1 1.what is reference frame A reference frame consists of (1) a coordinate system and (2) a set of synchronized clocks distributed throughout the coordinate grid and rest with respect to it. x z y § 25.1 reference frames and the classical Galilean relativity 2 § 25.1 reference frames and the classical Galilean relativity A reference frame has three spatial coordinates and one time coordinate (x, y, z, t). Four dimensional space-time synchronized clocks: l l A B O Inertial reference frames Reference frame noninertial reference frames § 25.1 reference frames and the classical Galilean relativity 3 ?Relativity is concerned with how an event described in one reference frame is related to its description in another reference frame. That is how the coordinates and times of events measured in one reference frame are related to the coordinate, time, and corresponding physical quantities in another reference frame. 2. Some fundamental concepts ?Event is something that happens at a particular place and instant. ?Observers belong to particular inertial frames of reference, they could be people, electronic instrument, or other suitable recorders. § 25.1 reference frames and the classical Galilean relativity ?The special theory of relativity is concerned with the relationship between events and physical quantities specified in different inertial reference frames. ?The general theory of relativity is concerned with the relationship between events and physical quantities specified in any reference frames. ?Transformation equations are that indicate how the four space and time coordinates specified in one reference frame are related to the corresponding quantities specified in another reference frame. ? ? ? ? ? ? ? ? ? ? ? ? ? ? ′ ′ ′ ′ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? t z y x t z y x § 25.1 reference frames and the classical Galilean relativity 4 ?Standard geometry we use for the special theory of relativity has two inertial reference frames called S and S’, with their x-and x’- coordinate axes collinear. Imagine collections of clocks distributed at rest throughout each respective frame; the clocks all are set to 0 s when the two origins coincide. Observer is at rest in frame S. § 25.1 reference frames and the classical Galilean relativity Observer is at rest in frame S’. § 25.1 reference frames and the classical Galilean relativity 5 3. Classical Galilean relativity 1the Galilean time transformation equation In classical physics time is a universal measure of the chronological ordering of events and the time interval between them. Watches in fast sports cars, airplanes, spacecraft, and oxcarts tick at the same rate as those at rest on the ground; the time interval between two events and the rate at which time passes are independent of the speed of the moving clock; they are same everywhere. tt ′ = § 25.1 reference frames and the classical Galilean relativity 2the Galilean spatial coordinate transformation equations zz yy vtxx = ′ = ′ ?= ′ tt ′ = vuu rrr O rrr rrr ?= ′ ?= ′ § 25.1 reference frames and the classical Galilean relativity 6 3the Galilean velocity component transformation equations vu t vtx t x u t x u xx x + ′ = + ′ == ′ ′ = ′ d )d( d d , d d zz yy xx uu uu vuu ′ = ′ = + ′ = The velocity components along a direction perpendicular to the motion are the same in two standard inertial reference frames. § 25.1 reference frames and the classical Galilean relativity 4the Galilean acceleration component transformation equations xx x a t vtx t x a t x a ′ = + ′ == ′ ′ = ′ 2 2 2 2 2 2 d )(d d d , d d zz yy xx aa aa aa = ′ = ′ = ′ The acceleration components are the same in the two inertial reference frames. § 25.1 reference frames and the classical Galilean relativity 7 FamamF mmaa ′ = ′′ == ′ = ′ = r rr r rr zz yy xx aa aa aa = ′ = ′ = ′ Descriptions of what happens as a result of the laws of mechanics may different from one inertial reference frame to another, but the laws of mechanics are the same. § 25.1 reference frames and the classical Galilean relativity § 25.2 the need for change and the postulates of the special theory 1. Why we need relativity theory? 