Special Relativity. Chris Prior. Trinity College Oxford. and

Transcription

1 Special Relativity Chris Prior ASTeC RAL Trinity College Oxford and 1

2 Overview The principle of special relativity Lorentz transformation and consequences Space-time 4-vectors: position, velocity, momentum, invariants, covariance. Derivation of E=mc 2 Examples of the use of 4-vectors Inter-relation between β and γ, momentum and energy An accelerator problem in relativity Photons and wave 4-vector Relativistic particle dynamics Lagrangian and Hamiltonian Formulation Radiation from an Accelerating Charge Motion faster than speed of light 2

3 Reading W. Rindler: Introduction to Special Relativity (OUP 1991) D.F. Lawden: An Introduction to Tensor Calculus and Relativity (Dover, 2003) N.M.J. Woodhouse: Special Relativity (Springer 2002) A.P. French: Special Relativity, MIT Introductory Physics Series (Nelson Thomes) C.Misner, K.Thorne and J.Wheeler: Relativity (Freeman, 1973) C.R. Prior: Special Relativity, CERN Accelerator School (Zeegse) 3

4 Historical Background Groundwork of Special Relativity laid by Lorentz in studies of electrodynamics, with crucial concepts contributed by Einstein to place the theory on a consistent footing. Maxwell s equations (1863) attempted to explain electromagnetism and optics through wave theory light propagates with speed c = m/s in ether but with different speeds in other frames the ether exists solely for the transport of e/m waves Maxwell s equations not invariant under Galilean transformations To avoid setting e/m apart from classical mechanics, assume light has speed c only in frames where source is at rest the ether has a small interaction with matter and is carried along with astronomical objects such as the Earth 4

5 Contradicted by Experiment Aberration of star light (small shift in apparent positions of distant stars) Fizeau s 1859 experiments on velocity of light in liquids Michelson-Morley 1907 experiment to detect motion of the earth through ether Suggestion: perhaps material objects contract in the direction of their motion L(v) =L 0 ( 1 v2 c 2 ) 1 2 This was the last gasp of the ether advocates and the germ of Special Relativity led by Lorentz, Minkowski and Einstein. 5

6 The Principle of Special Relativity 6

7 The Principle of Special Relativity A frame in which particles under no forces move with constant velocity is inertial. 6

8 The Principle of Special Relativity A frame in which particles under no forces move with constant velocity is inertial. Consider relations between inertial frames where measuring apparatus (rulers, clocks) can be transferred from one to another: related frames. 6

9 The Principle of Special Relativity A frame in which particles under no forces move with constant velocity is inertial. Consider relations between inertial frames where measuring apparatus (rulers, clocks) can be transferred from one to another: related frames. Assume: Behaviour of apparatus transferred from F to F' is independent of mode of transfer Apparatus transferred from F to F', then from F' to F'', agrees with apparatus transferred directly from F to F''. 6

10 The Principle of Special Relativity A frame in which particles under no forces move with constant velocity is inertial. Consider relations between inertial frames where measuring apparatus (rulers, clocks) can be transferred from one to another: related frames. Assume: Behaviour of apparatus transferred from F to F' is independent of mode of transfer Apparatus transferred from F to F', then from F' to F'', agrees with apparatus transferred directly from F to F''. The Principle of Special Relativity states that all physical laws take equivalent forms in related inertial frames, so that we cannot distinguish between the frames. 6

11 Simultaneity Two clocks A and B are synchronised if light rays emitted at the same time from A and B meet at the mid-point of AB Frame F A C B Frame F A C B A C B Frame F' moving with respect to F. Events simultaneous in F cannot be simultaneous in F'. Simultaneity is not absolute but frame dependent. 7

12 The Lorentz Transformation 8

13 The Lorentz Transformation Must be linear to agree with standard Galilean transformation in low velocity limit 8

14 The Lorentz Transformation Must be linear to agree with standard Galilean transformation in low velocity limit Preserves wave fronts of pulses of light, ct i.e. P = x 2 + y 2 + z 2 c 2 t 2 =0 whenever Q = x 2 + y 2 + z 2 c 2 t 2 =0 8

15 The Lorentz Transformation Must be linear to agree with standard Galilean transformation in low velocity limit Preserves wave fronts of pulses of light, ct i.e. P = x 2 + y 2 + z 2 c 2 t 2 =0 whenever Q = x 2 + y 2 + z 2 c 2 t 2 =0 Solution is the Lorentz transformation from frame F(t,x,y,z) to frame F'(t',x',y',z') moving with velocity v along the x-axis: 8

16 The Lorentz Transformation Must be linear to agree with standard Galilean transformation in low velocity limit Preserves wave fronts of pulses of light, ct i.e. P = x 2 + y 2 + z 2 c 2 t 2 =0 whenever Q = x 2 + y 2 + z 2 c 2 t 2 =0 Solution is the Lorentz transformation from frame F(t,x,y,z) to frame F'(t',x',y',z') moving with velocity v along the x-axis: ( t = γ t vx ) c 2 x = γ(x vt) y = y z = z where γ = 1 1 v2 c 2 8

17 The Lorentz Transformation Must be linear to agree with standard Galilean transformation in low velocity limit Preserves wave fronts of pulses of light, Lorentz ct i.e. P = x 2 + y 2 + z 2 c 2 t 2 =0 whenever Q = x 2 + y 2 + z 2 c 2 t 2 =0 Solution is the Lorentz transformation from frame F(t,x,y,z) to frame F'(t',x',y',z') moving with velocity v along the x-axis: ( t = γ t vx ) c 2 x = γ(x vt) y = y z = z where γ = 1 1 v2 c 2 8

18 The Lorentz Transformation Lorentz and the Zuidersee Must be linear to agree with standard Galilean transformation in low velocity limit First use of numerical Preserves techniques wave in fronts hydraulics of pulses of light, ct i.e. P = x 2 + y 2 + z 2 c 2 t 2 =0 whenever Q = x 2 + y 2 + z 2 c 2 t 2 =0 Lorentz Solution is the Lorentz transformation from frame F(t,x,y,z) to frame F'(t',x',y',z') moving with velocity v along the x-axis: ( t = γ t vx ) c 2 x = γ(x vt) y = y z = z where γ = 1 1 v2 c 2 8

19 The Lorentz Transformation Must be linear to agree with standard Galilean transformation in low velocity limit Preserves wave fronts of pulses of light, Lorentz ct i.e. P = x 2 + y 2 + z 2 c 2 t 2 =0 whenever Q = x 2 + y 2 + z 2 c 2 t 2 =0 Solution is the Lorentz transformation from frame F(t,x,y,z) to frame F'(t',x',y',z') moving with velocity v along the x-axis: ( t = γ t vx ) c 2 x = γ(x vt) y = y z = z where γ = 1 1 v2 c 2 8

