Introduction to Special Relativity

Topics: 􀂄 􀂄 􀂄 􀂄 Velocity transformation 8.022 (E&M) – Lecture 11 Introduction to Special Relativity Length contraction and Time dilation Lorentz transformations Special relativity 􀂄 Ready for the challenge? 􀂄 Special relativity seems easy but it’s not! 􀂄 A new way of thinking that often goes against intuition 􀂄 It will take some time to “digest it”, but believe me: it’s worth the effort! 􀂄 Why do we need it in 8.022? 􀂄 Weren't you frustrated last time when magnetic forces came out of nowhere? 􀂄 Special relativity naturally explains them in terms of electric forces seen from in a reference frame in motion 􀂄 This is important for everybody 􀂄 Physics majors: first of many iterations on a crucial topic 􀂄 Non Physics majors: chance to know what you are missing 􀂄 Don’t forget: you are still in time… G. Sciolla –MIT 8.022 – Lecture 11 2 1 3 The principles of special relativity 􀂄 Formulated in 1905 by A. Einstein 􀂄 Incredible but true: no Nobel Prize for this! 􀂄 􀂄 􀂄 􀂄 􀂄 (G. Sciolla – MIT 8.022 – Lecture 11 Based upon 2 postulates The laws of physics are the same for all reference frames The speed of light is the same (c) in all reference frames (Inertial) Reference frame System of coordinates in which the observer is non accelerating inertial = non accelerating) 4 Reference frames: examples 􀂄 Situation 􀂄 A train is moving with velocity v 􀂄 􀂄 􀂄 􀂄 􀂄 􀂄 􀂄 􀂄 Observers 1 and 2 􀂄 v G. Sciolla – MIT 8.022 – Lecture 11 w.r.t. to a station A table is anchored to the train A ball is falling from the table We can identify 3 systems of reference and 3 observers: Observer 1: sitting on a bench at the station Observer 2: sitting on the table on the train Observer 3: a bug sitting on top of the falling ball Who are the observers in an inertial reference frame? Observer 3 is not: the ball is falling with acceleration g 2G. Sciolla – MIT 8.022 – Lecture 11 5 Is time the same in all reference frames? 􀂄 ! 􀂄 Problem 􀂄 v //x axis 􀂄 􀂄 􀂄 􀂄 ight is h v G. Sciolla – MIT 8.022 – Lecture 11 These (apparently) innocent assumptions have amazing consequences such as time is not absoluteThe train is moving with velocity Observer 1: standing in the train Observer 2: at the station Observer 1 flashes a pulse of light vertically to a photosensor mounted on the floor of the train Both observers measure the time between when the lemitted and when the light reaches the sensor Will the 2 observers measure the same time? 6 Time in different reference frames 􀂄 􀂄 􀂄 1 ht t c ⎧ ⇒∆ =⎨ ⎩ ( ) 2 2 2 2 2 2 2 2 2 2 2 22 2 2 2 1 22 2 2 2 2 '(v t ) ' h+(v t )' t 1 1 1 ' ht t c h v vt t t t c c c c v c t tγ γ ⎧ ∆⎪ =⎨ ⎪⎩ ∆⎛ ⎞∆ = = + ∆ ⇒ ∆ −⎜⎝ ⎠ = − ⎟ ⇒ ∆= ∆ h h’ v Let’s calculate time measured by the 2 observers Train reference frame (observer 1) Station reference frame (observer 2) Distance traveled by light: h Velocity of light: c = ∆ Distance traveled by light: h'= h +Velocity of light: c Defining ∆= ∆ = ∆ = ∆ 37 Time