MASSA

CHUSE

TTSIN

STIT

UTE

OFTECHNOLOGY

Physics

Departm

entPhysics

8.286:The

Early

Universe

Novem

ber2,

2013Prof.

Alan

GuthR

EV

IEW

PR

OB

LEM

SFO

RQ

UIZ

2V

ersion2 ∗

QU

IZD

AT

E:Thursday,

Novem

ber7,

2013,duringthe

normal

classtim

e.

CO

VER

AG

E:Lecture

Notes

4and

5,and

pp.1–10

ofLecture

Notes

6;Prob-

lemSets

4,5,

and6;

Weinberg,

The

First

Three

Minutes,

Chapters

4–7;

InRyden’s

Introductionto

Cosm

ology,wehave

readChapters

4,5,

andSec.

6.1during

thisperiod.

These

chapters,how

ever,parallel

what

wehave

doneor

willbe

doingin

lecture,so

youshould

takethem

asan

aidto

learningthe

lec-ture

material;there

willbe

noquestions

onthis

quizexplicitly

basedon

thesesections

fromRyden.

But

wehave

alsoread

Chapters

10(N

ucleosynthesisand

theEarly

Universe)

and8(D

arkM

atter)in

Ryden,

andthese

willbe

includedon

thequiz,except

forSec.10.3

(Deuterium

Synthesis).Wewillreturn

todeu-

teriumsynthesis

laterin

thecourse.

Ryden’s

Eqs.

(10.11)and

(10.12)involve

similar

issuesfrom

statisticalmechanics,so

youshould

notworry

ifyoudo

notunderstand

theseequations.

(Infact,

youshould

worry

ifyou

dounderstand

them;as

wewill

discusslater,

theyare

spectacularlyincorrect.)

Eq.

(10.13),which

isobtained

bydividing

Eq.(10.11)

byEq.(10.12),is

nonethelesscorrect;

forthis

courseyou

neednot

worry

howto

derivethis

formula,

butyou

shouldassum

eit

andunderstand

itsconsequences,

asdescribed

byRyden

andalso

byWeinberg.

Chapters

4and

5of

Weinberg’s

bookare

packedwith

numbers;

youneed

notmem

orizethese

numbers,

butyou

shouldbe

familiar

with

theirorders

ofmagnitude.

Wewill

nottake

offfor

thespelling

ofnam

es,as

longas

theyare

vaguelyrecognizable.

Fordates

before1900,

itwill

besuffi

cientfor

youto

knowwhen

thingshappened

towithin

100years.

Fordates

after1900,

itwill

besuffi

cientifyou

canplace

eventswithin

10years.

You

shouldexpect

oneproblem

basedon

thereadings,and

severalcalculationalproblems.

One

ofth

eprob

lems

onth

equiz

will

be

takenverb

atim(or

atleast

alm

ost

verbatim

)from

either

the

hom

ework

assignm

ents,

orfrom

the

starredprob

lems

fromth

isset

ofR

eview

Prob

lems.

The

starredproblem

sare

theones

thatIrecom

mend

thatyou

reviewmost

carefully:Prob-

lems4,

5,6,

11,13,

15,17,

and19.

There

areonly

threereading

questions,Problem

s1,2,

and3.

PU

RP

OSE:These

reviewproblem

sare

notto

behanded

in,but

arebeing

made

availableto

helpyou

study.They

comemainly

fromquizzes

inprevious

years.

*Version

2,Novem

ber2,

2013(sam

edate

asoriginal).

The

document

datewas

correctedto

read2013

insteadof2011,and

crossreferences

within

Problem

s15,18,

19,and

theProblem

9solution

were

updated.

8.286Q

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2013p.2

Insom

ecases

thenum

berofpoints

assignedto

theproblem

onthe

quizislisted

—in

allsuch

casesitis

basedon

100points

forthe

fullquiz.

Inaddition

tothis

setof

problems,

youwill

findon

thecourse

web

pagethe

actualquizzes

thatwere

givenin

1994,1996,

1998,2000,

2002,2004,

2005,2007,

2009,and

2011.The

relevantproblem

sfrom

thosequizzes

havemostly

beenincorporated

intothese

reviewproblem

s,but

youstill

may

beinterested

inlooking

atthe

quizzes,just

tosee

howmuch

material

hasbeen

includedin

eachquiz.

The

coverageof

theupcom

ingquiz

will

notnecessarily

match

thecoverage

ofany

ofthe

quizzesfrom

previousyears.

The

coveragefor

eachquiz

inrecent

yearsisusually

describedat

thestart

ofthereview

problems,as

Idid

here.

REV

IEW

SESSIO

NA

ND

OFFIC

EH

OU

RS:T

ohelp

youstudy

forthe

quiz,Tingtao

Zhou

will

holdareview

sessionon

Monday,

Novem

ber4,

at7:30

pm,

inour

regularlecture

room,Room

34-101.Iwill

havemyusual

office

houron

Wednesday

evening,7:30

pm,in

Room

8-308.

INFO

RM

AT

ION

TO

BE

GIV

EN

ON

QU

IZ:

Each

quizin

thiscourse

will

haveasection

of“useful

information”

foryour

reference.For

thesecond

quiz,this

usefulinform

ationwillbe

thefollow

ing:

SP

EED

OF

LIG

HT

INC

OM

OV

ING

CO

OR

DIN

AT

ES:

vcoord

=c

a(t).

DO

PP

LER

SH

IFT

(For

motion

along

alin

e):

z=v/u

(nonrelativistic,sourcemoving)

z=

v/u

1−v/u

(nonrelativistic,observermoving)

z= √

1+β

1−β−

1(special

relativity,with

β=v/c)

CO

SM

OLO

GIC

AL

RED

SH

IFT

:

1+z≡

λobse

rved

λem

itted

=a(t

obse

rved )

a(tem

itted )

8.286Q

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2013p.3

SP

EC

IAL

RELA

TIV

ITY

:

Tim

eDilation

Factor:

γ≡1

√1−

β2,

β≡v/c

Lorentz-F

itzgeraldContraction

Factor:γ

Relativity

ofSim

ultaneity:Trailing

clockreads

laterby

anam

ountβ 0 /c.

Energy-M

omentum

Four-Vector:

pµ= (

Ec,p )

,p=γm

0 v,E

=γm

0 c2= √

(m0 c

2)2+|p| 2

c2,

p2≡|p| 2− (p

0 )2=|p| 2−

E2

c2

=−(m

0 c)2.

CO

SM

OLO

GIC

AL

EV

OLU

TIO

N:

H2= (

aa )2

=8π3Gρ−

kc2

a2,

a=−4π3G (

ρ+

3pc2 )

a,

ρm(t)

=a3(t

i )a3(t)

ρm(ti )

(matter),

ρr (t)

=a4(t

i )a4(t)

ρr (t

i )(radiation).

ρ=−3aa (

ρ+

pc2 )

,Ω≡ρ/ρc,

where

ρc=

3H

2

8πG

.

EV

OLU

TIO

NO

FA

MA

TT

ER

-DO

MIN

AT

ED

UN

IVER

SE:

Flat

(k=

0):a(t)∝

t2/3

Ω=

1.

Closed

(k>

0):ct

=α(θ−

sinθ)

,a√k

=α(1−

cosθ)

,

Ω=

21+

cosθ>

1,

where

α≡

4π3Gρ

c2 (

a√k )3

.

Open

(k<

0):ct

=α(sinh

θ−θ)

,a√κ

=α(cosh

θ−1)

,

Ω=

21+

coshθ<

1,

where

α≡

4π3Gρ

c2 (

a√κ )3

,

κ≡−k>

0.

8.286Q

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2013p.4

RO

BERT

SO

N-W

ALK

ER

MET

RIC

:

ds2=−c2dτ

2=−c2dt2+

a2(t)

dr2

1−kr2+r2 (d

θ2+sin

2θdφ

2 ) .

Alternatively,for

k>

0,wecan

definer=

sinψ

√k

,and

then

ds2=−c2dτ

2=−c2dt2+

a2(t)

dψ

2+sin

2ψ (d

θ2+sin

2θdφ

2 ),

where

a(t)=a(t)/ √

k.For

k<

0wecan

definer=

sinhψ

√−k,and

then

ds2=−c2dτ

2=−c2dt2+

a2(t)

dψ

2+sinh

2ψ (d

θ2+

sin2θdφ

2 ),

where

a(t)=a(t)/ √−

k.Note

thatacan

becalled

aifthere

isno

needto

relateitto

thea(t)

thatappears

inthe

firstequation

above.

HO

RIZ

ON

DIS

TA

NC

E:

p,h

oriz

on (t)

=a(t) ∫

t

0

c

a(t ′)dt ′

=3ct

(flat,matter-dom

inated).

SC

HW

AR

ZSC

HIL

DM

ET

RIC

:

ds2=−c2dτ

2=− (

1−2GM

rc2 )

c2dt2+ (

1−2GM

rc2 )

−1

dr2

+r2dθ2+r2sin

2θdφ

2,

GEO

DESIC

EQ

UA

TIO

N:

dds

gijdxj

ds

=12(∂i gk )dxk

ds

dx

ds

or:ddτ

gµνdxν

dτ

=12(∂µgλσ )

dxλ

dτ

dxσ

dτ

8.286Q

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PR

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LIS

T

1.Did

You

Dothe

Reading

(2000,2002)?.

...

...

...

6(Sol:

27)

2.Did

You

Dothe

Reading

(2007)?.

..

...

...

...

7(Sol:

28)

3.Did

You

Dothe

Reading

(2011)?.

..

...

...

...

11(Sol:

32)

*4.Evolution

ofan

Open

Universe

..

..

...

...

...

12(Sol:

34)

*5.Anticipating

aBig

Crunch

..

..

..

...

...

...

12(Sol:

35)

*6.Tracing

Light

Rays

inaClosed,

Matter-D

ominated

Universe

12(Sol:

36)

7.Lengths

andAreas

inaTwo-D

imensional

Metric

...

...

13(Sol:

38)

8.Geom

etryin

aClosed

Universe

..

..

...

...

...

14(Sol:

40)

9.The

General

SphericallySym

metric

Metric

..

...

...

15(Sol:

41)

10.Volum

esin

aRobertson-W

alkerUniverse

...

...

...

16(Sol:

42)

*11.The

Schwarzschild

Metric

..

...

..

...

...

...

16(Sol:

44)

12.Geodesics

..

..

..

...

...

..

...

...

...

17(Sol:

47)

*13.AnExercise

inTwo-D

imensional

Metrics

...

...

...

18(Sol:

49)

14.Geodesics

onthe

Surfaceof

aSphere

..

...

...

...

19(Sol:

52)

*15.Geodesics

inaClosed

Universe

..

..

...

...

...

19(Sol:

56)

16.A

Two-D

imensional

Curved

Space.

..

...

...

...

20(Sol:

59)

*17.Rotating

Frames

ofReference

...

..

...

...

...

22(Sol:

62)

18.The

Stabilityof

Schwarzschild

Orbits

..

...

...

...

24(Sol:

65)

*19.Pressure

andEnergy

Density

ofMysterious

Stuff.

..

...

26(Sol:

69)

8.286Q

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PR

OB

LEM

1:D

IDY

OU

DO

TH

ER

EA

DIN

G?

Parts

(a)-(c)of

thisproblem

come

fromQuiz

4,2000,

andparts

(d)and

(e)com

efrom

Quiz

3,2002.

(a)(5

points)Bywhat

factordoes

thelepton

number

percom

ovingvolum

eof

theuniverse

changebetw

eentem

peraturesofkT

=10

MeV

andkT

=0.1

MeV

?You

shouldassum

ethe

existenceof

thenorm

althree

speciesof

neutrinosfor

youransw

er.

(b)(5

points)Measurem

entsof

theprim

ordialdeuterium

abundancewould

givegood

constraintson

thebaryon

densityof

theuniverse.

How

ever,this

abun-dance

ishard

tomeasure

accurately.W

hichof

thefollow

ingis

NOT

areason

why

thisis

hardto

do?

(i)The

neutronin

adeuterium

nucleusdecays

onthe

timescale

of15minutes,

soalm

ostnone

ofthe

primordial

deuteriumproduced

inthe

Big

Bang

isstill

present.

(ii)The

deuteriumabundance

inthe

Earth’s

oceansis

biasedbecause,

beingheavier,less

deuteriumthan

hydrogenwould

haveescaped

fromthe

Earth’s

surface.

(iii)The

deuteriumabundance

inthe

Sunis

biasedbecause

nuclearreactions

tendto

destroyitby

convertingit

intohelium

-3.

