CHAPTER 2 Special Theory of Relativity

CHAPTER 2 Special Theory of Relativity Fall 2018 Prof. Sergio B. Mendes 1

Topics 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14 Inertial Frames of Reference Conceptual and Experimental Inconsistencies The Michelson-Morley Experiment Einstein s Postulates The Lorentz Transformation Time Dilation and Length Contraction Addition of Velocities Experimental Verification Twin Paradox Space-Time Doppler Effect Relativistic Momentum Relativistic Energy Computations in Modern Physics Electromagnetism and Relativity Fall 2018 Prof. Sergio B. Mendes 2

How to describe an event? The precise description of an event must be characterized by its location in space rr and its time tt. zz The set of (calibrated) rulers and (synchronized) clocks form a frame of reference that can be used to characterize events. Fall 2018 Prof. Sergio B. Mendes 3

What is an Inertial Frame of Reference? One in which Newton s laws are valid!! aa mmgg + TT = FF nnnnnn = mm aa FF nnnnnn = 00 vv = 00 aa = 00 FF nnnnnn = 00 vv = cccccccccccccccc aa = 00 Such a frame of reference is established when a body subjected to a null net external force, FF nnnnnn = 00, is observed to move with constant velocity (either rectilinear motion at constant speed or at rest). Fall 2018 Prof. Sergio B. Mendes 4

A non-inertial observer reaches conclusions that don t agree with Newton s laws aa = 0 aa FF nnnnnn 00 FF nnnnnn = 00 aa = aa aa aa FF nnnnnn = 00 aa = aa Fall 2018 Prof. Sergio B. Mendes 5

Therefore, we must use an inertial frame of reference to describe the laws of Mechanics Fall 2018 Prof. Sergio B. Mendes 6

Q: Well, how many inertial frames of reference are out there? A: An infinite number. Once we have found one inertial frame of reference (K), then any frame of reference moving at constant velocity vv oo with respect to K is also an inertial frame of reference. Fall 2018 Prof. Sergio B. Mendes 7

Proof Consider that K is at rest. Consider that K is moving with constant velocity vv oo with respect to K. Consider that the two frames of reference coincide at tt = 0. rr = rr vv oo tt K rr vv oo tt K rr xx = xx vv oo,xx tt yy = yy vv oo,yy tt zz = zz vv oo,xx tt vv = vv vv oo aa = aa tt = tt Fall 2018 Prof. Sergio B. Mendes 8

aa = aa mmm aa = mmm aa FF = FF mm = mm If Newton s laws are valid in one frame of reference (K), then they are also valid in another frame of reference (K ) moving at a uniform velocity relative to the first system. So, both are inertial frames of reference. Fall 2018 Prof. Sergio B. Mendes 9

Principle of (Classical) Relativity The laws of Mechanics (Newton s laws) are the same in all inertial frames of reference All inertial frames of reference are equivalent. Inertial frames of reference are related by: rr = rr vv oo tt tt = tt Galilean Transformation Fall 2018 Prof. Sergio B. Mendes 10

2.1 Conceptual and Experimental Inconsistencies Fall 2018 Prof. Sergio B. Mendes 11

Conceptual Inconsistencies: Although Newton s laws of motion had the same form under the Galilean transformation, Maxwell s equations did not. SS EE rr. ddaa = QQ iiiiiiiiiiii εε 0 CC EE rr. ddss = ddφ BB dddd SS BB rr. ddaa = 0 CC BB rr. ddss = μμ oo II + μμ oo εε oo ddφ EE dddd FF = qq EE + qq vv BB Fall 2018 Prof. Sergio B. Mendes 12

Experimental Inconsistencies: 2 EE xx 2 = εε 2 EE oo μμ oo tt 2 2 BB xx 2 = εε 2 BB oo μμ oo tt 2 In Maxwell s theory, the speed of light in terms of the permeability and permittivity of free space was given by: cc = 1 εε oo μμ oo Fall 2018 Prof. Sergio B. Mendes 13

The carrier medium for light, ether: Following the tradition of the time (that every wave has a medium to carry its propagation) the luminiferous ether was considered as the carrier medium for light propagation, where cc = 1 εε oo μμ oo Ether had to have such a low density that the planets could move through it without loss of energy It also had to have an elasticity to support the high velocity of light waves Fall 2018 Prof. Sergio B. Mendes 14

