2. 1 1 ST CHAPTER Special theory of relativity

3. 2 ZCT 104/3E ENGLISH TEACHING ENGLISH TEACHING 5 CHAPTERS WILL BE COVERED 5 CHAPTERS WILL BE COVERED 1. Special theory of relativity 2. Wave nature of particles 3. Particle nature of waves 4. Introduction to Quantum Mechanics 5. Atomic models

4. 3 LECTURE 1 Failure of Newtonian mechanics Newton

5. 4 Revision: Still remember Newton's 3 law of motion? 1. An object at rest will always be in the wrong place 2. An object in motion will always be headed in the wrong direction 3. For every action, there is an equal and opposite criticism =b, just joking

6. 5 Newtonian view of space and time Space and time are absolute Space and time are absolute Time flow independently of the state of motion of any physical system in 3-D space 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 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

7. 6 Inertial frames Inertial frames of reference is one in which an object subject to no forcess moves in straight line at constant speed 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 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 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

8. 7 Example of inertial frames of reference

9. 8 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 aeroplane with constant speed wrp to the ground, Newton second law takes the form of F’ = m a’ In the lab frame, Newton 2 nd law is F = ma In the lab frame, Newton 2 nd law is F = ma

10. 9 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 Newton’s law, conservation of energy and others physical principles Although the ball path is different in both inertial reference frames, both observers agree on the validity of Newton’s law, conservation of energy and others physical principles

11. 10 Galilean transformation It relates the kinematical quantities, such as position, velocity, acceleration between two inertial frames 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, z’=z, t’ = t Galilean addition law for velocity (in 1-D): v’x = vx - v Simply a daily experience

12. 11 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. 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.

13. 12Example Apply GT on th previous 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 The trajactories of the ball seen in the two frames, S’ (van) and S (ground observer) is related by GT as per

14. 13Example 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. 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’. Doraemiyan (S’) measure the end points of the stick at the same time, t’. Using x’ = x – vt’: Using x’ = x – vt’: x’(head) = x(head) – vt’ = a - vt’; x’(head) = x(head) – vt’ = a - vt’; x’(end) = x(end) – vt’ = (a+l) - vt’ x’(end) - x’(head) = x(end) - x(head) = l x(h) =a x(t)= l +a

15. 14 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’.

16. 15 Maxwell theory of light is inconsistent with Newton’s 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)

17. 16 Ether and Michelson-Morley Experiments In early 19 th century, it’s thought (incorrectly) that In early 19 th century, it’s thought (incorrectly) that there exist an omi-pervasive medium called Ether in which light propagates at a speed of 3x10 8 m/s (analogue to sound propagate in the mechanical medium of still air at speed 330m/s) there exist an omi-pervasive medium called Ether in which light propagates at a speed of 3x10 8 m/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 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 The effect of the ehter on speed of light can be experimentally measured

18. 17 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 -4 c) 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 -4 c) Consider a moving souce giving out two beams of light in different direction (say, 90 degree to each other) 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 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

19. 18 Experimental setup Both arms has same length L Both arms has same length L According to the ether wind concept: According to the ether wind concept: For arm 1, the speed of light to is c-v as it approaches M2, For arm 1, the speed of light to is c-v as it approaches M2, c+v as it is reflected from M2 c+v as it is reflected from M2 c’ = c - v v c’ = c + v c v

20. 19 Experimental setup For arm 2, the speed of light to-and-fro M1 is For arm 2, the speed of light to-and-fro M1 is v c c’= √ (c 2 - v 2 )

21. 20 The two light beams start out in phase. When they return and “recombined” by semi-transparent mirror M o interference pattern will be formed due to their difference in phase,  c  t /, where  t = t 1 - t 2  Lv 2 /c 3 is the time difference between the light beams when return to M o (figure a) The two light beams start out in phase. When they return and “recombined” by semi-transparent mirror M o interference pattern will be formed due to their difference in phase,  c  t /, where  t = t 1 - t 2  Lv 2 /c 3 is the time difference between the light beams when return to M o (figure a)

