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PDF Solutions Solutions to exercises Solutions to exercises Exercise 1.1 A‘stationary’ particle in anylaboratory on theEarth is actually subject to gravitationalforcesdue to theEarth andthe Sun. Thesehelp to ensure that theparticle moveswith thelaboratory.Ifstepsweretaken to counterbalance theseforcessothatthe particle wasreally not subject to anynet force, then the rotation of theEarth andthe Earth’sorbital motionaround theSun would carry thelaboratory away from theparticle, causing theforce-free particle to followacurving path through thelaboratory.Thiswouldclearly show that the particle didnot have constantvelocity in the laboratory (i.e.constantspeed in a fixed direction) andhence that aframe fixed in the laboratory is not an inertial frame.More realistically,anexperimentperformed usingthe kind of long, freely suspendedpendulum known as a Foucaultpendulum couldreveal the fact that a frame fixed on theEarth is rotating andthereforecannot be an inertial frame of reference. An even more practical demonstrationisprovidedbythe winds,which do not flowdirectly from areas of high pressure to areas of lowpressure because of theEarth’srotation. - Exercise 1.2 TheLorentzfactor is γ(V )=1/ 1−V2/c2. (a) If V =0.1c,then 1 γ = - =1.01 (to 3s.f.). 1 − (0.1c)2/c2 (b) If V =0.9c,then 1 γ = - =2.29 (to 3s.f.). 1 − (0.9c)2/c2 Notethatitisoften convenient to write speedsinterms of c instead of writingthe values in ms−1,because of thecancellation between factorsofc. ? @ AB Exercise 1.3 2 × 2 M = Theinverse of a matrix CDis ? @ 1 D −B M −1 = AD − BC −CA. Taking A = γ(V ), B = −γ(V )V/c, C = −γ(V)V/c and D = γ(V ),and noting that AD − BC =[γ(V)]2(1 − V 2/c2)=1,wehave ? @ γ(V )+γ(V)V/c [Λ]−1 = . +γ(V)V/cγ(V) This is thecorrect form of theinverse Lorentztransformation matrix. Exercise 1.4 First computethe Lorentzfactor: - γ(V )=1/ 1−V2/c2 - - =1/ 1−9/25 =1/ 16/25 =5/4. Thus themeasured lifetimeisΔT =5×2.2/4µs=2.8µs. Note that not all muons live for thesametime; rather,theyhavearangeoflifetimes.But alarge group of muons travelling with acommonspeed doeshaveawell-defined mean lifetime,and it is thedilationofthisquantity that is easily demonstrated experimentally. 279 Solutions to exercises Exercise 1.5 Thealternative definitionoflengthcan’t be used in the rest frame of therod as the rod does not move in its ownrestframe.The proper lengthis therefore defined as beforeand related to the positions of thetwo events as observedinthe restframe.