Presentation on theme: "Chapter 20 The special theory of relativity"— Presentation transcript:
1 Chapter 20 The special theory of relativity Albert Einstein ( 1879 ~ 1955 )
2 20-1 Troubles with classical physics The kinematics developed by Galileo and themechanics developed by Newton, which form thebasis of what we call “classical physics”, hadmany triumphs. However, a number ofexperimental phenomena can not be understoodwith these otherwise successful classical theories.Troubles with our ideas about timeThe pions ( or ) created at rest are observed
3 to decay ( to other particles ) with an average lifetime of only . In one particular experiment, pions were created in motion at a speed of In this case they were observed to travel in the laboratory an average distance ofbefore decaying, from which we conclude that they decay in a time given by , much larger than the lifetime measured for pions at rest.This effect, called “time dilation”, which cannot be explained by Newtonian physics. In Newtonian physics time is a universal coordinate having identical values for all observers.
4 2. Trouble with our ideas about length Suppose an observer in the above laboratory placed one marker at the location of the pion’s formation and another at the location of its decay.The distance between the markers is measured to be 17.4m. Now consider the observer who is traveling along with the pion at a speed of u=0.913c. This observer, to whom the pion appear to be at rest, measures its lifetime to be 26.0ns, and the distance between the markers isThus two observers measure different value for thesame length interval.3. Troubles with our ideas about light
5 20-2 The postulates of special relativity 1. Einstein offered two postulates that form thebasis of his special theory of relativity.(I) The principle of relativity: “The laws of physicsare the same in all inertial reference frames.”(II) The principle of the constancy of the speed oflight : “ The speed of light in free space has thesame value c in all inertial reference frames.”2. The first postulate declares that the laws ofphysics are absolute, universal, and same for allinertial observers.
6 The Second postulate is much more difficult to accept, because it violates our “ common sense”,which is firmly grounded in the Galilean kinematicsthat we have learned from everyday experiences.It implies that “it is impossible to accelerate aparticle to a speed greater than c”.
7 20-3 Consequences of Einstein’s postulates 1.The relativity of timeWe consider two observers: S is at rest on the ground, and S’ is in a train moving on a long straight track at constant speed u relative to S.The observers carry identical timing devices, illustrated in Fig 20-4, consisting of a flashing light bulb F attached to a detector D and separated by a distance from a mirror M.The bulb emits a flash of light that travels to the mirror, when the reflected light returns to D, the clock ticks and another flash is triggered.
8 MThe time interval between ticks is:(20-1)The interval is observed byeither S or S’ when theclock is at rest respect to thatobserver.FDFig 20-4
9 We now consider the situation when one observer looks at a clock carried by the other. Fig20-5 shows that S observes on the clock carriedby S’ on the moving train.BCAFig 20-5LLFDFDFDS
10 According to S, the flash is emitted at A, reflected at B, and detected at C.This interval is(20-20)Substituting for from Eq(20-1) and solvingEq(20-2) for gives(20-3)
11 The time interval measured by the observer (S’) relative to whom the clock is at rest is called the“proper time(正确时间) ”, andThat is, the observer relative to whom the clock is in motion measures a greater interval between ticks. This effect is called “time dilation”. All observer in motion relative to the clock measure “longer intervals”.Eq(20-3) is valid for any direction of the relative motion of S and S’.
12 2. The relativity of length Fig 20-6 shows the sequence of events asobserved by S for the moving clock which ison the train sideway, so that the light now travels along the direction of motion of the train.According to S the length of the clock is L, whichis different from the length measured by S’,relative to whom the clock is at rest.
14 In the process from A, B to C, the total time taken is (20-6)From Eq(20-3), setting(20-7)Setting Eqs(20-6) and (20-7) equal to one anotherand solving, we obtain(20-8)Eq(20-8) summarizes the effect known as “lengthcontraction”.
15 (a) The length measured by an observer who is at rest with respect to the object being measured iscalled the “rest length” or “proper length”.(b) All observers in motion relative to S’ measure ashorter length, but only for dimensions along thedirection of motion; length measurement transverse to the direction of motion are unaffected.(c) Under ordinary circumstances, and theeffects of length contraction are too small to beobserved.