22. 21 Now, when the whole set-up is rotated through 90 , arms 1 and 2 exchanges role 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 2  t As a result, the interefence pattern will be shifted as the time difference between the beams after rotation now becomes 2  t The number of interefence fringes shifted can be estimated via: The number of interefence fringes shifted can be estimated via: no. of fringe shift = 2c  t  2 Lv 2 / c 2  0.40 no. of fringe shift = 2c  t  2 Lv 2 / c 2  0.40 (taking v ≈ 10 -4 c, L = 11 m,  = 500nm) (taking v ≈ 10 -4 c, L = 11 m,  = 500nm) Very precise experiment Very precise experiment

23. 22 Fig. (b) shows expected fringe shift after a rotation of the interometer by 90 degree Fig. (b) shows expected fringe shift after a rotation of the interometer by 90 degree

24. 23 But MM sees only NULL result – no change in the interference pattern But MM sees only NULL result – no change in the interference pattern How to interprete the null result? 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 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 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 Einstein put it more strongly: the absolute frame of reference (i.e. the ether frame) has to be discarded NULL result

25. 24 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 onlyB. I,IIC. I, IIID. I, II, III E. Non of the above Ans: B

26. 25 Principle of special relativity Einstein believes that pure thought is sufficient to understand the world Einstein believes that pure thought is sufficient to understand the world The most incomprehensive thing in the universe is that the universe is comprehensible The most incomprehensive thing in the universe is that the universe is comprehensible

27. 26 Classical EM theory is inconsistent with Galileao transformation Their is inconsistency between EM and Newtonian view of absolute space and time 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: Einstein proposed SR to restore the inconsistency between the two based on two postulates:

28. 27 Postulates of SR 1. The laws of physics are the same in all inertial reference frames – a generalisation of Newton’s 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

29. 28 Postulate 2 simply means that Galilean transformation cannot be applied on light speed. It also explains the Null result of the MM experiment 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 doesn’t increase or reduced when the light source is moving Speed of light is always the same whether one is moving or stationary wrp to the source – its speed doesn’t increase or reduced when the light source is moving

30. 29  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 Einstein’s notion of space-time drastically revolutionarizes that of Newton’s

31. 30 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 IVSpecial relativity is inconsistent with the notion of the Ether frame A. III,IVB. I, II, IIIC. I, II, III,IV D. I, IIE. I, II,IV ANS: E, my own question

32. 31 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 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

33. 32 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. t A -t B, between the two lights arriving at O’? t is the time measured in the O frame.

34. 33 Time dilation as a consequence of Einstein’s postulate In frames that are moving wrp to the stationary frame, time runs slower In frames that are moving wrp to the stationary frame, time runs slower Gedanken experiment (thought experiment) Gedanken experiment (thought experiment)

35. 34 Gedanken Experiment Since light speed c is invariant (i.e. the same in all frames), it is used to measure time and space 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. 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. (b) Relative to a stationery observe on Earth, the mirror and O’ move with a speed v.

36. 35 Ligth triangle Consider the geometry of the triangle of the light Consider the geometry of the triangle of the light We can calculate the relationship between  t,  t’ and v We can calculate the relationship between  t,  t’ and v l 2 = (c  t/2 ) 2 l 2 = (c  t/2 ) 2 = d 2 + (u  t/2 ) 2 = d 2 + (u  t/2 ) 2

37. 36 Due to constancy of light postulate, both observer must agree on c: Due to constancy of light postulate, both observer must agree on c: Speed of light = total distance travelled divide by time taken Speed of light = total distance travelled divide by time taken For observer in O’, c = 2 d /  t’ For observer in O’, c = 2 d /  t’ For observer in O, c = 2 l /  t, where For observer in O, c = 2 l /  t, where l 2 = d 2 + (u  t/2 ) 2 Eliminating l and d,  t=  t’, where  u 2 /c 2   Lorentz factor, always > or equal 1, so that  t > =  t’ Lorentz factor, 

38. 37 Proper time Try to discriminate between two kinds of time interval: Try to discriminate between two kinds of time interval:  t’, 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)  t’, 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 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 Note that proper time is always ``shorter’’ compared to improper time

39. 38 The elapsed time  t between the same events in any other frame is dilated by a factor of  compared to the proper time interval  t’ The elapsed time  t between the same events in any other frame is dilated by a factor of  compared to the proper time interval  t’ In other words, according to a stationary observer, a moving clock runs slower than an identical stationary clock 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 天上方一日，人间已十年 1 day in the heaven = 10 years in the human plane 天上方一日，人间已十年

40. 39 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 bike’s 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.

