Special Relativity. Christopher R. Prior. Accelerator Science and Technology Centre Rutherford Appleton Laboratory, U.K.

Transcription

1 Special Relativity Christopher R. Prior Fellow and Tutor in Mathematics Trinity College, Oxford Accelerator Science and Technology Centre Rutherford Appleton Laboratory, U.K.

2 The principle of special relativity Lorentz transformation and consequences Space-time Overview 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 Relativistic particle dynamics Lagrangian and Hamiltonian Formulation Radiation from an Accelerating Charge Photons and wave 4-vector Motion faster than speed of light 2

3 Reading W. Rindler: Introduction to Special Relativity (OUP 1991) D. Lawden: An Introduction to Tensor Calculus and Relativity N.M.J. Woodhouse: Special Relativity (Springer 2002) A.P. French: Special Relativity, MIT Introductory Physics Series (Nelson Thomes) Misner, Thorne and Wheeler: Relativity C. 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 4 astronomical objects

5 Contradicted by: 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 v 2 c This was the last gasp of ether advocates and the germ of 5 Special Relativity led by Lorentz, Minkowski and Einstein.

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'. 7 Simultaneity is not absolute but frame dependent.

12 The Lorentz Transformation 8 i

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

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 i

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 i

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: i t = γ t vx c 2 ( ) x = γ x vt y = y z = z where γ = 1 v2 c

17 The Lorentz Transformation Lorentz 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: i t = γ t vx c 2 ( ) x = γ x vt y = y z = z where γ = 1 v2 c

18 The Lorentz Transformation Lorentz 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: i t = γ t vx c 2 ( ) x = γ x vt y = y z = z where γ = 1 v2 c Minkowski

19 The Lorentz Transformation Lorentz 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: Poincaré i t = γ t vx c 2 ( ) x = γ x vt y = y z = z where γ = 1 v2 c Minkowski

20 Set t = αt + β x Then x = γ x + δt y = εy z = ς z P = kq c 2 t 2 x 2 y 2 z 2 = k( c 2 t 2 x 2 y 2 z 2 ) c 2 ( αt + β x) 2 ( γ x + δt) 2 ε 2 y 2 ς 2 z 2 = k( c 2 t 2 x 2 y 2 z 2 ) Equate coefficients of x, y, z,t. Isotropy of space k = k( v ) = k( v ) = ±1 Outline of Derivation Apply some common sense (e.g. ε,ς,k = +1 and not -1) 9

21 General 3D form of Lorentz Transformation: x = x + v v x γt +(γ 1) v 2 t = γ t + v x c 2 10

22 Consequences: length contraction z Frame F z Frame F v Rod x Rod AB of length L' fixed in F' at x' A, x' B. What is its length measured in F? Must measure positions of ends in F at the same time, so events in F are (t,x A ) and (t,x B ). From Lorentz: A B x 11

23 Consequences: length contraction z Frame F z Frame F v Rod x Rod AB of length L' fixed in F' at x' A, x' B. What is its length measured in F? Must measure positions of ends in F at the same time, so events in F are (t,x A ) and (t,x B ). From Lorentz: A B x Moving objects appear contracted in the direction of the motion 11

24 Consequences: time dilation 12

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

26 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

27 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 So t = t B t A = γ t B t A t B vx c 2 = γ t > t 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 So t = t B t A = γ t B t A t B vx c 2 = γ t > t Moving clocks appear to run slow 12

29 Schematic Representation of the Lorentz Transformation t t Frame F t t Frame F Frame F Frame F L Δt Δt L x x x x Length contraction L<L Rod at rest in F. Measurement in F at fixed time t, along a line parallel to x-axis Time dilatation: Δt<Δt Clock at rest in F. Time difference in F from line parallel to x -axis 13

30 x = γ(x vt) t = γ t vx c 2 Schematic Representation of the Lorentz Transformation t t Frame F t t Frame F Frame F Frame F L Δt Δt L x x x x Length contraction L<L Rod at rest in F. Measurement in F at fixed time t, along a line parallel to x-axis Time dilatation: Δt<Δt Clock at rest in F. Time difference in F from line parallel to x -axis 13

