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MASSACHUSETTSINSTITUTEOFTECHNOLOGY
PhysicsDepartment
Physics8.286:TheEarlyUniverse
October28,2018
Prof.AlanGuth
REVIEW
PROBLEMSFOR
QUIZ2
RevisedVersion*
QUIZDATE:Monday,November5,2018,duringthenormalclasstime.
COVERAGE:LectureNotes4,5,andthroughthesectionon\DynamicsofaFlat
Radiation-DominatedUniverse"ofLectureNotes6;ProblemSets4,5,and6;Wein-
berg,TheFirstThreeMinutes,Chapters4{7;InRyden'sIntroductiontoCosmol-
ogy,wehavereadChapters4,5,andSec.6.1duringthisperiod.Thesechapters,
however,parallelwhatwehavedoneorwillbedoinginlecture,soyoushouldtake
themasanaidtolearningthelecturematerial;therewillbenoquestionsexplicitly
basedonthesesectionsfromRyden.ButwehavealsoreadChapters10(Nucleosyn-
thesisandtheEarlyUniverse)and8(DarkMatter)inRyden,andthesearerelevant
materialforthequiz,exceptforSec.10.3(DeuteriumSynthesis).Wewillreturnto
deuteriumsynthesislaterinthecourse.YoucanalsoignoreRyden'sEqs.(10.11),
(10.12),and(10.13)fornow.Chapters4and5ofWeinberg'sbookarepackedwith
numbers;youneednotmemorizethesenumbers,butyoushouldbefamiliarwith
theirordersofmagnitude.Wewillnottakeo�forthespellingofnames,aslongas
theyarevaguelyrecognizable.Fordatesbefore1900,itwillbesuÆcientforyouto
knowwhenthingshappenedtowithin100years.Fordatesafter1900,itwillbesuÆ-
cientifyoucanplaceeventswithin10years.Youshouldexpectone25-pointproblem
basedonthereadings,andseveralcalculationalproblems.Oneoftheproblems
onthequizwillbetakenverbatim
(oratleastalmostverbatim)from
eithertheproblem
setslistedabove(extracreditproblemsincluded),or
from
thestarredproblemsfrom
thissetofReview
Problems.Thestarred
problemsaretheonesthatIrecommendthatyoureviewmostcarefully:Problems
6,7,8,13,15,17,19,and21.
PURPOSE:Thesereviewproblemsarenottobehandedin,butarebeingmadeavail-
abletohelpyoustudy.Theycomemainlyfromquizzesinpreviousyears.Insome
casesthenumberofpointsassignedtotheproblemonthequizislisted|
inall
suchcasesitisbasedon100pointsforthefullquiz.
Inadditiontothissetofproblems,youwill�ndonthecoursewebpagetheactual
quizzesthatweregivenin1994,1996,1998,2000,2002,2004,2005,2007,2009,
2011,2013,and2016.Therelevantproblemsfromthosequizzeshavemostlybeen
incorporatedintothesereviewproblems,butyoustillmaybeinterestedinlooking
atthequizzes,justtoseehowmuchmaterialhasbeenincludedineachquiz.The
*RevisedNovember2,2018:Problem23referstoatableofintegrals,whichwasnot
includedintheoriginalversionofthereviewproblems.
8.286QUIZ2REVIEW
PROBLEMS,FALL2018
p.2
coverageoftheupcomingquizwillnotnecessarilymatchexactlythecoveragefrom
previousyears,butIbelievethatallthesereviewproblemswouldbefairproblems
fortheupcomingquiz.Thecoverageforeachquizinrecentyearsisusuallydescribed
atthestartofthereviewproblems,asIdidhere.In2016we�nishedWeinberg's
bookbythetimeofQuiz2,butotherwisethecoveragewasthesameasthisyear.
REVIEW
SESSION:Tohelpyoustudyforthequiz,HonggeunKimwillholdareview
session,atatimeandplacetobeannounced.
FUTUREQUIZ:Quiz3willbegivenonWednesday,December5,2018.
INFORMATION
TO
BEGIVEN
ON
QUIZ:
Eachquizinthiscoursewillhaveasectionof\usefulinformation"foryourreference.
Forthesecondquiz,thisusefulinformationwillbethefollowing:
DOPPLER
SHIFT(Formotionalongaline):
z=v=u
(nonrelativistic,sourcemoving)
z=
v=u
1�v=u
(nonrelativistic,observermoving)
z= s1+�
1���1
(specialrelativity,with�=v=c)
COSMOLOGICALREDSHIFT:
1+z��observed
�emitted
=a(tobserved )
a(temitted )
SPECIALRELATIVITY:
TimeDilationFactor:
�
1
p1��2
;
��v=c
Lorentz-FitzgeraldContractionFactor:
RelativityofSimultaneity:
Trailingclockreadslaterbyanamount�`0 =c.
Energy-MomentumFour-Vector:
p�= �Ec
;~p �;~p= m0 ~v;E= m0 c2= q(m0 c2)2+j~pj 2c2;
p2�j~pj 2� �p0 �2
=j~pj 2�E2
c2
=�(m0 c)2
:
8.286QUIZ2REVIEW
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KINEMATICSOFAHOMOGENEOUSLY
EXPANDING
UNI-
VERSE:
Hubble'sLaw:v=Hr,
wherev=
recessionvelocityofadistantobject,H
=
Hubble
expansionrate,andr=distancetothedistantobject.
PresentValueofHubbleExpansionRate(Planck2018):
H0=67:66�0:42km-s �1-Mpc �1
ScaleFactor:`p (t)=a(t)`c;
where`p (t)isthephysicaldistancebetweenanytwoobjects,a(t)
isthescalefactor,and`cisthecoordinatedistancebetweenthe
objects,alsocalledthecomovingdistance.
HubbleExpansionRate:H(t)=
1a(t)
da(t)
dt
.
LightRaysinComovingCoordinates:Lightraystravelinstraight
lineswithphysicalspeedcrelativetoanyobserver.InCartesian
coordinates,coordinatespeeddxd
t=
ca(t).Ingeneral,ds2
=
g�� dx�dx�=0:
HorizonDistance:
`p;horizon (t)=a(t) Z
t0
ca(t 0)dt 0
= �3ct
( at,matter-dominated),
2ct
( at,radiation-dominated).
COSMOLOGICALEVOLUTION:
H2= �_aa �2
=8�3
G��kc2
a2
;
�a=�4�3
G ��+3pc
2 �a;
�m(t)=a3(t
i )
a3(t)�m(ti )(matter);
�r (t)=a4(t
i )
a4(t)�r (ti )(radiation):
_�=�3_aa �
�+
pc2 �;��=�c;where�c=3H2
8�G
:
8.286QUIZ2REVIEW
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EVOLUTION
OFA
MATTER-DOMINATED
UNIVERSE:
Flat(k=0):
a(t)/t2=3
=1:
Closed(k>0):
ct=�(��sin�);
apk=�(1�cos�);
=
2
1+cos�>1;
where��4�3G�
c2 �apk �3
:
Open(k<0):
ct=�(sinh���);
ap�=�(cosh��1);
=
2
1+cosh�<1;
where��4�3G�
c2 �ap� �3
;
���k>0:
MINKOWSKIMETRIC(SpecialRelativity):
ds2��c2d�2=�c2dt2+dx2+dy2+dz2:
ROBERTSON-WALKER
METRIC:
ds2��c2d�2=�c2dt2+a2(t) �dr2
1�kr2+r2 �d�2+sin2�d�2 � �:
Alternatively,fork>0,wecande�ner=sin p
k,andthen
ds2��c2d�2��c2dt2+~a2(t) �d 2+sin2 �d�2+sin2�d�2 �;
where~a(t)=a(t)= pk.Fork<0wecande�ner=sinh
p�k,andthen
ds2��c2d�2=�c2dt2+~a2(t) �d 2+sinh2 �d�2+sin2�d�2 �;
where~a(t)=a(t)= p�k.Notethat~acanbecalledaifthereisnoneed
torelateittothea(t)thatappearsinthe�rstequationabove.
8.286QUIZ2REVIEW
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SCHWARZSCHILD
METRIC:
ds2��c2d�2=� �1�2GM
rc2 �c2dt2+ �1�2GM
rc2 �
�1
dr2
+r2d�2+r2sin2�d�2;
GEODESICEQUATION:
dds �gijdxj
ds �=12
(@i gk` )dxk
ds
dx`
ds
or:
dd� �g��dx�
d� �=12
(@�g��)dx�
d�
dx�
d�
8.286QUIZ2REVIEW
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p.6
PROBLEM
LIST
1.DidYouDotheReading(2000,2002)?.............7(Sol:34)
2.DidYouDotheReading(2007)?
...............8(Sol:35)
3.DidYouDotheReading(2011)?
...............12(Sol:39)
4.DidYouDotheReading(2013)?
...............13(Sol:41)
5.DidYouDotheReading(2016)?
...............13(Sol:42)
*6.EvolutionofanOpenUniverse
................16(Sol:45)
*7.AnticipatingaBigCrunch
..................16(Sol:45)
*8.TracingLightRaysinaClosed,Matter-DominatedUniverse
...16(Sol:46)
9.LengthsandAreasinaTwo-DimensionalMetric.........17(Sol:49)
10.GeometryinaClosedUniverse
................18(Sol:51)
11.TheGeneralSphericallySymmetricMetric
...........19(Sol:51)
12.VolumesinaRobertson-WalkerUniverse
............20(Sol:53)
*13.TheSchwarzschildMetric...................20(Sol:55)
14.Geodesics..........................21(Sol:58)
*15.AnExerciseinTwo-DimensionalMetrics............21(Sol:60)
16.GeodesicsontheSurfaceofaSphere..............22(Sol:62)
*17.GeodesicsinaClosedUniverse
................23(Sol:66)
18.ATwo-DimensionalCurvedSpace...............24(Sol:69)
*19.RotatingFramesofReference.................25(Sol:72)
20.TheStabilityofSchwarzschildOrbits..............27(Sol:75)
*21.PressureandEnergyDensityofMysteriousStu�.........28(Sol:79)
22.VolumeofaClosedThree-DimensionalSpace
..........29(Sol:80)
23.GravitationalBendingofLight
................31(Sol:81)
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PROBLEM
1:DID
YOU
DO
THEREADING
(2000,2002)
Parts(a)-(c)ofthisproblemcomefromQuiz4,2000,andparts(d)and(e)comefrom
Quiz3,2002.
(a)(5points)Bywhatfactordoestheleptonnumberpercomovingvolumeoftheuni-
versechangebetweentemperaturesofkT=10MeVandkT=0:1MeV?Youshould
assumetheexistenceofthenormalthreespeciesofneutrinosforyouranswer.
(b)(5points)Measurementsoftheprimordialdeuteriumabundancewouldgivegood
constraintsonthebaryondensityoftheuniverse.However,thisabundanceishard
tomeasureaccurately.WhichofthefollowingisNOTareasonwhythisishardto
do?
(i)Theneutroninadeuterium
nucleusdecaysonthetimescaleof15minutes,
soalmostnoneoftheprimordialdeuteriumproducedintheBigBangisstill
present.
(ii)ThedeuteriumabundanceintheEarth'soceansisbiasedbecause,beingheavier,
lessdeuteriumthanhydrogenwouldhaveescapedfromtheEarth'ssurface.
(iii)ThedeuteriumabundanceintheSunisbiasedbecausenuclearreactionstend
todestroyitbyconvertingitintohelium-3.
(iv)Thespectrallinesofdeuteriumarealmostidenticalwiththoseofhydrogen,so
deuteriumsignaturestendtogetwashedoutinspectraofprimordialgasclouds.
(v)Thedeuteriumabundanceissosmall(afewpartspermillion)thatitcanbe
easilychangedbyastrophysicalprocessesotherthanprimordialnucleosynthesis.
(c)(5points)Givethreeexamplesofhadrons.
(d)(6points)InChapter6ofTheFirstThreeMinutes,StevenWeinbergposedthe
question,\Whywastherenosystematicsearchforthis[cosmicbackground]radia-
tion,yearsbefore1965?"Indiscussingthisissue,hecontrasteditwiththehistory
oftwodi�erentelementaryparticles,eachofwhichwerepredictedapproximately20
yearsbeforetheywere�rstdetected.Nameoneofthesetwoelementaryparticles.
(Ifyounamethembothcorrectly,youwillget3pointsextracredit.However,one
rightandonewrongwillgetyou4pointsforthequestion,comparedto6pointsfor
justnamingoneparticleandgettingitright.)
Answer:
2ndAnswer(optional):
(e)(6points)InChapter6ofTheFirstThreeMinutes,StevenWeinbergdiscussesthree
reasonswhytheimportanceofasearchfora3 ÆKmicrowaveradiationbackground
wasnotgenerallyappreciatedinthe1950sandearly1960s.Choosethosethree
reasonsfromthefollowinglist.(2pointsforeachrightanswer,circleatmost3.)
8.286QUIZ2REVIEW
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(i)Theearliestcalculationserroneouslypredictedacosmicbackgroundtempera-
tureofonlyabout0:1 ÆK,andsuchabackgroundwouldbetooweaktodetect.
(ii)Therewasabreakdownincommunicationbetweentheoristsandexperimental-
ists.
(iii)Itwasnottechnologicallypossibletodetectasignalasweakasa3 ÆKmicrowave
backgrounduntilabout1965.
(iv)Sincealmostallphysicistsatthetimewerepersuadedbythesteadystatemodel,
thepredictionsofthebigbangmodelwerenottakenseriously.
(v)ItwasextraordinarilydiÆcultforphysiciststotakeseriouslyanytheoryofthe
earlyuniverse.
(vi)TheearlyworkonnucleosynthesisbyGamow,Alpher,Herman,andFollin,et
al.,hadattemptedtoexplaintheoriginofallcomplexnucleibyreactionsinthe
earlyuniverse.Thisprogramwasneververysuccessful,anditscredibilitywas
furtherunderminedasimprovementsweremadeinthealternativetheory,that
elementsaresynthesizedinstars.
PROBLEM
2:DID
YOU
DO
THEREADING
(2007)?(24points)
ThefollowingproblemwasProblem1ofQuiz2in2007.
(a)(6points)In1948RalphA.AlpherandRobertHermanwroteapaperpredicting
acosmicmicrowavebackgroundwithatemperatureof5K.Thepaperwasbased
onacosmologicalmodelthattheyhaddevelopedwithGeorgeGamow,inwhichthe
earlyuniversewasassumedtohavebeen�lledwithhotneutrons.Astheuniverse
expandedandcooledtheneutronsunderwentbetadecayintoprotons,electrons,and
antineutrinos,untilatsomepointtheuniversecooledenoughforlightelementsto
besynthesized.AlpherandHermanfoundthattoaccountfortheobservedpresent
abundancesoflightelements,theratioofphotonstonuclearparticlesmusthave
beenabout109.Althoughthepredictedtemperaturewasveryclosetotheactual
valueof2.7K,thetheorydi�eredfromourpresenttheoryintwoways.Circlethe
twocorrectstatementsinthefollowinglist.(3pointsforeachrightanswer;circleat
most2.)
8.286QUIZ2REVIEW
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(i)Gamow,Alpher,andHermanassumedthattheneutroncoulddecay,butnow
theneutronisthoughttobeabsolutelystable.
(ii)Inthecurrenttheory,theuniversestartedwithnearlyequaldensitiesofprotons
andneutrons,notallneutronsasGamow,Alpher,andHermanassumed.
(iii)Inthecurrenttheory,theuniversestartedwithmainlyalphaparticles,notall
neutronsasGamow,Alpher,andHermanassumed.(Note:analphaparticleis
thenucleusofaheliumatom,composedoftwoprotonsandtwoneutrons.)
(iv)Inthecurrenttheory,theconversionofneutronsintoprotons(andviceversa)
tookplacemainlythroughcollisionswithelectrons,positrons,neutrinos,and
antineutrinos,notthroughthedecayoftheneutrons.
(v)Theratioofphotonstonuclearparticlesintheearlyuniverseisnowbelieved
tohavebeenabout103,not109asAlpherandHermanconcluded.
(b)(6points)InWeinberg's\RecipeforaHotUniverse,"hedescribedtheprimordial
compositionoftheuniverseintermsofthreeconservedquantities:electriccharge,
baryonnumber,andleptonnumber.Ifelectricchargeismeasuredinunitsoftheelec-
troncharge,thenallthreequantitiesareintegersforwhichthenumberdensitycan
becomparedwiththenumberdensityofphotons.Foreachquantity,whichchoice
mostaccuratelydescribestheinitialratioofthenumberdensityofthisquantityto
thenumberdensityofphotons:
ElectricCharge:
(i)�109
(ii)�1000
(iii)�1
(iv)�10 �6
(v)eitherzeroornegligible
BaryonNumber:
(i)�10 �20
(ii)�10 �9
(iii)�10 �6
(iv)�1
(v)anywherefrom10 �5to1
LeptonNumber:
(i)�109
(ii)�1000
(iii)�1
(iv)�10 �6
(v)couldbeashighas�1,but
isassumedtobeverysmall
8.286QUIZ2REVIEW
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(c)(12points)The�gurebelowcomesfromWeinberg'sChapter5,andislabeledThe
ShiftingNeutron-ProtonBalance.
(i)(3points)Duringtheperiodlabeled\thermalequilibrium,"theneutronfraction
ischangingbecause(chooseone):
(A)Theneutronisunstable,anddecaysintoaproton,electron,andantineu-
trinowithalifetimeofabout1second.
(B)Theneutronisunstable,anddecaysintoaproton,electron,andantineu-
trinowithalifetimeofabout15seconds.
(C)Theneutronisunstable,anddecaysintoaproton,electron,andantineu-
trinowithalifetimeofabout15minutes.
