Relativity. Chapter 26. Quick Quizzes

Transcription

1 Chater 6 elatiity Quik Quizzes. (a). Less time will hae assed for you in your frame of referene than for your emloyer bak on Earth. Thus, to maximize your ayhek, you should hoose to hae your ay alulated aording to the elased time on a lok on Earth.. No. Your sleeing abin is at rest in your referene frame, thus, it will hae its roer length. There will be no hange in measured lengths of objets within your saeraft. Another obserer, on a saeraft traeling at a high seed relatie to yours, will laim that you are able to fit into a shorter sleeing abin (if your body is oriented in a diretion arallel to your diretion of trael relatie to the other obserer).. (a), (e). The outgoing roket will aear to hae a shorter length and a slower lok. The answers are the same for the inoming roket. Length ontration and time dilation deend only on the magnitude of the relatie eloity, not on the diretion. 4. (a) False (b) False () True (d) False A refleted hoton does exert a fore on the surfae. Although a hoton has zero mass, a hoton does arry momentum. When it reflets from a surfae, there is a hange in the momentum, just like the hange in momentum of a ball bouning off a wall. Aording to the momentum interretation of Newton s seond law, a hange in momentum results in a fore on the surfae. This onet is used in theoretial studies of sae sailing. These studies roose building non-owered saeraft with huge refletie sails oriented erendiularly to the rays from the Sun. The large number of hotons from the Sun refleting from the surfae of the sail will exert a fore whih, although small, will roide a ontinuous aeleration. This would allow the saeraft to trael to other lanets without fuel. 5. (a). The downstairs lok runs more slowly beause it is loser to Earth and hene exerienes a stronger graitational field than the ustairs lok does. 6

2 64 CHAPTE 6 Answers to Een Numbered Conetual Questions. To be stritly orret, the equation should be written as E γ m where γ. When the objet is at rest and γ, so the equation reerts to the oular form E m > KE ( γ ). When, E> E m and the differene aounts for the kineti energy of the moing mass. m aurately E E 4. The two obserers will agree on the seed of light and on the seed at whih they moe relatie to one another. 6. Seial relatiity desribes inertial referene frames -- that is, referene frames that are not aelerating. General relatiity desribes all referene frames.. You would see the same thing that you see when looking at a mirror when at rest. The theory of relatiity tells us that all exeriments will gie the same results in all inertial frames of referene.. The lok in orbit will run more slowly. The extra entrietal aeleration of the orbiting lok makes its history fundamentally different from that of the lok on Earth.. The lightyears reresents the roer length of a rod from Earth to Sirius, measured by an obserer seeing both the rod and Sirius nearly at rest. The astronaut sees Sirius oming toward her at. but also sees the distane ontrated to d ly ly. 5 ly d 5 ly 5 yr So, the trael time measured on her lok is t 6yr...

3 elatiity 65 Answers to Een Numbered Problems 5. (a).9 s (b). fringe shifts 4. (a). yr (b). lightyears 6. (a) 7 beats min (b) beats min. (a) 65. min (b).6 min. (a) a retangular box (b) sides erendiular to the motion are. m long, sides arallel to the motion are. m long. (a).97l (b). 4. (a) 7.4 m (b). with reset to the diretion of motion 6. (a). yr (b) 4.7 ly ().5 ly (d) 5.7 yr. (a) () 4.7 kg m s (b) 5.6 kg m s.6 kg m s. (a).4 (b) to the left Hz. (a) 99 MeV (b). GeV ().7 GeV..9. (a).5 MeV (b).45 MeV 4. m faster.5 kg, mslower.4 kg 6. (a).5 MeV (b).5 Hz..64 MeV,.4 kg m s for eah hoton 4. (a).5 MeV (b) 4.6 GeV 4. (a) 5. yr (b) 5. yr (). ly

4 66 CHAPTE m away from Earth 4. (a) ~ s (b) 5. (a) 4. J (b) ~ km 4. J 5. (b) (a) d t + (b) d t km 5. (a) $ (b) 9 $.