41. 40 To recap  t=  t’ ; proper time interval,  t’   t  t=  t’ ; proper time interval,  t’   t 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 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 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 One must be aware of the subtle different between which is the proper time and which is the improper one

42. 41 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 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 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

43. 42 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? 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. 13.3B. 2.88C.22.2 A. 13.3B. 2.88C.22.2 D. 1.77E. Non of the above D. 1.77E. Non of the above ANS: D, Cutnell, Q4, pg. 877

44. 43Solution 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? 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?  = 1/(1 – u 2 /c 2 ) 1/2 = 1/(1 – 0.975 2 ) 1/2 = 4.5  = 1/(1 – u 2 /c 2 ) 1/2 = 1/(1 – 0.975 2 ) 1/2 = 4.5 Use  t =  t’ Use  t =  t’ Given local interval between breaths  t’ = 1/8 = 0.125 min per breath (proper time interval) Given local interval between breaths  t’ = 1/8 = 0.125 min per breath (proper time interval)  t =  t =  4.5 x .125 = 0.563 min per breath  t =  t =  4.5 x .125 = 0.563 min per breath  t’ = 1.77 breath per min (as seen by the spaccraft observer)  t’ = 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 To an oberver on the spacecraft, you seem to breath at a slower rate

45. 44 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?

46. 45 Solution  t’ = 1.0 s is the proper time interval measured by the astronaut  t’ = 1.0 s is the proper time interval measured by the astronaut Earth observer measures a greater time interval,  t, than does the astronaut, who is at rest relative to the clock Earth observer measures a greater time interval,  t, than does the astronaut, who is at rest relative to the clock The Lorentz factor  u 2 /c 2  -1/2 =  0.92 2  -1/2 = 2.6 The Lorentz factor  u 2 /c 2  -1/2 =  0.92 2  -1/2 = 2.6 Hence,  t =  t’ = 2.6 x 1.0s = 2.6 s Hence,  t =  t’ = 2.6 x 1.0s = 2.6 s

47. 46 Example: Muon decay lifetime A muon is an unstable elementary particle which has a lifetime  0 = 2.2 microsecond (proper time, measured in the muon rest frame) and decays into lighter particles. Fast muons (say, travelling at v = 99%c) are created in the interactions of very high-energy particles as they enter the Earth's upper atmosphere. Assume v = 0.99c Assume v = 0.99c In the muon rest frame, the distance travelled by muon before decay is In the muon rest frame, the distance travelled by muon before decay is D’ = (0.99c)  0 D’ = (0.99c)  0 = 650 m = 650 m

48. 47 A muon travelling at 99% the speed of light. A muon travelling at 99% the speed of light. has a Lorentz factor  = 7.09 has a Lorentz factor  = 7.09 Hence, to an observer in the rest frame (e.g Earth) the lifetime of the muon is no longer  0 = 2.2  s but Hence, to an observer in the rest frame (e.g Earth) the lifetime of the muon is no longer  0 = 2.2  s but    x   = 7.09 x 2 microseconds = 15.6  s    x   = 7.09 x 2 microseconds = 15.6  s Thus the muon would appear to travel for 15.6 microseconds before it decays Thus the muon would appear to travel for 15.6 microseconds before it decays The distance it traversed as seen from Earth The distance it traversed as seen from Earth is D = (0.99c) x 15.6  s = 4,630 km (c.f. D’ = 650 m )

49. 48 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 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 Hence the dilated muon lifetime is confirmed experimentally

50. 49

51. 50 Length contraction Length measured differs from frame to frame – another consequence of relativistic effect Length measured differs from frame to frame – another consequence of relativistic effect Gedanken experiment again! Gedanken experiment again!

