2. PART TWO: RELATIVISTIC MECHANICS PHYS 141: Principles of Mechanics

3. I. Basic Principles A. Spacetime spacetime diagrams 1. Graphical Depiction: Before we look at how space and time are connected through special relativity, let’s establish how we can describe basic motion: spacetime diagrams. x t 1-D motion in time as measured from a Particular Reference Frame

4. I. Basic Principles A. Spacetime spacetime diagrams 1. Graphical Depiction: Before we look at how space and time are connected through special relativity, let’s establish how we can describe basic motion: spacetime diagrams. x t “PAST” “FUTURE” “PRESENT” 1-D motion in time as measured from a Particular Reference Frame

5. I. Basic Principles A. Spacetime spacetime diagrams 1. Graphical Depiction: Before we look at how space and time are connected through special relativity, let’s establish how we can describe basic motion: spacetime diagrams. x t “SOMEWHERE AHEAD” “HERE” “SOMEWHERE BEHIND” 1-D motion in time as measured from a Particular Reference Frame

6. I. Basic Principles A. Spacetime a.Event: a particular location and time. x t “event” (x,t) “OVER THERE, THEN” 1-D motion in time as measured from a Particular Reference Frame

7. I. Basic Principles A. Spacetime a.Event: a particular location and time. x t “event 1” (x 1,t 1 ) “event 2” (x 2,t 2 ) 1-D motion in time as measured from a Particular Reference Frame

8. I. Basic Principles A. Spacetime b.World Lines are Spacetime trajectories: How events are ordered. x t “event 1” (x 1,t 1 ) “event 2” (x 2,t 2 ) “Stay at x = x 1 for t 2 -t 1 seconds” 1-D motion in time as measured from a Particular Reference Frame

9. I. Basic Principles A. Spacetime b.World Lines: How events are ordered. x t “event 1” (x 1,t 1 ) “event 2” (x 2,t 2 ) “Instantaneously move From x 1 to x 2.” Note: this is impossible. 1-D motion in time as measured from a Particular Reference Frame

10. I. Basic Principles A. Spacetime c.Normalized Spacetime diagrams c.Normalized Spacetime diagrams. Let w = ct. Then the slope of a trajectory in spacetime is dw/dx = cdt/dx = c/v, with c = speed of light. x w = ct “event 1” (x 1,w 1 ) “event 2” (x 2,w 2 ) Speed from E1 to E2 is dw/dx = c/v =1, so v/c = 1. 1-D motion in time as measured from a Particular Reference Frame A

11. I. Basic Principles A. Spacetime c.Normalized Spacetime diagrams c.Normalized Spacetime diagrams. Let w = ct. Then the slope of a trajectory in spacetime is dw/dx = cdt/dx = c/v, with c = speed of light. x w = ct 1-D motion in time as measured from a Particular Reference Frame A

12. I. Basic Principles A. Spacetime Normalized Spacetime diagrams. Let w = ct. Then the slope of a trajectory in spacetime is dw/dx = cdt/dx = c/v, with c = speed of light. x w=ct v/c = 1dw/dx = c/v >1, so v/c <1 dw/dx = c/v <1, so v/c >1 1-D motion in time as measured from a Particular Reference Frame A

13. I. Basic Principles A. Spacetime Lightcones: d.Lightcones: Regions of spacetime in which events are connected by paths with speed v/c ≤ 1. x w “Future light cone” for A “Past light cone” for A v/c = 1 1-D motion in time as measured from a Particular Reference Frame A

14. I. Basic Principles A. Spacetime Lightcones: d.Lightcones: Regions of spacetime in which events are connected by paths with speed v/c ≤ 1. x w v/c = 1. “Future light cone” for A “Past light cone” for A Inaccessible to A according to relativity Inaccessible to A according to relativity v/c = 1.

15. Constant speed(x’,y’,z’,t’) I. Basic Principles B. Galilean Transformations Galilean Transformation For inertial systems, the Galilean Transformation allows us to translate between frames: all we have to do is (basically) subtract out the motion of primed frame (the train in this case).

16. I. Basic Principles B. Galilean Transformations

19. Galilean Transformation For inertial systems, the Galilean Transformation allows us to translate between frames: all we have to do is (basically) subtract out the motion of primed frame (the train in this case).