1troubles with our ideas about time The pion π + or π - created in a high-energy particle accelerator is a very unstable particle. Its lifetime at rest is 26.0 ns; when it moves at a speed of v= 0.913c, an average distance of D=17.4 m are observed before decaying in the laboratory. We can calculate the lifetime in this case by D/v=63.7 ns, it is much larger than the lifetime at rest. Such an effect cannot be explained by Newtoneian physics! 8 2troubles with our ideas about length Suppose an observer in the above laboratory placed one marker at location of pion’s formation and another at the location of its decay. The distance between the two markers is measured to be 17.4 m. Another observer who is traveling along with pion at a speed of u=0.913c . To this observer, the distance between the two markers showing the formation and decay of the pion is (0.913c)(26.0×10 -9s )=7.1 m. Two observers who are in relative motion measure different values for the same length interval. § 25.2 the need for change and the postulates of the special theory 3troubles with our ideas about speed If the observer A throws a ball at superluminal, speed, the result observed by observer O will be that the observer B get the ball before the observer A throwing it. § 25.2 the need for change and the postulates of the special theory 9 4troubles with our ideas about energy int E? radiation ?+ ee Electron-positron Electron and positron are initially toward one another at a very low speed. Electron and positron have annihilated one another and give a radiation. The walls of the container absorbed the radiation and increasing the internal energy of this system. § 25.2 the need for change and the postulates of the special theory 5troubles with our ideas about light Electromagnetism provides some subtle clues that all is not right with Glilean relativity. 2 2 00 2 2 2 2 00 2 2 or t B x B t E x E zz yy ? ? = ? ? ? ? = ? ? εμεμ m/s100.3 1 8 00 ×== εμ c This result is not dependent of the speed v of the source of the waves in the equation! § 25.2 the need for change and the postulates of the special theory 10 Experiment is performed in reference frame S: c c Experiment is performed in reference frame S’ which is moving at speed v in the same direction as light is moving: c v r c § 25.2 the need for change and the postulates of the special theory Therefore the Galilean transformation equations between two inertial reference frames need to be modified to accommodate the observed invariance of the speed of light. The new transformation equations , called the Lorentz transformation equations, together with their implications and consequences, constitute the special theory of relativity. 2. The postulates of special theory of relativity 1the speed of light in a vacuum has the same numerical value c in any inertial reference frame, independent of the motion of the source and observer. § 25.2 the need for change and the postulates of the special theory 11 It means that there is an ultimate speed c, the same in all directions and in all inertial reference frames. Light and any massless particles travel at this speed. No particle that have mass and carry energy or information can reach or exceed this limit, no matter how much or how long it is accelerated. This postulate is expression of the experimental result. (1)in a 1964 experiment, physicist at CERN. γγπ +→ 0 Rapid moving source Speed of the light was same as measured at rest in the laboratory. § 25.2 the need for change and the postulates of the special theory (2)W. Bertozzi experiment(1964) cv 95999999999.0= § 25.2 the need for change and the postulates of the special theory 12 2the fundamental laws of physics are the same for observers in all inertial reference frames. No frame is preferred. Descriptions of what happens as a result of the laws of physics may different from one inertial