(iv)The

spectrallinesofdeuterium

arealm

ostidenticalw

iththose

ofhydrogen,so

deuteriumsignatures

tendto

getwashed

outin

spectraof

primordial

gasclouds.

(v)The

deuteriumabundance

isso

small

(afew

partsper

million)

thatit

canbe

easilychanged

byastrophysical

processesother

thanprim

ordialnucleosynthesis.

(c)(5

points)Give

threeexam

plesof

hadrons.

(d)(6

points)In

Chapter

6ofT

heFirst

Three

Minutes,Steven

Weinberg

posedthe

question,“W

hywas

thereno

systematic

searchfor

this[cosm

icbackground]

radiation,years

before1965?”

Indiscussing

thisissue,

hecontrasted

itwith

thehistory

oftw

odifferent

elementary

particles,each

ofwhich

were

predictedapproxim

ately20

yearsbefore

theywere

firstdetected.

Nam

eone

ofthese

twoelem

entaryparticles.

(Ifyou

namethem

bothcorrectly,

youwill

get3

pointsextra

credit.How

ever,oneright

andone

wrong

willget

you4points

forthe

question,com

paredto

6points

forjust

naming

oneparticle

andgetting

itright.)A

nswer:

2ndAnsw

er(optional):

8.286Q

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(e)(6

points)In

Chapter

6ofT

heFirst

Three

Minutes,Steven

Weinberg

discussesthree

reasonswhy

theim

portanceof

asearch

fora3

Kmicrow

averadiation

backgroundwas

notgenerally

appreciatedin

the1950s

andearly

1960s.Choose

thosethree

reasonsfrom

thefollow

inglist.

(2points

foreach

rightansw

er,circleat

most

3.)

(i)The

earliestcalculations

erroneouslypredicted

acosm

icbackground

tem-

peratureof

onlyabout

0.1 K,and

suchabackground

would

betoo

weak

todetect.

(ii)There

was

abreakdow

nin

communication

between

theoristsand

experi-mentalists.

(iii)It

was

nottechnologically

possibleto

detectasignal

asweak

asa3

Kmicrow

avebackground

untilabout

1965.

(iv)Since

almost

allphysicistsat

thetim

ewere

persuadedby

thesteady

statemodel,

thepredictions

ofthe

bigbang

model

were

nottaken

seriously.

(v)It

was

extraordinarilydiffi

cultfor

physiciststo

takeseriously

anytheory

ofthe

earlyuniverse.

(vi)The

earlywork

onnucleosynthesis

byGam

ow,A

lpher,Herm

an,andFollin,

etal.,had

attempted

toexplain

theorigin

ofallcomplex

nucleibyreactions

inthe

earlyuniverse.

This

programwas

neververy

successful,and

itscredibility

was

furtherunderm

inedas

improvem

entswere

made

inthe

alternativetheory,that

elements

aresynthesized

instars.

PR

OB

LEM

2:D

IDY

OU

DO

TH

ER

EA

DIN

G?

(24points)

The

following

problemwas

Problem

1of

Quiz

2in

2007.

(a)(6

points)In

1948Ralph

A.A

lpherand

Robert

Herm

anwrote

apaper

predict-ing

acosm

icmicrow

avebackground

with

atem

peratureof5

K.The

paperwas

basedon

acosm

ologicalmodel

thatthey

haddeveloped

with

George

Gam

ow,

inwhich

theearly

universewas

assumed

tohave

beenfilled

with

hotneutrons.

Asthe

universeexpanded

andcooled

theneutrons

underwent

betadecay

intoprotons,

electrons,and

antineutrinos,until

atsom

epoint

theuniverse

cooledenough

forlight

elements

tobe

synthesized.Alpher

andHerm

anfound

thatto

accountfor

theobserved

presentabundances

oflightelem

ents,theratio

ofpho-tons

tonuclear

particlesmust

havebeen

about10

9.Although

thepredicted

temperature

was

veryclose

tothe

actualvalue

of2.7

K,the

theorydiffered

fromour

presenttheory

intw

oways.

Circle

thetw

ocorrect

statements

inthe

following

list.(3

pointsfor

eachright

answer;

circleat

most

2.)

(i)Gam

ow,Alpher,

andHerm

anassum

edthat

theneutron

coulddecay,

butnow

theneutron

isthought

tobe

absolutelystable.

8.286Q

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2013p.8

(ii)In

thecurrent

theory,the

universestarted

with

nearlyequal

densitiesof

protonsand

neutrons,not

allneutrons

asGam

ow,Alpher,

andHerm

anassum

ed.

(iii)In

thecurrent

theory,theuniverse

startedwith

mainly

alphaparticles,not

allneutrons

asGam

ow,Alpher,

andHerm

anassum

ed.(N

ote:an

alphaparticle

isthe

nucleusofa

heliumatom

,composed

oftwoprotons

andtw

oneutrons.)

(iv)In

thecurrent

theory,the

conversionof

neutronsinto

protons(and

viceversa)

tookplace

mainly

throughcollisions

with

electrons,positrons,neu-trinos,

andantineutrinos,

notthrough

thedecay

ofthe

neutrons.

(v)The

ratioof

photonsto

nuclearparticles

inthe

earlyuniverse

isnow

be-lieved

tohave

beenabout

103,

not10

9as

Alpher

andHerm

anconcluded.

(b)(6

points)In

Weinberg’s

“Recipe

foraHot

Universe,”

hedescribed

theprim

or-dialcom

positionofthe

universein

termsofthree

conservedquantities:

electriccharge,

baryonnum

ber,and

leptonnum

ber.If

electriccharge

ismeasured

inunits

ofthe

electroncharge,

thenall

threequantities

areintegers

forwhich

thenum

berdensity

canbe

compared

with

thenum

berdensity

ofphotons.

Foreach

quantity,which

choicemost

accuratelydescribes

theinitial

ratioof

thenum

berdensity

ofthis

quantityto

thenum

berdensity

ofphotons:

Electric

Charge:

(i)∼10

9(ii)∼

1000(iii)∼

1(iv)∼

10 −6

(v)either

zeroor

negligible

Baryon

Num

ber:(i)∼

10 −20

(ii)∼10 −

9(iii)∼

10 −6

(iv)∼1

(v)anyw

herefrom

10 −5to

1

Lepton

Num

ber:(i)∼

109

(ii)∼1000

(iii)∼1

(iv)∼10 −

6(v)

couldbe

ashigh

as∼1,

butis

assumed

tobe

verysm

all

8.286Q

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(c)(12

points)The

figurebelow

comes

fromWeinberg’s

Chapter

5,andislabeled

The

ShiftingNeutron-P

rotonBalance.

(i)(3

points)During

theperiod

labeled“therm

alequilibrium

,”the

neutronfraction

ischanging

because(choose

one):

(A)The

neutronis

unstable,and

decaysinto

aproton,

electron,and

an-tineutrino

with

alifetim

eof

about1second.

(B)The

neutronis

unstable,and

decaysinto

aproton,

electron,and

an-tineutrino

with

alifetim

eof

about15

seconds.

(C)The

neutronis

unstable,and

decaysinto

aproton,

electron,and

an-tineutrino

with

alifetim

eof

about15

minutes.

(D)Neutrons

andprotons

canbe

convertedfrom

oneinto

throughreac-

tionssuch

asantineutrino+proton←→

electron+

neutronneutrino

+neutron←→

positron+proton

.

(E)Neutrons

andprotons

canbe

convertedfrom

oneinto

theother

throughreactions

suchas

antineutrino+proton←→

positron+

neutronneutrino

+neutron←→

electron+proton

.

(F)Neutrons

andprotons

canbe

createdand

destroyedby

reactionssuch

asproton

+neutrino←→

positron+antineutrino

neutron+antineutrino←→

electron+

positron.

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(ii)(3

points)During

theperiod

labeled“neutron

decay,”the

neutronfraction

ischanging

because(choose

one):

(A)The

neutronis

unstable,and

decaysinto

aproton,

electron,and

an-tineutrino

with

alifetim

eof

about1second.

(B)The

neutronis

unstable,and

decaysinto

aproton,

electron,and

an-tineutrino

with

alifetim

eof

about15

seconds.

(C)The

neutronis

unstable,and

decaysinto

aproton,

electron,and

an-tineutrino

with

alifetim

eof

about15

minutes.

(D)Neutrons

andprotons

canbe

convertedfrom

oneinto

theother

throughreactions

suchas

antineutrino+proton←→

electron+

neutronneutrino

+neutron←→

positron+proton

.

(E)Neutrons

andprotons

canbe

convertedfrom

oneinto

theother

throughreactions

suchas

antineutrino+proton←→

positron+

neutronneutrino

+neutron←→

electron+proton

.

(F)Neutrons

andprotons

canbe

createdand

destroyedby

reactionssuch

asproton

+neutrino←→

positron+antineutrino

neutron+antineutrino←→

electron+

positron.

(iii)(3

points)The

masses

ofthe

neutronand

protonare

notexactly

equal,but

instead

(A)The

neutronis

more

massive

thanaproton

with

arest

energydiffer-

enceof

1.293GeV

(1GeV

=10

9eV

).

(B)The

neutronis

more

massive

thanaproton

with

arest

energydiffer-

enceof

1.293MeV

(1MeV

=10

6eV

).

(C)The

neutronis

more

massive

thanaproton

with

arest

energydiffer-

enceof

1.293KeV

(1KeV

=10

3eV

).

(D)The

protonismore

massive

thananeutron

with

arest

energydiffer-

enceof

1.293GeV

.

(E)The

protonismore

massive

thananeutron

with

arest

energydiffer-

enceof

1.293MeV

.

(F)The

protonismore

massive

thananeutron

with

arest

energydiffer-

enceof

1.293KeV

.

8.286Q

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PR

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LE

MS,FA

LL

2013p.11

(iv)(3

points)During

theperiod

labeled“era

ofnucleosynthesis,”(choose

one:)

(A)Essentially

allthe

neutronspresent

combine

with

protonsto

formhelium

nuclei,which

mostly

surviveuntil

thepresent

time.

(B)Essentially

allthe

neutronspresent

combine

with

protonsto

formdeuterium

nuclei,which

mostly

surviveuntil

thepresent

time.

(C)About

halftheneutrons

presentcom

binewith

protonsto

formhelium

nuclei,which

mostly

surviveuntilthe

presenttim

e,andthe

otherhalf

ofthe

neutronsrem

ainfree.

(D)About

halfthe

neutronspresent

combine

with

protonsto

formdeu-

teriumnuclei,

which

mostly

surviveuntil

thepresent

time,

andthe

otherhalf

ofthe

neutronsrem

ainfree.

(E)Essentially

allthe

protonspresent

combine

with

neutronsto

formhelium

nuclei,which

mostly

surviveuntil

thepresent

time.

(F)Essentially

allthe

protonspresent

combine

with

neutronsto

formdeuterium

nuclei,which

mostly

surviveuntil

thepresent

time.

PR

OB

LEM

3:D

IDY

OU

DO

TH

ER

EA

DIN

G?

(20points)

The

following

problemcom

esfrom

Quiz

2,2011.

(a)(8

points)During

nucleosynthesis,heavier

nucleiform

fromprotons

andneu-

tronsthrough

aseries

oftw

oparticle

reactions.

(i)In

The

First

Three

Minutes,

Weinberg

discussestw

ochains

ofreactions

that,starting

fromprotons

andneutrons,

endup

with

helium,He4.

De-

scribeat

leastone

ofthese

twochains.

(ii)Explain

brieflywhat

isthe

deuteriumbottleneck,and

what

isits

roleduring

nucleosynthesis.

(b)(12

points)In

Chapter

4of

The

First

Three

Minutes,

StevenWeinberg

makes

thefollow

ingstatem

entregarding

theradiation-dom

inatedphase

ofthe

earlyuniverse:

The

timethat

ittakes

forthe

universeto

coolfromone

temperature

toanother

isproportionalto

thediff

erenceof

theinverse

squaresof

thesetem

peratures.

Inthis

partofthe

problemyou

willexplore

more

quantitativelythis

statement.

(i)For

aradiation-dom

inateduniverse

thescale-factor

a(t)∝t1/2.

Find

thecosm

ictim

etas

afunction

ofthe

Hubble

expansionrate

H.

(ii)The

mass

densitystored

inradiation

ρrisproportionalto

thetem

peratureT

tothe

fourthpow

er:i.e.,

ρr

αT

4,for

someconstant

α.For

awide

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2013p.12

rangeof

temperatures

wecan

takeα

4.52×

10 −32kg·m

−3·K

−4.

