1ST CHAPTER Special theory of relativity

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1ST CHAPTERSpecial theory of relativity
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ZCT 104/3E

• ENGLISH TEACHING
• 5 CHAPTERS WILL BE COVERED
• Special theory of relativity
• Wave nature of particles
• Particle nature of waves
• Introduction to Quantum Mechanics
• Atomic models

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LECTURE 1

• Failure of Newtonian mechanics

Newton
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Revision Still remember Newton's 3 law of
motion?

• An object at rest will always be in the wrong
  place
• An object in motion will always be headed in the
  wrong direction
• For every action, there is an equal and opposite
  criticism
• b, just joking

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Newtonian view of space and time

• Space and time are absolute
• Time flow independently of the state of motion of
  any physical system in 3-D space
• In essence, time and space do not mix. The state
  of motion of a physical system does not affact
  the rate of time flow within the system

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Inertial frames

• Inertial frames of reference is one in which an
  object subject to no forcess moves in straight
  line at constant speed
• E.g. of inertial frames the lab frame and the
  constant-speed car frame
• Newtonian law of invariance (or called principle
  of Newtonian relativity) All inertial frames are
  equivalent, and the law of mechanics must be also
  take the same methematical form for all observers
  irrespective of their frame of references

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Example of inertial frames of reference
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Example of form invariance

• In the aeroplane with constant speed wrp to the
  ground, Newton second law takes the form of F
  m a
• In the lab frame, Newton 2nd law is F ma

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The laws of mechanics must be the same in all
inertial frames of reference

• Although the ball path is different in both
  inertial reference frames, both observers agree
  on the validity of Newtons law, conservation of
  energy and others physical principles

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Galilean transformation

• It relates the kinematical quantities, such as
  position, velocity, acceleration between two
  inertial frames
• S stationary frame (uses x,y,z,t as their
  coordinates)
• S moving wrp to S with constant speed u away
  from S (uses x,y,z,t as their coordinates)
• Galilean transformation for the coordinates (in
  1-D)
• x x vt, y y, zz, t t
• Galilean addition law for velocity (in 1-D)
• vx vx - v
• Simply a daily experience

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Galilean transformation and Newtonian view goes
hand-in-hand

• Galilean transformation assumes the notion of
  absolute space and time as hold by Newton, i.e.
  the length is independent of the state of motion.
  So is the flowing rate of time.

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Example

• Apply GT on th previous example
• The trajactories of the ball seen in the two
  frames, S (van) and S (ground observer) is
  related by GT as per

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Example

• Using Galilean transformation of corrdinates, one
  can show that observer in S and S measure
  different cordinates for the ends of a stick at
  rest in S, they agree on the length of the stick.
  Assume the stick has end coodintaes x a and x
  a l in S.

• Doraemiyan (S) measure the end points of the
  stick at the same time, t.
• Using x x vt
• x(head) x(head) vt a - vt
• x(end) x(end) vt (al) - vt
• x(end) - x(head) x(end) - x(head) l

x(h) a
x(t) l a
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Galilean transformation when applied on light
means speed of light is not constant
Frame S travel with velocity v relative to S. If
light travels with the same speed in all
directions relative too S, then (according to the
classical Galilean velocity-addition) it should
have different speeds as seen from S.
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Maxwell theory of light is inconsistent with
Newtons law of invariance

• Consider a gadanken case in an inertial frame
  moving at the speed of light, the electromagnetic
  (EM) wave is frozen and not waving anymore
• Maxwell theory of EM wave will fail in the
  light-speed frame of reference
• ? Galilean transformation is inconsistent with
  Maxwell theory of light
• ? Newtonian law of invariance fails for EM in the
  light-speed frame
• ? Galilean transformation is going to fail when v
  is approaching the speed of light it has to be
  supplanted by Lorentz transformation (to be
  learned later)

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Ether and Michelson-Morley Experiments

• In early 19th century, its thought (incorrectly)
  that
• there exist an omi-pervasive medium called
  Ether in which light propagates at a speed of
  3x108m/s (analogue to sound propagate in the
  mechanical medium of still air at speed 330m/s)
• Thought to be the absolute frame of reference
  that goes in accordance with Newtonian view of
  absolute space and time
• The effect of the ehter on speed of light can be
  experimentally measured

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Ether and Michelson-Morley Experiments