(D)Neutronsandprotonscanbeconvertedfrom
oneintothroughreactions
suchas
antineutrino+proton !electron+neutron
neutrino+neutron !positron+proton:
(E)Neutronsandprotonscanbeconvertedfromoneintotheotherthrough
reactionssuchas
antineutrino+proton !positron+neutron
neutrino+neutron !electron+proton:
(F)Neutronsandprotonscanbecreatedanddestroyedbyreactionssuchas
proton+neutrino !positron+antineutrino
neutron+antineutrino !electron+positron:
8.286QUIZ2REVIEW
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(ii)(3points)Duringtheperiodlabeled\neutrondecay,"theneutronfractionis
changingbecause(chooseone):
(A)Theneutronisunstable,anddecaysintoaproton,electron,andantineu-
trinowithalifetimeofabout1second.
(B)Theneutronisunstable,anddecaysintoaproton,electron,andantineu-
trinowithalifetimeofabout15seconds.
(C)Theneutronisunstable,anddecaysintoaproton,electron,andantineu-
trinowithalifetimeofabout15minutes.
(D)Neutronsandprotonscanbeconvertedfromoneintotheotherthrough
reactionssuchas
antineutrino+proton !electron+neutron
neutrino+neutron !positron+proton:
(E)Neutronsandprotonscanbeconvertedfromoneintotheotherthrough
reactionssuchas
antineutrino+proton !positron+neutron
neutrino+neutron !electron+proton:
(F)Neutronsandprotonscanbecreatedanddestroyedbyreactionssuchas
proton+neutrino !positron+antineutrino
neutron+antineutrino !electron+positron:
(iii)(3points)Themassesoftheneutronandprotonarenotexactlyequal,but
instead
(A)Theneutronismoremassivethanaprotonwitharestenergydi�erenceof
1.293GeV(1GeV=109eV).
(B)Theneutronismoremassivethanaprotonwitharestenergydi�erenceof
1.293MeV(1MeV=106eV).
(C)Theneutronismoremassivethanaprotonwitharestenergydi�erenceof
1.293KeV(1KeV=103eV).
(D)Theprotonismoremassivethananeutronwitharestenergydi�erenceof
1.293GeV.
(E)Theprotonismoremassivethananeutronwitharestenergydi�erenceof
1.293MeV.
(F)Theprotonismoremassivethananeutronwitharestenergydi�erenceof
1.293KeV.
8.286QUIZ2REVIEW
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(iv)(3points)Duringtheperiodlabeled\eraofnucleosynthesis,"(chooseone:)
(A)Essentiallyalltheneutronspresentcombinewithprotonstoformhelium
nuclei,whichmostlysurviveuntilthepresenttime.
(B)Essentiallyalltheneutronspresentcombinewithprotonstoformdeuterium
nuclei,whichmostlysurviveuntilthepresenttime.
(C)Abouthalftheneutronspresentcombinewithprotonstoformheliumnu-
clei,whichmostlysurviveuntilthepresenttime,andtheotherhalfofthe
neutronsremainfree.
(D)Abouthalftheneutronspresentcombinewithprotonstoformdeuterium
nuclei,whichmostlysurviveuntilthepresenttime,andtheotherhalfof
theneutronsremainfree.
(E)Essentiallyalltheprotonspresentcombinewithneutronstoformhelium
nuclei,whichmostlysurviveuntilthepresenttime.
(F)Essentiallyalltheprotonspresentcombinewithneutronstoformdeuterium
nuclei,whichmostlysurviveuntilthepresenttime.
PROBLEM
3:DID
YOU
DO
THEREADING
(2011)?(20points)
ThefollowingproblemcomesfromQuiz2,2011.
(a)(8points)Duringnucleosynthesis,heaviernucleiformfromprotonsandneutrons
throughaseriesoftwoparticlereactions.
(i)InTheFirstThreeMinutes,Weinbergdiscussestwochainsofreactionsthat,
startingfromprotonsandneutrons,endupwithhelium,He4.Describeatleast
oneofthesetwochains.
(ii)Explainbrie ywhatisthedeuterium
bottleneck,andwhatisitsroleduring
nucleosynthesis.
(b)(12points)InChapter4ofTheFirstThreeMinutes,StevenWeinbergmakesthe
followingstatementregardingtheradiation-dominatedphaseoftheearlyuniverse:
Thetimethatittakesfortheuniversetocoolfrom
onetemperaturetoanotheris
proportionaltothedi�erenceoftheinversesquaresofthesetemperatures.
Inthispartoftheproblemyouwillexploremorequantitativelythisstatement.
(i)Foraradiation-dominateduniversethescale-factora(t)/t1=2.Findthecosmic
timetasafunctionoftheHubbleexpansionrateH.
(ii)Themassdensitystoredinradiation�risproportionaltothetemperatureT
tothefourthpower:i.e.,�r '�T4,forsomeconstant�.Forawiderangeof
temperatureswecantake�'4:52�10 �32kg�m�3�K�4.Ifthetemperature
8.286QUIZ2REVIEW
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ismeasuredindegreesKelvin(K),then�rhasthestandardSIunits,[�r ]=
kg�m�3.UsetheFriedmannequationfora atuniverse(k=0)with�=�r
toexpresstheHubbleexpansionrateH
intermsofthetemperatureT.You
willneedtheSIvalueofthegravitationalconstantG'6:67�10 �11N�m2�
kg �2.WhatistheHubbleexpansionrate,ininverseseconds,atthestartof
nucleosynthesis,whenT=Tnucl '0:9�109K?
(iii)Usingtheresultsin(i)and(ii),expressthecosmictimetasafunctionofthe
temperature.YourresultshouldagreewithWeinberg'sclaim
above.Whatis
thecosmictime,inseconds,whenT=Tnucl ?
PROBLEM
4:DID
YOU
DO
THEREADING
(2013)?(25points)
ThefollowingproblemcomesfromQuiz2,2013.
(a)(6points)Theprimaryevidencefordarkmatteringalaxiescomesfrommeasuring
theirrotationcurves,i.e.,theorbitalvelocityvasafunctionofradiusR.Ifstars
contributedall,ormost,ofthemassinagalaxy,whatwouldweexpectforthe
behaviorofv(R)atlargeradii?
(b)(5points)Whatisactuallyfoundforthebehaviorofv(R)?
(c)(7points)Animportanttoolforestimatingthemassinagalaxyisthesteady-state
virialtheorem.Whatdoesthistheoremstate?
(d)(7points)AttheendofChapter10,Rydenwrites\Thus,theverystrongasymmetry
betweenbaryonsandantibaryonstodayandthelargenumberofphotonsperbaryon
arebothproductsofatinyasymmetrybetweenquarksandanitquarksintheearly
universe."Explaininoneorafewsentenceshowatinyasymmetrybetweenquarks
andanitquarksintheearlyuniverseresultsinastrongasymmetrybetweenbaryons
andantibaryonstoday.
PROBLEM
5:DID
YOU
DO
THEREADING
(2016)?(25points)
(a)(5points)InChapter8ofBarbaraRyden'sIntroductiontoCosmology,sheestimates
thecontributiontofromclustersofgalaxiesas
(i)0.01
(ii)0.05
(iii)0.20
(iv)0.60
(v)1.00
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(b)(4points)Onemethodofestimatingthetotalmassofaclusterofgalaxiesisbased
onthevirialtheorem.Withthismethod,oneestimatesthemassbymeasuring
(i)theradiuscontaininghalftheluminosityandalsothetemperatureoftheX-ray
emittinggasatthecenterofthegalaxy.
(ii)thevelocitydispersionperpendiculartothelineofsightandalsotheradius
containinghalfoftheluminosityofthecluster.
(iii)thevelocitydispersionalongthelineofsightandalsotheradiuscontaininghalf
oftheluminosityofthecluster.
(iv)thevelocitydispersionalongthelineofsightandalsotheredshiftofthecluster.
(v)thevelocitydispersionperpendiculartothelineofsightandalsotheredshiftof
thecluster.
(c)(4points)Anothermethodofestimatingthetotalmassofaclusterofgalaxiesisto
makedetailedmeasurementsofthex-raysemittedbythehotintraclustergas.
(i)Byassumingthatthisgasisthedominantcomponentofthemassofthecluster,
themassoftheclustercanbeestimated.
(ii)Byassumingthatthehotgascomprisesaboutathirdofthemassofthecluster,
thetotalmassoftheclustercanbeestimated.
(iii)Byassumingthatthegasisheatedbystarsandsupernovaethatmakeup
mostofthemassofthecluster,themassofthesestarsandsupernovaecanbe
estimated.
(iv)Byassumingthatthegasisheatedbyinteractionswithdarkmatter,which
dominatesthemassofthecluster,themassoftheclustercanbeestimated.
(v)Byassumingthatthisgasisinhydrostaticequilibrium,thetemperature,mass
density,andeventhechemicalcompositionoftheclustercanbemodeled.
8.286QUIZ2REVIEW
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(d)(6points)InChapter6ofTheFirstThreeMinutes,StevenWeinbergdiscussesthree
reasonswhytheimportanceofasearchfora3 ÆKmicrowaveradiationbackground
wasnotgenerallyappreciatedinthe1950sandearly1960s.Choosethosethree
reasonsfromthefollowinglist.(2pointsforeachrightanswer,circleatmost3.)
(i)Theearliestcalculationserroneouslypredictedacosmicbackgroundtempera-
tureofonlyabout0:1 ÆK,andsuchabackgroundwouldbetooweaktodetect.
(ii)Therewasabreakdownincommunicationbetweentheoristsandexperimental-
ists.
(iii)Itwasnottechnologicallypossibletodetectasignalasweakasa3 ÆKmicrowave
backgrounduntilabout1965.
(iv)Sincealmostallphysicistsatthetimewerepersuadedbythesteadystatemodel,
thepredictionsofthebigbangmodelwerenottakenseriously.
(v)ItwasextraordinarilydiÆcultforphysiciststotakeseriouslyanytheoryofthe
earlyuniverse.
(vi)TheearlyworkonnucleosynthesisbyGamow,Alpher,Herman,andFollin,et
al.,hadattemptedtoexplaintheoriginofallcomplexnucleibyreactionsinthe
earlyuniverse.Thisprogramwasneververysuccessful,anditscredibilitywas
furtherunderminedasimprovementsweremadeinthealternativetheory,that
elementsaresynthesizedinstars.
(e)(6points)In1948RalphA.AlpherandRobertHermanwroteapaperpredicting
acosmicmicrowavebackgroundwithatemperatureof5K.Thepaperwasbased
onacosmologicalmodelthattheyhaddevelopedwithGeorgeGamow,inwhichthe
earlyuniversewasassumedtohavebeen�lledwithhotneutrons.Astheuniverse
expandedandcooledtheneutronsunderwentbetadecayintoprotons,electrons,and
antineutrinos,untilatsomepointtheuniversecooledenoughforlightelementsto
besynthesized.AlpherandHermanfoundthattoaccountfortheobservedpresent
abundancesoflightelements,theratioofphotonstonuclearparticlesmusthave
beenabout109.Althoughthepredictedtemperaturewasveryclosetotheactual
valueof2.7K,thetheorydi�eredfromourpresenttheoryintwoways.Circlethe
twocorrectstatementsinthefollowinglist.(3pointsforeachrightanswer;circleat
most2.)
(i)Gamow,Alpher,andHermanassumedthattheneutroncoulddecay,butnow
theneutronisthoughttobeabsolutelystable.
(ii)Inthecurrenttheory,theuniversestartedwithnearlyequaldensitiesofprotons
andneutrons,notallneutronsasGamow,Alpher,andHermanassumed.
(iii)Inthecurrenttheory,theuniversestartedwithmainlyalphaparticles,notall
neutronsasGamow,Alpher,andHermanassumed.(Note:analphaparticleis
thenucleusofaheliumatom,composedoftwoprotonsandtwoneutrons.)
(iv)Inthecurrenttheory,theconversionofneutronsintoprotons(andviceversa)
tookplacemainlythroughcollisionswithelectrons,positrons,neutrinos,and
antineutrinos,notthroughthedecayoftheneutrons.
(v)Theratioofphotonstonuclearparticlesintheearlyuniverseisnowbelieved
tohavebeenabout103,not109asAlpherandHermanconcluded.
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�
PROBLEM
6:EVOLUTION
OFAN
OPEN
UNIVERSE
Thefollowingproblem
wastakenfrom
Quiz2,1990,whereitcounted10pointsoutof
100.Consideranopen,matter-dominateduniverse,asdescribedbytheevolutionequa-
tionsonthefrontofthequiz.Findthetimetatwhicha= p�=2�.
�
PROBLEM
7:ANTICIPATING
A
BIG
CRUNCH
Supposethatwelivedinaclosed,matter-dominateduniverse,asdescribedbythe
equationsonthefrontofthequiz.Supposefurtherthatwemeasuredthemassdensity
parametertobe0=2,andwemeasuredtheHubble\constant"tohavesomevalue
H0 .Howmuchtimewouldwehavebeforeouruniverseendedinabigcrunch,atwhich
timethescalefactora(t)wouldcollapseto0?
�
PROBLEM
8:
TRACING
LIGHT
RAYS
IN
A
CLOSED,MATTER-
DOMINATED
UNIVERSE(30points)
ThefollowingproblemwasProblem3,Quiz2,1998.
Thespacetimemetricforahomogeneous,isotropic,closeduniverseisgivenbythe
Robertson-Walkerformula:
ds2=�c2d�2=�c2dt2+a2(t) �dr2
1�r2+r2 �d�2+sin2�d�2 � �;
whereIhavetakenk=1.Todiscussmotionintheradialdirection,itismoreconvenient
toworkwithanalternativeradialcoordinate ,relatedtorby
r=sin :
Then
dr
p1�r2
=d ;
sothemetricsimpli�esto
ds2=�c2d�2=�c2dt2+a2(t) �d 2+sin2 �d�2+sin2�d�2 �:
(a)(7points)Alightpulsetravelsonanulltrajectory,whichmeansthatd�=0for
eachsegmentofthetrajectory.Consideralightpulsethatmovesalongaradialline,
so�=�=constant.Findanexpressionford =dtintermsofquantitiesthatappear
inthemetric.
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(b)(8points)Writeanexpressionforthephysicalhorizondistance`physattimet.You
shouldleaveyouranswerintheformofade�niteintegral.
Theform
ofa(t)dependsonthecontentoftheuniverse.Iftheuniverseismatter-
dominated(i.e.,dominatedbynonrelativisticmatter),thena(t)isdescribedbythe
parametricequations
ct=�(��sin�);
a=�(1�cos�);
where
��4�3G�a3
c2
:
Theseequationsareidenticaltothoseonthefrontoftheexam,exceptthatIhavechosen
k=1.
(c)(10points)Consideraradiallight-raymovingthroughamatter-dominatedclosed
universe,asdescribedbytheequationsabove.Findanexpressionford =d�,where
�istheparameterusedtodescribetheevolution.
(d)(5points)Supposethataphotonleavestheoriginofthecoordinatesystem( =0)
att=0.Howlongwillittakeforthephotontoreturntoitsstartingplace?Express
youranswerasafractionofthefulllifetimeoftheuniverse,frombigbangtobig
crunch.
PROBLEM
9:LENGTHSANDAREASINATWO-DIMENSIONALMET-
RIC(25points)
ThefollowingproblemwasProblem3,Quiz2,1994:
Supposeatwodimensionalspace,describedinpolarcoordinates(r;�),hasametric
givenby
ds2=(1+ar)2dr2+r2(1+br)2d�2;
whereaandbarepositiveconstants.Considerthepathinthisspacewhichisformedby
startingattheorigin,movingalongthe�=0linetor=r0 ,thenmovingat�xedrto
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�=�=2,andthenmovingbacktotheoriginat�xed�.Thepathisshownbelow:
a)(10points)Findthetotallengthofthispath.
b)(15points)Findtheareaenclosedbythispath.
PROBLEM
10:GEOMETRY
IN
A
CLOSED
UNIVERSE(25points)
ThefollowingproblemwasProblem4,Quiz2,1988:
ConsiderauniversedescribedbytheRobertson{Walkermetriconthe�rstpageof
thequiz,withk=1.Thequestionsbelowallpertaintosome�xedtimet,sothescale
factorcanbewrittensimplyasa,droppingitsexplicitt-dependence.
Asmallrodhasoneendatthepoint(r=h;�=0;�=0)andtheotherendatthe
point(r=h;�=��;�=0).Assumethat���1.
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(a)Findthephysicaldistance`pfromtheorigin(r=0)tothe�rstend(h;0;0)ofthe
rod.Youmay�ndoneofthefollowingintegralsuseful:
Zdr
p1�r2
=sin�1r
Zdr
1�r2
=12
ln �1+r
1�r �:
(b)Findthephysicallengthspoftherod.Expressyouranswerintermsofthescale
factora,andthecoordinateshand��.
(c)Notethat��istheanglesubtendedbytherod,asseenfromtheorigin.Writean
expressionforthisangleintermsofthephysicaldistance`p ,thephysicallengthsp ,
andthescalefactora.
PROBLEM
11:THEGENERALSPHERICALLY
SYMMETRIC
METRIC
(20points)
ThefollowingproblemwasProblem3,Quiz2,1986:
Themetricforagivenspacedependsofcourseonthecoordinatesystem
which
isusedtodescribeit.Itcanbeshownthatforanythreedimensionalspacewhichis
sphericallysymmetricaboutaparticularpoint,coordinatescanbefoundsothatthe
metrichastheform
ds2=dr2+�2(r) �d�2+sin2�d�2 �
forsomefunction�(r).Thecoordinates�and�havetheirusualranges:�variesbetween
0and�,and�variesfrom0to2�,where�=0and�=2�areidenti�ed.Giventhis
metric,considerthespherewhoseouterboundaryisde�nedbyr=r0 .
(a)Findthephysicalradiusaofthesphere.(By\radius",Imeanthephysicallength
ofaradiallinewhichextendsfromthecentertotheboundaryofthesphere.)
(b)Findthephysicalareaofthesurfaceofthesphere.