Ifthe

temperature

ismeasured

indegrees

Kelvin

(K),

thenρrhas

thestandard

SIunits,

[ρr ]=

kg·m−

3.Use

theFriedm

annequation

foraflat

universe(k

=0)

with

ρ=ρrto

expressthe

Hubble

expansionrate

Hin

termsofthe

temperature

T.You

will

needthe

SIvalue

ofthe

gravitationalconstant

G

6.67×

10 −11N·m

2·kg −2.

What

isthe

Hubble

expansionrate,in

inverseseconds,

atthe

startof

nucleosynthesis,when

T=T

nucl

0.9×

109K?

(iii)Using

theresults

in(i)

and(ii),express

thecosm

ictim

etas

afunction

ofthe

temperature.

Your

resultshould

agreewith

Weinberg’s

claimabove.

What

isthe

cosmic

time,

inseconds,

when

T=T

nucl ?

∗P

RO

BLEM

4:EV

OLU

TIO

NO

FA

NO

PEN

UN

IVER

SE

The

following

problemwas

takenfrom

Quiz

2,1990,

where

itcounted

10points

outof

100.

Consider

anopen,

matter-dom

inateduniverse,

asdescribed

bythe

evolutionequations

onthe

frontof

thequiz.

Find

thetim

etat

which

a/ √

κ=

2α.

∗P

RO

BLEM

5:A

NT

ICIP

AT

ING

AB

IGC

RU

NC

H

Supposethat

welived

inaclosed,

matter-dom

inateduniverse,

asdescribed

bythe

equationson

thefront

ofthe

quiz.Suppose

furtherthat

wemeasured

themass

densityparam

eterΩ

tobe

Ω0=

2,andwemeasured

theHubble

“constant”to

havesom

evalue

H0 .

How

much

timewould

wehave

beforeour

universeended

inabig

crunch,at

which

timethe

scalefactor

a(t)would

collapseto

0?

∗P

RO

BLEM

6:T

RA

CIN

GLIG

HT

RA

YS

INA

CLO

SED

,M

AT

TER

-D

OM

INA

TED

UN

IVER

SE

(30points)

The

following

problemwas

Problem

3,Quiz

2,1998.

The

spacetimemetric

forahom

ogeneous,isotropic,closeduniverse

isgiven

bythe

Robertson-W

alkerform

ula:

ds2=−c2dτ

2=−c2dt2+a2(t)

dr2

1−r2+r2 (d

θ2+

sin2θdφ

2 ) ,

where

Ihave

takenk=

1.Todiscuss

motion

inthe

radialdirection,

itis

more

convenientto

work

with

analternative

radialcoordinateψ,related

torby

r=

sinψ.

Then

dr

√1−

r2=dψ,

8.286Q

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2013p.13

sothe

metric

simplifies

to

ds2=−c2dτ

2=−c2dt2+a2(t)

dψ

2+

sin2ψ (d

θ2+

sin2θdφ

2 ).

(a)(7

points)A

lightpulse

travelson

anull

trajectory,which

means

thatdτ=

0for

eachsegm

entof

thetrajectory.

Consider

alight

pulsethat

moves

alongaradial

line,so

θ=φ=

constant.Find

anexpression

fordψ/dtin

termsof

quantitiesthat

appearin

themetric.

(b)(8

points)Write

anexpression

forthe

physicalhorizon

distance physat

time

t.You

shouldleave

youransw

erin

theform

ofadefinite

integral.

The

formofa(t)

dependson

thecontent

ofthe

universe.Ifthe

universeis

matter-

dominated

(i.e.,dominated

bynonrelativistic

matter),then

a(t)isdescribed

bythe

parametric

equationsct

=α(θ−

sinθ)

,

a=α(1−

cosθ)

,

where

α≡

4π3Gρa3

c2

.

These

equationsare

identicalto

thoseon

thefront

ofthe

exam,except

thatIhave

chosenk=

1.

(c)(10

points)Consider

aradial

light-raymoving

throughamatter-dom

inatedclosed

universe,as

describedby

theequations

above.Find

anexpression

fordψ/dθ,w

hereθis

theparam

eterused

todescribe

theevolution.

(d)(5

points)Suppose

thataphoton

leavesthe

originof

thecoordinate

system(ψ

=0)

att=

0.How

longwillit

takefor

thephoton

toreturn

toits

startingplace?

Express

youransw

eras

afraction

ofthe

fulllifetim

eof

theuniverse,

frombig

bangto

bigcrunch.

PR

OB

LEM

7:LEN

GT

HS

AN

DA

REA

SIN

AT

WO

-DIM

EN

SIO

NA

LM

ET

RIC

(25points)

The

following

problemwas

Problem

3,Quiz

2,1994:

Supposeatw

odim

ensionalspace,

describedin

polarcoordinates

(r,θ),has

ametric

givenby

ds2=

(1+ar)

2dr2+r2(1

+br)

2dθ2,

where

aand

bare

positiveconstants.

Consider

thepath

inthis

spacewhich

isform

edby

startingat

theorigin,

moving

alongthe

θ=

0line

tor=

r0 ,

then

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moving

atfixed

rto

θ=π/2,

andthen

moving

backto

theorigin

atfixed

θ.The

pathis

shownbelow

:

a)(10

points)Find

thetotallength

ofthis

path.

b)(15

points)Find

thearea

enclosedby

thispath.

PR

OB

LEM

8:G

EO

MET

RY

INA

CLO

SED

UN

IVER

SE

(25points)

The

following

problemwas

Problem

4,Quiz

2,1988:

Consider

auniverse

describedby

theRobertson–W

alkermetric

onthe

firstpage

ofthe

quiz,with

k=

1.The

questionsbelow

allpertain

tosom

efixed

timet,

so

thescale

factorcan

bewritten

simply

asa,dropping

itsexplicit

t-dependence.

Asm

allrod

hasone

endat

thepoint

(r=h,θ=

0,φ=

0)and

theother

end

atthe

point(r

=h,θ=

∆θ,φ=

0).Assum

ethat

∆θ

1.

8.286Q

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(a)Find

thephysical

distance pfrom

theorigin

(r=

0)to

thefirst

end(h,0,0)

ofthe

rod.You

may

findone

ofthe

following

integralsuseful:

∫dr

√1−

r2=

sin −1r

∫dr

1−r2=

12ln (

1+r

1−r )

.

(b)Find

thephysical

lengthspof

therod.

Express

youransw

erin

termsof

thescale

factora,

andthe

coordinateshand

∆θ.

(c)Note

that∆θisthe

anglesubtended

bythe

rod,asseen

fromthe

origin.Write

anexpression

forthis

anglein

termsof

thephysical

distance p ,

thephysical

lengthsp ,

andthe

scalefactor

a.

PR

OB

LEM

9:T

HE

GEN

ER

AL

SP

HER

ICA

LLY

SY

MM

ET

RIC

MET

-R

IC(20

points)

The

following

problemwas

Problem

3,Quiz

2,1986:

The

metric

foragiven

spacedepends

ofcourseon

thecoordinate

systemwhich

isused

todescribe

it.It

canbe

shownthat

forany

threedim

ensionalspace

which

isspherically

symmetric

aboutaparticular

point,coordinatescan

befound

sothat

themetric

hasthe

form

ds2=dr2+ρ2(r) [d

θ2+sin

2θdφ

2 ]

8.286Q

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LE

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LL

2013p.16

forsom

efunction

ρ(r).The

coordinatesθand

φhave

theirusual

ranges:θvaries

between

0and

π,and

φvaries

from0to

2π,w

hereφ=

0and

φ=

2πare

identified.Given

thismetric,

considerthe

spherewhose

outerboundary

isdefined

byr=r0 .

(a)Find

thephysical

radiusaof

thesphere.

(By“radius”,

Imean

thephysical

lengthof

aradial

linewhich

extendsfrom

thecenter

tothe

boundaryof

thesphere.)

(b)Find

thephysical

areaof

thesurface

ofthe

sphere.

(c)Find

anexplicit

expressionfor

thevolum

eof

thesphere.

Besure

toinclude

thelim

itsof

integrationfor

anyintegrals

which

occurin

youransw

er.

(d)Suppose

anew

radialcoordinateσis

introduced,where

σis

relatedto

rby

σ=r2.

Express

themetric

interm

sof

thisnew

variable.

PR

OB

LEM

10:V

OLU

MES

INA

RO

BERT

SO

N-W

ALK

ER

UN

IVER

SE

(20points)

The

following

problemwas

Problem

1,Quiz

3,1990:

The

metric

foraRobertson-W

alkeruniverse

isgiven

by

ds2=a2(t)

dr2

1−kr2+r2 (d

θ2+sin

2θdφ

2 ) .

Calculate

thevolum

eV(r

max )

ofthe

spheredescribed

by

r≤rm

ax.

You

shouldcarry

outany

angularintegrations

thatmay

benecessary,but

youmay

leaveyour

answer

inthe

formof

aradialintegralw

hichisnot

carriedout.

Besure,

however,

toclearly

indicatethe

limits

ofintegration.

∗P

RO

BLEM

11:T

HE

SC

HW

AR

ZSC

HIL

DM

ET

RIC

(25points)

The

followproblem

was

Problem

4,Quiz

3,1992:

The

spaceoutside

aspherically

symmetric

mass

Mis

describedby

theSchw

arzschildmetric,

givenat

thefront

ofthe

exam.Twoobservers,

designatedA

andB,are

locatedalong

thesam

eradialline,w

ithvalues

ofthe

coordinatergiven

byrAand

rB,respectively,

with

rA<rB.You

shouldassum

ethat

bothobservers

lieoutside

theSchw

arzschildhorizon.

8.286Q

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2R

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PR

OB

LE

MS,FA

LL

2013p.17

a)(5

points)Write

downthe

expressionfor

theSchw

arzschildhorizon

radiusR

S ,expressed

interm

sofM

andfundam

entalconstants.

b)(5

points)W

hatis

theproper

distancebetw

eenA

andB?

Itis

okayto

leavethe

answer

tothis

partin

theform

ofan

integralthat

youdo

notevaluate—

butbe

sureto

clearlyindicate

thelim

itsof

integration.

c)(5

points)Observer

Ahas

aclock

thatem

itsan

evenlyspaced

sequenceofticks,

with

propertim

eseparation

∆τA.W

hatwillbe

thecoordinate

timeseparation

∆tAbetw

eenthese

ticks?

d)(5

points)Ateach

tickofA’s

clock,alight

pulseis

transmitted.

Observer

B

receivesthese

pulses,andmeasures

thetim

eseparation

onhis

ownclock.

What

isthe

timeinterval

∆τB

measured

byB.

e)(5

points)Suppose

thatthe

object

creatingthe

gravitationalfield

isastatic

blackhole,

sothe

Schwarzschild

metric

isvalid

forall

r.Now

supposethat

oneconsiders

thecase

inwhich

observerA

lieson

theSchw

arzschildhorizon,

sorA≡R

S .Is

theproper

distancebetw

eenA

andB

finitefor

thiscase?

Does

thetim

einterval

ofthe

pulsesreceived

byB,∆

τB,diverge

inthis

case?

PR

OB

LEM

12:G

EO

DESIC

S(20

points)

The

following

problemwas

Problem

4,Quiz

2,1986:

Ordinary

Euclidean

two-dim

ensionalspace

canbe

describedin

polarcoordi-

natesby

themetric

ds2=dr2+r2dθ2.

(a)Suppose

thatr(λ)

andθ(λ)

describeageodesic

inthis

space,where

theparam

-eter

λis

thearc

lengthmeasured

alongthe

curve.Use

thegeneral

formula

onthe

frontof

theexam

toobtain

explicitdifferential

equationswhich

r(λ)and

θ(λ)must

obey.

(b)Now

introducethe

usualCartesian

coordinates,defined

by

x=rcos

θ,

y=rsin

θ.

Use

youransw

erto

(a)to

showthat

theline

y=

1is

ageodesic

curve.

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LE

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∗P

RO

BLEM

13:A

NEX

ER

CIS

EIN

TW

O-D

IMEN

SIO

NA

LM

ET

RIC

S(30

points)

(a)(8

points)Consider

firstatw

o-dimensional

spacewith

coordinatesrand

θ.The

metric

isgiven

by

ds2=

dr2+r2dθ2.

Consider

thecurve

describedby

r(θ)=

(1+εcos

2θ)r0,

where

εand

r0are

constants,and

θruns

fromθ1to

θ2 .

Write

anexpression,

inthe

formof

adefinite

integral,for

thelength

Sof

thiscurve.

(b)(5

points)Now

consideratw

o-dimensionalspace

with

thesam

etw

ocoordinates

rand

θ,butthis

timethe

metric

willbe

ds2= (1

+ra )

dr2+r2dθ2,

where

ais

aconstant.

