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CHAPTER 2 :
Special Theory of Relativity
Napatsakon SarapatSchool of Physics, Science and Technology, TRU
SC 4101307 Modern Physics, TRU
𝛾 =1
1 − 𝑣2
𝑐2
2.1 The Need for Ether2.2 The Michelson-Morley Experiment2.3 Einstein’s Postulates2.4 The Lorentz Transformation2.5 Time Dilation and Length Contraction2.6 Addition of Velocities2.7 Experimental Verification
2.8 Twin Paradox2.9 Spacetime2.10 Doppler Effect2.11 Relativistic Momentum2.12 Relativistic Energy2.13 Computations in Modern Physics2.14 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.
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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.
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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
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11
• Time (t) for all observers is a Fundamental invariant, i.e., the same for all inertial observers
𝑥′ = 𝑥 − 𝑢𝑡𝑦′ = 𝑦𝑧′ = 𝑧𝑡′ = 𝑡
The Inverse Relations
Step 1. Replace withStep 2. Replace “primed” quantities with
“unprimed” and “unprimed” with “primed”
12
𝑥 = 𝑥′ + 𝑢𝑡𝑦 = 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 ?
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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.
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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-
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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.
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𝑡
𝑡′
18
Ether
2.1 : The Need for Ether
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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
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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
𝜇0𝜀0
𝛻2 − 𝜇𝜀𝜕2
𝜕𝑡2E = 0
𝛻2 − 𝜇𝜀𝜕2
𝜕𝑡2B = 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.
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EM-wave EM-wave
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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
24
The M
ichels
on In
terfer
omete
r
Ether wind =v
25
Th
e M
ich
els
on
In
ter
fer
om
ete
r
26
The M
ichels
on In
terfer
omete
r
27
Rive
r𝑣
𝑐
𝑐+
𝑣
28
Rive
r
𝑣
𝑐
𝑐−
𝑣
29
Rive
r
𝑣
𝑐2 − 𝑣2
30
Rive
r
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+
𝑙1𝑐 − 𝑣
=2𝑐𝑙1
𝑐2 − 𝑣2=
2
𝑐
𝑙11 − 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−
𝑙11 − 𝑣2 𝑐2
32
The M
ichels
on In
terfer
omete
r 90o
Ether wind =v
Ether wind =v
Ether wind =v
The Analysis Rotating 90o
“Assuming the Galilean Transformation”
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+
𝑙2𝑐 − 𝑣
=2𝑐𝑙2
𝑐2 − 𝑣2=
2
𝑐
𝑙21 − 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
𝑙21 − 𝑣2 𝑐2
−𝑙1
1 − 𝑣2 𝑐2
34
The Analysis“Assuming the Galilean Transformation”
∆𝑡 =2
𝑐
𝑙2
1 − 𝑣2 𝑐2−
𝑙11 − 𝑣2 𝑐2
∆𝑡
∆𝑡′ =2
𝑐
𝑙21 − 𝑣2 𝑐2
−𝑙1
1 − 𝑣2 𝑐2
∆𝑡′
The Analysis ∆𝑡′ − ∆𝑡“Assuming the Galilean Transformation”
The binomial expansion 1 − 𝑣2 𝑐2 −1
2 is :
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The binomial expansion 1 − 𝑣2 𝑐2 −1 is :
The time difference is :
1 − 𝛽2 −1 = 1 + 𝛽2 + 𝛽4 + 𝛽6 + 𝛽8 ⋯
1 − 𝛽2 −12 = 1 +
1
2𝛽2 +
3
8𝛽4 +
5
16𝛽6 +
35
128𝛽8 ⋯
∆𝑡′ − ∆𝑡 =2
𝑐
𝑙1 + 𝑙21 − 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“Assuming the Galilean Transformation”
36
The time difference is :
∆𝑡′ − ∆𝑡 =2 𝑙1 + 𝑙2
𝑐1 + 𝛽2 + 𝛽4 + ⋯ − 1 +
1
2𝛽2 +
3
8𝛽4 + ⋯
∆𝑡′ − ∆𝑡 ≈𝑣2 𝑙1 + 𝑙2
𝑐3
37
Typical interferometer fringe pattern expectedwhen 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.”“สจัพจน์ข้อที ่ 1 ของไอน์สไตน์ ในกรอบเฉือ่ยทุกกรอบกฎเกณฑ์หรือสมการทางฟิสกิสจ์ะมรีปูเดยีวกนัเสมอ”
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.”“สจัพจน์ขอ้ที ่ 2 ของไอน์สไตน์ อตัราเร็วของแสงในสุญญากาศมคี่าคงที ่ ไม่ขึ้นกบัการเคลือ่นทีข่องแหล่งก าเนิดแสงหรือผู้สงัเกต”
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
Lightflash
Lightflash
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 :
Lightflash
48
Mary “sees” event B, then A.
