บทที่ 2 ทฤษฎีสัมพัทธภาพเฉพาะ

CHAPTER 2 :
Special Theory of Relativity
Napatsakon Sarapat
School of Physics, Science and Technology, TRU
SC 4101307 Modern Physics, TRU
𝛾 =
1
1 − 𝑣2
𝑐2

2.1 The Need for Ether
2.2 The Michelson-Morley
Experiment
2.3 Einstein’s Postulates
2.4 The Lorentz Transformation
2.5 Time Dilation and Length
Contraction
2.6 Addition of Velocities
2.7 Experimental Verification
2.8 Twin Paradox
2.9 Spacetime
2.10 Doppler Effect
2.11 Relativistic Momentum
2.12 Relativistic Energy
2.13 Computations in Modern
Physics
2.14 Electromagnetism and
Relativity
CHAPTER 2 :
Special Theory of Relativity

3
It was found that there was no
displacement of the interference fringes, so
that the result of the experiment was
negative and would, therefore, show that
there is still a difficulty in the theory
itself…
- Albert Michelson, 1907

4
Special Theory of Relativity ?
𝜔 𝑠′
𝜔 𝑠

5
Newtonian (Classical) Relativity
• Assumption :
• It is assumed that Newton’s laws of motion must be
measured with respect to (relative to) some reference
frame.

6
𝑥, 𝑦, 𝑧
𝑥′
, 𝑦′
, 𝑧′
• Assumption :
• It is assumed that Newton’s laws of motion must be
measured with respect to (relative to) some reference
frame.
Newtonian (Classical) Relativity

• A reference frame is called an inertial frame if
Newton laws are valid in that frame.
• Such a frame is established when a body, not
subjected to net external forces, is observed to
move in rectilinear motion at constant velocity.
7
Inertial Reference Frame
𝑢 ≡ 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑭 = 𝟎

Newtonian Principle of Relativity
• If Newton’s laws are valid in one reference
frame, then they are also valid in another
reference frame moving at a uniform velocity
relative to the first system.
• This is referred to as the Newtonian principle of
relativity or Galilean invariance.
8
𝑢′ ≡ 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑭′ = 𝟎

Inertial Frames 𝑂 and 𝑂′
• 𝑂 is at rest and 𝑂′ is moving with velocity 𝑢
• Axes are parallel
• 𝑂 and 𝑂′ are said to be INERTIAL COORDINATE
SYSTEMS 9
𝑃

The Galilean Transformation
For a point 𝑃
• In system 𝑂 ∶ 𝑃 = (𝑥, 𝑦, 𝑧, 𝑡)
• In system 𝑂 ′ ∶ 𝑃′ = (𝑥′, 𝑦′, 𝑧′, 𝑡′)
10
𝑃

Conditions of the Galilean Transformation
• Parallel axes
• 𝑂′ has a constant relative velocity in the x-direction with
respect to 𝑂
11
11
• Time (t) for all observers is a Fundamental invariant,
i.e., the same for all inertial observers
𝑥′
= 𝑥 − 𝑢𝑡
𝑦′
= 𝑦
𝑧′
= 𝑧
𝑡′
= 𝑡

The Inverse Relations
Step 1. Replace with
Step 2. Replace “primed” quantities with
“unprimed” and “unprimed” with “primed”
12
𝑥 = 𝑥′
+ 𝑢𝑡
𝑦 = 𝑦′
𝑧 = 𝑧′
𝑡 = 𝑡′

13
Ex-01 :
A passenger in a train moving at 30 m/s passes a man standing
on a station platform at 𝑡 = 𝑡′
= 0. Twenty seconds after the
train passes him, the man on the platform determines that a bird
flying along the tracks in the same direction as the train is 800
m away. What are the coordinates of the bird as determined by
the passenger ?

14
Ex-02 :
Refer to Ex-01. Five seconds after making the first coordinate
measurement, the man on the platform determines that the bird
is 850 m away. From these data find the velocity of the bird
(assumed constant) as determined by the man on the platform
and by the passenger on the train.

15
Ex-03 :
A sample of radioactive material, at rest in the laboratory, ejects
two electrons in opposite directions. One of the electrons has a
speed of 0.6c and the other has a speed of 0.7c, as measured by a
laboratory observer. According to classical velocity
transformations, what will be the speed of one electron as
measured from the other?
e
-
e
-

16
The Transition to Modern Relativity
• Although Newton’s laws of motion had
the same form under the Galilean
transformation, Maxwell’s equations did
not.

The Transition to Modern Relativity
• In 1905, Albert Einstein proposed a fundamental
connection between space and time and that Newton’s
laws are only an approximation. 17
𝑡
𝑡′

18
Ether
2.1 : The Need for Ether

19
2.1 : The Need for Ether
• The wave nature of light suggested that there
existed a propagation medium called the
luminiferous ether or just ether.
• Ether had to have such a low density that the planets
could move through it without loss of energy
• It also had to have an elasticity to support the high
velocity of light waves
EtherEM-wave EM-wave

20
Maxwell’s Equations
• In Maxwell’s theory the speed of
light, in terms of the permeability and
permittivity of free space, was given
by
• Thus the velocity of light between moving systems must
be a constant.
𝑐 =
1
𝜇0 𝜀0
𝛻2 − 𝜇𝜀
𝜕2
𝜕𝑡2
E = 0
𝛻2 − 𝜇𝜀
𝜕2
𝜕𝑡2
B = 0

Ether
An Absolute Reference System
• Ether was proposed as an absolute reference
system in which the speed of light was this
constant and from which other
measurements could be made.
• The Michelson-Morley experiment was an
attempt to show the existence of ether.
21
EM-wave EM-wave

22
Ether
2.2 : The Michelson-Morley Experiment
• Albert Michelson (1852–1931) was the first U.S.
citizen to receive the Nobel Prize for Physics (1907),
and built an extremely precise device called an
interferometer to measure the minute phase
difference between two light waves traveling in
mutually orthogonal directions.
EM-wave EM-wave