reference frame to another, but the underlying fundamental physical principles and laws are the same. Glileo assumed that the laws of mechanics were the same in all inertial reference frames. Einstein extended that idea to include all the laws of physics. § 25.2 the need for change and the postulates of the special theory § 25.3 consequences of Einstein’s postulates 1. Time dilation 0 l 0 τ s′ Event 1: emission of the pulse 0,0,0,0 1111 = ′ = ′ = ′ = ′ tzyx Event 2: reflection of the pulse cltzlyx 022022 ,0,,0 =′=′= ′ =′ Event 3: detection of the pulse cltzyx 03333 2,0,0,0 = ′ = ′ = ′ = ′ clt 00 2== ′ τ?Time interval: Proper time interval 13 Event 1: emission of the pulse 0,0,0,0 1111 ==== tzyx Event 2: reflection of the pulse 2,0 ,,2 22 022 τ τ == == tz lyvx Event 3: detection of the pulse τ τ == == 33 33 ,0 ,0, tz yvx s τ τv 0 l c vl 22 0 )2( 2 τ τ + = 22 0 22 0 11 2 cvcv cl ? = ? = τ τ § 25.3 consequences of Einstein’s postulates Conclusions: the moving clock has a greater time interval between its ticks than the clock that is at rest ; a moving clock runs slow. This is called time dilation. Time intervals in relativity are not absolute or universal but depend on whether the clock is moving or not: we say time is a relative not an absolute quantity. Time dilation has been confirmed by many experiments: § 25.3 consequences of Einstein’s postulates The proper time interval is the shortest. 14 Microscopic clock: Muons are unstable elementary particles with a (proper) lifetime of 2.2 μs. they are produced with very high speeds in the upper atmosphere when cosmic rays collide with air molecules. Take the height L 0 of the atmosphere to be 90 km in the reference frame of earth, if the average speed of the muons is 0.999978c, can the muons arrive the surface of the earth? § 25.3 consequences of Einstein’s postulates Solution: m660102.2100.3999978.0 68 =××××= ? L According to Newton’s mechanics The muons can not arrive the surface of the earth. According to relativity, the life-time of the muon is μs7.331 )999978.0(1 2.2 )(1 222 0 = ? = ? = cv τ τ The distance that muon can travel is m8.99507107.331103999978.0 68 =×××× ? The muon can arrive the surface of the earth. § 25.3 consequences of Einstein’s postulates 15 Example: your starship passes Earth with a relative speed of 0.9990c. After traveling 10.0 y(your time), you stop at lookout LP13, turn, and then travel back to earth with the same relative speed. The trip back takes another 10.0 y(your time). How long does the round trip take according to measurements made on Earth?(neglect any effect due to the accelerations) Solution: Event 1: start from Earth Event 2: stop at LP13 Proper time: y0.10 00 == t?τ § 25.3 consequences of Einstein’s postulates The Earth measurement of the time interval is y)(224 999.01 0.10 )(1 22 0 = ? = ? == cv t τ ?τ On the return trip, we have the same situation and the same data.thus the round trip requires y)(484 )(1 2 2 0 total = ? = cv t τ ? § 25.3 consequences of Einstein’s postulates 16 Macroscopic clocks: In October 1977, Joseph Hafele and Richard Keating flew four portable atomic clocks twice around the world on commercial airline, once in each direction. The prediction of the theory was verified within 10%. A few years later, physicist at the university of Maryland carried out a similar experiment with improved precision(1%). § 25.3 consequences of Einstein’s postulates 2. Length contraction 1lengths perpendicular to the direction of motion We can prove the sticks are the same length. 0 l x′ y′ o′ Measured using rulers at rest in S’ 0 l x y o Measured using rulers at rest in S § 25.3 consequences of Einstein’s postulates 17 Prove by contradiction: Make the hypothesis that a moving stick, oriented perpendicular to the direction of motion, is longer than stick at rest. An observer in reference frame S: 0 ll > x′ y′ o ′ Measured using rulers at rest in S’ 0 l x y o Measured using rulers at rest in S § 25.3 consequences of Einstein’s postulates An observer in reference frame S’: 0 ll > x′ y′ o′ Measured using rulers at rest in S’ 0 l x y o Measured using rulers at rest in S The two observer can compare their conclusions by communication, and find the results are contradictory, therefore, the hypothesis must be false. § 25.3 consequences of Einstein’s postulates 18 Conclusion: lengths measured perpendicular to the direction of motion are unaffected by the motion. 