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- 1. Theory of Relativity Albert Einstein Physics 100 Chapt 18
- 2. watching a light flash go by v c 2kThe man on earth sees c = √ κ (& agrees with Maxwell)
- 3. watching a light flash go by v cIf the man on the rocket sees c-v, he disagrees with Maxwell
- 4. Do Maxwell’s Eqns only work in one reference frame? If so, this would be the rest frame of the luminiferous Aether.
- 5. If so, the speed of light should change throughout the year upstream,downstream, light moves light moves slower faster “Aether wind”
- 6. Michelson-MorleyNo aether wind detected: 1907 Nobel Prize
- 7. Einstein’s hypotheses: 1. The laws of nature are equally valid in every inertial reference frame. Including Maxwell’s eqns2. The speed of light in empty space is same for all inertial observers, regard- less of their velocity or the velocity of the source of light.
- 8. All observers see light flashes go by them with the same speed vNo matter how fastthe guy on the rocketis moving!! c Both guys see the light flash travel with velocity = c
- 9. Even when the light flash istraveling in an opposite direction v c Both guys see the light flash travel past with velocity = c
- 10. Gunfight viewed by observer at rest He sees both shots fired simultaneously Bang ! Bang !
- 11. Viewed by a moving observer
- 12. Viewed by a moving observer He sees cowboy shoot 1st & cowgirl shoot later Bang ! Bang !
- 13. Viewed by an observer in the opposite direction
- 14. Viewed by a moving observer He sees cowgirl shoot 1st & cowboy shoot later Bang Bang ! !
- 15. Time depends of state of motion of the observer!!Events that occur simultaneously according to one observer can occur at different times for other observers
- 16. Light clock
- 17. Seen from the ground
- 18. Eventsy (x2,t2) (x1,t1) x x x1 x2 x t
- 19. Prior to Einstein, everyone agreed the distance between events depends Same events, different observers upon the observer, but not the time. y’ y’y (x2,t2) (x1,t1) x x (x1’,t1’) (x2’,t2’) t’ t’ x1’ x1’ dist’ x2’ x’ x’ x1 x2 x t dist
- 20. Time is the 4th dimension Einstein discovered that there is no “absolute” time, it too depends upon the state of motion of the observer Einstein Newton Space-Timecompletely Space different & 2 different aspects concepts Time of the same thing
- 21. How are the times seen by 2 different observers related? We can figure this out with simple HS-level math ( + a little effort)
- 22. Catch ball on a rocket ship Event 2: girl catches the ball w v= =4m/s t w=4m t=1s Event 1: boy throws the ball
- 23. Seen from earth V0=3m/s V0=3m/s Location of the 2events is different = 5m Elapsed time is m 2 ) (4 the same m )2 + w=4m √ (3 The ball appears d= v0t=3m to travel faster d t=1s v= = 5m/s t