52. 51 Two observers: O on Earth, O’ traveling to and fro from Earth and alpha centauri with speed u 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), = L 0 Total distance between Earth - alpha centauri – Earth, according to O (Earth observer), = L 0 O sees O’ return to Earth after  t 0 O sees O’ return to Earth after  t 0 Observer O’ in a spaceship is heading  C with speed u and returns to Earth after  t’ according to his clock Observer O’ in a spaceship is heading  C with speed u and returns to Earth after  t’ according to his clock

53. 52 Use some simple logics… In O: 2L 0 = u  t 0 In O: 2L 0 = u  t 0 In O’: 2L 0’ = u  t 0’ In O’: 2L 0’ = u  t 0’ Due to time dilation effect,  t 0’ is shorter than  t 0, i.e.  t 0 >  t 0’ Due to time dilation effect,  t 0’ is shorter than  t 0, i.e.  t 0 >  t 0’  t 0 is related to  t 0’ via a time dilation effect,  t 0’ =  t 0 / , hence  t 0 is related to  t 0’ via a time dilation effect,  t 0’ =  t 0 / , hence L 0’ / L 0 =  t 0’ /  t 0 = 1 / , or L 0’ / L 0 =  t 0’ /  t 0 = 1 / , or

54. 53 L 0’ = L 0 /  L 0 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  C) is at rest L 0 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  C) is at rest L 0’ is the length measured in the O’ frame, which is moving wrp to the “object” – here refer to the distance between E-  C L 0’ is the length measured in the O’ frame, which is moving wrp to the “object” – here refer to the distance between E-  C The length of a moving objecte is measured to be shorter than the proper length – length contraction The length of a moving objecte is measured to be shorter than the proper length – length contraction

55. 54 If an observer at rest wrp to an object measures its length to be L 0, 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 /  If an observer at rest wrp to an object measures its length to be L 0, 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 / 

56. 55 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, L p = 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/  Example of moving ruler

57. 56 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)?

58. 57Solution The observer sees the horizontal length of the ship to be contracted to a length of The observer sees the horizontal length of the ship to be contracted to a length of L = L p /  = 50 m√(1 – 0.950 2 ) = 15.6 m L = L p /  = 50 m√(1 – 0.950 2 ) = 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. The 25 m vertical height is unchanged because it is perpendicular to the direction of relative motion between the observer and the spaceship.

59. 58 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? 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? Example

60. 59 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 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

61. 60 Solution One can consider the altitude see by the stationary (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' /  = 435 m x [1- (0.97) 2 ] -1/2 = 106 m

62. 61 PYQ, KSCP 03/04 How fast does a rocket have to go for its length to be contracted to 99% of its rest length? 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 = L P = length of the rocket as seen by observer on the rocket itself Ans: Rest length = proper length = L P = length of the rocket as seen by observer on the rocket itself L improper length = length of the rocket as seen from Earth oberver L improper length = length of the rocket as seen from Earth oberver Always remember that proper length is longer than improper length Always remember that proper length is longer than improper length

63. 62 Lorentz Transformation All inertial frames are equivalent 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 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 A transformation law is required to related the space and time coordinates from one frame to another

64. 63 An event observed in two frames of reference must yield consistant results related by transformatin laws

65. 64 Different frame uses different notation for coordinates (because their clocks and ruler are different O' frame uses {x',y',z‘;t‘} to denote the coordinates of an event, whereas O frame uses {x,y,z;t} O' frame uses {x',y',z‘;t‘} to denote the coordinates of an event, whereas O frame uses {x,y,z;t} How to related {x',y',z',t‘} to {x,y,z;t}? How to related {x',y',z',t‘} to {x,y,z;t}? In Newtonian mechanics, we use Galilean transformation 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 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 The relativistic version of the transformation law is given by Lorentz transformation

66. 65 O’ O I see O’ moving with a velocity +u +u I measures the coordinates of M as {x,t} Object M Two observers in two inertial frames with relative motion I measures the coordinates of M as {x’,t’} I see O moving with a velocity -u -u-u

67. 66 Derivation of Lorentz transformation Our purpose is to find the transformation that relates {x,t} with {x’,t’} Our purpose is to find the transformation that relates {x,t} with {x’,t’}

68. 67 Consider a rocket moving with a speed u (O' frame) along the xx' direction wrp to the stationary O frame 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 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 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 After some times t, the origin of the wave centered at O has a radius r = ct, where r 2 = x 2 + y 2 + z 2

69. 68 From the view point of O', after some times t‘ the origin of the wave, centered at O' has a radius: 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) 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) 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 (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’ 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) 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) ; 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) ; Methematical details

70. 69 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) 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) Finally, the transformation obtained