(c)Findanexplicitexpressionforthevolumeofthesphere.Besuretoincludethe
limitsofintegrationforanyintegralswhichoccurinyouranswer.
(d)Supposeanewradialcoordinate�isintroduced,where�isrelatedtorby
�=r2:
Expressthemetricintermsofthisnewvariable.
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PROBLEM
12:VOLUMESIN
A
ROBERTSON-WALKER
UNIVERSE(20
points)
ThefollowingproblemwasProblem1,Quiz3,1990:
ThemetricforaRobertson-Walkeruniverseisgivenby
ds2=a2(t) �dr2
1�kr2+r2 �d�2+sin2�d�2 � �:
CalculatethevolumeV(rmax )ofthespheredescribedby
r�rmax:
Youshouldcarryoutanyangularintegrationsthatmaybenecessary,butyoumayleave
youranswerintheformofaradialintegralwhichisnotcarriedout.Besure,however,
toclearlyindicatethelimitsofintegration.
�
PROBLEM
13:THESCHWARZSCHILD
METRIC(25points)
ThefollowproblemwasProblem4,Quiz3,1992:
ThespaceoutsideasphericallysymmetricmassM
isdescribedbytheSchwarzschild
metric,givenatthefrontoftheexam.Twoobservers,designatedAandB,arelocated
alongthesameradialline,withvaluesofthecoordinatergivenbyrA
andrB,respectively,
withrA
<
rB.YoushouldassumethatbothobserverslieoutsidetheSchwarzschild
horizon.
a)(5points)WritedowntheexpressionfortheSchwarzschildhorizonradiusRS ,ex-
pressedintermsofM
andfundamentalconstants.
b)(5points)WhatistheproperdistancebetweenAandB?Itisokaytoleavethe
answertothispartintheformofanintegralthatyoudonotevaluate|
butbesure
toclearlyindicatethelimitsofintegration.
c)(5points)ObserverAhasaclockthatemitsanevenlyspacedsequenceofticks,
withpropertimeseparation��A.Whatwillbethecoordinatetimeseparation�tA
betweentheseticks?
d)(5points)AteachtickofA'sclock,alightpulseistransmitted.ObserverBreceives
thesepulses,andmeasuresthetimeseparationonhisownclock.Whatisthetime
interval��B
measuredbyB.
e)(5points)Supposethattheobjectcreatingthegravitational�eldisastaticblack
hole,sotheSchwarzschildmetricisvalidforallr.Nowsupposethatoneconsiders
thecaseinwhichobserverAliesontheSchwarzschildhorizon,sorA
�RS .Isthe
properdistancebetweenAandB�niteforthiscase?Doesthetimeintervalofthe
pulsesreceivedbyB,��B,divergeinthiscase?
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PROBLEM
14:GEODESICS(20points)
ThefollowingproblemwasProblem4,Quiz2,1986:
OrdinaryEuclideantwo-dimensionalspacecanbedescribedinpolarcoordinatesby
themetric
ds2=dr2+r2d�2:
(a)Supposethatr(�)and�(�)describeageodesicinthisspace,wheretheparameter
�isthearclengthmeasuredalongthecurve.Usethegeneralformulaonthefront
oftheexamtoobtainexplicitdi�erentialequationswhichr(�)and�(�)mustobey.
(b)NowintroducetheusualCartesiancoordinates,de�nedby
x=rcos�;
y=rsin�:
Useyouranswerto(a)toshowthattheliney=1isageodesiccurve.
�
PROBLEM
15:AN
EXERCISEIN
TWO-DIMENSIONALMETRICS(30
points)
(a)(8points)Consider�rstatwo-dimensionalspacewithcoordinatesrand�.The
metricisgivenby
ds2=dr2+r2d�2:
Considerthecurvedescribedbyr(�)
=(1+�cos2�)r0;
where�andr0areconstants,and�runsfrom�1to�2 .Writeanexpression,inthe
formofade�niteintegral,forthelengthSofthiscurve.
(b)(5points)Nowconsideratwo-dimensionalspacewiththesametwocoordinatesr
and�,butthistimethemetricwillbe
ds2= �1+ra �
dr2+r2d�2;
whereaisaconstant.�isaperiodic(angular)variable,witharangeof0to2�,with
2�identi�edwith0.WhatisthelengthRofthepathfromtheorigin(r=0)to
thepointr=r0 ;�=0,alongthepathforwhich�=0everywherealongthepath?
Youcanleaveyouranswerintheformofade�niteintegral.(Besure,however,to
specifythelimitsofintegration.)
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(c)(7points)Forthespacedescribedinpart(b),whatisthetotalareacontainedwithin
theregionr<r0 .Againyoucanleaveyouranswerintheformofade�niteintegral,
makingsuretospecifythelimitsofintegration.
(d)(10points)Againforthespacedescribedinpart(b),considerageodesicdescribed
bytheusualgeodesicequation,
dds �gijdxj
ds �=12
(@i gk` )dxk
ds
dx`
ds:
Thegeodesicisdescribedbyfunctionsr(s)and�(s),wheresisthearclengthalong
thecurve.Writeexplicitlyboth(i.e.,fori=1=randi=2=�)geodesicequations.
PROBLEM
16:GEODESICSON
THESURFACEOFA
SPHERE
Inthisproblemwewilltestthegeodesicequationbycomputingthegeodesiccurves
onthesurfaceofasphere.WewilldescribethesphereasinLectureNotes5,withmetric
givenby
ds2=a2 �d�2+sin2�d�2 �:
(a)Clearlyonegeodesiconthesphereistheequator,whichcanbeparametrizedby
�=�=2and�= ,where isaparameterwhichrunsfrom0to2�.Showthatif
theequatorisrotatedbyanangle�aboutthex-axis,thentheequationsbecome:
cos�=sin sin�
tan�=tan cos�
:
(b)Usingthegenericformofthegeodesicequationonthefrontoftheexam,derivethe
di�erentialequationwhichdescribesgeodesicsinthisspace.
(c)Showthattheexpressionsin(a)satisfythedi�erentialequationforthegeodesic.
Hint:Thealgebraonthiscanbemessy,butIfoundthingswerereasonablysimple
ifIwrotethederivativesinthefollowingway:
d�
d =�
cos sin�
p1�sin2 sin2�
;
d�
d =
cos�
1�sin2 sin2�
:
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�
PROBLEM
17:GEODESICSIN
A
CLOSED
UNIVERSE
ThefollowingproblemwasProblem3,Quiz3,2000,whereitwasworth40pointsplus5
pointsextracredit.
ConsiderthecaseofclosedRobertson-Walkeruniverse.Takingk=1,thespacetime
metriccanbewrittenintheform
ds2=�c2d�2=�c2dt2+a2(t) �dr2
1�r2+r2 �d�2+sin2�d�2 � �:
Wewillassumethatthismetricisgiven,andthata(t)hasbeenspeci�ed.Whilegalaxies
areapproximatelystationaryinthecomovingcoordinatesystemdescribedbythismetric,
wecanstillconsideranobjectthatmovesinthissystem.Inparticular,inthisproblem
wewillconsideranobjectthatismovingintheradialdirection(r-direction),underthe
in uenceofnoforcesotherthangravity.Hencetheobjectwilltravelonageodesic.
(a)(7points)Expressd�=dtintermsofdr=dt.
(b)(3points)Expressdt=d�intermsofdr=dt.
(c)(10points)Iftheobjecttravelsonatrajectorygivenbythefunctionrp (t)between
sometimet1andsomelatertimet2 ,writeanintegralwhichgivesthetotalamount
oftimethataclockattachedtotheobjectwouldrecordforthisjourney.
(d)(10points)Duringatimeintervaldt,theobjectwillmoveacoordinatedistance
dr=drd
tdt:
Letd`denotethephysicaldistancethattheobjectmovesduringthistime.By
\physicaldistance,"Imeanthedistancethatwouldbemeasuredbyacomovingob-
server(anobserverstationarywithrespecttothecoordinatesystem)whoislocated
atthesamepoint.Thequantityd`=dtcanberegardedasthephysicalspeedvphys
oftheobject,sinceitisthespeedthatwouldbemeasuredbyacomovingobserver.
Writeanexpressionforvphysasafunctionofdr=dtandr.
(e)(10points)Usingtheformulasatthefrontoftheexam,derivethegeodesicequation
ofmotionforthecoordinateroftheobject.Speci�cally,youshouldderivean
equationoftheform
dd� �Adr
d� �=B �dt
d� �
2+C �dr
d� �
2+D �d�
d� �
2+E �d�
d� �
2
;
whereA,B,C,D,andEarefunctionsofthecoordinates,someofwhichmightbe
zero.
(f)(5pointsEXTRACREDIT)OnProblem1ofProblemSet6welearnedthatina
atRobertson-Walkermetric,therelativisticallyde�nedmomentumofaparticle,
p=
mvphys
q1�v2p
hys
c2
;
fallso�as1=a(t).Usethegeodesicequationderivedinpart(e)toshowthatthe
sameistrueinacloseduniverse.
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PROBLEM
18:A
TWO-DIMENSIONALCURVED
SPACE(40points)
ThefollowingproblemwasProblem3,Quiz2,2002.
Consideratwo-dimensionalcurvedspacedescribedbypolar
coordinatesuand�,where0�u�aand0���2�,and
�=2�isasusualidenti�edwith�=0.Themetricisgivenby
ds2=
adu2
4u(a�u)+ud�2:
Adiagramofthespaceisshownattheright,butyoushouldof
coursekeepinmindthatthediagramdoesnotaccuratelyre ect
thedistancesde�nedbythemetric.
(a)(6points)FindtheradiusR
ofthespace,de�nedasthe
lengthofaradial(i.e.,�=constant)line.Youmayexpress
youranswerasade�niteintegral,whichyouneednoteval-
uate.Besure,however,tospecifythelimitsofintegration.
(b)(6points)FindthecircumferenceSofthespace,de�nedas
thelengthoftheboundaryofthespaceatu=a.
(c)(7points)Consideranannularregionasshown,consisting
ofallpointswithau-coordinateintherangeu0
�u�
u0+du.FindthephysicalareadAofthisregion,to�rst
orderindu.
(d)(3points)Usingyouranswertopart(c),writeanexpressionforthetotalareaof
thespace.
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(e)(10points)Considerageodesiccurveinthisspace,describedbythefunctionsu(s)
and�(s),wheretheparametersischosentobethearclengthalongthecurve.Find
thegeodesicequationforu(s),whichshouldhavetheform
dds �F(u;�)du
ds �=:::;
whereF(u;�)isafunctionthatyouwill�nd.(NotethatbywritingFasafunction
ofuand�,wearesayingthatitcoulddependoneitherorbothofthem,butweare
notsayingthatitnecessarilydependsonthem.)Youneednotsimplifytheleft-hand
sideoftheequation.
(f)(8points)Similarly,�ndthegeodesicequationfor�(s),whichshouldhavetheform
dds �G(u;�)d�
ds �=:::;
whereG(u;�)isafunctionthatyouwill�nd.Again,youneednotsimplifythe
left-handsideoftheequation.
�
PROBLEM
19:ROTATING
FRAMESOFREFERENCE(35points)
ThefollowingproblemwasProblem3,Quiz2,2004.
Inthisproblemwewillusetheformalismofgeneralrelativityandgeodesicstoderive
therelativisticdescriptionofarotatingframeofreference.
Theproblemwillconcerntheconsequencesofthemetric
ds2=�c2d�2=�c2dt2+ hdr2+r2(d�+!dt)2+dz2 i;
(P19.1)
whichcorrespondstoacoordinatesystem
rotatingaboutthez-axis,where�isthe
azimuthalanglearoundthez-axis.Thecoordinateshavetheusualrangeforcylindrical
coordinates:�1<t<1,0�r<1,�1<z<1,and0��<2�,where�=2�is
identi�edwith�=0.
EXTRAINFORMATION
Toworktheproblem,youdonotneedtoknowanythingaboutwherethismetric
camefrom.However,itmight(ormightnot!)helpyourintuitiontoknowthat
Eq.(P19.1)wasobtainedbystartingwithaMinkowskimetricincylindrical
coordinates�t,�r,��,and�z,
c2d�2=c2d�t2� �d�r2+�r2d��2+d�z2 �;
andthenintroducingnewcoordinatest,r,�,andzthatarerelatedby
�t=t;
�r=r;
��=�+!t;
�z=z;
sod�t=dt,d�r=dr,d��=d�+!dt,andd�z=dz.
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(a)(8points)Themetriccanbewritteninmatrixformbyusingthestandardde�nition
ds2=�c2d�2�g��dx�dx�;
wherex0�t,x1�r,x2��,andx3�z.Then,forexample,g11(whichcanalsobe
calledgrr )isequalto1.Findexplicitexpressionstocompletethelistofthenonzero
entriesinthematrixg�� :
g11 �grr=1
g00 �gtt=?
g20 �g02 �g�t �gt�=?
g22 �g��=?
g33 �gzz=?
(P19.2)
Ifyoucannotanswerpart(a),youcanintroduceunspeci�edfunctionsf1 (r),f2 (r),f3 (r),
andf4 (r),with
g11 �grr=1
g00 �gtt=f1 (r)
g20 �g02 �g�t �gt�=f1 (r)
g22 �g��=f3 (r)
g33 �gzz=f4 (r);
(P19.3)
andyoucanthenexpressyouranswerstothesubsequentpartsintermsoftheseunspec-
i�edfunctions.
(b)(10points)Usingthegeodesicequationsfromthefrontofthequiz,
dd� �g��dx�
d� �=12
(@�g��)dx�
d�
dx�
d�
;
explicitlywritetheequationthatresultswhenthefreeindex�isequalto1,corre-
spondingtothecoordinater.
(c)(7points)Explicitlywritetheequationthatresultswhenthefreeindex�isequal
to2,correspondingtothecoordinate�.
(d)(10points)Usethemetricto�ndanexpressionfordt=d�intermsofdr=dt,d�=dt,
anddz=dt.Theexpressionmayalsodependontheconstantscand!.Besureto
notethatyouranswershoulddependonthederivativesoft,�,andzwithrespect
tot,not�.(Hint:�rst�ndanexpressionford�=dt,intermsofthequantities
indicated,andthenaskyourselfhowthisresultcanbeusedto�nddt=d�.)
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PROBLEM
20:
THE
STABILITY
OF
SCHWARZSCHILD
ORBITS
(30
points)
ThisproblemwasProblem4,Quiz2in2007.Ihavemodi�edthereferencetothe
homeworkproblemtocorrespondtothecurrent(2016)context,whereitisProblem3of
ProblemSet6.In2007ithadalsobeenahomeworkproblempriortothequiz.
ThisproblemisanelaborationoftheProblem3ofProblemSet6,forwhichboth
thestatementandthesolutionarereproducedattheendofthisquiz.Thismaterialis
reproducedforyourreference,butyoushouldbeawarethatthesolutiontothepresent
problemhasimportantdi�erences.Youcancopyfromthismaterial,buttoallowthe
gradertoassessyourunderstanding,youareexpectedtopresentalogical,self-contained
answertothisquestion.
Inthesolutiontothathomeworkproblem,itwasstatedthatfurtheranalysisofthe
orbitsinaSchwarzschildgeometryshowsthatthesmalleststablecircularorbitoccurs
forr=3RS.Circularorbitsarepossiblefor32RS
<r<3RS
,buttheyarenotstable.
Inthisproblemwewillexplorethecalculationsbehindthisstatement.
Wewillconsiderabodywhichundergoessmalloscillationsaboutacircularorbitat
r(t)=r0 ,�=�=2,wherer0isaconstant.Thecoordinate�willthereforebe�xed,but
alltheothercoordinateswillvaryasthebodyfollowsitsorbit.
(a)(12points)The�rststep,sincer(�)willnotbeaconstantinthissolution,willbe
toderivetheequationofmotionforr(�).Thatis,fortheSchwarzschildmetric
ds2=�c2d�2=�h(r)c2dt2+h(r) �1dr2+r2d�2+r2sin2�d�2;
(P20.1)
where
h(r)�1�RSr
;
workouttheexplicitformofthegeodesicequation
dd� �g��dx�
d� �=12@g��
@x�
dx�
d�
dx�
d�
;
(P20.2)
forthecase�=r.Youshouldusethisresultto�ndanexplicitexpressionfor
d2r
d�2
:
Youmayallowyouranswertocontainh(r),itsderivativeh0(r)withrespecttor,
andthederivativewithrespectto�ofanycoordinate,includingdt=d�.
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(b)(6points)Itisusefultoconsiderrand�tobetheindependentvariables,while
treatingtasadependentvariable.Findanexpressionfor
�dt
d� �
2
intermsofr,dr=d�,d�=d�,h(r),andc.Usethisequationtosimplifytheexpression
ford2r=d�2obtainedinpart(a).Thegoalistoobtainanexpressionoftheform
d2r
d�2=f0 (r)+f1 (r) �d�
d� �
2
:
(P20.3)
wherethefunctionsf0 (r)andf1 (r)mightdependonRSorc,andmightbepositive,
negative,orzero.Notethattheintermediatestepsinthecalculationinvolveaterm
proportionalto(dr=d�)2,butthenetcoeÆcientforthistermvanishes.
(c)(7points)Tounderstandtheorbitwewillalsoneedtheequationofmotionfor�.
Evaluatethegeodesicequation(P20.2)for�=�,andwritetheresultintermsof
thequantityL,de�nedby
L�r2d�
d�:
(P20.4)
(d)(5points)Finally,wecometothequestionofstability.SubstitutingEq.(P20.4)
intoEq.(P20.3),theequationofmotionforrcanbewrittenas
d2r
d�2=f0 (r)+f1 (r)L2
r4
:
Nowconsiderasmallperturbationaboutthecircularorbitatr=r0 ,andwritean
equationthatdeterminesthestabilityoftheorbit.(Thatis,ifsomeexternalforce
givestheorbitingbodyasmallkickintheradialdirection,howcanyoudetermine
whethertheperturbationwillleadtostableoscillations,orwhetheritwillstartto
grow?)Youshouldexpressthestabilityrequirementintermsoftheunspeci�ed
functionsf0 (r)andf1 (r).YouareNOTaskedtocarryoutthealgebraofinserting
theexplicitformsthatyouhavefoundforthesefunctions.