Ether wind =v
23
1. AC is parallel to the motion of
the Earth inducing an “ether
wind”
2. Light from source S is split by
mirror A and travels to mirrors
C and D in mutually
perpendicular directions
3. After reflection the beams
recombine at A slightly out of
phase due to the “ether wind” as
viewed by telescope E.
The Michelson Interferometer

Ether wind =v
25
TheMichelson
Interferometer

28
River
𝑣
𝑐
𝑐−𝑣

30
River
𝑐
𝑣
𝑐2 − 𝑣2

The Analysis
“Assuming the Galilean Transformation”
Time t1 from A to C and
back :
31
Time t2 from A to D and
back :
So that the change in time is :
𝑡1 =
𝑙1
𝑐 + 𝑣
+
𝑙1
𝑐 − 𝑣
=
2𝑐𝑙1
𝑐2 − 𝑣2
=
2
𝑐
𝑙1
1 − 𝑣2 𝑐2
𝑡2 =
𝑙2
𝑐2 − 𝑣2
+
𝑙2
𝑐2 − 𝑣2
=
2𝑙2
𝑐2 − 𝑣2
=
2
𝑐
𝑙2
1 − 𝑣2 𝑐2
∆𝑡 = 𝑡2 − 𝑡1 =
2
𝑐
𝑙2
1 − 𝑣2 𝑐2
−
𝑙1
1 − 𝑣2 𝑐2

32
TheMichelsonInterferometer
90o
Ether wind =v
Ether wind =v
Ether wind =v

The Analysis  Rotating 90o
Time t1 from A to C and
back :
33
Time t2 from A to D and
back :
So that different change in time is :
𝑡′2 =
𝑙2
𝑐 + 𝑣
+
𝑙2
𝑐 − 𝑣
=
2𝑐𝑙2
𝑐2 − 𝑣2
=
2
𝑐
𝑙2
1 − 𝑣2 𝑐2
𝑡′1 =
𝑙1
𝑐2 − 𝑣2
+
𝑙1
𝑐2 − 𝑣2
=
2𝑙1
𝑐2 − 𝑣2
=
2
𝑐
𝑙1
1 − 𝑣2 𝑐2
∆𝑡′
= 𝑡′2 − 𝑡′1 =
2
𝑐
𝑙2
1 − 𝑣2 𝑐2
−
𝑙1
1 − 𝑣2 𝑐2

34
The Analysis
∆𝑡 =
2
𝑐
𝑙2
1 − 𝑣2 𝑐2
−
𝑙1
1 − 𝑣2 𝑐2
∆𝑡
∆𝑡′ =
2
𝑐
𝑙2
1 − 𝑣2 𝑐2
−
𝑙1
1 − 𝑣2 𝑐2
∆𝑡′

The Analysis  ∆𝑡′ − ∆𝑡
The binomial expansion 1 − 𝑣2
𝑐2 −
1
2 is :
35
The binomial expansion 1 − 𝑣2 𝑐2 −1 is :
The time difference is :
1 − 𝛽2 −1
= 1 + 𝛽2
+ 𝛽4
+ 𝛽6
+ 𝛽8
⋯
1 − 𝛽2 −
1
2 = 1 +
1
2
𝛽2
+
3
8
𝛽4
+
5
16
𝛽6
+
35
128
𝛽8
⋯
∆𝑡′ − ∆𝑡 =
2
𝑐
𝑙1 + 𝑙2
1 − 𝑣2 𝑐2
−
𝑙1 + 𝑙2
1 − 𝑣2 𝑐2
=
2 𝑙1 + 𝑙2
𝑐
1
1 − 𝑣2 𝑐2
−
1
1 − 𝑣2 𝑐2
; 𝛽 =
𝑣
𝑐
; 𝛽 =
𝑣
𝑐

The Analysis  ∆𝑡′ − ∆𝑡
36
The time difference is :
∆𝑡′
− ∆𝑡 =
2 𝑙1 + 𝑙2
𝑐
1 + 𝛽2
+ 𝛽4
+ ⋯ − 1 +
1
2
𝛽2
+
3
8
𝛽4
+ ⋯
∆𝑡′
− ∆𝑡 ≈
𝑣2
𝑙1 + 𝑙2
𝑐3

37
Typical interferometer fringe pattern expected
when the system is rotated by 90°

38
Results
• Using the Earth’s orbital speed as:
o together with
o So that the time difference becomes
• Although a very small number, it was within the
experimental range of measurement for light
waves.
𝑣 = 3 × 104
𝑚/𝑠
𝑙1 ≈ 𝑙2 = 1.2 𝑚
∆𝑡′
− ∆𝑡 ≈
𝑣2 𝑙1 + 𝑙2
𝑐2
= 8 × 10−17
𝑠

39
Michelson’s Conclusion
• Michelson noted that he should be able to
detect a phase shift of light due to the time
difference between path lengths but found
none.
• He thus concluded that the hypothesis of the
stationary ether must be incorrect.
• After several repeats and refinements with
assistance from Edward Morley (1893-1923),
again a null result.
• Thus, ether does not seem to exist!

40
Possible Explanations
• Many explanations were proposed but the most
popular was the ether drag hypothesis.
• This hypothesis suggested that the Earth somehow
“dragged” the ether along as it rotates on its axis and
revolves about the sun.
• This was contradicted by stellar abberation wherein
telescopes had to be tilted to observe starlight due to
the Earth’s motion. If ether was dragged along, this
tilting would not exist.

The Lorentz-Fitz Gerald Contraction
• Another hypothesis proposed independently by both
H. A. Lorentz and G. F. FitzGerald suggested that the
length ℓ1, in the direction of the motion was
contracted by a factor of
41
1 − 𝑣2 𝑐2
…thus making the path lengths equal to account
for the zero phase shift.
o This, however, was an ad hoc assumption
that could not be experimentally tested.