- 24. Flash a light on a rocket ship Event 2: light flash reaches the girl w c= t0 w t0 Event 1: boy flashes the light
- 25. Seen from earth V VSpeed has toBe the same 2 2+ w )Dist is longer √( vt w d=Time must be vt longer d =√ (vt) +w 2 2 c= t=? t t
- 26. How is t related to t0?t= time on Earth clock t0 = time on moving clock w c =√ (vt)2+w2 c = t0 t ct = √ (vt)2+w2 ct0 = w (ct)2 = (vt)2+w2 (ct)2 = (vt)2+(ct0)2 (ct)2-(vt)2= (ct0)2 (c2-v2)t2= c2t02 c2 1 t = 2 2 t 02 t2 = t2 2 c – v 2 1 – v /c 0 2 1 t= t0 √1 – v2/c2 t = γ t0this is called γ
- 27. Properties of γ = 1 √1 – v2/c2Suppose v = 0.01c (i.e. 1% of c) 1 1 γ = √1 – (0.01c)2/c2 = √1 – (0.01)2c2/c2 1 1 1 γ = √1 – (0.01)2 = = √1 – 0.0001 √0.9999 γ = 1.00005
- 28. Properties 1 of γ = √1 – v(cont’d) 2 /c2Suppose v = 0.1c (i.e. 10% of c) 1 1 γ = √1 – (0.1c)2/c2 = √1 – (0.1)2c2/c2 1 1 1 γ = √1 – (0.1)2 = = √1 – 0.01 √0.99 γ = 1.005
- 29. Let’s make a chart v γ =1/√(1-v2/c2)0.01 c 1.00005 0.1 c 1.005
- 30. Other values of 1 γ = √1 – v2/c2Suppose v = 0.5c (i.e. 50% of c) 1 1 γ = √1 – (0.5c)2/c2 = √1 – (0.5)2c2/c2 1 1 1 γ = √1 – (0.5)2 = = √1 – (0.25) √0.75 γ = 1.15
- 31. Enter into chart v γ =1/√(1-v2/c2)0.01 c 1.00005 0.1 c 1.005 0.5c 1.15
- 32. Other values of 1 γ = √1 – v2/c2Suppose v = 0.6c (i.e. 60% of c) 1 1 γ =√1 – (0.6c)2/c2 = √1 – (0.6)2c2/c2 1 1 1 γ = √1 – (0.6)2 = = √1 – (0.36) √0.64 γ = 1.25
- 33. Back to the chart v γ =1/√(1-v2/c2)0.01 c 1.00005 0.1 c 1.005 0.5c 1.15 0.6c 1.25
- 34. Other values of 1 γ = √1 – v2/c2Suppose v = 0.8c (i.e. 80% of c) 1 1 γ = √1 – (0.8c)2/c2 = √1 – (0.8)2c2/c2 1 1 1 γ = √1 – (0.8)2 = = √1 – (0.64) √0.36 γ = 1.67
- 35. Enter into the chart v γ =1/√(1-v2/c2)0.01 c 1.00005 0.1 c 1.005 0.5c 1.15 0.6c 1.25 0.8c 1.67
- 36. Other values of 1 γ = √1 – v2/c2Suppose v = 0.9c (i.e.90% of c) 1 1 γ = √1 – (0.9c)2/c2 = √1 – (0.9)2c2/c2 1 1 1 γ = √1 – (0.9)2 = = √1 – 0.81 √0.19 γ = 2.29
- 37. update chart v γ =1/√(1-v2/c2)0.01 c 1.00005 0.1 c 1.005 0.5c 1.15 0.6c 1.25 0.8c 1.67 0.9c 2.29
- 38. Other values of 1 γ = √1 – v2/c2Suppose v = 0.99c (i.e.99% of c) 1 1 γ =√1 – (0.99c)2/c2 = √1 – (0.99)2c2/c2 1 1 1 γ = √1 – (0.99)2 = = √1 – 0.98 √0.02 γ = 7.07
- 39. Enter into chart v γ =1/√(1-v2/c2)0.01 c 1.00005 0.1 c 1.005 0.5c 1.15 0.6c 1.25 0.8c 1.67 0.9c 2.290.99c 7.07
- 40. Other values of 1 γ = √1 – v2/c2Suppose v = c 1 1 γ = √1 – (c)2/c2 = √1 – c2/c2 1 1 1 γ = = = √1 – 12 √0 0 γ = ∞ Infinity!!!
- 41. update chart v γ =1/√(1-v2/c2)0.01 c 1.00005 0.1 c 1.005 0.5c 1.15 0.6c 1.25 0.8c 1.67 0.9c 2.290.99c 7.071.00c ∞
- 42. Other values of 1 γ = √1 – v2/c2Suppose v = 1.1c 1 1 γ = √1 – (1.1c)2/c2 = √1 – (1.1)2c2/c2 1 1 1 γ = √1 – (1.1)2 = = √1-1.21 √ -0.21 γ = ??? Imaginary number!!!