θis

aperiodic

(angular)variable,

with

arange

of0to

2π,w

ith2πidentified

with

0.W

hatisthe

lengthR

ofthepath

fromthe

origin(r

=0)

tothe

pointr=r0 ,θ

=0,

alongthe

pathfor

which

θ=

0everyw

herealong

thepath?

You

canleave

youransw

erin

theform

ofadefinite

integral.(B

esure,

however,

tospecify

thelim

itsof

integration.)

(c)(7

points)For

thespace

describedin

part(b),w

hatisthe

totalareacontained

within

theregion

r<r0 .

Again

youcan

leaveyour

answer

inthe

formof

adefinite

integral,making

sureto

specifythe

limits

ofintegration.

(d)(10

points)Again

forthe

spacedescribed

inpart

(b),consider

ageodesic

de-scribed

bythe

usualgeodesic

equation,

dds

gij d

xj

ds

=12(∂i gk )

dxk

ds

dx

ds.

The

geodesicisdescribed

byfunctions

r(s)and

θ(s),where

sisthe

arclength

alongthe

curve.Write

explicitlyboth

(i.e.,for

i=1=

rand

i=2=

θ)geodesic

equations.

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PR

OB

LEM

14:G

EO

DESIC

SO

NT

HE

SU

RFA

CE

OF

ASP

HER

E

Inthis

problemwewill

testthe

geodesicequation

bycom

putingthe

geodesiccurves

onthe

surfaceof

asphere.

Wewill

describethe

sphereas

inLecture

Notes

5,with

metric

givenby

ds2=a2 (d

θ2+

sin2θdφ

2 ).

(a)Clearly

onegeodesic

onthe

sphereis

theequator,

which

canbe

parametrized

byθ=

π/2

andφ

=ψ,where

ψis

aparam

eterwhich

runsfrom

0to

2π.

Showthat

ifthe

equatoris

rotatedby

anangle

αabout

thex-axis,

thenthe

equationsbecom

e:cos

θ=

sinψsin

α

tanφ=

tanψcos

α.

(b)Using

thegeneric

formofthe

geodesicequation

onthe

frontofthe

exam,derive

thedifferential

equationwhich

describesgeodesics

inthis

space.

(c)Show

thatthe

expressionsin

(a)satisfy

thedifferential

equationfor

thegeodesic.

Hint:

The

algebraon

thiscan

bemessy,

butIfound

thingswere

reasonablysim

pleifIwrote

thederivatives

inthe

following

way:

dθ

dψ

=−

cosψsin

α√

1−sin

2ψsin

2α

,dφ

dψ

=cos

α

1−sin

2ψsin

2α

.

∗P

RO

BLEM

15:G

EO

DESIC

SIN

AC

LO

SED

UN

IVER

SE

The

following

problemwas

Problem

3,Quiz

3,2000,

where

itwas

worth

40points

plus5

pointsextra

credit.

Consider

thecase

ofclosed

Robertson-W

alkeruniverse.

Taking

k=

1,the

spacetimemetric

canbe

written

inthe

form

ds2=−c2dτ

2=−c2dt2+a2(t)

dr2

1−r2+r2 (d

θ2+

sin2θdφ

2 ) .

Wewill

assumethat

thismetric

isgiven,

andthat

a(t)has

beenspecified.

While

galaxiesare

approximately

stationaryin

thecom

ovingcoordinate

systemdescribed

bythis

metric,w

ecan

stillconsideran

object

thatmoves

inthis

system.In

particu-lar,in

thisproblem

wewillconsider

anob

jectthat

ismoving

inthe

radialdirection(r-direction),under

theinfluence

ofno

forcesother

thangravity.

Hence

theob

jectwilltravel

onageodesic.

(a)(7

points)Express

dτ/dtin

termsofdr/dt.

8.286Q

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2013p.20

(b)(3

points)Express

dt/dτin

termsofdr/dt.

(c)(10

points)If

theob

jecttravels

onatrajectory

givenby

thefunction

rp (t)

between

sometim

et1and

somelater

timet2 ,w

ritean

integralwhich

givesthe

totalamount

oftim

ethat

aclock

attachedto

theob

jectwould

recordfor

thisjourney.

(d)(10

points)During

atim

einterval

dt,the

object

willm

oveacoordinate

distance

dr=drdtdt.

Letd denote

thephysicaldistance

thatthe

object

moves

duringthis

time.

By

“physicaldistance,”Im

eanthe

distancethat

would

bemeasured

byacom

ovingobserver

(anobserver

stationarywith

respectto

thecoordinate

system)who

islocated

atthe

samepoint.

The

quantityd /dtcan

beregarded

asthe

physicalspeed

vphysof

theob

ject,since

itis

thespeed

thatwould

bemeasured

bya

comoving

observer.Write

anexpression

forvphysas

afunction

ofdr/dtand

r.

(e)(10

points)Using

theform

ulasat

thefront

ofthe

exam,derive

thegeodesic

equationof

motion

forthe

coordinaterof

theob

ject.Specifically,

youshould

derivean

equationof

theform

ddτ [

Adr

dτ ]

=B (

dt

dτ )

2

+C (

dr

dτ )

2

+D (

dθ

dτ )

2

+E (

dφ

dτ )

2

,

where

A,B,C,D,and

Eare

functionsofthe

coordinates,someofw

hichmight

bezero.

(f)(5

pointsEXTRA

CREDIT

)OnProblem

1of

Problem

Set6welearned

thatin

aflat

Robertson-W

alkermetric,

therelativistically

definedmom

entumof

aparticle,

p=

mvphys

√1−

v2phys

c2

,

fallsoff

as1/a(t).

Use

thegeodesic

equationderived

inpart

(e)to

showthat

thesam

eis

truein

aclosed

universe.

PR

OB

LEM

16:A

TW

O-D

IMEN

SIO

NA

LC

URV

ED

SPA

CE

(40points)

The

following

problemwas

Problem

3,Quiz

2,2002.

Consider

atw

o-dimensional

curvedspace

describedby

polarcoordinates

uand

θ,where

0≤u≤

aand

0≤θ≤

2π,

andθ=

2πis

asusual

identifiedwith

θ=

0.The

metric

isgiven

by

ds2=

adu

2

4u(a−

u)+udθ2.

Adiagram

ofthespace

isshow

nat

theright,but

youshould

ofcoursekeep

inmind

thatthe

diagramdoes

notaccurately

reflectthe

distancesdefined

bythe

metric.

8.286Q

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(a)(6

points)Find

theradius

Rof

thespace,

definedas

thelength

ofaradial

(i.e.,θ=

constant)line.

You

may

expressyour

answer

asadefinite

integral,which

youneed

notevaluate.

Besure,how

ever,tospecify

thelim

itsof

integration.

(b)(6

points)Find

thecircum

ferenceS

ofthe

space,de-

finedas

thelength

ofthe

boundaryof

thespace

atu=a.

(c)(7

points)Consider

anannular

regionas

shown,

con-sisting

ofall

pointswith

au-coordinate

inthe

rangeu

0 ≤u≤u

0+

du.

Find

thephysical

areadA

ofthis

region,to

firstorder

indu.

(d)(3

points)Using

youransw

erto

part(c),w

ritean

expressionfor

thetotalarea

ofthe

space.

(e)(10

points)Consider

ageodesic

curvein

thisspace,

describedby

thefunctions

u(s)and

θ(s),where

theparam

etersis

chosento

bethe

arclength

alongthe

curve.Find

thegeodesic

equationfor

u(s),which

shouldhave

theform

dds [

F(u,θ)

du

ds ]

=...

,

where

F(u,θ)

isafunction

thatyou

will

find.(N

otethat

bywriting

Fas

afunction

ofuand

θ,weare

sayingthat

itcould

dependon

eitheror

bothof

them,but

weare

notsaying

thatit

necessarilydepends

onthem

.)You

neednot

simplify

theleft-hand

sideof

theequation.

(f)(8

points)Sim

ilarly,findthe

geodesicequation

forθ(s),w

hichshould

havethe

formdds [

G(u,θ)

dθ

ds ]

=...

,

8.286Q

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2013p.22

where

G(u,θ)

isafunction

thatyou

willfind.

Again,you

neednot

simplify

theleft-hand

sideof

theequation.

∗P

RO

BLEM

17:R

OTA

TIN

GFR

AM

ES

OF

REFER

EN

CE

(35points)

The

following

problemwas

Problem

3,Quiz

2,2004.

Inthis

problemwewilluse

theform

alismofgeneralrelativity

andgeodesics

toderive

therelativistic

descriptionof

arotating

frameof

reference.

The

problemwillconcern

theconsequences

ofthe

metric

ds2=−c2dτ

2=−c2dt2+ [d

r2+r2(dφ+ωdt)

2+

dz2 ]

,(P

17.1)

which

correspondsto

acoordinate

systemrotating

aboutthe

z-axis,where

φis

theazim

uthalangle

aroundthe

z-axis.The

coordinateshave

theusual

rangefor

cylindricalcoordinates:−∞

<t<∞

,0≤

r<∞

,−∞<z<∞

,and

0≤φ<

2π,

where

φ=

2πis

identifiedwith

φ=

0.

EXTRA

INFORM

ATIO

N

To

work

theproblem

,you

donot

needto

knowanything

aboutwhere

thismetric

came

from.How

ever,it

might

(ormight

not!)help

yourintuition

toknow

thatEq.(P

17.1)was

obtainedby

startingwith

aM

inkowskim

etricin

cylindricalcoordinates

t,r,φ,

andz,

c2dτ

2=c2dt2− [d

r2+r2dφ

2+

dz2 ]

,

andthen

introducingnew

coordinatest,r,φ,

andz

thatare

relatedby

t=t,

r=r,

φ=φ+ωt,

z=z,

sodt=

dt,

dr=

dr,

dφ=

dφ+ωdt,

anddz=

dz.

(a)(8

points)The

metric

canbe

written

inmatrix

formby

usingthe

standarddefinition

ds2=−c2dτ

2≡gµνdxµdxν,

where

x0≡

t,x

1≡r,x

2≡φ,

andx

3≡z.

Then,

forexam

ple,g11(w

hichcan

alsobe

calledgrr )

isequal

to1.

Find

explicitexpressions

tocom

pletethe

list

8.286Q

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PR

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2013p.23

ofthe

nonzeroentries

inthe

matrix

gµν :

g11 ≡

grr=

1

g00 ≡

gtt=

?

g20 ≡

g02 ≡

gφt ≡

gtφ

=?

g22 ≡

gφφ=

?

g33 ≡

gzz=

?

(P17.2)

Ifyou

cannotansw

erpart

(a),youcan

introduceunspecified

functionsf1 (r),

f2 (r),

f3 (r),

andf4 (r),w

ith

g11 ≡

grr=

1

g00 ≡

gtt=f1 (r)

g20 ≡

g02 ≡

gφt ≡

gtφ

=f1 (r)

g22 ≡

gφφ=f3 (r)

g33 ≡

gzz=f4 (r)

,

(P17.3)

andyou

canthen

expressyour

answers

tothe

subsequentparts

interm

sof

theseunspecified

functions.

(b)(10

points)Using

thegeodesic

equationsfrom

thefront

ofthe

quiz,

ddτ

gµνdxν

dτ

=12(∂µgλσ )

dxλ

dτ

dxσ

dτ

,

explicitlywrite

theequation

thatresults

when

thefree

indexµis

equalto

1,corresponding

tothe

coordinater.

(c)(7

points)Explicitly

write

theequation

thatresults

when

thefree

indexµis

equalto

2,corresponding

tothe

coordinateφ.

(d)(10

points)Use

themetric

tofind

anexpression

fordt/d

τin

termsof

dr/d

t,dφ/dt,and

dz/dt.

The

expressionmay

alsodepend

onthe

constantscand

ω.

Besure

tonote

thatyour

answer

shoulddepend

onthe

derivativesoft,φ,and

zwith

respectto

t,not

τ.(H

int:first

findan

expressionfor

dτ/dt,

interm

sof

thequantities

indicated,and

thenask

yourselfhow

thisresult

canbe

usedto

finddt/d

τ.)

8.286Q

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PR

OB

LEM

18:T

HE

STA

BIL

ITY

OF

SC

HW

AR

ZSC

HIL

DO

RB

ITS

(30points)

This

problemwas

Problem

4,Quiz

2in

2007.I

havemodified

thereference

tothe

homew

orkproblem

tocorrespond

tothe

current(2013)

context,where

itis

Problem

3of

Problem

Set6.

In2007

ithad

alsobeen

ahom

ework

problemprior

tothe

quiz.

This

problemis

anelaboration

ofthe

Problem

3of

Problem

Set6,

forwhich

boththe

statement

andthe

solutionare

reproducedat

theend

ofthis

quiz.This

materialis

reproducedfor

yourreference,but

youshould

beaw

arethat

thesolution

tothe

presentproblem

hasim

portantdifferences.

