Special Theory Relativity
Inertial Frame of Reference: A frame of reference in which Newton’s laws of motion are valid is known as an
inertial frame of reference.
Non-Inertial Frame of Reference: A frame of reference in which Newton’s Laws are not valid is known as a Non-
inertial frame of reference.
Galilean Transformation
𝑍 𝑆 𝑍′ 𝑆 ′
𝑣
𝑥 𝑃
𝑣𝑡 𝑥′
𝑋′
0 0′
𝑋
𝑌 𝑌′
Consider two inertial frames of reference 𝑆, 𝑆 ′ . Frame 𝑆 ′ is moving with a velocity 𝑣 along the positive 𝑋-axis
relative to the frame 𝑆. Let the two frames of reference 𝑆, 𝑆 ′ coincide at time 𝑡 = 0.
Let be the co-ordinates of the point 𝑃 with respect to the frames 𝑆, 𝑆 ′ are 𝑥, 𝑦, 𝑧, 𝑡 and 𝑥 ′ , 𝑦 ′ , 𝑧 ′ , 𝑡 ′ .
From Figure,
𝑥 = 𝑥 ′ + 𝑣𝑡
𝑥 ′ = 𝑥 − 𝑣𝑡
Similarly 𝑦 ′ = 𝑦,
𝑧 ′ = 𝑧,
𝑡′ = 𝑡
The above equations are called Galilean Transformation equations.
Inverse Galilean Transformation equations are
𝑥 = 𝑥 ′ + 𝑣𝑡 ′
𝑦 = 𝑦′
𝑧 = 𝑧′
𝑡 = 𝑡′
Space interval is invariant under Galilean transformation
Time interval is invariant under Galilean Transformation.
Laws of mechanics are invariant under Galilean Transformation.
Lorentz Transformation
Consider two inertial frames of reference 𝑆, 𝑆 ′ . Frame 𝑆 ′ is moving with a velocity 𝑣 along the positive 𝑋-axis
relative to the frame 𝑆. Let the two frames of reference 𝑆, 𝑆 ′ coincide at time𝑡 = 0. Let be the co-ordinates of the
point 𝑃 with respect to the frames 𝑆, 𝑆 ′ are 𝑥, 𝑦, 𝑧, 𝑡 and 𝑥 ′ , 𝑦 ′ , 𝑧 ′ , 𝑡 ′ .
Let a beam of light is emitted from the origin 𝑂 at time 𝑡 = 𝑜.The beam of light reaches the point 𝑃 after a time.
K.V.Ganesh Kumar, GDC Ganapavaram[Type text] Page 1
𝑍 𝑆 𝑍′ 𝑆 ′
𝑣
𝑋′
0 0′
𝑋
𝑌 𝑌′
Distance 𝑥 2 +𝑦 2 +𝑧 2