42
2.3 : Einstein’s Postulates
• Albert Einstein (1879–1955) was only two years
old when Michelson reported his first null
measurement for the existence of the ether.
• At the age of 16 Einstein began thinking about
the form of Maxwell’s equations in moving
inertial systems.
• In 1905, at the age of 26, he published his
startling proposal about the principle of
relativity, which he believed to be fundamental.

43
Einstein’s Two Postulates
• With the belief that Maxwell’s equations must be
valid in all inertial frames, Einstein proposes the
following postulates:
1 “The principle of relativity : The
laws of physics are the same in all
inertial systems. There is no way to
detect absolute motion, and no
preferred inertial system exists.”
“สัจพจน์ข้อที่ 1 ของไอน์สไตน์ ในกรอบ
เฉื่อยทุกกรอบกฎเกณฑ์หรือสมการทาง
ฟิสิกส์จะมีรูปเดียวกันเสมอ”

44
Einstein’s Two Postulates
• With the belief that Maxwell’s equations must be
valid in all inertial frames, Einstein proposes the
following postulates:
2“The constancy of the speed of
light : Observers in all inertial systems
measure the same value for the speed
of light in a vacuum.”
“สัจพจน์ข้อที่ 2 ของไอน์สไตน์ อัตราเร็ว
ของแสงในสุญญากาศมีค่าคงที่ ไม่ขึ้นกับ
การเคลื่อนที่ของแหล่งกาเนิดแสงหรือผู้

Re-evaluation of Time
𝑡𝑡′
• In Newtonian physics we previously assumed that t = t’45
Thus with “synchronized” clocks,
events in K and K’ can be considered
simultaneous

Re-evaluation of Time
• Einstein realized that each system must have its own
observers with their own clocks and meter sticks 46
𝑡
Thus events considered
simultaneous in K may not be in K’
𝑡′

The Problem of Simultaneity
Frank at rest is equidistant from events A and B :
Frank
Light
flash
Light
flash
47
Frank “sees” both flashbulbs go off simultaneously.
0 +10m+10m
K
A B

The Problem of Simultaneity
Mary, moving to the right with speed v, observes
events A and B in different order :
Light
flash
48
Mary “sees” event B, then A.
Mary
0 +10m+10m
K’
𝑣
Light
flash
A B

This suggests that each coordinate system has
its own observers with “clocks” that are synchronized… 49
We thus observe…
• Two events that are simultaneous in one reference
frame (K) are not necessarily simultaneous in
another reference frame (K’) moving with respect
to the first frame.
Light
flash
Mary
0 +1m+1m
K’
𝑣
Light
flash
A BFrank
Light
flash
Light
flash
0 +1m+1m
K
A B
𝑡 𝑡′

50
Synchronization of Clocks
Step 1: Place observers with clocks
throughout a given system.
Step 2: In that system bring all
the clocks together at
one
location.
Step 3: Compare the clock readings.
If all of the clocks agree,
then the clocks are said to be synchronized.
K’

51
A method to synchronize…
• One way is to have one clock at the origin set to t
= 0 and advance each clock by a time (d/c) with d
being the distance of the clock from the origin.
• Allow each of these clocks to begin timing when a
light signal arrives from the origin.
𝑑
𝑡 =
𝑑
𝑐
𝑡 = 0
𝑑
𝑡 =
𝑑
𝑐

52
The Lorentz Transformations
known as the Lorentz transformation equations
2) account for the problem of
simultaneity between these
observers
1) preserve the constancy of the speed
of light (c) between inertial
observers;
The special set of linear transformations that :
𝐾
𝑣
𝐾′
𝑐𝑐

𝐾
53
Derivation
• Use the fixed system K and the moving system K’
• At t = 0 the origins and axes of both systems are
coincident with system K’ moving to the right along the
x axis.
• A flashbulb goes off at the origins when t = 0.
• According to postulate 2, the speed of light will be c in
both systems and the wavefronts observed in both
𝑣
𝐾′

Spherical wavefronts in 𝐾:
54
Derivation
𝐾
𝑥
𝑦
𝑧
𝑥2
+ 𝑦2
+ 𝑧2
= 𝑐2
𝑡2
𝑥 = 𝑥′
+ 𝑣𝑡, 𝑦 = 𝑦′, 𝑧 = 𝑧′ and 𝑡 = 𝑡′
𝑣
𝐾′
𝑥′
𝑦′
𝑧′
𝑥′2 + 𝑦′2 + 𝑧′2 = 𝑐2 𝑡′2
Spherical wavefronts in 𝐾′:
Note: these are not preserved in the classical transformations with

1) Let 𝑥′ = 𝛾 𝑥 − 𝑣𝑡 so that 𝑥 = 𝛾′ 𝑥′ + 𝑣𝑡′
2) By Einstein’s first postulate : 𝛾 = 𝛾′
3) The wavefront along the 𝑥, 𝑥′- axis must
satisfy :
𝑥 = 𝑐𝑡and 𝑥′ = 𝑐𝑡′
4) Thus 𝑐𝑡′
= 𝛾 𝑐𝑡 − 𝑣𝑥 𝑐 and
𝑐𝑡 = 𝛾 𝑐𝑡′ + 𝑣𝑥′ 𝑐
5) Solving the first one above for t’ and
substituting into the second... 55
Derivation

𝑥′
=
𝑥 − 𝑣𝑡
1 − 𝑣2 𝑐2
𝑥 =
𝑥′ + 𝑣𝑡′
1 − 𝑣2 𝑐2
𝑦′
= 𝑦 𝑦 = 𝑦′
𝑧′
= 𝑧 𝑧 = 𝑧′
𝑡′
=
𝑡 − 𝑣𝑥 𝑐2
1 − 𝑣2 𝑐2
𝑡 =
𝑡′
+ 𝑣𝑥′ 𝑐2
1 − 𝑣2 𝑐2
56
Thus the complete Lorentz Transformation
𝑣
𝐾′ 𝐾