You

cancopy

fromthis

material,

butto

allowthe

graderto

assessyour

understanding,you

areexpected

topresent

alogical,self-contained

answer

tothis

question.

Inthe

solutionto

thathom

ework

problem,it

was

statedthat

furtheranalysis

ofthe

orbitsin

aSchw

arzschildgeom

etryshow

sthat

thesm

alleststable

circularorbit

occursfor

r=

3RS .

Circular

orbitsare

possiblefor

32RS<r<

3RS,but

theyare

notstable.

Inthis

problemwewill

explorethe

calculationsbehind

thisstatem

ent.

Wewill

considerabody

which

undergoessm

alloscillations

aboutacircular

orbitat

r(t)=r0 ,θ=π/2,w

herer0is

aconstant.

The

coordinateθwilltherefore

befixed,

butallthe

othercoordinates

will

varyas

thebody

followsits

orbit.

(a)(12

points)The

firststep,since

r(τ)willnot

beaconstant

inthis

solution,will

beto

derivethe

equationof

motion

forr(τ).

That

is,for

theSchw

arzschildmetric

ds2=−c2dτ

2=−h(r)c

2dt2+h(r) −

1dr2+r2dθ2+r2sin

2θdφ

2,

(P18.1)

where

h(r)≡1−

RSr,

work

outthe

explicitform

ofthe

geodesicequation

ddτ [

gµνdxν

dτ ]

=12∂gλσ

∂xµ

dxλ

dτ

dxσ

dτ

,(P

18.2)

forthe

caseµ=r.

You

shoulduse

thisresult

tofind

anexplicit

expressionfor

d2r

dτ

2.

You

may

allowyour

answer

tocontain

h(r),itsderivative

h ′(r)with

respectto

r,and

thederivative

with

respectto

τof

anycoordinate,including

dt/dτ.

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(b)(6

points)It

isusefulto

considerrand

φto

bethe

independentvariables,w

hiletreating

tas

adependent

variable.Find

anexpression

for

(dt

dτ )

2

interm

sof

r,dr/dτ,

dφ/dτ,

h(r),and

c.Use

thisequation

tosim

plifythe

expressionfor

d2r/

dτ

2obtained

inpart

(a).The

goalisto

obtainan

expressionof

theform

d2r

dτ

2=f0 (r)

+f1 (r) (

dφ

dτ )

2

.(P

18.3)

where

thefunctions

f0 (r)

andf1 (r)

might

dependon

RSor

c,and

might

bepositive,negative,

orzero.

Note

thatthe

intermediate

stepsin

thecalculation

involveaterm

proportionalto

(dr/dτ)

2,but

thenet

coefficient

forthis

termvanishes.

(c)(7

points)Tounderstand

theorbit

wewillalso

needthe

equationofm

otionfor

φ.Evaluate

thegeodesic

equation(P

18.2)for

µ=φ,

andwrite

theresult

interm

sof

thequantity

L,defined

byL≡r2dφ

dτ.

(P18.4)

(d)(5

points)Finally,w

ecom

eto

thequestion

ofstability.Substituting

Eq.(P

18.4)into

Eq.

(P18.3),the

equationof

motion

forrcan

bewritten

as

d2r

dτ

2=f0 (r)+

f1 (r)

L2

r4.

Now

considerasm

allperturbation

aboutthe

circularorbit

atr=

r0 ,

andwrite

anequation

thatdeterm

inesthe

stabilityof

theorbit.

(That

is,ifsom

eexternal

forcegives

theorbiting

bodyasm

allkickin

theradialdirection,

howcan

youdeterm

inewhether

theperturbation

willlead

tostable

oscillations,orwhether

itwill

startto

grow?)

You

shouldexpress

thestability

requirement

interm

sof

theunspecified

functionsf0 (r)

andf1 (r).

You

areNOT

askedto

carryout

thealgebra

ofinserting

theexplicit

formsthat

youhave

foundfor

thesefunctions.

8.286Q

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2013p.26

∗P

RO

BLEM

19:P

RESSU

RE

AN

DEN

ER

GY

DEN

SIT

YO

FM

YST

E-

RIO

US

ST

UFF

(25points)

The

following

problemwas

Problem

3,Quiz

3,2002.

InLecture

Notes

6,with

furthercalculations

inProblem

4of

Problem

Set6,

athought

experiment

involvingapiston

was

usedto

showthat

p=

13ρc2for

radiation.In

thisproblem

youwill

applythe

sametechnique

tocalculate

thepressure

ofm

yste

rious

stuff,which

hasthe

propertythat

theenergy

densityfalls

offin

proportionto

1/ √V

asthe

volumeV

isincreased.

Iftheinitialenergy

densityofthe

mysterious

stuffisu

0=ρ0 c

2,thenthe

initialconfiguration

ofthe

pistoncan

bedraw

nas

The

pistonis

thenpulled

outward,

sothat

itsinitial

volumeV

isincreased

toV

+∆V.You

may

consider∆V

tobe

infinitesimal,

so∆V

2can

beneglected.

(a)(15

points)Using

thefact

thatthe

energydensity

ofmysterious

stufffalls

offas

1/ √

V,find

theam

ount∆U

bywhich

theenergy

insidethe

pistonchanges

when

thevolum

eis

enlargedby

∆V.Define

∆U

tobe

positiveifthe

energyincreases.

(b)(5

points)If

the(unknow

n)pressure

ofthe

mysterious

stuffis

calledp,

howmuch

work

∆W

isdone

bythe

agentthat

pullsout

thepiston?

(c)(5

points)Use

yourresults

from(a)

and(b)

toexpress

thepressure

pof

themysterious

stuffin

termsof

itsenergy

densityu.

(Ifyou

didnot

answer

parts(a)

and/or(b),explain

asbest

youcan

howyou

would

determine

thepressure

ifyou

knewthe

answers

tothese

twoquestions.)

8.286Q

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2013p.27

SO

LU

TIO

NS

PR

OB

LEM

1:D

IDY

OU

DO

TH

ER

EA

DIN

G?

(a)This

isatotal

trickquestion.

Lepton

number

is,of

course,conserved,

sothe

factoris

just1.

SeeWeinberg

chapter4,

pages91-4.

(b)The

correctansw

eris

(i).The

othersare

allreal

reasonswhy

it’shard

tomeasure,

althoughWeinberg’s

bookem

phasizesreason

(v)abit

more

thanmodern

astrophysicistsdo:

astrophysicistshave

beenlooking

forother

ways

thatdeuterium

might

beproduced,

butno

significantmechanism

hasbeen

found.See

Weinberg

chapter5,

pages114-7.

(c)The

most

obviousansw

erswould

beproton,

neutron,and

pimeson.

How

ever,there

aremany

otherpossibilities,including

many

thatwere

notmentioned

byWeinberg.

SeeWeinberg

chapter7,

pages136-8.

(d)The

correctansw

erswere

theneutrino

andthe

antiproton.The

neutrinowas

firsthypothesized

byWolfgang

Pauliin

1932(in

orderto

explainthe

kine-matics

ofbeta

decay),and

firstdetected

inthe

1950s.After

thepositron

was

discoveredin

1932,theantiproton

was

thoughtlikely

toexist,and

theBevatron

inBerkeley

was

builtto

lookfor

antiprotons.It

made

thefirst

detectionin

the1950s.

(e)The

correctansw

erswere

(ii),(v)and

(vi).The

otherswere

incorrectfor

thefollow

ingreasons:

(i)the

earliestprediction

ofthe

CMB

temperature,

byAlpher

andHerm

anin

1948,was

5degrees,

not0.1

degrees.

(iii)Weinberg

quoteshis

experimentalcolleagues

assaying

thatthe

3 K

radi-ation

couldhave

beenobserved

“longbefore

1965,probably

inthe

mid-

1950sand

perhapseven

inthe

mid-1940s.”

ToWeinberg,

however,

thehistorically

interestingquestion

isnot

when

theradiation

couldhave

beenobserved,but

why

radioastronom

ersdid

notknow

thatthey

oughtto

try.

(iv)Weinberg

arguesthat

physicistsat

thetim

edid

notpay

attentionto

eitherthe

steadystate

modelor

thebig

bangmodel,as

indicatedby

thesentence

initem

(v)which

isadirect

quotefrom

thebook:

“Itwas

extraordinarilydiffi

cultfor

physiciststo

takeseriously

anytheory

ofthe

earlyuniverse”.

8.286Q

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PR

OB

LEM

2:D

IDY

OU

DO

TH

ER

EA

DIN

G?

(24points)

(a)(6

points)In

1948Ralph

A.A

lpherand

Robert

Herm

anwrote

apaper

predict-ing

acosm

icmicrow

avebackground

with

atem

peratureof5

K.The

paperwas

basedon

acosm

ologicalmodel

thatthey

haddeveloped

with

George

Gam

ow,

inwhich

theearly

universewas

assumed

tohave

beenfilled

with

hotneutrons.

Asthe

universeexpanded

andcooled

theneutrons

underwent

betadecay

intoprotons,

electrons,and

antineutrinos,until

atsom

epoint

theuniverse

cooledenough

forlight

elements

tobe

synthesized.Alpher

andHerm

anfound

thatto

accountfor

theobserved

presentabundances

oflightelem

ents,theratio

ofpho-tons

tonuclear

particlesmust

havebeen

about10

9.Although

thepredicted

temperature

was

veryclose

tothe

actualvalue

of2.7

K,the

theorydiffered

fromour

presenttheory

intw

oways.

Circle

thetw

ocorrect

statements

inthe

following

list.(3

pointsfor

eachright

answer;

circleat

most

2.)

(i)Gam

ow,Alpher,

andHerm

anassum

edthat

theneutron

coulddecay,

butnow

theneutron

isthought

tobe

absolutelystable.

(ii)In

thecurrent

theory,the

universestarted

with

nearlyequal

densitiesof

protonsand

neutrons,not

allneutrons

asGam

ow,Alpher,

andHerm

anassum

ed.

(iii)In

thecurrent

theory,theuniverse

startedwith

mainly

alphaparticles,not

allneutrons

asGam

ow,Alpher,

andHerm

anassum

ed.(N

ote:an

alphaparticle

isthe

nucleusofa

heliumatom

,composed

oftwoprotons

andtw

oneutrons.)

(iv)In

thecurrent

theory,the

conversionof

neutronsinto

protons(and

viceversa)

tookplace

mainly

throughcollisions

with

electrons,positrons,neu-trinos,

andantineutrinos,

notthrough

thedecay

ofthe

neutrons.

(v)The

ratioof

photonsto

nuclearparticles

inthe

earlyuniverse

isnow

be-lieved

tohave

beenabout

103,

not10

9as

Alpher

andHerm

anconcluded.

(b)(6

points)In

Weinberg’s

“Recipe

foraHot

Universe,”

hedescribed

theprim

or-dialcom

positionofthe

universein

termsofthree

conservedquantities:

electriccharge,

baryonnum

ber,and

leptonnum

ber.If

electriccharge

ismeasured

inunits

ofthe

electroncharge,

thenall

threequantities

areintegers

forwhich

thenum

berdensity

canbe

compared

with

thenum

berdensity

ofphotons.

Foreach

quantity,which

choicemost

accuratelydescribes

theinitial

ratioof

thenum

berdensity

ofthis

quantityto

thenum

berdensity

ofphotons:

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Electric

Charge:

(i)∼10

9(ii)∼

1000(iii)∼

1(iv)∼

10 −6

(v)either

zeroor

negligible

Baryon

Num

ber:(i)∼

10 −20

(ii)∼10 −

9(iii)∼

10 −6

(iv)∼1

(v)anyw

herefrom

10 −5to

1

Lepton

Num

ber:(i)∼

109

(ii)∼1000

(iii)∼1

(iv)∼10 −

6(v)

couldbe

ashigh

as∼1,

butis

assumed

tobe

verysm

all

(c)(12

points)The

figurebelow

comes

fromWeinberg’s

Chapter

5,andislabeled

The

ShiftingNeutron-P

rotonBalance.

(i)(3

points)During

theperiod

labeled“therm

alequilibrium

,”the

neutronfraction

ischanging

because(choose

one):

(A)The

neutronis

unstable,and

decaysinto

aproton,

electron,and

an-tineutrino

with

alifetim

eof

about1second.

(B)The

neutronis

unstable,and

decaysinto

aproton,

electron,and

an-tineutrino

with

alifetim

eof

about15

seconds.

(C)The

neutronis

unstable,and

decaysinto

aproton,

electron,and

an-tineutrino

with

alifetim

eof

about15

minutes.