20. 1. Coordinate Transformations: transformations between two reference frames. Consider a primed reference in motion relative to an unprimed one (let v be along the positive x- direction) x y a)Galilean transformations: coordinates for a point P in the moving primed frame as seen from the (unmoving) unprimed frame (and vice versa). x = x’ + vt’.(I.B.1-4) y = y’. z = z’. t = t’. I. Basic Principles B. Galilean Transformations

21. x y x = x’ + vt’. y = y’. z = z’. t = t’. x’ y’ v P x x’ I. Basic Principles B. Galilean Transformations 1. Coordinate Transformations: transformations between two reference frames. Consider a primed reference in motion relative to an unprimed one (let v be along the positive x- direction) a)Galilean transformations: coordinates for a point P in the moving primed frame as seen from the (unmoving) unprimed frame (and vice versa).

22. x y x = x’ + vt’. y = y’. z = z’. t = t’. x’ y’ v P x vt’ x’ I. Basic Principles B. Galilean Transformations 1. Coordinate Transformations: transformations between two reference frames. Consider a primed reference in motion relative to an unprimed one (let v be along the positive x- direction) a)Galilean transformations: coordinates for a point P in the moving primed frame as seen from the (unmoving) unprimed frame (and vice versa).

23. x y x = x’ + vt’. y = y’. z = z’. t = t’. x’ y’ v P x vt’ x’ I. Basic Principles B. Galilean Transformations 1. Coordinate Transformations: transformations between two reference frames. Consider a primed reference in motion relative to an unprimed one (let v be along the positive x- direction) a)Galilean transformations: coordinates for a point P in the moving primed frame as seen from the (unmoving) unprimed frame (and vice versa).

24. x y x = x’ + vt’. y = y’. z = z’. t = t’. x’ y’ v P x vt’ x’ I. Basic Principles B. Galilean Transformations 1. Coordinate Transformations: transformations between two reference frames. Consider a primed reference in motion relative to an unprimed one (let v be along the positive x- direction) a)Galilean transformations: coordinates for a point P in the moving primed frame as seen from the (unmoving) unprimed frame (and vice versa).

25. x y x = x’ + vt’. y = y’. z = z’. t = t’. x’ y’ v P x vt’ x’ I. Basic Principles B. Galilean Transformations 1. Coordinate Transformations: transformations between two reference frames. Consider a primed reference in motion relative to an unprimed one (let v be along the positive x- direction) a)Galilean transformations: coordinates for a point P in the moving primed frame as seen from the (unmoving) unprimed frame (and vice versa).

26. II. Simultaneity A. Set Up 1. Consider two observers, A&B, stationary with respect to each other and reference frame (w,x). How do we calibrate their identical clocks? x w AB

27. II. Simultaneity A. Set Up 1. Consider two observers, A&B, stationary with respect to each other and reference frame (w,x). How do we calibrate their identical clocks? x w AB wBwB Events are simultaneous (clocks calibrated) By backtracking To the equidistant Spacetime position Between light paths. Events W A and W B are simultaneous in this frame. wAwA Note: dotted line indicates same TIME for unprimed frame

28. II. Simultaneity A. Set Up 2. What happens to the axes for a Galilean Transformation? x w (x,t)

29. II. Simultaneity A. Set Up 2. What happens to the axes for a Galilean Transformation? Now consider a frame moving with speed v wrt the original frame. x t (x,t) x=(x’ + vt’)

30. II. Simultaneity A. Set Up 2. What happens to the axes for a Galilean Transformation? Now consider a frame moving with speed v wrt the original frame. x t (x,t) x=(x’ + vt’) x=(x’ + v(2t’))

31. II. Simultaneity A. Set Up 2. What happens to the axes for a Galilean Transformation? Now consider a frame moving with speed v wrt the original frame. x t (x,t) x=(x’ + v(3t’)) x=(x’ + vt’) x=(x’ + v(2t’))

32. II. Simultaneity A. Set Up 2. What happens to the axes for a Galilean Transformation? Now consider a frame moving with speed v wrt the original frame. x t (x,t) x=(x’ + v(3t’)) x=(x’ + vt’) x=(x’ + v(2t’)) x’

33. II. Simultaneity A. Set Up 2. What happens to the axes for a Galilean Transformation? Now consider a frame moving with speed v wrt the original frame. x t (x,t) x’