71. 70 the constant k is identified as the Lorentz factor,  the constant k is identified as the Lorentz factor,  Note that, now, the length and time interval measured become dependent of the state of motion (in terms of  ) – in contrast to Newton’s viewpoint Note that, now, the length and time interval measured become dependent of the state of motion (in terms of  ) – in contrast to Newton’s viewpoint Lorentz transformation reduces to Galilean transformation when u << c (show this yourself) Lorentz transformation reduces to Galilean transformation when u << c (show this yourself) i.e. LT  GT in the limit v<<c i.e. LT  GT in the limit v<<c Space and time now becomes state-of-motion dependent (via  )

72. 71 How to express {x, t} in terms of {x’, t’} We have related {x',t'} in terms of {x,t } as per We have related {x',t'} in terms of {x,t } as per Now, how do we express {x, t } in terms of {x’, t’}

73. 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

74. 73 Length contraction can be recovered from the LT Consider the rest length of a ruler as measured in frame O’ is L’ =  x’ = x’ 2 - x’ 1 (proper length) measured at the same instant in that frame, hence t’ 2 = t’ 1 Consider the rest length of a ruler as measured in frame O’ is L’ =  x’ = x’ 2 - x’ 1 (proper length) measured at the same instant in that frame, hence t’ 2 = t’ 1 What is the length of the rule as measured by O? What is the length of the rule as measured by O? The length in O, according the LT is The length in O, according the LT is L’  x’ = x’ 2 - x’ 1 =  (x 2 - x 1 ) – u(t 2 -t 1 )] The length of the ruler in O is simply the distance btw x 2 and x 1 measured at the same instant in that frame, hence t 2 = t 1, hence L’ =  L The length of the ruler in O is simply the distance btw x 2 and x 1 measured at the same instant in that frame, hence t 2 = t 1, hence L’ =  L

75. 74 Similarly, how would you recover time dilation from the LT? Do it as homework

76. 75 Lorentz velocity transformation O’ M is moving with a velocity +u x from my point of view Object M I see M moving with a velocity +u x ’ I see O’ moving with a velocity +u +u O Object M O How to relate the velocity of the object M as seen in the O’ (u’ x ) frame to that seen in the O frame (u x )?

77. 76 Derivation By definition, u x = dx/dt, u’ x = dx’/dt’ The velocity in the O’ frame can be obtained by taking the differentials of the Lorentz transformation,

78. 77Combining where we have made used of the definition u x = dx/dt

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

80. 79 Please try to understand the definition of u x, u’ x, u so that you wont get confused Please try to understand the definition of u x, u’ x, u so that you wont get confused

81. 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 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. u x = c Say object M is moving with speed of light as seen by O, i.e. u x = c According to LT, the speed of M as seen by O’ is According to LT, the speed of M as seen by O’ is

82. 81 That is, in either frame, both observers agree that the speed of light they measure is the same, c = 3 x 10 8 m/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

83. 82 To recap the LT given in the previous analysis relates u’ x to u x in which O’ is moving with +u wrp to O, the LT given in the previous analysis relates u’ x to u x in which O’ is moving with +u wrp to O,

84. 83 From the view point of O’ To express u x in terms of u’ x simply perform the similar derivation from the view point of O’ such that O is moving in the –u direction:. To express u x in terms of u’ x simply perform the similar derivation from the view point of O’ such that O is moving in the –u direction:.

85. 84 Recap: Lorentz transformation relates {x’,t’}  {x,t}; u’ x  u x

86. 85Example 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). 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 t1t1 The two slapping occurs at t 1, t 2 such that  t = t 2 - t 1 = 1 s, and  x =0. t2t2 To his new girlfriend in a car moving with speed u on the road (we call this moving frame O’), what is the time interval between the two slapping? O’ u

87. 86 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’,  t’ in terms of  t is simply This is nothing but just the time dilation effect calculated using LT approach

88. 87 Example (relativistic velocity addition) Rocket 1 is approaching rocket 2 on a 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?

89. 88 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: Diagramatical translation of the question in text

90. 89 Choose the observer in the middle as in the stationary frame, O Choose the observer in the middle as in the stationary frame, O Choose rocket 1 as the moving frame O‘ Choose rocket 1 as the moving frame O‘ Call the velocity of rocket 2 as seen from rocket 1 u’ x. This is the quantity we are interested in Call the velocity of rocket 2 as seen from rocket 1 u’ x. This is the quantity we are interested in Frame O' is moving in the +ve direction as seen in O, so u = +4c/5 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 The velocity of rocket 2 as seen from O is in the -ve direction, so u x = - 4c/5 -ve direction, so u x = - 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) Now, what is the velocity of rocket 2 as seen from frame O', u ’ x = ? (intuitively, u ’ x must be in the negative direction)

91. 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.

