Solution Manual for Modern Physics, 2nd Edition

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INSTRUCTORSOLUTIONSMANUALMODERN PHYSICSS E C O N DE D I T I O NRANDY HARRISUniversity of California, Davis

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iiiContents2.Special Relativity13.Waves and Particles I: Electromagnetic Radiation Behaving as Particles314.Waves and Particles II: Matter Behaving as Waves415.Bound States: Simple Cases556.Unbound States: Obstacles, Tunneling and Particle-Wave Propagation767.Quantum Mechanics in Three Dimensions and the Hydrogen Atom928.Spin and Atomic Physics1159.Statistical Mechanics13210.Bonding: Molecules and Solids15111.Nuclear Physics16612.Fundamental Particles and Interactions178

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1CHAPTER 2Special Relativity2.1According to Anna, if clocks at the train’s ends sent light signals when they read noon, the signals would reachAnna together somewhat after noon. According to Bob outside, the signals must start at different times, so thatboth reach themovingAnna together. The clocks read noon at different times according to Bob.2.2She decides to turn on the lights simultaneously. According to her, light signals leave her brain together andreach her hands simultaneously, so her hands act simultaneously. According to Bob outside, the light signalheading toward her trailing hand will reach that hand first, for the hand moves toward the signal. This hand actsfirst. The signal heading toward her front hand has to catch up with that hand, taking more time and causing thathand to act later.2.3By symmetry, if an observer in S sees the origin of frame S´ moving at speedv, an observer in S´ must see theorigin of S moving at the same speed.2.4(a)Slower.(b)Later.(c)Ground observers must see my clock running slowly, so their clock at Y must be ahead. I don’t see clocksX and Y as synchronized, so when I pass by X, the clock at Y certainly doesnotread zero, so even thoughit does run slowly according to me, it might (must) nevertheless be ahead of mine when I pass it.2.5If it passes through in one frame, it must do so in all others. Moving parallel to the ground is an issue ofsimultaneity. If you are “on the disk,” the plate has bothx-andy-components of velocity and it will not be equaldistances from the ground at the same time—it will be slanted. If slanted, the disk can pass through it withoutcollision.2.6Physical laws arenotthe same if you are not in an inertial frame. If you are in an accelerating frame, you know it,no matter what others may be doing. Objects in your frame would appear to change velocity without cause. Thephysical laws are always obeyed for the observer in the inertial frame, but not for the twin who turns around.There is an asymmetry.2.7No, clocks run at different rates no matter how low the relative velocity, though considerable total travel timemight be required before a significant effect is noticed.2.8I am inertial, so I must see this moving clock running slowly. Each time the space alien passes in front of me Iwill see his clock getting further and further behind. Passing observers always agree on the readings of localclocks, so he will agree that his clock is getting further and further behind—that my clock is running faster thanhis. He is in a frame constantly accelerating toward me, so according to him, my clock is continuously “jumpingahead,” as in the Twin Paradox.2.9Sound involves a medium, and three speeds—wave, source, and observer—relative to that. Light has no medium.The only speeds are the relative velocity of the source and observer, and of course the speed of light.

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Chapter 2 Special Relativity22.10First, for a massive object,E = mc2is incomplete. If something is moving, it isE =umc2, which wouldinvalidate the argument. Second, some things have energy and no mass, so if any of these are generated, theenergy and mass of what remains could change.2.11No, for if we consider a frame in which the initial object is at rest, kinetic energy clearly increases, so masswould have to decrease.2.12No, for if we consider a frame in which the final object is at rest, kinetic energy clearly decreases, so mass wouldhave to increase.2.13Yes, of course. We could consider a frame in which it is either at rest or moving, giving different values for KE.No, for mass/internal energy is the same in any frame of reference, so the change in mass is the same, and so thecorresponding opposite change in kinetic energy must be the same.2.14Yes, if it loses internal energy—in whatever form it may be taken away—it loses mass.2.15No, the light was emitted in a frame that, relative to me, is moving away, so I see a longer wavelength (a veryslight change). The observer at the front waits longer to see the light, so a greater relative velocity is involved,and thus a larger red shift. Simply turn the bus up on end and let gravity accomplish the same thing.2.16Noelectrontravels from one side of the screen to the other. Nothing that can have any information about the left-hand-side of the screen moves to the right-hand-side. The events (electrons hitting phosphors) are all plannedahead of time, and have no effect upon one another.2.17v≥ 1. Classical mechanics applies whenv<<c, andv= 1. At what speed willvbe 1.01?2211−vc= 1.010.14c2.18Insertingx´=–vt´andx= 0 into (2-4) yields–vt´= 0 +Btandt´= 0 +Dt. Dividing these equations gives–v=B/D. Insertingx´=ct´andx=ctinto (2-4) yieldsct´=Act+Bt=At(c–v) ort´=At(1–v/c) andt´=Cct+Dt=t(Cc+A). Setting these two equations equal givesA(1–v/c) =Cc+AorC=–v/c2A.2.19If I am in frame S and say that the post in S´ is shorter, then my saw will not slice off any of the post in frame S´,but the saw in frame S´ will slice off some of the post in my frame S. An observer in frame S´, where the relativespeed(if not direction) is the same, would also have to say that the post in the “other frame,” frame S, is shorter.Therefore, the saw in frame S will slice off some of the post in frame S´. This is a contradiction. Thus, the postscannot be contracted in either frame. The same argument would work if we supposed that the post in the “otherframe” were elongated rather than contracted.2.20Your time is longer.tyou=vtCarl→60s =211(0.5)−tCarltCarl=52s2.21L=L0221−vc→35m =L021(0.6)−L0=43.75m2.22Let Anna be S, Bob S. It is simplest to lett=t= 0 be the time when Anna passes Earth.According to Bob, theexplosion event occurs atx= 5ly,t= 2yr.

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Instructor Solutions Manual for Harris,Modern Physics, Second Edition3According to Anna,x=v(x−v t) =()21(5ly)(0.8 )(2yr)1(0.8)−−c=5.67ly,andt=v2−+v xtc=2210.8(5ly)2yr1(0.8 )−+−ccc=−3.33yr.Negative!? How can this be? Togain some appreciation of the Lorentz transformation, let’s see how involved itis to solve the problem without it.According to Bob: The planet explodes att= 2yr. At this time, Anna will havemoved (0.8c) (2yr) = 1.6ly. Bob sees a distance between Anna and Planet Y of 3.4ly and a relative velocitybetween Anna and the light from the explosion of 1.8c. So the light from the explosion will reach Anna inanothert= 3.4ly/1.8c= 1.89yr—at the time, according to Bob, of 2 + 1.89 = 3.889yr. The distance he now seesbetween Anna and Planet Y (or the center of the debris) is 5ly−(0.8c) (3.889yr) = 1.889ly. Meanwhile he willhave seen less time go by on Anna’s clock:21(0.8)−3.889yr = 2.33yr.According to Anna: The distance PlanetY is from herself when she receives the bad news will be shorter:21(0.8)−1.889ly = 1.133ly. The bigquestion: If Planet Y moves toward Anna at 0.8c, and the light from the explosion moves toward Anna atc, howlong will it take for the light to get 1.133ly in front of the planet? The relative velocity between Planet Y and thelight from the explosion is 0.2c, and 0.2ct= 1.133lyt= 5.667yr. If Anna’s clock reads 2.33yr when shegets the bad news, and it took 5.667yr for the news to reach her, the explosion must have occurred att=−3.33yr.The problem is a good example of two events that occur in a different order in two frames of reference. Theexplosion (one event) occurs before Anna and Bob cross (another event) according to Anna, but after accordingto Bob.2.23Let Anna be S, Bob S. The flash of the flashbulb is the event in question here. We seek Bob’s time.We knowthat it is 100ns27ns, i.e., either 127ns or 73ns.2vvtxtc=+=2210.6100ns1(0.6 )c xcc +−=0.75cx+ 125ns.Ifxis positive (Anna’s arm is stretched out in the positive direction), then it must be 127ns.Later.We might be tempted to use time dilation to answer this, but care is necessary. There is no doubt that thewristwatch on Anna’s hand reads100nswhen the flash goes off. ButBobwill not agree that it reads zero whenAnna passed him. It is set back a bit according to Bob (like the front clocks in Example 2.4). Thus, Bob seesslightly more than 100ns go by on the wristwatch. Time dilation would then give the time Bob sees go by on hisown: slightly more than0.6(100ns) = 125ns.(b)Now knowing thattmust be 127ns, we can solve forx, the location in Anna’s frame where the flash occurs(i.e., her hand). 12710−9s =80.75310 m/sx+ 12510−9sx=0.8m. (A reasonable arm length.)2.24Bob, standing in the barn/S, sees a lengthL=221/−vcL. 10ft =221/(16ft)−vcv =0.78c.(b)The observer stationary in the barn.Pole−vaulter Anna, S, will never agree that the pole fits in all atonce; the barn is shorter than10ft!(c)Weseekt,knowingthat, accordingtoabarnobserver,theends ofthe pole areatthedoorssimultaneously,i.e.,t=0,andthatthedistancebetweentheseeventsis10ft=3.048m.2121212()()−=−−+−vvttxxttc=2210.78(3.08m)01(0.78 )−+−ccc=−1.2710−8s.