31 Schematic Representation of the Lorentz Transformation t t Frame F t t Frame F Frame F Frame F L Δt Δt L x x x x Length contraction L<L Rod at rest in F. Measurement in F at fixed time t, along a line parallel to x-axis Time dilatation: Δt<Δt Clock at rest in F. Time difference in F from line parallel to x -axis 13

32 Example: High Speed Train All clocks synchronised. A s clock and driver s clock read 0 as front of train emerges from tunnel. 14

33 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 14

34 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 =100 1 v = = 50m γ c

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 100 =100 1 v = = 50m γ 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 14

36 Question 1 A s clock (and the driver's clock) reads zero as the driver exits tunnel. What does B s clock read when the guard goes in? 15

37 Question 1 A s clock (and the driver's clock) reads 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 15

38 Question 1 A s clock (and the driver's clock) reads 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 15

39 Question 2 What does the guard s clock read as he goes in? 16

40 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 16

41 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 16

42 Question 3 Where is the guard when his clock reads 0? 17

43 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. 17

44 Repeat within framework of Lorentz transformation Question 1 A s clock (and the driver's clock) reads zero as the driver exits tunnel. What does B s clock read when the guard goes in? 18

45 Repeat within framework of Lorentz transformation Question 1 A s clock (and the driver's clock) reads 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. 18

46 Repeat within framework of Lorentz transformation Question 1 A s clock (and the driver's clock) reads 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. x = γ(x vt ) x = γ(x + vt) t = γ t vx t = γ c 2 t + vx c 2 18

47 Repeat within framework of Lorentz transformation Question 1 A s clock (and the driver's clock) reads 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. x = γ(x vt ) x = γ(x + vt) t = γ t vx t = γ c 2 t + vx c 2 18

48 Question 2 What does the guard s clock read as he goes in? 19

49 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. 19

50 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. x = γ(x vt ) t = γ t vx c 2 x = γ(x + vt) t = γ t + vx c 2 19

51 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. x = γ(x vt ) t = γ t vx x = γ(x + vt) t = γ c 2 t + vx c 2 x B = 100, x G = 100 = t G = 100γ 1 γv = 50 v 19

52 Question 3 Where is the guard when his clock reads 0? 20

53 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. 20

54 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. x = γ(x vt ) t = γ t vx c 2 x = γ(x + vt) t = γ t + vx c 2 20

55 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. x = γ(x vt ) t = γ t vx x = γ(x + vt) t = γ c 2 t + vx c 2 x G = 100, t G =0 = x = γx = 200 m Or 100m from the entrance to the tunnel 20

56 Question 4 Where was the driver when his clock reads the same as the guard s when he enters the tunnel? 21

57 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. 21

58 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. x = γ(x vt ) t = γ t vx c 2 x = γ(x + vt) t = γ t + vx c 2 21

59 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. x D = 0, t D = 50 x = γ(x vt ) t = γ t vx v x = γ(x + vt) t = γ c 2 t + vx c 2 = x = γvt D = 100 or 100 m beyond the tunnel exit 21

60 Example: Cosmic Rays 22

61 Example: Cosmic Rays µ-mesons are created in the upper atmosphere, 90km from earth. Their half life is τ=2 µs, so they can travel at most c=600m before decaying. So how do more than 50% reach the earth s surface? 22

62 Example: Cosmic Rays µ-mesons are created in the upper atmosphere, 90km from earth. Their half life is τ=2 µs, so they can travel at most c=600m before decaying. So how do more than 50% reach the earth s surface? Mesons see distance contracted by γ, so vτ 90 γ km 22

63 Example: Cosmic Rays µ-mesons are created in the upper atmosphere, 90km from earth. Their half life is τ=2 µs, so they can travel at most c=600m before decaying. So how do more than 50% reach the earth s surface? Mesons see distance contracted by γ, so vτ 90 γ km Earthlings say mesons clocks run slow so their half-life is γτ and v(γτ) 90 km 22

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

65 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): s 2 = c 2 t 2 x 2 y 2 z 2 = c 2 t 2 1 x2 + y 2 + z 2 = c 2 t 2 1 v2 c 2 c 2 t 2 is called proper time, time in instantaneous rest frame, an invariant. Δs=cΔτ is called the separation between 23 two events = c 2 t γ 2