Relative to frame 𝑆, 𝐶= =
Time 𝑡
2 2 2
Distance 𝑥 ′ +𝑦 ′ +𝑧 ′
′
Relative to Frame 𝑆 , 𝐶= =
Time 𝑡
𝑥2 + 𝑦2 + 𝑧2
𝐶=
𝑡
𝑐2 𝑡2 = 𝑥2 + 𝑦2 + 𝑧2
𝑥2 + 𝑦2 + 𝑧2 − 𝑐2 𝑡2 = 0 … … … … … … … … … … … 1
𝑥′ 2 + 𝑦′ 2 + 𝑧′ 2
𝐶=
𝑡′
2 2 2 2
𝑐2 𝑡′ = 𝑥′ + 𝑦′ + 𝑧′
2 2 2 2
𝑥 ′ + 𝑦 ′ + 𝑧 ′ − 𝑐 2 𝑡 ′ = 0 … … … … … … … … … . .2
From equations 1 and 2
2 2 2 2
𝑥2 + 𝑦2 + 𝑧2 − 𝑐2 𝑡2 = 𝑥′ + 𝑦′ + 𝑧′ − 𝑐2 𝑡′
𝑦 ′ = 𝑦,
𝑧′ = 𝑧
2 2
𝑥 2 − 𝑐 2 𝑡 2 = 𝑥 ′ − 𝑐 2 𝑡 ′ ……………………………………….3
From Galilean Transformation
𝑥 ′ = 𝑥 − 𝑣𝑡
𝐿𝑒𝑡 𝑥 ′ = 𝑘 𝑥 − 𝑣𝑡 ……………………….4
Inverse Galilean Transformation
𝑥 = 𝑘 𝑥 ′ + 𝑣𝑡 ′
𝑥 = 𝑘 𝑘 𝑥 − 𝑣𝑡 + 𝑣𝑡 ′
𝑥
= 𝑘 𝑥 − 𝑣𝑡 + 𝑣𝑡 ′
𝑘
𝑥
𝑣𝑡 ′ = − 𝑘 𝑥 − 𝑣𝑡
𝑘
𝑥
𝑣𝑡 ′ = − 𝑘𝑥 + 𝑘𝑣𝑡
𝑘
𝑥 𝑘𝑥
𝑡′ = − + 𝑘𝑡
𝑘𝑣 𝑣
K.V.Ganesh Kumar, GDC Ganapavaram[Type text] Page 2
𝑥 𝑘𝑥
𝑡 ′ = 𝑘𝑡 + −
𝑘𝑣 𝑣
𝑥 1
𝑡 ′ = 𝑘𝑡 − 𝑘 −
𝑣 𝑘
𝑘𝑥 1
𝑡 ′ = 𝑘𝑡 − 1− 2
𝑣 𝑘
𝑥 1
𝑡 ′ = 𝑘 𝑡 − 1 − 2 … … … … … … … … . .5
𝑣 𝑘
2 2 2 ′2 2 ′2
From equation 3, 𝑥 −𝑐 𝑡 =𝑥 −𝑐 𝑡
2
𝑥 1
𝑥 2 − 𝑐 2 𝑡 2 = 𝑘 2 𝑥 − 𝑣𝑡 2
− 𝑐2 𝑘2 𝑡 − 1− 2
𝑣 𝑘
Comparing the coefficients of 𝑡 2 on both sides,
−𝑐 2 = 𝑘 2 𝑣 2 − 𝑐 2 𝑘 2
𝑐2 = 𝑐2 𝑘2 − 𝑘2𝑣 2
𝑐2 = 𝑘2 𝑐2 − 𝑣2
𝑐2 1
𝑘2 = 2 2
= 2
𝑐 −𝑣 1−𝑣
𝑐2
1
𝑘=
2
1−𝑣
𝑐2
From equation 4,
𝑥 ′ = 𝑘 𝑥 − 𝑣𝑡
𝑥 − 𝑣𝑡
𝑥′ =
2
1 − 𝑣 𝑐2
From equation 5,
𝑥 1 1 𝑥 𝑐2 − 𝑣2
𝑡′ = 𝑘 𝑡 − 1− 2 = 𝑡− 1−
𝑣 𝑘 2 𝑣 𝑐2
1−𝑣
𝑐2
2 𝑡− 𝑣𝑥
1 𝑥 𝑐 𝑐2
= 𝑡− =
2 𝑣 𝑣2 2
1−𝑣 1−𝑣
𝑐2 𝑐2
𝑡− 𝑣𝑥
𝑡′ = 𝑐2
2
1−𝑣 2
𝑐
Lorentz Transformation Equations
𝑥 − 𝑣𝑡
𝑥′ =
2
1−𝑣
𝑐2
′
𝑦 = 𝑦,
𝑧 ′ = 𝑧,
𝑡 − 𝑣𝑥 2
′
𝑡 = 𝑐
2
1−𝑣 2
𝑐
Length Contraction or Lorentz-Fitzgerald Contraction
Consider two inertial frames of reference 𝑆, 𝑆 ′ . Frame 𝑆 ′ is moving with a velocity 𝑣 along the positive 𝑋-axis relative to the frame
𝑆. Let the two frames of reference 𝑆, 𝑆 ′ coincide at time 𝑡 = 0.