2length oriented along the direction of motion x y x ′ y ′ O O ′ Measured by us in lab Measure by Fairy godmother 0 l τ l 0 τ 0 00 γττ ll v == 0 τ l v = )(1 22 0 0 cvl l l ?== γ § 25.3 consequences of Einstein’s postulates l 0 is the proper length. Length contraction only occurs for those lengths (or components of lengths ) oriented along the direction of motion. Conclusion: The length l is shorter than l 0 when measured in a frames in which it is moving. § 25.3 consequences of Einstein’s postulates The proper length is longest in all measurement. 19 Example 1: Sally (at point A) and Sam’s spaceship (of proper length L 0 =230 m) pass each other with constant relative speed v. Sally measures a time interval of 3.75 μs for the ship to pass her (from the passage of point B to the passage of point C). In terms of c, what is the relative speed v between Sally and the ship? § 25.3 consequences of Einstein’s postulates Solution: t cvL t L t L v ?? γ ? 2 00 )(1? === c Ltc cL v 210.0 )230()1057.3)(10998.2( 230 )( 268 2 0 2 0 = +×× = + = ? ? Solving this equation for v leads us to § 25.3 consequences of Einstein’s postulates 20 Example 2:(a)Can a person, in principle, travel from Earth to the galactic center (which is about 23000 ly distant) in a normal lifetime?Explain, using either time-dilation or length-contraction arguments. (b)What constant speed is needed to make the trip in 30 y (proper time)? Solution: (a) Do not consider the important problem such as fuel requirements, stresses to the human body due to the accelerations, the answer is yes. § 25.3 consequences of Einstein’s postulates (b) d=23000 ly=23000c (y) c c dct c v v d cv t t 99999915.0 299792203 000017013.01 299792458 )2300030(1)(1 )(1 22 0 2 0 = = + = + = + = = ? = ? ? ? § 25.3 consequences of Einstein’s postulates 21 Einstein’s train: s s ′ Station frame: Train frame: In S frame: the two light signals arrive C simultaneously. 3. Simultaneity is relative . .. A C B A′ C ′ B′ u ),,,( tzyxI AAA ),,,( tzyxII BBB Sframe S′frame . § 25.3 consequences of Einstein’s postulates ... A C B . A′ C ′ B′ u ),,,( 2 tzyxII BBB ′′′′),,,( 1 tzyxI AAA ′′′′ . . .. A C B A′ C ′ B′ u ),,,( tzyxI AAA ),,,( tzyxII BBB Sframe S′frame Sframe S′frame § 25.3 consequences of Einstein’s postulates 22 Conclusions: If two events are simultaneous in one inertial reference frame, they may not be simultaneous to any other inertial reference frame moving with respect to the first. § 25.3 consequences of Einstein’s postulates In frame S’： the two light signals do not arrive C’ simultaneously. § 25.4 The Lorentz transformation equations Modify the Galilean relativity: the new transformation equations must approach the Galilean equations in the limit of small speed. 1. The transformation of coordinates vtx = x x′ O O′ y y′ S S’ BtAxx += ′ For O’ 0= ′ x AvB BtAvtx ?= =+= ′ 0 )( vtxAAvtAxx ?=?= ′ 23 )2()( )1()( vtxAx vtxAx + ′ = ?= ′ vtx = x x′ O O′ y y′ S S’ 1 x′ 2 x′ )()( 1212120 ttAvxxAxxl ???= ′ ? ′ = 0, 1212 =?=? ttlxx All = 0 γ γ =∴ =?= A l lcvl 0 0 22 1Q )2()( )1()( vtxx vtxx + ′ = ?= ′ γ γ § 25.4 The Lorentz transformation equations Eliminate the x’ of equations (1) and (2) )4()( )3()( 2 2 cxvtt cvxtt ′ + ′ = ?= ′ γ γ )( )( 2 x c v tt zz yy vtxx ?= ′ = ′ = ′ ?= ′ γ γ Transformation equations )( )( 2 x c v tt zz yy tvxx ′ + ′ = ′ = ′ = ′ + ′ = γ γ Inverse equations § 25.4 The Lorentz transformation equations 24 2. Important deductions Notice that 1, →<< γcv 1time dilation if )( )( 2 022 2 011 cxvtt cxvtt ′ + ′ = ′ + ′ = γ γ 012 00 2 12 12 )( )()( γτγ γ τ = ′ ? ′ = ′ ? ′ + ′ ? ′ = ?= tt xx c v tt tt then tt zz yy vtxx = ′ = ′ = ′ ?