�
PROBLEM
21:PRESSURE
AND
ENERGY
DENSITY
OFMYSTERI-
OUSSTUFF(25points)
ThefollowingproblemwasProblem3,Quiz3,2002.
InLectureNotes6,withfurthercalculationsinProblem
4ofProblem
Set6,a
thoughtexperimentinvolvingapistonwasusedtoshowthatp=13�c2forradiation.In
thisproblemyouwillapplythesametechniquetocalculatethepressureofmysterious
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stu�,whichhasthepropertythattheenergydensityfallso�inproportionto1= pVas
thevolumeVisincreased.
Iftheinitialenergydensityofthemysteriousstu�isu0
=�0 c2,thentheinitial
con�gurationofthepistoncanbedrawnas
Thepistonisthenpulledoutward,sothatitsinitialvolumeVisincreasedtoV+�V.
Youmayconsider�Vtobein�nitesimal,so�V2canbeneglected.
(a)(15points)Usingthefactthattheenergydensityofmysteriousstu�fallso�as
1= pV,�ndtheamount�Ubywhichtheenergyinsidethepistonchangeswhenthe
volumeisenlargedby�V.De�ne�Utobepositiveiftheenergyincreases.
(b)(5points)Ifthe(unknown)pressureofthemysteriousstu�iscalledp,howmuch
work�W
isdonebytheagentthatpullsoutthepiston?
(c)(5points)Useyourresultsfrom(a)and(b)toexpressthepressurepofthemysterious
stu�intermsofitsenergydensityu.(Ifyoudidnotanswerparts(a)and/or(b),
explainasbestyoucanhowyouwoulddeterminethepressureifyouknewthe
answerstothesetwoquestions.)
PROBLEM
22:VOLUMEOFACLOSEDTHREE-DIMENSIONALSPACE
(15points)
ThisproblemisageneralizationofProblem2ofProblemSet5.
Recallthatthespatialpartofthemetricforacloseduniversecanbewrittenas
ds2=R2 �d 2+sin2 �d�2+sin2�d�2 ��:
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Inthisproblem
wewillconsideramoregeneralmetric,whichalsodescribesaclosed
three-dimensionalspace,butonethatisnothomogeneous.Themetricwillbegivenby
ds2=R2 �d 2+f2( ) �d�2+sin2�d�2 ��;
wheref( )issomeunspeci�edfunction.Thecoordinates�and�havetheusualrange,
0����,and0���2�,and variesintherange0� ��.
Writeanintegralexpressionforthevolumeofthisspace.Theintegralshouldbe
overasinglevariableonly.Hint:asinProblem2ofProblemSet5,youcanbreakthe
volumeupintosphericalshellsofin�nitesimalthickness,extendingfrom to +d :
8.286QUIZ2REVIEW
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PROBLEM
23:GRAVITATIONALBENDING
OFLIGHT(30points)
Whenalightraypassesbyamassiveobject,generalrelativitypredictsthatitwill
bebent.Sincemostcelestialobjectsarenearlyspherical,wecanusetheSchwarzschild
metrictocalculatethebending.Furthermore,sinceweareusuallyinterestedinobjects
thatarenotblackholesoranywherenearlyasdense,wecanobtainanaccurateanswer
bycarryingoutthecalculationinaweak-�eldapproximation.Foraphotonthatgrazes
theSun,forexample,thevalueofRSch =R�,theSchwarzschildradiusovertheradiusof
theSun,isabout4�10 �6.
StartingwiththeSchwarzschildmetric,
ds2=� �1�RSch
r �c2dt2+ �1�RSch
r ��1
dr2+r2(d�2+sin2�d�2);
(P23.1)
whereRSch=2GM=c2,wecanexpandinpowersofRSch =randkeeponlythe�rstorder
terms:d
s2=� �1�RSch
r �c2dt2+ �1+RSch
r �dr2+r2(d�2+sin2�d�2):
(P23.2)
ForthisproblemitisusefultoswitchtoCartesian-likecoordinates,de�nedintermsof
r,�,and�bytheusualCartesianformulas,
x=rsin�cos�;
y=rsin�sin�;
z=rcos�:
(P23.3)
Generalrelativityallowsustomakeanycoordinaterede�nitionsthatwemightwant,as
longaswecalculatethemetricintermsofthenewcoordinates.Itisusefultocontinue
tousethequantityr,butnowitwillbethoughtofasafunctionofthecoordinatesx,y,
andz:
r=(x2+y2+z2)1=2:
(P23.4)
ThemetriccanthenberewrittenastheMinkowskimetricofspecialrelativity,plussmall
corrections:d
s2=�c2dt2+dx2+dy2+dz2+RSch
r
c2dt2+RSch
r
(dr)2;
(P23.5)
wherefromEq.(4)onecanseethat
dr=1r
(xdx+ydy+zdz):
(P23.6)
(a)(6points)ForthemetricasapproximatedbyEqs.(P23.5)and(P23.6),writethe
expressionsforgtt ,gxx ,andgxy .
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Thetrajectoryofthephotonislightlike,sowecannotuse�toparameterizethe
trajectory,becausepropertimeintervalsalongalightliketrajectoryarezero.Nonetheless,
itcanbeshownthatonecanusean\aÆneparameter"�,forwhichthegeodesicequation
hastheusualform:
dd� �g��dx�
d� �=12
[@�g�� ]dx�
d�
dx�
d�
:
(P23.7)
Toobtainananswerthatisaccurateto�rstorderinG,webeginbyconsideringthe
unperturbedphotontrajectory|
thetrajectoryitwouldhaveifGweretakenaszero,so
RSch=2GM=c2=0.Thiswouldbeastraightlineinthe(x;y;z)coordinates,asshown
inthediagrambelow:
Herebiscalledtheimpactparameter.Wecanparameterizethispathby
x(�)=�;
y(�)=b;
z(�)=0;
t(�)=�=c:
(P23.8)
Wewillcalculatethede ection(to�rstorderinG)byassumingthatthephotonpathis
accuratelydescribedbyEq.(P23.8),andwewillcalculatethey-velocitythatthephoton
acquiresduetothegravitationalattractionoftheSun.
(b)(9points)Withthegoalofcalculatingd2y=d�2,weevaluatethegeodesicequation
for�=y.Startherebyevaluatingtheleft-handsideofEq.(P23.7)for�=y,to
�rstorderinG.Expandthederivativewithrespectto�usingtheproductrule,
workingoutexplicitlythederivativesoftherelevantg��withrespectto�.Inparts
(b)and(c),youmayassumethatx(�),y(�),z(�),andt(�),aswellasdx=d�,
dy=d�,dz=d�,anddt=d�,areallgiventosuÆcientaccuracybyEq.(P23.8)andits
derivativeswithrespectto�.(Becareful:itislikelythattherearemoretermsthan
youwillat�rstnotice.)
(c)(9points)Evaluatetheright-handsideofEq.(P23.7)for�=y,to�rstorderinG.
Carryoutallderivativesexplicitly.(Italwayspaystobecareful.)
(d)(2points)Useyouranswerstoparts(c)and(d)to�ndanequationford2y=d�2.
(e)(4points)Ifthephotonstartsoutontheunperturbedtrajectory,itsinitialvalueof
dy=d�willbezero.The�nalvalueofdy=d�willthenbe
dy
d� �����nal= Z1�
1d2y
d�2d�:
(P23.9)
Usethisfacttoexpressthede ectionangle�,to�rstorderinG,asanexplicit
integral.Youneednotcarryouttheintegral,butyoumaywishtousethetableof
8.286QUIZ2REVIEW
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integralsgivenbelowtocarryitoutsothatyoucancheckyouranswer.Thecorrect
�nalansweris
�=4GM
c2b:
(P23.10)
TABLEOFINTEGRALS:
Z1�
1
1
(x2+b2)dx=�b
Z1�
1
1
(x2+b2)3=2dx=
2b2
Z1�
1
1
(x2+b2)2dx=
�2b3
Z1�
1
x2
(x2+b2)2dx=
�2b
Z1�
1
x2
(x2+b2)5=2dx=
23b2
Z1�
1
x2
(x2+b2)3dx=
�8b3
8.286QUIZ2REVIEW
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SOLUTIONS
PROBLEM
1:DID
YOU
DO
THEREADING?
(a)Thisisatotaltrickquestion.Leptonnumberis,ofcourse,conserved,sothefactor
isjust1.SeeWeinbergchapter4,pages91-4.
(b)Thecorrectansweris(i).Theothersareallrealreasonswhyit'shardtomeasure,
althoughWeinberg'sbookemphasizesreason(v)abitmorethanmodernastrophysi-
cistsdo:astrophysicistshavebeenlookingforotherwaysthatdeuteriummightbe
produced,butnosigni�cantmechanismhasbeenfound.SeeWeinbergchapter5,
pages114-7.
(c)Themostobviousanswerswouldbeproton,neutron,andpimeson.However,there
aremanyotherpossibilities,includingmanythatwerenotmentionedbyWeinberg.
SeeWeinbergchapter7,pages136-8.
(d)Thecorrectanswersweretheneutrinoandtheantiproton.Theneutrinowas�rst
hypothesizedbyWolfgangPauliin1932(inordertoexplainthekinematicsofbeta
decay),and�rstdetectedinthe1950s.Afterthepositronwasdiscoveredin1932,
theantiprotonwasthoughtlikelytoexist,andtheBevatroninBerkeleywasbuilt
tolookforantiprotons.Itmadethe�rstdetectioninthe1950s.
(e)Thecorrectanswerswere(ii),(v)and(vi).Theotherswereincorrectforthefol-
lowingreasons:
(i)theearliestpredictionoftheCMBtemperature,byAlpherandHermanin1948,
was5degrees,not0.1degrees.
(iii)Weinbergquoteshisexperimentalcolleaguesassayingthatthe3 ÆKradiation
couldhavebeenobserved\longbefore1965,probablyinthemid-1950sandper-
hapseveninthemid-1940s."ToWeinberg,however,thehistoricallyinteresting
questionisnotwhentheradiationcouldhavebeenobserved,butwhyradio
astronomersdidnotknowthattheyoughttotry.
(iv)Weinbergarguesthatphysicistsatthetimedidnotpayattentiontoeitherthe
steadystatemodelorthebigbangmodel,asindicatedbythesentenceinitem
(v)whichisadirectquotefromthebook:\ItwasextraordinarilydiÆcultfor
physiciststotakeseriouslyanytheoryoftheearlyuniverse".
8.286QUIZ2REVIEW
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PROBLEM
2:DID
YOU
DO
THEREADING?(24points)
(a)(6points)In1948RalphA.AlpherandRobertHermanwroteapaperpredicting
acosmicmicrowavebackgroundwithatemperatureof5K.Thepaperwasbased
onacosmologicalmodelthattheyhaddevelopedwithGeorgeGamow,inwhichthe
earlyuniversewasassumedtohavebeen�lledwithhotneutrons.Astheuniverse
expandedandcooledtheneutronsunderwentbetadecayintoprotons,electrons,and
antineutrinos,untilatsomepointtheuniversecooledenoughforlightelementsto
besynthesized.AlpherandHermanfoundthattoaccountfortheobservedpresent
abundancesoflightelements,theratioofphotonstonuclearparticlesmusthave
beenabout109.Althoughthepredictedtemperaturewasveryclosetotheactual
valueof2.7K,thetheorydi�eredfromourpresenttheoryintwoways.Circlethe
twocorrectstatementsinthefollowinglist.(3pointsforeachrightanswer;circleat
most2.)
(i)Gamow,Alpher,andHermanassumedthattheneutroncoulddecay,butnow
theneutronisthoughttobeabsolutelystable.
(ii)Inthecurrenttheory,theuniversestartedwithnearlyequaldensitiesofprotons
andneutrons,notallneutronsasGamow,Alpher,andHermanassumed.
(iii)Inthecurrenttheory,theuniversestartedwithmainlyalphaparticles,notall
neutronsasGamow,Alpher,andHermanassumed.(Note:analphaparticleis
thenucleusofaheliumatom,composedoftwoprotonsandtwoneutrons.)
(iv)Inthecurrenttheory,theconversionofneutronsintoprotons(andviceversa)
tookplacemainlythroughcollisionswithelectrons,positrons,neutrinos,and
antineutrinos,notthroughthedecayoftheneutrons.
(v)Theratioofphotonstonuclearparticlesintheearlyuniverseisnowbelieved
tohavebeenabout103,not109asAlpherandHermanconcluded.
(b)(6points)InWeinberg's\RecipeforaHotUniverse,"hedescribedtheprimordial
compositionoftheuniverseintermsofthreeconservedquantities:electriccharge,
baryonnumber,andleptonnumber.Ifelectricchargeismeasuredinunitsoftheelec-
troncharge,thenallthreequantitiesareintegersforwhichthenumberdensitycan
becomparedwiththenumberdensityofphotons.Foreachquantity,whichchoice
mostaccuratelydescribestheinitialratioofthenumberdensityofthisquantityto
thenumberdensityofphotons:
8.286QUIZ2REVIEW
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ElectricCharge:
(i)�109
(ii)�1000
(iii)�1
(iv)�10 �6
(v)eitherzeroornegligible
BaryonNumber:
(i)�10 �20
(ii)�10 �9
(iii)�10 �6
(iv)�1
(v)anywherefrom10 �5to1
LeptonNumber:
(i)�109
(ii)�1000
(iii)�1
(iv)�10 �6
(v)couldbeashighas�1,but
isassumedtobeverysmall
(c)(12points)The�gurebelowcomesfromWeinberg'sChapter5,andislabeledThe
ShiftingNeutron-ProtonBalance.
(i)(3points)Duringtheperiodlabeled\thermalequilibrium,"theneutronfraction
ischangingbecause(chooseone):
(A)Theneutronisunstable,anddecaysintoaproton,electron,andantineu-
trinowithalifetimeofabout1second.
(B)Theneutronisunstable,anddecaysintoaproton,electron,andantineu-
trinowithalifetimeofabout15seconds.
(C)Theneutronisunstable,anddecaysintoaproton,electron,andantineu-
trinowithalifetimeofabout15minutes.
(D)Neutronsandprotonscanbeconvertedfrom
oneintothroughreactions
suchas
8.286QUIZ2REVIEW
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antineutrino+proton !electron+neutron
neutrino+neutron !positron+proton:
(E)Neutronsandprotonscanbeconvertedfromoneintotheotherthrough
reactionssuchas
antineutrino+proton !positron+neutron
neutrino+neutron !electron+proton:
(F)Neutronsandprotonscanbecreatedanddestroyedbyreactionssuchas
proton+neutrino !positron+antineutrino
neutron+antineutrino !electron+positron:
(ii)(3points)Duringtheperiodlabeled\neutrondecay,"theneutronfractionis
changingbecause(chooseone):
(A)Theneutronisunstable,anddecaysintoaproton,electron,andantineu-
trinowithalifetimeofabout1second.
(B)Theneutronisunstable,anddecaysintoaproton,electron,andantineu-
trinowithalifetimeofabout15seconds.
(C)Theneutronisunstable,anddecaysintoaproton,electron,andantineu-
trinowithalifetimeofabout15minutes.
(D)Neutronsandprotonscanbeconvertedfromoneintotheotherthrough
reactionssuchas
antineutrino+proton !electron+neutron
neutrino+neutron !positron+proton:
(E)Neutronsandprotonscanbeconvertedfromoneintotheotherthrough
reactionssuchas
antineutrino+proton !positron+neutron
neutrino+neutron !electron+proton:
(F)Neutronsandprotonscanbecreatedanddestroyedbyreactionssuchas
proton+neutrino !positron+antineutrino
neutron+antineutrino !electron+positron:
8.286QUIZ2REVIEW
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(iii)(3points)Themassesoftheneutronandprotonarenotexactlyequal,but
instead
(A)Theneutronismoremassivethanaprotonwitharestenergydi�erenceof
1.293GeV(1GeV=109eV).
(B)Theneutronismoremassivethanaprotonwitharestenergydi�erenceof
1.293MeV(1MeV=106eV).
(C)Theneutronismoremassivethanaprotonwitharestenergydi�erenceof
1.293KeV(1KeV=103eV).
(D)Theprotonismoremassivethananeutronwitharestenergydi�erenceof
1.293GeV.
(E)Theprotonismoremassivethananeutronwitharestenergydi�erenceof
1.293MeV.
(F)Theprotonismoremassivethananeutronwitharestenergydi�erenceof
1.293KeV.
(iv)(3points)Duringtheperiodlabeled\eraofnucleosynthesis,"(chooseone:)
(A)Essentiallyalltheneutronspresentcombinewithprotonstoformhelium
nuclei,whichmostlysurviveuntilthepresenttime.
(B)Essentiallyalltheneutronspresentcombinewithprotonstoformdeuterium
nuclei,whichmostlysurviveuntilthepresenttime.
(C)Abouthalftheneutronspresentcombinewithprotonstoformheliumnu-
clei,whichmostlysurviveuntilthepresenttime,andtheotherhalfofthe
neutronsremainfree.
(D)Abouthalftheneutronspresentcombinewithprotonstoformdeuterium
nuclei,whichmostlysurviveuntilthepresenttime,andtheotherhalfof
theneutronsremainfree.
(E)Essentiallyalltheprotonspresentcombinewithneutronstoformhelium
nuclei,whichmostlysurviveuntilthepresenttime.
(F)Essentiallyalltheprotonspresentcombinewithneutronstoformdeuterium
nuclei,whichmostlysurviveuntilthepresenttime.