An Absolute Inertial Frame of Reference Ether was proposed as an absolute reference system in which the speed of light was this constant and from which other measurements could be made. The Michelson-Morley experiment was an attempt to show the existence of ether. Fall 2018 Prof. Sergio B. Mendes 15

2.2 The Michelson-Morley Experiment Albert Michelson (1852 1931) was the first U.S. citizen to receive the Nobel Prize for Physics (1907). He built an extremely precise device called an interferometer to measure the minute phase difference between two light waves traveling in mutually orthogonal directions. Fall 2018 Prof. Sergio B. Mendes 16

The Michelson Interferometer Fall 2018 Prof. Sergio B. Mendes 17

How does it work? 1. AC is parallel to the motion of the Earth inducing an ether wind 2. Light from source S is split by mirror A and travels to mirrors C and D in mutually perpendicular directions 3. After reflection the beams recombine at A slightly out of phase due to the ether wind as viewed by telescope E. Fall 2018 Prof. Sergio B. Mendes 18

A Typical Interference Pattern Fall 2018 Prof. Sergio B. Mendes 19

The Analysis assuming the Galilean Transformation!! Time tt 1 from A to C and back: tt 1 = ll 1 cc + vv + ll 1 cc vv Time tt 2 from A to D and back: tt 2 = = 2 cc ll 1 cc 2 vv 2 ll 2 cc 2 vv 2 + ll 2 cc 2 vv 2 So that the change in time is: tt = tt 2 tt 1 = 2 ll 2 cc = 2 ll 2 cc = 2 ll 1 cc Fall 2018 Prof. Sergio B. Mendes 20 1 1 1 vv2 cc 2 1 1 vv2 cc 2 1 vv2 cc 2 2 ll 1 cc 1 1 vv2 cc 2

The Analysis (continued) tt = tt 2 tt 1 = 2 ll 2 cc 1 1 vv2 cc 2 2 ll 1 cc 1 1 vv2 cc 2 Upon rotating the apparatus by 90º, the optical path lengths ll 1 and ll 2 are interchanged producing a different change in time: ttt = ttt 2 tt 1 = 2 ll 2 cc 1 1 vv2 cc 2 2 ll 1 cc 1 1 vv2 cc 2 Fall 2018 Prof. Sergio B. Mendes 21

Difference in times upon rotation: tt tt = 2 ll 1 + ll 2 cc 1 1 vv2 cc 2 1 1 vv2 cc 2 ll 1 + ll 2 vv 2 cc 3 Fall 2018 Prof. Sergio B. Mendes 22

Increasing the pathlength: Fall 2018 Prof. Sergio B. Mendes 23

Crunching the numbers: vv = 3 10 4 mm/ss cc = 3 10 8 mm/ss ll 1 ll 2 = 11 mm tt tt = ll 1 + ll 2 vv 2 cc 3 = 7 10 16 ss λλ = 589 10 9 mm cc = 3 10 8 mm/ss TT = λλ cc = 589 10 9 mm 3 10 8 mm/ss = 2 10 15 ss ffffffffffffffff oooo aaaa iiiiiiiiiiiiiiiiiiiiiiii ffffffffffff = 7 10 16 ss 2 10 15 ss 0.4 iiiiiiiiiiiiiiiiiiii rrrrrrrrrrrrrrrrrrrr 0.01 Fall 2018 Prof. Sergio B. Mendes 24

The Experiments on the relative motion of the earth and ether have been completed and the result decidedly negative. The expected deviation of the interference fringes from the zero should have been 0.40 of a fringe the maximum displacement was 0.02 and the average much less than 0.01 and then not in the right place. As displacement is proportional to squares of the relative velocities it follows that if the ether does slip past the relative velocity is less than one sixth of the earth s velocity. Albert Abraham Michelson, 1887 Fall 2018 Prof. Sergio B. Mendes 25

2.3 Einstein s Two Postulates 1. The principle of relativity: The laws of physics are the same in all inertial frames of reference. There is no way to detect absolute motion and no preferred inertial system exists. 2. The constancy of the speed of light: Observers in all inertial frames of reference measure the same value for the speed of light when propagating in vacuum. Fall 2018 Prof. Sergio B. Mendes 26