16 3 The relativistic addition of velocities Let us now modify our timing device, as shown in Fig20-7. The flashing bulb F is moved to the mirror end and is replaced by a device P that emits particles at a speed , as measured by an observer at rest with respect to the device.particlePlightFD
17 The time interval measured by an observer (such as S’) who is at rest with respect to the device is: (20-9)What’s the velocity of the particles, measured by the observer S on the ground?(20-12)Eq(20-12) gives one form of the velocity addition law.
18 (a) According to Galileo and Newton, a projectile fired forward at speed in a train that is movingat speed u should have a speed relative to an observer on the ground.This clearly permits speeds in excess of c to be realized.
19 (b) The Eq(20-12) prevents the relative speed from ever exceeding c. IfThus, Eq(20-12) is consistent with Einstein’ssecond postulate
20 Sample Problem 20-2A spaceship is moving away from the Earth at a speed of 0.80c when it fires a missile parallel to the direction of motion of the ship. The missile moves at a peed of 0.60c relative to the ship. What would be the speed of the missile as measured by an observer on the Earth?Compare with the predictions of Galilean kinematics.
21 20-4 The Lorentz transformation We wish to calculate the coordinates x’ , y’ ,Z’, t’ of an event as observed by S’ from thecoordinates of x, y, z, t of the same eventaccording to S.We simplify this problem somewhat, withoutlosing generality by always choosing the xand x’ axes to be along the direction of .Namely the velocity of O’ is , respective to O.
22 The Lorentz transformation equations are (20-14)Where the factor is(20-15)
23 It is convenient in relativity equation to introduce the speed parameter , defined as(20-16)The inverse Lorentz transformation:–u u(20-17)
24 Sample problem 20-3In inertial frame S, a red light and a blue light areseparated by a distance , with the redlight at the larger value of x. The blue light flashes,and later the red light flashes. Frame S’ ismoving in the direction of increasing x with aspeed of What is the distance betweenthe two flashes and the time between them asmeasured in S’ ?
25 The velocity in x direction of O’ is u, respective to O. Table 20-2The velocity in x direction of O’ is u, respective to O.Lorentz transformationInverse transformationInterval transformationInverse Interval transformation
27 20-5* Measuring the space-time coordinates of an event
28 20-6 The transformation of velocities is the velocity of a particle in S referenceis the velocity of S’ reference relative to S in x directionis the velocity of the particle in S’ reference, then
29 (20-18)In similar fashion,(20-19)Eqs(20-18) and (20-19) give the Lorentz velocitytransformation.
30 1 We now show directly that Lorentz velocity transformation gives the result demanded byEinstein’s Second postulate ( the constancy of thespeed of light ).Suppose that the common event being observedby S and S’ is the passage of a light beam alongthe x direction. Observer S measures and
31 .Using Eqs (20-18) and (20-19)Thus the speed of light is indeed the same for allobservers or all frames.
32 2. When ( or equivalently, when ), Eqs(20-18) and (20-19) reduce toand (20-20)which are the Galilean results.
33 Sample Problem 20-4A particle is accelerated from rest in the lab until its velocity is 0.60c. As viewed from a frame that is moving with the particle at a speed of 0.60c relative to the laboratory, the particle is then given an additional increment of velocity amounting to 0.60c. Find the final velocity of the particle as measured in the lab frame.
34 20-7 Consequences of the Lorentz transformation 1.The relativity of timeFig20-15 shows a different view of the time dilationeffect.(a)S……(b)……Fig 20-15
35 (20-21), the is a proper time ( )Note that: the time dilation effect is completelysymmetric. If a clock C at rest in S is observed byS’, then S’ concludes that clock C is running slow.
36 (a) The relativity of simultaneity Suppose S’ has two clocks at rest, located atand A flash of light emitted from a pointmidway between the clocks reaches the two clocksIf and , then
37 * * According to S’ ( see Fig 20-16a), . According to S, the light signal reaches clock 1 before it reaches clock 2, and thus the arrival of the light signals at the locations of the two clocks is not simultaneous to S.1212cc**(a)S(b)SFig 20-16(b) The twin paradox2. The relativity of length
38 Sample problem 20-5 An observer S is standing on a platform of length on a space station. A rocket passes at arelative speed of 0.8c moving parallel to the edgeof the platform. The observer S notes that the frontand back of the rocket simultaneously line up withthe ends of the platform at a particular instant (Fig20-29a)According to S, what is the time necessary forthe rocket to pass a particular point on the platform?