𝛾 = 1/ 1 − 𝛽2 𝛽 = 𝑣 𝑐
𝑥′ = 𝛾 𝑥 − 𝑣𝑡 𝑥 = 𝛾 𝑥′ + 𝑣𝑡′
𝑦′
= 𝑦 𝑦 = 𝑦′
𝑧′
= 𝑧 𝑧 = 𝑧′
𝑡′ = 𝛾 𝑡 − 𝛽𝑥 𝑐 𝑡 = 𝛾 𝑡′ + 𝛽𝑥′ 𝑐
57
Thus the complete Lorentz Transformation
𝑣
𝐾′ 𝐾

0
1
2
3
4
5
6
7
8
9
10
0.00 0.20 0.40 0.60 0.80 1.00 1.20
Relativisticfactorγ
v/c
Properties of γ 58
• Recall β = v/c < 1 for all observers.
1) 𝛾 ≥ 1 equals 1 only when v = 0.
2)Graphofβ:notev≠c)
𝛾 =
1
1 − 𝛽2

Remarks
1) If v << c, i.e., β ≈ 0 and 𝛾 ≈ 1, we see these
equations reduce to the familiar Galilean
transformation.
2) Space and time are now not separated.
3) For non-imaginary transformations, the frame
velocity cannot exceed c.
59

60
Ex-04 :
As measured by 𝑂, a flashbulb goes off at 𝑥 = 100𝑘𝑚, 𝑦 =
10𝑘𝑚, 𝑧 = 1𝑘𝑚 at 𝑡 = 5 × 10−4
𝑠. What are the coordinates
𝑥’, 𝑦’, 𝑧’, and t’ of this event as determined by a second observer.
𝑂’, moving relative to 𝑂 at −0.8𝑐 along the common 𝑥 − 𝑥’ axis?
𝑂′(𝑥’, 𝑦’, 𝑧’)
𝑂(𝑥, 𝑦, 𝑧)

61
Ex-05 :
Suppose that a particle moves relative to 𝑂’ with a constant
velocity of 𝑐/2 in the 𝑥’𝑦’ − 𝑝𝑙𝑎𝑛𝑒 such that its trajectory makes
an angle of 60° with the 𝑥’ − 𝑎𝑥𝑖𝑠. If the velocity of 𝑂’ with
respect to 𝑂 is 0.6𝑐 along the 𝑥 − 𝑥’ 𝑎𝑥𝑖𝑠, find the equation of
motion of the particle as determined by 𝑂.
𝑂′(𝑥’, 𝑦’, 𝑧’) 𝑥′
𝑥𝑂(𝑥, 𝑦, 𝑧)

𝐴’ 𝐵’
62
Ex-06 :
A train ½ mile long (as measured by an observer on the train) is
traveling at a speed of 100 mi/hr. Two lightning bolts strike the
ends of the train simultaneously as determined by an observer on
the ground. What is the time separation as measured by an
observer on the train?
𝐴 𝐵

63
Ex-07 :
Observer 𝑂 notes that two events are separated in space and
time by 600 m and 8 × 10−7
s. How fast must an observer 𝑂’ be
moving relative to 𝑂 in order that the events be simultaneous to
𝑂’?
𝑂′(𝑥’, 𝑦’, 𝑧’)
𝑡’1
𝑡’2
𝑡1
𝑡2

64
2.5: Time Dilation and Length Contraction
• Time Dilation :
Clocks in K’ run slow with respect to stationary clocks in
K.
• Length Contraction :
Lengths in K’ are contracted with respect to the same
lengths stationary in K.
Consequences of the Lorentz Transformation:

65
Time Dilation
To understand time dilation the idea of proper time must
be understood :
• The term proper time ,T0 , is the time difference between
two events occurring at the same position in a system as
measured by a clock at that position.
Same location

66
Time Dilation
Not Proper Time
Beginning and ending of the event occur at
different positions

Mary
𝑣
K’
Melinda
𝑥′1𝑥′2
𝑡′1𝑡′2
67
Frank’s clock is at the same position in system K when the sparkler is lit
in (a) and when it goes out in (b). Mary, in the moving system K’, is
beside the sparkler at (a). Melinda then moves into the position where
and when the sparkler extinguishes at (b). Thus, Melinda, at the new
position, measures the time in system K’ when the sparkler goes out in
Time Dilation
K
Frank
𝑥1, 𝑥2
𝑡1𝑡2

68
According to Mary and Melinda…
𝑡′2 − 𝑡′1 =
𝑡2 − 𝑡1 𝑡 − 𝑣 𝑐2 𝑥2 − 𝑥1
1 − 𝑣2 𝑐2
• Mary and Melinda measure the two times for the sparkler
to be lit and to go out in system K’ as times t’1 and t’2 so that
by the Lorentz transformation:
• Note here that Frank records 𝒙 𝟐 − 𝒙 𝟏 = 𝟎 in K
with a proper time: 𝑻 𝟎 = 𝒕 𝟐 − 𝒕 𝟏 or
with 𝑇′ = 𝒕′ 𝟐 − 𝒕′ 𝟏
𝑇′ =
𝑇0
1 − 𝑣2 𝑐2
= 𝛾𝑇0

69
Time Dilation
1) T ’ > T0 or the time measured between two
events at different positions is greater than
the time between the same events at one
position: time dilation.
2) The events do not occur at the same space and
time coordinates in the two system
3) System K requires 1 clock and K’ requires 2
clocks.