Let a rod of length 𝑙 is placed in the reference frame 𝑆 ′ with its length parallel to 𝑋-axis. Co-ordinates of the ends of the rod with
respect to the frames 𝑆, 𝑆 ′ are 𝑥1 , 𝑥2 and 𝑥1′ , 𝑥2′ .
K.V.Ganesh Kumar, GDC Ganapavaram[Type text] Page 3
𝑍 𝑆 𝑍′ 𝑆 ′
𝑣
𝑥1 ′
𝑥2 ′
𝑋′
0 0′
𝑋
𝑥1
𝑥2
𝑌 𝑌′
Length of the rod in frame 𝑆
𝑙 = 𝑥2 − 𝑥1
Length of the rod in frame 𝑆 ′
𝑙 ′ = 𝑥2′ − 𝑥1′
From Lorentz transformation
𝑥 − 𝑣𝑡
𝑥′ =
2
1−𝑣
𝑐2
𝑥2 − 𝑣𝑡 𝑥1 − 𝑣𝑡 𝑥2 − 𝑥1
𝑙 ′ = 𝑥2′ − 𝑥1′ = − =
2 2 2
1−𝑣 1−𝑣 1−𝑣
𝑐2 𝑐2 𝑐2
𝑙
𝑙′ =
2
1−𝑣
𝑐2
2
𝑙 = 𝑙′ 1 − 𝑣
𝑐2
Hence a moving rod appears to be contracted for a stationary observer.
𝑣2
Case (i): 𝑊ℎ𝑒𝑛 𝑣 ≪ 𝑐 𝑐2
~0
∴ 𝑙 = 𝑙′ 1 = 𝑙′
𝑙 = 𝑙′
2
Case(ii): 𝑊ℎ𝑒𝑛 𝑣 𝑖𝑠 comparable to 𝑐, 𝑙 = 𝑙 ′ 1 − 𝑣
𝑐2
∴ 𝑙 < 𝑙′
Moving rod appears to be contracted for a stationary observer
𝑣2
Case(iii): 𝑊ℎ𝑒𝑛 𝑣 = 𝑐, 𝑐2
=1
𝑙 = 𝑙′ 0 = 0
∴𝑙=0
𝑊ℎ𝑒𝑛 𝑣 > 𝑐, 𝑙 = 𝐶𝑜𝑚𝑝𝑙𝑒𝑥 𝑁𝑢𝑚𝑏𝑒𝑟
Hence no object can travel faster than the speed of light.
K.V.Ganesh Kumar, GDC Ganapavaram[Type text] Page 4
Time Dilation
Consider two inertial frames of reference 𝑆, 𝑆 . Frame 𝑆 ′ is moving with a velocity 𝑣 along the positive 𝑋-axis
′
relative to the frame 𝑆. Let the two frames of reference 𝑆, 𝑆 ′ coincide at time 𝑡 = 0.
Let a clock be placed in the frame𝑆.
Time interval in frame 𝑆 is ∆𝑡 = 𝑡2 − 𝑡1
Time interval in frame 𝑆 ′ is ∆𝑡 ′ = 𝑡2 ′ − 𝑡1 1
From Lorentz Transformation
𝑡 − 𝑣𝑥 𝑐 2
′
𝑡 =
2
1 − 𝑣 𝑐2
𝑡2 − 𝑣𝑥 𝑐 2 𝑡1 − 𝑣𝑥 𝑐 2 𝑡2 − 𝑡1 ∆𝑡
∆𝑡 ′ = 𝑡2 ′ − 𝑡1 1 = = = =
2 2 2 2
1 − 𝑣 𝑐2 1 − 𝑣 𝑐2 1 − 𝑣 𝑐2 1 − 𝑣 𝑐2
∆𝑡
∴ ∆𝑡 ′ = 2 = 𝑘∆𝑡
1−𝑣 𝑐2
𝑣2
Case (i): 𝑊ℎ𝑒𝑛 𝑣 ≪ 𝑐 𝑐2
~0
∴ ∆𝑡 ′ = ∆𝑡
Case(ii): When 𝑣is comparable to 𝑐
∆𝑡 ′ > ∆𝑡
Hence the time interval of a moving observer is more than the time interval of a stationery observer.