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PROBLEM
3:DID
YOU
DO
THEREADING?(20points) y
(a)(8points)
(i)(4points)WewillusethenotationXA
toindicateanucleus,*whereX
is
thesymbolfortheelementwhichindicatesthenumberofprotons,whileA
isthemassnumber,namelythetotalnumberofprotonsandneutrons.With
thisnotationH1,H2,H3,He3andHe4standforhydrogen,deuterium,tritium,
helium-3andhelium-4nuclei,respectively.StevenWeinberg,inTheFirstThree
Minutes,chapterV,page108,describestwochainsofreactionsthatproduce
helium,startingfromprotonsandneutrons.Theycanbewrittenas:
p+n!H2+
H2+n!H3+
H3+p!He4+ ;
p+n!H2+
H2+p!He3+
He3+n!He4+ :
ThesearethetwoexamplesgivenbyWeinberg.However,di�erentchainsof
twoparticlereactionscantakeplace(ingeneralwithdi�erentprobabilities).
Forexample:
p+n!H2+
H2+H2!He4+ ;
p+n!H2+
H2+n!H3+
H3+H2!He4+n;
p+n!H2+
H2+p!He3+
He3+H2!He4+p;
:::
Studentswhodescribedchainsdi�erentfromthoseofWeinberg,butthatcan
stilltakeplace,gotfullcreditforthispart.Also,noticethatphotonsinthe
reactionsabovecarrytheadditionalenergyreleased.However,sincethemain
pointwastodescribethenuclearreactions,studentswhodidn'tincludethe
photonsstillreceivedfullcredit.
(ii)(4points)ThedeuteriumbottleneckisdiscussedbyWeinberginTheFirstThree
Minutes,chapterV,pages109-110.Thekeypointisthatfrompart(i)itshould
beclearthatdeuterium(H2)playsacrucialroleinnucleosynthesis,sinceitisthe
startingpointforallthechains.However,thedeuteriumnucleusisextremely
looselyboundcomparedtoH3,He3,orespeciallyHe4.So,therewillbea
*Noticethatsomestudentstalkedaboutatoms,whilewearetalkingaboutnuclei
formation.Duringnucleosynthesisthetemperatureiswaytoohightoallowelectrons
andnucleitobindtogethertoform
atoms.Thishappensmuchlater,intheprocess
calledrecombination.
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rangeoftemperatureswhicharelowenoughforH3,He3,andHe4nucleitobe
bound,buttoohightoallowthedeuteriumnucleustobestable.Thisisthe
temperaturerangewherethedeuteriumbottleneckisinaction:evenifH3,He3,
andHe4nucleicouldinprinciplebestableatthosetemperatures,theydonot
formbecausedeuterium,whichisthestartingpointfortheirformation,cannot
beformedyet.Nucleosynthesiscannotproceedatasigni�cantrateuntilthe
temperatureislowenoughsothatdeuteriumnucleiarestable;atthispointthe
deuteriumbottleneckhasbeenpassed.
(b)(12points)
(i)(3points)Ifwetakea(t)=bt1=2,forsomeconstantb,wegetfortheHubble
expansionrate:
H=
_aa=
12t
=)
t=
12H:
(ii)(6points)ByusingtheFriedmannequationwithk=0and�=�r=�T4,we
�nd:
H2=8�3
G�r=8�3
G�T4
=)
H=T2 r8�3
G�:
IfwesubstitutethegivennumericalvaluesG'6:67�10 �11N�m2�kg �2and
�'4:52�10 �32kg�m�3�K�4weget:
H'T2�5:03�10 �21s �1�K�2:
NoticethattheunitscorrectlycombinetogiveH
inunitsofs �1ifthetemper-
atureisexpressedindegreesKelvin(K).Indetail,wesee:
[G�] 1=2=(N�m2�kg �2�kg�m�3�K�4)1=2=s �1�K�2;
whereweusedthefactthat1N=1kg�m�s �2.AtT=Tnucl '0:9�109Kwe
get:
H'4:07�10 �3s �1:
(iii)(3points)Usingtheresultsinparts(i)and(ii),weget
t=
12H
' �9:95�1019
T2
�s�K2:
8.286QUIZ2REVIEW
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Togoodaccuracy,thenumeratorintheexpressionabovecanberoundedto
1020.TheaboveequationagreeswithWeinberg'sclaim
that,foraradiation
dominateduniverse,timeisproportionaltotheinversesquareofthetempera-
ture.InparticularforT=Tnuclweget:
tnucl '123s�2min:
ySolutionwrittenbyDanieleBertolini.
PROBLEM
4:DID
YOU
DO
THEREADING?(25points)
(a)(6points)Theprimaryevidencefordarkmatteringalaxiescomesfrommeasuring
theirrotationcurves,i.e.,theorbitalvelocityvasafunctionofradiusR.Ifstars
contributedall,ormost,ofthemassinagalaxy,whatwouldweexpectforthe
behaviorofv(R)atlargeradii?
Answer:Ifstarscontributedmostofthemass,thenatlargeradiithemasswould
appeartobeconcentratedasasphericallumpatthecenter,andtheorbitsofthe
starswouldbe\Keplerian,"i.e.,orbitsina1=r2gravitational�eld.Then~F=m~a
impliesthat
1R2
/v2
R
=)
v/1p
R:
(b)(5points)Whatisactuallyfoundforthebehaviorofv(R)?
Answer:v(R)looksnearly atatlargeradii.
(c)(7points)Animportanttoolforestimatingthemassinagalaxyisthesteady-state
virialtheorem.Whatdoesthistheoremstate?
Answer:Foragravitationallyboundsysteminequilibrium,
Kineticenergy=�12
(Gravitationalpotentialenergy):
(Theequalityholdswhenever�I�0,whereIisthemomentofinertia.)
(d)(7points)AttheendofChapter10,Rydenwrites\Thus,theverystrongasymmetry
betweenbaryonsandantibaryonstodayandthelargenumberofphotonsperbaryon
arebothproductsofatinyasymmetrybetweenquarksandanitquarksintheearly
universe."Explaininoneorafewsentenceshowatinyasymmetrybetweenquarks
andanitquarksintheearlyuniverseresultsinastrongasymmetrybetweenbaryons
andantibaryonstoday.
Answer:WhenkTwaslargecomparedto150MeV,theexcessofquarksoveranti-
quarkswastiny:onlyabout3extraquarksforevery109antiquarks.Buttherewas
massivequark-antiquarkannihilationaskTfellbelow150MeV,sothattodaywe
seetheexcessquarks,boundintobaryons,andalmostnosignofantiquarks.
8.286QUIZ2REVIEW
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PROBLEM
5:DID
YOU
DO
THEREADING
(2016)?(25points)
(a)(5points)InChapter8ofBarbaraRyden'sIntroductiontoCosmology,sheestimates
thecontributiontofromclustersofgalaxiesas
(i)0.01
(ii)0.05
(iii)0.20
(iv)0.60
(v)1.00
(b)(4points)Onemethodofestimatingthetotalmassofaclusterofgalaxiesisbased
onthevirialtheorem.Withthismethod,oneestimatesthemassbymeasuring
(i)theradiuscontaininghalftheluminosityandalsothetemperatureoftheX-ray
emittinggasatthecenterofthegalaxy.
(ii)thevelocitydispersionperpendiculartothelineofsightandalsotheradius
containinghalfoftheluminosityofthecluster.
(iii)thevelocitydispersionalongthelineofsightandalsotheradiuscontaininghalf
oftheluminosityofthecluster.
(iv)thevelocitydispersionalongthelineofsightandalsotheredshiftofthecluster.
(v)thevelocitydispersionperpendiculartothelineofsightandalsotheredshiftof
thecluster.
Explanation:Thevirialtheoremrelatesthekineticenergytothepotentialenergy.The
keyrelationshipis
12M<v2>=�2GM2
rh
;
whereM
isthemassofthecluster,<v2>
istheaveragesquaredvelocityofits
galaxies,andrh
istheradiuscontaininghalfthetotalmass,whichisestimatedby
theradiuscontaininghalftheluminosity.�isanumericalfactordependingonthe
structureofthecluster,estimatedat0.4basedonobservedclusters.Velocitiesalong
thelineofsightaremeasuredbythespreadinDopplershifts,whilevelocitiesperpen-
diculartothelineofsightareessentiallyimpossibletomeasure,eliminatinganswers
(ii)and(v).Sincerhisneeded,neither(i)nor(iv)includeenoughinformation.(iii)
isexactlyright.
(c)(4points)Anothermethodofestimatingthetotalmassofaclusterofgalaxiesisto
makedetailedmeasurementsofthex-raysemittedbythehotintraclustergas.
(i)Byassumingthatthisgasisthedominantcomponentofthemassofthecluster,
themassoftheclustercanbeestimated.
(ii)Byassumingthatthehotgascomprisesaboutathirdofthemassofthecluster,
thetotalmassoftheclustercanbeestimated.
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(iii)Byassumingthatthegasisheatedbystarsandsupernovaethatmakeup
mostofthemassofthecluster,themassofthesestarsandsupernovaecanbe
estimated.
(iv)Byassumingthatthegasisheatedbyinteractionswithdarkmatter,which
dominatesthemassofthecluster,themassoftheclustercanbeestimated.
(v)Byassumingthatthisgasisinhydrostaticequilibrium,thetemperature,mass
density,andeventhechemicalcompositionoftheclustercanbemodeled.
Explanation:Thedominantcomponentofthemassisapparentlydarkmatter,sothe
hotintraclustergasisonlyasmallfraction,andwehavenodirectwayofknowing
whatfraction.Butthegassettlesintoastateofhydrostaticequilibrium
which
isdeterminedbypressuresandgravitationalforces.Thegascanbemappedby
measuringitsx-rays,whichallowsastronomerstoestimatethegravitationalforces,
andhencethemass.
(d)(6points)InChapter6ofTheFirstThreeMinutes,StevenWeinbergdiscussesthree
reasonswhytheimportanceofasearchfora3 ÆKmicrowaveradiationbackground
wasnotgenerallyappreciatedinthe1950sandearly1960s.Choosethosethree
reasonsfromthefollowinglist.(2pointsforeachrightanswer,circleatmost3.)
(i)Theearliestcalculationserroneouslypredictedacosmicbackgroundtempera-
tureofonlyabout0:1 ÆK,andsuchabackgroundwouldbetooweaktodetect.
(ii)Therewasabreakdownincommunicationbetweentheoristsandexperimental-
ists.
(iii)Itwasnottechnologicallypossibletodetectasignalasweakasa3 ÆKmicrowave
backgrounduntilabout1965.
(iv)Sincealmostallphysicistsatthetimewerepersuadedbythesteadystatemodel,
thepredictionsofthebigbangmodelwerenottakenseriously.
(v)ItwasextraordinarilydiÆcultforphysiciststotakeseriouslyanytheoryofthe
earlyuniverse.
(vi)TheearlyworkonnucleosynthesisbyGamow,Alpher,Herman,andFollin,et
al.,hadattemptedtoexplaintheoriginofallcomplexnucleibyreactionsinthe
earlyuniverse.Thisprogramwasneververysuccessful,anditscredibilitywas
furtherunderminedasimprovementsweremadeinthealternativetheory,that
elementsaresynthesizedinstars.
Answer:Thecorrectanswerswere(ii),(v)and(vi).Theotherswereincorrectforthe
followingreasons:
(i)theearliestpredictionoftheCMBtemperature,byAlpherandHermanin1948,
was5degrees,not0.1degrees.
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(iii)Weinbergquoteshisexperimentalcolleaguesassayingthatthe3 ÆKradiation
couldhavebeenobserved\longbefore1965,probablyinthemid-1950sandper-
hapseveninthemid-1940s."ToWeinberg,however,thehistoricallyinteresting
questionisnotwhentheradiationcouldhavebeenobserved,butwhyradio
astronomersdidnotknowthattheyoughttotry.
(iv)Weinbergarguesthatphysicistsatthetimedidnotpayattentiontoeitherthe
steadystatemodelorthebigbangmodel,asindicatedbythesentenceinitem
(v)whichisadirectquotefromthebook:\ItwasextraordinarilydiÆcultfor
physiciststotakeseriouslyanytheoryoftheearlyuniverse".
(e)(6points)In1948RalphA.AlpherandRobertHermanwroteapaperpredicting
acosmicmicrowavebackgroundwithatemperatureof5K.Thepaperwasbased
onacosmologicalmodelthattheyhaddevelopedwithGeorgeGamow,inwhichthe
earlyuniversewasassumedtohavebeen�lledwithhotneutrons.Astheuniverse
expandedandcooledtheneutronsunderwentbetadecayintoprotons,electrons,and
antineutrinos,untilatsomepointtheuniversecooledenoughforlightelementsto
besynthesized.AlpherandHermanfoundthattoaccountfortheobservedpresent
abundancesoflightelements,theratioofphotonstonuclearparticlesmusthave
beenabout109.Althoughthepredictedtemperaturewasveryclosetotheactual
valueof2.7K,thetheorydi�eredfromourpresenttheoryintwoways.Circlethe
twocorrectstatementsinthefollowinglist.(3pointsforeachrightanswer;circleat
most2.)
(i)Gamow,Alpher,andHermanassumedthattheneutroncoulddecay,butnow
theneutronisthoughttobeabsolutelystable.
(ii)Inthecurrenttheory,theuniversestartedwithnearlyequaldensitiesofprotons
andneutrons,notallneutronsasGamow,Alpher,andHermanassumed.
(iii)Inthecurrenttheory,theuniversestartedwithmainlyalphaparticles,notall
neutronsasGamow,Alpher,andHermanassumed.(Note:analphaparticleis
thenucleusofaheliumatom,composedoftwoprotonsandtwoneutrons.)
(iv)Inthecurrenttheory,theconversionofneutronsintoprotons(andviceversa)
tookplacemainlythroughcollisionswithelectrons,positrons,neutrinos,and
antineutrinos,notthroughthedecayoftheneutrons.
(v)Theratioofphotonstonuclearparticlesintheearlyuniverseisnowbelieved
tohavebeenabout103,not109asAlpherandHermanconcluded.
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PROBLEM
6:EVOLUTION
OFAN
OPEN
UNIVERSE
Theevolutionofanopen,matter-dominateduniverseisdescribedbythefollowing
parametricequations:
ct=�(sinh���)
ap�=�(cosh��1):
Evaluatingthesecondoftheseequationsata= p�=2�yieldsasolutionfor�:
2�=�(cosh��1)
=)
cosh�=3
=)
�=cosh�1(3):
Wecanusetheseresultsinthe�rstequationtosolvefort.Notingthat
sinh�= pcosh2��1=p
8=2 p2;
wehave
t=�c h2 p
2�cosh�1(3) i:
Numerically,t�1:06567�=c.
PROBLEM
7:ANTICIPATING
A
BIG
CRUNCH
Thecriticaldensityisgivenby
�c=3H20
8�G
;
sothemassdensityisgivenby
�=0 �c=2�c=3H20
4�G
:
(S5.1)
Substitutingthisrelationinto
H20=8�3
G��kc2
a2
;
we�nd
H20=2H20 �kc2
a2
;
fromwhichitfollowsthat
apk=
cH0
:
(S5.2)
8.286QUIZ2REVIEW
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Nowuse
�=4�3G�a3
k3=2c2
:
SubstitutingthevalueswehavefromEqs.(S5.1)and(S5.2)for�anda= pk,wehave
�=
cH0
:
(S5.3)
Todeterminethevalueoftheparameter�,use
apk=�(1�cos�);
whichwhencombinedwithEqs.(S5.2)and(S5.3)impliesthatcos�=0:Theequation
cos�=0hasmultiplesolutions,butweknowthatthe�-parameterforaclosedmatter-
dominateduniversevariesbetween0and�duringtheexpansionphaseoftheuniverse.
Withinthisrange,cos�=0impliesthat�=�=2.Thus,theageoftheuniverseatthe
timethesemeasurementsaremadeisgivenby
t=�c
(��sin�)
=
1H0 �
�2�1 �:
Thetotallifetimeofthecloseduniversecorrespondsto�=2�,or
t�nal=2��c
=
2�
H0
;
sothetimeremainingbeforethebigcrunchisgivenby
t�nal �t=
1H0 h2�� ��2�1 �i=
�3�2
+1 �1H
0
:
PROBLEM
8:
TRACING
LIGHT
RAYS
IN
A
CLOSED,MATTER-
DOMINATED
UNIVERSE
(a)Since�=�=constant,d�=d�=0,andforlightraysonealwayshasd�=0.The
lineelementthereforereducesto
0=�c2dt2+a2(t)d 2:
8.286QUIZ2REVIEW
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Rearranginggives
�d d
t �2
=
c2
a2(t);
whichimpliesthat
d d
t=�c
a(t):
Theplussigndescribesoutwardradialmotion,whiletheminussigndescribesinward
motion.
(b)Themaximumvalueofthe coordinatethatcanbereachedbytimetisfoundby
integratingitsrateofchange:
hor= Z
t0
ca(t 0)dt 0:
Thephysicalhorizondistanceistheproperlengthoftheshortestlinedrawnatthe
timetfromtheoriginto = hor ,whichaccordingtothemetricisgivenby
`phys (t)= Z
= hor
=0
ds= Z
hor
0
a(t)d =
a(t) Z
t0
ca(t 0)dt 0:
(c)Frompart(a),
d d
t=
ca(t):
Bydi�erentiatingtheequationct=�(��sin�)statedintheproblem,one�nds
dt
d�=�c
(1�cos�):
Then
d d
�=d d
tdt
d�=�(1�cos�)
a(t)
:
Thenusinga=�(1�cos�),asstatedintheproblem,onehastheverysimpleresult
d d
�=1:
8.286QUIZ2REVIEW
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(d)Thispartisverysimpleifoneknowsthat mustchangeby2�beforethephoton
returnstoitsstartingpoint.Sinced =d�=1,thismeansthat�mustalsochange
by2�.Froma=�(1�cos�),onecanseethatareturnstozeroat�=2�,sothis
isexactlythelifetimeoftheuniverse.So,
Timeforphotontoreturn
Lifetimeofuniverse
=1:
Ifitisnotclearwhy mustchangeby2�forthephotontoreturntoitsstarting
point,thenrecalltheconstructionofthecloseduniversethatwasusedinLecture
Notes5.Thecloseduniverseisdescribedasthe3-dimensionalsurfaceofaspherein
afour-dimensionalEuclideanspacewithcoordinates(x;y;z;w):
x2+y2+z2+w2=a2;
whereaistheradiusofthesphere.TheRobertson-Walkercoordinatesystem
is
constructedonthe3-dimensionalsurfaceofthesphere,takingthepoint(0;0;0;1)
asthecenterofthecoordinatesystem.Ifwede�nethew-directionas\north,"
thenthepoint(0;0;0;1)canbecalledthenorthpole.Eachpoint(x;y;z;w)onthe
surfaceofthesphereisassignedacoordinate ,de�nedtobetheanglebetweenthe
positivewaxisandthevector(x;y;z;w).Thus =0atthenorthpole,and =�
fortheantipodalpoint,(0;0;0;�1),whichcanbecalledthesouthpole.Inmaking
theroundtripthephotonmusttravelfromthenorthpoletothesouthpoleand
back,foratotalrangeof2�.