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Chapter 2 Special Relativity4Why negative? Since we chosex2−x1as positive, it must be that the front of the pole reaching its door isEvent 2, the back reaching its door Event 1. The answer shows that the front time is a smaller time; itoccurs earlier, as it must. According to the pole−vaulter/S,front leaves before back enters.2.250.8c=211(0.8)−=5 .3Bob sees Anna’s ship contracted to 100m/v= 100m/53= 60m, so Bob Jr. will have to beatx=60m.(b)We seekt, knowingx,x, andt.2 =−+vvtxtc=850.8(60m)03310 m/s−+=−2.6710−7s.2.26Calling the front light Event 2, Anna frame S,()()2121212−=−+−vvttxxttc=2(60m)0+vvc. Sincethis is positive, the front time is the larger (later), soback light must go on first.(b)4010−9s =2221(60m)1/−vcvc→()2291/(4010s)−−vc2c2= (v2/c2) (60m)2→(4010−9s)2(3108m/s)2=()22928(60m)(4010s) (310 m/s−+v2/c2v/c=0.1962.27Bob and Bob Jr. see a 25m−long plane.L=L0/v→25m = 40m221/−vcvc=0.781(b)If zero on Anna’s clock is thefrontof her spaceplane (where Bob is), then the time at the back occurs atx=−40m, as well as atx=−25m andt= 0.2 =−+vvtxtc=8210.781( 25m)0310 m/s1(0.781)−−+−=1.0410−7s.A positive number. Sensible. Back enters after front leaves.(c)We know thatt= 104ns (“atthistime”) andx= 0 (Bob’s location), and seekx.x=v(x+v t)→()87210(0.781310 m/s)(1.0410s)1(0.781)−= +−xx=−24.375m. The front of her ship is theorigin, so24.375msticks out. Note: “at this time”, observers at the back (x=−40m) and atx=−24.375m,i.e., 15.625m apart in Anna’s frame, see the ends of Bob’s hangar. This is correct: Anna should see a hangarlength ofL= 25m/v= 25m21(0.781)−= 15.625m.

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Instructor Solutions Manual for Harris,Modern Physics, Second Edition52.28Since Bob moves to the right, let’s make his frame S. Thus, we seek times in frame S.v= 2→2211/−vc= 232=vc. The clockat the back of Anna’s ship is presently atx=12L0. (It is also atx=L0.) Bob looks at thisclock att= 0 on his own clock. Call this Event a. Withxa=12L0,ta= 0, we may findta, the time in Anna’sframe.2=+avaavtxtc03 2 1202=+Lc=32cL0.The clock at the center of Anna’s ship is atx=14L0(rather than12L0) in Bob’s frame. The time on thisclock,tb, will therefore simply be half the previous result.tb=34cL0.2.29If it takes light 40ns to travel half the length of Bob’s ship, then:12L0= (3108m/s)(4010−9s)L0=24m.(b)The origins are atthe center. We know that the event atx= 12m (i.e., the front of Anna’s ship) andt=−24ns occurs at t = 0.2= + vvtxtc→920(12m)( 2410s)−=+ −vvcvc=0.6c.(c)0.8c=21541(0.8)=−. Logically, at the initial instant, if the front clock says−24ns, the back has to say+24ns. Now consider Anna’s origin at t = 20ns.2= + vvtxtc→2010−9s =850.604310/−+ tm st= 1610−9s. Agrees. Another route: If 20ns passes for Bob, he must see less time pass on the clock’s inAnna’s frame. Sincev=5140.8=, he must see 80% as much time pass, and 80% of 20ns is 16ns. Similarly,Bob must see 16ns pass on all of Anna’s clocksas diagram shows. Att= 40ns, Bob mustsee that 80% of40ns, 32ns, has passed on Anna’s clocksas shown. Finally, 80ns for Bob must show 80% of 80ns, or 64ns,pass on Anna’s clocks.2.30It is tricky to use time dilation to relate readings on twodifferentclocks moving relative to you. You would neveragree that the gas station clocks were synchronized. However, gas station observers (S) watch asingleclock(yours, S) and must observe time dilation. When you get to the second station, observers there must see yourclock as having registered less time than in their frame:the gas station clock must be ahead.

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Chapter 2 Special Relativity6(b)In the frame of the gas stations,t= 900m/20m/s = 45s.t=vt→45s =821(20 / 310 ) −tt=()18221(20 / 310 )(45s)−8211(20 / 310 )(45s)2−= 45s−10−13s.Yours is behind by10−13s. Anotherroute: Time zero is you (S) passing first station. What would an observer in your frame and abreast thesecond station att= 0 (it’s a long bus) see on the second station’s clock?x= 900m,t= 0t= ?20=−+vv xtc→8220m/s0(900m)(310 m/s)=−+vtt= 210−13s. According to you, then, thesecond station’s clock is ahead of the first’s by 210−13s. Now, you do see less time pass on those clocksthan on your own—10−13s less—but this will still leave the second station’s clockaheadof yours by 10−13swhen you get there. Yet another route: Whattwill your clock read when you pass the second station, atx=900m,t= 45s?2 =−+vvtxtc=8282120m/s(900m)45s(310 m/s)1(20 / 310 )−+−()821311(20 / 310 )210s45s2−+− += 45s−210−13s + 110−13s−210−26s45s−110−13s, i.e., 10−13s behind.2.31According to those on the ground, this is simple classical mechanics: a speed and distance according to a groundobserver are used to find a time according to a ground observer. Let’s call the ground frame S.t=68410 m0.8310 m/s= 1.6710−2s. There are two ways to find the time on the spaceplane—time dilation, or lengthcontraction.Time dilation: The ground observer will see more time pass on his clock than on the plane. His ist,the plane’st. (Note:0.8c= 1.67)t=vt→1.6710−2s = (1.67)tt= 0.0100s.Lengthcontraction:Theplaneobserverseesadifferentcoast−to−coastdistance.HeseesL=261(0.8) (410 m)−= 2.4106m. How much time will it take a 2,400m object to pass at 0.8c? Classicalmechanics:t=682.410 m0.8310 m/s= 0.0100s. The clock seen on the plane from the ground will be 0.0167s−0.0100s =0.0067s behind. But how can it be behind if the people on the plane must see time ongroundclockspassing more slowly? Should they not see thegroundclocks behind? Assuming the plane and ground clocks readzero as the plane started across the country, the plane clock will indeed be behind the ground clock on the othercoast, just as we calculated. The people on the plane willby no meansagree that the clock on thefarcoast wassynchronized, that it read zero when the plane started across country (relative simultaneity). In fact, they will saythat it was significantly ahead, so that, even though they “see” less time pass on this far−off clock, they will agreethat it is ahead when they pass it.2.32Passengers see a 5km-long object shrunk toL=LRest/v=25km 1(0.8)−= 3km. If the ends of this object are atthe train’s ends simultaneously (according to observers on the train), the train must be3kmlong.(b)Look at it from the ground’s point of view, the train is not 5km long; it’s not even 3km long, butL=LRest/v=23km 1(0.8)−= 1.8km. Surely the back end will pass its station before the front end passes its, so thefront station is the more current information.Noreason to slow down.