5 elatiity 67 Problem Solutions 6. (a) As the lane flies from O to B along ath I, ground air + wind gies m m ground +. s s t OB L m s ground m s, and m.67 s ur air ur ground ur wind For the lane following ath II from O to A, ground air + wind with erendiular to eah other. The Pythagorean theorem then yields ground and wind and m m m ground air wind. 9. s s s t OA L 9. m s ground m.4 s ur ur wind air ur ground (b) For the return flight along ath I, ground air + wind gies so m m ground.. s s t BO ground m s ur air ur ur L m wind ground.5 s. m s As the lane flies from A to O along ath II, ground and erendiular to eah other. The Pythagorean theorem gies wind are again 9. s m ground air wind ur air ur ground and t AO L ground.4 s ur wind

6 6 CHAPTE 6 () The total times of flight are t t t I OB + BO.67 s +.5 s 4.7 s and t t t II OA + AO.4 s 4. s The differene in total flight times is t ti tii 9 s 6. (a) L tnet 4 ( ) (. m s) m. ms 5.9 s (b) Sine a fringe shift ours for eery half-waelength hange made in the otial ath length, the number of fringe shifts exeted is ( ) N d net t net λ λ 5 ( )( ). m s.9 s 9 55 m. fringe shifts 6. t. s t γ t -(.) 5. s 6.4 (a) Obserers on Earth measure the time for the astronauts to reah Alha Centauri as 4.4 yr. But these obserers are moing relatie to the astronaut s internal t E biologial lok and hene, exeriene a dilated ersion of the roer time interal t measured on that lok. From te γ t, we find ( ) (.4 yr) t t γ t yr E E (b) The astronauts are moing relatie to the san of sae searating Earth and Alha Centauri. Hene, they measure a length ontrated ersion of the roer distane, L 4. ly. The distane measured by the astronauts is L L γ L ( ) 4. ly.95. ly

7 elatiity (a) Sine your shi is idential to his, the roer length of your shi (that is, the length you measure for your shi at rest relatie to you) must be the same as the roer length that he measured for his own shi. Thus, both shis must be m meters long when measured by obserers at rest relatie to them. (b) His shi moes relatie to you at the same seed your shi moes relatie to him. Thus, you obsere the same length ontration of his shi as he obsered for your shi and measure his shi to be 9 m in length. () From γ L L L with L m and L 9 m, we find ( ) L L 9 m m. 6.6 (a) The time for 7 beats, as measured by the astronaut and any obserer at rest with reset to the astronaut, is. min. The obserer in the shi then measures a rate of 7 beats min. t (b) The obserer on Earth moes at.9 relatie to the astronaut and measures the time for 7 beats as t t γ t. min -(.9). min This obserer then measures a beat rate of 7 beats. min beats min 6.7 (a) t.6 s t γ t -(.9) 7. s 7 (b) d ( t).9. m s. s m () d ( t ).9. m s.6 s 7.6 m

8 7 CHAPTE 6 6. (a) As measured by obserers in the shi (that is, at rest relatie to the astronaut), the time required for 75. ulses is. min. t The time interal required for 75. ulses as measured by the Earth obserer is. min t γ t, so the Earth obserer measures a ulse rate of rate t. min 65. min (b) If.99, then t γ t. min.99 and the ulse rate obsered on Earth is rate t. min.6 min That is, the life san of the astronaut (rekoned by the total number of his heartbeats) is muh longer as measured by an Earth lok than by a lok aboard the sae ehile. 6.9 As seen by an Earth based obserer, the time for the muon to trael 4.6 km is d 4.6 m t.99. m s ( ) (a) In the rest frame of the muon, this time (the roer lifetime) is t 4.6 m 6 t (.99). s. µ s γ.99(. m s) (b) The muon is at rest in its frame and thus traels zero distane as measured in this frame. Howeer, during this time interal, the muon sees Earth moe toward it by a distane of (.99. m s) (. 6 s) d t 6.5 m.65 km