= ′ § 25.4 The Lorentz transformation equations Event 1: Musician 1 begins to play s0 m0 1 1 = = t x Event 2: Musician 2 begins to play s0 m 2 02 = = t lx 0 l x y o S In S reference frame 0, 12012 =?==?= tttlxxx ?? 2the relativity of simultaneity § 25.4 The Lorentz transformation equations 25 )( )( 2 x c v tt zz yy vtxx ?= ′ = ′ = ′ ?= ′ γ γ )( )( 2 x c v tt zz yy tvxx ′ + ′ = ′ = ′ = ′ + ′ = γ γ )( )( 2 x c v tt tvxx ??γ? ??γ? ?= ′ ?= ′ )( )( 2 x c v tt tvxx ′ + ′ = ′ + ′ = ??γ? ??γ? We can get Using the Lorentz transformation equations § 25.4 The Lorentz transformation equations In S’ reference frame Event 1: Musician 1 begins to play s0 m0 1 1 = ′ = ′ t x Event 2: Musician 2 begins to play 2 02 02 / cvltt lxx γ? γ? ?= ′ = ′ = ′ = ′ Conclusions: If two events are simultaneous in one inertial reference frame, they will not be simultaneous to any other inertial reference frame moving at non speed v with respect to the first. Only when 0,0 == xt ?? 0= ′ t?then § 25.4 The Lorentz transformation equations 26 3. Relativistic velocity addition 1velocity parallel or antiparallel to the direction of motion of the two inertial reference frames )dd(dand)dd(d )(and)( 2 2 x c v tttvxx x c v tttvxx ′ + ′ = ′ + ′ = ′ + ′ = ′ + ′ = γγ γγ from We have )d(d )dd( d d 2 x c v t tvx t x u x ′ + ′ ′ + ′ == γ γ § 25.4 The Lorentz transformation equations 22 1 d d 1 d d c uv vu t x c v v t x u x x x ′ + + ′ = ′ ′ + + ′ ′ = 2velocity perpendicular to the direction of motion of the two inertial reference frames t y u t y u yy d d , d d = ′ ′ =′ )dd(danddd )(and 2 2 x c v ttyy x c v ttyy ′ + ′ = ′ = ′ + ′ = ′ = γ γfrom We have § 25.4 The Lorentz transformation equations 27 )d/d1( dd )dd( d d d 22 tx c v ty x c v t y t y u y ′′ + ′ = ′ + ′ ′ == γγ )1( 2 c uv u u x y y ′ + ′ = γ like manner )1( 2 c uv u u x z z ′ + ′ = γ In special relativity, even though the y-and y’- coordinates in the two frames are the same , the velocity components are not same because of the relativity of time( ).tt ′≠ § 25.4 The Lorentz transformation equations Example 1: The distance of two lamps A and B on a train is 25 m, the train is moving at the speed of v=20.0 m/s with respect to the ground. If the observer on the train declare that the two lamps is turn on simultaneously, then what is the conclusion for the observer on the ground? v r A B S S’ § 25.4 The Lorentz transformation equations 28 Solution: If s/m107.2 8 ×=v then s107.1 1 7 2 2 2 ? ×= ? ′ + ′ =?= c v x c v t ttt BA ?? ? The effect of the relativity is not in evidence. s106.5 1 15 2 2 2 ? ×= ? ′ + ′ =?= c v x c v t ttt BA ?? ? According to Lorentz’ transformation The effect of the relativity is in evidence. § 25.4 The Lorentz transformation equations Example 2: Two events happened simultaneously at two points of distance 1000 m on x axis in an inertial reference frame K. The observer in another inertial reference frame K’ which was moving along the x axis measured the distance of the two events is 2000 m. Find the time interval of the two events in K’ frame. § 25.4 The Lorentz transformation equations Solution: ()tvxx ??γ? ?= ′ ? ? ? ? ? ? ???= ′ ? x c v tt 2 γ According to Lorentz’ transformation tv ′ →→ ?γAnalysis: 29 then )s(1077.5 103 1000 2 3 2 )( 6 8 22 12 ? ×?= × ××?= ?=?= ′ ? ′ = ′ c xv x c v tttt ?γ ??γ? cv c v 2 3 2 )(1 1 from =? ? =γ What does the “-” means? One can get 2= ′ = x x ? ? γ m2000,0,m1000 = ′ == xtx ??? from § 25.4 The Lorentz transformation equations Example 3: An Earth starship has been sent to check an Earth outpost on the planet P1407, whose moon houses a battle group of the often hostile Reptulians. As the ship follows a straight line course first past the planet and then past the moon, it detects a high energy microwave burst at the Reptulian moon base and then, 1.10s later, an explosion at the Earth outpost, which is 4.00×10 8 m from the Reptulian base as measured from the ship’s reference frame. The Reptulians have obviously attacked the Earth outpost, so the starship begins to prepare for a confrontation with them. § 25.4 The Lorentz transformation equations 30 (a)The speed of the ship relative to the planet and its moon is 0.980c. What are the distance and time interval between the burst and the explosion as measured in the planet –moon inertial frame? Solution: Event 1: burst; event 2: explosion S frame: starship; S’ frame: planet-moon § 25.4 The Lorentz transformation equations s10.1 m1000.4 e 8 =?