2.7:ExperimentalVerification
Figure 2.18: The number of muons detected with speeds near 0.98c is much
different (a) on top of a mountain than (b) at sea level, because of the muon’s
decay. The experimental result agrees with our time dilation equation.
Time Dilation and Muon Decay
• At 2000m, we detect 1000
muons in period t0 traveling
at speed near 0.98c.
• At sea level, we detect only
542 muons in the same time
period t0 traveling at speed
near 0.98c.
A
B

71
Atomic Clock Measurement
Figure 2.20: Two airplanes took off (at different times) from Washington, D.C.,
where the U.S. Naval Observatory is located. The airplanes traveled east and
west around Earth as it rotated. Atomic clocks on the airplanes were compared
with similar clocks kept at the observatory to show that the moving clocks in
the airplanes ran slower.

72
2.8: Twin Paradox
The Set-up
Twins Mary and Frank at age 30 decide on two career paths: Mary decides to
become an astronaut and to leave on a trip 8 lightyears (ly) from the Earth at
a great speed and to return; Frank decides to reside on the Earth.
The Problem
Upon Mary’s return, Frank reasons that her clocks measuring her age must
run slow. As such, she will return younger. However, Mary claims that it is
Frank who is moving and consequently his clocks must run slow.
The Paradox
Who is younger upon Mary’s return?

73
Ex-8 :
The average lifetime of μ − 𝑚𝑒𝑠𝑜𝑛𝑠 with a speed of 0.95𝑐 is
measured to be 6 × 10−6
𝑠. Compute the average lifetime of
μ − 𝑚𝑒𝑠𝑜𝑛𝑠 in a system in which they are at rest.
𝑥𝑂(𝑥, 𝑦, 𝑧, 𝑡)
μ 𝑥′
𝑂′(𝑥′, 𝑦′, 𝑧′, 𝑡′)

74
Ex-9:
An airplane is moving with respect to the earth with a speed of
600 m/s. As determined by earth clocks, how long will it take for
the airplane’s clock to fall behind by two microseconds?

75
Length Contraction
To understand length contraction the idea
of proper length must be understood :
• Let an observer in each system K and K’ have a
meter stick at rest in their own system such that
each measure the same length at rest.
• The length as measured at rest is called the
proper length.

76
Length Contraction
K
𝑥′
𝑣K’
𝑥
𝑥′ =
𝑥 − 𝑣𝑡
1 − 𝑣2 𝑐2
𝑥′ =
𝑥 − 𝑣𝑡
1 − 𝑣2 𝑐2

77
What Frank and Mary see…
Each observer lays the stick down along
his or her respective x axis, putting the left end at
𝑥1 (or 𝑥′1) and the right end at 𝑥2 (or 𝑥′2).
•Thus, in system K, Frank measures his stick
to be :
•Similarly, in system K’, Mary measures her
stick at rest to be :
𝐿 = 𝑥2 − 𝑥1
𝐿′ = 𝑥′2 − 𝑥′1

78
What Frank and Mary measure
• Where both ends of the stick must be measured
simultaneously, i.e, 𝑡2 = 𝑡1
• Here Mary’s proper length is 𝐿0 = 𝑥′2 − 𝑥′
1
• and Frank’s measured length is 𝐿 = 𝑥2 − 𝑥1
Frank in his rest frame measures the
moving length in Mary’s frame moving with
velocity.
• Thus using the Lorentz transformations Frank
measures the length of the stick in K’ as:
𝑥′2 − 𝑥′
1 =
𝑥2 − 𝑥1 − 𝑣 𝑡2 − 𝑡1
1 − 𝑣2 𝑐2

79
Frank’s measurement
So Frank measures the moving length as 𝐿 given
by
but since both Mary and Frank in their respective
frames measure 𝐿′0 = 𝐿0
and 𝐿0 > 𝐿, i.e. the moving stick shrinks.
𝐿0 =
𝐿
1 − 𝑣2 𝑐2
= 𝛾𝐿
𝐿 = 𝐿0 1 − 𝑣2 𝑐2 =
𝐿0
𝛾

80
Ex-10 :
How fast does a rocket ship have to go for its length to 99% of its
rest length?
𝑂′(𝑥’, 𝑦’, 𝑧’)

81
Ex-11 :
Calculate the Lorentz contraction of the earth’s diameter as
measured by an observer 𝑂’ who is stationary with respect to
sun.
Orbital velocity of Earth 3 x 104 m/s
𝑂′(𝑥’, 𝑦’, 𝑧’)
Diameter of the Earth as 7920 mi

82
Ex-12 :
A meterstick makes an angle of 30° with respect to the 𝑥’ −
𝑎𝑥𝑖𝑠 of 𝑂’. What must be the value of 𝑣 if the meterstick makes
an angle of 45° with respect to the 𝑥 − 𝑎𝑥𝑖𝑠 of 𝑂?
𝑥𝑂(𝑥, 𝑦, 𝑧)
𝑦′
𝑦

𝑥′
𝑂′(𝑥′, 𝑦′, 𝑧′)
83
Ex-13 :
A cube has a (proper) volume of 1000 𝑐𝑚3
. Find the volume as
determined by an observer 𝑂’ who moves at a velocity of 0.8𝑐
relative to the cube in a direction parallel to one edge.
𝑥

84
2.6 : Addition of Velocities
• Taking differentials of the Lorentz
transformation, relative velocities may be
calculated :
𝑥′ =
𝑥 − 𝑣𝑡
1 − 𝑣2 𝑐2
Takingdifferentialsofthe
Lorentztransformation
𝑑𝑥′ =
𝑑𝑥 − 𝑣𝑑𝑡
1 − 𝑣2 𝑐2
𝑦′ = 𝑦 𝑑𝑦′ = 𝑑𝑦
𝑧′ = 𝑧 𝑑𝑧′ = 𝑑𝑧
𝑡′ =
𝑡 − 𝑣𝑥 𝑐2
1 − 𝑣2 𝑐2
𝑑𝑡′ =
𝑑𝑡 − 𝑣 𝑐2 𝑑𝑥
1 − 𝑣2 𝑐2