9 elatiity 7 6. Length ontration ours only in the dimension arallel to the motion. (a) The sides labeled L and L in the figure at the right are unaffeted, but the side labeled L will aear ontrated giing the box a retangular shae. (b) The dimensions of the box, as measured by the obserer moing at. relatie to it, are est Frame of Box. L L L L. m, L L. m, and L L L. m.. m 6. The roer length of the faster shi is three times that of the slower shi ( Lf Ls ) they both aear to hae the same ontrated length, L. Thus, ( ), or ( ) 9 9( ) L L L s s s f s f, yet This gies f ( s ) (a) Obserer A measures the roer length, LA L, of the rod that is at rest relatie to her. Howeer, obserer B is moing at.955 relatie to this rod and will measure the ontrated length L A for it, where L (. ) L L γ L L A A

10 7 CHAPTE 6 (b) Obserer A is moing at.955 relatie to the rod at rest in B s referene frame, and she measures the ontrated length LB L for the length of this rod. This rod s roer length L B (the length measured by obserer B who is at rest relatie to this rod) is found from L L as B B L B LB L (.955).7L Thus, the ratio of the length of A s rod to the length of B s rod, as measured by obserer B, is ratio L L.97L.7L A B. 6. The trakside obserer sees the suertrain length-ontrated as L L m.95 m The suertrain aears to fit in the tunnel with 5 m m 9 m to sare 6.4 Note: Exess digits are retained in some stes gien below to more learly illustrate the method of solution. Obserer s rest frame rod s rest frame.995 We are gien that L. m, and θ. (both measured in the obserer s rest frame). The omonents of the rod s length as measured in the obserer s rest frame are L y L y L q L Losθ. m os..7 m x L x L x and L Lsinθ. m sin.. m y The omonent of length arallel to the motion has been ontrated, but the omonent erendiular to the motion is unaltered. Thus, L L. m and L x Lx.7 m 7.4 m (.995) y y

11 elatiity 7 (a) The roer length of the rod is then x y L L + L 7.4 m +. m 7.4 m (b) The orientation angle in the rod s rest frame is L y. m θ tan tan. L x 7.4 m 6.5 The entrietal aeleration is roided by the graitational fore, so m r GMm, giing Cooer s seed as r / GM GM r ( E + h) / or 4 ( 6.67 N m kg )( 5.9 kg) 6 6 ( ) m 7. m s Then the time eriod of one orbit is, 6 π ( ) 6.54 m E + h π T 7. m s (a) The time differene for orbits is 5.6 s / t t ( γ ) t T t t + T 7. m/s ( 5.6 s) 9. µ s. m/s (b) For one orbit, 9. µ s t t.7 µ s The ress reort is aurate to one digit

12 74 CHAPTE (a) t 5. yr t γ t (.7). yr (b) d ( t) [ ].7. yr.7. ly yr. yr 4.7 ly () The astronauts see Earth flying out the bak window at.7 : ( ) [ ] ( 5. yr ) d t.7 5. yr.7. ly yr.5 ly (d) Mission ontrol gets signals for. yr while the battery is oerating and then for 4.7 yr after the battery stos owering the transmitter, 4.7 ly away:. yr yr 5.7 yr 6.7 The momentum of the eletron is ( 9. kg)(.9) (.9 kg) e γ m e.9 If the roton has the same momentum, then ( 7.67 kg γ m (.9 kg) ) whih redues to.9 and yields ( ) ( )( )... m s. 5 m s 6. The momentum of the eletron is e γ m e ( 9. kg). (a) When., e ( kg)(.)(. m s) kg m s (b) If.5, e.6 kg m s, and