• If exists, from the viewpoint of the light
  source, the Ether wind appears to drift with a
  relative speed of u wrp to Earth (One assumes
  that ether frame is fixed wrp to the Sun, hence
  one expects u 10-4c)
• Consider a moving souce giving out two beams of
  light in different direction (say, 90 degree to
  each other)
• Since the light source is moving through the
  omi-perasive ehter medium, the different
  directions of the two beams of light would mean
  that these two beams will move with different
  velocities when viewed in the frame of moving
  source

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Experimental setup

• Both arms has same length L
• According to the ether wind concept
• For arm 1, the speed of light to is c-v as it
  approaches M2,
• cv as it is reflected from M2

v
c c - v
v
c
c c v
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Experimental setup

• For arm 2, the speed of light to-and-fro M1 is

v
c v (c2 - v2)
c
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• The two light beams start out in phase. When they
  return and recombined by semi-transparent
  mirror Mo interference pattern will be formed
  due to their difference in phase, Df cDt /l,
  where Dt t1 - t2 Lv2/c3 is the time
  difference between the light beams when return to
  Mo (figure a)

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• Now, when the whole set-up is rotated through
  90?, arms 1 and 2 exchanges role
• As a result, the interefence pattern will be
  shifted as the time difference between the beams
  after rotation now becomes 2Dt
• The number of interefence fringes shifted can be
  estimated via
• no. of fringe shift 2cDt /l 2 Lv2/lc2 0.40
• (taking v 10-4c, L 11 m, l 500nm)
• Very precise experiment

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• Fig. (b) shows expected fringe shift after a
  rotation of the interometer by 90 degree

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NULL result

• But MM sees only NULL result no change in the
  interference pattern
• How to interprete the null result?
• If Maxwell theory of light is right (as EM wave)
  the notion of ether as an medium in which light
  is propagating has to be discarded
• Put simply ether is not shown to exist
• Einstein put it more strongly the absolute frame
  of reference (i.e. the ether frame) has to be
  discarded

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PYQ (past year question), KSCP 2003/04

• What were the consequences of the negative result
  of the Michelson-Morley experiment?
• I. It render untenable the hypothesis of the
  ether
• II. It suggests the speed of light in the free
  space is the same everywhere, regardless of any
  motion of source or observer
• III. It implies the existence of a unique frame
  of reference in which the speed of light in this
  frame is equal to c
• A. III only B. I,II C. I, III D. I, II, III
• E. Non of the above

Ans B
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Principle of special relativity

• Einstein believes that pure thought is sufficient
  to understand the world
• The most incomprehensive thing in the universe is
  that the universe is comprehensible

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Classical EM theory is inconsistent with Galileao
transformation

• Their is inconsistency between EM and Newtonian
  view of absolute space and time
• Einstein proposed SR to restore the
  inconsistency between the two based on two
  postulates

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Postulates of SR

1. The laws of physics are the same in all inertial
   reference frames a generalisation of Newtons
   relativity
2. The speed of light in vacuum is the same for all
   observers independent of the motion of the source
   constancy of the speed of light

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• Postulate 2 simply means that Galilean
  transformation cannot be applied on light speed.
  It also explains the Null result of the MM
  experiment
• Speed of light is always the same whether one is
  moving or stationary wrp to the source its
  speed doesnt increase or reduced when the light
  source is moving

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Einsteins notion of space-time drastically
revolutionarizes that of Newtons

• The notion of absolute frame of reference is
• discarded
• The Newton notion that time is absolute and
• flows independently of the state of motion (or
• the frame of reference chosen) is radically
• modified the rate of time flow does depends
• on the frame of reference (or equivalently, the
• state of motion).
• This being so due to the logical consequence
• of the constancy of the speed of light in all
  
• inertial frame

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PYQ (past year question), Final 2003/04

• Which of the following statement(s) is (are)
  true?
• I. The assumption of the Ether frame is
  inconsistent with the experimental observation
• II. The speed of light is constant
• III. Maxwell theory of electromagnetic radiation
  is inconsistent with the notion of the Ether
  frame
• IV Special relativity is inconsistent with the
  notion of the Ether frame
• A. III,IV B. I, II, III C. I, II, III,IV
• D. I, II E. I, II,IV
• ANS E, my own question

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Simultaneity is not an absolute concept but frame
dependent

• Simultaneity in one frame is not guaranteed in
  another frame of reference (due to postulate 2)
• Two lightning bolts strike the ends of a moving
  boxcar. (a) The events appear to be simultaneous
  to the stationary observer at O but (b) for the
  observer at O, the front of the train is struck
  before the rear

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Try to calculate it yourself

• The breakdown of simultaneity means that the two
  lights from A and B are not arriving at O at
  the same time. Can you calculate what is the time
  lag, i.e. tA-tB, between the two lights arriving
  at O? t is the time measured in the O frame.