34. II. Simultaneity A. Set Up 2. What happens to the axes for a Galilean Transformation? Now consider a frame moving with speed v wrt the original frame. x t (x,t) x’ t’=t

35. II. Simultaneity A. Set Up 2. What happens to the axes for a Galilean Transformation? Now consider a frame moving with speed v wrt the original frame. x t x’

36. II. Simultaneity A. Set Up 3.Why? Newtonian addition of speeds. x t x’ t’

37. II. Simultaneity A. Set Up 2. What happens to the axes for a Galilean Transformation? Now consider a frame moving with speed v wrt the original frame. x t + =

38. I. Basic Principles C. The ‘Principle of Relativity’ “The Laws of Mechanics are the same in every inertial frame, and the Galilean Transformation is valid.” Problem: Electromagnetism (1850)

39. I. Basic Principles D. Postulates of Special Relativity 1. The speed of light in vacuum is a constant, independent of the motion of the source, the observer, or both. 2. The Laws of Physics are everywhere the same for inertial frames, and the connection between frames is the Lorentz Transformation. 3. When v/c is small, then the LT reduces to the GT: x’ = (x - vt) & t’ = t. x’x v P x’ = (x - vt)/{1 - (v/c) 2 } 1/2. y’ = y. z’ = z. t’ =  t - vx/c 2 )/{1 - (v/c) 2 } 1/2. (I.D.1-4)

40. II. Simultaneity A. Set Up 2. What happens to the axes for a Lorentz Transformation? x w (x,w)

41. II. Simultaneity A. Set Up x w (x’,w’)

42. II. Simultaneity A. Set Up x w

43. II. Simultaneity B. Result 1. Now what is measured by another observer moving with respect to the original frame with speed (v’)? x w AB w’

44. 1. Now what is measured by another observer moving with respect to the original frame with speed (v’)? x w AB w’ x’ II. Simultaneity B. Result

45. II. Simultaneity B. Results 1. Now what is measured by another observer moving with respect to the original frame with speed (v’)? x w A B w’ x’ wB’wB’

46. II. Simultaneity B. Results 1. Now what is measured by another observer moving with respect to the original frame with speed (v’)? x w A B w’ wA’wA’ Simultaneity Is frame dependent wB’wB’ Events w A ’ and w B ’ are simultaneous in the red frame. However, now the events are NOT simultaneous in the unprimed frame. Note: dotted line indicates same TIME for primed frame

47. II. Simultaneity B. Results 1. Now what is measured by another observer moving with respect to the original frame with speed (v)? x w t3t3 t2t2 t1t1

48. II. Simultaneity B. Results 1. Now what is measured by another observer moving with respect to the original frame with speed (v)? x w w’ x’ t3’t3’

49. II. Simultaneity B. Results 1. Now what is measured by another observer moving with respect to the original frame with speed (v)? x w w’ x’ t3’t3’ Example: x = 1 lys, t=0s, v/c = 0.5 x’= (1 lys - 0.5c(0s))/{1-0.5 2 } 1/2 = 1.15*(5.5) = 1.15 lys t’ = (0 s - 0.5c(1 ly-s)/c 2 )/{1-0.5 2 } 1/2 =1.15*(-.5) = -0.6 second.

50. II. Simultaneity B. Results 1. Now what is measured by another observer moving with respect to the original frame with speed (v)? x w w’ x’ t3’t3’ Example: x = 5 lys, t=0s, v/c = 0.5 x’= (5 lys – 0.5c(0s))/{1-0.5 2 } 1/2 = 1.15*(5) = 5.8 lys t’ = (0 s – 0.5c(5 lys)/c 2 )/{1-0.5 2 } 1/2 =1.15*(-2.5) = -2.9 second.

51. II. Simultaneity B. Results 1. Now what is measured by another observer moving with respect to the original frame with speed (v’)? x w w’ x’

52. II. Simultaneity B. Results 1. Now what is measured by another observer moving with respect to the original frame with speed (v’)? x w w’ x’

53. (II.C.1) Invariant: independent of frame that is measured What is the space time interval on a lightcone? II. Simultaneity C. Spacetime Interval Consider a light pulse that travels in a sphere.