(This works, because event 1 andevent 2 still occurat theend-pointsofthe rod andthe rod nevermovesinthe restframe S%.) As before,itishelpful to write down all the intervals that areknown in atable. EventS(laboratory) S% (restframe) % % 2(t2,0) (t2,x2) % % 1(t1,0) (t1,x1) % % % % Intervals (t2 − t1, 0) (t2 − t1,x2 −x1) ≡(Δt, Δx) ≡ (Δt%, Δx%) Relation to intervals (L/V,0) (?,LP) By examiningthe intervals,itcan be seen that Δx, Δt and Δx% areknown. From theintervaltransformation rules,onlyEquation1.33 relates thethree known intervals.Substitutingthe known intervals into that equationgives LP = γ(V )(0 − V (L/V )).Inthisway,lengthcontractionispredicted as before: L = LP/γ(V ). Exercise 1.6 Thereceivedwavelength is lessthanthe emitted wavelength. This meansthatthe jet is approaching. We can therefore useEquation1.42 providedthatwec-hange thesignofV.Combining it with theformula fλ = c showsthat λ% = λ (c − V )/(c + V ).Squaring bothsides andrearranginggives (λ%/λ)2 =(c−V)/(c+V). Fromthisitfollows that (λ%/λ)2(c + V )=(c−V), so V (1 +(λ%/λ)2)=c(1 − (λ%/λ)2), thus V = c(1 − (λ%/λ)2)/(1 +(λ%/λ)2). Substituting λ% =4483 × 10−10 mand λ =5850 × 10−10 m, thespeed is found to be v =0.26c (to 2s.f.). Exercise1.7 Letthe spacestation be theoriginofframe S, andthe nearer of the spacecraftthe originofframe S%,which therefore moveswith speed V = c/2 as measured in S. Letthese twoframesbeinstandard configuration. Thevelocity of thefurtherofthe twospacecraft, as observedinS,isthen v =(3c/4, 0, 0).It follows from thevelocity transformation that the velocity of thefurtherspacecraft % % as observedfrom thenearer will be v =(vx,0,0),where v − V 3c/4 − c/2 v% = x = =2c/5. x 2 2 1 − vxV/c 1 − (3c/4)(c/2)/c Exercise 1.8 Δx =(5−7) m = −2 mand c Δt =(5−3) m =2m. Sincethe spacetimeseparation is (Δs)2 =(cΔt)2−(Δx)2 in this case, it follows that 280 Solutions to exercises (Δs)2 =(2m)2−(2 m)2 =0.The value (Δs)2 =0is permitted;itdescribes situations in which thetwo events couldbelinkedbyalight signal. In fact, any such separationissaid to be light-like. Exercise 1.9 Startwith (Δs%)2 =(cΔt%)2−(Δx%)2.The aim is to show that (Δs%)2 =(Δs)2. Substitute Δx% = γ(Δx − V Δt) and c Δt% = γ(c Δt − V Δx/c) so that 1 ) (Δs%)2 = γ2 c2(Δt)2 − 2V ΔxΔt + V 2(Δx)2/c2 1 ) − γ2 (Δx)2 − 2V ΔxΔt + V 2(Δt)2 . Crossterms involving Δx Δt cancel. Collectingcommonterms in c2(Δt)2 and (Δx)2 gives (Δs%)2 = γ2c2(Δt)2(1 − V 2/c2) − γ2(Δx)2(1 − V 2/c2). Finally,notingthat γ2 =[1−V2/c2]−1,there is acancellation of terms, giving (Δs%)2 = c2(Δt)2 − (Δx)2 =(Δs)2, thus showingthat (Δs%)2 =(Δs)2. Exercise 1.10 Since (Δs)2 =(cΔt)2−(Δl)2,and (Δs)2 is invariant, it follows that all inertial observers will find (c Δt)2 =(Δs)2+(Δl)2,where (Δl)2 cannot be negative.Since (Δl)2 =0in theframe in which theproper timeis measured, it follows that no otherinertial observercan findasmaller valuefor the timebetween the events. Exercise 1.11 In Terra’s frame,Stella’sshiphas velocity (vx,vy,vz)=(−V,0,0).Itfollows from thevelocity transformation that % in Astra’s frame,the velocity of Stella’sshipwill be (vx, 0, 0),where % 2 vx =(vx−V)/(1 − vxV/c ).Taking vx = −V gives (−V − V ) −2V v% = = . x (1 − (−V )V/c2) 1+V2/c2 Taking themagnitude of this singlenon-zero velocity component givesthe speed of approach, 2V/(1 + V 2/c2),asrequired. Exercise 1.12 In Terra’s frame,the signals wouldhaveanemitted