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Time dilation as a consequence of Einsteins
postulate

• In frames that are moving wrp to the stationary
  frame, time runs slower
• Gedanken experiment (thought experiment)

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Gedanken Experiment

• Since light speed c is invariant (i.e. the same
  in all frames), it is used to measure time and
  space
• A mirror is fixed to a moving vehicle, and a
  light pulse leaves O at rest in the vehicle.
• (b) Relative to a stationery observe on Earth,
  the mirror and O move with a speed v.

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Ligth triangle

• Consider the geometry of the triangle of the
  light
• We can calculate the relationship between Dt, Dt
  and v
• l 2 (cDt/2 )2
• d 2 (u Dt/2 )2

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Lorentz factor, g

• Due to constancy of light postulate, both
  observer must agree on c
• Speed of light total distance travelled divide
  by time taken
• For observer in O, c 2 d /Dt
• For observer in O, c 2 l / Dt, where
• l 2 d 2 (u Dt/2 )2
• Eliminating l and d, Dt gDt, where
• g (1 - u2/c2)-1/2
• Lorentz factor, always gt or equal 1, so that
• Dt gt Dt

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Proper time

• Try to discriminate between two kinds of time
  interval
• Dt , proper time that measures the time interval
  of the two events at the same point in space
  (e.g. light emitted and received at the same
  point in the vehicle)
• Proper time is the time measured by a clock that
  is stationary wrp to the events that it measures
• Note that proper time is always shorter
  compared to improper time

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• The elapsed time Dt between the same events in
  any other frame is dilated by a factor of g
  compared to the proper time interval Dt
• In other words, according to a stationary
  observer, a moving clock runs slower than an
  identical stationary clock
• Chinese proverb
• 1 day in the heaven 10 years in the human plane
  ?????,?????

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Example

• The watch of a student in the class is running
  at a rate different than that of a student
  ponteng class to lumba motosikal haram.
• The time of the student on the bikes is running
  at a slower rate compared to that of the student
  in the class
• Onc can imagine that when the watch on the arms
  of the motocyclist ticks once in a second (as is
  concluded by the local, or rest, observer, i,e,
  the motocyclist), the student in the class
  (non-local observer) find the watch of the
  motocyclist ticks at 1.000001 second per second.

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To recap

• Dt gDt proper time interval, Dt Dt
• The rate of time flowing in one frame is
  different from the others (frames that are moving
  with a constant speed relative to a give frame)
• The relationship between the time intervals of
  the two frames moving at an non-zero relatively
  velocity are given by the time dilation formula
• One must be aware of the subtle different between
  which is the proper time and which is the
  improper one

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Example

• When you are measuring the time interval between
  your heartbeats (on your bed in you bedroom)
  using your watch, you are measuring the proper
  time interval
• Say a doctor who is in a car traveling at some
  constant speed with recpect to you is monitoring
  your heartbeat by some remote device. The time
  interval between the heartbeat measured by him,
  is improper time because he is moving wrp to you

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PYQ, Semester Test I, 2003/04

• Suppose that you are travelling on board a
  spacecraft that is moving with respect to the
  Earth at a speed of 0.975c. You are breathing at
  a rate of 8.0 breaths per minute. As monitored on
  Earth, what is your breathing rate?
• A. 13.3 B. 2.88 C.22.2
• D. 1.77 E. Non of the above
• ANS D, Cutnell, Q4, pg. 877

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Solution

• Suppose that you are travelling on board a
  spacecraft that is moving with respect to the
  Earth at a speed of 0.975c. You are breathing at
  a rate of 8.0 breaths per minute. As monitored on
  Earth, what is your breathing rate?
• g 1/(1 u2/c2)1/2 1/(1 0.9752)1/2 4.5
• Use Dt gDt
• Given local interval between breaths Dt 1/8
  0.125 min per breath (proper time interval)
• ? Dt gDt 4.5 x 0.125 0.563 min per breath
• ? 1/ Dt 1.77 breath per min (as seen by the
  spaccraft observer)
• To an oberver on the spacecraft, you seem to
  breath at a slower rate

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Example (read it yourself)

• A spacecraft is moving past the Earth at a
  constant speed 0.92c. The astronaut measures the
  time interval between successive ticks'' of the
  spacecraft clock to be 1.0 s. What is the time
  interval that an Earth observer measures between
  ticks'' of the astronaut's clock?