13 elatiity 75 () When.9, e 5.6 kg m s 6.9 Momentum must be onsered, so the momenta of the two fragments must add to zero. Thus, their magnitudes must be equal, or (.5 kg)(.9) γ m.9 ( 4.96 kg) For the heaier fragment, ( 7.67 kg ) ( 4.96 kg) whih redues to.7 and yields.5 6. Using the relatiisti form, m γ m we find the differene from the lassial momentum, m : γ m m ( γ )m (a) The differene is.% when ( γ ) m. γ m : γ or.4 (b) The differene is.% when ( γ ) m.γ m : γ.9.9 or Taking toward the right as the ositie diretion, with L eloity of roton relatie to laboratory, eloity of roton relatie to eletron, and el eloity of eletron e relatie to laboratory, the relatiisti eloity addition equation gies L e + el.7+.9 e el (.7)(.9)

14 76 CHAPTE 6 6. Taking toward the right as ositie, with L eloity of L relatie to, LE eloity of L relatie to Earth, E eloity of relatie to Earth, and E E eloity of Earth relatie to, the relatiisti eloity addition equation gies LE E L LE E.7+ (. 7) (.7)(.7) EO eloity of Enterrise relatie to Obserers on Earth. EO.9 KO. KO eloity of Klingon shi relatie to Obserers on Earth. OK KO eloity of Obserers on Earth relatie to Klingon shi The relatiisti eloity addition equation yields + + ( ) (.9 )(. ).9. EO OK EK + EO OK With S eloity of roket relatie to shi, E eloity of roket relatie to Earth, SE eloity of shi relatie to Earth, and ES SE eloity of Earth relatie to shi, the relatiisti eloity addition equation gies + ( ) (.95)(.75) S E ES + E ES Taking toward the right as ositie, with BA eloity of B relatie to A, A eloity of roket relatie to A, B eloity of roket relatie to B, and B B eloity of B relatie to roket, the relatiisti eloity addition equation gies B A BA B A (.95) (.9) (.95) (.9)

15 elatiity First, determine the eloity of the ulsar relatie to the roket. Taking toward Earth as ositie, with P eloity of ulsar relatie to roket, PE eloity of ulsar relatie to Earth, E eloity of roket relatie to Earth, and E E eloity of Earth relatie to roket, the relatiisti eloity addition equation gies + (.995) (. 95 )(.995 ) +.95 PE E P PE E The eriod of the ulsar in its own referene frame is roket s frame of referene is. s, and its eriod in the t t. s t γ ( t ) 6. s ( P ) (.9997) and the frequeny in the roket s frame is f t 6. s.6 Hz 6.7 The instrutors measure a roer time of t 5 min on their lok. (a) First determine the eloity of the students relatie to the instrutors (and hene relatie to the offiial lok). Taking toward the right in Figure P6.7 as the ositie diretion, with SI eloity of students relatie to instrutors, SE eloity of students relatie to Earth, IE eloity of instrutors relatie to Earth, and eloity of Earth relatie to instrutors, we find EI IE + + ( ) (.6)(.) SE EI SI SE EI The elased time on the student s lok is then t t γ t (b) The elased time on Earth is 5 min ( ) (.5 SI ) 54 min t 5 min t γ t. ( EI ) ( ) 5 min

16 7 CHAPTE 6 MeV E m.67 kg. m s 99 MeV -.6 J 7 6. (a) (b) E γm γ E E 99 MeV.95. MeV. GeV () KE E E. MeV 99 MeV.7 MeV.7 GeV 6.9 If KE E, then giing γ ( γ ) KE E E E E Therefore, γ 4 ( ) 6. The work done aelerating a free artile from rest to seed equals the kineti energy gien that artile. Thus, KE E E γ E E 75 MeV 7 For a roton, E (.67 kg) (. ) m MeV 99 MeV -.6 J Thus, 75 MeV γ.99 giing 99 MeV γ 4.99 or

17 elatiity The nonrelatiisti exression for kineti energy is KE m, while the relatiisti exression is KE E E ( γ ) E ( γ ) m where γ. Thus, when the relatiisti kineti energy is twie the redited nonrelatiisti alue, we hae m m or + Squaring both sides of the last result and simlifying gies 4 + Ignoring the triial solution, we must hae 4 + This is a quadrati equation of the form x x with x ( ) quadrati formula gies Sine x ± 5 x, we ignore the negatie solution and find +. Alying the + 5 x.6 whih yields The energy inut to the eletron will be W E E ( γ γ ) E f i f i or W E ( ) ( ) where E.5 MeV f i (a) If f.9 and i.5, then W (.5 Me).5 MeV (.9 ) (.5)