54. x w=ct ds 2 = 0, lightlike ds 2 >0, timelike, Causally connected ds 2 <0, Spacelike, Causally unconnected A

55. 1. What is measured by another observer moving with respect to the original frame with speed (v’)? x w w’ w* We know that w = 3 w* = 2 both measured from the black Reference frame clock. What is w’ measured from red Reference Frame clock? III.Time Dilation A. Definition

56. 1. Now what is measured by another observer moving with respect to the original frame with speed (v’)? x w w’ w* w’ =  w* w=  w’, and so w =   w*. III.Time Dilation A. Definition

57. 1. Now what is measured by another observer moving with respect to the original frame with speed (v’)? x w w’ w* w’ =  w* w=  w’, and so w =   w*. w’ = w/  w’ < w  Time divisions in the primed frame are different than in the unprimed frames III.Time Dilation A. Definition

58. 1. Clocks moving relative to an observer are measured by that observer to run slow compared to a clock at rest. 2. Another thought experiment: Albert on a train w/speed v. t’ = 0: light leaves Note: only one clock is needed to measure emission and return of light. III.Time Dilation B. Derivation

59. Light travels a total distance equal to 2d. Light requires t’ = 2d/c seconds to reach the floor.

60. 2.And now Albert’s view from the platform. Now we’ll need two synchronized clocks to measure both events:

61. Clock 1 Clock 2

62. Clock 1 Clock 2

63. Clock 1 Clock 2

64. Clock 1 Clock 2

65. Clock 1 Clock 2

66. Clock 1 Clock 2

67. Clock 1 Clock 2

68. Clock 1 Clock 2

69. Clock 1 Clock 2

70. Clock 1 Clock 2

71. 3.Mathematical formulation (pseudo-classical): L = vt. d = ct’. D = (d 2 + L 2 ) 1/2. c = D/t = (d 2 + L 2 ) 1/2 /t; c 2 = (d 2 + L 2 )/t 2. t 2 = (d/c) 2 + (vt/c) 2 ; t 2 (1- (v/c) 2 ) = (d/c) 2. However, d = ct’, and so t = t’/{1- (v/c) 2 } 1/2. t’ = t{1- (v/c) 2 } 1/2. Measured from platform Measured on train

72. t = t’/{1- (v/c) 2 } 1/2. (III.B.1) t’ = t{1- (v/c) 2 } 1/2. (III.B.2) Note: a)t is the time in the train as measured by the platform observer. b)t’ is the time in the train as measured by the train observer. c)t > t’: moving clocks run slow as measured in the platform frame. III.Time Dilation

73. III. Time Dilation C. Implications Let  = 1/{1- (v/c) 2 } 1/2 (III.B.3) t =  t’. (III.B.4) t’ = t/ . (III.B.5)

74. 1.Verifications & Implications a)Muon lifetime Particle decay time longer as measured in lab (rest) frame b)Atomic clock experiments c)The “Twin Paradox” III. Time Dilation C. Implications

75. III. Time Dilation Twin “Paradox” x w w’ w* Rocket and Earth Clocks synchronized At w’ and w*

76. III. Time Dilation Twin “Paradox” x w** w’ w* On return, Rocket and Earth Clocks synchronized At w’ and w**

77. III. Time Dilation Twin “Paradox” x w** w’ w* On return, Rocket and Earth Clocks synchronized At w’ and w**

78. III. Time Dilation Twin “Paradox” x w’ x’

79. III. Time Dilation Twin “Paradox” x w’ x’

80. III. Time Dilation Twin “Paradox” x w’ x’

81. III. Time Dilation Twin “Paradox” x w’ x’

82. III. Time Dilation Twin “Paradox” x w’

83. III. Time Dilation Twin “Paradox” x w’

84. III. Time Dilation Twin “Paradox” x w’

85. 2.Example: An astronaut travels to a star 100 lys away with a speed equal to 0.99c. How much time elapses on the ship as measured by someone in the ship?  The time on the ship is given by B.2  t’ =  t{1- (v/c) 2 } 1/2. Note:  t, the time elapsed on the Earth is {100 ly/0.99 ly/y} = 101 y, since v = 0.99c. Thus,  t 0 =  t’ =  t{1- (v/c) 2 } 1/2 ~ (101 yrs)(1 -.99 2 ) 1/2 = 14 yrs. “proper time” III. Time Dilation D. Implications