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Solution

• Dt 1.0 s is the proper time interval measured
  by the astronaut
• Earth observer measures a greater time interval,
  Dt, than does the astronaut, who is at rest
  relative to the clock
• The Lorentz factor g 1/(1- u2/c2) -1/2 1/(1-
  0.922)-1/2 2.6
• Hence, Dt gDt 2.6 x 1.0s 2.6 s

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Example Muon decay lifetime

• A muon is an unstable elementary particle which
  has a lifetime t0 2.2 microsecond (proper
  time, measured in the muon rest frame) and decays
  into lighter particles.
• Fast muons (say, travelling at v 99c) are
  created in the interactions of very high-energy
  particles as they enter the Earth's upper
  atmosphere.

• Assume v 0.99c
• In the muon rest frame, the distance travelled by
  muon before decay is
• D (0.99c)t0
• 650 m

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• A muon travelling at 99 the speed of light.
• has a Lorentz factor g 7.09
• Hence, to an observer in the rest frame (e.g
  Earth) the lifetime of the muon is no longer t0
  2.2 ms but
• t g x t0 7.09 x 2 microseconds 15.6 ms
• Thus the muon would appear to travel for 15.6
  microseconds before it decays
• The distance it traversed as seen from Earth
• is D (0.99c) x 15.6 ms 4,630 km (c.f. D
  650 m )

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Muon are detected at a much lower altitude

• Observation has verified the relativistic effect
  of time dilation muons are detected at a
  distance of 4700 m below the atmospheric level in
  which they are produced
• Hence the dilated muon lifetime is confirmed
  experimentally

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(No Transcript)
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Length contraction

• Length measured differs from frame to frame
  another consequence of relativistic effect
• Gedanken experiment again!

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• Two observers O on Earth, O traveling to and
  fro from Earth and alpha centauri with speed u
• Total distance between Earth - alpha centauri
  Earth, according to O (Earth observer), L0
• O sees O return to Earth after Dt0
• Observer O in a spaceship is heading aC with
  speed u and returns to Earth after Dt according
  to his clock

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Use some simple logics

• In O 2L0 uDt0
• In O 2L0 uDt0
• Due to time dilation effect, Dt0 is shorter
  than Dt0 , i.e. Dt0 gt Dt0
• Dt0 is related to Dt0 via a time dilation
  effect, Dt0 Dt0 /g , hence
• L0 / L0 Dt0 /Dt0 1 / g , or

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• L0 L0 / g
• L0 is defined as the proper length length of
  object measured in the frame in which the object
  (in this case, the distance btw Earth and aC) is
  at rest
• L0 is the length measured in the O frame, which
  is moving wrp to the object here refer to the
  distance between E- aC
• The length of a moving objecte is measured to be
  shorter than the proper length length
  contraction

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• If an observer at rest wrp to an object measures
  its length to be L0 , an observer moving with a
  relative speed u wrp to the object will find the
  object to be shorter than its rest length by a
  foctor 1 / g .

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Example of moving ruler

• A stick moves to the right with a speed v (as
  seen in a rest frame, O)
• (a) The stick as viewed by a frame attached to
  it (O frame, Lp proper length)
• (b) The stick as seen by an observer in a frame
  O. The length measured in the O frame (L) is
  shorter than the proper length by a factor 1/ g

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Length contraction only happens along the
direction of motion

• Example A spaceship in the form of a triangle
  flies by an oberver at rest wrp to the ship (see
  fig (a)), the distance x and y are found to be
  50.0 m and 25.0 m respectively. What is the shape
  of the ship as seen by an observer who sees the
  ship in motion along the direction shown in fig
  (b)?

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Solution

• The observer sees the horizontal length of the
  ship to be contracted to a length of
• L Lp/g 50 mv(1 0.9502) 15.6 m
• The 25 m vertical height is unchanged because it
  is perpendicular to the direction of relative
  motion between the observer and the spaceship.

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Example

• An observer on Earth sees a spaceship at an
  altitude of 435 moving downward toward the Earth
  with a speed of 0.97c. What is the altitude of
  the spaceship as measured by an observer in the
  spaceship?