Mary
0 +10m+10m
K’
𝑣Lightflash
A B
This suggests that each coordinate system hasits 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.
Lightflash
Mary
0 +1m+1m
K’𝑣
Lightflash
A BFrank
Lightflash
Lightflash
0 +1m+1m
K
A B
𝑡 𝑡′
50
Synchronization of ClocksStep 1: Place observers with clocks
throughout a given system.
Step 2: In that system bring allthe 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 systems must be spherical.
𝑣
𝐾′
Spherical wavefronts in 𝐾:
54
Derivation
𝐾
𝑥
𝑦
𝑧
𝑥2 + 𝑦2 + 𝑧2 = 𝑐2𝑡2
𝑥 = 𝑥′ + 𝑣𝑡, 𝑦 = 𝑦′, 𝑧 = 𝑧′ and 𝑡 = 𝑡′
𝑣
𝐾′
𝑥′
𝑦′
𝑧′
𝑥′2 + 𝑦′2 + 𝑧′2 = 𝑐2𝑡′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 𝑐2𝑥 =
𝑥′ + 𝑣𝑡′
1 − 𝑣2 𝑐2
𝑦′ = 𝑦 𝑦 = 𝑦′
𝑧′ = 𝑧 𝑧 = 𝑧′
𝑡′ =𝑡 − 𝑣𝑥 𝑐2
1 − 𝑣2 𝑐2𝑡 =
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
Rel
ativ
isti
c fa
cto
r γ
v/c
Properties of γ 58
• Recall β = v/c < 1 for all observers.
1) 𝛾 ≥ 1 equals 1 only when v = 0.
2)
Gra
ph
of β
: n
ote
v ≠
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’)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 DilationTo 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 (b).
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 𝒙𝟐 − 𝒙𝟏 = 𝟎 in K with a proper time: 𝑻𝟎 = 𝒕𝟐 − 𝒕𝟏 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: Experimental Verification
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’sdecay. 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 ParadoxThe Set-upTwins 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 ProblemUpon 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 ParadoxWho 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’)
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’)
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 0.8𝑐relative to the cube in a direction parallel to one edge.
𝑥𝑂(𝑥, 𝑦, 𝑧)
84
2.6 : Addition of Velocities
• Taking differentials of the Lorentz transformation, relative velocities may be calculated :
𝑥′ =𝑥 − 𝑣𝑡
1 − 𝑣2 𝑐2
Taki
ng di
ffere
ntial
s of t
he
Lore
ntz t
rans
form
ation 𝑑𝑥′ =
𝑑𝑥 − 𝑣𝑑𝑡
1 − 𝑣2 𝑐2
𝑦′ = 𝑦 𝑑𝑦′ = 𝑑𝑦
𝑧′ = 𝑧 𝑑𝑧′ = 𝑑𝑧
𝑡′ =𝑡 − 𝑣𝑥 𝑐2
1 − 𝑣2 𝑐2𝑑𝑡′ =
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′=
𝑢𝑥 − 𝑣
1 − 𝑣 𝑐2 𝑢𝑥
𝑢′𝑦 =𝑑𝑦′
𝑑𝑡′=
𝑢𝑦
𝛾 1 − 𝑣 𝑐2 𝑢𝑥
𝑢′𝑧 =𝑑𝑧′
𝑑𝑡′=
𝑢𝑧
𝛾 1 − 𝑣 𝑐2 𝑢𝑥
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
Taki
ng di
ffere
ntial
s of t
he
Lore
ntz t
rans
form
ation 𝑑𝑥 =
𝑑𝑥′ + 𝑣𝑑𝑡′
1 − 𝑣2 𝑐2
𝑦 = 𝑦′ 𝑑𝑦 = 𝑑𝑦′
𝑧 = 𝑧′ 𝑑𝑧 = 𝑑𝑧′
𝑡 =𝑡′ + 𝑣 𝑐2 𝑥′
1 − 𝑣2 𝑐2𝑑𝑡 =
𝑑𝑡′ + 𝑣 𝑐2 𝑑𝑥′
1 − 𝑣2 𝑐2
So that…
87