Consequences: Source and Observer at rest: c Observer in motion with respect to the Source: c Source in motion with respect to the Observer: c -v v Fall 2018 Prof. Sergio B. Mendes 27

Consequences In Newtonian physics, we previously assumed that tt = tt. Therefore K and K would always agree if two events happen at the same time (simultaneous) or not. Einstein realized that events considered simultaneous in K may not be in K. Fall 2018 Prof. Sergio B. Mendes 28

Because speed of light is absolute then simultaneity is relative K K K K Fall 2018 Prof. Sergio B. Mendes 29

Two events that are simultaneous in one reference frame (K) are not necessarily simultaneous in another reference frame (K ) moving with respect to the first frame. This suggests that each coordinate system must have its own set of observers with their own set of synchronized clocks. Fall 2018 Prof. Sergio B. Mendes 30

2.4 Lorentz Transformations The special set of linear transformations that preserve the constancy of the speed of light between inertial observers. Fall 2018 Prof. Sergio B. Mendes 31

Two inertial frames of reference KK and KK 1. The axes along (xx, yy, zz) are parallel to the corresponding axes along (xxx, yyy, zzz). 2. The relative motion between the two inertial frames of reference is along the x-axis (and x -axis). 3. Consider that the origins OO and OO of the two systems coincide at tt = ttt = 0 As seen by KK As seen by KK OO OOO vv oo vv oo OO OOO Fall 2018 Prof. Sergio B. Mendes 32

As seen by KK KK: (xx, tt) vv oo vv oo vv oo KKK: (xxx, ttt) vv oo vv oo Fall 2018 Prof. Sergio B. Mendes 33

As seen by KK : KK: (xx, tt) vv oo vv oo vv oo KKK: (xxx, ttt) vv oo vv oo Fall 2018 Prof. Sergio B. Mendes 34

As seen by KK: A flashbulb goes off at the origins when tt = ttt = 0. vv oo Fall 2018 Prof. Sergio B. Mendes 35

As seen by KKK: A flashbulb goes off at the origins when tt = ttt = 00. vv oo Fall 2018 Prof. Sergio B. Mendes 36

According to Postulate 2, the speed of light will be c in both systems!! The wavefronts observed in both systems must be spherical with respect to their own coordinates and time. KK OO KKK OOO xx 2 + yy 2 + zz 2 = cc 2 tt 2 xxx 2 + yyy 2 + zzz 2 = cc 2 ttt 2 Fall 2018 Prof. Sergio B. Mendes 37

xx 2 + yy 2 + zz 2 cc 2 tt 2 = 0 0 = xxx 2 + yyy 2 + zzz 2 cc 2 ttt 2 yy = yyy zz = zzz as in Galilean transformation xx 2 cc 2 tt 2 = xxx 2 cc 2 ttt 2 Fall 2018 Prof. Sergio B. Mendes 38

xx = γγ xx vv oo tt linear modification in Galilean transformation due to symmetry xx = γγ xxx + vv oo ttt = γγ γγ xx vv oo tt + vv oo ttt solve for: tt = 1 γγ vv oo 1 γγ 2 xx + γγ tt Fall 2018 Prof. Sergio B. Mendes 39

xx 2 cc 2 tt 2 = xxx 2 cc 2 ttt 2 xx = γγ xx vv oo tt tt = 1 γγ vv oo 1 γγ 2 xx + γγ tt Fall 2018 Prof. Sergio B. Mendes 40

xx 2 : 1 = γ 2 cc2 γγ 2 2 vv 1 γγ2 2 1 oo 1 γγ 2 = cc2 γγ 2 vv oo 2 1 γγ 2 1 = vv oo cc 2 2 tt 2 : cc 2 = γγ 2 2 1 vv oo cc 2 γγ 2 γγ 2 = 1 vv 2 oo cc 2 2 xx tt: 0 = γγ 2 vv oo cc 2 1 vv oo 1 γγ 2 γγ 2 vv oo = cc 2 1 vv oo 1 γγ 2 vv oo 2 cc 2 = 1 γγ 2 1 γγ = 1 1 ββ 2 ββ vv oo cc Fall 2018 Prof. Sergio B. Mendes 41