20 The Lorentz Transformation Must be linear to agree with standard Galilean transformation in low velocity limit Preserves wave fronts of pulses of light, Lorentz ct i.e. P = x 2 + y 2 + z 2 c 2 t 2 =0 whenever Q = x 2 + y 2 + z 2 c 2 t 2 =0 Solution is the Lorentz transformation from frame F(t,x,y,z) to frame F'(t',x',y',z') moving with velocity v along the x-axis: ( t = γ t vx ) c 2 x = γ(x vt) y = y z = z where γ = Minkowski 1 1 v2 c 2 8

21 The Lorentz Transformation Must be linear to agree with standard Galilean transformation in low velocity limit Preserves wave fronts of pulses of light, Lorentz ct i.e. P = x 2 + y 2 + z 2 c 2 t 2 =0 whenever Q = x 2 + y 2 + z 2 c 2 t 2 =0 Solution is the Lorentz transformation from frame F(t,x,y,z) to frame F'(t',x',y',z') moving with velocity v along the x-axis: ( t = γ t vx ) c 2 Poincaré x = γ(x vt) y = y z = z where γ = Minkowski 1 1 v2 c 2 8

22 Outline of Derivation Set t =αt + βx x =γx + δt y =ɛy z =ζz where α, β, γ, δ, ɛ, ζ are constants kp = Q k(c 2 t 2 x 2 y 2 z 2 )=c 2 t 2 x 2 y 2 z 2 = c 2 (αt + βx) 2 (γx + δt) 2 ɛ 2 y 2 ζ 2 z 2 Equate coefficients of x, y, z, t Impose isotropy of space = k = k( v) =k( v ) =±1 Apply some common sense (e.g. ɛ, ζ, k = +1 and not -1) 9

23 3D form of Lorentz Transformation ( x v x = x v γt (γ 1) ( t = γ t ) v x c 2 v 2 ) 10

24 Consequences: length contraction ct Frame F Frame F ct v x Rod A B ArodAB of length L,fixedinframeF at x A, x B. What is its length measured in F? Must measure position of ends in F at same time, so events in F are (ct, x A )andct, x B ). By Lorentz: x A = γ(x } L A vt) = x B x A = = γ(x B x A ) x B = γ(x B vt) = γl > L x 11

25 Consequences: length contraction ct Frame F Frame F ct v x Rod A B ArodAB of length L,fixedinframeF at x A, x B. What is its length measured in F? Must measure position of ends in F at same time, so events in F are (ct, x A )andct, x B ). By Lorentz: x A = γ(x } L A vt) = x B x A = = γ(x B x A ) x B = γ(x B vt) = γl > L Moving objects appear contracted in the direction of the motion x 11

26 Consequences: time dilation 12

27 Consequences: time dilation Clock in frame F at point with coordinates (x,y,z) at different times t A and t B 12

28 Consequences: time dilation Clock in frame F at point with coordinates (x,y,z) at different times t A and t B In frame F' moving with speed v, Lorentz transformation gives t A = γ ( t A vx c 2 ) t B = γ ( t B vx c 2 ) 12

29 Consequences: time dilation Clock in frame F at point with coordinates (x,y,z) at different times t A and t B In frame F' moving with speed v, Lorentz transformation gives t A = γ ( t A vx c 2 ) t B = γ ( t B vx c 2 ) So ) Δt = t B t A = γ (t B t A = γδt >Δt 12

30 Consequences: time dilation Clock in frame F at point with coordinates (x,y,z) at different times t A and t B In frame F' moving with speed v, Lorentz transformation gives t A = γ ( t A vx c 2 ) t B = γ ( t B vx c 2 ) So ) Δt = t B t A = γ (t B t A = γδt >Δt Moving clocks appear to run slow 12

31 Schematic Representation ct Frame F ct ct Frame F ct Frame F Frame F L t t L Length contraction L<L x x Rod at rest in F.MeasurementsinF at a fixed time t, along a line parallel to x-axis Time dilation Δt <Δt x x Clock at rest in F. Time difference in F from line parallel to t -axis 13

32 Schematic t = γ Representation x = γ(x vt) ( t vx ) c 2 ct Frame F ct ct Frame F ct Frame F Frame F L t t L Length contraction L<L x x Rod at rest in F.MeasurementsinF at a fixed time t, along a line parallel to x-axis Time dilation Δt <Δt x x Clock at rest in F. Time difference in F from line parallel to t -axis 13

33 Schematic Representation ct Frame F ct ct Frame F ct Frame F Frame F L t t L Length contraction L<L x x Rod at rest in F.MeasurementsinF at a fixed time t, along a line parallel to x-axis Time dilation Δt <Δt x x Clock at rest in F. Time difference in F from line parallel to t -axis 13

34 Example: High Speed Train All clocks synchronised. A s clock and driver s clock read 0 as front of train emerges from tunnel. 15

35 Example: High Speed Train All clocks synchronised. A s clock and driver s clock read 0 as front of train emerges from tunnel. Observers A and B at exit and entrance of tunnel say the train is moving, has contracted and has length 15

36 Example: High Speed Train All clocks synchronised. A s clock and driver s clock read 0 as front of train emerges from tunnel. Observers A and B at exit and entrance of tunnel say the train is moving, has contracted and has length 100 γ =100 1 v = = 50m c

37 Example: High Speed Train All clocks synchronised. A s clock and driver s clock read 0 as front of train emerges from tunnel. Observers A and B at exit and entrance of tunnel say the train is moving, has contracted and has length 100 γ =100 1 v = = 50m c 2 4 But the tunnel is moving relative to the driver and guard on the train and they say the train is 100 m in length but the tunnel has contracted to 50 m 15

38 Question 1 A s clock and the driver's clock read zero as the driver exits tunnel. What does B s clock read when the guard goes in? 16

39 Question 1 A s clock and the driver's clock read zero as the driver exits tunnel. What does B s clock read when the guard goes in? 16

40 Question 1 A s clock and the driver's clock read zero as the driver exits tunnel. What does B s clock read when the guard goes in? Moving train length 50m, so driver has still 50m to travel before he exits and his clock reads 0. A's clock and B's clock are synchronised. Hence the reading on B's clock is 16

41 Question 1 A s clock and the driver's clock read zero as the driver exits tunnel. What does B s clock read when the guard goes in? Moving train length 50m, so driver has still 50m to travel before he exits and his clock reads 0. A's clock and B's clock are synchronised. Hence the reading on B's clock is 50 v = 100 3c 200ns 16

42 Question 2 What does the guard s clock read as he goes in? 17

43 Question 2 What does the guard s clock read as he goes in? 17

44 Question 2 What does the guard s clock read as he goes in? To the guard, tunnel is only 50m long, so driver is 50m past the exit as guard goes in. Hence clock reading is 17