86. 1. Definition: the length of an object L is measured to be shorter when it is moving relative to the observer than when it is at rest. 2. Mathematical formulation: L = L 0 {1- (v/c) 2 } 1/2. (IV.D.1) 3. Example: in the previous example, if the ship has a length of 100m as measured in its rest frame, what is its length as seen from the Earth? L = L 0 {1- (v/c) 2 } 1/2 = (100m)(1 -.99 2 ) 1/2 = 14 m. IV. Length Contraction A. Formulation

87. 1. Relative velocity is not computed by simply adding (or subtracting velocity vectors. 2. Mathematical Formulation: u = (v ± u’)/(1 ± vu’/c 2 ).(V.A.1) v u’ V. Velocity Addition A. Formulation

88. 3. Example: two particles collide with individual speeds of.9c and.5c, respectively as measured in the lab. What is the relative speed of the second particle as seen by the first particle? Classically, we would just add these speeds and get 1.4c, which is clearly wrong if SR is valid. u = (v + u’)/(1 + vu’/c 2 ) = (1.4c)/(1 +.45c/c) = 0.97c. vu’ V. Velocity Addition A. Formulation

89. V. Velocity Addition B. Spacetime 1. How does velocity addition appear in ST diagrams? a)Galilean addition x w=ct v/c = 1  = arctan(.5) = 27 o Let a train move at v/c = 0.5 wrt the platform. Let’s consider An object at rest on the train as it speeds by the platform. x’ v’/c = 0.5

90. V. Velocity Addition B. Spacetime 1. How does velocity addition appear in ST diagrams? a)Galilean addition x w=ct v/c = 1  = arctan(.5) = 27 o Let a train move at v/c = 0.5 wrt the platform. Let’s consider An object at rest on the train as it speeds by the platform. Now consider someone running Down the train at u’/c = 0.5 x’ v’/c = 0.5 u’/c = 0.5 Galileo: combined velocity = c. D’oh!

91. V. Velocity Addition B. Spacetime 1. How does velocity addition appear in ST diagrams? a)Lorentz addition x w=ct v/c = 1  = arctan(.5) = 27 o Let a train move at v/c = 0.5 wrt the platform. Let’s consider An object at rest on the train as it speeds by the platform. Now consider someone running down the train at u’/c = 0.5 x’ v’/c = 0.5 u’/c = 0.5 Einstein: combined velocity =.8c.

92. A. Momentum p = mv/{1 -  2 } 1/2,=  mv =  mc. (VI.A.1) Equation A.1 => v for a massive object can never be ≥ c. B.Energy 1. Relativistic Kinetic Energy: K =  mc 2 - mc 2 = (  - 1)mc 2, or(VI.B.1) K = mc 2 ( 1/{1 -  2 } 1/2 - 1). Note: for  << 1, then 1/{1 -  2 } 1/2 ~ 1 + (1/2)  2, and K~ mc 2 (1 + (1/2)  2 - 1) = (1/2)mv 2, (VI.B.2) VI. Momentum & Energy

93. 2. Total Energy E =  mc 2 = mc 2 + K, where(B.3) The quantity mc 2 is the rest energy of the object. Using the momentum equation (A.1), we can also write: E 2 = p 2 c 2 + (mc 2 ) 2.(B.4) The consequences of (B.3, 4) are that mass and energy are equivalent, and that they can be converted into each other. VI. Momentum & Energy

94. 3. Example: Suppose our astronaut has a mass of 70kg. Compare her classical and relativistic KE. K(class.) = 1/2mv 2 = 3.2 x 10 18 J. K(rel.)=  mc 2 - mc 2 = 3.8 x 10 19 J. 4. Example: How much energy would be released if our astronaut were converted completely into energy? E(rest) = mc 2 = 6.3 x 10 18 J. E(tot) = E(rest) + K ~ 4 x 10 19 J ~ 10,000 Megatons of TNT. VI. Momentum & Energy

95. 5. Another example: An electron has a total energy of 10.0 “Mega-electron volts” (MeV) as measured in a particular laboratory frame. What is the value of its K, v? Note: E electron (rest) = m electron c 2 = 8.2 x 10 -14 J = 0.511 MeV. Thus, K = E – mc 2 = 9.5 MeV. E =  mc 2 ;  = E/mc 2 = 19.6;  2 = {1 –  2 } -1 = 384 =>  2 = 1 - 1/384 = 0.997.  = 0.999 VI. Momentum & Energy