So that…
85
defining velocities as : 𝑢′ 𝑥 = 𝑑𝑥′ 𝑑𝑡′, 𝑢′ 𝑦 =
𝑑𝑦′ 𝑑𝑡′, 𝑢′ 𝑧 = 𝑑𝑧′ 𝑑𝑡′, etc. it is easily shown that :
𝑢′ 𝑥 =
𝑑𝑥′
𝑑𝑡′
=
𝑢 𝑥 − 𝑣
1 − 𝑣 𝑐2 𝑢 𝑥
𝑢′ 𝑦 =
𝑑𝑦′
𝑑𝑡′
=
𝑢 𝑦
𝛾 1 − 𝑣 𝑐2 𝑢 𝑥
𝑢′ 𝑧 =
𝑑𝑧′
𝑑𝑡′
=
𝑢 𝑧
𝛾 1 − 𝑣 𝑐2 𝑢 𝑥
1
2
3

86
2.6 : Addition of Velocities
• Taking differentials of the Lorentz
transformation, relative velocities may be
calculated :
𝑥 =
𝑥′ + 𝑣𝑡′
1 − 𝑣2 𝑐2
Takingdifferentialsofthe
Lorentztransformation
𝑑𝑥 =
𝑑𝑥′ + 𝑣𝑑𝑡′
1 − 𝑣2 𝑐2
𝑦 = 𝑦′ 𝑑𝑦 = 𝑑𝑦′
𝑧 = 𝑧′ 𝑑𝑧 = 𝑑𝑧′
𝑡 =
𝑡′ + 𝑣 𝑐2 𝑥′
1 − 𝑣2 𝑐2
𝑑𝑡 =
𝑑𝑡′ + 𝑣 𝑐2 𝑑𝑥′
1 − 𝑣2 𝑐2

So that…
87
defining velocities as : 𝑢 𝑥 = 𝑑𝑥 𝑑𝑡, 𝑢 𝑦 = 𝑑𝑦 𝑑𝑡,
𝑢 𝑧 = 𝑑𝑧 𝑑𝑡, etc. it is easily shown that :
𝑢 𝑥 =
𝑑𝑥
𝑑𝑡
=
𝑢′ 𝑥 + 𝑣
1 + 𝑣 𝑐2 𝑢′ 𝑥
𝑢 𝑦 =
𝑑𝑦
𝑑𝑡
=
𝑢′ 𝑦
𝛾 1 + 𝑣 𝑐2 𝑢′ 𝑥
𝑢 𝑧 =
𝑑𝑧
𝑑𝑡
=
𝑢′ 𝑧
𝛾 1 + 𝑣 𝑐2 𝑢′ 𝑥
1
2
3

88
The Resolution
1) Frank’s clock is in an inertial system during the entire trip;
however, Mary’s clock is not. As long as Mary is traveling at
constant speed away from Frank, both of them can argue that
the other twin is aging less rapidly.
2) When Mary slows down to turn around, she leaves her original
inertial system and eventually returns in a completely different
inertial system.
3) Mary’s claim is no longer valid, because she does not remain in
the same inertial system. There is also no doubt as to who is in
the inertial system. Frank feels no acceleration during Mary’s
entire trip, but Mary does.

89
Ex-14 :
Rocket A travels to the right and rocket B travels to the left, with
velocities 0.8c and 0.6c, respectively, relative to the earth. What
is the velocity of rocket A measured from rocket B ?
𝐴𝐵

90
Ex-15 :
Repeat Ex-14: If rocket A travels with a velocity of 0.8𝑐 in the
+ 𝑦 − 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 relative to the earth.(Rocket B stills travels in
the – 𝑥 − 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛)
𝐵
𝐴

91
Ex-16 :
A particle moves with a speed of 0.8𝑐 at an angle of 30o to the
𝑥 − 𝑎𝑥𝑖𝑠, as determined by 𝑂. What is the velocity of the particle
as determined by a second observer, 𝑂’, moving with a speed of
− 0.6𝑐 along the common 𝑥 − 𝑥’ 𝑎𝑥𝑖𝑠 ?
𝑂’
𝑂
𝑂′

92
2.9: Spacetime
• When describing events in relativity, it is convenient to
represent events on a spacetime diagram.
• In this diagram one spatial coordinate x, to specify position,
is used and instead of time t, ct is used as the other
coordinate so that both coordinates will have dimensions of
length.
• Spacetime diagrams were first used by H. Minkowski in 1908
and are often called Minkowski diagrams. Paths in
Minkowski spacetime are called worldlines.

93
Spacetime Diagram
Figure 2.21 : A spacetime diagram is used to specify events. The
worldline denoting the path from event A to event B is shown.
𝑐𝑡
𝑥
𝑐𝑡 𝐵
𝑐𝑡 𝐴
𝐵(𝑥 𝐵, 𝑐𝑡 𝐵)
0
World line
𝐴(𝑥 𝐴, 𝑐𝑡 𝐴)

94
Particular Worldlines
𝑐𝑡
𝑥
0
Spaceship
Light signal
𝑣 𝑐
Figure 2.22 : A light signal has the slope of 45° on a spacetime
diagram. A spaceship moving along the x axis with speed v is a
straight line on the spacetime diagram with a slope c/v.