- 43. Complete the chart v γ =1/√(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005 0.5c 1.15 0.6c 1.25 0.8c 1.67 0.9c 2.29 0.99c 7.07 1.00c ∞Larger than c Imaginary number
- 44. Plot results: ∞ Never-never land 1γ = √1 – v2/c2 x x x x x v=c
- 45. Moving clocks run slower v t0 t= 1 t 0 √1 – v2/c2 t t = γ t0 γ >1 t > t0
- 46. Length contraction v L0 time=t L0 = vt te r! or Shman on Time = t0 =t/γrocket Length = vt0 =vt/γ =L0/γ
- 47. Moving objects appear shorter Length measured when object is at rest L = L0/γ γ >1 L < L0 V=0.9999c V=0.86c V=0.1c V=0.99c
- 48. Length contraction
- 49. mass: change in v F=m0a = m0 time t0 a time=t0 m0 Ft0 change in v = m0 Ft0 m0 = change in v mass Ft γ Ft0m= = = γ m0 increases!! change in v change in v m = γ m0 t=γt0 by a factor γ
- 50. Relativistic mass increase m0 = mass of an object when it is at rest “rest mass”mass of a moving γobject increases as vc, m∞ m = γ m0 as an object moves faster, it gets harder & harder to accelerate by the γ factor v=c
- 51. summary• Moving clocks run slow γ o f o r c t• Moving objects appear shorter f a a y• Moving object’s mass increases B
- 52. Plot results: ∞ Never-never land 1γ = √1 – v2/c2 x x x x x v=c
- 53. α- Twin paradox centauri rs y ea ht Twin brother lig & sister 4.3 She will travel to α -centauri (a near- by star on a specialHe will stay home rocket ship v = 0.9c& study Phys 100
- 54. Light yeardistance light travels in 1 year dist = v x time = c yr 1cyr = 3x108m/s x 3.2x107 s = 9.6 x 1015 m We will just use cyr units & not worry about meters
- 55. Time on the boy’s clock r cy 0.9c =4 . 3 v= d0 0.9c v= According to the boy & his clock on Earth: d0 4.3 cyr = 4.8 yrs tout = = 0.9c v d0 4.3 cyr = 4.8 yrs tback = = 0.9c v ttotal = tout+tback = 9.6yrs
- 56. What does the boy see on her clock? yr 0.9c 4. 3c v= d= 0.9c v= According to the boy her clock runs slower tout 4.8 yrs t = out = 2.3 = 2.1 yrs γ tback 4.8 yr tback = γ = = 2.1 yrs 2.3 ttotal = tout+tback = 4.2yrs
- 57. So, according to the boy: yr 0.9c 4. 3c v= d= 0.9c v= his clock her clock out: 4.8yrs 2.1yrs back: 4.8yrs 2.1yrs ges She a total: 9.6yrs 4.2yrs less
- 58. But, according to the girl, the boy’s clock ismoving &, so, it must be 0.9c running slower v= According to her, the boy’s clock on Earth says: tout 2.1 yrs tout = γ = = 0.9 yrs 2.3 tback 2.1 yrs = 0.9 yrs tback = = 2.3 .9c γ v=0 ttotal = tout+tback = 1.8yrs
- 59. Her clock advances 4.2 yrs& she sees his clock advance only 1.8 yrs, contradict ion??AShe should think he has aged less than her!!
- 60. Events in the boy’s life: As seen by him As seen by her She leaves 4.8 yrs 0.9 yrs She arrives & starts turn short time ???? Finishes turn & heads home 4.8 yrs 0.9 yrs She returns 9.6+ yrs 1.8 + ??? yrs
- 61. turning around as seen by her According to her, these 2 events occur very,very far apart from each other He sees herHe sees her finish turningstart to turn Time interval between 2 events depends on the state of motion of the observer
- 62. Gunfight viewed by observer at rest He sees both shots fired simultaneously Bang ! Bang !
- 63. Viewed by a moving observer
- 64. Viewed by a moving observer He sees cowboy shoot 1st & cowgirl shoot later Bang ! Bang !
- 65. In fact, ???? = 7.8+ years as seen by him as seen by her She leaves 4.8 yrs 0.9 yrs She arrives & starts turn short time 7.8+ yrs ??? Finishes turn & heads home 4.8 yrs 0.9 yrs She returns 9.6+ yrs 1.8 + ???yrs 9.6+ yrs
- 66. No paradox: both twins agree The twin that “turned around” is younger
- 67. Ladder & Barn Door paradoxStan & Ollie puzzle over howto get a 2m long ladder thrua 1m wide barn door ??? 1m 2m ladder
- 68. Ollie remembers Phys 100 & the theory of relativityStan, pick upthe ladder &run very fast 1m 2m tree ladder
- 69. View from Ollie’s ref. frame 1m Push, Stan! 2m/γ V=0.9cOllie Stan (γ=2.3)
- 70. View from Stan’s ref. frame But it 1m/ doesn’t fit, γ Ollie!!V=0.9c(γ=2.3) 2mOllie Stan
- 71. If Stan pushes both ends of theladder simultaneously, Ollie sees the two ends move at different times: Too late 1m Too Stan! soon nk Stan! clu clan k V=0.9c Ollie Stan Stan (γ=2.3)
- 72. Fermilab proton accelerator V=0.9999995c γ =1000 2km
- 73. Stanford electron acceleratorv=0.99999999995 c 3k mγ=100,000
- 74. statusEinstein’s theory of “special relativity” has been carefully tested in many very precise experiments and found to be valid.Time is truly the 4th dimension of space & time.
- 75. testγ=29.3

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