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Instructor Solutions Manual for Harris,Modern Physics, Second Edition7(c)From the ground point of view, the train travels 5km–1.8km = 3.2km from the time the employee at theback sees his sign to the time the conductor sees his: t = dist/speed = 3.2103m / (0.83108m/s) =13.3μs.2.33According to the muons, the distance from the mountaintop to sea level is21910m 1(0.9952)187m.−=According to a muon, this distance would pass in78187m6.2610s0.626 s0.9952310 m/s−==.0.626 / 2.200.75.−==NeNAnd3950.75.527=2.34We may “work in either frame”.Muon’s frame:= 2.2s. Distance to Earth is shorter: 4km21(0.93)−= 1.47km.t=8dist1, 470mspeed0.93310 m/s== 5.2710−6s.0NN= e−(5.27/2.2)= e−2.4or9.1%.Earth frame: Distance to Earth is 4km. Lifetime is longer.22.21(0.93)−s= 5.99s.t=8dist4, 000mspeed0.93310 m/s== 1.4310−5s.0NN= e−(14.3/5.99)= e−2.4or 9.1%(b)= 2.2s.t=8dist4, 000mspeed0.93310 m/s== 1.4310−5s.0NN= e−(14.3/2.2)= e−6.5or0.14%2.35According to observers on the plane, these two events occur at the same location. Their times differ byt0.0221/ =→−ttvc()128201(420 / 310 )(10s)=−t()82121(420 / 310 )(10s)−= 10s−9.810−12s.The plane’s clock will read9.8ps earlier.May also solve in plane frame, where 4.2 km is contracted.2.36L=221/−vcL→50m−0.110−9m =221/(50m)−vc. Must use the approximation, sincev/cis apparentlysmall.()()12221+ −vc1−2212vc.50m−0.110−9m =2211(50m)2−vc0.110−9m =2212vc(50m)v= 210−6c=600m/s.2.37There are two equivalent routes.First: Anna doesn’t see a distance of 20ly, but rather a length−contracted one of20ly/, and it passes in twenty years of her life, so its speed isv=20ly /20yrv=vc. Solving,vc=221−vcor22vc=1−22vcorv=12c.

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Chapter 2 Special Relativity8Second: From Bob’s perspective, Planet Y is 20ly away. But Bob, seeing Anna age twenty years, must, owing totime−dilation, see more time go by on his own clock:v20yr. Anna’s speed is thusv=20ly20yrv=vc. Sameequation.2.38The plank has moved to a different frame, frame S. The length along thex−axis isLx=Locos0, and along they−axis it isLy=Losino. An observer in S will see thex−leg contracted, but not they.Lx=Lx/v=Locoso221/,−vcandLy=Ly=Losino.L=22+xyLL=()22222200cos1sin.−+ooLvcLBut sin2o= 1−cos2o.L=Lo()()2222cos11cos−+−oovc=Lo2221()cos.−ovcAs the angleogoes from zero to 90º, thelength of the plank to an observer in S goes fromLo221−vc(simple length contraction) to merelyLo(nocontraction perpendicular to axes of relative motion). Now, tan=yxLL=22sincos1−ooooLLvc= tano2211−vc=tano. As the speed increases, the angleseen by an observer in S increases. This makes sense in view ofthe fact that the faster the relative motion, the shorter thex−component of the plank will be. They−componentdoesn’t change, so the angle is larger.2.39If it can pass through according to one observerit mustbe able to pass through according to the other! Therecannot be a collision in one frame and not in the other! It’s just a matter ofwhen. One argument is that itsdimensions perpendicular to the direction of relative motion don’t change—its height is the same in either frame,so it cannot hit. Another argument is that while Bob sees a plank and doorway at the same angle, Anna, at restwith respect to the plank, sees a plank at a smaller angle,0. She also sees a wall moving toward her, contractedalong the direction of relative motion and therefore at a larger angle than theseen by Bob. The top will passthrough first! Let’s say that the bottom passes through at time zero on both clocks. According to Bob, the top willalso (simultaneously) pass through at time zero:t= 0. It can’t according to Anna! The top passing through (anevent) occurs att= 0 and atx=L0cos0. Findt.2=+vtxtc→0020cos=+ vv Ltct=−2vcL0cos0. According to Anna, in frame S, thetop passes through2vcL0cos0earlier. This adds up! We know thatAnna sees a smaller angle than Bob. If it passes through at once according to Bob, the top must pass through firstaccording to Anna. If it does not go through all at once according to Anna, it can indeed be larger than the “hole”.

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Instructor Solutions Manual for Harris,Modern Physics, Second Edition92.40We have a speed and time according to the lab and wish to know a distance according to that frame.distance = (0.943108m/s)(0.03210−6s) =9.02m.(b)If the experimenter sees 0.032s pass on his own clock, he will see less pass on the clock glued to theparticle.t=21(0.94) −t→0.032s =21(0.94) −tt=0.011s.(c)Length contraction. According to the particle, the lab is21(0.94) (9.02m)−=3.08mlong. Let’s see, if3.08m of lab passes by in 0.011s, how fast is the lab moving?63.08m0.01110s−= .94c. It all fits!2.41According to Earth observer, muon “lives” longer.t=0221/−tvc=211(0.92)−2.2s = 2.55(2.2s) = 5.61s.Dist = speedtime = (0.923108m/s)(5.6110−6s) =1,549m.2.42According to an observer in the lab, the pion survivest=2226ns.1−vcButmoving a distance of 13m at speedv,this time must also bet=13mv. Thus:2226ns1/−vc=13mv→v82.610s13m−=221/−vc→8510 m/sv=221/−vc→282(510 )v= 1−282(310 )vv=2.57108m/s.2.43Calling to the right positive and Anna frame S, we know that this tick of the clock/event occurs att= 0 andx=12according to Bob. We seekt.t=2−+vv xtc.v=21112− =2.Thust=122(30m)0−+c=−830m310 m/s=−10−7s =−100ns.(b)We have a distance traveled according to Bob, 30m, and a speed, and so find a time according to Bob viaclassical mechanics.t=30m2c=830m2310 m/s= 1.4110−7s =141ns.(c)If Anna’s back clock has moved to the middle of Bob’s ship, Anna’s front clock, formerly at Bob’s middle,will now be at Bob’s front,x= 60m. Thus the locations, according to Bob, arex= 30 andx= 60m. Thetimes are both 141ns.For Anna’s back clock:t=9122(30m)141 10s−−+c=−830m310 m/s+214110−9s =100ns.For Anna’s front:t=91/22(60m)141 10sc−−+=−860 m310 m / s+214110−9s =0.