66 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. Conditional present Absolute future t Fundamental invariant (preservation of speed of Absolute past light): s 2 = c 2 t 2 x 2 y 2 z 2 = c 2 t 2 1 x2 + y 2 + z 2 = c 2 t 2 1 v2 c 2 c 2 t 2 = c 2 t γ 2 x is called proper time, time in instantaneous rest frame, an invariant. Δs=cΔτ is called the separation between 23 two events

67 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 Lorentz matrix L ct x y z 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 24

68 Invariants Basic invariant c 2 t 2 x 2 y 2 z 2 =(ct, x, y, z) ct x y z = XT gx = X X Inner product of two 4-vectors A =(a 0,a), B =(b 0, b) A A = A T ga = a 0 b 0 a 1 b 1 a 2 b 2 a 3 b 3 = a 0 b 0 a b Invariance: A A =(LA) T g(la) =A T (L T gl)a = A T ga = A A Similarly A B = A B 25

69 4-Vectors in S.R. Mechanics 26

70 4-Vectors in S.R. Mechanics Velocity: V = dx dτ = γ dx dt = γ d (ct, x) =γ(c, v) dt 26

71 4-Vectors in S.R. Mechanics Velocity: V = dx dτ = γ dx dt = γ d (ct, x) =γ(c, v) dt Note invariant V V = γ 2 (c 2 v 2 )= c2 v 2 1 v 2 /c 2 = c2 26

72 4-Vectors in S.R. Mechanics Velocity: V = dx dτ = γ dx dt = γ d (ct, x) =γ(c, v) dt Note invariant Momentum: 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 γv = mv is the relativistic 3-momentum 26

73 Example of Transformation: Addition of Velocities 27

74 Example of Transformation: Addition of Velocities A particle moves with velocity in frame F, so has 4- velocity V = γ u (c, u) 27

75 Example of Transformation: Addition of Velocities A particle moves with velocity in frame F, so has 4- velocity Add velocity by transforming to frame F to get new velocity V = γ u (c, u) w = (w x,w y,w z ) corresponding to 4-vector ( ) W = γ w c, w 27

76 Example of Transformation: Addition of Velocities A particle moves with velocity in frame F, so has 4- velocity Add velocity by transforming to frame F to get new velocity V = γ u (c, u) w = (w x,w y,w z ) corresponding to 4-vector ( ) W = γ w c, w Lorentz transformation gives x = γ v ( x + vt) t = γ v t + vx c 2 t γ u, x γ uu, t γ w, x γ ww 27

77 Example of Transformation: Addition of Velocities A particle moves with velocity in frame F, so has 4- velocity Add velocity by transforming to frame F to get new velocity Lorentz transformation gives γ w = γ v γ u + vγ uu x γ w w x = γ v γ w w y = γ u u y γ w w z = γ u u z V = γ u (c, u) w = (w x,w y,w z ) c 2 ( ) γ u u x + vγ u corresponding to 4-vector x = γ v ( x + vt) t = γ v u y w x = u x + v 1 + vu w y = x γ c 2 v 1 + vu x c 2 t + vx c 2 w z = ( ) W = γ w c, w t γ u, x γ uu, t γ w, x γ ww 27 u z γ v 1 + vu x c 2

78 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, dp dt 28

79 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, dp dt Note: 3-force equation: f = dp dt = m 0 d dt (γv) 28

80 Einstein s Relation 29

81 Einstein s Relation Momentum invariant P P = m 2 0V V = m 2 0c 2 29

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

83 Einstein s Relation Momentum invariant P P = m 2 0V V = m 2 0c 2 Differentiate P dp dτ 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

84 Einstein s Relation Momentum invariant P P = m 2 0V V = m 2 0c 2 Differentiate P dp dτ 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

85 Einstein s Relation Momentum invariant P P = m 2 0V V = m 2 0c 2 Differentiate P dp dτ 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

86 Einstein s Relation Momentum invariant P P = m 2 0V V = m 2 0c 2 Differentiate P dp dτ 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