92. 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

93. 92 Relativistic Dynamics By Einstein’s postulate, the onservational law of linear momentum must also hold true in all frames of reference By Einstein’s postulate, the onservational law of linear momentum must also hold true in all frames of reference m1u1m1u1 m2u2m2u2 m1v1m1v1 m2v2m2v2 Conservation of linear momentum classically means m 1 u 1 +m 2 u 2 = m 1 v 1 +m 2 v 2

94. 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 Newton’s mechanics 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 Newton’s 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. 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 it’s at rest, m 0 That is, the mass of an moving object, m, is different from its value when it’s at rest, m 0

95. 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 p classical = mu, has to be modified to 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 p classical = mu, has to be modified to p sr = mu =  m 0 u (where the relativisitic mass m =  m 0 is not the same the rest mass m 0 Read up the text for a more rigorous illustration why the definition of classical momentum is inconsistent with LT Read up the text for a more rigorous illustration why the definition of classical momentum is inconsistent with LT

96. 95 Grafically… O O’ v M I see M is at rest. Its mass is m 0, momentum, p’ = 0 I see the momentum of M as p = mv=m 0  v

97. 96 Two kinds of mass Differentiate two kinds of mass: rest mass and relativistic mass m 0 = 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 =  m 0 of an object changes depends on its speed

98. 97 Behaviour of p SR as compared to p classic Classical momentum is constant in mass, p classic = m 0 v Classical momentum is constant in mass, p classic = m 0 v Relativisitic momentum is p SR = m 0  v Relativisitic momentum is p SR = m 0  v p SR / p classic =    as v  c p SR / p classic =    as v  c In the other limit, p SR / p classic =  as v << c In the other limit, p SR / p classic =  as v << c

99. 98 Example The rest mass of an electron is m 0 = 9.11 x 10 -31 kg. p = m 0  u Compare it with that calculated with classical definition. If it moves with u = 0.75 c, what is its relativistic momentum? m0m0

100. 99 Solution The Lorentz factor is  = [1-(u/c) 2 ] -1/2 = [1-(0.75c/c) 2 ] -1/2 =1.51 The Lorentz factor is  = [1-(u/c) 2 ] -1/2 = [1-(0.75c/c) 2 ] -1/2 =1.51 Hence the relativistic momentum is simply Hence the relativistic momentum is simply p =  x m 0 x 0.75c = 1.51 x 9.11 x 10 -31 kg x 0.75 x 3 x 10 8 m/s = 3.1 x 10 -22 kg m/s = Ns In comparison, classical momentum gives p classical = m 0 x 0.75c = 2.5 x 10 -22 Ns – about 34% lesser than the relativistic value

101. 100 Work-Kinetic energy theorem Recall the law of conservation of mechanical energy: 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,  K

102. 101 K1K1 F K2 F s  = K 2 - K 1 W = F s Conservation of mechanical energy: W =  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 K 1 = 0, then E = K 2. But in general, U also needs to be taken into account for E.

103. 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 m class (cosnt)  m SR = m 0  p class = m class u  p SR = (m 0  )u We will like to derive K in SR in the following slides E.g, in classical mechanics, K = p 2 /2m = 2 mu 2 /2. However, this relationship has to be supplanted by the relativistic version K = mu 2 /2  K = E – m 0 c 2 = mc 2 - m 0 c 2 we must also modify the relation btw work and energy so that the law conservation of energy is consistent with SR

104. 103 Force, work and kinetic energy When a force is acting on an object with rest mass m 0, it will get accelerated (say from rest) to some speed (say u) and increase in kinetic energy from 0 to K When a force is acting on an object with rest mass m 0, 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: Force,F = dp/dt, work done, W = F dx, and conservation of mechanical energy,  K = W

105. 104 Derivation of relativistic kinetic energy Force = rate change of momentum Chain rule in calculus where, by definition, is the velocity of the object