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Chapter 2 Special Relativity10(d)In the diagrams, Event 1 shows an observer in Anna’s ship, att=−100ns by her wristwatch, looking at aclock at the middle of Bob’s, which reads zero. Event 2 shows another observer in Anna’s ship, att= +100ns by her wristwatch, looking atthe same clockin Bob’s, which reads 141ns. The time elapsing betweenthese events in Anna’s frame is +100−(−100) = 200ns, but they see this single clock at a fixed location inBob’s ship mark off only 141ns, which is less than 200ns by the usual factor:v. That is, 200ns =2141ns. Events 3 and 4 show observers in Anna’s frame, whoare less than 60m apart (not both being at thevery ends of their own ship) viewingat the same time(t= 0) the very ends of Bob’s ship. It is less than60m long according to them. How much less? According to Bob, the two clocks on Anna’s ship are 30mapart, but this is a length−contracted observation, so they must be more than 30m apart in their own frame:230m = 42.4m. As noted, this is how long Bob’s ship appears from Anna’s frame. This fits! Annashould see the length of Bob’s ship as60mv=60m2=230m = 42.4m.2.44Whenvc,vapproaches 1, so equations (2-12) becomex´ =x−vtandt´ =t−2vcx, while equations (2-13)becomex=x´ +vt´ andt=t´ +2vcx´. So long as neitherxnorx´ is large, the two time equations becomeequivalent,t´ =t, and then the two position equations are also equivalent,x´ =x–vt. These are equations (2-1).If, however, we are talking about events at such largexthat2vcxis not negligible, then the correspondence fails.2.45Bob will waitt=12ly0.6c= 20yr for Anna to get there and 20yr for her to return. Bob ages 40yr. Bob, always in aninertial frame, will observe Anna’s aging slowly the whole way.v40yr=240yr1/1(0.6)−= 32yr.Anna “sees” Bob age slower than herself on the way out and on the way back, but “sees” Bob age horribly duringthe interval in which she accelerates. We need not calculate from Anna’s perspective. Both must agree on theirrespective ages when they reunite. Bob is 20 + 40 =60yr. Anna is 20 + 32 =52yr.2.46A distance of 30ly is traveled at 0.9c.Time in Bob’s frame =distance in Bob's framespeed=30ly0.9c=33.3yr.(b)Anna will see Bob age less than herself, but how much does Anna age? She sees a distance of21(0.9)−30ly = 13.08ly pass by her window at 0.9c.Time in Anna’s frame =distance in Anna's framespeed=13.08ly0.9c= 14.5yr.This is Anna’s age according to herself, reckoned via length contraction. It is also Anna’s age according toBob, for Bob ages 33.3yr, but will determine that Anna ages less (events transpire less rapidly):21(0.9)−33.3yr = 14.5yr.

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Instructor Solutions Manual for Harris,Modern Physics, Second Edition11Anna, aging 14.5yr, will determine thatBobages less than herself, by the same factor.21(0.9)−14.5yr=6.33yr. (c) and (d) already answered:14.5yr. Yes, each says,correctly, that the other is younger.2.47For Anna in S´, where the planets are 24ly apart, (2-12b) evaluated for two events is21212212()()−=+−+−vvttxxttc=50.8 (24ly)(0)3++=c32yr.(b)Relative to Earth, Carl simply has avthat is the opposite of Anna’s, so in this case the time interval is–32yr.(c)Bob waits 50yr, so the Planet X clock (synchronized in his frame at least) reads 50yr. All observers agreethat the clock on Planet X reads 50yr when they pass. According to Anna, the clock on Planet X (event 2) is32 yr ahead of that on Earth, so she says that the Earth clock is chiming 18yr now. Carl says that the PlanetX clock is 32yrbehind, so the Earth clock is right now chiming 82yr.2.48Calling to the right positive and Anna frame S, we know that this tick of the clock/event occurs att= 0 andx=−24m according to this observer in Bob’s frame. We seekt.t=2−+vv xtcandv=211(0.6)−= 1.25.Thus,t=0.61.25( 24m)0−−+c=818m310 m/s= 610−8s =60ns.(b)Before your friend steps on, Anna, at the center of her ship, is seen from Bob’s frame to be age zero.Afterward, your friend is in Anna’s frame, where the clock reads 60ns. But once in Anna’s frame, all clocksaresynchronizedin that frame. Specifically, the clock right at Anna’s location now also must read 60ns.Anna’s agejumps forward by 60ns.(c)x would be +24m rather than−24m, so the new time would be−60nsand Anna’s age, starting at zero,jumps backward by 60ns.(d)In part (b) your friend is accelerating toward Anna, and Anna’s agejumps forward. In part (c) your friendaccelerates away from Anna (before the ship-changing, Anna was approaching your friend, but afterward isno longer doing so), and Anna’s agejumps backward.If you think about this with relative simultaneity inmind, it makes sense. If your two friends jump off your ship at the same time (t1=t2= 0), they cannotpossibly arrive on Anna’s ship at the same time (t1≠t2). They must necessarily arrive there at two differenttimes (ages) of Anna.2.49Thewholeideahereisthatyouarejumpingintoanewframe,andtheclocksinthenowmovingEarth−Centaurus A frame are unsynchronized. How out of synch is the one on Centaurus A? Let us call theEarth−Centaurus A system frame S, with their separation given the symbol W. You are initially in frame S, but atthe passing of the origins you instantly jump into frame S, moving at 5m/s toward Centaurus A. We know thatyour time is t= 0, and we seek t, the time on a clock in Centaurus A atx= + W in frame S.t=2 −vvtxc→0 =2W−vvtc, ort=2vcW.This being positive, the clocks in front of you will all be ahead.2vc=16225m/s910m /s= 5.5610−17s/m.Thus,t= W5.5610−17s/m

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Chapter 2 Special Relativity12(a)If W = 21023m then clockjumps aheadbyt= 1.11107s =128days.(b)If W = 4.5109m thent=250ns.(c)Need only reverse the sign ofx. Clocks will bebehind by same amounts.(d)If the traveler is moving away from Earth, then decelerates to a stop (stoppingbeing the reverse ofstartingto jogawayfrom a planet), he moves to a frame in which clocks back on Earth are immediately advanced.If he furthermore turns around and jumps to a frame moving back toward Earth, he moves to a frame inwhich clocks back on Earth are again immediately advanced. Acceleration toward Earth causes clocks thereto advance. The effect depends not only on the speed involved, but also on the distance away; as we see, bymerely jogging this way and that at ordinary speeds, we move through frames in which clocks on heavenlybodiesveryfar away change by a great deal. Readings on local clocks, those around the solar system, don’tvary much.2.50I am S´. Att´ = 0, I seekton a clock at the 100cm mark, i.e.,x= 50cm (with origin at meterstick center).t=2−+vv xtc→=250cm−+vvtcort=vc250cm.The clock at 0cm, orx=–50cm, would readvc2–50cm.(b)Changing direction would simply change the sign on each value calculated.(c)When I turn around, the clock at the 0cm mark jumps forward byvc21m, as I accelerate toward it, just asBob’s clock on Earth jumps forward as Anna accelerates toward it.2.51Toward is= 180º, cos=−1.fobs=fsource221/1/−−vcvc=fsource(1/)(1/)(1/)(1/)−+−−v cv cv cv c=fsource1/1/+−v cv c2.52fobs=fsource1/1/−+v cv c→obsc=source1/1/−+cv cv cobs=1/1/+−v cv csource=10.910.9+−source=4.36source2.53If we see a shorter wavelength, a higher frequency, it must be movingtoward.obssourceff=1/1/+−v cv c→obssource//cc=1/1/+−v cv c→532412=1/1/+−v cv c→1.667 =1/1/+−v cv cvc=0.25.(b)Away:obsc=source1/1/−+cv cv c→obs1=110.2553210.25−+obs=687nm.(c)If=90º,fobs=fsource221/−vc→obsc=21(0.25)532nm−cobs=549nm.A period in the observer’s frame is longer than for 532nm light, due solely to time dilation.

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Instructor Solutions Manual for Harris,Modern Physics, Second Edition132.54Since the observed frequency is larger(wavelength smaller) than the source, the galaxy must be movingtowardEarth.fobs=fsource1/1/+−v cv c→obssourceff=1/1/+−v cv c→obssource//cc=1/1/+−v cv c→396.85396.58=1/1/+−v cv c→1.00136 =1/1/+−vcvcvc=0.000681. The wavelength shift is about one part in a thousand, and the speed is roughlyone−thousandth that of light.2.55Yes—movementpartiallytoward must be compensating for time dilation effect.fobs=fsource221/1(/) cos−+vcv cbutfobs=fsource1 + (v/c) cos=221/−vc→1+ 0.8 cos=21(0.8)−= 120º. This fits, for the movement is partiallytoward.2.56Toward:obsc=source1/1/+−cv cv c→540nmc=source1/1/+−cv cv cAway:obsc=source1/1/−+cv cv c→650nmc=source1/1/−+cv cv cDivide first equation by second:650540=11+−v cv cvc=0.0924Plug back in:540nmc=source10.092410.0924+−csource=592nm(Yellow.)2.57When moving away,fobs=fsource1/1/−+v cv corobsc=sourcec1/1/−+v cv corobs=source(1+v/c)1/2(1−v/c)−1/2source(1+12v/c)(1+12v/c)source+source(v/c). Moving toward would just change the sign. The total range is thus 2sourcev/c=/B3k T mc2.(b)vrms=234273(1.3810J/K)(510 K) /1.6710kg−−= 3.5104m/s.Thus,=483.5102656 nm310=0.15nm.2.58The car moving away would detect a frequencyfobs=fsource1/1/−+v cv c=fsource(1−v/c)1/2(1 +v/c)−1/2fsource(1−12v/c) (1−12v/c)fsource(1−v/c). It then becomes a source of this frequency, moving away from the finalobserver (i.e., the radar gun). The final observed frequency would bef'obs=f'source1/1/−+v cv c=1/(1/)1/−−+sourcev cfv cv c(1/ ) (1/ )−−sourcefv cv cfsource(1−2v/c) = 900MHz(1−2830310)= 900MHz−180Hz.The beat frequency (the difference) is180Hz.