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Draw the diagram yourself

• As a useful strategy to solve physics problem you
  should always try to translate the problems from
  text into diagramatical form with all the correct
  labelling

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Solution

• One can consider the altitude see by
  thestationary (Earth) observer as the proper
  length (say, L'). The observer in the spaceship
  should sees a contracted length, L, as compared
  to the proper length. Hence the moving observer
  in the ship finds the altitude to be
• L L' / g 435 m x 1- (0.97)2-1/2 106 m

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PYQ, KSCP 03/04

• How fast does a rocket have to go for its length
  to be contracted to 99 of its rest length?
• Ans Rest length proper length LP length of
  the rocket as seen by observer on the rocket
  itself
• L improper length length of the rocket as seen
  from Earth oberver
• Always remember that proper length is longer than
  improper length

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Lorentz Transformation

• All inertial frames are equivalent
• Hence all physical processes analysed in one
  frame can also be analysed in other inertial
  frame and yield consistent results
• A transformation law is required to related the
  space and time coordinates from one frame to
  another

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An event observed in two frames of reference must
yield consistant results related by transformatin
laws
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Different frame uses different notation for
coordinates (because their clocks and ruler are
different

• O' frame uses x',y',zt to denote the
  coordinates of an event, whereas O frame uses
  x,y,zt
• How to related x',y',z',t to x,y,zt?
• In Newtonian mechanics, we use Galilean
  transformation
• However, as discussed, GT fails when u ? c
  because the GT is not consistent with the
  constancy of the light speed postulate
• The relativistic version of the transformation
  law is given by Lorentz transformation

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Two observers in two inertial frames with
relative motion

I measures the coordinates of M as x,t
O
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Derivation of Lorentz transformation

• Our purpose is to find the transformation that
  relates x,t with x,t

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• Consider a rocket moving with a speed u (O'
  frame) along the xx' direction wrp to the
  stationary O frame
• A light pulse is emitted at the instant t' t 0
• when the two origins of the two reference frames
  coincide
• The light signal travels as a spherical wave at a
  constant speed c in both frames
• After some times t, the origin of the wave
  centered at O has a radius r ct, where
• r 2 x2 y2 z2

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Methematical details

• From the view point of O', after some times t
  the origin of the wave, centered at O' has a
  radius
• r' ct' , (r )2 (x)2 (y )2 (z )2
• y'y, z' z (because the motion of O' is along
  the xx) axis no change for y,z coordinates
  (condition A)
• The transformation from x to x (and vice versa)
  must be linear, i.e. x ? x (condition B)
• Boundary condition (1) In the limit of v ?c,
  from the viewpoint of O, the origin of O is
  located on the wavefront (to the right of O)? x
  0 must correspond to x ct
• Boundary condition (2) In the same limit, from
  the viewpoint of O, the origin of O is located
  on the wavefront (to the left of O) ? x 0
  corresponds to x -ct
• Putting everything together we assume the form x
  k(x - ct) to relate x to x,t as this is the
  form that fulfill all the conditions (A,B) and
  boundary consdition (1) (k some proportional
  constant to be determined)
• Likewise, we assume the form x k(x ct ) to
  relate x to x ,t as this is the form that
  fulfill all the conditions (A,B) and boundary
  consdition (2)

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Finally, the transformation obtained

• Hence, with r ct , r ct , x k(x ct ),
  x k(x - ct) we solve for x',t' in terms of
  x,t to obtain the desired transformation law
  (do it as an exercise)

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Space and time now becomes state-of-motion
dependent (via g)

• the constant k is identified as the Lorentz
  factor, g
• Note that, now, the length and time interval
  measured become dependent of the state of motion
  (in terms of g) in contrast to Newtons
  viewpoint
• Lorentz transformation reduces to Galilean
  transformation when u ltlt c (show this yourself)
• i.e. LT ? GT in the limit vltltc

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How to express x, t in terms of x, t

• We have related x',t' in terms of x,t as per
  
• Now, how do we express x, t in terms of x,
  t

72
O moving to the right with velocity u is
equivalent to O moving to the left with velocity
-u
The two transformations above are equivalent use
which is appropriate in a given question
73
Length contraction can be recovered from the LT

• Consider the rest length of a ruler as measured
  in frame O is L Dx x2 - x1 (proper
  length) measured at the same instant in that
  frame, hence t2 t1
• What is the length of the rule as measured by O?
• The length in O, according the LT is
• L Dx x2 - x1 g (x2 - x1) u(t2
  -t1)
• The length of the ruler in O is simply the
  distance btw x2 and x1 measured at the same
  instant in that frame, hence t2 t1, hence L
  g L

74

• Similarly, how would you recover time dilation
  from the LT?
• Do it as homework

75
Lorentz velocity transformation

• How to relate the velocity of the object M as
  seen in the O (ux) frame to that seen in the O
  frame (ux)?