45 Question 2 What does the guard s clock read as he goes in? To the guard, tunnel is only 50m long, so driver is 50m past the exit as guard goes in. Hence clock reading is 17

46 Question 3 Where is the guard when his clock reads 0? 18

47 Question 3 Where is the guard when his clock reads 0? 18

48 Question 3 Where is the guard when his clock reads 0? Guard s clock reads 0 when driver s clock reads 0, which is as driver exits the tunnel. To guard and driver, tunnel is 50m, so guard is 50m from the entrance in the train s frame, or 100m in tunnel frame. So the guard is 100m from the entrance to the tunnel when his clock reads 0. 18

49 Repeat within framework of Lorentz transformation Question 1 A s clock and the driver's clock read zero as the driver exits tunnel. What does B s clock read when the guard goes in? 19

50 Repeat within framework of Lorentz transformation Question 1 A s clock and the driver's clock read zero as the driver exits tunnel. What does B s clock read when the guard goes in? F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard. 19

51 Repeat within framework of Lorentz transformation Question 1 A s clock and the driver's clock read zero as the driver exits tunnel. What does B s clock read when the guard goes in? F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard. We know x A =0,x B = 100, x D =0,x G = 100. when G and B coincide. We want t B 19

52 Repeat within framework of Lorentz transformation Question 1 A s clock and the driver's clock read zero as the driver exits tunnel. What does B s clock read when the guard goes in? F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard. We know x A =0,x B = 100, x D =0,x G = 100. when G and B coincide. ) x = γ(x vt ) t = γ (t vx x = γ(x + vt) t = γ c 2 ( t + vx c 2 ) We want t B 19

53 Repeat within framework of Lorentz transformation Question 1 A s clock and the driver's clock read zero as the driver exits tunnel. What does B s clock read when the guard goes in? x = γ(x + vt) F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard. We know x A =0,x B = 100, x D =0,x G = 100. We want t B when G and B coincide. ) x = γ(x vt ) t = γ (t vx 100 = γ(100 + vt B ) c 2 = t B = γ γv t = γ ( t + vx c 2 ) = 50 v 19

54 Question 2 What does the guard s clock read as he goes in? 20

55 Question 2 What does the guard s clock read as he goes in? F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard. 20

56 Question 2 What does the guard s clock read as he goes in? F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard. We know x A =0,x B = 100, x D =0,x G = 100. when B and G coincide. We want t G 20

57 Question 2 What does the guard s clock read as he goes in? F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard. We know x A =0,x B = 100, x D =0,x G = 100. when B and G coincide. ) x = γ(x vt ) t = γ (t vx x = γ(x + vt) t = γ c 2 ( t + vx c 2 ) We want t G 20

58 Question 2 What does the guard s clock read as he goes in? F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard. We know x A =0,x B = 100, x D =0,x G = 100. We want t G when B and G coincide. ) x = γ(x vt ) t = γ (t vx 100 = γ(100 vt G) c 2 ( x = γ(x + vt) t = γ t + vx ) = t G = 100 γ 1 =+ 50 γv v c 2 20

59 Question 3 Where is the guard when his clock reads 0? 21

60 Question 3 Where is the guard when his clock reads 0? F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard. 21

61 Question 3 Where is the guard when his clock reads 0? F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard. We know x A =0,x B = 100, x D =0,x G = 100. We want x corresponding to x G = 100 given t G =0. 21

62 Question 3 Where is the guard when his clock reads 0? We know x A =0,x B = 100, x D =0,x G = 100. We want x corresponding to x G = 100 given t G =0. ) x = γ(x vt ) t = γ (t vx x = γ(x + vt) F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard. t = γ c 2 ( t + vx c 2 ) 21

63 Question 3 Where is the guard when his clock reads 0? F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard. We know x A =0,x B = 100, x D =0,x G = 100. We want x corresponding to x G = 100 given t G =0. ) x = γ(x vt ) t = γ (t vx c 2 x = γ(100 v 0) = 200 m ( x = γ(x + vt) t = γ t + vx ) c 2 Or 100m from the entrance to the tunnel 21

64 Question 4 Where was the driver when his clock reads the same as the guard s when he enters the tunnel? 22

65 Question 4 Where was the driver when his clock reads the same as the guard s when he enters the tunnel? F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard. 22

66 Question 4 Where was the driver when his clock reads the same as the guard s when he enters the tunnel? F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard. We know x A =0,x B = 100, x D =0,x G = 100. when t D = 50 v. We want x D 22

67 Question 4 Where was the driver when his clock reads the same as the guard s when he enters the tunnel? F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard. We know x A =0,x B = 100, x D =0,x G = 100. when t D = 50 v. x = γ(x vt ) x = γ(x + vt) t = γ t = γ ) (t vx c 2 ( t + vx c 2 ) We want x D 22

68 Question 4 Where was the driver when his clock reads the same as the guard s when he enters the tunnel? We know x A =0,x B = 100, x D =0,x G = 100. when t D = 50 v. x = γ(x vt ) x = γ(x + vt) F(t,x) is frame of A and B, F'(t',x') is frame of driver and guard. t = γ t = γ ) (t vx c 2 ( t + vx c 2 ) x =γ(x D vt D) We want x D = 50γ = 100 m The driver is already out of the tunnel by 100m 22

69 Example: Cosmic Rays Muons 23

70 Example: Cosmic Rays Muons Muons are created in the upper atmosphere, 90km from earth. Their half life is τ=2 μs, so they can travel at most c=600 m before decaying. So how do more than 50% reach the earth s surface? 23

71 Example: Cosmic Rays Muons Muons are created in the upper atmosphere, 90km from earth. Their half life is τ=2 μs, so they can travel at most c=600 m before decaying. So how do more than 50% reach the earth s surface? Muons see distance contracted by γ, so vτ 90 γ km 23

72 Example: Cosmic Rays Muons Muons are created in the upper atmosphere, 90km from earth. Their half life is τ=2 μs, so they can travel at most c=600 m before decaying. So how do more than 50% reach the earth s surface? Muons see distance contracted by γ, so vτ 90 γ km Earthlings say muons clocks run slow so their halflife is γτ and v(γτ) 90 km 23

73 Example: Cosmic Rays Muons Muons are created in the upper atmosphere, 90km from earth. Their half life is τ=2 μs, so they can travel at most c=600 m before decaying. So how do more than 50% reach the earth s surface? Muons see distance contracted by γ, so vτ 90 γ km Earthlings say muons clocks run slow so their halflife is γτ and Both give v(γτ) 90 km γv c = 90 km cτ = 150, v c, γ

74 Space-time 24

75 Space-time An invariant is a quantity that has the same value in all inertial frames. 24

76 Space-time An invariant is a quantity that has the same value in all inertial frames. Lorentz transformation is based on invariance of c 2 t 2 (x 2 + y 2 + z 2 )=(ct) 2 x 2 24

