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Physics_100_chapt_18.ppt
- 1. Theory of Relativity Physics100 Chapt 18 Albert Einstein
- 2. watching a lightflash go by c v The man on earth sees c = (& agrees with Maxwell) 2k k
- 3. watching a lightflash go by c v If the man on the rocket sees c-v, he disagrees with Maxwell
- 4. Do Maxwell’s Eqnsonly work in one reference frame? If so, this would be the rest frame of the luminiferous Aether.
- 5. If so, thespeed of light should change throughout the year upstream, light moves slower downstream, light moves faster “Aether wind”
- 6. Michelson-Morley No aether winddetected: 1907 Nobel Prize
- 7. Einstein’s hypotheses: 1. Thelaws of nature are equally valid in every inertial reference frame. Including Maxwell’s eqns 2. 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 seelight flashes go by them with the same speed c v Both guys see the light flash travel with velocity = c No matter how fast the guy on the rocket is moving!!
- 9. Even when thelight flash is traveling in an opposite direction c v Both guys see the light flash travel past with velocity = c
- 10. Gunfight viewed byobserver at rest Bang ! Bang ! He sees both shots fired simultaneously
- 11. Viewed by amoving observer
- 12. Viewed by amoving observer Bang ! Bang ! He sees cowboy shoot 1st & cowgirl shoot later
- 13. Viewed by anobserver in the opposite direction
- 14. Viewed by amoving observer Bang ! Bang ! He sees cowgirl shoot 1st & cowboy shoot later
- 15. Time depends ofstate of motion of the observer!! Events that occur simultaneously according to one observer can occur at different times for other observers
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- 17. Seen from theground
- 18.
- 19. Same events, differentobservers x y x t (x1,t1) x (x2,t2) x1 x2 x’ y’ x1’ (x1’,t1’) y’ x’ x1’ x2’ (x2’,t2’) t’ t’ Prior to Einstein, everyone agreed the distance between events depends upon the observer, but not the time. dist’ dist
- 20. Time is the4th dimension Einstein discovered that there is no “absolute” time, it too depends upon the state of motion of the observer Newton Space & Time Einstein Space-Time completely different concepts 2 different aspects of the same thing
- 21. How are thetimes seen by 2 different observers related? We can figure this out with simple HS-level math ( + a little effort)
- 22. Catch ball ona rocket ship w=4m t=1s v= =4m/s w t Event 1: boy throws the ball Event 2: girl catches the ball
- 23. Seen from earth w=4m v0t=3m v== 5m/s d t t=1s V0=3m/s V0=3m/s Location of the 2 events is different Elapsed time is the same The ball appears to travel faster
- 24. Flash a lighton a rocket ship w t0 c= w t0 Event 1: boy flashes the light Event 2: light flash reaches the girl
- 25. Seen from earth w vt c== d t t=? V V Speed has to Be the same Dist is longer Time must be longer (vt)2+w2 t
- 26. How is trelated to t0? c = (vt)2+w2 t t= time on Earth clock c = w t0 t0 = time on moving clock ct = (vt)2+w2 (ct)2 = (vt)2+w2 ct0 = w (ct)2 = (vt)2+(ct0)2 (ct)2-(vt)2= (ct0)2 (c2-v2)t2= c2t0 2 t2 = t0 2 c2 c2 – v2 t2 = t0 2 1 1 – v2/c2 t = t0 1 1 – v2/c2 this is called g t = g t0
- 27. Properties of g= 1 1 – v2/c2 1 1 – (0.01c)2/c2 g = Suppose v = 0.01c (i.e. 1% of c) 1 1 – (0.01)2c2/c2 = 1 1 – (0.01)2 g = 1 1 – 0.0001 = 1 0.9999 = g = 1.00005
- 28. Properties of g= (cont’d) 1 1 – v2/c2 1 1 – (0.1c)2/c2 g = Suppose v = 0.1c (i.e. 10% of c) 1 1 – (0.1)2c2/c2 = 1 1 – (0.1)2 g = 1 1 – 0.01 = 1 0.99 = g = 1.005
- 29. Let’s make achart v g =1/(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005
- 30. Other values ofg = 1 1 – v2/c2 1 1 – (0.5c)2/c2 g = Suppose v = 0.5c (i.e. 50% of c) 1 1 – (0.5)2c2/c2 = 1 1 – (0.5)2 g = 1 1 – (0.25) = 1 0.75 = g = 1.15
- 31. Enter into chart vg =1/(1-v2/c2) 0.01 c 1.00005 0.1 c 1.005 0.5c 1.15
- 32. Other values ofg = 1 1 – v2/c2 1 1 – (0.6c)2/c2 g = Suppose v = 0.6c (i.e. 60% of c) 1 1 – (0.6)2c2/c2 = 1 1 – (0.6)2 g = 1 1 – (0.36) = 1 0.64 = g = 1.25
- 33. Back to thechart v g =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 ofg = 1 1 – v2/c2 1 1 – (0.8c)2/c2 g = Suppose v = 0.8c (i.e. 80% of c) 1 1 – (0.8)2c2/c2 = 1 1 – (0.8)2 g = 1 1 – (0.64) = 1 0.36 = g = 1.67
- 35. Enter into thechart v g =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 ofg = 1 1 – v2/c2 1 1 – (0.9c)2/c2 g = Suppose v = 0.9c (i.e.90% of c) 1 1 – (0.9)2c2/c2 = 1 1 – (0.9)2 g = 1 1 – 0.81 = 1 0.19 = g = 2.29
- 37. update chart v g=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 ofg = 1 1 – v2/c2 1 1 – (0.99c)2/c2 g = Suppose v = 0.99c (i.e.99% of c) 1 1 – (0.99)2c2/c2 = 1 1 – (0.99)2 g = 1 1 – 0.98 = 1 0.02 = g = 7.07
- 39. Enter into chart vg =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
- 40. Other values ofg = 1 1 – v2/c2 1 1 – (c)2/c2 g = Suppose v = c 1 1 – c2/c2 = 1 1 – 12 g = 1 0 = 1 0 = g = Infinity!!!