39 (b) What is the rest length of the rocket? (c) According to S’ on the rocket what is the length D of the platform?(d) According to S’, how long does it take for observer S to pass the entire length of the rocket?(e) According to S, the ends of the rocket simultaneously line up with the ends of the platform. Are these events simultaneous to S’?
41 (d)(e) According to S’, the rocket has a rest length ofand the platform has a contracted length of From Fig 20-19b and 20-19c, the time intervalor
42 20-8 Relativistic momentum Here we discuss the relativistic view of linearmomentum. Consider the collision shown in Fig20-20a, viewed from the S frame. Two particles,each of mass m, move with equal and oppositevelocity v and –v along the x axis. They collide atthe origin, and the distance between their lines ofapproach has been adjusted so that after thecollision the particles move along the y axis withequal and opposite final velocities (Fig 20-20b).
43 (a) (b) Fig 20-20 (c) (d) y S S Frame S S’ Frame S’ S’ 1 1 2 2 Before collisionAfter the collisionvS’1Frame S’12S’2vFig 20-20(c)(d)
44 The collision to be perfectly elastic, in the S frame Initial :Final :The momentum is conserved in the S frame.According to S’ which moves relative to the Sframe with speed (Fig 20-20c) , particle 2is at rest before the collision. Using Eqs (20-18)and (20-19) we can find the transformed x’ and y’component of the initial and final velocities.
45 Thus in the S’ frame:,momentum is not conserved. Therefore,if we are to retain the conservation of momentum
46 as a general law consistent with Einstein’s first postulate, we must find a new definition ofmomentum, that is(20-23)In terms of components,and (20-24)There the speed v in the denominator of theseexpressions is always the speed of the particle as
47 measured in particular inertial frame. It is not the speed of an inertial frame.This new definition restores conservation ofmomentum in the collision. In the S frame, thevelocities before and after are equal and opposite,and thus Eq(20-23) again gives zero for the initialand final momenta. In the S’ frame,(20-25)
48 20-9 Relativistic energy Using the velocity shown in Figs 20-20c and 20-20d, you can show that, with , the totalinitial and final kinetic energies are(20-26)Thus is not equal to According to SThis situation violates the relativity postulate,we require a new definition of kinetic energy if we
49 are to preserve the law of conservation of energy and the relativity postulate.(b) The classical expression for kinetic energy alsoviolates the Second relativity postulate by allowingspeed in excess of the speed of light. These is nolimit ( in either classical or relativistic dynamic ) tothe energy we can give to a particle.(c) Relativistic kinetic energy is(20-27)
50 Using Eq(20-27), we can show that kinetic energy is conserved in the S’ frame of the collision of Fig20-20.2. Energy and mass in special relativityWe can also express Eq(20-27) as(20-30)where the total relativistic energy E is defined as(20-31)
51 and rest energy(20-32)The rest energy can be regarded as the internalenergy of a particle at rest.According to Eq(20-32), whenever we addenergy to a object that remain at rest, weincrease its mass by an amountIf we compress a spring and increase its potentialenergy by an mount , then its mass increasesby
52 (b) The total relativistic energy must be conserved in any interaction. The sun radiates an energy ofevery second, and the correspondingchange in the mass is
53 Sample problem 20-8 Two 35g putty balls are thrown toward each other, each with a speed of 1.7m/s. The balls strike eachother head-on and stick together. By how muchdoes the mass of the combined ball differ from thesum of the masses of the two original balls?Solution:We treat the two putty balls as an isolated system.No external work is done on this isolated system.
54 With , whereWe haveThe corresponding increase in mass isSuch a tiny increase in mass is beyond our ability to measure.
55 3. Conservation of total relativistic energy Eq(20-30) can be written asManipulation of Eqs(20-23) and (20-31) gives auseful relationship among the total energy,momentum, and rest energy
56 Sample problem 20-10 A certain accelerator produces a beam of neutral Kaons ( ) with kinetic energy 325Mev. Consider a Kaon that decays in flight twopion( ) . Find the kinetic energy ofeach pion in the special case in which the pionstravel parallel or antiparallel to the direction of theKaon beam.Solution:From Eq(20-33), the initial total relativistic energyis
57 The initial momentum is The total energy of the final system consisting ofthe two pions is(20-35)Thus we have one equation in the two unknownsand , with conservation of momentum.