18 CHAPTE 6 (b) When f.99 and i.9, we hae W (.5 Me).45 MeV ( 9) ( ) When a ubial box moes at seed, the dimension arallel to the motion is length ontrated and other dimensions are unaffeted. If all edges of the box had length L when at rest, the olume of the moing box is ( ) V L L L V The relatiisti density is m m. g ρ V V (. m) (.9).4 g m Note that E m m ρ as suggested in the roblem statement. V V V 6.4 Let m be the mass of the fragment moing at.6, and m be the mass moing at.97. From onseration of mass-energy, m E + m (.6) (.97) m giing.m () m.4 kg The momenta of the two fragments must add to zero, so the magnitudes must be equal, giing or γ m γ m. This yields ( ). m.6 6.m.97, or.5m () Substituting equation () into () gies 7.7 +, or 7 6. m.4 kg m Equation () then yields.5(.5 ) m.5 kg m kg.4 kg

19 elatiity 6.5 The total kineti energy equals the energy of the original hoton minus the total rest energy of the air, or KE E E. MeV.5 MeV.9 MeV γ 6.6 (a) The energy of the original hoton must equal the total energy (total kineti energy lus total rest energy) of the air. Thus, E KE E +.5 MeV + γ total.5 MeV.5 MeV (b) Eγ.5 MeV.6 J f 4 h 6.6 J s MeV.5 Hz 6.7 To rodue minimum energy hotons, both members of the roton-antiroton air should be at rest (that is, hae zero kineti energy) just before annihilation. Then, the total momentum is zero both before and after annihilation. This means that the two hotons must hae equal magnitude but oositely direted momenta, and hene, equal energies. From onseration of energy, ( ) energy of a roton is E E γ E + or Eγ E for eah hoton. The rest ( ev.6 ) 7 m.67 kg. m s M J 99 MeV f Eγ 99 MeV.6 J 4 h 6.6 J s MeV.7 Hz and. m s λ f.7 Hz 5. m. fm

20 CHAPTE 6 6. The total momentum is zero both before and after the annihilation. Thus, the momenta of the two hotons must hae equal magnitudes and be oositely direted. Sine E, the hoton energies are also equal and onseration of energy gies γ γ Eγ KE+ E E E or E E γ ( ) E.5 MeV E γ (.6).64 MeV γ Eγ Me.64 (.64 Me)(.6 J MeV). m s.4 kg m s 6.9 ( γ ) KE E E E so KE γ + E giing ( + KE E ) (a) When q( V ) e KE 5 V 5 ev, and E 99 MeV, this yields ev ( 99 ev) 5.. m s (b) When KE q( V ) e 5. V 5 MeV.75 ( + 5 MeV 99 MeV) 6.4 From E + with E 5E E, we find that E 4 (a) For an eletron,.5 MeV 4.5 MeV (b) For a roton, 99 MeV 4 MeV GeV

21 elatiity 6.4 E KE E. MeV.5 MeV.5 MeV, so E + E gies + + E E.5 MeV.5 MeV.4 MeV 6.4 (a) The astronomer on Earth measured both the seed and distane in his frame of referene. Thus, the time to imat on his lok is L. ly. yr t.. 5. yr () The obserer on the meteoroid sees Earth rushing at him at seed. but sees a length ontrated distane of searation gien by L L. ly.. ly (b) The time to imat as omuted by the obserer on the meteoroid is L. ly. yr t.. 5. yr 6.4 (a) Sine Ted and Mary are in the same frame of referene, they measure the same seed for the ball, namely u.. (b) L t u. m.. m s ( ) 7.5 s () The distane of searation, as measured by Jim, is L L. m.6.4 m Taking toward the right in Figure P6.4 as ositie, with relatie to Jim, BJ BT eloity of ball relatie to Ted, and TJ relatie to Jim, the relatiisti eloity addition equation gies eloity of ball eloity of Ted BT TJ BJ BT TJ (.)(.6 + ) +. Thus, aording to Jim, the ball moes with a seed of.