(D)Neutrons

andprotons

canbe

convertedfrom

oneinto

throughreac-

tionssuch

as

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antineutrino+proton←→

electron+

neutronneutrino

+neutron←→

positron+proton

.

(E)Neutrons

andprotons

canbe

convertedfrom

oneinto

theother

throughreactions

suchas

antineutrino+proton←→

positron+

neutronneutrino

+neutron←→

electron+proton

.

(F)Neutrons

andprotons

canbe

createdand

destroyedby

reactionssuch

asproton

+neutrino←→

positron+antineutrino

neutron+antineutrino←→

electron+

positron.

(ii)(3

points)During

theperiod

labeled“neutron

decay,”the

neutronfraction

ischanging

because(choose

one):

(A)The

neutronis

unstable,and

decaysinto

aproton,

electron,and

an-tineutrino

with

alifetim

eof

about1second.

(B)The

neutronis

unstable,and

decaysinto

aproton,

electron,and

an-tineutrino

with

alifetim

eof

about15

seconds.

(C)The

neutronis

unstable,and

decaysinto

aproton,

electron,and

an-tineutrino

with

alifetim

eof

about15

minutes.

(D)Neutrons

andprotons

canbe

convertedfrom

oneinto

theother

throughreactions

suchas

antineutrino+proton←→

electron+

neutronneutrino

+neutron←→

positron+proton

.

(E)Neutrons

andprotons

canbe

convertedfrom

oneinto

theother

throughreactions

suchas

antineutrino+proton←→

positron+

neutronneutrino

+neutron←→

electron+proton

.

(F)Neutrons

andprotons

canbe

createdand

destroyedby

reactionssuch

asproton

+neutrino←→

positron+antineutrino

neutron+antineutrino←→

electron+

positron.

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(iii)(3

points)The

masses

ofthe

neutronand

protonare

notexactly

equal,but

instead

(A)The

neutronis

more

massive

thanaproton

with

arest

energydiffer-

enceof

1.293GeV

(1GeV

=10

9eV

).

(B)The

neutronis

more

massive

thanaproton

with

arest

energydiffer-

enceof

1.293MeV

(1MeV

=10

6eV

).

(C)The

neutronis

more

massive

thanaproton

with

arest

energydiffer-

enceof

1.293KeV

(1KeV

=10

3eV

).

(D)The

protonismore

massive

thananeutron

with

arest

energydiffer-

enceof

1.293GeV

.

(E)The

protonismore

massive

thananeutron

with

arest

energydiffer-

enceof

1.293MeV

.

(F)The

protonismore

massive

thananeutron

with

arest

energydiffer-

enceof

1.293KeV

.

(iv)(3

points)During

theperiod

labeled“era

ofnucleosynthesis,”(choose

one:)

(A)Essentially

allthe

neutronspresent

combine

with

protonsto

formhelium

nuclei,which

mostly

surviveuntil

thepresent

time.

(B)Essentially

allthe

neutronspresent

combine

with

protonsto

formdeuterium

nuclei,which

mostly

surviveuntil

thepresent

time.

(C)About

halftheneutrons

presentcom

binewith

protonsto

formhelium

nuclei,which

mostly

surviveuntilthe

presenttim

e,andthe

otherhalf

ofthe

neutronsrem

ainfree.

(D)About

halfthe

neutronspresent

combine

with

protonsto

formdeu-

teriumnuclei,

which

mostly

surviveuntil

thepresent

time,

andthe

otherhalf

ofthe

neutronsrem

ainfree.

(E)Essentially

allthe

protonspresent

combine

with

neutronsto

formhelium

nuclei,which

mostly

surviveuntil

thepresent

time.

(F)Essentially

allthe

protonspresent

combine

with

neutronsto

formdeuterium

nuclei,which

mostly

surviveuntil

thepresent

time.

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PR

OB

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3:D

IDY

OU

DO

TH

ER

EA

DIN

G?

(20points) †

(a)(8

points)

(i)(4

points)Wewill

usethe

notationXA

toindicate

anucleus,*

where

Xis

thesym

bolfor

theelem

entwhich

indicatesthe

number

ofprotons,

while

Aisthe

mass

number,nam

elythe

totalnumber

ofprotonsand

neu-trons.

With

thisnotation

H1,H

2,H

3,He3and

He4stand

forhydrogen,

deuterium,tritium

,helium

-3and

helium-4

nuclei,respectively.

StevenWeinberg,

inThe

First

Three

Minutes,

chapterV,page

108,describes

twochains

ofreactions

thatproduce

helium,starting

fromprotons

andneutrons.

They

canbe

written

as:

p+n→

H2+γ

H2+n→

H3+γ

H3+p→

He4+γ,

p+n→

H2+γ

H2+p→

He3+γ

He3+n→

He4+γ.

These

arethe

twoexam

plesgiven

byWeinberg.

How

ever,differentchains

oftw

oparticle

reactionscan

takeplace

(ingeneral

with

differentproba-

bilities).For

example:

p+n→

H2+γ

H2+H

2→He4+γ,

p+n→

H2+γ

H2+n→

H3+γ

H3+H

2→He4+n,

p+n→

H2+γ

H2+p→

He3+γ

He3+H

2→He4+p,

...

Studentswho

describedchains

differentfrom

thoseof

Weinberg,

butthat

canstilltake

place,got

fullcreditfor

thispart.

Also,notice

thatphotons

inthe

reactionsabove

carrythe

additionalenergyreleased.

How

ever,sincethe

main

pointwas

todescribe

thenuclear

reactions,studentswho

didn’tinclude

thephotons

stillreceivedfull

credit.

(ii)(4

points)The

deuteriumbottleneck

isdiscussed

byWeinberg

inThe

First

Three

Minutes,

chapterV,pages

109-110.The

keypoint

isthat

frompart

(i)it

shouldbe

clearthat

deuterium(H

2)plays

acrucial

rolein

*Notice

thatsom

estudents

talkedabout

atoms,

while

weare

talkingabout

nucleiform

ation.During

nucleosynthesisthe

temperature

isway

toohigh

toallow

electronsand

nucleito

bindtogether

toform

atoms.

This

happensmuch

later,in

theprocess

calledrecom

bination.

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nucleosynthesis,since

itis

thestarting

pointfor

allthe

chains.How

ever,the

deuteriumnucleus

isextrem

elyloosely

boundcom

paredto

H3,He3,

orespecially

He4.

So,there

will

bearange

oftem

peratureswhich

arelow

enoughfor

H3,He3,

andHe4nuclei

tobe

bound,but

toohigh

toallow

thedeuterium

nucleusto

bestable.

This

isthe

temperature

rangewhere

thedeuterium

bottleneckis

inaction:

evenifH

3,He3,

andHe4

nucleicouldin

principlebe

stableat

thosetem

peratures,theydo

notform

becausedeuterium

,which

isthe

startingpoint

fortheir

formation,cannot

beform

edyet.

Nucleosynthesis

cannotproceed

atasignificant

rateuntil

thetem

peratureislow

enoughso

thatdeuterium

nucleiarestable;at

thispoint

thedeuterium

bottleneckhas

beenpassed.

(b)(12

points)

(i)(3

points)If

wetake

a(t)=

bt1/2,

forsom

econstant

b,weget

forthe

Hubble

expansionrate:

H=aa=

12t

=⇒t=

12H.

(ii)(6

points)Byusing

theFriedm

annequation

with

k=

0and

ρ=ρr=αT

4,wefind:H

2=

8π3Gρr=

8π3GαT

4=⇒

H=T

2 √8π3Gα.

Ifwesubstitute

thegiven

numericalvalues

G

6.67×

10 −11N·m

2·kg −2

andα

4.52×

10 −32kg·m

−3·K

−4weget:

HT

2×5.03×

10 −21s −

1·K−

2.

Notice

thatthe

unitscorrectly

combine

togive

Hin

unitsof

s −1if

thetem

peratureis

expressedin

degreesKelvin

(K).

Indetail,w

esee:

[Gα] 1/2=

(N·m

2·kg −2·kg·m

−3·K

−4)

1/2=

s −1·K

−2,

where

weused

thefact

that1N

=1kg·m

·s −2.

AtT

=T

nucl

0.9×

109K

weget:

H

4.07×

10 −3s −

1.

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(iii)(3

points)Using

theresults

inparts

(i)and

(ii),weget

t=

12H (

9.95×

1019

T2

)s·K

2.

Togood

accuracy,the

numerator

inthe

expressionabove

canbe

roundedto

1020.

The

aboveequation

agreeswith

Weinberg’s

claimthat,

fora

radiationdom

inateduniverse,

timeis

proportionalto

theinverse

squareof

thetem

perature.In

particularfor

T=T

nuclweget:

tnucl

123s≈

2min.

†Solutionwritten

byDaniele

Bertolini.

PR

OB

LEM

4:EV

OLU

TIO

NO

FA

NO

PEN

UN

IVER

SE

The

evolutionof

anopen,

matter-dom

inateduniverse

isdescribed

bythe

fol-low

ingparam

etricequations:

ct=α(sinh

θ−θ)

a√κ=α(cosh

θ−1)

.

Evaluating

thesecond

ofthese

equationsat

a/ √

κ=

2αyields

asolution

forθ:

2α=α(cosh

θ−1)

=⇒cosh

θ=

3=⇒

θ=

cosh −1(3)

.

Wecan

usethese

resultsin

thefirst

equationto

solvefor

t.Noting

that

sinhθ= √

cosh2θ−

1=√8=

2 √2,

wehave

t=αc [2 √

2−cosh −

1(3) ].

Num

erically,t≈

1.06567

α/c.

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PR

OB

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5:A

NT

ICIP

AT

ING

AB

IGC

RU

NC

H

The

criticaldensityis

givenby

ρc=

3H

20

8πG

,

sothe

mass

densityis

givenbyρ=

Ω0 ρ

c=

2ρc=

3H

20

4πG

.(S5.1)

Substitutingthis

relationinto

H20=

8π3Gρ−

kc2

a2,

wefind

H20=

2H

20 −kc2

a2,

fromwhich

itfollow

sthat

a√k=

cH0.

(S5.2)

Now

use

α=

4π3Gρa3

k3/2c

2.

Substitutingthe

valueswehave

fromEqs.

(S5.1)and

(S5.2)for

ρand

a/ √

k,we

haveα=

cH0.

(S5.3)

Todeterm

inethe

valueof

theparam

eterθ,

use

a√k=α(1−

cosθ)

,

which

when

combined

with

Eqs.

(S5.2)and

(S5.3)im

pliesthat

cosθ

=0.The

equationcos

θ=

0has

multiple

solutions,but

weknow

thatthe

θ-parameter

foraclosed

matter-dom

inateduniverse

variesbetw

een0and

πduring

theexpansion

phaseof

theuniverse.

Within

thisrange,cos

θ=

0im

pliesthat

θ=π/2.

Thus,the

ageof

theuniverse

atthe

timethese

measurem

entsare

made

isgiven

by

t=αc(θ−

sinθ)

=1H0 (

π2−

1 ).

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The

totallifetimeof

theclosed

universecorresponds

toθ=

2π,or

tfinal =

2παc

=2π

H0,

sothe

timerem

ainingbefore

thebig

crunchis

givenby

tfinal −

t=

1H0 [2

π− (π2−

1 )]=

(3π2+

1 )1H0.

PR

OB

LEM

6:T

RA

CIN

GLIG

HT

RA

YS

INA

CLO

SED

,M

AT

TER

-D

OM

INA

TED

UN

IVER

SE

(a)Since

θ=φ=

constant,dθ=dφ=

0,andfor

lightrays

onealw

ayshas

dτ=

0.The

lineelem

enttherefore

reducesto

0=−c2dt2+a2(t)d

ψ2.

Rearranging

gives(dψdt )

2

=c2

a2(t)

,

which

implies

that

dψdt=±

c

a(t).

The

plussign

describesoutw

ardradialm

otion,while

theminus

signdescribes

inward

motion.

(b)The

maxim

umvalue

oftheψcoordinate

thatcan

bereached

bytim

etisfound

byintegrating

itsrate

ofchange:

ψhor= ∫

t

0

c

a(t ′)dt ′.

The

physicalhorizondistance

isthe

properlength

oftheshortest

linedraw

nat

thetim

etfrom

theorigin

toψ=ψ

hor ,which

accordingto

themetric

isgiven

by

phys (t)

= ∫ψ

=ψ

hor

ψ=

0

ds= ∫

ψhor

0

a(t)dψ=

a(t) ∫t

0

c

a(t ′)dt ′.

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(c)From

part(a),

dψdt=

c

a(t).

Bydifferentiating

theequation

ct=

α(θ−

sinθ)

statedin

theproblem

,one

findsdt

dθ=αc(1−

cosθ)

.