77 Space-time An invariant is a quantity that has the same value in all inertial frames. Lorentz transformation is based on invariance of c 2 t 2 (x 2 + y 2 + z 2 )=(ct) 2 x 2 4D-space with coordinates (t,x,y,z) is called space-time and the point (t,x,y,z)=(t,x) is called an event. 24

78 Space-time An invariant is a quantity that has the same value in all inertial frames. Lorentz transformation is based on invariance of c 2 t 2 (x 2 + y 2 + z 2 )=(ct) 2 x 2 4D-space with coordinates (t,x,y,z) is called space-time and the point (t,x,y,z)=(t,x) is called an event. Fundamental invariant (preservation of speed of light): ( Δs 2 = c 2 Δt 2 Δx 2 Δy 2 Δz 2 = c 2 Δt 2 1 Δx2 +Δy 2 +Δz 2 ) = c 2 Δt 2 ( 1 v2 c 2 ) c 2 Δt 2 = c 2 ( Δt γ ) 2 24

79 Space-time An invariant is a quantity that has the same value in all inertial frames. Lorentz transformation is based on invariance of c 2 t 2 (x 2 + y 2 + z 2 )=(ct) 2 x 2 4D-space with coordinates (t,x,y,z) is called space-time and the point (t,x,y,z)=(t,x) is called an event. Fundamental invariant (preservation of speed of light): ( Δs 2 = c 2 Δt 2 Δx 2 Δy 2 Δz 2 = c 2 Δt 2 1 Δx2 +Δy 2 +Δz 2 ) = c 2 Δt 2 ( 1 v2 c 2 ) c 2 Δt 2 = c 2 ( Δt γ ) 2 τ = dt γ is called proper time, the time in the instantaneous rest-frame and an invariant. s is called the separation between two events. 24

80 Space-time An invariant is a quantity that has the same value in all inertial frames. Lorentz transformation is based on invariance of c 2 t 2 (x 2 + y 2 + z 2 )=(ct) 2 x 2 Absolute future Conditional 4D-space with coordinates (t,x,y,z) is called space-time present and the point (t,x,y,z)=(t,x) is called an event. t x Fundamental invariant (preservation of speed of light): ( Δs 2 = c 2 Δt 2 Δx 2 Δy 2 Δz 2 = c 2 Δt 2 1 Δx2 +Δy 2 +Δz 2 ) c 2 Δt 2 Absolute ( ) ( ) = c 2 Δt 2 1 v2 Δt past 2 c 2 = c 2 γ τ = dt γ is called proper time, the time in the instantaneous rest-frame and an invariant. s is called the separation between two events. 24

81 4-Vectors The Lorentz transformation can be written in matrix form as ( t = γ t vx ) c 2 x = γ(x vt) y = y z = z = ct x y z = γ c 0 0 c γ γv γv ct x y z 25

82 4-Vectors The Lorentz transformation can be written in matrix form as ( t = γ t vx ) c 2 x = γ(x vt) y = y z = z = ct x y z = γ c 0 0 c γ γv γv ct x y z Lorentz matrix L 25

83 4-Vectors The Lorentz transformation can be written in matrix form as ( t = γ t vx ) c 2 x = γ(x vt) y = y z = z = ct x y z = γ c 0 0 c γ γv γv ct x y z Lorentz matrix L Position 4-vector X 25

84 4-Vectors The Lorentz transformation can be written in matrix form as ( t = γ t vx ) c 2 x = γ(x vt) y = y z = z = ct x y z = γ c 0 0 c γ γv γv ct x y z Lorentz matrix L Position 4-vector X 25

85 4-Vectors The Lorentz transformation can be written in matrix form as ( t = γ t vx ) c 2 x = γ(x vt) y = y z = z = ct x y z = γ c 0 0 c γ γv γv ct x y z Lorentz matrix L An object made up of 4 elements which transforms like X is called a 4-vector (analogous to the 3-vector of classical mechanics) Position 4-vector X 25

86 4-Vector Invariants Basic invariant: ct c 2 t 2 x 2 y 2 z 2 =(ct,x,y,z) x y z = X T gx = X X 26

87 4-Vector Invariants Basic invariant: ct c 2 t 2 x 2 y 2 z 2 =(ct,x,y,z) x y z = X T gx = X X 26

88 4-Vector Invariants Basic invariant: ct c 2 t 2 x 2 y 2 z 2 =(ct,x,y,z) x y z Inner product of two 4-vectors A = ( ) a0, B = a ( b0 b ) A B = A T gb = a 0 b 0 a 1 b 1 a 2 b 2 a 3 b 3 = a 0 b 0 a b In particular A A = a 2 0 a 2. = X T gx = X X is Invariance: A B =(LA) T g(lb) =A T (L T gl)b = A T gb = A B since L T gl = L (exercise: check) 26

89 4-Vectors in S.R. Mechanics 27

90 4-Vectors in S.R. Mechanics Velocity: V = dx dτ = γ dx dt = γ d (ct, x) =γ(c, v) dt 27

91 4-Vectors in S.R. Mechanics Velocity: Note invariant: V = dx dτ = γ dx dt = γ d (ct, x) =γ(c, v) dt V V = γ 2 (c 2 v 2 )= c2 v 2 1 v 2 /c 2 = c2 27

92 4-Vectors in S.R. Mechanics Velocity: Note invariant: Momentum: V = dx dτ = γ dx dt = γ d (ct, x) =γ(c, v) dt V V = γ 2 (c 2 v 2 )= c2 v 2 1 v 2 /c 2 = c2 P = m 0 V = m 0 γ(c, v) =(mc, p) 27

93 4-Vectors in S.R. Mechanics Velocity: Note invariant: Momentum: V = dx dτ = γ dx dt = γ d (ct, x) =γ(c, v) dt V V = γ 2 (c 2 v 2 )= c2 v 2 1 v 2 /c 2 = c2 P = m 0 V = m 0 γ(c, v) =(mc, p) m = m 0 γ is the relativistic mass p = m 0 γ v = m v is the relativistic 3-momentum 27

94 4-Force From Newton s 2 nd Law expect 4-Force given by F = dp dτ = γ dp dt = γ d ( dt (mc, p) =γ = γ ( c dm dt, f ) c dm dt, d p ) dt 28

95 4-Force From Newton s 2 nd Law expect 4-Force given by F = dp dτ = γ dp dt = γ d ( dt (mc, p) =γ = γ ( c dm dt, f ) c dm dt, d p ) dt Note: 3-force equation: f = d p dt = m 0 d dt (γ v) 28