106. 105 Explicitly, p =  m 0 u, Hence, dp/du = d/du(  m 0 u) = m 0 [u (d  /du) +  ]  = m 0  + (u 2 /c 2 )  3 ] = m 0 (1-u 2 /c 2 ) -3/2 in which we have inserted the relation integrate

107. 106 E = mc 2 is the total relativistic energy of an moving object E = mc 2 is the total relativistic energy of an moving object The relativisitic kinetic energy of an object of rest mass m 0 travelling at speed u E 0 = m 0 c 2 is called the rest energy of the object. Its value is a constant for a given object Any object has non-zero rest mass contains energy as per E 0 = m 0 c 2 One can imagine that masses are ‘energies frozen in the form of masses’ as per E 0 = m 0 c 2

108. 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 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 m 0 due to the contribution from its relativistic kinetic energy – this is a pure relativistic effect not possible in classical mechanics E = mc 2 relates the mass of an object to the total energy released when the object is converted into pure energy

109. 108 Example, 10 kg of mass, if converted into pure energy, it will be equivalent to E = mc 2 = 10 x (3 x10 8 ) 2 J = 9 x10 17 J – equivalent to a few tons of TNT explosive

110. 109 PYQ, KSCP 2003/04 (i)What is the rest mass of a proton in terms of MeV? (i)What is the rest mass of a proton in terms of MeV? Ans: Ans: (i)1.67 x 10 -27 kg x (3x10 8 m/s) 2 = 1.503x10 -10 J = (1.503 x10 -10 /1.6x10 -19 ) eV = 939.4 MeV (i)1.67 x 10 -27 kg x (3x10 8 m/s) 2 = 1.503x10 -10 J = (1.503 x10 -10 /1.6x10 -19 ) eV = 939.4 MeV

111. 110 PYQ, KSCP 2003/04 What is the relativistic mass of a proton whose kinetic energy is 1 GeV? What is the relativistic mass of a proton whose kinetic energy is 1 GeV? Ans: rest mass of proton, m p = 939.4 MeV Ans: rest mass of proton, m p = 939.4 MeV

112. 111 Reduction of relativistic kinetic energy to the classical limit The expression of the relativistic kinetic energy 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.

113. 112 Expand  with binomial expansion For u << c, we can always expand  in terms of (u/c) 2 as For u << c, we can always expand  in terms of (u/c) 2 as i.e., the relativistic kinetic energy reduces to classical expression in the u << c limit

114. 113Example An electron moves with speed u = 0.85c. Find its total energy and kinetic energy in eV. An electron moves with speed u = 0.85c. Find its total energy and kinetic energy in eV. CERN’s picture: the circular accelerator accelerates electron almost the speed of light CERN’s picture: the circular accelerator accelerates electron almost the speed of light

115. 114 Solution Due to mass-energy equivalence, sometimes we express the mass of an object in unit of energy Electron has rest mass m 0 = 9.1 x 10 -31 kg The rest mass of the electron can be expressed as energy equivalent, via m 0 c 2 = 9.1 x 10 -31 kg x (3 x 10 8 m/s)2 = 8.19 x 10 -14 J = 8.19 x 10 -14 x (1.6x10 -19 ) -1 eV = 511.88 x 10 3 eV = 0.511 MeV

116. 115Solution First, find the Lorentz factor,  = 1.89 First, find the Lorentz factor,  = 1.89 The rest mass of electron, m 0 c 2, is The rest mass of electron, m 0 c 2, is 0.5 MeV Hence the total energy is Hence the total energy is E = mc 2 =  m 0 c 2 )= 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 -  m 0 c 2 = (0.97 – 0.51)MeV = 0.46 MeV Kinetic energy is the difference between the total relativistic energy and the rest mass, K = E -  m 0 c 2 = (0.97 – 0.51)MeV = 0.46 MeV

117. 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 m 0 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. Calculate (i) the kinetic energy of the system and (ii) mass increase for a completely inelastic head-on of two balls (with rest mass m 0 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. m0m0m0m0 m0m0m0m0 M u u