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Chapter 2 Special Relativity142.59The “object” here is the projectile. Let’s choose away from Earth as positive. Therefore Bob, on Earth, is frameS, while Anna (moving in the positive direction) is frame S. We are given that the velocity isv= 0.6cbetweenthe frames, and that the object moves in the positive (away from Earth) direction atu= 0.8c relative to frameS/Bob.u=21−−uvuvc=0.80.61(0.8)(0.6)−−cc=0.385c.(b)The relativistic velocity transformation works for light just as for a massive object. Thus,u=c, andu=21−−uvuvc=0.61(1)(0.6)−−cc=c. This had better be the case! Light traveling atcin both frames is built in to theLorentz Transformation equations, from which the velocity transformation is derived.2.60The “object” here is Carl. Let’s choose Anna, on Earth, as frame S and Bob as frame S. (Since the Sframe bydefinition moves in the positive direction relative to S, we’ve chosen toward Earth as positive.) We are given thatthe velocity isv= 0.8cbetween the frames, and that the object Carl moves in the positive (toward Earth)direction atu= 0.9crelative to frame S/Anna. We seek the velocityuof the object/Carl relative to frame S/Bob.u=21−−uvuvc=0.90.81(0.8)(0.9)−−cc=0.357c. Positive meanstoward Earth.(b)Bob sees Carl moving at 0.357cin one direction and Earth moving at 0.8cin the other (i.e., toward Bob).These are both velocitiesaccording to Bob. They may be added using the classical expression to find arelative velocityaccording to Bob.1.157c. Note: This is a velocity of Carl relative to Earthaccording toBob. It isnotsaid that an observer sees something else moving at greater than crelative to him or herself.2.61(c−u)(c−v) > 0c2−(u+v)c+uv> 0→c2+uv> (u+v)c→c>21 ++uvu vc=u2.62The lab is S; Particle 2 is S, moving atv= + 0.99crelative to the lab; and Particle 1 is the object, which moves atu=−0.99cthrough the lab.u=21−−uvuvc=0.990.991( 0.99)(0.99)−−− −cc=−0.9995c2.63uy=.22xcu−(b)xu=21−−xxuvu vc=221−−xxxuuuc=0andyu=21−yxvuu vc=22221−−xxvcuuc=222222111−−−xxxcuucuc=c.The light beam has nox-component, and its speed overall must bec.2.64In frame S, the velocity components of the light beam areux=ccosanduy=csin. Equations (2-20) apply.xu=21−−xxuvu vcandyu=21−yxvuu vc.

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Instructor Solutions Manual for Harris,Modern Physics, Second Edition15Plugging in:xu=−−coscos1cvvc,yu=−vsincos1cvc22+xyuu=22( cos)cos1−−cvvc+2222( sin)1cos1− −vccvc=()22222( cos)1/( sin)cos1−+−−cvvccvc.Multiplying out the numerator,c2cos2−2c vcos+v2+c2sin2−v2sin2=c2cos2−2c vcos+v2+c2(1−cos2)−v2(1−cos2) =−2c vcos+c2+v2cos2. Multiplying out the denominator, 1−2 (v/c) cos+(v2/c2) cos2. The numerator isc2times the denominator. Conclusion: Though the components may be different,the light beam moves atcin both frames of reference.2.65ux=–ccos 60° =–c/2.uy=csin 60° =3 2.cxu=112211222145−−= −−−ccccccandyu=21−yxvuu vc=112222123 2111()−−−ccccc=.c35´= tan-1(–3/4) = 36.9° north of west.u´ =22(0.8 )(0.6 )+cc=c, as it must be.(b)Would change only the sign ofv. Thus,xu=112211222()1−+−−−ccccc=0, andyu=112222123 2()111()−−−−ccccc=c. Again, speed isc, but direction is alongy´.2.66Whenxu=21−−xxuvu vcis zero, it divides positivex´−components from negative ones according to an observer inS´. This occurs whenux=vorucos=v. But the “object” moving in frame S here is light, for whichu=c. Thusccos=vor= cos−1(v/c).(b)Atv= 0, this is 90°, which makes sense. The beacon and Anna are in the same frame, and light emitted onthe +xside would be on the +xside according to both. Atv=c, the angle is 0°. Only the light emitted by thebeacon directly along the horizontal axis would appear to Anna to be moving in the positive direction. Allthe rest would appear to have a negative component according to Anna.(c)According to Anna, the beacon shines essentially all of its light in front of it, in the direction it is movingrelative to Anna.2.67u2211−uc=222211++−xyzuuuc=()()()2222222111111−−++−−−yxzxvxvxuuvucu v cu v cu v c

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Chapter 2 Special Relativity16=()2222222222211111−−−−+−+−xxxyzu vcu vvvuvuucccc=222222222242222221112211−−+−+−−−+−xxxxxyzu vcu vu vuu vvvvuucccccccc=222222222222221111111−−−−−−+−xxyzu vcvvvvuuucccccc=22222221111−++−−xxyzu vcvuuucc=21−xu vcvu2.68We suppress the indexifor clarity.21 =−xuu vmucvum21−−xxuvu vc=vumux–vvum. Summing this over allparticles of the system would reproduce equation (2-23).2.69umumu=u. This becomes significant whenunearsc.2.70p=umu=27821(1.6710kg)(0.8310 m/s)1(0.8)−−=6.6810−19kgm/s.(b)E=umc2=278221(1.6710kg)(310 m/s)1(0.8)−−=2.5110−10J.(c)KE =()1−umc2=2782211 (1.6710kg)(310 m/s)1(0.8)−−−=1.0010−10J.2.71Einternal=mc2= (1kg) (3108m/s)2=91016J(Huge!);KE = (u−1)mc2= (53−1)(1kg) (3108m/s)2=61016J;Etotal=umc2=Einternal+KE =1.51017J.2.72To melt ice, energy (heat) must be added. This increases the internal thermal energy, hence the mass. It takes3.33105J/kg to change ice at 0ºC to water at 0ºC. Water is 18g/mol, so, with one mole, we have 18g.3.33105Jkg0.018kg = 6103J.E=m c2m=2Ec=31622610 J910m /s= 6.710−14kg. The mass of the iceis less, by67pg. Not much!2.7312kx2=12(18N/m)(0.5m)2= 2.25J. If it gains this much internal energy, its mass increases correspondingly:Einternal=mc2m= (2.25J)/91016m2/s2=2.510−17kg.