Then

dψdθ=dψdt

dt

dθ=α(1−

cosθ)

a(t).

Then

usinga=α(1−

cosθ),as

statedin

theproblem

,onehas

thevery

simple

result

dψdθ=

1.

(d)This

partis

verysim

pleif

oneknow

sthat

ψmust

changeby

2π

beforethe

photonreturns

toits

startingpoint.

Sincedψ/dθ=

1,thismeans

thatθmust

alsochange

by2π.From

a=α(1−

cosθ),

onecan

seethat

areturns

tozero

atθ=

2π,so

thisis

exactlythe

lifetimeof

theuniverse.

So,

Tim

efor

photonto

returnLifetim

eof

universe=

1.

Ifit

isnot

clearwhy

ψmust

changeby

2π

forthe

photonto

returnto

itsstarting

point,then

recallthe

constructionof

theclosed

universethat

was

usedin

Lecture

Notes

5.The

closeduniverse

isdescribed

asthe

3-dimensional

surfaceof

asphere

inafour-dim

ensionalEuclidean

spacewith

coordinates(x,y,z,w

):x

2+y2+z2+w

2=a2,

where

ais

theradius

ofthe

sphere.The

Robertson-W

alkercoordinate

systemis

constructedon

the3-dim

ensionalsurface

ofthe

sphere,taking

thepoint

(0,0,0,1)

asthe

centerof

thecoordinate

system.If

wedefine

thew-direction

as“north,”

thenthe

point(0,0

,0,1)

canbe

calledthe

northpole.

Each

point(x,y,z,w

)on

thesurface

ofthesphere

isassigned

acoordinate

ψ,defined

tobe

theangle

between

thepositive

waxis

andthe

vector(x,y,z,w

).Thus

ψ=

0at

thenorth

pole,andψ=πfor

theantipodalpoint,(0,0

,0,−

1),which

canbe

calledthe

southpole.

Inmaking

theround

tripthe

photonmust

travelfrom

thenorth

poleto

thesouth

poleand

back,for

atotalrange

of2π

.

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Discussion:

Somestudents

answered

thatthe

photonwould

returnin

thelife-

timeofthe

universe,butreached

thisconclusion

without

consideringthe

detailsofthe

motion.

The

argument

was

simply

that,atthe

bigcrunch

when

thescale

factorreturns

tozero,alldistances

would

returnto

zero,includingthe

distancebetw

eenthe

photonand

itsstarting

place.This

statement

iscorrect,but

itdoes

notquite

answer

thequestion.

First,the

statement

inno

way

rulesout

thepos-

sibilitythat

thephoton

might

returnto

itsstarting

pointbefore

thebig

crunch.Second,

ifweuse

thedelicate

butwell-m

otivateddefinitions

thatgeneral

rel-ativists

use,it

isnot

necessarilytrue

thatthe

photonreturns

toits

startingpoint

atthe

bigcrunch.

Tobe

concrete,letmeconsider

aradiation-dom

inatedclosed

universe—ahypothetical

universefor

which

theonly

“matter”

presentconsists

ofmassless

particlessuch

asphotons

orneutrinos.

Inthat

case(you

cancheck

mycalculations)

aphoton

thatleaves

thenorth

poleat

t=

0just

reachesthe

southpole

atthe

bigcrunch.

Itmight

seemthat

reachingthe

southpole

atthe

bigcrunch

isnot

anydifferent

fromcom

pletingthe

roundtrip

backto

thenorth

pole,sincethe

distancebetw

eenthe

northpole

andthe

southpole

iszero

att=tC

runch ,

thetim

eof

thebig

crunch.How

ever,suppose

weadopt

theprinciple

thatthe

instantof

theinitial

singularityand

theinstant

ofthe

finalcrunch

areboth

toosingular

tobe

consideredpart

ofthe

spacetime.

We

will

allowourselves

tomathem

aticallyconsider

times

rangingfrom

t=

εto

t=

tC

runch −

ε,where

εis

arbitrarilysm

all,but

wewill

nottry

todescribe

what

happensexactly

att=

0or

t=tC

runch .

Thus,w

enow

consideraphoton

thatstarts

itsjourney

att=ε,

andwefollow

ituntil

t=tC

runch −

ε.For

thecase

ofthe

matter-dom

inatedclosed

universe,such

aphoton

would

traverseafraction

ofthe

fullcircle

thatwould

bealm

ost1,

andwould

approach1as

ε→0.

Bycontrast,

forthe

radiation-dominated

closeduniverse,

thephoton

would

traverseafraction

ofthe

fullcircle

thatis

almost

1/2,and

itwould

approach1/2

asε→

0.Thus,

fromthis

pointof

viewthe

twocases

lookvery

different.In

theradiation-dom

inatedcase,

onewould

saythat

thephoton

hascom

eonly

half-way

backto

itsstarting

point.

PR

OB

LEM

7:LEN

GT

HS

AN

DA

REA

SIN

AT

WO

-DIM

EN

-SIO

NA

LM

ET

RIC

a)Along

thefirst

segment

dθ=

0,so

ds2=

(1+ar)

2dr2,

ords=

(1+ar)dr.

Integrating,the

lengthof

thefirst

segment

isfound

tobe

S1= ∫

r0

0

(1+ar)dr=r0+

12ar20.

Along

thesecond

segment

dr=

0,so

ds=r(1

+br)

dθ,

where

r=r0 .

Sothe

lengthof

thesecond

segment

is

S2= ∫

π/2

0

r0 (1

+br

0 )dθ=π2r0 (1

+br

0 ).

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Finally,the

thirdsegm

entisidenticalto

thefirst,so

S3=S

1 .The

totallengthis

then

S=

2S

1+S

2=

2 (r0+

12ar20 )

+π2r0 (1

+br

0 )

=(2

+π2 )

r0+

12(2a+πb)r

20.

b)Tofind

thearea,it

isbest

todivide

theregion

intoconcentric

stripsas

shown:

Note

thatthe

striphas

acoordinate

width

ofdr,

butthe

distanceacross

thewidth

ofthe

stripis

determined

bythe

metric

tobe

dh=

(1+ar)dr.

The

lengthof

thestrip

iscalculated

thesam

eway

asS

2in

part(a):

s(r)=π2r(1

+br)

.

The

areais

then

dA

=s(r)

dh,

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soA

= ∫r0

0

s(r)dh

= ∫r0

0

π2r(1

+br)(1

+ar)dr

=π2 ∫

r0

0

[r+(a

+b)r

2+abr

3]dr

=π2 [

12r20+

13(a

+b)r

30+

14abr

40 ]

PR

OB

LEM

8:G

EO

MET

RY

INA

CLO

SED

UN

IVER

SE

(a)Asone

moves

alongaline

fromthe

originto

(h,0,0),there

isno

variationinθ

orφ.

Sodθ=dφ=

0,and

ds=

adr

√1−

r2.

So

p= ∫

h

0

adr

√1−

r2=asin −

1h.

(b)In

thiscase

itis

onlyθthat

varies,so

dr=dφ=

0.So

ds=ardθ,

so

sp=ah∆θ.

(c)From

part(a),one

hash=

sin( p /a)

.

Insertingthis

expressioninto

theansw

erto

(b),and

thensolving

for∆θ,

onehas

∆θ=

sp

asin(

p /a)

.

Note

thatas

a→∞

,thisapproaches

theEuclidean

result,∆θ=sp / p .

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PR

OB

LEM

9:T

HE

GEN

ER

AL

SP

HER

ICA

LLY

SY

MM

ET

RIC

MET

-R

IC

(a)The

metric

isgiven

byds2=dr2+ρ2(r) [d

θ2+

sin2θdφ

2 ].

The

radiusais

definedas

thephysical

lengthof

aradial

linewhich

extendsfrom

thecenter

tothe

boundaryofthe

sphere.The

lengthofa

pathisjust

theintegral

ofds,so

a= ∫

radialpath

fromorigin

tor0

ds.

The

radialpath

isat

aconstant

valueofθand

φ,so

dθ=dφ=

0,and

thends=dr.

So

a= ∫

r0

0

dr=

r0.

(b)Onthe

surfacer=r0 ,

sodr≡

0.Then

ds2=ρ2(r

0 ) [dθ2+

sin2θdφ

2 ].

Tofind

thearea

element,

considerfirst

apath

obtainedby

varyingonly

θ.Then

ds=ρ(r

0 )dθ.

Similarly,

apath

obtainedby

varyingonly

φhas

lengthds=

ρ(r0 )sin

θdφ.

Furthermore,

thesetw

opaths

areperpendicular

toeach

other,afact

thatis

incorporatedinto

themetric

bythe

absenceof

adrdθ

term.Thus,

thearea

ofasm

allrectangle

constructedfrom

thesetw

opaths

isgiven

bythe

productof

theirlengths,

so

dA

=ρ2(r

0 )sinθdθdφ.

The

areais

thenobtained

byintegrating

overthe

rangeof

thecoordinate

variables:

A=ρ2(r

0 ) ∫2π

0

dφ ∫

π

0

sinθdθ

=ρ2(r

0 )(2π) (−

cosθ ∣∣∣ π0 )

=⇒A

=4πρ2(r

0 ).

Asacheck,

noticethat

ifρ(r)

=r,

thenthe

metric

becomes

themetric

ofEuclidean

space,in

sphericalpolar

coordinates.In

thiscase

theansw

erabove

becomes

thewell-know

nform

ulafor

thearea

ofaEuclidean

sphere,4πr2.

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(c)Asin

Problem

2ofP

roblemSet

5,wecan

imagine

breakingup

thevolum

einto

sphericalshellsofinfinitesim

althickness,with

agiven

shellextendingfrom

rto

r+dr.

Bythe

previouscalculation,the

areaof

suchashellis

A(r)

=4πρ2(r).

(Inthe

previouspart

weconsidered

onlythe

caser=r0 ,but

thesam

eargum

entapplies

forany

valueofr.)

The

thicknessofthe

shellisjust

thepath

lengthds

ofaradial

pathcorresponding

tothe

coordinateinterval

dr.

Forradial

pathsthe

metric

reducesto

ds2=dr2,

sothe

thicknessof

theshell

isds=dr.

The

volumeof

theshell

isthen

dV

=4πρ2(r)

dr.

The

totalvolumeis

thenobtained

byintegration:

V=

4π ∫

r0

0

ρ2(r)

dr.

Checking

theansw

erfor

theEuclidean

case,ρ(r)

=r,

onesees

thatit

givesV

=(4π/3)r

30 ,as

expected.

(d)Ifrisreplaced

byanew

coordinateσ≡r2,then

theinfinitesim

alvariationsof

thetw

ocoordinates

arerelated

by

dσdr=

2r=

2 √σ,

so

dr2=dσ

2

4σ

.

The

functionρ(r)

canthen

bewritten

asρ( √

σ),

so

ds2=dσ

2

4σ

+ρ2( √

σ) [d

θ2+

sin2θdφ

2 ].

PR

OB

LEM

10:V

OLU

MES

INA

RO

BERT

SO

N-W

ALK

ER

UN

IVER

SE

The

productof

differentiallength

elements

correspondingto

infinitesimal

changesin

thecoordinates

r,θand

φequals

thedifferential

volumeelem

entdV.

Therefore

dV

=a(t)

dr

√1−

kr2 ×

a(t)rdθ×

a(t)rsin

θdφ

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The

totalvolumeis

then

V= ∫

dV

=a3(t) ∫

rm

ax

0

dr ∫

π

0

dθ ∫

2π

0

dφ

r2sin

θ√1−

kr2

Wecan

dothe

angularintegrations

immediately:

V=

4πa3(t) ∫

rm

ax

0

r2dr

√1−

kr2.

[PedagogicalN

ote:Ifyou

don’tsee

throughthe

solutionsabove,then

notethat

thevolum

eof

thesphere

canbe

determined

byintegration,after

firstbreaking

thevolum

einto

infinitesimal

cells.A

genericcell

isshow

nin

thediagram

below:

The

cellincludesthe

volumelying

between

rand

r+dr,betw

eenθand

θ+dθ,

andbetw

eenφand

φ+dφ.

Inthe

limit

asdr,dθ,

anddφall

approachzero,

thecell

approachesarectangular

solidwith

sidesof

length:

ds1=a(t)

dr

√1−

kr2

ds2=a(t)r

dθ

ds3=a(t)r

sinθdθ.

Here

eachdsiscalculated

byusing

themetric

tofind

ds2,in

eachcase

allowing

onlyone

ofthequantities

dr,dθ,or

dφto

benonzero.