Lorentz Transformations: xx = γγ xx γγ vv oo tt tt = γγ vv oo cc 2 xx + γγ tt γγ = 1 1 ββ 2 ββ vv oo cc Fall 2018 Prof. Sergio B. Mendes 42

Inverse Lorentz Transformations: xx = γγ xx + γγ vv oo tt tt = γγ vv oo cc 2 xx + γγ ttt γγ = 1 1 ββ 2 ββ vv oo cc Fall 2018 Prof. Sergio B. Mendes 43

Relativistic Factor γγ ββ = Fall 2018 Prof. Sergio B. Mendes 44

2.5 Time Dilation and Length Contraction Fall 2018 Prof. Sergio B. Mendes 45

Time Dilation: vv oo Fall 2018 Prof. Sergio B. Mendes 46

Time Dilation: tt = γγ vv oo cc 2 xx + γγ tt ttt 1 xx 1 = xx 2 ttt 1 = γγ vv oo cc 2 xx 1 + γγ tt 1 ttt 2 = γγ vv oo cc 2 xx 2 + γγ tt 2 Fall 2018 Prof. Sergio B. Mendes 47

Time Dilation & Proper Time: ttt 1 ttt 2 ttt 2 tt 1 = γγ tt 2 tt 1 tt 2 tt 1 = proper time, time duration measured at same location Fall 2018 Prof. Sergio B. Mendes 48

Length Contraction: vv oo Fall 2018 Prof. Sergio B. Mendes 49

Length Contraction: xx = γγ xx + γγ vv oo tt xx 1 = γγ xxx 1 + γγ vv oo tt 1 xx 2 = γγ xxx 2 + γγ vv oo tt 2 tt 2 = ttt 1 xxx 2 xxx 1 = 1 γγ xx 2 xx 1 xx 2 xx 1 = proper length, length measured at rest Fall 2018 Prof. Sergio B. Mendes 50

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2.6 Addition of Velocities Fall 2018 Prof. Sergio B. Mendes 53

Taking the differentials: tt = γγ vv oo cc 2 xx + γγ tt ddtt = γγ vv oo cc 2 ddxx + γγ dddd xx = γγ xx γγ vv oo tt ddxx = γγ dddd γγ vv oo ddtt yy = yy zz = zz ddyy = ddyy ddzz = dddd Fall 2018 Prof. Sergio B. Mendes 54

Along x-axis ddxx = γγ dddd γγ vv oo ddtt ddtt = γγ vv oo cc 2 ddxx + γγ dddd ddxx ddtt = γγ dddd γγ vv oo dddd γγ vv oo dddd + γγ dddd cc 2 uuu xx ddxx ddtt = uu xx vv oo 1 vv oo uu xx cc 2 Fall 2018 Prof. Sergio B. Mendes 55

ddyy Along y-axis ddyy = ddyy ddtt = γγ vv oo cc 2 ddtt = γγ vv oo cc 2 ddxx + γγ dddd dddd dddd + γγ dddd uuu yy ddyy ddtt = uu yy γγ 1 vv oo uu xx cc 2 Fall 2018 Prof. Sergio B. Mendes 56

Along z-axis ddzz = dddd ddzz ddtt = γγ vv oo cc 2 ddtt = γγ vv oo cc 2 ddxx + γγ dddd dddd dddd + γγ dddd uuu zz ddzz ddtt = uu zz γγ 1 vv oo uu xx cc 2 Fall 2018 Prof. Sergio B. Mendes 57

In Summary, Addition of Velocities: uuu xx = uu xx vv oo 1 vv oo uu xx cc 2 uuu yy = uu yy γγ 1 vv oo uu xx cc 2 uuu zz = uu zz γγ 1 vv oo uu xx cc 2 Fall 2018 Prof. Sergio B. Mendes 58

Inverted Relations: uu xx = uuu xx + vv oo 1 + vv oo uuu xx cc 2 uu yy = uuu yy γγ 1 + vv oo uuu xx cc 2 uu zz = uuu zz γγ 1 + vv oo uuu xx cc 2 Fall 2018 Prof. Sergio B. Mendes 59

Example uu xx = +0.990 cc vv oo = 0.600 cc uuu xx = = uu xx vv oo 1 vv oo uu xx cc 2 0.990 cc 0.600 cc 0.600 cc 0.990 cc 1 cc 2 = 0.997 cc Fall 2018 Prof. Sergio B. Mendes 60