- 41. update chart v g=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
- 42. Other values ofg = 1 1 – v2/c2 1 1 – (1.1c)2/c2 g = Suppose v = 1.1c 1 1 – (1.1)2c2/c2 = 1 1 – (1.1)2 g = 1 1-1.21 = 1 -0.21 = g = ??? Imaginary number!!!
- 43. Complete the chart vg =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: 1 1 –v2/c2 g = v=c x x x x x Never-never land
- 45. Moving clocks runslower t t0 t = g t0 t = t0 1 1 – v2/c2 v g >1 t > t0
- 46. Length contraction man on rocket Time= t0 =t/g Length = vt0 L0 v time=t L0 = vt =vt/g =L0/g
- 47. Moving objects appearshorter L = L0/g g >1 L < L0 Length measured when object is at rest V=0.1c V=0.86c V=0.99c V=0.9999c
- 48.
- 49. mass: m0 a F=m0a change in v time Ft0 m0 changein v = time=t0 t=gt0 Ft change in v m = = m0 mass increases!! t0 = gm0 Ft0 change in v m0 = gFt0 change in v = m = g m0 by a factor g
- 50. Relativistic mass increase m0= mass of an object when it is at rest “rest mass” m = g m0 mass of a moving object increases by the g factor as vc, m as an object moves faster, it gets harder & harder to accelerate g v=c
- 51. summary • Moving clocksrun slow • Moving objects appear shorter • Moving object’s mass increases
- 52. Plot results: 1 1 –v2/c2 g = v=c x x x x x Never-never land
- 53. Twin paradox Twin brother &sister She will travel to a-centauri (a near- by star on a special rocket ship v = 0.9c He will stay home & study Phys 100 a-centauri
- 54. Light year distance lighttravels in 1 year dist = v x time 1cyr = 3x108m/s x 3.2x107 s = 9.6 x 1015 m We will just use cyr units & not worry about meters = c yr
- 55. Time on theboy’s clock tout = d0 v 4.3 cyr 0.9c = = 4.8 yrs According to the boy & his clock on Earth: tback = d0 v 4.3 cyr 0.9c = = 4.8 yrs ttotal = tout+tback = 9.6yrs
- 56. What does theboy see on her clock? tout = tout g 4.8 yrs 2.3 = = 2.1 yrs According to the boy her clock runs slower tback = tback g 4.8 yr 2.3 = = 2.1 yrs ttotal = tout+tback = 4.2yrs
- 57. So, according tothe boy: his clock her clock out: 4.8yrs 2.1yrs back: 4.8yrs 2.1yrs total: 9.6yrs 4.2yrs
- 58. But, according tothe girl, the boy’s clock is moving &, so, it must be running slower tout = tout g 2.1 yrs 2.3 = = 0.9 yrs According to her, the boy’s clock on Earth says: tback = tback g 2.1 yrs 2.3 = = 0.9 yrs ttotal = tout+tback = 1.8yrs
- 59. Her clock advances4.2 yrs & she sees his clock advance only 1.8 yrs, She should think he has aged less than her!!
- 60. As seen byhim Events in the boy’s life: As seen by her She leaves She arrives & starts turn Finishes turn & heads home She returns 4.8 yrs 4.8 yrs short time 9.6+ yrs 0.9 yrs ???? 0.9 yrs 1.8 + ??? yrs
- 61. turning around asseen by her He sees her start to turn He sees her finish turning According to her, these 2 events occur very,very far apart from each other Time interval between 2 events depends on the state of motion of the observer
- 62. Gunfight viewed byobserver at rest Bang ! Bang ! He sees both shots fired simultaneously
- 63. Viewed by amoving observer
- 64. Viewed by amoving observer Bang ! Bang ! He sees cowboy shoot 1st & cowgirl shoot later
- 65. as seen byhim In fact, ???? = 7.8+ years as seen by her She leaves She arrives & starts turn Finishes turn & heads home She returns 4.8 yrs 4.8 yrs short time 9.6+ yrs 0.9 yrs ??? 0.9 yrs 1.8 + ???yrs 7.8+ yrs 9.6+ yrs
- 66. No paradox: bothtwins agree The twin that “turned around” is younger
- 67. Ladder & BarnDoor paradox 1m 2m ??? ladder Stan & Ollie puzzle over how to get a 2m long ladder thru a 1m wide barn door
- 68. Ollie remembers Phys100 & the theory of relativity 1m 2m ladder Stan, pick up the ladder & run very fast tree
- 69. View from Ollie’sref. frame 1m 2m/g Push, Stan! V=0.9c (g=2.3) Ollie Stan
- 70. View from Stan’sref. frame 2m 1m/g V=0.9c (g=2.3) Ollie Stan But it doesn’t fit, Ollie!!
- 71. If Stan pushesboth ends of the ladder simultaneously, Ollie sees the two ends move at different times: 1m Too soon Stan! V=0.9c (g=2.3) Ollie Stan Stan Too late Stan!
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- 74. status Einstein’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.