 10–24 Kg is moving with a speed of 1.8 × 108
m/s. Calculate its relativistic mass. (Given C = 3 × 108 m/s).
10.The total energy of a particle is exactly twice its rest energy.
Calculate its speed.
6.A particle is moving with a speed of 0.5 C. Calculate the ratio of
the rest mass and the mass while in motion.
11.A particle is moving with a speed of 0.6 C. Calculate the
ratio of the rest mass and the mass while in motion.
7.If 1 kg of a substance is fully converted into energy, how much
energy is produced ? (Given C = 3 × 108 m/s).
8.Draw a graph showing variation of mass of body with its speed
Assignment B:
P
H
Y
S
I
C
S

48. phy.kasarla@iscnagpur.ac.in
B.Sc. Sem.-VI Physics-I Unit-I
Relativity
12.If one gram of a substance is fully converted into energy in one second
find out how many calories of heat will be produced.
14.Two space crafts A and B are moving away from earth along the same
line with speed 0.8 c and 0.6 c respectively. Find the velocity of B with
respect to A.
15.Calculate the speed of an electron at which its relativistic energy is 1.25
times the rest energy.
16.What would be the speed of a particle so that its mass is double of its
rest mass ?
17.Calculate the velocity at which electron mass is 3 times the rest mass.
13.A rocket moving with speed 108 m/s ejects a projectile in its direction of
motion with speed relative to rocket 2 × 108 m/s. Find the speed as
measured by an observer on earth.
Assignment B contd..:
P
H
Y
S
I
C
S