76
Derivation

• By definition, ux dx/dt, ux dx/dt
• The velocity in the O frame can be obtained by
  taking the differentials of the Lorentz
  transformation,

77
Combining

• where we have made used of the definition ux
  dx/dt

78
Comparing the LT of velocity with that of GT
Lorentz transformation of velocity
Galilean transformation of velocity
GT reduces to LT in the limit u ltlt c
79

• Please try to understand the definition of ux ,
  ux , u so that you wont get confused

80
LT is consistent with the constancy of speed of
light

• in either O or O frame, the speed of light seen
  must be the same, c
• Say object M is moving with speed of light as
  seen by O, i.e. ux c
• According to LT, the speed of M as seen by O is

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• That is, in either frame, both observers agree
  that the speed of light they measure is the same,
  c 3 x 108m/s
• In contrast, according to GT, the speed of light
  seen by O would be

Which is inconsistent with constancy of speed of
light postulate
82
To recap

• the LT given in the previous analysis relates ux
  to ux in which O is moving with u wrp to O,

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From the view point of O

• To express ux in terms of ux simply perform the
  similar derivation from the view point of O such
  that O is moving in the u direction .

84
Recap Lorentz transformation relates x,t ??
x,t ux ??ux
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Example

• A boy is slapped twice on the face by his old
  girlfriend. This is happening in a hotel room (a
  rest frame we call O).

O
The two slapping occurs at t1 , t2 such that Dt
t2- t1 1 s, and Dx 0.
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The time t as seen by O in terms of t is simply
related by
Hence the time interval as measured by his new
girlfriend in O, Dt in terms of Dt is simply
This is nothing but just the time dilation effect
calculated using LT approach
87
Example (relativistic velocity addition)

• Rocket 1 is approaching rocket 2 on a
• head-on collision course. Each is moving at
  velocity 4c/5 relative to an independent observer
  midway between the two. With what velocity does
  rocket 2 approaches rocket 1?

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Diagramatical translation of the question in text

• C.f. In GT, their relative speed would just be
  4c/5 4c/5 1.6 c which violates constancy of
  speed of light postulate. See how LT handle this
  situation

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• Choose the observer in the middle as in the
  stationary frame, O
• Choose rocket 1 as the moving frame O
• Call the velocity of rocket 2 as seen from rocket
  1 ux. This is the quantity we are interested in
• Frame O' is moving in the ve direction as seen
  in O, so u 4c/5
• The velocity of rocket 2 as seen from O is in the
  
• -ve direction, so ux - 4c/5
• Now, what is the velocity of rocket 2 as seen
  from frame O', u x ? (intuitively, u x must
  be in the negative direction)

90

• Use the LT

i.e. the velocity of rocket 2 as seen from rocket
1 (the moving frame, O) is 40c/41, which means
that O sees rocket 2 moving in the ve direction
(to the left in the picture), as expected.
91
PYQ, KSCP 2003/04

• A man in a spaceship moving at a velocity of 0.9c
  with respect to the Earth shines a light beam in
  the same direction in which the spaceship is
  travelling. Compute the velocity of the light
  beam relative to Earth using (i) Galilean
  approach (ii) special relativity approach 6
  marks. Please define clearly all the symbols
  used in
• your working.
• Ans
• O is the moving frame travelling at v 0.9c
  with respect to the Earth. Speed of the light
  beam as seen in the frame O is u c. O is the
  Earth frame. We wish to find the speed of the
  light beam as seen from frame O, u.
• (i) According to Galilean transformation, u u
  v c 0.9c 1.9c.(ii) Use

92
Relativistic Dynamics

• By Einsteins postulate, the onservational law of
  linear momentum must also hold true in all frames
  of reference

Conservation of linear momentum classically means
m1u1 m2u2 m1v1 m2v2
93
Modification of expression of linear momentum

• Classically, p mu. In the other frame, p m
  u the mass m (as seen in frame O) is the
  same as m (as seen in O frame) this is
  according to Newtons mechanics
• However, simple consideration will reveal that in
  order to preserver the consistency between
  conservation of momentum and the LT, the
  definition of momentum has to be modified such
  that m is not equal to m.
• That is, the mass of an moving object, m, is
  different from its value when its at rest, m0

94
In other words

• In order to preserve the consistency between
  Lorentz transformation of velocity and
  conservation of linear momentum, the definition
  of 1-D linear momentum, classically defined as
  pclassical mu, has to be modified to
• psr mu gm0u (where the relativisitic mass m
  gm0 is not the same the rest mass m0
• Read up the text for a more rigorous illustration
  why the definition of classical momentum is
  inconsistent with LT

95
Grafically
96
Two kinds of mass

• Differentiate two kinds of mass rest mass and
  relativistic mass
• m0 rest mass the mass measured in a frame
  where the object is at rest. The rest mass of an
  object must be the same in all frames (not only
  in its rest frame).
• Relativisitic mass m g m0 of an object changes
  depends on its speed

97
Behaviour of pSR as compared to pclassic

• Classical momentum is constant in mass, pclassic
  m0v
• Relativisitic momentum is pSR m0gv
• pSR / pclassic g ? ? as v ? c
• In the other limit, pSR / pclassic 1 as v ltlt c

98
Example

• The rest mass of an electron is m0 9.11 x
  10-31kg.