96 Einstein s Relation: Energy and Mass 29

97 Einstein s Relation: Energy and Mass Momentum invariant P P = m 2 0V V = m 2 0c 2 29

98 Einstein s Relation: Energy and Mass Momentum invariant Differentiate P dp dτ P P = m 2 0V V = m 2 0c 2 dp =0 = V =0 = V F =0 dτ ( = γ(c, v) γ c dm ) dt, f =0 = d dt (mc2 ) v f =0 29

99 Einstein s Relation: Energy and Mass Momentum invariant Differentiate P dp dτ P P = m 2 0V V = m 2 0c 2 dp =0 = V =0 = V F =0 dτ ( = γ(c, v) γ c dm ) dt, f =0 = d dt (mc2 ) v f =0 v f = rate at which force does work = rate of change of kinetic energy 29

100 Einstein s Relation: Energy and Mass Momentum invariant Differentiate P dp dτ P P = m 2 0V V = m 2 0c 2 dp =0 = V =0 = V F =0 dτ ( = γ(c, v) γ c dm ) dt, f =0 = d dt (mc2 ) v f =0 v f = rate at which force does work = rate of change of kinetic energy Therefore kinetic energy is T = mc 2 + constant = m 0 c 2 (γ 1) 29

101 Einstein s Relation: Energy and Mass Momentum invariant Differentiate P dp dτ P P = m 2 0V V = m 2 0c 2 dp =0 = V =0 = V F =0 dτ ( = γ(c, v) γ c dm ) dt, f =0 = d dt (mc2 ) v f =0 v f = rate at which force does work = rate of change of kinetic energy Therefore kinetic energy is T = mc 2 + constant = m 0 c 2 (γ 1) E=mc 2 is total energy 29

102 Einstein s Relation: Energy and Mass Momentum invariant Differentiate P dp dτ P P = m 2 0V V = m 2 0c 2 dp =0 = V =0 = V F =0 dτ ( = γ(c, v) γ c dm ) dt, f =0 = d dt (mc2 ) v f =0 v f = rate at which force does work = rate of change of kinetic energy Therefore kinetic energy is T = mc 2 + constant = m 0 c 2 (γ 1) E=mc 2 is total energy 29

103 Summary of 4-Vectors Position X = (ct, x) Velocity V = γ(c, v) Momentum P = m 0 V = m(c, v) = Force F = γ ( c dm ) dt, f = γ ( ) E c, p ( 1 c de dt, f ) 30

104 Example: Addition of Velocities An object has velocity u =(u x,u y )inframef, which moves with velocity v =(v, 0) with respect to frame F. 31

105 Example: Addition of Velocities An object has velocity u =(u x,u y )inframef, which moves with velocity v =(v, 0) with respect to frame F. The 4-velocity U = γ u (c, u x,u y ) has to be Lorentz transformed to F, resulting in a 4-velocity W = γ w (c, w x,w y ): y Frame F Frame F y w =(w x,w y ) v u =(u x,u y ) x x 31

106 Example: Addition of Velocities An object has velocity u =(u x,u y )inframef, which moves with velocity v =(v, 0) with respect to frame F. The 4-velocity U = γ u (c, u x,u y ) has to be Lorentz transformed to F, resulting in a 4-velocity W = γ w (c, w x,w y ): cγ w γ w w x γ w w y = γv/c γ 0 γ γv/c cγ u γ u u x γ u u y 31

107 Example: Addition of Velocities An object has velocity u =(u x,u y )inframef, which moves with velocity v =(v, 0) with respect to frame F. The 4-velocity U = γ u (c, u x,u y ) has to be Lorentz transformed to F, resulting in a 4-velocity W = γ w (c, w x,w y ): cγ w γ w w x γ w w y = γv/c γ 0 γ γv/c cγ u γ u u x γ u u y γ w =γγ u ( 1+ vu x c 2 ) γ w w x =γγ u (v + u x ) γ w w y =γ u u y 31

108 Example: Addition of Velocities An object has velocity u =(u x,u y )inframef, which moves with velocity v =(v, 0) with respect to frame F. The 4-velocity U = γ u (c, u x,u y ) has to be Lorentz transformed to F, resulting in a 4-velocity W = γ w (c, w x,w y ): cγ w γ w w x γ w w y = γv/c γ 0 γ γv/c cγ u γ u u x γ u u y γ w =γγ u ( 1+ vu x c 2 ) γ w w x =γγ u (v + u x ) γ w w y =γ u u y w x = v + u x ( 1+ vu ) x c 2 w y = γ u y ( 1+ vu x c 2 ) 31

109 Using Invariants A neater way of finding the speed of the particle as measured in frame F : An observer in F has 4-velocity V and the object has 4-velocity U U V is invariant Evaluated in frame F : U = γ u (c, u x,u y ), V = γ(c, v, 0) = U V = γγ u (c 2 + vu x ) Evaluated in frame F : U =(c, 0, 0), V = γ w (c, w x,w y ) = U V = c 2 γ w Hence γ w = γγ u ( 1+ vu x c 2 ) 32

110 Basic Quantities used in Accelerator Calculations Relative velocity Velocity Momentum β = v c v = βc p = mv = m 0 γv = m 0 γβc Kinetic energy T = mc 2 m 0 c 2 =(γ 1)m 0 c 2 =(γ 1)E 0 γ = (1 v2 c 2 ) 1 2 = ( 1 β 2) 1 2 = (βγ) 2 = γ2 v 2 c 2 = γ 2 1 = β 2 = v2 c 2 =1 1 γ 2

111 Velocity and Energy T =m 0 c 2 (γ 1) = E 0 (γ 1) γ =1 + β = T m 0 c 2 =1+ T E γ 2 p =m 0 βγc = E 0 c βγ 34

112 Velocity and Energy T =m 0 c 2 (γ 1) = E 0 (γ 1) γ =1 + β = T m 0 c 2 =1+ T E γ 2 p =m 0 βγc = E 0 c βγ For v c, γ = (1 v2 c 2 ) v 2 c v 4 c so T = m 0 c 2 (γ 1) 1 2 m 0v 2 34

113 Energy-Momentum Important invariant: 35

114 Energy-Momentum Important invariant: P =(E/c, p) = P P = E 2 /c 2 p 2 and P = m 0 V = P P = m 2 0V V = m 2 0c 2 = E 2 0/c 2 35

115 Energy-Momentum Important invariant: P =(E/c, p) = P P = E 2 /c 2 p 2 and P = m 0 V = P P = m 2 0V V = m 2 0c 2 = E 2 0/c 2 E 2 c 2 = p 2 + m 2 0c 2 or E 2 = p 2 c 2 + E

116 Energy-Momentum Important invariant: P =(E/c, p) = P P = E 2 /c 2 p 2 and P = m 0 V = P P = m 2 0V V = m 2 0c 2 = E 2 0/c 2 E 2 c 2 = p 2 + m 2 0c 2 or E 2 = p 2 c 2 + E 2 0 = p 2 c 2 = E 2 E 2 0 =(E E 0 )(E + E 0 )=T (T +2E 0 ) 35