Discussion:Somestudentsansweredthatthephotonwouldreturninthelifetime
oftheuniverse,butreachedthisconclusionwithoutconsideringthedetailsofthe
motion.Theargumentwassimplythat,atthebigcrunchwhenthescalefactor
returnstozero,alldistanceswouldreturntozero,includingthedistancebetween
thephotonanditsstartingplace.Thisstatementiscorrect,butitdoesnotquite
answerthequestion.First,thestatementinnowayrulesoutthepossibilitythat
thephotonmightreturntoitsstartingpointbeforethebigcrunch.Second,ifwe
usethedelicatebutwell-motivatedde�nitionsthatgeneralrelativistsuse,itisnot
necessarilytruethatthephotonreturnstoitsstartingpointatthebigcrunch.To
beconcrete,letmeconsideraradiation-dominatedcloseduniverse|ahypothetical
universeforwhichtheonly\matter"presentconsistsofmasslessparticlessuchas
photonsorneutrinos.Inthatcase(youcancheckmycalculations)aphotonthat
leavesthenorthpoleatt=0justreachesthesouthpoleatthebigcrunch.It
mightseemthatreachingthesouthpoleatthebigcrunchisnotanydi�erentfrom
completingtheroundtripbacktothenorthpole,sincethedistancebetweenthe
northpoleandthesouthpoleiszeroatt=tCrunch ,thetimeofthebigcrunch.
However,supposeweadopttheprinciplethattheinstantoftheinitialsingularity
8.286QUIZ2REVIEW
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andtheinstantofthe�nalcruncharebothtoosingulartobeconsideredpartofthe
spacetime.Wewillallowourselvestomathematicallyconsidertimesrangingfrom
t=�tot=tCrunch ��,where�isarbitrarilysmall,butwewillnottrytodescribe
whathappensexactlyatt=0ort=tCrunch .Thus,wenowconsideraphotonthat
startsitsjourneyatt=�,andwefollowituntilt=tCrunch ��.Forthecaseofthe
matter-dominatedcloseduniverse,suchaphotonwouldtraverseafractionofthe
fullcirclethatwouldbealmost1,andwouldapproach1as�!0.Bycontrast,for
theradiation-dominatedcloseduniverse,thephotonwouldtraverseafractionofthe
fullcirclethatisalmost1/2,anditwouldapproach1/2as�!0.Thus,fromthis
pointofviewthetwocaseslookverydi�erent.Intheradiation-dominatedcase,one
wouldsaythatthephotonhascomeonlyhalf-waybacktoitsstartingpoint.
PROBLEM
9:
LENGTHS
AND
AREAS
IN
A
TWO-DIMEN-
SIONALMETRIC
a)Alongthe�rstsegmentd�=0,sods2=(1+ar)2dr2,ords=(1+ar)dr.Integrating,
thelengthofthe�rstsegmentisfoundtobe
S1= Z
r0
0
(1+ar)dr=r0+12
ar20:
Alongthesecondsegmentdr=0,sods=r(1+br)d�,wherer=r0 .Sothelength
ofthesecondsegmentis
S2= Z
�=2
0
r0 (1+br0 )d�=�2
r0 (1+br0 ):
Finally,thethirdsegmentisidenticaltothe�rst,soS3=S1 .Thetotallengthis
then
S=2S1+S2=2 �r0+12
ar20 �+�2
r0 (1+br0 )
=
�2+�2 �
r0+12
(2a+�b)r20:
8.286QUIZ2REVIEW
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b)To�ndthearea,itisbesttodividetheregionintoconcentricstripsasshown:
Notethatthestriphasacoordinatewidthofdr,butthedistanceacrossthewidth
ofthestripisdeterminedbythemetrictobe
dh=(1+ar)dr:
ThelengthofthestripiscalculatedthesamewayasS2inpart(a):
s(r)=�2
r(1+br):
Theareaisthen
dA=s(r)dh;
so
A= Z
r0
0
s(r)dh
= Zr0
0
�2r(1+br)(1+ar)dr
=�2 Zr0
0
[r+(a+b)r2+abr3]dr
=
�2 �12
r20+13
(a+b)r30+14
abr40 �
8.286QUIZ2REVIEW
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PROBLEM
10:GEOMETRY
IN
A
CLOSED
UNIVERSE
(a)Asonemovesalongalinefromtheoriginto(h;0;0),thereisnovariationin�or�.
Sod�=d�=0,and
ds=
adr
p1�r2
:
So
`p= Z
h0
adr
p1�r2
=asin�1h:
(b)Inthiscaseitisonly�thatvaries,sodr=d�=0.So
ds=ard�;
so
sp=ah��:
(c)Frompart(a),onehas
h=sin(`p =a):
Insertingthisexpressionintotheanswerto(b),andthensolvingfor��,onehas
��=
sp
asin(`p =a):
Notethatasa!1,thisapproachestheEuclideanresult,��=sp =`p .
PROBLEM
11:THEGENERALSPHERICALLY
SYMMETRICMETRIC
(a)Themetricisgivenby
ds2=dr2+�2(r) �d�2+sin2�d�2 �:
Theradiusaisde�nedasthephysicallengthofaradiallinewhichextendsfromthe
centertotheboundaryofthesphere.Thelengthofapathisjusttheintegralofds,
so
a= Z
radialpathfrom
origintor0
ds:
8.286QUIZ2REVIEW
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Theradialpathisataconstantvalueof�and�,sod�=d�=0,andthends=dr.
So
a= Z
r0
0
dr=
r0:
(b)Onthesurfacer=r0 ,sodr�0.Then
ds2=�2(r
0 ) �d�2+sin2�d�2 �:
To�ndtheareaelement,consider�rstapathobtainedbyvaryingonly�.Thends=
�(r0 )d�.Similarly,apathobtainedbyvaryingonly�haslengthds=�(r0 )sin�d�.
Furthermore,thesetwopathsareperpendiculartoeachother,afactthatisincor-
poratedintothemetricbytheabsenceofadrd�term.Thus,theareaofasmall
rectangleconstructedfromthesetwopathsisgivenbytheproductoftheirlengths,
so
dA=�2(r
0 )sin�d�d�:
Theareaisthenobtainedbyintegratingovertherangeofthecoordinatevariables:
A=�2(r
0 ) Z2�
0
d� Z
�0
sin�d�
=�2(r
0 )(2�) ��cos� ����0 �
=)
A=4��2(r
0 ):
Asacheck,noticethatif�(r)=r,thenthemetricbecomesthemetricofEuclidean
space,insphericalpolarcoordinates.Inthiscasetheanswerabovebecomesthe
well-knownformulafortheareaofaEuclideansphere,4�r2.
(c)AsinProblem
2ofProblem
Set5,wecanimaginebreakingupthevolumeinto
sphericalshellsofin�nitesimalthickness,withagivenshellextendingfrom
rto
r+dr.Bythepreviouscalculation,theareaofsuchashellisA(r)=4��2(r).(In
thepreviouspartweconsideredonlythecaser=r0 ,butthesameargumentapplies
foranyvalueofr.)Thethicknessoftheshellisjustthepathlengthdsofaradial
pathcorrespondingtothecoordinateintervaldr.Forradialpathsthemetricreduces
tods2=dr2,sothethicknessoftheshellisds=dr.Thevolumeoftheshellisthen
dV=4��2(r)dr:
Thetotalvolumeisthenobtainedbyintegration:
V=4� Z
r0
0
�2(r)dr:
8.286QUIZ2REVIEW
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CheckingtheanswerfortheEuclideancase,�(r)=r,oneseesthatitgivesV=
(4�=3)r30 ,asexpected.
(d)Ifrisreplacedbyanewcoordinate��r2,thenthein�nitesimalvariationsofthe
twocoordinatesarerelatedby
d�d
r=2r=2 p�;
so
dr2=d�2
4�
:
Thefunction�(r)canthenbewrittenas�( p�),so
ds2=d�2
4�+�2( p�) �d�2+sin2�d�2 �:
PROBLEM
12:VOLUMESIN
A
ROBERTSON-WALKER
UNIVERSE
Theproductofdi�erentiallengthelementscorrespondingtoin�nitesimalchangesin
thecoordinatesr;�and�equalsthedi�erentialvolumeelementdV.Therefore
dV=a(t)
dr
p1�kr2 �a(t)rd��a(t)rsin�d�
Thetotalvolumeisthen
V= ZdV=a3(t) Zrmax
0
dr Z
�0
d� Z
2�
0
d�r2sin�
p1�kr2
Wecandotheangularintegrationsimmediately:
V=4�a3(t) Zrm
ax
0
r2dr
p1�kr2
:
[PedagogicalNote:Ifyoudon'tseethroughthesolutionsabove,thennotethatthevolume
ofthespherecanbedeterminedbyintegration,after�rstbreakingthevolumeinto
in�nitesimalcells.Agenericcellisshowninthediagrambelow:
8.286QUIZ2REVIEW
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Thecellincludesthevolumelyingbetweenrandr+dr,between�and�+d�,and
between�and�+d�.Inthelimitasdr,d�,andd�allapproachzero,thecell
approachesarectangularsolidwithsidesoflength:
ds1=a(t)
dr
p1�kr2
ds2=a(t)rd�
ds3=a(t)rsin�d�:
Hereeachdsiscalculatedbyusingthemetricto�ndds2,ineachcaseallowingonly
oneofthequantitiesdr,d�,ord�tobenonzero.Thein�nitesimalvolumeelement
isthendV=ds1 ds2 ds3 ,resultingintheanswerabove.Thederivationreliesonthe
orthogonalityofthedr,d�,andd�directions;theorthogonalityisimpliedbythe
metric,whichotherwisewouldcontaincrosstermssuchasdrd�.]
[Extension:Theintegralcaninfactbecarriedout,usingthesubstitution
pkr=sin
(ifk>0)
p�kr=sinh
(ifk>0).
Theansweris
V= 8>>>>>><>>>>>>:
2�a3(t) 24sin�1 �pkrmax �
k3=2
� p1�kr2m
ax
k
35(ifk>0)
2�a3(t) "p1�kr2m
ax
(�k)
�sinh�1 �p�krmax �
(�k)3=2
#(ifk<0)
.]
8.286QUIZ2REVIEW
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PROBLEM
13:THESCHWARZSCHILD
METRIC
a)TheSchwarzschildhorizonisthevalueofrforwhichthemetricbecomessingular.
Sincethemetriccontainsthefactor�
1�2GM
rc2 �;
itbecomessingularat
RS=2GM
c2
:
b)TheseparationbetweenAandB
ispurelyintheradialdirection,sotheproper
lengthofasegmentalongthepathjoiningthemisgivenby
ds2= �1�2GM
rc2 �
�1
dr2;
so
ds=
dr
q1�2GM
rc2
:
TheproperdistancefromAtoBisobtainedbyaddingtheproperlengthsofallthe
segmentsalongthepath,so
sAB
= ZrB
rA
dr
q1�2GM
rc2
:
EXTENSION:Theintegrationcanbecarriedoutexplicitly.Firstusetheexpression
fortheSchwarzschildradiustorewritetheexpressionforsAB
as
sAB
= ZrB
rA
prdr
pr�RS
:
Thenintroducethehyperbolictrigonometricsubstitution
r=RScosh2u:
Onethenhas
pr�RS= pRS
sinhu
8.286QUIZ2REVIEW
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dr=2RScoshusinhudu;
andtheinde�niteintegralbecomes
Zp
rdr
pr�RS
=2RS Zcosh2udu
=RS Z(1+cosh2u)du
=RS �u+12
sinh2u �
=RS(u+sinhucoshu)
=RSsinh�1 �rrR
S
�1 �+ pr(r�RS):
Thus,
sAB
=RS �sinh�1 �rrB
RS
�1 ��sinh �1 �rrA
RS
�1 ��
+ prB(rB
�RS)� prA(rA
�RS):
c)Atickoftheclockandthefollowingtickaretwoeventsthatdi�eronlyintheirtime
coordinates.Thus,themetricreducesto
�c2d�2=� �1�2GM
rc2 �c2dt2;
so
d�= r1�2GM
rc2
dt:
Thereadingontheobserver'sclockcorrespondstothepropertimeintervald�,so
thecorrespondingintervalofthecoordinatetisgivenby
�tA
=
��A
q1�2GM
rA
c2
:
d)SincetheSchwarzschildmetricdoesnotchangewithtime,eachpulseleavingAwill
takethesamelengthoftimetoreachB.Thus,thepulsesemittedbyAwillarrive
atBwithatimecoordinatespacing
�tB
=�tA
=
��A
q1�2GM
rA
c2
:
8.286QUIZ2REVIEW
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SOLUTIONS,FALL2018
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TheclockatB,however,willreadthepropertimeandnotthecoordinatetime.
Thus,
��B
= r1�2GM
rBc2
�tB
=
vuut1�2GM
rB
c2
1�2GM
rA
c2
��A
:
e)Fromparts(a)and(b),theproperdistancebetweenAandBcanberewrittenas
sAB
= ZrB
RS
prdr
pr�RS
:
Thepotentiallydivergentpartoftheintegralcomesfromtherangeofintegrationin
theimmediatevicinityofr=RS ,sayRS<r<RS+�.Forthisrangethequantity
printhenumeratorcanbeapproximatedby pRS,sothecontributionhastheform
pRS Z
RS
+�
RS
dr
pr�RS
:
Changingtheintegrationvariabletou�r�RS,thecontributioncanbeeasily
evaluated:
pRS Z
RS
+�
RS
dr
pr�RS
= pRS Z
�0
dup
u=2 pRS �<1:
So,althoughtheintegrandisin�niteatr=RS,theintegralisstill�nite.
TheproperdistancebetweenAandBdoesnotdiverge.
Lookingattheanswertopart(d),however,onecanseethatwhenrA
=RS,
Thetimeinterval��B
diverges.
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.58
PROBLEM
14:GEODESICS
Thegeodesicequationforacurvexi(�),wheretheparameter�isthearclength
alongthecurve,canbewrittenas
dd� �gijdxj
d� �=12
(@i gk` )dxk
d�
dx`
d�:
Heretheindicesj,k,and`aresummedfrom1tothedimensionofthespace,sothere
isoneequationforeachvalueofi.
(a)Themetricisgivenby
ds2=gij dxidxj=dr2+r2d�2;
so
grr=1;
g��=r2;
gr�=g�r=0:
Firsttakingi=r,thenonvanishingtermsinthegeodesicequationbecome
dd� �grrdr
d� �=12
(@r g�� )d�
d�
d�
d�;
whichcanbewrittenexplicitlyas
dd� �
dr
d� �=12 �@
r r2 � �d�
d� �
2
;
or
d2r
d�2=r �d�
d� �
2
:
Fori=�,onehasthesimpli�cationthatgijisindependentof�forall(i;j).So
dd� �r2d�
d� �=0:
(b)The�rststepistoparameterizethecurve,whichmeanstoimaginemovingalong
thecurve,andexpressingthecoordinatesasafunctionofthedistancetraveled.(I
amcallingthelocusy=1acurveratherthanaline,sincethetechniquesthatare
usedhereareusuallyappliedtocurves.Sincealineisaspecialcaseofacurve,there
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.59
isnothingwrongwithtreatingthelineasacurve.)InCartesiancoordinates,the
curvey=1canbeparameterizedas
x(�)=�;
y(�)=1:
(Theparameterizationisnotunique,becauseonecanchoose�=0torepresentany
pointalongthecurve.)Convertingtothedesiredpolarcoordinates,
r(�)= px2(�)+y2(�)= p�2+1;
�(�)=tan�1y(�)
x(�)=tan�1(1=�):
Calculatingtheneededderivatives,*
dr
d�=
�
p�2+1
d2r
d�2=
1
p�2+1 �
�2
(�2+1)3=2
=
1
(�2+1)3=2
=
1r3
d�
d�=�
1
1+ �1� �2
1�2=�1r
2
:
Then,substitutingintothegeodesicequationfori=r,
d2r
d�2=r �d�
d� �
2()1r
3
=r ��1r
2 �2
;
whichchecks.Substitutingintothegeodesicequationfori=�,
dd� �r2d�
d� �=0()dd
� �r2 ��1r
2 ��=0;
whichalsochecks.