2.7 Experimental Verification of Special Relativity Fall 2018 Prof. Sergio B. Mendes 61

Cosmic Rays and Muon Decay NN NN oo = ee llll 2 tt ττ ττ = 1.52 μμμμ Fall 2018 Prof. Sergio B. Mendes 62

(Incorrect) Classical Calculation h = 2000 mm vv = 0.98 cc tt = h vv = 6.80 μμμμ NN NN oo = ee llll 2 tt ττ ττ = 1.52 μμμμ llll 2 6.80 μμμμ = ee 1.52 μμμμ = 4.5% Don t agree with experiment Fall 2018 Prof. Sergio B. Mendes 63

Relativistic Calculation vv = 0.98 cc γγ = 1 1 0.98 2 5 tt = tt γγ 6.80 μμμμ = 5 1.36 μμμμ NN NN oo = ee llll 2 tt ττ llll 2 1.36 μμμμ = ee 1.52 μμμμ = 54% Agrees with experiment Fall 2018 Prof. Sergio B. Mendes 64

Atomic Clock Measurement Fall 2018 Prof. Sergio B. Mendes 65

Velocity Addition ππ 0 γγ + γγ vv oo = 0.99975 cc uu xx = uuu xx + vv oo 1 + vv oo uuu xx cc 2 uuu xx = cc vv xx = cc Fall 2018 Prof. Sergio B. Mendes 66

2.10 Relativistic Doppler effect for light waves in vacuum Although light velocity in vacuum is always constant c, the frequency will change for a relative motion between source and observer Source- Observer approaching Higher Frequency Source- Observer receding Lower Frequency Fall 2018 Prof. Sergio B. Mendes 67

Relativistic Doppler effect for light waves in vacuum cc vv vv TT cc TT lllllllllll oooo ttttt wwwwwwww tttttttttt = cc TT vv TT nn = cc TT vv TT λλ ff = cc λλ = cc nn cc TT vv TT Fall 2018 Prof. Sergio B. Mendes 68

nn = pppppppppppp tttttttt oooo ttttt ssssssssssss tttttttt pppppppppppp oooo ttttt ssssssssssss = TTT oo 1 fffoo TT = γγ TTT oo nn = TT fff oo γγ ff = cc nn cc TT vv TT = cc γγ cc vv fff oo = 1 + ββ 1 ββ fff oo Fall 2018 Prof. Sergio B. Mendes 69

Relativistic Doppler effect for light propagating in vacuum Source- Observer approaching Source- Observer receding ff = 1 + ββ 1 ββ fff oo ff = 1 ββ 1 + ββ fff oo blueshifted redshifted Fall 2018 Prof. Sergio B. Mendes 70

Rotation of Venus Fall 2018 Prof. Sergio B. Mendes 71

Laser Cooling Fall 2018 Prof. Sergio B. Mendes 72

Laser Radar Technology Fall 2018 Prof. Sergio B. Mendes 73

2.11 Relativistic Linear Momentum Fall 2018 Prof. Sergio B. Mendes 74

Classical Expressions from Galileo and Newton: pp mm uu Linear Momentum dd pp FF = dddd Newton s law Newton s law is invariant under Galilean transformations, but not under Lorentz transformations. Fall 2018 Prof. Sergio B. Mendes 75

Relativistic Linear Momentum Invariance of Newton s law under Lorentz transformations will be shown to lead to: pp = ΓΓ mm uu ΓΓ 1 1 uu cc 2 Fall 2018 Prof. Sergio B. Mendes 76

Conservation of Linear Momentum in an Elastic Collision: uu xx = 0 uu xx = uuu xx + vv 1 + vv uuu xx cc 2 uu xx = vv uu xx = 0 uu yy = uu oo uu yy = uuu yy γγ 1 + vv uuu xx cc 2 uu yy = uu oo γγ uu yy = 2 uu oo γγ mm mm pp = mm uu + uu 0 mm uu xx = 0 uu yy = ± uu oo uu xx = 0 uu yy = 2 uu oo Fall 2018 Prof. Sergio B. Mendes 77