117 Energy-Momentum Important invariant: P =(E/c, p) = P P = E 2 /c 2 p 2 and P = m 0 V = P P = m 2 0V V = m 2 0c 2 = E 2 0/c 2 E 2 c 2 = p 2 + m 2 0c 2 or E 2 = p 2 c 2 + E 2 0 = p 2 c 2 = E 2 E0 2 =(E E 0 )(E + E 0 )=T(T +2E 0 ) Example: ISIS at RAL accelerates protons (E 0 = 938 MeV) to 800 MeV = pc = 800 ( ) MeV =1.463 GeV 35

118 Energy-Momentum Important invariant: P =(E/c, p) = P P = E 2 /c 2 p 2 and P = m 0 V = P P = m 2 0V V = m 2 0c 2 = E 2 0/c 2 E 2 c 2 = p 2 + m 2 0c 2 or E 2 = p 2 c 2 + E 2 0 = p 2 c 2 = E 2 E 2 0 =(E E 0 )(E + E 0 )=T (T +2E 0 ) Example: ISIS at RAL accelerates protons (E 0 = 938 MeV) to 800 MeV βγ = m 0βγc 2 m 0 c 2 = pc E 0 =1.56 = pc = 800 ( ) MeV γ 2 = (βγ) 2 +1= γ =1.85 =1.463 GeV β = βγ γ =

119 Relationships between small variations in parameters ΔE, ΔT, Δp, Δβ, Δγ

120 Relationships between small variations in parameters ΔE, ΔT, Δp, Δβ, Δγ (βγ) 2 = γ 2 1 = βγδ(βγ) = γδγ = βδ(βγ) = Δγ (1)

121 Relationships between small variations in parameters ΔE, ΔT, Δp, Δβ, Δγ (βγ) 2 = γ 2 1 = βγδ(βγ) = γδγ = βδ(βγ) = Δγ (1) 1 γ 2 = 1 β 2 = 1 Δγ = βδβ (2) γ3

122 Relationships between small variations in parameters ΔE, ΔT, Δp, Δβ, Δγ (βγ) 2 = γ 2 1 = βγδ(βγ) = γδγ = βδ(βγ) = Δγ (1) 1 γ 2 = 1 β 2 = 1 Δγ = βδβ (2) γ3 Δp p = Δ(m 0βγc) m 0 βγc = Δ(βγ) βγ = 1 Δγ β 2 γ = 1 ΔE β 2 E = γ 2 Δβ β = γ γ +1 ΔT T (exercise)

123 Relationships between small variations in parameters ΔE, ΔT, Δp, Δβ, Δγ (βγ) 2 = γ 2 1 = βγδ(βγ) = γδγ = βδ(βγ) = Δγ (1) 1 γ 2 = 1 β 2 = 1 Δγ = βδβ (2) γ3 Δp p = Δ(m 0βγc) m 0 βγc = Δ(βγ) βγ = 1 Δγ β 2 γ = 1 ΔE β 2 E = γ 2 Δβ β = γ γ +1 ΔT T (exercise) Note: valid to first order only

124 Δβ β Δp p ΔT T ΔE E Δγ γ = Δβ β Δβ β = γ 2 Δβ β = γ(γ +1) Δβ β = (βγ)2 Δβ β = (γ 2 1) Δβ β Δp p 1 Δp γ 2 p Δp p Δγ γ Δp p ( 1+ 1 ) Δp γ p β 2 Δp p Δp p Δβ β ΔT T 1 γ(γ +1) ΔT T γ ΔT γ +1 T ΔT T ( 1 1 ) ΔT γ T ΔE E = Δγ γ 1 Δγ β 2 γ 2 γ 1 Δγ γ 2 1 γ 1 Δγ β 2 γ γ Δγ γ 1 γ Δγ γ Table 1: Incremental relationships between energy, velocity and momentum. 37

125 4-Momentum Conservation 38

126 4-Momentum Conservation Equivalent expression for 4-momentum P = m 0 γ(c, v) =(mc, p) = ( ) E c, p 38

127 4-Momentum Conservation Equivalent expression for 4-momentum P = m 0 γ(c, v) =(mc, p) = ( ) E c, p Invariant m 2 0c 2 = P P = E2 c 2 p 2 = E2 c 2 = p 2 + m 2 0c 2 38

128 4-Momentum Conservation Equivalent expression for 4-momentum P = m 0 γ(c, v) =(mc, p) = ( ) E c, p Invariant m 2 0c 2 = P P = E2 c 2 p 2 = E2 c 2 = p 2 + m 2 0c 2 Classical conservation laws: conservation of total mass conservation of total 3-momentum 38

129 4-Momentum Conservation Equivalent expression for 4-momentum P = m 0 γ(c, v) =(mc, p) = ( ) E c, p Invariant m 2 0c 2 = P P = E2 c 2 p 2 = E2 c 2 = p 2 + m 2 0c 2 Classical conservation laws: conservation of total mass conservation of total 3-momentum In relativity, mass equivalent to energy (incl. rest energy) 38

130 4-Momentum Conservation Equivalent expression for 4-momentum P = m 0 γ(c, v) =(mc, p) = ( ) E c, p Invariant m 2 0c 2 = P P = E2 c 2 p 2 = E2 c 2 = p 2 + m 2 0c 2 Classical conservation laws: conservation of total mass conservation of total 3-momentum In relativity, mass equivalent to energy (incl. rest energy) = and p i constant particles i E i = total 4-momentum particles i particles i P i constant 38

131 Problem

132 Problem M 1 A body of mass M disintegrates while at rest into two parts of rest masses M 1 and M 2. M 2

133 Problem M 1 A body of mass M disintegrates while at rest into two parts of rest masses M 1 and M 2. M 2 Show that the energies of the parts are given by

134 Solution Before: P = ( ) Mc, 0 P 2 = After: ( ) E2 c, p P 1 = ( ) E1 c, p 40

135 Solution Before: P = ( ) Mc, 0 P 2 = After: ( ) E2 c, p P 1 = ( ) E1 c, p Conservation of 4-momentum: 40

136 Solution Before: P = ( ) Mc, 0 P 2 = After: ( ) E2 c, p P 1 = ( ) E1 c, p Conservation of 4-momentum: P = P 1 + P 2 P P 1 = P 2 ( P P 1 ) ( P P 1 )= P 2 P 2 P P 2P P 1 + P 1 P 1 = P 2 P 2 M 2 c 2 2ME 1 + M 2 1 c 2 = M 2 2 c 2 E 1 = M 2 + M M 2 2M c 2 40

137 Example of use of invariants 41

138 Example of use of invariants Two particles have equal rest mass m 0. 41

139 Example of use of invariants Two particles have equal rest mass m 0. Frame 1: one particle at rest, total energy is E 1. Frame 2: centre of mass frame where velocities are equal and opposite, total energy is E 2. 41