87 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

88 Basic Quantities used in Accelerator Calculations Relative velocity β = v c Velocity v = βc Momentum p = mv = m 0 γβc Kinetic energy T = (m m 0 )c 2 = m 0 c 2 (γ 1) γ = 1 v2 c = 1 β = (βγ) 2 = γ2 v 2 c 2 = γ 2 1 = β 2 = v2 c 2 =1 1 γ 2 31

89 Velocity v. Energy T = m 0 (γ 1)c 2 γ = 1+ T m 0 c 2 β = 1 1 γ 2 p = m 0 βγc For v c, γ = 1 v2 c v 2 c v 4 c so T = m 0 c 2 (γ 1) 1 2 m 0v 2 32

90 Energy-Momentum Invariant P P = m 2 0 V V = m 2 0c 2 and E P = c, p E 2 c 2 p2 = m 2 0c 2 = 1 c 2 E2 0 where E 0 is rest energy = p 2 c 2 = E 2 E0 2 = (E E 0 )(E + E 0 ) = T (T +2E 0 ) 33

91 Energy-Momentum Invariant P P = m 2 0 V V = m 2 0c 2 and E P = c, p E 2 c 2 p2 = m 2 0c 2 = 1 c 2 E2 0 where E 0 is rest energy = p 2 c 2 = E 2 E0 2 = (E E 0 )(E + E 0 ) = T (T +2E 0 ) Example: ISIS 800 MeV protons (E 0 =938 MeV) => pc=1.463 GeV 33

92 Energy-Momentum Invariant P P = m 2 0 V V = m 2 0c 2 and E P = c, p E 2 c 2 p2 = m 2 0c 2 = 1 c 2 E2 0 where E 0 is rest energy = p 2 c 2 = E 2 E0 2 Example: ISIS 800 MeV protons (E 0 =938 MeV) => pc=1.463 GeV = (E E 0 )(E + E 0 ) = T (T +2E 0 ) βγ = m 0βγc 2 m 0 c 2 = pc E 0 =1.56 γ 2 = (βγ) 2 +1= γ =1.85 β = βγ γ =

93 Relationships between small variations in parameters ΔE, ΔT, Δp, Δβ, Δγ (βγ) 2 = γ 2 1 = βγ (βγ) = γ γ = β (βγ) = γ (1) 1 γ 2 = 1 β 2 = 1 γ = β β (2) γ3 p p = (m 0βγc) m 0 βγc = (βγ) βγ = 1 γ β 2 γ = 1 E β 2 E = γ 2 β β = γ γ +1 T T (exercise) 34

94 Relationships between small variations in parameters ΔE, ΔT, Δp, Δβ, Δγ (βγ) 2 = γ 2 1 = βγ (βγ) = γ γ = β (βγ) = γ (1) 1 γ 2 = 1 β 2 = 1 γ = β β (2) γ3 Note: valid to first order only p p = (m 0βγc) m 0 βγc = (βγ) βγ = 1 γ β 2 γ = 1 E β 2 E = γ 2 β β = γ γ +1 T T 34 (exercise)

95 β β p p T T E E γ γ = β β β β = γ 2 β β = γ(γ + 1) β β = (βγ)2 β β = (γ 2 1) β β p p 1 p γ 2 p p p γ γ p p 1+ 1 p γ p β 2 p p p p β β T T 1 γ(γ + 1) T T γ T γ +1 T T T 1 1 T γ T E E = γ γ 1 γ β 2 γ 2 γ 1 γ γ 2 1 γ 1 γ β 2 γ γ γ γ 1 γ γ γ Table 1: Incremental relationships between energy, velocity and momentum. 35

96 4-Momentum Conservation 36

97 4-Momentum Conservation Equivalent expression for 4-momentum P = m 0γ(c, v ) = (mc, p ) = E c, p ( ) 36

98 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 36

99 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 Classical momentum conservation laws conservation of 4- momentum. Total 3- momentum and total energy are conserved. 36

100 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 Classical momentum conservation laws conservation of 4- momentum. Total 3- momentum and total energy are conserved. 36