*Ifyoudonotrememberhowtodi�erentiate�=tan �1(z),thenyoushouldknow
howtoderiveit.Writez=tan�=sin�=cos�,so
dz= �cos�
cos�+sin2�
cos2� �d�=(1+tan2�)d�:
Then
d�d
z=
1
1+tan2�=
11+z2
:
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.60
PROBLEM
15:AN
EXERCISE
IN
TWO-DIMENSIONALMETRICS(30
points)
(a)Since
r(�)=(1+�cos2�)r0;
astheangularcoordinate�changesbyd�,rchangesby
dr=dr
d�d�=�2�r0cos�sin�d�:
ds2isthengivenby
ds2=dr2+r2d�2
=4�2r20cos2�sin2�d�2+(1+�cos2�)2r20d�2
= �4�2cos2�sin2�+(1+�cos2�)2 �r20d�2;
so
ds=r0 q4�2cos2�sin2�+(1+�cos2�)2d�:
Since�runsfrom�1to�2asthecurveissweptout,
S=r0 Z
�2
�1 q4�2cos2�sin2�+(1+�cos2�)2d�:
(b)Since�doesnotvaryalongthispath,
ds= r1+ra
dr;
andso
R= Z
r0
0 r1+ra
dr:
(c)Sincethemetricdoesnotcontainatermindrd�,therand�directionsareorthog-
onal.Thus,ifoneconsidersasmallregioninwhichrisintheintervalr 0tor 0+dr 0,
and�isintheinterval� 0to� 0+d� 0,thentheregioncanbetreatedasarectangle.
Thesidealongwhichrvarieshaslengthdsr= p1+(r 0=a)dr 0,whilethesidealong
which�varieshaslengthds�=r 0d� 0.Theareaisthen
dA=dsrds�=r 0 p1+(r 0=a)dr 0d� 0:
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.61
Tocovertheareaforwhichr<r0 ,r 0mustbeintegratedfrom0tor0 ,and� 0must
beintegratedfrom0to2�:
A= Z
r0
0
dr 0 Z
2�
0
d� 0r 0 p1+(r 0=a):
But
Z2�
0
d� 0=2�;
so
A=2� Z
r0
0
dr 0r 0 p1+(r 0=a):
Youwerenotaskedtocarryouttheintegration,butitcanbedonebyusingthe
substitutionu=1+(r 0=a),sodu=(1=a)dr 0,andr 0=a(u�1).Theresultis
A=4�a2
15 �2+ �3r20
a2
+r0a�2 � r1+r0a �:
(d)ThenonzerometriccoeÆcientsaregivenby
grr=1+ra
;
g��=r2;
sothemetricisdiagonal.Fori=1=r,thegeodesicequationbecomes
dds �grrdr
ds �=12@grr
@r
dr
ds
dr
ds+12@g��
@r
d�
ds
d�
ds;
soifwesubstitutethevaluesfromabove,wehave
dds ��1+ra �dr
ds �=12@@
r �1+ra � �dr
ds �
2+12@r2
@r �
d�
ds �
2
:
Simplifyingslightly,dd
s ��1+ra �dr
ds �=
12a �
dr
ds �
2+r �d�
ds �
2
:
Theansweraboveisperfectlyacceptable,butonemightwanttoexpandtheleft-hand
side:
dds ��1+ra �dr
ds �=1a �dr
ds �
2+ �1+ra �d2r
ds2
:
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.62
Insertingthisexpansionintotheboxedequationabove,the�rsttermcanbebrought
totheright-handside,giving
�1+ra �d2r
ds2
=�12
a �dr
ds �
2+r �d�
ds �
2
:
Thei=2=�equationissimpler,becausenoneofthegijcoeÆcientsdependon�,
sotheright-handsideofthegeodesicequationvanishes.Onehassimply
dds �r2d�
ds �=0:
Formostpurposesthisisthebestwaytowritetheequation,sinceitleadsimmedi-
atelytor2(d�=ds)=const:However,itispossibletoexpandthederivative,giving
thealternativeform
r2d2�
ds2+2rdr
ds
d�
ds=0:
PROBLEM
16:GEODESICSON
THESURFACEOFA
SPHERE
(a)RotationsareeasytounderstandinCartesiancoordinates.Therelationshipbetween
thepolarandCartesiancoordinatesisgivenby
x=rsin�cos�
y=rsin�sin�
z=rcos�:
Theequatoristhendescribedby�=�=2,and�= ,where isaparameter
runningfrom0to2�.Thus,theequatorisdescribedbythecurvexi( ),where
x1=x=rcos
x2=y=rsin
x3=z=0:
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.63
Nowintroduceaprimedcoordinatesystemthatisrelatedtotheoriginalsystemby
arotationinthey-zplanebyanangle�:
x=x0
y=y 0cos��z 0sin�
z=z 0cos�+y 0sin�:
Therotatedequator,whichweseektodescribe,isjustthestandardequatorinthe
primedcoordinates:
x0=rcos ;
y 0=rsin ;
z 0=0:
Usingtherelationbetweenthetwocoordinatesystemsgivenabove,
x=rcos
y=rsin cos�
z=rsin sin�:
UsingagaintherelationsbetweenpolarandCartesiancoordinates,
cos�=zr
=sin sin�
tan�=yx
=tan cos�:
(b)Asegmentoftheequatorcorrespondingtoanintervald haslengthad ,sothe
parameter isproportionaltothearclength.Expressedintermsofthemetric,this
relationshipbecomes
ds2=gijdxi
d
dxj
d d 2=a2d 2:
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.64
Thusthequantity
A�gijdxi
d
dxj
d
isequaltoa2,sothegeodesicequation(5.50)reducestothesimplerform
of
Eq.(5.52).(NotethatwearefollowingthenotationofLectureNotes5,except
thatthevariableusedtoparameterizethepathiscalled ,ratherthan�ors.Al-
thoughAisnotequalto1asweassumedinLectureNotes5,itiseasilyseenthat
Eq.(5.52)followsfrom(5.50)providedonlythatA=constant.)Thus,
dd �gijdxj
d �=12
(@i gk` )dxk
d
dx`
d
:
Forthisproblemthemetrichasonlytwononzerocomponents:
g��=a2;
g��=a2sin2�:
Takingi=�inthegeodesicequation,
dd �g��d�
d �=12
@� g��d�
d
d�
d
=)
d2�
d 2=sin�cos� �d�
d �
2
:
Takingi=�,
dd �a2sin2�d�
d �=0
=)
dd �sin2�d�
d �=0:
(c)Thispartismainlyalgebra.Takingthederivativeof
cos�=sin sin�
implies
�sin�d�=cos sin�d :
Then,usingthetrigonometricidentitysin�=p
1�cos2�,one�nds
sin�= q1�sin2 sin2�;
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.65
so
d�
d =�
cos sin�
p1�sin2 sin2�:
Similarly
tan�=tan cos�
=)
sec2�d�=sec2 d cos�:
Then
sec2�=tan2�+1=tan2 cos2�+1
=
1cos2 [sin2 cos2�+cos2 ]
=sec2 [sin2 (1�sin2�)+cos2 ]
=sec2 [1�sin2 sin2�];
So
d�
d =
cos�
1�sin2 sin2�:
Toverifythegeodesicequationsofpart(b),itiseasiesttocheckthesecondone
�rst:
sin2�d�
d =(1�sin2 sin2�)
cos�
1�sin2 sin2�
=cos�;
soclearly
dd �sin2�d�
d �=
dd (cos�)=0:
Toverifythe�rstgeodesicequationfrompart(b),�rstcalculatetheleft-handside,
d2�=d 2,usingourresultford�=d :
d2�
d 2
=
dd �
d�
d �=
dd (�
cos sin�
p1�sin2 sin2� ):
Aftersomestraightforwardalgebra,one�nds
d2�
d 2=
sin sin�cos2�
�1�sin2 sin2� �3=2
:
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.66
Theright-handsideofthe�rstgeodesicequationcanbeevaluatedusingtheexpres-
sionfoundaboveford�=d ,giving
sin�cos� �d�
d �
2= q1�sin2 sin2�sin sin�
cos2�
�1�sin2 sin2� �2
=
sin sin�cos2�
�1�sin2 sin2� �3=2
:
Sotheleft-andright-handsidesareequal.
PROBLEM
17:GEODESICSIN
A
CLOSED
UNIVERSE
(a)(7points)Forpurelyradialmotion,d�=d�=0,sothelineelementreducesdo
�c2d�2=�c2dt2+a2(t) �dr2
1�r2 �:
Dividingbydt2,
�c2 �d�d
t �2
=�c2+
a2(t)
1�r2 �drd
t �2
:
Rearranging,
d�d
t= s1�
a2(t)
c2(1�r2) �drd
t �2
:
(b)(3points)
dt
d�=
1d�d
t=
1
s1�
a2(t)
c2(1�r2) �drd
t �2
:
(c)(10points)Duringanyintervalofclocktimedt,thepropertimethatwouldbe
measuredbyaclockmovingwiththeobjectisgivenbyd�,asgivenbythemetric.
Usingtheanswerfrompart(a),
d�=d�d
tdt= s1�
a2(t)
c2(1�r2p ) �drp
dt �
2dt:
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.67
Integratingto�ndthetotalpropertime,
�= Z
t2
t1 s1�
a2(t)
c2(1�r2p ) �drp
dt �
2dt:
(d)(10points)Thephysicaldistanced`thattheobjectmovesduringagiventime
intervalisrelatedtothecoordinatedistancedrbythespatialpartofthemetric:
d`2=ds2=a2(t) �dr2
1�r2 �=)
d`=
a(t)
p1�r2dr:
Thus
vphys=d`
dt=
a(t)
p1�r2
drd
t:
Discussion:Acommonmistakewastoinclude�c2dt2intheexpressionford`2.To
understandwhythisisnotcorrect,weshouldthinkabouthowanobserverwould
measured`,thedistancetobeusedincalculatingthevelocityofapassingobject.
Theobserverwouldplaceameterstickalongthepathoftheobject,andshewould
marko�thepositionoftheobjectatthebeginningandendofatimeintervaldtmeas .
Thenshewouldreadthedistancebysubtractingthetworeadingsonthemeterstick.
Thissubtractionisequaltothephysicaldistancebetweenthetwomarks,measured
atthesametimet.Thus,whenwecomputethedistancebetweenthetwomarks,
wesetdt=0.Tocomputethespeedshewouldthendividethedistancebydtmeas ,
whichisnonzero.
(e)(10points)Westartwiththestandardformulaforageodesic,aswrittenonthe
frontoftheexam:
dd� �g��dx�
d� �=12
(@�g��)dx�
d�
dx�
d�
:
Thisformulaistrueforeachpossiblevalueof�,whiletheEinsteinsummation
conventionimpliesthattheindices�,�,and�aresummed.Wearetryingtoderive
theequationforr,soweset�=r.Sincethemetricisdiagonal,theonlycontribution
ontheleft-handsidewillbe�=r.Ontheright-handside,thediagonalnatureof
themetricimpliesthatnonzerocontributionsariseonlywhen�=�.Thetermwill
vanishunlessdx�=d�isnonzero,so�mustbeeitherrort(i.e.,thereisnomotion
inthe�or�directions).However,theright-handsideisproportionalto
@g��
@r
:
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.68
Sincegtt=�c2,thederivativewithrespecttorwillvanish.Thus,theonlynonzero
contributionontheright-handsidearisesfrom�=�=r.Using
grr=
a2(t)
1�r2
;
thegeodesicequationbecomes
dd� �grrdr
d� �=12
(@r grr )dr
d�
dr
d�;
or
dd� �
a2
1�r2dr
d� �=12 �
@r �a2
1�r2 ��dr
d�
dr
d�;
or�nally
dd� �
a2
1�r2dr
d� �=a2
r
(1�r2)2 �dr
d� �
2
:
Thismatchestheformshowninthequestion,with
A=
a2
1�r2
;andC=a2
r
(1�r2)2
;
withB=D=E=0.
(f)(5pointsEXTRACREDIT)Thealgebraherecangetmessy,butitisnottoobad
ifonedoesthecalculationinaneÆcientway.Onegoodwaytostartistosimplify
theexpressionforp.Usingtheanswerfrom(d),
p=
mvphys
q1�v2p
hys
c2
=
m
a(t)
p1�r2
dr
dt
q1�
a2
c2(1�r2) �dr
dt �
2
:
Usingtheanswerfrom(b),thissimpli�esto
p=m
a(t)
p1�r2
drd
tdt
d�=m
a(t)
p1�r2
dr
d�:
Multiplythegeodesicequationbym,andthenusetheaboveresulttorewriteitas
dd� �
ap
p1�r2 �=ma2
r
(1�r2)2 �dr
d� �
2
:
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.69
Expandingtheleft-handside,
LHS=
dd� �
ap
p1�r2 �=
1
p1�r2
dd� fapg+ap
r
(1�r2)3=2
dr
d�
=
1
p1�r2
dd� fapg+ma2
r
(1�r2)2 �dr
d� �
2
:
Insertingthisexpressionbackintoleft-handsideoftheoriginalequation,onesees
thatthesecondtermcancelstheexpressionontheright-handside,leaving
1
p1�r2
dd� fapg=0:
Multiplyingby p1�r2,onehasthedesiredresult:
dd� fapg=0
=)
p/1
a(t):
PROBLEM
18:A
TWO-DIMENSIONALCURVED
SPACE(40points)
(a)For�=constant,theexpressionforthemetricreducesto
ds2=
adu2
4u(a�u)
=)
ds=12 ra
u(a�u)du:
To�ndthelengthoftheradiallineshown,onemustin-
tegrate
this
expression
from
the
value
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.70
ofuatthecenter,whichis0,tothevalueofuattheouteredge,whichisa.So
R=12 Za
0 ra
u(a�u)du:
Youwerenotexpectedtodoit,buttheintegralcanbecarriedout,givingR=
(�=2) pa.
(b)Foru=constant,theexpressionforthemetricreducesto
ds2=ud�2
=)
ds=p
ud�:
Since�runsfrom0to2�,andu=aforthecircumference
ofthespace,
S= Z
2�
0
pad�=2� pa:
(c)Toevaluatetheanswerto�rstorderindumeanstoneglect
anytermsthatwouldbeproportionaltodu2orhigherpow-
ers.Thismeansthatwecantreattheannulusasifitwere
arbitrarilythin,inwhichcasewecanimaginebendingit
intoarectanglewithoutchangingitsarea.Theareaisthen
equaltothecircumferencetimesthewidth.Boththecir-
cumferenceandthewidthmustbecalculatedbyusingthe
metric:
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.71
dA=circumference�width
=[2� pu0]� �12 ra
u0 (a�u0 )du �
=
� ra
(a�u0 )du:
(d)Wecan�ndthetotalareabyimaginingthatitisbrokenupintoannuluses,where
asingleannulusstartsatradialcoordinateuandextendstou+du.Asinpart(a),
thisexpressionmustbeintegratedfromthevalueofuatthecenter,whichis0,to
thevalueofuattheouteredge,whichisa.
A=� Z
a0 r
a
(a�u)du:
Youdidnotneedtocarryoutthisintegration,buttheanswerwouldbeA=2�a.
(e)Fromthelistatthefrontoftheexam,thegeneralformulaforageodesiciswritten
as
dds �gijdxj
ds �=12@gk`
@xidxk
ds
dx`
ds:
Themetriccomponentsgijarerelatedtods2by
ds2=gijdxidxj;
wheretheEinsteinsummationconvention(sumoverrepeatedindices)isassumed.
Inthiscase
g11 �guu=
a
4u(a�u)
g22 �g��=u
g12=g21=0;
whereIhavechosenx1=uandx2=�.Theequationwithdu=dsontheleft-hand
sideisfoundbylookingatthegeodesicequationsfori=1.Ofcoursej,k,and`
mustallbesummed,buttheonlynonzerocontributionsarisewhenj=1,andk
and`areeitherbothequalto1orbothequalto2:
dds �guudu
ds �=12@guu
@u �
du
ds �
2+12@g��
@u �
d�
ds �
2
:
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.72
dds �
a
4u(a�u)
du
ds �=12 �dd
u �a
4u(a�u) ���
du
ds �
2+12 �dd
u(u) ��d�
ds �
2
=12 �a
4u(a�u)2 �
a
4u2(a�u) ��
du
ds �
2+12 �d�
ds �
2
=
18a(2u�a)
u2(a�u)2 �du
ds �
2+12 �d�
ds �
2
:
(f)Thispartissolvedbythesamemethod,butitissimpler.Hereweconsiderthe
geodesicequationwithi=2.Theonlytermthatcontributesontheleft-handside
isj=2.Ontheright-handsideone�ndsnontrivialexpressionswhenkand`are
eitherbothequalto1orbothequalto2.However,thetermsontheright-handside
bothinvolvethederivativeofthemetricwithrespecttox2=�,andthesederivatives
allvanish.So
dds �g��d�
ds �=12@guu
@� �
du
ds �
2+12@g��
@� �
d�
ds �
2
;
whichreducesto
dds �ud�
ds �=0:
PROBLEM
19:ROTATING
FRAMESOFREFERENCE(35points)
(a)Themetricwasgivenas
�c2d�2=�c2dt2+ hdr2+r2(d�+!dt)2+dz2 i;
andthemetriccoeÆcientsarethenjustreado�fromthisexpression:
g11 �grr=1
g00 �gtt=coeÆcientofdt2=�c2+r2!2
g20 �g02 �g�t �gt�=12 �
coeÆcientofd�dt=r2!2
g22 �g��=coeÆcientofd�2=r2
g33 �gzz=coeÆcientofdz2=1:
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.73
Notethattheo�-diagonaltermg�tmustbemultipliedby1/2,becausetheexpression
3X�
=0
3X�
=0g��dx�dx�
includesthetwoequaltermsg20d�dt+g02dtd�,whereg20 �g02 .