140 Example of use of invariants Two particles have equal rest mass m 0. Frame 1: one particle at rest, total energy is E 1. Frame 2: centre of mass frame where velocities are equal and opposite, total energy is E 2. Problem: Relate E 1 to E 2 41

141 P 1 = ( E1 m 0 c 2 c ), p P 2 = ( ) m 0 c, 0 Total energy E 1 (Fixed target experiment) P 1 = ( ) E2 2c, p P 2 = ( ) E2 2c, p Total energy E 2 (Colliding beams experiment) 42

142 P 1 = ( E1 m 0 c 2 c ), p P 2 = ( ) m 0 c, 0 Total energy E 1 (Fixed target experiment) P 1 = ( ) E2 2c, p P 2 = ( ) E2 2c, p Total energy E 2 (Colliding beams experiment) Invariant: P 2 (P 1 + P 2 ) 42

143 P 1 = ( E1 m 0 c 2 c ), p P 2 = ( ) m 0 c, 0 Total energy E 1 (Fixed target experiment) P 1 = ( ) E2 2c, p P 2 = ( ) E2 2c, p Total energy E 2 (Colliding beams experiment) Invariant: P 2 (P 1 + P 2 ) 42

144 Collider Problem 43

145 Collider Problem In an accelerator, a proton p 1 with rest mass m 0 collides with an anti-proton p 2 (with the same rest mass), producing two particles W 1 and W 2 with equal rest mass M 0 =100m 0 43

146 Collider Problem In an accelerator, a proton p 1 with rest mass m 0 collides with an anti-proton p 2 (with the same rest mass), producing two particles W 1 and W 2 with equal rest mass M 0 =100m 0 Experiment 1: p 1 and p 2 have equal and opposite velocities in the lab frame. Find the minimum energy of p 2 in order for W 1 and W 2 to be produced. 43

147 Collider Problem In an accelerator, a proton p 1 with rest mass m 0 collides with an anti-proton p 2 (with the same rest mass), producing two particles W 1 and W 2 with equal rest mass M 0 =100m 0 Experiment 1: p 1 and p 2 have equal and opposite velocities in the lab frame. Find the minimum energy of p 2 in order for W 1 and W 2 to be produced. Experiment 2: in the rest frame of p 1, find the minimum energy E' of p 2 in order for W 1 and W 2 to be produced. 43

148 Experiment 1 W 1 p 1 p 2 W 2

149 Experiment 1 E 2 c 2 = p 2 + m 2 0c 2 = Particles with same rest-mass and same momentum have same energies. W 1 p 1 p 2 W 2

150 Experiment 1 p 1 E 2 c 2 = p 2 + m 2 0c 2 = W 1 p 2 Particles with same rest-mass and same momentum have same energies. Total 3-momentum is zero before collision, so is zero afterwards W 2

151 Experiment 1 p 1 E 2 c 2 = p 2 + m 2 0c 2 = W 1 p 2 Particles with same rest-mass and same momentum have same energies. Total 3-momentum is zero before collision, so is zero afterwards 4-momenta before collision: ( ) ( ) E E P 1 = c, p P 2 = c, p W 2

152 Experiment 1 p 1 W 2 E 2 c 2 = p 2 + m 2 0c 2 = W 1 p 2 Particles with same rest-mass and same momentum have same energies. Total 3-momentum is zero before collision, so is zero afterwards 4-momenta before collision: ( ) ( ) E E P 1 = c, p P 2 = c, p 4-momenta after collision: ( ) ( ) E E P 1 = c, q P 2 = c, q

153 Experiment 1 p 1 W 2 E 2 c 2 = p 2 + m 2 0c 2 = W 1 p 2 Particles with same rest-mass and same momentum have same energies. Total 3-momentum is zero before collision, so is zero afterwards 4-momenta before collision: ( ) ( ) E E P 1 = c, p P 2 = c, p 4-momenta after collision: ( ) ( ) E E P 1 = c, q P 2 = c, q Total energy is conserved = 2E =2E = E = E > rest energy = M 0 c 2 = 100m 0 c 2

154 Experiment 1 p 1 W 2 E 2 c 2 = p 2 + m 2 0c 2 = W 1 p 2 Particles with same rest-mass and same momentum have same energies. Total 3-momentum is zero before collision, so is zero afterwards 4-momenta before collision: ( ) ( ) E E P 1 = c, p P 2 = c, p 4-momenta after collision: ( ) ( ) E E P 1 = c, q P 2 = c, q Total energy is conserved = 2E =2E = E = E > rest energy = M 0 c 2 = 100m 0 c 2

155 Experiment 2 Before collision P 1 =(m 0 c, 0), P 2 = ( ) E c, p W 1 p 1 TotalenergyisE 1 = E + m 0 c 2 p 2 W 2 Use previous result 2m 0 c 2 E 1 = E2 2 E 2 in the centre of mass frame to relate E 1 to total energy 2m 0 c 2 E 1 =E 2 2 = 2m 0 c 2( E + m 0 c 2) =(2E) 2 > (200m 0 c 2 ) 2 = E > ( )m 0 c 2 20, 000 m 0 c 2

156 Photons and Wave 4-Vectors Monochromatic plane wave: sin(ωt k x) k is the wave vector, k = 2π λ ; ω is the angular frequency, ω =2πν The phase 1 2π (ωt k x) is the number of wave crests passing an observer Invariant: ωt ( ω ) k x =(ct, x) c, k 4-momentum P = Position 4-vector ( ) E c,p ( ω ) = c,k = K Wave 4-vector 46

157 Relativistic Doppler Shift For light rays ω = c ( ω ) k so K = c, k is anullvector and can be written K = ω c (1, n) where n =1. 47

158 Relativistic Doppler Shift For light rays ω = c ( ω ) k so K = c, k is anullvector and can be written K = ω c (1, n) where n =1. y Frame F Frame F v θ θ x y x 47

159 Relativistic Doppler Shift For light rays ω = c ( ω ) k so K = c, k is anullvector and can be written K = ω c (1, n) where n =1. y Frame F Frame F v θ θ x y x In F, K = ω ( 1, cos θ, sin θ) c For F, use Lorentz transformation: K = LK 47

160 Relativistic Doppler Shift For light rays ω = c ( ω ) k so K = c, k is anullvector and can be written K = ω c (1, n) where n =1. y Frame F Frame F v θ θ x ω /c (ω /c)cosθ (ω /c)sinθ y = x In F, K = ω ( 1, cos θ, sin θ) c For F, use Lorentz transformation: K = LK γ γv/c 0 γv/c γ ω/c (ω/c)cosθ (ω/c)sinθ 47