101 Problem A body of mass M disintegrates while at rest into two parts of rest masses M 1 and M 2. Show that the energies of the parts are given by 37

102 Solution Before: P = Mc,0 P 2 = After: E2 c, p P 1 = E1 c, p 38

103 Solution Before: P = Mc,0 P 2 = Conservation of 4-momentum: After: E2 c, p P 1 = E1 c, p 38

104 Solution Before: P = Mc,0 P 2 = After: E2 c, p Conservation of 4-momentum: P = P 1 + P 2 P P 1 = P 2 ( ) ( P P 1 ) = P 2 P 2 P P 1 P P 2P P 1 + P 1 P 1 = P 2 P 2 M 2 c 2 2ME 1 + M 1 2 c 2 = M 2 2 c 2 E 1 = M 2 + M 1 2 M 2 2 2M c 2 P 1 = 38 E1 c, p

105 Example of use of invariants 39

106 Example of use of invariants Two particles have equal rest mass m 0. 39

107 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. 39

108 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 39

109 Total energy E 1 P 1 = E1 m 0 c 2 c, p P 2 = m 0 c,0 (Fixed target experiment) Total energy E 2 P 1 = E2 2c, p P 2 = E2 2c, p (Colliding beams expt) Invariant: P 2 (P 1 + P 2 ) 40

110 Total energy E 1 P 1 = E1 m 0 c 2 c, p P 2 = m 0 c,0 (Fixed target experiment) Total energy E 2 P 1 = E2 2c, p P 2 = E2 2c, p (Colliding beams expt) Invariant: P 2 (P 1 + P 2 ) 40

111 Collider Problem 41

112 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 41

113 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 Expt 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. 41

114 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 Expt 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. Expt 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. 41

115 Experiment 1 Note: same m 0, same p mean same E. p 1 W 1 p 2 Total 3-momentum is zero before collision and so is zero afterwards. P 1 = 4-momenta before collision: ( E c, p ) P 2 = E c, p ( ) W 2 4-momenta after collision: ( ) P 2 = ( q ) P 1 = E c, q E c, Energy conservation E=E > rest energy = M 0 c 2 = 100 m 0 c 2 42

116 Before collision: P 1 = ( ) P 2 = m 0 c, 0 ( ) E c, p W 1 Experiment 2 p 1 Total energy is E 1 = E + m 0 c 2 Use previous result C.O.M frame W 2 to relate E 1 to total energy E 2 in 2m 0 c 2 E 1 = E 2 2 2m 0 c 2 ( E + m 0 c 2 ) = (2E) 2 > ( 200 m 0 c 2 ) 2 E > 2m 0 c 2 E 1 = E 2 2 ( ) m 0 c 2 20,000m 0 c 2 43 p 2

117 4-Acceleration 4-Acceleration=rate of change of 4-Velocity A = dv dτ = γ d γc, γv dt Use 1 v v =1 γ2 c 2 = 1 dγ γ 3 dt A = γ 3 v a γ, γa + 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 44

118 Radiation from an accelerating charged particle 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 2a2 Rearranged: R = 2e2 3c 3 γ6 a 2 (a v)2 c 2 Relativistic Larmor Formula 45

119 Motion under constant acceleration; world lines Introduce rapidity ρ defined by β = v c = tanh ρ = γ = 1 = cosh ρ 1 β 2 Then And V = γ(c, v) =c(cosh ρ, sinh ρ) 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 46

120 c sinh ρ = c sinh aτ c = x = x 0 + c2 a dx = γv = γ dt = dx dτ cosh aτ c 1 = Particle Paths cosh 2 ρ sinh 2 ρ =1 x x 0 + c2 a 2 c 2 t 2 = c4 a 2 cosh ρ = cosh aτ c = γ = dt dτ = t = c a sinh aτ c x = x 0 + c a 2 τ 2 a 2 c x aτ 2 t c a aτ c = τ 47

121 c sinh ρ = c sinh aτ c = x = x 0 + c2 a dx = γv = γ dt = dx dτ cosh aτ c 1 = Particle Paths cosh 2 ρ sinh 2 ρ =1 x x 0 + c2 a 2 c 2 t 2 = c4 a 2 cosh ρ = cosh aτ c = γ = dt dτ = t = c a sinh aτ c x = x 0 + c2 a a 2 τ 2 c x aτ 2 Non-relativistic paths are parabolic t c a aτ c = τ 47