118. 117Solution (i) K = 2mc 2 - 2m 0 c 2 = 2(  m 0 c 2 (i) K = 2mc 2 - 2m 0 c 2 = 2(  m 0 c 2 (ii) E before = E after  2  m 0 c 2 =  Mc 2  M = 2  m 0 (ii) E before = E after  2  m 0 c 2 =  Mc 2  M = 2  m 0 Mass increase  M = M - 2m 0 = 2(  m 0 Mass increase  M = M - 2m 0 = 2(  m 0 Approximation: u/c = …=1.5x10 -6   1 + ½ u 2 /c 2 (binomail expansion)  M  2  1 + ½ u 2 /c 2  m 0 Approximation: u/c = …=1.5x10 -6   1 + ½ u 2 /c 2 (binomail expansion)  M  2  1 + ½ u 2 /c 2  m 0 Mass increase  M = M - 2m 0 Mass increase  M = M - 2m 0  (u 2 /c 2  m 0 = 1.5x10 -6 m 0  (u 2 /c 2  m 0 = 1.5x10 -6 m 0 Comparing K with  Mc 2 : 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 Comparing K with  Mc 2 : 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 In contrast, in classical mechanics, momentum is conserved but kinetic energy is not in an inelastic collision

119. 118 Conservation of energy-momentum In terms of relativistic momentum, the relativistic total energy can be expressed as followed Relativistic momentum and relativisitc Energy

120. 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)

121. 120 Example: measuring pion mass using conservation of momentum- energy pi meson decays into a muon + massless neutrino pi meson decays into a muon + massless neutrino If the mass of the muon is known to be 106 MeV/c 2, and the kinetik energy of the muon is measured to be 4.6 MeV, find the mass of the pion If the mass of the muon is known to be 106 MeV/c 2, and the kinetik energy of the muon is measured to be 4.6 MeV, find the mass of the pion

122. 121Solution

123. 122

124. 123 Binding energy The nucleus of a deuterium comprises of one neutron and one proton. Both nucleons are bounded within the deuterium nucleus The nucleus of a deuterium comprises of one neutron and one proton. Both nucleons are bounded within the deuterium nucleus Neutron, m n proton, m p Deuterium, m d Nuclear fusion Initially, the total Energy = (m n + m n )c 2 After fusion, the total energy = m d c 2 + U U Analogous to exothermic process in chemistry

125. 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 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 U is called the binding energy

126. 125 U can be explained in terms energy-mass equivalence relation, as followed U can be explained in terms energy-mass equivalence relation, as followed For the following argument, we will ignore KE for simplicity sake For the following argument, we will ignore KE for simplicity sake Experimentally, we finds that m n + m p > m d Experimentally, we finds that m n + m p > m d By conservation of energy-momentum, By conservation of energy-momentum, E(before) = E(after) E(before) = E(after) m n c 2 + m p c 2 + 0 = m d c 2 + U m n c 2 + m p c 2 + 0 = m d c 2 + U Hence, U = (m p + m n )c 2 - m d c 2 =  mc 2 Hence, U = (m p + m n )c 2 - m d c 2 =  mc 2 The difference in mass between deuterium and the sum of (m n + m n )c 2 is converted into the binding energy that binds the proton to the neutron together The difference in mass between deuterium and the sum of (m n + m n )c 2 is converted into the binding energy that binds the proton to the neutron together

127. 126Example m n = 1.008665u; m p = 1.007276u; m n = 1.008665u; m p = 1.007276u; m d = 2.013553u; m d = 2.013553u; u = standard atomic unit = mass of 1/12 of the mass of a 12 C nucleus u = standard atomic unit = mass of 1/12 of the mass of a 12 C nucleus = 1.66 x 10 -27 kg = 1.66 x 10 -27 x c 2 J = 1.494 x 10 -10 J = 1.494 x 10 -10 /(1.6x10 -19 ) eV = 933.75 x 10 6 eV = 933.75 x MeV Hence the binding energy U  mc  = (m p + m n )c 2 - m d c 2 =0.002388u = 2.23 MeV

128. 127 Fission Such as Such as The reverse of nuclear fusion is nuclear fission The reverse of nuclear fusion is nuclear fission An parent nuclide M disintegrates into daughter nuclides such that their total mass  m i < M. An parent nuclide M disintegrates into daughter nuclides such that their total mass  m i < M. The energy of the mass deficit equivalent, The energy of the mass deficit equivalent, Q = (M -  m i )c 2 =  mc 2 called disintegration energy will be released

129. 128 SR finishes here…