22 4 CHAPTE The lok, at rest in the shi s frame of referene, will measure a roer time of h before sounding. Obserers on Earth moe at.75 relatie to the lok t and measure an elased time of t h t γ ( t ) 5 h -(.75) The obserers on Earth see the lok moing away at traeled before the alarm sounds as.75 and omute the distane d t 6 s h..75. m s 5 h m 6.45 With nl eloity of nuleus relatie to laboratory, el eloity of eletron relatie to laboratory, en eloity of eletron relatie to nuleus, and ne en eloity of nuleus relatie to eletron, the relatiisti eloity addition equation gies ne el nl +.7 ne el + (.7)(.5) Taking away from Earth as ositie, with JE eloity of jet of material relatie to Earth, eloity of jet relatie to quasar, and JQ Earth, the relatiisti eloity addition equation gies JE JQ QE + JQ QE (.55)(. 7 + ) + QE eloity of quasar relatie to 6.47 (a) Obserers on Earth measure the distane to Andromeda to be d 6 6. ly. yr referene, is. The time for the tri, in Earth s frame of t γ t. yr

23 elatiity 5 The required seed is then 6 (. yr) d t. yr - 5 (.5 ) whih gies Squaring both sides of this equation and soling for yields (b) ( γ ) KE m, and γ Thus, ( ) KE () ost KE rate kg. m s 6. J kwh J $. kwh $.7.6 J ( 6. ) The solution to this roblem is easier if we start with art (b). First, omute the seed of the rotons. KE MeV KE ( γ ) E, so γ + +.6, or γ ~ E 99 MeV Thus, γ ~

24 6 CHAPTE 6 (b) The diameter of the galaxy, as seen in the roton s frame of referene, is 5 L ly 5 L L ( ) ~ ly γ 7 Sine 6 ly yr.56 s. m s m 6 5 m km L ly ~ km ly m (a) The roton sees the galaxy rushing by at. The time, in the roton s frame of referene for the galaxy to ass is or ( 5 yr) 5 7 L ly 5.56 s t ~ yr 6 s yr t ~ s L L. In 6.49 The length of the sae shi, as measured by obserers on Earth, is Earth s frame of referene, the time required for the shi to ass oerhead is Thus, ( ) L L L t 6 t m m.75 s s L (. m s) or m s s. m s 7.74 m.4.

25 elatiity 7 7. kg 6 m s 4. KE m J 6.5 (a) Classially, (b) 6 m s 7 γ.5. m s Using the aroximation x + for x<< gies γ + 6.4, so ( x) ( γ ) m ( KE ) 4. J kg. m s In the limit <<, the lassial and relatiisti equations yield the same results. 6.5 (a) Taking toward Earth as ositie, with LE eloity of lander relatie to Earth, LS eloity of lander relatie to shi, and SE eloity of shi relatie to Earth, the relatiisti eloity addition equation gies LS SE LE +.94 LS SE + (.)(.6) + (b) The length ontrated distane of searation as measured in the shi s frame of referene is 6 L L. ly.6.6 ly () The aliens obsere the.6-ly distane losing beause the robe nibbles into it from one end at., and Earth redues it from the other end at.6. Thus, (.6 yr). time 6 ly yr (d) KE ( γ ) m (. 946) 5 m ( 4. kg). 7.5 J s

26 CHAPTE The kineti energy gained by the eletron will equal the loss of otential energy, so KE q V e. MV. MeV (a) If Newtonian mehanis remained alid, then would be KE m, and the seed attained ( ) KE. MeV.6 J MeV m m (b) KE ( γ ) kg s KE. MeV E, so γ + +. E.5 MeV The atual seed attained is γ (a) When at rest, muons hae a mean lifetime of t. µ s. In a frame of referene where they moe at.95, the dilated mean lifetime of the muons will be t. µ s τ γ ( t ) 7. µ s (.95) (b) In a frame of referene where the muons trael at.95, the time required to trael. km is d. m t.95. m s ( ) 5.5 s.5 µs If 4 N 5. muons started the. km tri, the number remaining at the end is.5 s 7. s N Ne 5. e. t τ 4 µ µ 4