122 c sinh ρ = c sinh aτ c = x = x 0 + c2 a dx = γv = γ dt = dx dτ cosh aτ c 1 cosh ρ = cosh aτ c = γ = dt dτ = Particle Paths cosh 2 ρ sinh 2 ρ =1 x x 0 + c2 a 2 c 2 t 2 = c4 Relativistic paths are hyperbolic a 2 = t = c a sinh aτ c x = x 0 + c2 a a 2 τ 2 c x aτ 2 Non-relativistic paths are parabolic t c a aτ c = τ 47

123 c sinh ρ = c sinh aτ c = x = x 0 + c2 a dx = γv = γ dt = dx dτ cosh aτ c 1 cosh ρ = cosh aτ c = γ = dt dτ = Particle Paths cosh 2 ρ sinh 2 ρ =1 x x 0 + c2 a 2 c 2 t 2 = c4 Relativistic paths are hyperbolic a 2 = t = c a sinh aτ c x = x 0 + c2 a a 2 τ 2 c x aτ 2 Non-relativistic paths are parabolic t c a aτ c = τ 47

124 3-force eqn of motion under potential V: Standard Lagrangian formalism: Since v 2 = x 2 + Relativistic Lagrangian and Hamiltonian Formulation y 2 + z 2 d dt f = d p dt L x, deduce = L x m 0 d dt etc 1 L = m 0 c 2 1 v2 2 c 2 x ( 1 v 2 /c 2 ) 1/2 = V x L x = etc m 0 x ( 1 v 2 /c 2 ), L 1/2 x = V x V = m 0c 2 γ V Relativistic Lagrangian 48

125 Hamiltonian H = x L L = x m 0 1 v2 c 2 ( 1 x 2 + y 2 + z 2 ) L 2 = m 0 γ v 2 + m 0 c2 γ = E + V, + V = m 0 γ c 2 + V total energy Since H = c( p 2 + m 2 0 c 2 ) 12 + V Hamilton s equations of motion 49

126 Photons and Wave 4-Vectors Monochromatic plane wave: k = wave vector, k = 2π λ ; sin(ωt k x) ω = angular frequency = 2πν 1 Phase ( is the number of wave crests passing 2π ωt k x ) an observer, an invariant. ωt ω k x =(ct, x) c, k Position 4-vector, X Wave 4-vector, K 50

127 For light rays, phase velocity is K = ω (1, n) c Relativistic Doppler Shift c = ω k n = (cos θ, sin θ, 0) So where is a unit vector F F' Lorentz transform ct ω c, x ω c n θ θ' v 51

128 For light rays, phase velocity is K = ω (1, n) c Relativistic Doppler Shift c = ω k n = (cos θ, sin θ, 0) So where is a unit vector F θ F' θ' v Lorentz transform ct = γ ct ω c, x ω c n ct vx c x = γ(x vt) y = y 51

129 For light rays, phase velocity is K = ω (1, n) c Relativistic Doppler Shift c = ω k n = (cos θ, sin θ, 0) So where is a unit vector F θ F' θ' v Lorentz transform ω = γ ω cosθ = γ ct ω c, x ω c n ω vω cos θ c ω cos θ v ω c ω sin θ = ω sin θ 51

130 For light rays, phase velocity is K = ω (1, n) c Relativistic Doppler Shift c = ω k n = (cos θ, sin θ, 0) So where is a unit vector F θ F' θ' v Lorentz transform ω = γω tan θ = ct ω c, x ω c n 1 v c cos θ sin θ γ cos θ v c 51

131 For light rays, phase velocity is K = ω (1, n) c Relativistic Doppler Shift So where is a unit vector F θ F' θ' v Note: transverse Doppler effect even when θ=½π c = ω k n = (cos θ, sin θ, 0) Lorentz transform ω = γω tan θ = ct ω c, x ω c n 1 v c cos θ sin θ γ cos θ v c 51

132 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. c P x d α O 52

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