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Instructor Solutions Manual for Harris,Modern Physics, Second Edition172.74500103W3,600s = 1.8109J.E=mc2→m=916221.810 J910m /s=210−8kgor 20g.2.75p=umu=3142481(9.11 10kg)(2.410 m/s)2.410 m/s1310 m/s−− = (1.000000003) (9.1110−31kg)(2.4104m/s)=2.1910−26kgm/s.(b)p =umu=3162881(9.11 10kg)(2.410 m/s)2.410 m/s1310 m/s−− = (1.667) (9.1110−31kg)(2.4108m/s) =3.6410−22kgm/s.(c)% error =−classicalcorrectcorrectppp100%=−uumumumu100% =11−u100%. In the first case,111.000000003−100%= (0.999999997−1)100% =−310−7% or310−7% low.In the second case,111.667−100% = (0.6−1)100% =−40% or40% low.Simply put, the classical expression is good so long asuis not significantly different from 1.2.76Before: ptotal=0.6(16)(0.6c) +0.8(9)(−0.8c) =54(16)(0.6c) +53(9)(0.8c) = 0(b)v= 0.6cand we seeku, given variousu.u=21−−uvuvc.Givenu= +0.6c:u=0.60.61(0.6)(0.6)−−cc=0. Givenu=−0.8c:u=0.80.61( 0.8)(0.6)−−− −cc=−0.946c.Givenu=−0.6c:u=0.60.61( 0.6)(0.6)−−− −cc=−0.882c. Givenu= +0.8c:u=0.80.61(0.8)(0.6)−−cc=0.385c.(c)Before:ptotal=1(16)(0)10−+21(9)( .946 )1(0.946)−−c=−26.25c.After:ptotal=21(16)( 0.882 )1(0.882)−−c+21(9)(0.385)1(0.385)−=−26.25c.2.77Momentum can be arbitrarily large.p=umu=m cuu=cu=c221/−ucu=c/22.78(u−1)mc2=()1222(1/)1−−−ucmc2()2212[1()/]1− −−ucmc2=12mu2

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Chapter 2 Special Relativity182.79The area that Earth presents to (and that absorbs) the incoming sunlight is simply a circle whose radius is that ofEarth. area =RE2=(6.37106m)2= 1.271014m2. The power absorbed is thus power =powerareaarea =(1.53W/m2)(1.271014m2) = 1.911017W.mt=int2/Etc=1716221.91 10W910m / s= 2.1kg/s or1.835kg/day.2.80In orbit:F=ma→earth2GMmr=m2vrorv2=earthGMr. Ifr=Rearth(i.e., 6.37106m) thenv=1122246(6.6710N m /kg )(5.9810kg)6.3710 m−= 7913m/s. The Empire State Building is of mass 365103ton8890Nton219.8m/s= 3.31108kg. Thus its KE needs to be12(3.31108kg) (7913)2= 11016J. But 1kg convertsto (1kg)(3108m/s)2= 91016J. More than enough.2.81IntensityPowerArea=2Power4RPower = Intensity4R2= (1.53W/m2)1124(1.510m)= 4.241026W.So every second, 4.241026J of energy is put out by the sun.m=int2Ec=2616224.2410J/s910m /s=4.71109kg persecond.2.82If one kilogram explodes, 106J is released. But how much mass must actually be converted to produce suchenergy?m=int2Ec=6162210 J910m /s= 1.1110−11kg.111.11 10kg1kg−=1.1110−11(b)Suppose we have one kilogram. If one part in ten−thousand is converted,1kg10,000= 0.0001kg is converted.How much energy is released?Eint=mc2= (0.0001kg) (91016m2/s2) = 91012J. Explosive yield:91012J/kg. A much greater percent is converted, so it is much more powerful.2.83(u−1)mc2=mc2u= 2→211(/)−uc= 2u=c3/2. Fast! Internal energy is large.2.84KE =()1−fumec2−()1−iumec2=()−fiuumec2=3116222211(9.11 10kg)(910m /s )1(0.6)1(0.3)−−−−=1.6510−14J(b)3116222211(9.11 10kg)(910m /s )1(0.9)1(0.6)−−−−=8.5610−14JIf12mu2were correct, the second one would require only 67% more energy. Here we see it is more thanfive times as much. This is due to the steep rise in KE nearc.

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Instructor Solutions Manual for Harris,Modern Physics, Second Edition192.85KE = KEf=311622211 (9.11 10kg)(910m /s )1(0.9998)−−−= 4.0210−12J = 25.1MeV, requiring anaccelerating potential of25.1MV.2.86First find the kinetic energy fromqV=KE; then either (1) solve foru, then findp, or (2) calculateEfromKE, then use (2−28).Let’s do (1).qV=KE→qV= (u−1)mpc2→(1.610−19C)(109V)=271622211 (1.6710kg)(910m /s )1(/)−−−ucu= 0.875c.p=umu=27821(1.6710kg)(0.875310 m/s)1(0.875)−−=9.0510−19kgm/s.2.87It acquires a KE of 500MeV = 810−11J. KE = (u−1)mc2→810−11J =2716222211 (1.6610kg)(910m /s )1/−−−ucu=0.759c(b)4810−11J =2721 (1.6610kg)2−uu= 6.21108m/s =2.07c(c)4()11810J−=2716222211 (1.6610kg)(910m /s )1/−−−ucu=0.948c2.88Momentum conserved:0.6m0(0.6c) =0.8(0.66m0)(0.8c) +um uEnergy conserved:0.6m0c2=0.8(0.66m0)c2+um c2. The physics is done; the rest is math. We wish to solveforu. Divide energy equation byc2, and rearrange the equations:(1)0.6m0(0.6c)−0.8(0.66m0)(0.8c) =um u(2)0.6m0−0.8(0.66m0) =umDivide (1) by (2): u =0.600.800.600.80(0.6 )(0.66)(0.8 )(0.66)−−mcmcmm=(5 / 4)(0.6 )(5 / 3)(0.66)(0.8 )(5 / 4)(5 / 3)(0.66)−−cc=−0.867c.It moves at0.867c in the opposite direction. Plug back into either (1) or (2) to findm.Using (2):0.6m0−0.8(0.66m0) =um→(5/4)m0−(5/3)(0.66m0) =211(0.867)−mm=0.0748m0. Much mass is lost, because there is a significant increase in kinetic energy.2.89Since carbon−14 is “slow”,1.Pf=Pi→mCuC+emeue= 0.Ef=Ei→mCc2+emec2=mBc2→(13.99995) +22e1(0.00055)1−uc= (14.02266)ue= 0.99971c. Plug back in to momentum equation:(13.99995)uC+21(0.00055)(0.99971 )1(0.99971)−c= 0uC=−1.6210−3c(~1600the electron’s speed).

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Chapter 2 Special Relativity20KEe= (−1)mc2=311622211 (9.11 10kg)(910m /c )1(0.99971)−−−= 3.310−12J =20.6MeV.KEC=12mu2=12(13.99995u1.6610−27kg/u) (1.6210−33108m/s)2= 2.7510−15J =17.1keV(about11, 000of KEe).2.90Let’s usem0for an “atomic mass unit”, rather thanu(for obvious reasons).Momentum:0.8(3m0) (+0.8c) +0.6(4m0)(−0.6c) =0(6m0)(0) +um u→1m0c=um uEnergy:0.8(3m0)c2+0.6(4m0)c2= +10m0c2=0(6m0)c2+umc2→+4m0=umDivide the two:u=c/4. Plug back in: 4m0=0.25mm=3.87u(b)Since mass/internal energy increases, KE must decrease. The long way: KEfinal= (0.25−1) (3.87m0)c2=0.127m0c2. KEinitial= (0.8−1) (3m0)c2+ (0.6−1) (4m0)c2= 3m0c2KE =−2.87m0c2. The short way:KE =−mc2=−((3.87+6)−(3+4))m0c2=−2.87m0c2=−2.87u1.6610−27kg/u(3108m/s)2=−4.2910−10J2.91Momentum:0.6(10kg)(0.6c) =0.6m1(−0.6c) +0.8m2(0.8c)→7.5kg =−0.75m1+1.33m2Energy:0.6(10kg)c2=0.6m1c2+0.8m2c2→12.5kg = 1.25m1+ 1.67m2Solve: MultiplyE−equation by 0.6, then add: 15kg = 2.33m2m2= 6.43kg. Reinsert:m1= 1.43kg(b)KE =−mc2=−(6.43kg + 1.43kg−10kg) (91016m2/c2) =1.931017J2.92Momentum:0.8m1(0.8c) +0.6m2(−0.6c) = 01.33m1= 0.75m2m2=1.78m1(b)Energy:0.8m1c2+0.6m2c2=0mfc2→1.67m1+ 1.25m2=mf→1.67m1+ 1.25(1.78m1) =mfmf=3.89m1.(c)KE =−mc2=−(3.89m1−m1−1.78m1)c2=−1.11m1c22.93(0.9−1)m0c2=2111(0.9)−−m0c2=1.29m0c2(b)2(u−1)m0c2= 1.29m0c2u=1.292+ 1 = 1.647.211(/)−uc= 1.647u=0.795c.(c)Experiment A: Momentum:0.9m0(0.9c) =fumuf. Energy:0.9m0c2+m0c2=fum c2Divide equations:uf=0.900.900(0.9 )+mcmm=221(0.9 )1(0.9)111(0.9)−+−c= 0.627c. Plug back in to momentum equation:211(0.9)−m0(0.9c) =211(0.627)−m(0.627c)m =2.57m0