27 elatiity (a) An obserer at rest relatie to the mirror sees the light trael a distane D d x, where x t is the distane the shi moes toward the mirror in time t. Sine this obserer agrees that the seed of light is, the time for it to trael distane D is D d t t. Soling for t yields t d + (b) The obserer in the roket sees a length-ontrated initial distane to the mirror of L d and the mirror moing toward the shi at seed. Thus, he measures D L y, where y t is the distane the the distane the light traels as mirror moes toward the shi before the light reflets off it. This obserer also measures the seed of light to be, so the time for it to trael D t distane D is t d Soling for t and simlifying yields t d The students are to measure a roer time of t T on the lok at rest relatie to them. The rofessor will measure a dilated time gien by ( ) t γ t T+t, or T T + t () where T is the time she should wait before sending the light signal, and t is the transit time for the signal, both measured in Earth s frame of referene. The distane, measured in Earth s frame, the signal must trael to reah the reeding d T + t, and the transit time is students is d t ( T t) + or T t ()

28 9 CHAPTE 6 Substituting equation () into () yields T T T + Thus, ( ) T T T ( )( + ) or T T The work required equals the inrease in the graitational otential energy, or GMSunm W. If this is to equal the rest energy of the mass remoed, then g m g GM m GM or g Sun g Sun ( 6.67 N m kg )(.99 kg) (. m s).47 m.47 km 6.57 (a) The omonents of length, measured in the frame moing with the rod, are L Losθ x and Ly L sin θ The stationary obserer will see a length ontrated omonent in the diretion arallel to the motion, with the other omonent unaffeted. Therefore, ( ) L L x Lx osθ and Ly Ly Lsinθ The length of the rod, as measured by the stationary obserer, is θ θ θ L L x + Ly L os os + sin or os L L θ (b) The orientation angle seen by the stationary obserer is gien by Ly Lsinθ tanθ, or tanθ γ tanθ L x L osθ

29 elatiity The seed of light in a auum, exressed in km h, is m km 6 s 9.. km h s m h At the seed limit, the momentum is m ( ) m( 9. km h) 9 lim lim 9. km h. km h lim Similarly, at 9 km h, m( 9 km h) exess momentum Fine k ( ) k lim lim m ( 9. km h), and with the fine roortional to the with onstant, we find that Fine $. $. k m 9 km h 9. km h m km h (a) When 9 km h, ( ) m( 9 km h) 9 m 9 km h. km h and the fine is ( 9 km h) $. Fine k ( lim ) m( 9 km h 9. km h ) $ ( m) km h (b) When 9 km h ( 9 km h) m m 9 km h. km h and the fine is 9 (.65 9 km h) $. Fine m(.65 km h 9. km h ) $. ( m) km h 9 9

30 9 CHAPTE Aording to Earth-based obserers, the times required for the two tris are L. yr For Seedo: T S.5 yr.95 S L. yr For Goslo: T G 6.67 yr.75 G Thus, after Seedo lands, he must wait and age at the same rate as lanet-based T T T yr 5.64 yr before Goslo obserers, for an additional arries. S G S The time required for the tri aording to Seedo s internal biologial lok (whih measures the roer time for his aging roess during the tri) is T T γ S S TS ( S ).5 yr yr When Goslo arries, Seedo has aged a total of ts T S + TS yr yr The time required for the tri aording to Goslo s internal biologial lok (and hene the amount he ages) is TG tg T G TG ( G ) ( 6.67 yr) (.75) 7.64 yr γ Thus, we see that when he arries, Goslo is older than Seedo, haing aged an additional t t 7.64 yr.9 yr 5.45 yr G S