frequency fem =1Hz.InAstra’s frame,the Doppler effect tells us that thesignals wouldbe receivedwith adifferent frequency frec.Onthe outward legofthe journey, the signals wouldberedshifted andthe receivedfrequency wouldbe - frec = fem (c − V )/(c + V ). On thereturnleg of thejourney, thesignals wouldbeblueshiftedand thereceived frequency wouldbe - frec = fem (c + V )/(c − V ). Exercise 2.1 TheLorentzfactor is - - - γ =1/ 1−v2/c2 =1/ 1−16c2/25c2 =1/ 9/25 =5/3. 281 Solutions to exercises Theelectron hasmass m =9.11 × 10−31 kg. Thus themagnitude of theelectron’s momentum is p =5/3×4c/5×m =(5/3)×(4×3.00×108 ms−1/5)×9.11×10−31 kg =3.6×10−22 kg ms−1. 2 Exercise 2.2 Thekinetic energy is EK =(γ−1)mc .Taking the speed to be 9c/10,the Lorentzfactor is - - γ =1/ 1−v2/c2 =1/ 1−(9/10)2 =2.29. Notingthat m =1.88 × 10−28 kg, thekinetic energy is −28 8 −1 2 −11 EK =(2.29 − 1) × 1.88 × 10 kg × (3.00 × 10 ms ) =2.2×10 j. Exercise 2.3 v =3c/5 corresponds to aLorentzfactor - - γ(v)=1/ 1−v2/c2 =1/ 1−9/25 =5/4. −27 Theproton hasmass mP =1.67 × 10 kg, therefore thetotal energy is E = γ(v)mc2 =(5/4)×1.67×10−27 kg×(3.00×108 ms−1)2 =1.88×10−10 j. 2 Exercise 2.4 Sincethe total energy is E = γmc ,itisclear that th-etotal energy is twice themassenergywhen γ =2.Thismeansthat 2=1/ 1−v2/c2. 2 2 2 2 Squaring andinverting boths√ides, 1/4=1−v /c ,sov /c =3/4.Taking the positive square root, v/c = 3/2. Exercise 2.5 (a) TheenergydifferenceisΔE=Δmc2,where Δm =3.08 × 10−28 kg. Thus ΔE =3.08 × 10−28 kg × (3.00 × 108 ms−1)2 =2.77 × 10−11 j. Convertingtoelectronvolts,thisis 2.77 × 10−11 j/1.60 × 10−19 JeV−1 =1.73 × 108 eV =173 MeV. (b) From ΔE =Δmc2,the mass difference is Δm =ΔE/c2.Now, ΔE =13.6eV or,convertingtojoules, ΔE =13.6eV × 1.60 × 10−19 JeV−1 =2.18 × 10−18 j. Therefore Δm =2.18 × 10−18 j/(3.00 × 108 ms−1)2 =2.42 × 10−35 kg. Note that themassesofthe electron andproton are 9.11 × 10−31 kg and 1.67 × 10−27 kg, respectively,sothe mass difference from chemical binding is small enough to be negligible in most cases.However,mass–energy equivalence is not unique to nuclear reactions. % Exercise 2.6 Thetransformationsare E = γ(V )(E − Vpx√)and % 2 2 2 px = γ(V )(px − VE/c ).Inthiscase, E =3mec and px = 8mec .For relative-speed V =4c/5 between the twoframes, theLorentzfactor is γ =1/ 1−(4/5)2 =5/3.Substitutingthe values, √ % 2 2 E =5/3(3mec − 4c/5 × 8mec)=1.23mec 282 Solutions to exercises and √ % 2 2 p =5/3( 8mec − 4c/5 × 3mec /c )=0.714mec. Exercise 2.7 (a) Foraphoton m =0,so 6.63 × 10−34 Js×5.00 × 1014 s−1 p=E/c = hf/c = =1.11×10−27 kg ms−1. 3.00 × 108 ms−1 (b) Usingthe Newtonian relationthatthe force is equaltothe rateofchange of momentum (weshall have more to sayabout this later), themagnitude of the forceonthe sail will be F = np,where n is therate at which photons are absorbedbythe sail (number of photons persecond). Thus n = F/p =10N/1.11 × 10−27 kg ms−1 =9.0×1027 s−1.
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