161 Relativistic Doppler Shift For light rays ω = c ( ω ) k so K = c, k is anullvector and can be written K = ω c (1, n) where n =1. y Frame F Frame F v θ θ x ω /c (ω (/c)cosθ (ω /c)sinθ y ( ) ω vω cos θ = γ ω c ω cosθ = γ ω cos θ v ω c ) = x In F, K = ω ( 1, cos θ, sin θ) c For F, use Lorentz transformation: K = LK γ γv/c 0 γv/c γ ω/c (ω/c)cosθ (ω/c)sinθ ω sin θ = ω sin θ 47

162 Relativistic Doppler Shift For light rays ω = c ( ω ) k so K = c, k is anullvector and can be written K = ω c (1, n) where n =1. y Frame F Frame F v θ θ x ω /c (ω (/c)cosθ (ω /c)sinθ y ( ) ω vω cos θ = γ ω c ω cosθ = γ ω sin θ = ω sin θ ω cos θ v ω c ) = x In F, K = ω ( 1, cos θ, sin θ) c For F, use Lorentz transformation: K = LK ω = γω (1 v ) c cos θ γ γv/c 0 γv/c γ 0 sin θ 0 ( 0 1 tan θ = γ cos θ v c ) ω/c (ω/c)cosθ (ω/c)sinθ 47

163 Relativistic Doppler Shift For light rays ω = c ( ω ) k so K = c, k is anullvector and can be written K = ω c (1, n) where n =1. y Frame F Frame F v θ θ x ω /c (ω (/c)cosθ (ω /c)sinθ y ( ) ω vω cos θ = γ ω c ω cosθ = γ ω sin θ = ω sin θ ω cos θ v ω c ) = x In F, K = ω ( 1, cos θ, sin θ) c For F, use Lorentz transformation: K = LK ω = γω (1 v ) c cos θ γ γv/c 0 γv/c γ 0 sin θ 0 ( 0 1 tan θ = γ cos θ v c ) Note ω/c there is a (ω/c)cosθ transverse Doppler (ω/c)sinθ effect even when θ = 1 2 π 47

164 4-Acceleration 4-Acceleration=rate of change of 4-Velocity A = dv dτ = γ d ( ) γc,γ v dt Use 1 v v =1 γ2 c 2 = 1 dγ γ 3 dt A = γ ( 3 v a γ,γ a + γ3 c = v v c 2 ( ) v a c 2 = ) v v a c 2 In instantaneous rest-frame A =(0, a), A A = a 2 48

165 Radiation from an accelerating charge Rate of radiation, R, known to be invariant and proportional to a 2 in instantaneous rest frame. But in instantaneous rest-frame A A = a 2 ( ( v ) ) 2 a Deduce R A A = γ c γ 2 a2 Rearranged: R = e2 6πɛ 0 c 3 γ6 [ a 2 ( a v)2 c 2 ] Relativistic Larmor Formula 49

166 Radiation from an accelerating charge Rate of radiation, R, known to be invariant and proportional to a 2 in instantaneous rest frame. But in instantaneous rest-frame A A = a 2 ( ( v ) ) 2 a Deduce R A A = γ c γ 2 a2 Rearranged: R = e2 6πɛ 0 c 3 γ6 [ a 2 ( a v)2 c 2 ] Relativistic Larmor Formula If a v, R γ 6, but if a v, R γ 4 49

167 Motion under constant acceleration; world lines Introduce rapidity ρ defined by β = v c = tanh ρ = γ = 1 1 β 2 = cosh ρ Then V = γ(c, v) =c(cosh ρ, sinh ρ) And A = dv dτ = c(sinh ρ, cosh ρ) dρ dτ So constant acceleration satisfies a 2 = a 2 = A A = c 2 ( dρ dτ ) 2 = dρ dτ = a c, so ρ = aτ c 50

168 dx dx =γ dτ dt = γv = c sinh ρ = c sinh aτ c = x =x 0 + (cosh c2 aτ ) a c 1 dt dτ =γ = cosh ρ = cosh aτ c = t = c a sinh aτ c Particle Paths 51

169 Particle Paths dx dx =γ dτ dt = γv = c sinh ρ = c sinh aτ c = x =x 0 + (cosh c2 aτ ) a c 1 dt dτ =γ = cosh ρ = cosh aτ c = t = c a sinh aτ c = cosh 2 ρ sinh 2 ρ =1 ( x x 0 + c2 a ) 2 c 2 t 2 = c4 Relativistic paths are hyperbolic a 2 51

170 Particle Paths dx dx =γ dτ dt = γv = c sinh ρ = c sinh aτ c = x =x 0 + (cosh c2 aτ ) a c 1 dt dτ =γ = cosh ρ = cosh aτ c = t = c a sinh aτ c = cosh 2 ρ sinh 2 ρ =1 ( x x 0 + c2 a ) 2 c 2 t 2 = c4 Relativistic paths are hyperbolic a 2 x = x 0 + c2 a x aτ 2 t c a aτ c = τ ( a 2 τ 2 ) c Non-relativistic paths are parabolic 51

171 Particle Paths dx dx =γ dτ dt = γv = c sinh ρ = c sinh aτ c = x =x 0 + (cosh c2 aτ ) a c 1 dt dτ =γ = cosh ρ = cosh aτ c = t = c a sinh aτ c = cosh 2 ρ sinh 2 ρ =1 ( x x 0 + c2 a ) 2 c 2 t 2 = c4 Relativistic paths are hyperbolic a 2 x = x 0 + c2 a x aτ 2 t c a aτ c = τ ( a 2 τ 2 ) c Non-relativistic paths are parabolic 51

172 Relativistic Lagrangian and Hamiltonian Formulation 4-force equation of motion under a potential V : f = V = d p dt = m d 0 dt (γ v) ( ) d ẋ = m 0 = V etc., where v 2 =ẋ 2 +ẏ 2 +ż 2 dt 1 v2 /c 2 x Compare with standard Lagrangian formulation: d dt ( ) L ẋ = L x. Deduce Relativistic Langangian: ( ) 1 L = m 0 c 2 1 v2 2 m 0 c 2 V = c 2 γ V. Note: L T-V 52

173 Hamiltonian H, bydefinition: H = x,y,z ẋ L ẋ L= m 0 1 v2 /c 2 (ẋ2 +ẏ 2 +ż 2) L =m 0 γv 2 + m 0c 2 γ + V = m 0 γc 2 + V =T + V, Total Energy Since E 2 = p 2 c 2 + m 2 0c 4, H = c ( p 2 + m 2 0c 2) V Then Hamilton s equations of motion are: p = H x, x = H p 53

174 Motion faster than light 1. Two rods sliding over each other. Speed of intersection point is v/sinα, which can be made greater than c. 2. Explosion of planetary nebula. Observer sees bright spot spreading out. Light from P arrives t=dα 2 /2c later. ct P x d α O 54

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