If it moves with u 0.75 c, what is its
relativistic momentum?
Compare it with that calculated with classical
definition.
99
Solution

• The Lorentz factor is g 1-(u/c)2 -1/2
  1-(0.75c/c)2 -1/21.51
• Hence the relativistic momentum is simply
• p g x m0 x 0.75c
• 1.51 x 9.11 x 10-31kg x 0.75 x 3 x 108 m/s
  3.1 x 10-22 kg m/s Ns

In comparison, classical momentum gives
pclassical m0 x 0.75c 2.5 x 10-22 Ns about
34 lesser than the relativistic value
100
Work-Kinetic energy theorem

• Recall the law of conservation of mechanical
  energy

Work done by external force on a system, W
the change in kinetic energy of the system, DK
101
Conservation of mechanical energy W DK
The total energy of the object, E K U.
Ignoring potential energy, E of the object is
solely in the form of kinetic energy. If K1 0,
then E K2. But in general, U also needs to be
taken into account for E.
102

• In classical mechanics, mechanical energy
  (kinetic potential) of an object is closely
  related to its momentum and mass
• Since in SR we have redefined the classical mass
  and momentum to that of relativistic version
• mclass(cosnt) ? mSR m0g
• pclass mclass u ? pSR (m0g)u

• we must also modify the relation btw work and
  energy so that the law conservation of energy is
  consistent with SR

• E.g, in classical mechanics, K p2/2m 2mu2/2.
  However, this relationship has to be supplanted
  by the relativistic version K mu2/2 ? K E
  m0c2 mc2 - m0c2

• We will like to derive K in SR in the following
  slides

103
Force, work and kinetic energy

• When a force is acting on an object with rest
  mass m0, it will get accelerated (say from rest)
  to some speed (say u) and increase in kinetic
  energy from 0 to K

K as a function of u can be derived from first
principle based on the definition of
work done, W F dx,
and conservation of mechanical energy, DK W
104
Derivation of relativistic kinetic energy
where, by definition,
is the velocity of the object
105
Explicitly, p gm0u,

• Hence, dp/du d/du(gm0u)
• m0 u (dg/du) g
  
• m0 g (u2/c2) g3 m0 (1-u2/c2)-3/2

in which we have inserted the relation
106

• E mc2 is the total relativistic energy of an
  moving object

107

• Or in other words, the total relativistic energy
  of a moving object is the sum of its rest energy
  and its relativistic kinetic energy

• The mass of an moving object m is larger than its
  rest mass m0 due to the contribution from its
  relativistic kinetic energy this is a pure
  relativistic effect not possible in classical
  mechanics
• E mc2 relates the mass of an object to the
  total energy released when the object is
  converted into pure energy

108

• Example, 10 kg of mass, if converted into pure
  energy, it will be equivalent to
• E mc2 10 x (3 x108) 2 J 9 x1017J
• equivalent to a few tons of TNT explosive

109
PYQ, KSCP 2003/04

• (i) What is the rest mass of a proton in terms of
  MeV?
• Ans
• (i) 1.67 x 10-27kg x (3x108 m/s)2 1.503x10-10J
  (1.503 x10-10/1.6x10-19) eV 939.4 MeV

110
PYQ, KSCP 2003/04

• What is the relativistic mass of a proton whose
  kinetic energy is 1 GeV?
• Ans rest mass of proton, mp 939.4 MeV

111
Reduction of relativistic kinetic energy to the
classical limit

• The expression of the relativistic kinetic energy

• must reduce to that of classical one in the limit
  u ? 0 when compared with c, i.e.