dilation 􀂄 We just deri􀂄 Gamma factor: 􀂄 Since ∆t’=γ∆t 􀃎 ∆∆t 􀂄 ∆i􀂄 ∆the clock 􀂄 Conclusion: 't tγ∆ =∆ 2 2 2 1 1 1 1 1 v v cv c γ β β β ≡ = > ≡ ≡ −− G. Sciolla – MIT 8.022 – Lecture 11 ved a very important result! t’ is always larger than t’ = tme measured by the observer in the station who sees the clock in motion t = time measured by the observer on the train, at rest wrt Clocks in motion run slower (time dilation) with 8 Length in different reference frames 􀂄 Problem 2 􀂄 of the train 􀂄 v G. Sciolla – MIT 8.022 – Lecture 11 Now observer 1 flashes a pulse of light horizontally from left end The light is reflected by a mirror on the right end wall and detected by a photosensor on the left wall What is the length of the train measured by each observer? 48.022 – Lecture 11 G. Sciolla – MIT 8.022 – Lecture 11 9 Length in train reference frames 􀂄 For observer in train reference frame 􀂄 Events we are interested in: emission and reception of light 􀂄 Time in between the two: ∆t= ∆ttrain 􀂄 Length of the train: v L 2L = 2 ct Lt c ∆ ⇒∆ = G. Sciolla – MIT 10 Length in the station reference frame 􀂄 Calculate separately ∆x1 (L􀃆R) and ∆x2 (R􀃆L) 􀂄 ∆t1 is shorter because train and light move in opposite directions 􀂄 ∆t2 is longer because train and light move in the same direction 􀂄 L’ (t’) = length (time) measured from station reference frame 􀂄 Rearrange terms: v 1 1 2 2 ' ( ' ' ) /' ( ' ' ) /t L vt c t L vt c ∆ = − ∆ ∆ = + ∆ 1 2 ' ' ' ' Lt c v Lt c v ⎧ ∆ =⎪⎪+⎨ ⎪ ∆ = ⎪ −⎩ 511 Length contraction 􀂄 ∆t’1 and ∆t’2: 􀂄 ∆t’=γ∆t 􀃆 􀂄 Since γ>1 􀃆 Moving objects 1 2 2 22 2 2 2 ' ' ' = ' + ' 2 2 2 ' ' ' (1 ) L Lt t t c c c LL L v c c c γ ∆ ∆ ∆ = + = − + = = = − − 22 ' 2' 'L Lt t c c LLγ γ γ γ=∆ = ∆ = ⇒ = G. Sciolla – MIT 8.022 – Lecture 11 Total time in the station reference frame = sum of Remember how time dilates: appear contracted (length contraction) v c v c v 12 Summary so far 􀂄 􀂄 􀂄 􀂄 Consequences: 􀂄 􀂄 􀂄 Length contraction 􀂄 i􀂄 REALLY??? Can we check this experimentally??? ' LL γ = ' γ∆ =∆ G. Sciolla – MIT 8.022 – Lecture 11 Assume Special Relativity postulates hold: The laws of physics are the same for all reference frames The speed of light is the same (c) in all reference frames Time dilation clocks in motion run slower movng objects appear contracted t t 613 Cosmic Ray Muons 􀂄 Cosmic ray muons: 􀂄 􀂄 􀂄 µ µs) 􀂄 Question: Can muons produced i􀂄 Input: 􀂄 􀂄 Atmosphere ~20 Km thick G. Sciolla – MIT 8.022 – Lecture 11 Application: Cosmic rays are energetic particles (mainly protons) coming from somewhere in the Universe When they hit the atmosphere they will produce showers of particles are of particular interest because they are very penetrating and have a long lifetime (2.2 n the upper atmosphere reach the ground? Muon’s velocity = 99.99% of velocity of light c 14 Cosmic Ray Muons (2) 􀂄 Inputs: 􀂄 vµ 􀂄 􀂄 ∆l = 0.9999 c ∆􀂄 􀂄 γ = 1/sqrt(1-v2/c2) ~ 71 􀂄 Approach 1: our perspective 􀂄 τµ = 2.2 µ􀂄 τ’ /71 x 2.2 µµs 􀃆 ∆l = 42 