𝑣2
Case(iii): 𝑊ℎ𝑒𝑛 𝑣 = 𝑐, 𝑐 2 = 1
∆𝑡 ′ = ∞
′
𝑊ℎ𝑒𝑛 𝑣 > 𝑐, ∆𝑡 =Complex Number
Hence no object can travel faster than the speed of light.
Einstein’s Mass-Energy Equivalence
From Newton’s Second law
𝑑𝑃 𝑑 𝑑𝑣 𝑑𝑚
𝐹= = 𝑚𝑣 = 𝑚 +𝑣
𝑑𝑡 𝑑𝑡 𝑑𝑡 𝑑𝑡
By Work-Energy theorem, work done is equal to the change in kinetic energy.
𝑊 = 𝐹. 𝑑𝑥 = 𝑑𝐾
𝑑𝑣 𝑑𝑚
𝑑𝐾 = 𝐹. 𝑑𝑥 = 𝑚 +𝑣 𝑑𝑥
𝑑𝑡 𝑑𝑡
𝑑𝑣 𝑑𝑚
= 𝑚 𝑑𝑥 + 𝑣 𝑑𝑥
𝑑𝑡 𝑑𝑡
𝑑𝑥 𝑑𝑥
= 𝑚 𝑑𝑣 + 𝑣 𝑑𝑚
𝑑𝑡 𝑑𝑡
2
= 𝑚 𝑣 𝑑𝑣 + 𝑣 𝑑𝑚
∴ 𝑑𝐾 = 𝑚 𝑣 𝑑𝑣 + 𝑣 2 𝑑𝑚
Relativistic mass
𝑚0
𝑚=
2
1 − 𝑣 𝑐2
K.V.Ganesh Kumar, GDC Ganapavaram[Type text] Page 5
2
𝑚02 𝑚02 𝑚02 𝑐 2
𝑚 = 2 = 2 =
1−𝑣 𝑐 − 𝑣2 𝑐2 − 𝑣2
𝑐 2
𝑐2
𝑚2 𝑐 2 − 𝑣 2 = 𝑚02 𝑐 2
𝑚2 𝑐 2 − 𝑚2 𝑣 2 = 𝑚02 𝑐 2
2𝑚𝑐 2 𝑑𝑚 − 𝑚2 2𝑣 𝑑𝑣 + 𝑣 2 2𝑚 𝑑𝑚 = 0
2𝑚𝑐 2 𝑑𝑚 = 𝑚2 2𝑣 𝑑𝑣 + 𝑣 2 2𝑚 𝑑𝑚
𝑐 2 𝑑𝑚 = 𝑚 𝑣 𝑑𝑣 + 𝑣 2 𝑑𝑚
𝑑𝐾 = 𝑐 2 𝑑𝑚
𝑚
2
𝑑𝐾 = 𝑐 𝑑𝑚
𝑚0
𝐾 = 𝑐2 𝑚 𝑚
𝑚0
2
𝐾 = 𝑐 𝑚 − 𝑚0
The above equation gives the relativistic kinetic energy of a moving body.
Energy at rest is given by
𝑚0 𝑐 2
Total energy
𝐸 = 𝑐 2 𝑚 − 𝑚0 + 𝑚0 𝑐 2 = 𝑚𝑐 2
The above equation gives Einstein’s mass-energy equivalence.
Hence Mass and Energy are not two different physical quantities. Mass can be converted in to energy and vice-
versa.