= ×=?= b be ttt xxx ? ? According to the transformation equation )( )( 2 x c v tt tvxx ??γ? ??γ? ?= ′ ?= ′ 0252.5 )980.0(1 1 1 1 222 = ? = ? = cv γ s04.1 m1086.3 8 ?= ′ ×= ′ t x ? ? § 25.4 The Lorentz transformation equations 31 (b)What is the meaning of the minus sign for ?t’? bb ttttt ′ < ′ ??= ′ ? ′ = ′ ee s04.1? It means that the burst occurred 1.04 s after the explosion in the moon-planet reference frame. (c)Did the burst cause the explosion, or vice versa? If there is a causal relationship between the two events, information must travel from the location of one event to the location of the other to cause by it. The required speed of the information in the ship frame is m/s1064.310.11000.4 88 info ×=×== txv ?? They are unrelated events! § 25.4 The Lorentz transformation equations Example 4: An observer S sees a big flash of light 1200 m from his position and a small flash of light 720 m closer to him directly in line with the big flash. He determines that the time interval between the flashes is 5.00 μs and the big flash occurs first. (a) what is the relative velocity of a second observer S’for whom these flashes occur at the same place in the S’ reference frame? (b) what time interval between them does S’ measure?(c) From the point view of S’, which flash occur first? v r § 25.4 The Lorentz transformation equations 32 Solution: (a) Event 1: big flash; Event 2: small flash m7201200)7201200( 12 ?=??=?= xxx? )m/s(48.01044.1/ 0)( 0 8 ctxv tvxx x ?=×?== =?= ′ ∴ = ′ ?? ??γ? ?Q m1200 m720 x x ′ O S S′ big small O′ v r § 25.4 The Lorentz transformation equations (b) what time interval between them does S’ measure? 0s1039.4 (0.48)-1 )10998.2( )720(1044.1 1000.5 )( 6 2 28 8 6 2 12 >×= × ?××? ?× = ?= ′ ? ′ = ′ ? ? c xv tttt ? ?γ? (c) the time interval in the s’ frame is positive The order of the flashes is the same in the S’ frame as it is in the S frame. § 25.4 The Lorentz transformation equations 33 Example 5: Two galaxies are speeding away from the Earth along a line in opposite directions, each with a speed of 0.75c with respect to the planet. At what speed are they moving apart with respect to each other? Solution: Earth ? x u ′ v S S’ O O’ c cc c uv vu u x x x 96.0 )75.0(1 75.075.0 1 2 2 = + + = ′ + + ′ = § 25.4 The Lorentz transformation equations Example 6: An armada of spaceships that is 1.00 ly long(in its frame) moves with speed 0.800c relative to ground station S. A messenger travels from the rear of the armada to the front with a speed of 0.950c relative to S. How long does the trip take as measured (a) in the messenger’s rest frame, (b) in the armada’s rest frame, and (c) by an observer in frame S? Solution: (a) in the messenger’s rest frame S m ’ c cc cvu vu u mx mx mx 625.0 95.080.01 95.080.0 1 2 ?= ×? ? = ? ? =′ The velocity of the armada is § 25.4 The Lorentz transformation equations 34 The length of the armada as measured in S’is )ly(781.0)625.0(1)ly(0.1 2 0 =?== m m l l γ The length of the trip is )y(25.1 625.0 )ly(781.0 == ′ = ′ cu l t mx m ? (b)in armada’s frame S a ’, the velocity of the messenger is c cc cvu vu u aax aax ax 625.0 80.095.01 80.095.0 1 2 = ×? ? = ? ? = ′ § 25.4 The Lorentz transformation equations The length of the trip is )y(6.1 625.0 )ly(0.1 0 == ′ = ′ cu l t ax a ? (c) in system S, the length of the armada is )ly(60.0)80.0(1)ly(0.1 2 0 =?