95
Worldlines and Time
Figure 2.23 : Clocks positioned at 𝑥1 and 𝑥2 can be synchronized by sending a
light signal from a position 𝑥3 halfway between. The light signals intercept
the worldlines of 𝑥1 and 𝑥2 at the same time t.
𝑐𝑡
𝑥
0
𝑐𝑡
Light Light
𝑥1 𝑥3𝑥2
𝑡 𝑡

96
Worldlines and Time
Figure 2.23 : If the positions 𝑥1(= 𝑥’1) and 𝑥2(= 𝑥’2) of the previous figure are on a
moving system 𝐾’ when the flashbulb goes off, the times will not appear
simultaneously in system 𝐾, because the worldlines for 𝑥’1 and 𝑥’2 are slanted.
𝑐𝑡
𝑥
0
Light
Light
𝑥1 𝑥3𝑥2
𝑡2𝑡1
𝑐𝑡1
𝑐𝑡2
𝑣 𝑣

97
The Light Cone
𝑐𝑡
𝑥
Future Present
Elsewhere
Past
𝑐𝑡
𝑥
Future
Present
Elsewhere
Past
𝑦𝐴
𝐵
(𝑎) (𝑏)

98
Spacetime Interval
• Since all observers “see” the same speed of light,
then all observers, regardless of their velocities,
must see spherical wave fronts.
𝑠2
= 𝑥2
− 𝑐2
𝑡2
= 𝑥′ 2
− 𝑐2
𝑡′ 2
= 𝑠′ 2
𝑐𝑡
𝑥
𝑐𝑡
𝑐𝑡′
𝑥′
𝑐𝑡′

99
• If we consider two events, we can determine the
quantity Δs2 between the two events, and we find
that it is invariant in any inertial frame. The
quantity Δs is known as the spacetime interval
between two events.
Spacetime Interval

100
There are three possibilities for the invariant quantity Δs2:
1) Δs2 = 0 : Δx2 = c2 Δt2, and the two events can be
connected only by a light signal. The events are said to
have a lightlike separation.
2) Δs2 > 0 : Δx2 > c2 Δt2, and no signal can travel fast
enough to connect the two events. The events are not
causally connected and are said to have a spacelike
separation.
3) Δs2 < 0 : Δx2 < c2 Δt2, and the two events can be causally
connected. The interval is said to be timelike.
Spacetime Interval

101
2.10: The Doppler Effect
• The Doppler effect of sound in introductory physics is represented
by an increased frequency of sound as a source such as a train (with
whistle blowing) approaches a receiver (our eardrum) and a decreased
frequency as the source recedes.
• Also, the same change in sound frequency occurs when the source is
fixed and the receiver is moving. The change in frequency of the
sound wave depends on whether the source or receiver is moving.
• On first thought it seems that the Doppler effect in sound violates
the principle of relativity, until we realize that there is in fact a
special frame for sound waves. Sound waves depend on media such
as air, water, or a steel plate in order to propagate; however, light
does not!

103
The Relativistic Doppler Effect
Consider a source of light (for example, a star) and a receiver
(an astronomer) approaching one another with a relative velocity v.
1) Consider the receiver in system K and the light source in
system K’ moving toward the receiver with velocity v.
2) The source emits n waves during the time interval T.
3) Because the speed of light is always c and the source is moving
with velocity v, the total distance between the front and rear of
the wave transmitted during the time interval T is :
Length of wave train = cT − vT

104
The Relativistic Doppler Effect :
Source and Receiver Approaching
𝑣
𝜆 =
𝑐𝑇 − 𝑣𝑇
𝑛
• Because there are n waves, the wavelength is
given by
𝑐
𝒗𝑻
𝒄𝑻
𝒄𝑻 − 𝒗𝑻
∗ 𝑝𝑎𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 𝑛𝜆 ∗ (𝑠𝑎𝑚𝑒 𝑔𝑟𝑎𝑡𝑖𝑛𝑔 𝑑𝑠𝑖𝑛𝜃 = 𝑛𝜆)

105
• And the resulting frequency is (form 𝑣 = 𝑓𝜆)
𝑓 =
𝑐𝑛
𝑣 𝑐
o In this frame : 𝑓0 =
𝑛
𝑇0
and 𝑇0 =
𝑇
𝛾
𝑓 =
𝑐𝑓0 𝑇/𝛾
Thus : 𝑓 =
1
1 − 𝑣/𝑐
𝑓0
𝛾
=
1 − 𝑣2/𝑐2
1 − 𝑣/𝑐
𝑓0

106
• With β = v / c the resulting frequency is given by :
𝑓 =
1 + 𝛽
1 − 𝛽
𝑓0
𝑣 𝑐
(source and receiver
approaching)

107
Source and Receiver Receding
𝑣
𝜆 =
𝑐𝑇 + 𝑣𝑇
𝑛
• Because there are n waves, the wavelength is
given by
𝒗𝑻 𝒄𝑻
𝒄𝑻 + 𝒗𝑻
∗ 𝑝𝑎𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 𝑛𝜆 ∗ (𝑠𝑎𝑚𝑒 𝑔𝑟𝑎𝑡𝑖𝑛𝑔 𝑑𝑠𝑖𝑛𝜃 = 𝑛𝜆)
𝑐

108
• In a similar manner, it is found that :
Source and Receiver Receding
𝑓 =
1 − 𝛽
1 + 𝛽
𝑓0
(source and receiver receding)
𝑣 𝑐

109
The Relativistic Doppler Effect
• If we agree to use a +
sign for β (+v/c) when
the source and
receiver are
approaching
• If we agree to use a –
sign for β (–v/c) when
they are receding
𝑓 =
1 + 𝛽
1 − 𝛽
𝑓0𝑓 =
1 − 𝛽
1 + 𝛽
𝑓0

110
2.11: Relativistic Momentum
Because physicists believe that the
conservation of momentum is fundamental, we
begin by considering collisions where there do
not exist external forces and
𝑑 𝑝
𝑑𝑡
= 𝐹𝑒𝑥𝑡 = 0𝐹 𝐴 𝐹 𝐵
Newton’s first law (law of inertia) : An object in motion with a constant
velocity will continue in motion unless acted upon by some net external
force.