Addition of Velocities or Transformation of Velocities
Consider two inertial frames of reference 𝑆, 𝑆 ′ . Frame 𝑆 ′ is moving with a velocity 𝑣 along the positive 𝑋-axis
relative to the frame 𝑆. Let the two frames of reference 𝑆, 𝑆 ′ coincide at time 𝑡 = 0.
In reference frame 𝑆, an object moves a distance 𝑑𝑥 in time 𝑑𝑡. Similarly in reference frame 𝑆 ′ , the object moves
a distance 𝑑𝑥 ′ in time 𝑑𝑡 ′ .
𝑑𝑥
Velocity in Reference frame 𝑆 𝑢 =
𝑑𝑡
𝑑𝑥 ′
Velocity in Reference frame 𝑆 ′ 𝑢′ =
𝑑𝑡 ′
From Lorentz Transformation
𝑥 ′ = 𝑘 𝑥 − 𝑣𝑡
𝑡 ′ = 𝑘 𝑡 − 𝑣𝑥 𝑐 2
From Inverse Lorentz Transformation
′
𝑥 = 𝑘 𝑥 ′ + 𝑣𝑡 ′ , 𝑡 = 𝑘 𝑡 ′ + 𝑣𝑥 𝑐 2
′
𝑑𝑥 = 𝑘 𝑑𝑥 ′ + 𝑣 𝑑𝑡 ′ , 𝑑𝑡 = 𝑘 𝑑𝑡 ′ + 𝑣𝑑𝑥 𝑐 2
𝑑𝑥 𝑘 𝑑𝑥 ′ + 𝑣 𝑑𝑡 ′ 𝑑𝑥 ′ + 𝑣 𝑑𝑡 ′
𝑢= = =
𝑑𝑡 𝑘 𝑑𝑡 ′ + 𝑣𝑑𝑥 ′ ′
𝑑𝑡 ′ + 𝑣𝑑𝑥 𝑐 2
𝑐 2
𝑑𝑥 ′
′ +𝑣
= 𝑑𝑡
𝑣 𝑑𝑥 ′
1+ 2 ′
𝑐 𝑑𝑡
𝑢′ + 𝑣
𝑢= ′
1 + 𝑢 𝑣 𝑐2
The above equation represents the relativistic law of addition of velocities.
Case(i) : When 𝑢′ ≪ 𝑐 , 𝑣 ≪ 𝑐
𝑢′ 𝑣
~0
𝑐2
K.V.Ganesh Kumar, GDC Ganapavaram[Type text] Page 6
∴ 𝑢 = 𝑢′ + 𝑣
Case(ii): When 𝑢′ = 𝑐 , 𝑣 = 𝑐
𝑐+𝑐 2𝑐
𝑢= = =𝑐
1 + 𝑐. 𝑐 𝑐 2 2
𝑢=𝑐
Hence addition of velocity of to the velocity of light reproduces the velocity of light.
Michelson-Morley Experiment
M1
𝐴 𝐴′
Ray 1 𝒍
L
S
M2
′
𝐺 𝐷 𝐺 𝐵 𝐵′
M Ray 2
Aim: Aim of Michelson-Morley experiment is to determine the velocity of Earth relative to Ether.
Michelson-Morley Interferometer is shown in figure. Light emitted from the monochromatic source 𝑆 falls
on the half silvered glass plate 𝐺. The glass plate 𝐺 is oriented at an angle of 450 to the incident light. Hence the
light incident on the glass plate 𝐺 is divided in to two perpendicular beams of light. The two beams of light are
reflected back from the two mirrors 𝑀1 , 𝑀2 and meet at 𝐺 to produce interference pattern. The interference
pattern can be observed through the telescope𝑇.
Since the apparatus is moving with a velocity 𝑣 along with the Earth, the optical paths of two beams are not equal.
The two beams are reflected at the points 𝐴′ , 𝐵′ instead of 𝐴, 𝐵 and interfere at 𝐺 ′ .