Relativistic Linear Momentum pp = mm ΓΓ uu pp = mm ΓΓ uu + uu = 0 ΓΓ 1 1 uu cc 2 uu xx = vv uu yy = uu oo γγ ΓΓ uu xx = 0 ΓΓ uu yy = ΓΓ 2 uu oo γγ = 2 uu oo 1 uu oo 2 cc 2 ΓΓ = 1 1 vv 2 + uu oo 2 γγ 2 cc 2 uu xx = 0 ΓΓ uu xx = 0 uu yy = ± uu oo ΓΓ uu yy = 2 uu oo 1 uu oo 2 cc 2 ΓΓ = 1 1 uu oo 2 cc 2 Fall 2018 Prof. Sergio B. Mendes 78

Relativistic Linear Momentum pp = mm ΓΓ uu ΓΓ 1 1 uu cc 2 Fall 2018 Prof. Sergio B. Mendes 79

2.12 Relativistic Energy FF = dd pp dddd 2 2 dd pp WW 12 = KK 2 KK 1 = FF dd rr = 1 1 dddd dd rr Fall 2018 Prof. Sergio B. Mendes 80

pp = mm ΓΓ uu dd pp = mm dd ΓΓ uu uu = dd rr dddd dd rr = uu dddd 2 dd pp KK 2 KK 1 = 1 dddd dd rr 2 dd ΓΓ uu = mm 1 dddd uu dddd 2 = mm uu dd ΓΓ uu 1 Fall 2018 Prof. Sergio B. Mendes 81

Solving the integral: ΓΓ 1 1 uu cc 2 2 KK = mm uu dd ΓΓ uu 1 uu = mm 0 uu 1 uu cc 2 3 2 dddd = mm cc 2 1 1 uu cc 2 1 Fall 2018 Prof. Sergio B. Mendes 82

Kinetic Energy KK = mm cc 2 1 1 uu cc 2 1 uu cc 2 1 1 2 1 + 1 2 uu cc 2 1 uu cc KK mm uu2 2 Fall 2018 Prof. Sergio B. Mendes 83

Comparison of Classical and Relativistic Kinetic Energy Fall 2018 Prof. Sergio B. Mendes 84

Experimental Results Fall 2018 Prof. Sergio B. Mendes 85

Total Relativistic Energy KK = mm cc 2 1 1 uu cc 2 1 KK + mm cc 2 mm cc 2 Total Energy: EE tttttttttt = = = ΓΓ mm cc 2 1 uu cc 2 Rest Energy: EE 0 = mm cc 2 Fall 2018 Prof. Sergio B. Mendes 86

Hydrogen Fusion and Solar Energy 1 HH + 1 HH 2 HH + ee + + νν ee + 0.42 MeV 2 HH + 1 HH 3 HHee + γγ + 5.49 MeV 3 HHHH + 3 HHHH 4 HHee + 2 1 HH + γγ + 12.86 MeV Fall 2018 Prof. Sergio B. Mendes 87

Total Energy and Linear Momentum EE tttttttttt = mm cc 2 1 uu cc 2 EE tttttttttt 2 = mm2 cc 4 1 uu cc 2 pp = mm uu 1 uu cc 2 pp 2 cc 2 = mm2 uu 2 cc 2 1 uu cc 2 EE tttttttttt 2 pp 2 cc 2 = mm 2 cc 4 Fall 2018 Prof. Sergio B. Mendes 88

Massless Particles: EE tttttttttt 2 pp 2 cc 2 = mm 2 cc 4 mm = 0 EE tttttttttt = pp cc uu = cc Fall 2018 Prof. Sergio B. Mendes 89

Units of Energy: WW = qq VV ee = 1.6022 10 19 CC VV = 1 VV WW = ee VV = 1.6022 10 19 JJ 1 eeee Fall 2018 Prof. Sergio B. Mendes 90

Units of Mass: EE 0 = mm cc 2 mm = EE 0 cc 2 1 kkkk = 8.987 10 16 JJ/cc 2 1 uu 1 12 mmmmmmmm oooo nnnnnnnnnnnnnn 12 CC = 1.66054 10 27 kkkk = 931.494 MMMMMM/cc 2 Particle Mass (MeV) / c 2 Mass (u) electron 0.511 0.54858 10-3 proton 938.27 1.007276 neutron 939.57 1.008665 Higgs boson 125,090 134.3 Fall 2018 Prof. Sergio B. Mendes 91