𝑣
111
• To see how the classical form p = mu fails and to
determine the correct relativistic definition of p,
consider the case of an inelastic collision
Relativistic Momentum
𝑥′𝑥
𝑣
𝑚1
𝑢1
𝑚 2
−𝑢 2
𝑚1 𝑚 2
𝑚′1
𝑢′1
𝑚′ 2
−𝑢′ 2
𝑚1 𝑚 2𝑆′𝑆
Before
After
Before
After

112
• Rather than abandon the conservation of linear
momentum, let us look for a modification of the
definition of linear momentum that preserves both it and
Newton’s second law.
• To do so requires reexamining mass to conclude that :
𝛾 =
1
1−𝛽2
and 𝛽 = 𝑣/𝑐
𝑝 = 𝑚 𝑣 = 𝛾𝑚0 𝑣 : Relativistic momentum

113
 Some physicists like to refer to the mass in Equation (2.48)
as the rest mass m0 and call the term m = γm0 the relativistic
mass. In this manner the classical form of momentum, m, is
retained. The mass is then imagined to increase at high
speeds.
 Most physicists prefer to keep the concept of mass as an
invariant, intrinsic property of an object. We adopt this
latter approach and will use the term mass exclusively to
mean rest mass. Although we may use the terms mass and
rest mass synonymously, we will not use the term relativistic
mass. The use of relativistic mass to often leads the student
into mistakenly inserting the term into classical expressions
where it does not apply.

114
2.12 : Relativistic Energy
• Due to the new idea of relativistic mass, we must now
redefine the concepts of work and energy.
• Therefore, we modify Newton’s second law to include our new
definition of linear momentum, and force becomes:
𝐹 =
𝑑 𝑝
𝑑𝑡
=
𝑑
𝑑𝑡
𝛾𝑚0 𝑣
𝑊 = 0
𝑥
𝐹 𝑑 𝑥

115
Momentum and Energy (continued)
𝑝2
𝑐2
= 𝐸2
− 𝐸0
2
• The first term on the right-hand side is just E2, and the
second term is E0
2. The last equation becomes
• We rearrange this last equation to find the result we are
seeking, a relation between energy and momentum.
𝐸2
= 𝑝2
𝑐2
+ 𝐸0
2
𝐸2
= 𝑝2
𝑐2
+ 𝑚0
2
𝑐4
• Equation is a useful result to relate the total energy of a particle with
its momentum. The quantities (E2 – p2c2) and m are invariant
quantities. Note that when a particle’s velocity is zero and it has no
momentum, Equation correctly gives E0 as the particle’s total energy.

116
2.13 : Computations in Modern Physics
• We were taught in introductory physics that the
international system of units is preferable when doing
calculations in science and engineering.
• In modern physics a somewhat different, more
convenient set of units is often used.
• The smallness of quantities often used in modern physics
suggests some practical changes.

117
Units of Work and Energy
• Recall that the work done in accelerating a charge
through a potential difference is given by W = qV.
• For a proton, with the charge e = 1.602 × 10−19 C being
accelerated across a potential difference of 1 V, the work
done is
W = (1.602 × 10−19)(1 V) = 1.602 × 10−19 J

118
The Electron Volt (eV)
• The work done to accelerate the proton across a potential
difference of 1 V could also be written as
W = (1 e)(1 V) = 1 eV
• Thus eV, pronounced “electron volt,” is also a unit of
energy. It is related to the SI (Système International) unit joule by
the 2 previous equations.
1 eV = 1.602 × 10−19 J

Other Units
1) Rest energy of a particle:
Example: E0 (proton)
2) Atomic mass unit (amu):
Example: carbon-12
119
Mass (12C atom)
Mass (12C atom)

120
Ex-17 :
A star is receding from the earth at speed of 5 x 10-3 c .
What is the wavelength shift for the sodium D2 line (5890
A)?

121
Ex-18 :
Suppose that the Doppler shift in the sodium D2 line
(5890 A) is 100 A when the light is observed from a
distant star. Determine the star’s velocity of recession.

122
Ex-19 :
A man in a rocket ship moving with a speed of 0.6c
away from a space platform shines a light of
wavelength 5000 A toward the platform. What is the
frequency of the light as seen by an observer on the
platform.

123
Ex-22 :
โปรตอนเคลื่อนที่ด้วยความเร็วเท่าใด มวลของโปรตอนจึงเป็นสองเท่าของมวลนิ่งจากสมการ
𝑚 = 𝛾𝑚0 =
𝑚0
1−(𝑣/𝑐)2
𝑚0 𝑣
ก่อน หลัง
𝑚

124
Ex-23 :
From the rest masses listed in the Appendix calculate the
rest energy of an electron in joules and electron-volts.

125
Ex-24 :
A body at rest spontaneously breaks up into two part which
move in opposite directions. The part have rest masses of 3
kg and 5.33 kg and respective speeds of 0.8c and 0.6c .
Find the rest mass of the original body.

126
Ex-25 :
What is the speed of an electron that is accelerated through
a potential difference of 105 V?

127
Ex-26 :
Calculate the momentum of 1 MeV electron.

128
Ex-27 :
Calculate the kinetic energy of an electron whose
momentum is 2 MeV/c.

129
Ex-28 :
Calculate the velocity of a electron whose kinetic
energy is 2 MeV.

130
Ex-29 :
Calculate the velocity of a electron whose kinetic
energy is 2 MeV.

131
Ex-29 :
Calculate the momentum of an electron whose velocity
is 0.8c.

132
Ex-30 :
Compute the effective mass of a 5000 A photon.

Binding Energy
• The equivalence of mass and energy becomes apparent when we
study the binding energy of systems like atoms and nuclei that
are formed from individual particles.
• The potential energy associated with the force keeping the system
together is called the binding energy EB.
133

Binding EnergyThe binding energy is the difference between the rest energy of the
individual particles and the rest energy of the combined bound system.
134

Electromagnetism and Relativity
• Einstein was convinced that magnetic fields appeared as
electric fields observed in another inertial frame. That
conclusion is the key to electromagnetism and relativity.
• Einstein’s belief that Maxwell’s equations describe
electromagnetism in any inertial frame was the key that led
Einstein to the Lorentz transformations.
• Maxwell’s assertion that all electromagnetic waves travel at
the speed of light and Einstein’s postulate that the speed of
light is invariant in all inertial frames seem intimately
connected.
135

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