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Instructor Solutions Manual for Harris,Modern Physics, Second Edition21Experiment B: Momentum:0.795m0(−0.795c) +0.795m0(+0.795c) =fumuf.Energy: 20.795m0c2=fumc2. Fromp-equation,uf= 0.E-equation becomes: 20.795m0= 1mm= 2211(0.795)−m0=3.29m0. Though mass increases in the both completely inelastic collisions,Experiment B, the collider, with the same initial kinetic energy input, yields more mass, simply becauseuf= 0. There is no final kinetic energy.2.94Energy:u(8.8710−28kg)c2=0.9(2.4910−28kg)c2+0.8(2.4910−28kg)c2211(/)−uc=(2.294)(2.49)(1.67)(2.49)8.87+u=0.437c.Momentumx:0.437(8.87)(0.437c) =0.9(2.49)(0.9ccos1) +0.8(2.49)(0.8ccos2)Momentumy: 0 =0.9(2.49)(0.9csin1)−0.8(2.49)(0.8csin2)Or: 4.312−5.141 cos1= 3.32 cos2and 5.141 sin1= 3.32 sin2. Square both:18.591−44.335 cos1+ 26.432 cos21= 11.022 cos22and 26.432 sin21= 11.022 sin22.Add: 18.591−44.335 cos1+ 26.432 = 11.0221=39.9º(the 0.9c+)Plug back in: 5.141 sin 39.9º = 3.32 sin22=83.6º(the 0.8c−)Mass decreases, so KE must increase. As is often the case, this can be seen another way: There is a frame ofreference moving with the kaon in which the process is simply a stationary object (i.e., the kaon) exploding intotwo parts. KEi= 0, KEf> 0.2.95In the new frame, the initial particle moves left at 0.6cand the right-hand fragment is stationary. The left-handfragment moves atu=21−−uvuvc=20.60.6( 0.6 )(0.6 )1−−−−ccccc=−0.882c.In the new frame,Pinitial=0.6m0(–0.6c) =–1.25m00.6c=–0.75m0cPfinal=0.882m(–0.882c) =–2.125m0.882c. Can relatem0andmvia energy conservation, and using originalframe is easiest:m0c2= 2(1.25)mc2m= 0.4m0. Plug back in:Pfinal=–2.125 (0.4m0)0.882c=–0.75m0c2.96()2221= −+ iiuiuiim cm cm c.E= 0→()2210 −+  =iuiim cm cor()221 −= − iuiim cm c2.97E=umc2andp=umu. Squaring both:E2=211(/)−ucm2c4andp2=211(/)−ucm2u2.E2−p2c2=211(/)−ucm2c4−211(/)−ucm2u2c2=m2c2222()1(/)−−cuuc=m2c42.98In the product22112211()()(2)(). . .()(2)()()−++−−+f rf rdrf rdrf rdrf rdrf rdrf rdrf r, all terms but21()()f rf rcancel.

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Chapter 2 Special Relativity22(b)In multiplying out the product11122222()()(2)()111. . . 1++−−−−−g r drg rdr drg rdr drg rdr drcccc, anyterm with a fractionsquaredor higher order will be negligible. Thus, only the leading 1 and terms with afraction to the first power survive.11122222()()(2)()1. . .++−−−−−−g r drg rdr drg rdr drg rdr drcccc. Theterms beyond the 1 define the integral ofg(r) fromr1tor2.(c)mg(r) =2GMmrg(r) =2GMr. So,212222112()111111()=−=−−rrf rGM drGMf rrrcrcor21212111()() 1=−−f rf rGMrrc. This is analogous to equation (2-29), where the satellite is the higherpointr2andEarthisthelowerpointr1.Thus,analogousto(2-30),wehave12212111( )() 1= −−t rt rGMrrcorEarthsatellite2Earthsatellite1111= −−ttGMrrc2.99Although the satelliteappearsnot to move in the sky, it is moving, and the point on Earth’s equator is alsomoving. For geosynchronous orbit,22=→=vGMmGMmvrrrandT=286, 400s.=rv112224(6.6710N m /kg )(5.9810kg)86, 400s / 2−=vv. Solving this forvgivesv= 3.07103m/s. Subtracting the tangentialspeed of 460m/s at Earth’s surface gives a relative speed of 2.61103m/s. The orbit radius is given by3286, 400s3.0710 m/s=r, orr= 4.23107m/s. Now, as in the GPS example, for the speed-dependent part,()1/ 211SatelliteEarthSatellite23817.6102.61 101310−−==−− ttt()11Satellite13.810−+tand for the gravitational part,1432EarthSatellite82674.010m /s111(310 m/s)6.410 m4.2310 m −−tt()10Satellite15.8910−= −t.Accountingforboth (5.89–0.385.5),()10EarthSatellite15.510− −tt. The difference in a day is (86,400s) (5.510−10)= 4.810−5s =48μs.2.100Radius is of dimensions [L], mass [M],c[L][T], andG(Nm2/kg2)32[L][T] [M]r=MaGbcd→[L] = [M]a3bd2bbd[L][L][T][M][T]. Considering mass gives 0 = a−b, or a = bConsidering time gives 0 =−2b−d, or d =−2b. Considering length gives 1 = 3b + d. But since d =−2b thisbecomes 1 = 3b−2b, b = 1. This in turns gives d =−2 and a = 1. Thusr=MG/c22.101KE + PE =E+2(/)−GM Ecr= 0. The pulse’s energy cancels, leavingr=2GMc.

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Instructor Solutions Manual for Harris,Modern Physics, Second Edition232.102Those of velocity +1m/s will be at 1m, of velocity +2m/s at 2m, of velocity−1m/s at−1m, etc.(b)If the observer jumps tovmeters from the origin, on a particle moving atvmeters per second, he will findparticles moving atv+1 meters per second one meter further away, and particles moving atv−1 meters persecond one meter closer to the origin. But if moving atvmeters per second himself, he will see relativevelocities, respectively, of +1m/s and−1m/s, just as does the person at the origin.2.103000320000210101222222 =−+−++vuuuvcvcvctx uccuccuccuor()02200012122=++uvccuucccu.(b)Ifu0<c, then()22021+uc> 1. Thus, for3tto be negative means that0uvcc, but, as noted,vis notallowed to exceedu0.(c)22200000221021012−→−+→+uuuuucccccor0220211+ucuc2.1042=− + vvtxtc. Divide both sides byt:21=−+vtvxttc. If the time intervals are of oppositesign, then210−+vxtcor2.xcctvThe speed needed to travel thexin timetis greater thanc. Usingthe complementary Lorentz transformation equation gives210++vxtcor2 −xctv. This too implies aspeedgreater thanc.2.105Asvc→0,v→1, so the matrix in (1−15) becomes:−−10001000010001vcvc(b)00xct=1000100000100001−     − vcvcxct=−−xvt00ctxv c.In the limitvc→0, the termxv/ccan be ignored, leavingx=x−vtandt=t.