defining velocities as : 𝑢𝑥 = 𝑑𝑥 𝑑𝑡, 𝑢𝑦 = 𝑑𝑦 𝑑𝑡, 𝑢𝑧 = 𝑑𝑧 𝑑𝑡, etc. it is easily shown that :
𝑢𝑥 =𝑑𝑥
𝑑𝑡=
𝑢′𝑥 + 𝑣
1 + 𝑣 𝑐2 𝑢′𝑥
𝑢𝑦 =𝑑𝑦
𝑑𝑡=
𝑢′𝑦𝛾 1 + 𝑣 𝑐2 𝑢′𝑥
𝑢𝑧 =𝑑𝑧
𝑑𝑡=
𝑢′𝑧𝛾 1 + 𝑣 𝑐2 𝑢′𝑥
1
2
3
88
The Resolution1) 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
𝑐𝑡𝐵
𝑐𝑡𝐴
𝐵(𝑥𝐵 , 𝑐𝑡𝐵)
0
World line𝐴(𝑥𝐴, 𝑐𝑡𝐴)
94
Particular Worldlines
𝑐𝑡
𝑥0
Spaceship
Light signal
𝑣 𝑐
Figure 2.22 : A light signal has the slope of 45° on a spacetimediagram. 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
𝑥0
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
𝑥0
Light
Light
𝑥1 𝑥3𝑥2
𝑡2𝑡1
𝑐𝑡1𝑐𝑡2
𝑣 𝑣
97
The Light Cone𝑐𝑡
𝑥
Future PresentElsewhere
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 − 𝑐2𝑡2 = 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 intervalbetween 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 spacelikeseparation.
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!
102Recall the Doppler Effect
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
𝑐𝑇 − 𝑣𝑇
The Relativistic Doppler Effect :Source and Receiver Approaching
𝑣 𝑐
o In this frame : 𝑓0 =𝑛
𝑇0and 𝑇0 =
𝑇
𝛾
𝑓 =𝑐𝑓0𝑇/𝛾
𝑐𝑇 − 𝑣𝑇
Thus : 𝑓 =1
1 − 𝑣/𝑐
𝑓0𝛾
=1 − 𝑣2/𝑐2
1 − 𝑣/𝑐𝑓0
106
• With β = v / c the resulting frequency is given by :
The Relativistic Doppler Effect :Source and Receiver Approaching
𝑓 =1 + 𝛽
1 − 𝛽𝑓0
𝑣 𝑐
(source and receiver approaching)
107
The Relativistic Doppler Effect :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 :
The Relativistic Doppler Effect :Source and Receiver Receding
𝑓 =1 − 𝛽
1 + 𝛽𝑓0
(source and receiver receding)
𝑣 𝑐
109The 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 − 𝛽𝑓0𝑓 =
1 − 𝛽
1 + 𝛽𝑓0
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 = 0
𝐹 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
𝑚 1
𝑢 1𝑚 2
−𝑢 2
𝑚 1 𝑚 2
𝑚′ 1𝑢′ 1
𝑚′ 2−𝑢′ 2
𝑚 1 𝑚 2𝑆′𝑆
BeforeAfter
BeforeAfter
112
Relativistic Momentum
• 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−𝛽2and 𝛽 = 𝑣/𝑐
𝑝 = 𝑚 𝑣 = 𝛾𝑚0 𝑣 : Relativistic momentum
113
Relativistic Momentum
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
𝑑𝑡𝛾𝑚0 𝑣
𝑊 = 0𝑥 𝐹 𝑑 𝑥
115
Momentum and Energy (continued)
𝑝2𝑐2 = 𝐸2 − 𝐸02
• 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 = 𝑝2𝑐2 + 𝐸02
𝐸2 = 𝑝2𝑐2 + 𝑚02𝑐4
• 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 Units1) 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 :โปรตอนเคล่ือนท่ีดว้ยความเร็วเท่าใด มวลของโปรตอนจึงเป็นสองเท่าของมวลน่ิงจากสมการ
𝑚 = 𝛾𝑚0 =𝑚0
1−(𝑣/𝑐)2
𝑚0 𝑣
ก่อน หลงั
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
A Conducting Wire
136