Km: 􀂄 􀂄 The ∆lµ 􀂄 ∆l = ∆l’/γ τ=2.2 µs: OK! G. Sciolla – MIT 8.022 – Lecture 11 Application: = 99.99% of velocity of light c, atmosphere ~ 20 Km Non relativistic approach: t = 0.6 Km < 20 Km: NO, they cannot reach the ground Relativistic approach s in muon’s reference frame In our reference frame: = τγ = s = 156 Now muon can travel: OK! Approach 2: muons’ perspective ’ = 20 Km of atmosphere appear contracted to a relativistic = 20Km/71 ~ 0.3 Km that can be traveled with Relativity: same phyiscs in all reference frames! 715 More on Cosmic Ray Muons 􀂄 􀂄 􀂄 􀂄 β = 0.9999 􀃆 2/c2) ~ 71 􀂄 􀂄 ll0exp(-t/τ) 􀂄 􀂄 τ’ µ = 156 µ􀂄 At sea level: 􀂄 L=20Km 􀃆 T=66 µs 􀃆 Nsea =N01560 􀂄 On Mount Everest: 􀂄 L=12Km 􀃆 T=40 µs 􀃆 N=N01560 􀃆 µ N(t) t G. Sciolla – MIT 8.022 – Lecture 11 The number of cosmic muons detected at sea level and on the top of Mount Everest are different. By how much? Hypotheses: Muons are produced in the upper atmosphere: ~ 20 Km γ = 1/sqrt(1-vMount Everest ~ 8 Km Muons decay exponentiay N(t) = NChoose 1 RF and stay with it s in our R.F. exp(-66/)=0.65 NEverest exp(-40/)=0.77 NAt sea level expect ~15% less cosmic than on Mount Everest: OK! How do lengths perpendicular to v transform? 􀂄 Thought experiment 􀂄 Train moving towards a tunnel with velocity v=0.9c􀂄 Height of train in train’s RF: htra= 3.5 m in􀂄 Height of tunnel in tunnel’s RF: h’tunnel = 4.0 m 􀂄 If we have Lorentz contractions: L’=L/γ 􀂄 γ = 1/sqrt(1-0.92)=2.29􀂄 In tunnel’s reference frame: the train moves with β=0.9 􀃆 h’train= htrain/γ = 3.5/2.29 = 1.5m 􀃆 no problem: it will fit!􀂄 In train’s reference frame: tunnel moves with velocity β=0.9 􀃆 htunnel= h’tunnel/γ = 4/2.29 = 1.7m < htrain 􀃆 they will smash!􀃆 Different observers come to different conclusions 􀃆 against relativity principle! 􀃆 Lorentz contraction cannot happen G. Sciolla –MIT 8.022 – Lecture 11 16 8 17 Lorentz transformation 􀂄 􀂄 􀂄 O’ is moving w.r.t. O with velocity v //x axis where 􀂄 inate in the O 􀂄 iO’ reference frame 􀂄 􀂄 􀂄 x y z x’ y’ z’ O O’ v G. Sciolla – MIT 8.022 – Lecture 11 “Time dilation” and “Length contraction” are consequences of the so called “Lorentz transformation” Consider 2 inertial reference frames: O and O’ (x,y,z,t) the coordreference frame (x’,y’,z’,t’) the coordnate in the Lorentz transformation: Linear transformation that relates the coordinate in the 2 R.F. Why linear? Because reference frames are inertial 18 Lorentz transformation (2) 􀂄 i􀂄 v //x 􀃆 i􀂄 Goal: calcul􀂄 􀂄 􀂄 ility v 􀃆 􀂄 iv 􀃆 x y z x’ y’ z’ O O’ v ' (1) ' ' ' (2) x y y z z t = +⎧ ⎪ =⎪⎨ =⎪ ⎪ = +⎩ ' ( ) (3)(1) 0 ' ) -(4 xAvt t CB A x v Dt = −⎧ ⇒ = + =⇒ ⇒ ⎨ = +⎩ (3) : ' ( ) -. (4): ' ' ( ) ' ( ) (3) ' (5) x A x vt x t DA = − = + − = −⎧ ⇒ + = ⇒ ⎨ =⎩ G. Sciolla – MIT 8.022 – Lecture 11 The most general form for a lnear transformation: z and y do not change because ignore themn the following ate coefficients A,B,C,D First requirement: O and O’ overlap at t=0: At t=t’=0, x=x’=0 For O, the orign of O’ moves away with veocFor O’, the origin of O moves away wth velocity -Ax Bt Cx Dt Substitute in A x vt Bt Substitute in Fromx vt Avt v Cx Dt vDt A x vt Cx At = − = − 919 Lorentz transformation (3) 􀂄 Second requirement: 􀂄 Send a li􀂄 i(3) and use (5): x y z x’ y’ z’ O O’ v 2 ' ' ( ) ( ) ( ) ( )' ( ) = ( ) x A vt ct vC A c == − = −⎧ ⇒ + = − ⇒⎨ =− = + +⎩ 2 ' ( ) (3) ' (6) x vt x c = −⎧ ⎪ ⇒ ⎨ ⎛ ⎞ = −⎜ ⎟⎪ ⎝ ⎠⎩ G. Sciolla – MIT 8.022 – Lecture 11 ght pulse along the x direction at t=0 After a time t the coordnates of the light pulse are x=ct and x’=ct’. Substitute in ct x vt A ct cCct At Act vt c Cx At c Cc tAt A x vt A t 20 Lorentz transformation (4) 􀂄 Third requirement: 􀂄 Send a li􀂄 ithe total displ2+y’22x y z x’ y’ z’ O O’ v () 2 2 2 2 2 2 2 2 2 2 22 2 2 2 ' ' ( ') ( ) ( ) 1 1 x y vAx y x c x ct A v c γ + = ⎛ ⎞− + = −⎜ ⎟⎝ ⎠ = = ⇒ + = ⇒= ≡ ⎛⎞−⎜⎟⎝⎠ 2 ' ( ) ' x x vt t x c γ γ = −⎧ ⎪⎨ ⎛ ⎞ = −⎜ ⎟⎪ ⎝ ⎠⎩ ⇒ G. Sciolla – MIT 8.022 – Lecture 11 ght pulse along the y direction at t=0 After a time t the coordnates of the light pulse are (x=0; y=ct) in O; in O’ acement is: x’= (ct’). Substitute (3) and (6): 2 2 Since 0 and ct vt c A t y ct A vt c At vt 1021 Lorentz transformation: summary Summarizing: is 􀂄 To go from O (at rest) to O’ (in motion): 􀂄 􀂄 x y z x’ y’ z’ O O’ v 2 ' ( ) ' x x vt t x c γ γ = −⎧ ⎪⎨ ⎛ ⎞ = −⎜ ⎟⎪ ⎝ ⎠⎩ 2 ( ' ') ' ' x x vt t x c γ γ = +⎧ ⎪⎨ ⎛ ⎞ = +⎜ ⎟⎪ ⎝ ⎠⎩ G. Sciolla – MIT 8.022 – Lecture 11 when O’ moves wrt O with velocity +v//x axTo go from O’ (in motion) to O(at rest), just change the sign of the velocity: The other coordinates (y and z) are not affected vt vt 22 Transformation of velocity 􀂄 􀂄 x //+x axis 􀂄 x 􀂄 Conclusion: x y z x’ y’ z’ O O’ v 22 2 2 (( ' ')) ' ' ' '( ' ' ) ''/' ' 1 '/' 1 x x x d x vt u vvdt dt dxd t x cc u vdx dt v v dx dt c c γ γ + + = = = ⎛ ⎞ ++⎜ ⎟⎝ ⎠ ++ = = + + 2 2' ' ' 1 1 x x x x x x u v u v u u vu c c + − = = + − G. Sciolla – MIT 8.022 – Lecture 11 Consequence of Lorentz transformations Observer in motion O’ shoots a bullet with velocity u’What is the velocity of the bullet umeasured by O? dx dx vdt vu and vu 1123 Velocity not //to v 􀂄 􀂄 y perpendicular to v 􀂄 x 􀂄 Conclusion: x y z x’ y’ z’ O O’ v 22 2 2 ' ' ( ' ')( ' ' ) ''/' ' ( ' ') /' (1 ) y y x dy dyu vvdt dtd t x cc udy dt vdt dx c c γγ γ γ = = = ⎛ ⎞ ++⎜ ⎟⎝ ⎠ = = + + 2 2 an ' ' ' (1 ) d (1 ) y y y y x x u u u u vu c c γ γ = = + − G. Sciolla – MIT 8.022 – Lecture 11 How do we sum velocity not //to the relative motion of the 2 R.F.? Observer in motion O’ shoots a bullet with velocity u’What is the velocity of the bullet umeasured by O? dy dx vu dt vu 24 Summary and outlook 􀂄 Today: 􀂄 􀂄 􀂄 􀂄 􀂄 Next time: 􀂄 􀂄 How to transform electric fiel􀂄 G. Sciolla – MIT 8.022 – Lecture 11 Principle of Special Relativity and its amazing consequences Length contraction and Time dilation Lorentz transformations Velocity transformation (v always < c) More on Relativity: ds and forces Prove that E and B are intimately connected 12