Units of Linear Momentum EE tttttttttt 2 pp 2 cc 2 = mm 2 cc 4 lliiiiiiiiii mmmmmmmmmmmmmmmm = eeeeeeeeeeee cc Fall 2018 Prof. Sergio B. Mendes 92

Binding Energy: 2 2 2 EE 0, pp + 2 EE 0, nn EE 0, HHHH = EE BB,HHHH 2 1.007276 uu + 2 1.008665 uu 4.001505 u cc 2 = 0.0304 uu cc 2 = 28.3 MMMMMM = EE BB,HHHH Fall 2018 Prof. Sergio B. Mendes 93

2.9 Spacetime Representation Fall 2018 Prof. Sergio B. Mendes 94

Conventional Representation: (tt, xx) Fall 2018 Prof. Sergio B. Mendes 95

Spacetime Representation (xx, cccc) When describing events in relativity, it is convenient to represent events on a spacetime diagram. In this diagram one spatial coordinate x, to specify position, is used and instead of time t, ct is used as the other coordinate so that both coordinates will have dimensions of length. Spacetime diagrams were first used by H. Minkowski in 1908 and are often called Minkowski diagrams. Paths in Minkowski spacetime are called worldlines. Fall 2018 Prof. Sergio B. Mendes 96

Worldline Fall 2018 Prof. Sergio B. Mendes 97

tan θθ = xx cc tt = uu cc = ββ xx cc tt uu tan θθ = uu cc = ββ 1 θθ θθ 45 Fall 2018 Prof. Sergio B. Mendes 98

Light Cone Fall 2018 Prof. Sergio B. Mendes 99

Spacetime and Inertial Frames of Reference xx = γγ xx γγ vv oo tt tt = γγ vv oo cc 2 xx + γγ tt cc tt θθ cc tt θθ xx xx tan θθ = vv oo cc Fall 2018 Prof. Sergio B. Mendes 100

Back to the Lorentz Transformations: ddtt = γγ vv oo cc 2 ddxx = γγ dddd γγ vv oo ddtt ddyy = ddyy ddzz = dddd ddxx + γγ dddd xx yy zz cccc four-vector ddxx 2 + ddyy 2 + ddzz 2 cc 2 ddtt 2 = ddxxx 2 + ddyy 2 + ddzz 2 cc 2 ddttt 2 Invariant: its value does not change among inertial frames of reference Fall 2018 Prof. Sergio B. Mendes 101

Example 2.13 KK = 2.00 GGGGGG KK = 2.00 GGGGGG EE tttttttttt =?? = KK + mm cc 2 = 2.00 GGGGGG + 0.93827 GeV = 2.94 GGGGGG Fall 2018 Prof. Sergio B. Mendes 102

Example 2.13 KK = 2.00 GGGGGG KK = 2.00 GGGGGG pp =?? EE tttttttttt 2 pp 2 cc 2 = mm 2 cc 4 pp = 1 cc EE tttttttttt 2 mm cc 2 2 = 1 cc 2.938 GGGGGG 2 0.93827GGGGGG 2 = 2.78 GGGGGG cc Fall 2018 Prof. Sergio B. Mendes 103

Example 2.13 KK = 2.00 GGGGGG KK = 2.00 GGGGGG uu =?? ββ =?? ΓΓ =?? EE tttttttttt = KK + mm cc 2 = mm cc 2 2 = ΓΓ mm cc2 1 uu cc ΓΓ = EE tttttttttt mm cc 2 = 2.938 GGGGGG 0.93827 GGGGGG = 3.13 ββ = ΓΓ2 1 ΓΓ 2 = 0.948 Fall 2018 Prof. Sergio B. Mendes 104

Topics 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13 2.14 Inertial Frames of Reference Conceptual and Experimental Inconsistencies The Michelson-Morley Experiment Einstein s Postulates The Lorentz Transformation Time Dilation and Length Contraction Addition of Velocities Experimental Verification Twin Paradox Space-Time Doppler Effect Relativistic Momentum Relativistic Energy Computations in Modern Physics Electromagnetism and Relativity Fall 2018 Prof. Sergio B. Mendes 105