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Chapter 2 Special Relativity242.106The matrix to find (x,t) values from (x,t) is of the form:xyxct=000100001000−     −vcvcvvvvxyxct0.8c=53and530.8 =43, so this matrix is−−00010000100054335433. The space−time point is5ly002yrcin Bob’s frame S.Thus, in Anna’s frame S:54335433005ly0100000100002yr−     − c=(17 / 3)ly00(10 / 3)yr−c. The position is 5.67ly and the time−3.33yr,in agreement with Exercise 22.2.107 =−xvxtvAAAc =yyAA =zzAA =−tvtxvAAAcSquaring,2222222+=−xvxxttvvAAA AAccand2222222+ =−tvtxtxvvAAA AAccSubtracting, the middle terms cancel:22222222211−=−−−txvtxvvAAAAcc.Butv2=2211−vc, so this becomes22−txAA, and since =yyAAand =zzAA, equation (2-36) follows.2.108p=0.8m(0.8c) =53(1kg)(0.8c) =43(1kg) c =4108kgm/s.E=53(1kg)c2=1.51017JThe matrix to find(,)xpEfrom(,)xpEis of the formxyzpppEc=000100001000−     −vcvcvvxyzvvpppE cSince0.8c=53and530.8 =43, the matrix is−−00010000100054335433. Momentum−energy in S is43253(1kg)00(1kg)cccandmatrix multiplication gives55443 33 35 54 43 33 3(1kg)(1kg)00000(1kg)(1kg)(1kg)−=−+cccccfor momentum−energy in S. In its rest frame,its momentum iszero; andE/c= (1kg)cE=(1kg)c2. Its energy is internal only; no KE.

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Instructor Solutions Manual for Harris,Modern Physics, Second Edition252.109yp=21 −xvuu vcm21−yxvuu vc=umuy=py. Thez−component follows similarly.2.110 =−xvxvEppcc. If1,vcthenv1 andEmc2. Thus,xpbecomespx–mv, essentially equation(2-21).=−vxEEv pccc. In the same limit, this becomesm´c=mc, simply confirming that the mass is thesame in both frames.2.111p=0.83m0(0.8c) =533m0(0.8c) =4m0c.E=533m0c2=5m0c2.(b)xu=21−−xxuvu vc=20.80.50.5 .(0.8 )(0.5 )1−=−ccccccp´ =0.53m0(0.5c) =233m0(0.5c) =3m0c.E´ =233m0c2=23m0c2. =−xvxvEppcc=00024(0.5)53.3−=m cm cm c=−vxEEv pccc0002=5(0.5) 423.3m cm cm c−=2.112The frame in which the final single object is at rest is simplest, in which case the invariant is just (Mc2/c)2=M2c2.(b)In the lab frame there are two terms in the total energy. Inserting both and using the relationship suggestedto eliminatePtotal, gives2222222totaltotal(/)(/)(/)iiEcPEcmcEcm c−=+−−=22222222/2/iiiEcE mm cEcm c++−+= 2(mEi+m2c2). The invariant is the same no matter which frame isconsidered, soM2c2= 2(mEi+m2c2)M=2222imEcm+.(c)Momentum and energy conservation areifuiufmuMu=and222ifuumcmcMc+=.If we square bothsides of both we have222222ifuiufm uM u=and22424242242iifuuum cm cm cM c++=. If we now subtractc2times the former from the latter we have2242222242422422222iiiffuuiuuufm cm u cm cm cM cM u c−++=−.Theidentifycannowbeusedoneachside,yielding242424242ium cm cm cM c++=,or22222222ium cm cM c+=. Noting that2iumcisEi, this is the same result as before.

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Chapter 2 Special Relativity262.113 =xxduadt=22() 1−−−+xxvu vduvcv dxdtc=22222()11−+−−−+xxxxxvvuvduducu vu vccv dxdtc=22222221()11−−+−−−+xxxxxvu vvuvccduu vu vccv dxdtc=2222211()−−−+xxvu vvduccv dxdtc. Dividing top and bottom bydtyields22222111−− =−+xxxvduu vvdtccav dxdtcBut=xxduadt, and 1−22vc=21vand−2v dxdtc+1 = 1−2xvuc, so3321 =−xxxvaau vc =yyduadt=221−−+xyvvu vducv dxdtc=22222211+ −−−+yxyxxvvududucu vu vccv dxdtc.Dividing top and bottom bydt,222222111+ −− =−+xyyxxyvduvduudtcdtu vu vccav dxdtcBut=xxduadt,=yyduadtand−2v dxdtc+1 = 1−2xvuc, so223222211 =+−−yxyyxxvvu vaacau vu vcc.2.1140.5c=211(0.5)−=23andvcvis therefore13. Thus, the matrix is−−00010000100021331233.

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Instructor Solutions Manual for Harris,Modern Physics, Second Edition27Now multiplying,2121333312123333000001000100001000100000−−     −−  =−−00010000100054335433. The upper−left element isv, which in this case is53.2211/−vc=53v=0.8c. This is as it should be: If Smoves at 0.5c relative to S, which moves at 0.5crelative to S, the velocity addition formula gives a velocity of20.50.5(0.5 )(0.5 )1++ccccc= 0.8cfor Srelative to S.2.115In the new frame, the right-moving stream is stationary, so the distance between the charges is larger byvandthe charge density thus smaller byv.right=/v=21(1 3)−=8 3. This is the charge density in the restframe of the charges in the stream. The charges in the left-moving stream move atc/3 relative to the “lab,” andthe lab moves atc/3 relative to the stationary stream, so the speed of the left-moving stream relative to thestationary stream, by the relativistic velocity transformation, is233(3)(3)1++ccccc= 0.6c. Relative to the stationarystream, the left-moving charges are close together, so the density is higher by0.6.Thus,left=0.6right=548 3=58 12.(b)The streams push in opposite directions on the point charge and are the same distance away, so the netelectricforcewilldependonlyonthedifferencebetweenthechargedensities.()058 128 32==−EFqEqr.08 24=qrFor the magnetic force, only the left-movingstream is indeed moving, thus producing a magnetic field. The current is the charge per distance times thedistance per unit time.()58 12 0.68 4.==IccIn the new frame, the point charge is moving at speedc/3, so it experiences a magnetic force()2point charge00(3)8 42()8 24.===BFqvBq ccrcqrNoting that2001, =cwe see that this is the same as the electric force.Because the point charge wouldnot experience a net force in the “lab” frame, it must not experience one in the new frame. The electric andmagnetic forces must be equal and opposite.2.116W=0fuu dp=02211/−fuu dmuuc=2223/ 2221/ 20(/)1(1/)(1/)+−−fuu cduu muduucuc222223/ 2223/ 20/1/(1/)(1/)−+−−fuu cucu muduucuc=223 / 201(1/)−fuu mduuc=22201/−fumcuc=20fuumc=2(1)−fumc2.117F=dpdt=221/ 21(1/)−dmudtuc=2223 / 2221/ 2/1(1/)(1/)+−−ucdudumudtdtucuc=222223/ 2223/ 2(/)1/(1/)(1/)−+−−u cuucdumdtucuc=223 / 21(1/)−dumdtuc=3udum dt.

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Chapter 2 Special Relativity28(b)F=dum dtonly whenuis essentially unity, atspeeds much less than c.(c)F=223 / 21(1/)−dumdtuc→(F/m)dt=223/ 2(1/)−duuc→(F/m)223 / 2(1/)=−dudtuc→(F/m)t=221/ 2(1/)−uuc→221/ 2()(1/)−=F m tucu→2222() (1/)−=Ft mucu()222()1()=+Ft mFt mcuu=211()+F tmFt mc.(d)u→c.2.118t=820.99310 m/s9.8m/s= 3.03107s =0.96yr.(b)u=211(/)+FmFtmct→0.993108m/s =222819.8/9.8m/s1 +310 m/sm stmtm→(0.993108m/s)22289.8m/s1310 m/s+ t= (9.8m/s2)2t2t=8220.99310 m/s9.8m/s1(0.99)−= 2.15108s =6.8yr. Because whenuapproachescthe momentum begins togrow much more rapidly with speed than classically, force must be applied for a much greater time.2.119x=0211(/)=+ftFu dtt dtmFtmc=2202(/)1(/)+ftmcFmct dtFFtmc=2201(/)+ftmcFtmcF=()221(/)1+−mcFtmcF2.120dt=221/−ucdt=2221111(/)−+gtcgtcdt=2221(/)(/)1(/)+−+gtcgtcgtcdt=21(/)+dtgtc.Integrating both sides,t=201(/)+tdtgtc=1sinh−cgtgcort=sinhcgtgc

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