== s s l l γ The length of the trip is )y(0.4 80.095.0 )ly(60.0 = ? = ′ ? ′ = ′ ccuu l t axmx s S ? § 25.4 The Lorentz transformation equations 35 § 25.5 The Doppler effect 1. The longitudinal Doppler effect How the frequency of a light source is affected by the relative motion of the source and observer? Consider a firefly, the proper frequency is 0 0 1 τ ν = Event 0: 0 0 0 0 = ′ = ′ t x Event 2: 02 2 2 0 τ= ′ = ′ t x Event 1: 01 1 0 τ= ′ = ′ t x In frame S’(firefly) Event 0: Event 1: Event 2: 0 0 0 0 = = t x 02 02 2 2 γτ τγ = = t vx 01 01 γτ τγ = = t vx In frame S(ground) According to the transformation equation )( )( 2 x c v tt tvxx ′ + ′ = ′ + ′ = γ γ The time interval between the zero flash and the first flash seen by your eyes at origin O, when the firefly is receding from you § 25.5 The Doppler effect 36 The time interval between the first flash and the second flash c v t 0 0 222 τγ γττ? +== The frequency of the flashes seen at the origin )1()1( 11 0 0 0 0 c v c v c v + = + = + = γ ν γτ τγ γτ ν The receding source is given by c v 0 0 τγ γττ += § 25.5 The Doppler effect c v c v c v c v + ? = + ? = 1 1 )1( 1 0 2 2 0 recede νν ν The frequency seen by your eyes at origin O, when the firefly is approaching from you c v c v ? + = 1 1 0 appro ν ν § 25.5 The Doppler effect 37 2. The transverse Doppler effect If the firefly is moving transverse to the line of sight, The proper period of the firefly is τ 0 , the proper frequency in its frame S’ is ν 0 and is very large. The frequency of the firefly in frame S is 2 2 0 0 1 11 c v ?=== ν γττ ν 1The transverse Doppler effect is always a red shift; 2 this effect has no classical analog; 3 it is always smaller than longitudinal Doppler effect. § 25.5 The Doppler effect § 25.6 Relativistic dynamics 1. Relativistic momentum Implications of relativity for the dynamics of a single particle, the effect of forces, the work done by these forces, and the consequent changes in momentum and kinetic energy of the particle. ump rr = class Classical momentum If we accept the postulates of the special theory of relativity, the classical definition of the momentum cannot be valid at high speeds. Because it leads that the law of momentum conservation is violate. Then ?=p r 38 O O′ x′ x y′ S′ y S m m ju ′? ? 0 v r ju ? 0 S frame: according to the definition of the classical momentum, consider y-axis direction jmu ? 0 j u m ? 0 γ ? Before collision jmu ? 0 ? j u m ? 0 γ After collision Total momentum is not conserved! § 25.6 Relativistic dynamics We define the relativistic momentum of a particle 2 2 1 1 c u ump ? =≡ γγ rr m is called rest mass, u is the speed of the particle. rest mass Now we can verify that the total momentum is conservative before and after collision of the two particles. We consider only the direction of y-axis. § 25.6 Relativistic dynamics 39 S frame: according to new definition of the momentum Before collision: ju c u m jmup ? 1 ? 0 2 2 0 01 ? ==γ r For m in S frame For m in S’ frame 2 2 0 0 1, c v u u uvu yx ??=?== γ [] ? ? ? ? ? ? ? ? ?? +? = 2 2 0 21 222 2 1 )(1 1 c v u cuu mp yx § 25.6 Relativistic dynamics 1 2 2 0 0 2 1 p c u u mp ?= ? ?= Substitution of u x and u y , after collision: following the similar process ju c u m p ? 1 0 2 2 0 1 ? ?= r For m in S frame For m in S’ frame 1 2 2 0 0 2 1 p c u u mp ?= ? = § 25.6 Relativistic dynamics 40 Conclusion: 1The total y-component of the momentum of the two particles before and after the collision is zero. 2when u<<c, γ→1, the relativistic momentum reduce to the classical expression. 3when u →c, γ→∞, γm →∞, t u m t um t um t p F d d d d d )d( d d r r rr r γ γγ +=== For constant force, we can not accelerate a particle infinitely. § 25.6 Relativistic dynamics 2. The CWE theorem revisited We shall require that the relativistic total energy E satisfy two conditions: 1the total energy E of any isolated system is conserved. 2W will approach the classical value when u/c approaches zero. t vm t p F d )(d d d rr r γ == The force in relativity § 25.6 Relativistic dynamics 41 The kinetic energy KE is the work done by the net force in accelerating a particle from rest to some velocity. For one dimension ∫∫∫ === vvv mvvx t mv xFKE 000 )(dd d )(d d γ γ 2 2 2 21 2 2 2 2 1 )2()1( 2 1 1 d )(d d d c v c v c v mv c v m v mv v p ? ? ? ? ? ? ? ???? = = ? γ § 25.6 Relativistic dynamics 23 2 2 )1( d )(d d d c v m v mv v p ? == γ 22 2 2 2 0 23 2 2 0 )1 1 1 ( d)1()(d mcmc c v mc vv c v mmvvKE vv ?