Ph - MITweb.mit.edu/8.286/www/quiz18/q2rp-euf18-2up.pdf · ANDING UNI-VERSE: Hubble's La w: v = Hr, where v = recession v elo cit y of a distan t ob ject, H = Hubble expansion rate,

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

PROBLEMS,FALL2018

p.3

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

PROBLEMS,FALL2018

p.4

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

PROBLEMS,FALL2018

p.5

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

PROBLEMS,FALL2018

p.6

PROBLEM

LIST

1.DidYouDotheReading(2000,2002)?.............7(Sol:34)

2.DidYouDotheReading(2007)?

...............8(Sol:35)


...............12(Sol:39)


...............13(Sol:41)


...............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)

8.286QUIZ2REVIEW

PROBLEMS,FALL2018

p.7

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

PROBLEMS,FALL2018

p.8

(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

PROBLEMS,FALL2018

p.9

(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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p.10

(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

PROBLEMS,FALL2018

p.11

(ii)(3points)Duringtheperiodlabeled\neutrondecay,"theneutronfractionis

changingbecause(chooseone):







(D)Neutronsandprotonscanbeconvertedfromoneintotheotherthrough

reactionssuchas




reactionssuchas






(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

PROBLEMS,FALL2018

p.12

(iv)(3points)Duringtheperiodlabeled\eraofnucleosynthesis,"(chooseone:)

(A)Essentiallyalltheneutronspresentcombinewithprotonstoformhelium

nuclei,whichmostlysurviveuntilthepresenttime.

(B)Essentiallyalltheneutronspresentcombinewithprotonstoformdeuterium


(C)Abouthalftheneutronspresentcombinewithprotonstoformheliumnu-

clei,whichmostlysurviveuntilthepresenttime,andtheotherhalfofthe

neutronsremainfree.

(D)Abouthalftheneutronspresentcombinewithprotonstoformdeuterium

nuclei,whichmostlysurviveuntilthepresenttime,andtheotherhalfof

theneutronsremainfree.

(E)Essentiallyalltheprotonspresentcombinewithneutronstoformhelium


(F)Essentiallyalltheprotonspresentcombinewithneutronstoformdeuterium


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

PROBLEMS,FALL2018

p.13

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

8.286QUIZ2REVIEW

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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







ists.






earlyuniverse.






(e)(6points)In1948RalphA.AlpherandRobertHermanwroteapaperpredicting











most2.)













8.286QUIZ2REVIEW

PROBLEMS,FALL2018

p.16

�

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)


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)


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)


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)


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)


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)


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,


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


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)


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 :

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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:


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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p.34

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











most2.)













(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):







(D)Neutronsandprotonscanbeconvertedfrom

oneintothroughreactions

suchas

8.286QUIZ2REVIEW

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reactionssuchas






(ii)(3points)Duringtheperiodlabeled\neutrondecay,"theneutronfractionis

changingbecause(chooseone):







(D)Neutronsandprotonscanbeconvertedfromoneintotheotherthrough

reactionssuchas




reactionssuchas






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


(B)Essentiallyalltheneutronspresentcombinewithprotonstoformdeuterium


(C)Abouthalftheneutronspresentcombinewithprotonstoformheliumnu-

clei,whichmostlysurviveuntilthepresenttime,andtheotherhalfofthe

neutronsremainfree.

(D)Abouthalftheneutronspresentcombinewithprotonstoformdeuterium

nuclei,whichmostlysurviveuntilthepresenttime,andtheotherhalfof

theneutronsremainfree.

(E)Essentiallyalltheprotonspresentcombinewithneutronstoformhelium


(F)Essentiallyalltheprotonspresentcombinewithneutronstoformdeuterium


8.286QUIZ2REVIEW

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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.

8.286QUIZ2REVIEW

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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.

8.286QUIZ2REVIEW

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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







ists.






earlyuniverse.






Answer:Thecorrectanswerswere(ii),(v)and(vi).Theotherswereincorrectforthe

followingreasons:

(i)theearliestpredictionoftheCMBtemperature,byAlpherandHermanin1948,

was5degrees,not0.1degrees.

8.286QUIZ2REVIEW

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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











most2.)













8.286QUIZ2REVIEW

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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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SOLUTIONS,FALL2018

p.47

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

PROBLEM

SOLUTIONS,FALL2018

p.48

(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

PROBLEM

SOLUTIONS,FALL2018

p.49

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

PROBLEM

SOLUTIONS,FALL2018

p.50

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

PROBLEM

SOLUTIONS,FALL2018

p.51

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

PROBLEM

SOLUTIONS,FALL2018

p.52

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

PROBLEM

SOLUTIONS,FALL2018

p.53

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

PROBLEM

SOLUTIONS,FALL2018

p.54

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

PROBLEM

SOLUTIONS,FALL2018

p.55

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

PROBLEM

SOLUTIONS,FALL2018

p.56

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

PROBLEM

SOLUTIONS,FALL2018

p.57

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,andàresummedfrom1tothedimensionofthespace,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:


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

andàreeitherbothequalto1orbothequalto2:

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�ndsnontrivialexpressionswhenkandàre

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


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�=�,


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


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:


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.

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