(b)Startingwiththegeneralexpression
dd� �g��dx�
d� �=12
(@�g��)dx�
d�
dx�
d�
;
weset�=r:
dd� �gr�dx�
d� �=12
(@r g��)dx�
d�
dx�
d�
:
Whenwesum
over�ontheleft-handside,theonlyvalueforwhichgr�
6=0is
�=1�r.Thus,theleft-handsideissimply
LHS=
dd� �grrdx1
d� �=
dd� �
dr
d� �=d2r
d�2
:
TheRHSincludeseverycombinationof�and�forwhichg��dependsonr,sothat
@rg�� 6=0.Thismeansgtt ,g��,andg�t .So,
RHS=12
@r (�c2+r2!2) �dt
d� �
2+12
@r (r2) �d�
d� �
2+@r (r2!)d�
d�
dt
d�
=r!2 �dt
d� �
2+r �d�
d� �
2+2r!d�
d�
dt
d�
=r �d�
d�+!dt
d� �
2
:
Notethatthe�nalterminthe�rstlineisreallythesumofthecontributionsfrom
g�tandgt�,wherethetwotermswerecombinedtocancelthefactorof1/2inthe
generalexpression.Finally,
d2r
d�2=r �d�
d�+!dt
d� �
2
:
IfoneexpandstheRHSas
d2r
d�2=r �d�
d� �
2+r!2 �dt
d� �
2+2r!d�
d�
dt
d�;
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.74
thenonecanidentifythetermproportionalto!2asthecentrifugalforce,andthe
termproportionalto!astheCoriolisforce.
(c)Substituting�=�,
dd� �g��dx�
d� �=12
(@�g��)dx�
d�
dx�
d�
:
ButnoneofthemetriccoeÆcientsdependon�,sotheright-handsideiszero.The
left-handsidereceivescontributionsfrom�=�and�=t:
dd� �g��d�
d�+g�tdt
d� �=
dd� �r2d�
d�+r2!dt
d� �=0;
so
dd� �r2d�
d�+r2!dt
d� �=0:
Notethatonecannot\factorout"r2,sincercandependon�.Ifthisequation
isexpandedtogiveanequationford2�=d�2,theterm
proportionalto!would
beidenti�edastheCoriolisforce.Thereisnotermproportionalto!2,sincethe
centrifugalforcehasnocomponentinthe�direction.
(d)IfEq.(P19.1)oftheproblemisdividedbyc2dt2,oneobtains
�d�d
t �2
=1�1c
2 "�dr
dt �
2+r2 �d�d
t+! �
2+ �dzd
t �2 #
:
Thenusing
dt
d�=
1�
d�
dt �;
onehas
dt
d�=
1
vuut1�1c
2 "�dr
dt �
2+r2 �d�d
t+! �
2+ �dzd
t �2 #:
Notethatthisequationisreallyjust
dt
d�=
1
p1�v2=c2
;
adaptedtotherotatingcylindricalcoordinatesystem.
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.75
PROBLEM
20:THE
STABILITY
OF
SCHWARZSCHILD
ORBITS�
(30
points)
Fromthemetric:
ds2=�c2d�2=�h(r)c2dt2+h(r) �1dr2+r2d�2+r2sin2�d�2;
(S20.1)
andtheconventionds2=g��dx�dx�wereadthenonvanishingmetriccomponents:
gtt=�h(r)c2;grr=
1h(r);g��=r2;g��=r2sin2�:
(S20.2)
Wearetoldthattheorbithas�=�=2,soontheorbitd�=0andtherelevantmetric
andmetriccomponentsare:
ds2=�c2d�2=�h(r)c2dt2+h(r) �1dr2+r2d�2;
(S20.3)
gtt=�h(r)c2;grr=
1h(r);g��=r2:
(S20.4)
Wealsoknowthat
h(r)=1�RSr:
(S20.5)
(a)Thegeodesicequation
dd� �g��dx�
d� �=12@g��
@x�
dx�
d�
dx�
d�;
(S20.6)
fortheindexvalue�=rtakestheform
dd� �grrdr
d� �=12@g��
@r
dx�
d�
dx�
d�
:
Expandingout
dd� �
1hdr
d� �=12@gtt
@r �
dt
d� �
2+12@grr
@r �
dr
d� �
2+12@g��
@r �
d�
d� �
2:
Usingthevaluesin(S20.4)toevaluatetheright-handsideandtakingthederivativeson
theleft-handside:
�h0
h2 �dr
d� �
2+1hd2r
d�2=�12
c2h0 �dt
d� �
2�12h0
h2 �dr
d� �
2+r �d�
d� �
2:
*SolutionbyBartonZwiebach.
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.76
Hereh0�dh
drandwehavesupressedtheargumentsofhandh0toavoidclutter.Collecting
theunderlinedtermstotherightandmultiplyingbyh,we�nd
d2r
d�2=�12
h0hc2 �dt
d� �
2+12h0h �dr
d� �
2+rh �d�
d� �
2:
(S20.7)
(b)Dividingtheexpression(S20.3)forthemetricbyd�2wereadily�nd
�c2=�hc2 �dt
d� �
2+1h �dr
d� �
2+r2 �d�
d� �
2;
andrearranging,
hc2 �dt
d� �
2=c2+1h �dr
d� �
2+r2 �d�
d� �
2:
(S20.8)
Thisisthemostusefulformoftheanswer.Ofcourse,wealsohave
�dt
d� �
2=1h
+
1h2c2 �dr
d� �
2+
r2
hc2 �d�
d� �
2:
(S20.9)
Weusenow(S20.8)tosimplify(S20.7):
d2r
d�2=�12
h0 c2+1h �dr
d� �
2+r2 �d�
d� �
2 !+12h0h �dr
d� �
2+rh �d�
d� �
2:
Expandingout,thetermswith(dr
d�)2cancelandwe�nd
d2r
d�2=�12
h0c2+ �rh�12
h0r2 ��d�
d� �
2:
(S20.10)
Thisisanacceptableanswer.Onecansimplify(S20.10)furtherbynotingthath0=
RS=r2andrh=r�RS:d
2rd�2=�12RSc2
r2
+ �r�32
RS ��d�
d� �
2:
(S20.11)
Inthenotationoftheproblemstatement,wehave
f0 (r)=�12RSc2
r2
;
f1 (r)=r�32
RS:
(S20.12)
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.77
(c)Thegeodesicequation(S20.6)for�=�gives
dd� �g��d�
d� �=12@g��
@�
dx�
d�
dx�
d�:
Sincenometriccomponentdependson�,theright-handsidevanishesandweget:
dd� �r2d�
d� �=0
!
dd�L=0;where
L�r2d�
d�:
(S20.13)
ThequantityLisaconstantofthemotion,namely,itisanumberindependentof�.
(d)Using(S20.13)thesecond-orderdi�erentialequation(S20.11)forr(�)takestheform
statedintheproblem:
d2r
d�2=f0 (r)+f1 (r)
r4
L2�H(r);
(S20.14)
wherewehaveintroducedthefunctionH(r)(recallthatLisaconstant!).Thedi�erential
equationthentakestheform
d2r
d�2=H(r):
(S20.15)
Sincewearetoldthatacircularorbitwithradiusr0
exists,thefunctionr(�)=
r0
mustsolvethisequation.Beingtheconstantfunction,theleft-handsidevanishesand,
consequently,theright-handsidemustalsovanish:
H(r0 )=f0 (r0 )+f1 (r0 )
r40
L2=0:
(S20.16)
ToinvestigatestabilityweconsiderasmallperturbationÆr(�)oftheorbit:
r(�)=r0+Ær(�);with
Ær(�)�r0
atsomeinitial�:
Substitutingthisinto(S20.15)weget,to�rstnontrivialapproximation
d2Ær
d�2
=H(r0+Ær)'H(r0 )+ÆrH0(r
0 )=ÆrH0(r
0 );
whereH0(r)=dH(r)
dr
andweusedH(r0 )=0from(S20.16).Theresultingequation
d2Ær(�)
d�2
=H0(r
0 )Ær(�);
(S20.17)
isfamiliarbecauseH0(r
0 )isjustanumber.Theconditionofstabilityisthatthisnumber
isnegative:H0(r
0 )<0.Indeed,inthiscase(S20.17)istheharmonicoscillatorequation
d2x
dt2
=�!2x;withreplacements
x$Ær;t$�;�!2$H0(r
0 );
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.78
andthesolutiondescribesboundedoscillations.Sostabilityrequires:
StabilityCondition:H0(r
0 )=
ddr �f0 (r)+f1 (r)
r4
L2 �
r=r0
<0:
(S20.18)
Thisistheanswertopart(d).
������������������������������
Forstudentsinterestedingettingthefamousresultthatorbitsarestableforr>3RSwe
completethispartoftheanalysisbelow.FirstweevaluateH0(r
0 )in(S20.18)usingthe
valuesoff0andf1in(S20.12):
H0(r
0 )=
ddr ��
12RSc2
r2
+ �1r
3 �3RS
2r4 �L2 �
r=r0
=RSc2
r30
�3L2
r50
(r0 �2RS):
Theinequalityin(S20.18)thengivesus
RSc2�3L2
r20
(r0 �2RS)<0;
(S20.19)
wherewemultipliedbyr30>0.TocompletethecalculationweneedthevalueofL2for
theorbitwithradiusr0 .ThisvalueisdeterminedbythevanishingofH(r0 ):
�12RSc2
r20
+(r0 �32
RS)L2
r40
=0
!
L2
r20
=12
RSc2
(r0 �32RS):
Note,incidentally,thattheequalitytotherightdemandsthatforacircularorbitr0>
32RS.SubstitutingtheabovevalueofL2=r20in(S20.19)weget:
RSc2�32
RSc2
(r0 �32RS) (r0 �2RS)<0:
CancellingthecommonfactorsofRSc2we�nd
1�32(r0 �2RS)
(r0 �32RS)<0;
whichisequivalentto
32(r0 �2RS)
(r0 �32RS)>1:
Forr0>32RS,weget3(r
0 �2RS)>2(r0 �32
RS)
!
r0>3RS:
(S20.20)
ThisisthedesiredconditionforstableorbitsintheSchwarzschildgeometry.
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.79
PROBLEM
21:PRESSUREAND
ENERGYDENSITYOFMYSTERIOUS
STUFF
(a)Ifu/1= pV,thenonecanwrite
u(V+�V)=u0 rV
V+�V
:
(Theaboveexpressionisproportionalto1= pV+�V,andreducestou=u0when
�V=0.)Expandingto�rstorderin�V,
u=
u0
q1+�VV
=
u0
1+12�VV
=u0 �1�12�VV �
:
Thetotalenergyistheenergydensitytimesthevolume,so
U=u(V+�V)=u0 �1�12�VV �
V �1+�VV �
=U0 �1+12�VV �
;
whereU0=u0 V:Then
�U=12�VV
U0:
(b)Theworkdonebytheagentmustbethenegativeoftheworkdonebythegas,which
isp�V.So
�W
=�p�V
:
(c)Theagentmustsupplythefullchangeinenergy,so
�W
=�U=12�VV
U0:
Combiningthiswiththeexpressionfor�W
frompart(b),oneseesimmediately
that
p=�12U0
V
=
�12
u0:
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.80
PROBLEM
22:VOLUMEOFACLOSEDTHREE-DIMENSIONALSPACE
(15points)
Themetricforthespacethatweareconsideringis
ds2=R2 �d 2+f2( ) �d�2+sin2�d�2 ��;
Forcomparison,themetricforthesurfaceofasphereofradiusRisgivenby
ds2=R2 �d�2+sin2�d�2 �:
Bycomparingthesetwo,oneseesthatthesetofpointsdescribedby =constant
(varying�and�)hasthesamemetricasasphereofradiusr=Rf( ).Wecan
saveourselvessometroubleincalculatingbyrememberingthattheareaofsucha
sphericalsurfaceofradiusris4�r2=4�R2f2( ).
Thevolumeofthesphericalshellshownintheproblemisjusttheareatimesthe
thickness.Thethicknessisnotd ,since isonlyacoordinate|
rememberthat
incurvedspaceacoordinateandadistancearetwodi�erentthings.Thedistanceis
givenbythemetric.Considerinthiscasearadiallineextendingfrom to +d ,
atconstant�and�.Then
ds2=R2d 2
;
andsothelengthofthelinesegmentisds=Rd .
Thevolumeofthesphericalshellisthengivenby
dV= �4�R2f2( ) �Rd
:
Wemustnowintegrateovertherangeof ,for0to�.So,
V=4�R3 Z�
0
f2( )d
:
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.81
PROBLEM
23:GRAVITATIONALBENDING
OFLIGHT(30points)
(a)(6points)Notethat
dr2=
1r2(xdx+ydy+zdz)2
=
1r2 �x2dx2+y2dy2+z2dz2+2xydxdy+2xzdxdz+2yzdydz �:
(S23.1)
Byusingthisexpressionfor(dr)2inEq.(P23.5),wehavethefullexpressionfords2
writtenout,fromwhichwecanreado�thecomponentsofg�� :
gtt=coeÆcientofdt2=�c2 �1�RSch
r �
gxx=coeÆcientofdx2=1+RSch
r3
x2
gxy=12
ofcoeÆcientofdxdy=RSch
r3
xy:
(S23.2)
Anumberofpeoplemissedthefactorof1/2inthevalueofgxy .Itarisesbecause
thegeneralformulaiswrittenasds2=g��dx�dx�,whichwhenexpandedbecomes
ds2=gxx dx2+gyy dy2+gzz dz2+gtt dt2+gxy dxdy+gyxdydx+::::
Sincedxdy=dydx,thecoeÆcientofdxdyisgxy+gyx=2gxy .
(b)(9points)Itwillbeusefultoknowthederivativesofr:
@r
@x=
@@x(x2+y2+z2)1=2
=12
(x2+y2+z2) �1=2
@@x(x2+y2+z2)=xr
:
(S23.3)
Similarly,
@r
@y=yr
and
@r
@z=zr
;
(S23.4)
and
dr
d�=@r
@x
dx
d�+@r
@y
dy
d�+@r
@z
dz
d�
=xr
:
(S23.5)
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.82
Inthe2ndlineIusedthevalueof@r=@xfromEq.(S23.3),andthederivatives
dx
d�=1;
dy
d�=dz
d�=0
(S23.6)
thatcanbefoundfromEq.(P23.8).
Now,toexpandtheleft-handsideofthegeodesicequation:
dd� �g��dx�
d� �=
dd� �gyydy
d�+gyxdx
d� �
=
dd� ��1+RSch
r3
y2 �dy
d�+RSch
r3
xydx
d� �
=d2y
d�2 �3RSch
r4
xrxydx
d�+RSch
r3
dx
d�ydx
d�
=
d2y
d�2 �3RSch b
r5
x2+RSch b
r3
:
(S23.7)
NotethatIdroppedaterm
RSch y2
r3
d2y
d�2
andaterm
RSch
r3
xyd2x
d�2
;
whichisjusti�edbecausetheaccelerationd2y
d�2
willbeproportionaltoG,andRSch
is
proportionaltoG,sothistermis2ndorderinG.Theproblemstatedthatweareto
workto�rstorderinG.Nopointsweretakeno�,however,fromstudentswhoretained
theseorothernegligibleterms.
Note,however,thatd2yd�2isnotnegligible,andappearsintheanswer.Thisis
becausedy=d�isnotactuallyzero,butisoforderG.dy=d�iszerofortheunperturbed
path,butinrealitythephotonpicksupasmallvelocityinthey-direction,causedby
thegravitationalattractionoftheSunandproportionaltoG.d2y=d�2
willalsobe
proportionaltoG.Whendy=d�multipliesafactorproportionaltoRSch ,theproductis
oforderG2andhencenegligible.Butd2y=d�2byitselfisoforderGandisnotnegligible.
Noteonpropagationoferrors:Inormallydonottakeo�pointsforpropagatingerrors,
soforexampleastudentwhoforgotthefactorof1/2indetermininggxywouldgetfull
creditonpart(b),eventhoughtheanswerwouldcontaintermsthatarewrongbyafactor
of1/2.However,itseemsrighttometomakeanexceptiontothisruleincaseswhere
anerroronpart(a)causestheconsequentansweronalaterparttobecometrivial.For
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.83
example,ifastudentdescribedametricinpart(a)whichhadnodependenceonr,then
manyofthetermsinparts(b)and(c)wouldnotbepresent.InsuchcasesIstilltook
o�pointsinparts(b)and(c),becauseitdidn'tseemfairtometogivesuchastudent
creditforcalculatingtheseterms,whenthestudentexhibitednosuchcapability.
(c)(9points)
12@@
y(g��)dx�
d�
dx�
d�=12@@
y(gxx) �dx
d� �
2+12@@
y(gtt ) �dt
d� �
2
=12@@
y �1+RSch
r3
x2 ��12
c2@@
y �1�RSch
r ��1c
2 �
=�12 �
3RSch
r4
yrx2 ��12 �RSch
r2
yr �
=
�32RSch b
r5
x2�12RSch b
r3
:
(S23.8)
(d)(2points)CombiningEqs.(S23.7)and(S23.8),we�nd
d2y
d�2=�32RSch b
r5
x2�12RSch b
r3
+3RSch b
r5
x2�RSch b
r3
=
32RSch b �x2
r5
�1r
3 �:
(S23.9)
(e)(4points)The�nalvalueofdy=d�isgivenbyEq.(P23.9),whilethe�nalvalueof
dx=d�willbeequalto1,atleastuptopossiblecorrectionsproportionaltoG.Thus,
the�nalvelocitywillmakeanangle�relativetothehorizontal,where
tan�=
dy=d�j�nal
dx=d�j�nal
= Z1�
1d2y
d�2d�:
Sincetan�willbeproportionaltoG,thesmallangleapproximationtan�=�will
apply,and
�� Z
1�1
d2y
d�2d�:
(S23.10)
8.286QUIZ2REVIEW
PROBLEM
SOLUTIONS,FALL2018
p.84
Then,usingEqs.(S23.9)andcombiningwithEqs.(P23.4)and(P23.8),
�=32
RSch b Z
1�1 �
x2
r5
�1r
3 �d�
=
32RSch b Z
1�1 �
�2
(�2+b2)5=2 �
1
(�2+b2)3=2 �d�:
(S23.11)
Youwerenotaskedtocarryouttheseintegrals,butusingthetableofintegralsgiven
withtheproblem,one�nds
�=32
RSch b �23
b2 �2b
2 �=�2RSch
b
=�4GM
c2b
:
(S23.12)
Theminussignindicatesthatthede ectionisdownward,asonewouldexpect.