=? ? = ?== ∫∫ ? γ γ Expanding γ by the binomial theorem yields L++≈?= ? 2 2 21 2 2 2 1 1)1( c v c v γ Rest energy E 0 § 25.6 Relativistic dynamics 42 1When v<<c, the expression for the kinetic energy reduce to the classical expression. class 2 2 2 2 2 22 2 1 ) 2 1 1( KEmv mc c v mc mcmcKE == ?+≈ ?=γ Conclusion: 2the relativistic total energy E is then defined as the sum of the kinetic energy and the rest energy. 0 22 EKEmcKEmcE +=+==γ § 25.6 Relativistic dynamics 3the speed of light is an unreachable upper bound on the speed of a particle (with nonzero mass) in special relativity. If we want v→c, we need to do an infinite amount of work on the particle to accomplish this feat! § 25.6 Relativistic dynamics 43 3. The relation of relativistic momentum and the total relativistic energy 2 and, mcEmvp γγ == from We find E c v p 2 = from )1( 2 2 422 2222422222 c v cm vmccmpcE ?= ?=? γ γγ We find 2 0 42222 EcmpcE ==? § 25.6 Relativistic dynamics E 2 mc pc Example 1: A particular object is observed to move through the laboratory at high speed. Its total energy and the components of its momentum are measured by lab workers to be E=4.51017 J, p x =3.8×10 8 kg · m/s, p y =3.0 × 10 8 kg · m/s, and p z =2.0 × 10 8 kg · m/s. What is the object’s rest mass? Solution: 42222 cmpcE =? [ ] kg6.4 )1074.1( 1074.1 )100.3()100.3()108.3()105.4( 2 2135 35 228282821742 = × = ×= ×+×+×?×= c m ccm § 25.6 Relativistic dynamics 44 Example 2: A proton is moving at speed v=0.900c. a. find its total energy E; b. Find its kinetic energy; c. Determine the magnitude p of its relativistic momentum. Solution: a. The total energy MeV102.15J1044.3 )100.3(1067.129.2 9.01 1 1 1 310 2827 2 2 2 2 2 2 ×=×= ××××= ? = ? == ? ? mcmc c v mcE γ § 25.6 Relativistic dynamics b. Find its kinetic energy MeV1021.1 J1094.1 )100.3)(1067.1)(129.2( )1( 3 10 2827 2 ×= ×= ××?= ?= ? ? mcKE γ c. Determine the magnitude p of its relativistic momentum m/skg1003.1 )1039.0)(1067.1(29.2 18 827 2 ?×= ×××= == ? ? E c v mvp γ § 25.6 Relativistic dynamics 45 Example 3: A proton of mass 938.3 MeV/c 2 is accelerated across a potential difference of 202.2 MeV. Determine its total energy (in MeV) and momentum (in MeV/c). What is the speed of the particle? Solution: MeV 1140 MeV0.202MeV3.938 2 = += += KEmcE ccEEp MeV/5.647/ 2 0 2 =?= ccEEcv cvEEE 5683.011403.98311 1/ 2222 0 22 00 =?=?= ?==γ § 25.6 Relativistic dynamics 4. Implications of the equivalence between mass and energy 22 mcKEmcE +==γ According to 1 A particle can only have kinetic energy KE; 2 A particle can only have rest energy mc 2 ; 3 A particle can have both kinetic energy KE and rest energy mc 2 ; 222 mcmcmcKE ?γ =?= It is possible to convert an isolated system of particles with mass to a system of particle with less mass, even zero mass, and –remarkably – vice versa. § 25.6 Relativistic dynamics 46 Example 4: There are several fusion reactions that convert mass directly into energy and can power stars and drive hydrogen bombs. One such process fuses two nuclei of heavy hydrogen(deuterium) together, resulting in a still heavier hydrogen(tritium) nucleus, an ordinary hydrogen nucleus(proton), and the KE they fly off with. It is customary to write such a reaction in terms of the neutral atoms involved(neglecting the tiny amounts of energy holding the electrons to each atom, ≈10 eV). thus KE++→+ HHHH 1 1 3 1 2 1 2 1 Determine the energy liberated per fusion. § 25.6 Relativistic dynamics Solution: The mass of Deuterium is 1876.12 MeV; The mass of Tritium is 2809.43 MeV ; The mass of Hydrogen is 938.783 MeV. KEcmcmcmcm HTDD ++=+ 2222 J1045.6MeV03.4 MeV783.938MeV41.2809 MeV12.1876MeV12.187 13? ×== ++= + KE KE § 25.6 Relativistic dynamics 47 “mass is equivalent to energy” “mass is converted into energy” “energy is converted into mass” § 25.6 Relativistic dynamics