112
Expand g with binomial expansion

• For u ltlt c, we can always expand g in terms of
  (u/c)2 as

i.e., the relativistic kinetic energy reduces to
classical expression in the u ltlt c limit
113
Example

• An electron moves with speed u 0.85c. Find its
  total energy and kinetic energy in eV.
• CERNs picture the circular accelerator
  accelerates electron almost the speed of light

114
Solution

• Due to mass-energy equivalence, sometimes we
  express the mass of an object in unit of energy
• Electron has rest mass m0 9.1 x 10-31kg
• The rest mass of the electron can be expressed as
  energy equivalent, via
• m0c2 9.1 x 10-31kg x (3 x 108m/s)2
• 8.19 x 10-14 J
• 8.19 x 10-14 x (1.6x10-19)-1 eV
• 511.88 x 103 eV 0.511 MeV

115
Solution

• First, find the Lorentz factor, g 1.89
• The rest mass of electron, m0c2, is
• 0.5 MeV
• Hence the total energy is
• E mc2 g (m0c2) 1.89 x 0.5 MeV 0.97 MeV
• Kinetic energy is the difference between the
  total relativistic energy and the rest mass, K
  E - m0c2 (0.97 0.51)MeV 0.46 MeV

116
Conservation of Kinetic energy in relativistic
collision

• Calculate (i) the kinetic energy of the system
  and (ii) mass increase for a completely inelastic
  head-on of two balls (with rest mass m0 each)
  moving toward the other at speed u/c 1.5x10-6
  (the speed of a jet plane). M is the resultant
  mass after collision, assumed at rest.

M
u
u
m0
m0
117
Solution

• (i) K 2mc2 - 2m0c2 2(g-1)m0c2
• (ii) Ebefore Eafter ? 2g m0c2 Mc2 ? M 2g
  m0
• Mass increase DM M - 2m0 2(g -1)m0
• Approximation u/c 1.5x10-6 ? g 1 ½ u2/c2
  (binomail expansion) ? M 2(1 ½ u2/c2)m0
• Mass increase DM M - 2m0
• 
  (u2/c2)m0 1.5x10-6m0
• Comparing K with DMc2 the kinetic energy is not
  lost in relativistic inelastic collision but is
  converted into the mass of the final composite
  object, i.e. kinetic energy is conserved
• In contrast, in classical mechanics, momentum is
  conserved but kinetic energy is not in an
  inelastic collision

118
In terms of relativistic momentum, the
relativistic total energy can be expressed as
followed
Relativistic momentum and relativisitc Energy
119
In SR, both relativistic mass-energy and momentum
are always conserved in a collision (in contrast
to classical mechanics in which KE is not
conserved in inelastic collision)
120
Example measuring pion mass using conservation
of momentum-energy

• pi meson decays into a muon massless neutrino
• If the mass of the muon is known to be 106
  MeV/c2, and the kinetik energy of the muon is
  measured to be 4.6 MeV, find the mass of the pion

121
Solution
122
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123
Binding energy

• The nucleus of a deuterium comprises of one
  neutron and one proton. Both nucleons are bounded
  within the deuterium nucleus

Initially, the total Energy (mn mn)c2
Nuclear fusion
After fusion, the total energy mdc2 U
Analogous to exothermic process in chemistry
124

• U is the energy that will be released when a
  proton and a neutron is fused in a nuclear
  reaction. The same amount of energy is required
  if we want to separate the proton from the
  neutron in a deuterium nucleus
• U is called the binding energy

125

• U can be explained in terms energy-mass
  equivalence relation, as followed
• For the following argument, we will ignore KE for
  simplicity sake
• Experimentally, we finds that mn mp gt md
• By conservation of energy-momentum,
• E(before) E(after)
• mnc2 mpc2 0 mdc2 U
• Hence, U (mp mn)c2 - mdc2 Dmc2
• The difference in mass between deuterium and the
  sum of (mn mn)c2 is converted into the binding
  energy that binds the proton to the neutron
  together

126
Example

• mn 1.008665u mp 1.007276u
• md 2.013553u
• u standard atomic unit mass of 1/12 of the
  mass of a 12C nucleus
• 1.66 x 10-27kg
• 1.66 x 10-27 x c2 J 1.494 x 10-10 J
• 1.494 x 10-10/(1.6x10-19) eV
• 933.75 x 106 eV 933.75 x MeV

• Hence the binding energy
• U Dmc2 (mp mn)c2 - mdc2
• 0.002388u 2.23 MeV

127
Fission

• Such as
• The reverse of nuclear fusion is nuclear fission
• An parent nuclide M disintegrates into daughter
  nuclides such that their total mass ?mi lt M.
• The energy of the mass deficit equivalent,
• Q (M - ?mi)c2 Dmc2 called disintegration
  energy will be released

128
SR finishes here