From ∆ 𝐺𝐴′ 𝐷 𝐺𝐴′ 2 = 𝐴′ 𝐷 2 + 𝐺𝐷 2
𝑐 2 𝑡 2 = 𝑙2 + 𝑣 2 𝑡 2
𝑙2 = 𝑐 2 − 𝑣 2 𝑡 2
𝑙2
𝑡2 = 2
𝑐 − 𝑣2
𝑙 𝑙
𝑡= =
𝑐2 − 𝑣2 2
𝑐 1 − 𝑣 𝑐2
−1/2
𝑙 2 𝑙 2
= 1 − 𝑣 𝑐2 = 1 + 𝑣 2𝑐 2
𝑐 𝑐
Hence the time taken by the light beam 1 to reach 𝐺 ′
2𝑙 2
𝑡1 = 2𝑡 = 1 + 𝑣 2𝑐 2
𝑐
Let be 𝑡2 the time taken by the light beam 2 to reach the glass plate 𝐺 ′ .
Velocity of light beam from 𝐺 to 𝐵′ is 𝑐 − 𝑣 and from 𝐵′ to 𝐺 ′ is 𝑐 + 𝑣 .
K.V.Ganesh Kumar, GDC Ganapavaram[Type text] Page 7
𝑙 𝑙 1 1 2𝑐
𝑡2 = + =𝑙 + =𝑙 2
𝑐−𝑣 𝑐+𝑣 𝑐−𝑣 𝑐+𝑣 𝑐 − 𝑣2
−1
𝑙 2𝑐 2𝑙 2 2𝑙 2
= 1 − 𝑣 𝑐2 = 1 + 𝑣 𝑐2
𝑐 𝑐
𝑣 2
𝑐2 1− 𝑐2
2𝑙 2
𝑡2 = 1 + 𝑣 𝑐2
𝑐
Time lag between the two beams
∆𝑡 = 𝑡2 − 𝑡1
2𝑙 2 2𝑙 2
= 1 + 𝑣 𝑐2 − 1 + 𝑣 2𝑐 2
𝑐 𝑐
2𝑙 2 2
= 1 + 𝑣 𝑐 2 − 1 − 𝑣 2𝑐 2
𝑐
2𝑙 𝑣 2
=
𝑐 2𝑐 2
𝑙𝑣 2
= 3
𝑐
𝑙𝑣 2
∆𝑡 = 3
𝑐
𝑙𝑣 2 𝑙𝑣 2
Path difference = 𝑐. ∆𝑡 = 𝑐. =
𝑐3 𝑐2
𝑙𝑣 2
Path difference in terms of Wavelength = 2
𝜆𝑐
Mirrors 𝑀1 , 𝑀2 are interchanged by rotating the apparatus by 900
𝑙𝑣 2
Path difference = −
𝜆𝑐 2
𝑙𝑣 2 𝑙𝑣 2 2𝑙𝑣 2
Resultant Path difference = 2
− − =
𝜆𝑐 𝜆𝑐 2 𝜆𝑐 2
2𝑙𝑣 2
Hence Fringe Shift ∆𝑛 = 2
𝜆𝑐
In Michelson-Morley Experiment, 𝑙 = 10𝑚, 𝑣 = 3 × 104 𝑚 𝑠 , 𝜆 = 5000 × 10−10 𝑚, 𝑐 = 3 × 108 𝑚/𝑠
2 × 10 × 3 × 104 2
∴ ∆𝑛 = = 0.4
5000 × 10−10 × 3 × 108 2
Hence a fringe shift of 0.4 was expected. But Michelson-Morley observed a fringe shift of only 0.001. This is known
as Null Result.
Significance of Null Result:
It is impossible to measure the speed of Earth relative to Ether. Hence the concept of Ether is rejected.
Speed of light in vacuum is the same for all observers.
K.V.Ganesh Kumar, GDC Ganapavaram[Type text] Page 8