The

infinitesimalvolum

eelem

entisthen

dV

=ds1 ds2 ds3 ,resulting

inthe

answer

above.The

derivation

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relieson

theorthogonality

ofthe

dr,dθ,

anddφdirections;

theorthogonality

isim

pliedby

themetric,

which

otherwise

would

containcross

termssuch

asdrdθ.]

[Extension:

The

integralcan

infact

becarried

out,usingthe

substitution

√kr=

sinψ

(ifk>

0)√−

kr=

sinhψ

(ifk>

0).

The

answer

is

V=

2πa3(t)

sin −1 (√

krm

ax )

k3/2

− √1−

kr2m

ax

k

(if

k>

0)

2πa3(t) [√

1−kr2m

ax

(−k)

−sinh −

1 (√−krm

ax )

(−k)

3/2

](if

k<

0).]

PR

OB

LEM

11:T

HE

SC

HW

AR

ZSC

HIL

DM

ET

RIC

a)The

Schwarzschild

horizonis

thevalue

ofrfor

which

themetric

becomes

sin-gular.

Sincethe

metric

containsthe

factor

(1−

2GM

rc2 )

,

itbecom

essingular

at

RS=

2GM

c2

.

b)The

separationbetw

eenA

andB

ispurely

inthe

radialdirection,sothe

properlength

ofasegm

entalong

thepath

joiningthem

isgiven

by

ds2= (

1−2GM

rc2 )

−1

dr2,

so

ds=

dr

√1−

2GM

rc2

.

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The

properdistance

fromA

toB

isobtained

byadding

theproper

lengthsof

allthe

segments

alongthe

path,so

sAB= ∫

rB

rA

dr

√1−

2GM

rc2

.

EX

TE

NSIO

N:The

integrationcan

becarried

outexplicitly.

First

usethe

expressionfor

theSchw

arzschildradius

torew

ritethe

expressionfor

sAB

as

sAB= ∫

rB

rA

√rdr

√r−

RS

.

Then

introducethe

hyperbolictrigonom

etricsubstitution

r=RScosh

2u.

One

thenhas

√r−

RS= √

RSsinh

u

dr=

2RScosh

usinh

udu,

andthe

indefiniteintegral

becomes

∫√rdr

√r−

RS

=2RS ∫

cosh2udu

=RS ∫

(1+cosh

2u)d

u

=RS (

u+

12sinh

2u )

=RS (u

+sinh

ucosh

u)

=RSsinh −

1 (√rRS−

1 )+ √

r(r−RS )

.

Thus,

sAB=RS [sinh −

1 (√rB

RS−

1 )−sinh −

1 (√rA

RS−1 )]

+ √rB(rB−RS )− √

rA(rA −

RS )

.

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c)A

tickofthe

clockand

thefollow

ingtick

aretw

oevents

thatdiffer

onlyin

theirtim

ecoordinates.

Thus,

themetric

reducesto

−c2dτ

2=− (

1−2GM

rc2 )

c2dt2,

so

dτ= √

1−2GM

rc2

dt.

The

readingon

theobserver’s

clockcorresponds

tothe

propertim

einterval

dτ,

sothe

correspondinginterval

ofthe

coordinatetis

givenby

∆tA=

∆τA

√1−

2GM

rAc2

.

d)Since

theSchw

arzschildmetric

doesnot

changewith

time,

eachpulse

leavingA

willtake

thesam

elength

oftim

eto

reachB.Thus,the

pulsesem

ittedby

Awillarrive

atB

with

atim

ecoordinate

spacing

∆tB=

∆tA=

∆τA

√1−

2GM

rAc2

.

The

clockat

B,how

ever,will

readthe

propertim

eand

notthe

coordinatetim

e.Thus,

∆τB= √

1−2GM

rBc2∆tB

=

√√√√1−

2GM

rBc2

1−2GM

rAc2

∆τA.

e)From

parts(a)

and(b),the

properdistance

between

Aand

Bcan

berew

rittenas

sAB= ∫

rB

RS

√rdr

√r−

RS

.

The

potentiallydivergent

partof

theintegral

comes

fromthe

rangeof

inte-gration

inthe

immediate

vicinityofr=R

S ,say

RS<r<RS+ε.

Forthis

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rangethe

quantity √rin

thenum

eratorcan

beapproxim

atedby √

RS ,

sothe

contributionhas

theform√

RS ∫

RS+ε

RS

dr

√r−

RS

.

Changing

theintegration

variabletou≡

r−RS ,the

contributioncan

beeasily

evaluated:√RS ∫

RS+ε

RS

dr

√r−

RS

= √R

S ∫ε

0

du√u=

2 √R

S ε<∞

.

So,although

theintegrand

isinfinite

atr=RS ,

theintegral

isstillfinite.

The

properdistance

between

Aand

Bdoes

notdiverge.

Looking

atthe

answer

topart

(d),how

ever,one

cansee

thatwhen

rA=RS ,

The

timeinterval

∆τB

diverges.

PR

OB

LEM

12:G

EO

DESIC

S

The

geodesicequation

foracurve

xi(λ),

where

theparam

eterλ

isthe

arclength

alongthe

curve,can

bewritten

as

ddλ

gijdxj

dλ

=12(∂i gk )dxk

dλ

dx

dλ

.

Here

theindices

j,k,

and are

summed

from1to

thedim

ensionof

thespace,

sothere

isone

equationfor

eachvalue

ofi.

(a)The

metric

isgiven

by

ds2=gij d

xidxj=dr2+r2dθ2,

sogrr=

1,

gθθ=r2,

grθ=gθr=

0.

First

takingi=r,the

nonvanishingterm

sin

thegeodesic

equationbecom

e

ddλ

grrdr

dλ

=12(∂r gθθ )dθ

dλ

dθ

dλ,

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which

canbe

written

explicitlyas

ddλ

dr

dλ

=12 (∂

r r2 ) (

dθ

dλ )

2

,

or

d2r

dλ

2=r (

dθ

dλ )

2

.

Fori=θ,

onehas

thesim

plificationthat

gij

isindependent

ofθfor

all(i,j).

So

ddλ

r2dθ

dλ

=0.

(b)The

firststep

isto

parameterize

thecurve,

which

means

toim

aginemoving

alongthe

curve,and

expressingthe

coordinatesas

afunction

ofthe

distancetraveled.

(Iam

callingthe

locusy=

1acurve

ratherthan

aline,

sincethe

techniquesthat

areused

hereare

usuallyapplied

tocurves.

Sincealine

isa

specialcaseofa

curve,thereisnothing

wrong

with

treatingthe

lineas

acurve.)

InCartesian

coordinates,the

curvey=

1can

beparam

eterizedas

x(λ)=λ,

y(λ)=

1.

(The

parameterization

isnot

unique,becauseone

canchoose

λ=

0to

representany

pointalong

thecurve.)

Converting

tothe

desiredpolar

coordinates,

r(λ)= √

x2(λ)

+y2(λ)

= √λ

2+

1,

θ(λ)=

tan −1y(λ)x(λ)

=tan −

1(1/λ)

.

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Calculating

theneeded

derivatives,*

dr

dλ=

λ√λ

2+1

d2r

dλ

2=

1√λ

2+1 −

λ2

(λ2+1)

3/2=

1

(λ2+

1)3/2=

1r3

dθ

dλ=−

1

1+ (

1λ )2

1λ2=−

1r2.

Then,

substitutinginto

thegeodesic

equationfor

i=r,

d2r

dλ

2=r (

dθ

dλ )

2⇐⇒1r3=r (−

1r2 )

2

,

which

checks.Substituting

intothe

geodesicequation

fori=θ,

ddλ

r2dθ

dλ

=0⇐⇒

ddλ

r2 (−

1r2 )

=0,

which

alsochecks.

PR

OB

LEM

13:A

NEX

ER

CIS

EIN

TW

O-D

IMEN

SIO

NA

LM

ET

RIC

S(30

points)

(a)Since

r(θ)=

(1+εcos

2θ)r0,

asthe

angularcoordinate

θchanges

bydθ,rchanges

by

dr=

dr

dθdθ=−2εr

0cos

θsin

θdθ.

*If

youdo

notrem

ember

howto

differentiateφ

=tan −

1(z),then

youshould

knowhow

toderive

it.Write

z=

tanφ=

sinφ/cos

φ,so

dz= (

cosφ

cosφ+

sin2φ

cos2φ )

dφ=

(1+tan

2φ)d

φ.

Then

dφdz=

11+tan

2φ=

11+z2.

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ds2is

thengiven

by

ds2=

dr2+r2dθ2

=4ε2r

20cos

2θsin

2θdθ2+(1

+εcos

2θ)

2r20dθ2

= [4ε2cos

2θsin

2θ+(1

+εcos

2θ)

2 ]r20dθ2,

sods=r0 √

4ε2cos

2θsin

2θ+

(1+εcos

2θ)

2dθ.

Sinceθruns

fromθ1to

θ2as

thecurve

issw

eptout,

S=r0 ∫

θ2

θ1 √

4ε2cos

2θsin

2θ+

(1+εcos

2θ)

2dθ.

(b)Since

θdoes

notvary

alongthis

path,

ds= √

1+radr,

andso

R= ∫

r0

0 √1+radr.

(c)Since

themetric

doesnot

containaterm

indrdθ,

therand

θdirections

areorthogonal.

Thus,ifone

considersasm

allregionin

which

risin

theinterval

r ′

tor ′+

dr ′,and

θisin

theinterval

θ ′toθ ′+

dθ ′,then

theregion

canbe

treatedas

arectangle.

The

sidealong

which

rvaries

haslength

dsr= √

1+

(r ′/a)d

r ′,while

theside

alongwhich

θvaries

haslength

dsθ=r ′d

θ ′.The

areais

then

dA

=dsrdsθ=r ′ √

1+

(r ′/a)d

r ′dθ ′.

Tocover

thearea

forwhich

r<r0 ,r ′

must

beintegrated

from0to

r0 ,

andθ ′

must

beintegrated

from0to

2π:

A= ∫

r0

0

dr ′ ∫

2π

0

dθ ′r ′ √

1+

(r ′/a)

.

But

∫2π

0

dθ ′=

2π,

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so

A=

2π ∫

r0

0

dr ′r ′ √

1+

(r ′/a)

.

You

were

notasked

tocarry

outthe

integration,but

itcan

bedone

byusing

thesubstitution

u=

1+

(r ′/a),

sodu=

(1/a)d

r ′,and

r ′=

a(u−1).

The

resultis

A=

4πa2

15 [2+ (

3r20

a2

+r0a−2 ) √

1+r0a ]

.

(d)The

nonzerometric

coefficients

aregiven

by

grr=

1+ra,

gθθ=r2,

sothe

metric

isdiagonal.

Fori=

1=r,

thegeodesic

equationbecom

es

dds

grr dr

ds

=12∂grr

∂r

dr

ds

dr

ds+

12∂gθθ

∂r

dθ

ds

dθ

ds,

soifwesubstitute

thevalues

fromabove,

wehave

dds (1

+ra )

dr

ds

=12∂∂r (1

+ra ) (

dr

ds )

2

+12∂r2

∂r (

dθ

ds )

2

.

Simplifying

slightly,

dds (1

+ra )

dr

ds

=12a (

dr

ds )

2

+r (

dθ

ds )

2

.

The

answer

aboveis

perfectlyacceptable,

butone

might

want

toexpand

theleft-hand

side:

dds (1

+ra )

dr

ds

=1a (

dr

ds )

2

+ (1+ra )

d2r

ds2.

Insertingthis

expansioninto

theboxed

equationabove,

thefirst

termcan

bebrought

tothe

right-handside,

giving

(1+ra )

d2r

ds2=−

12a (

dr

ds )

2

+r (

dθ

ds )

2

.

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The

i=

2=θequation

issim

pler,because

noneof

thegij

coefficients

dependon

θ,sothe

right-handside

ofthe

geodesicequation

vanishes.One

hassim

ply

dds

r2 dθ

ds

=0.

Formost

purposesthis

isthe

bestway

towrite

theequation,since

itleads

im-

mediately

tor2(d

θ/ds)

=con

st.How

ever,it

ispossible

toexpand

thederiva-

tive,giving

thealternative

form

r2 d

2θ

ds2+

2r dr

ds

dθ

ds=

0.

PR

OB

LEM

14:G

EO

DESIC

SO

NT

HE

SU

RFA

CE

OF

ASP

HER

E

(a)Rotations

areeasy

tounderstand

inCartesian

coordinates.The

relationshipbetw

eenthe

polarand

Cartesian

coordinatesis

givenby

x=rsin

θcos

φ

y=rsin

θsin

φ

z=rcos

θ.

The

equatoristhen

describedby

θ=π/2,and

φ=ψ,w

hereψ

isaparam

eterrunning

from0to

2π.Thus,the

equatorisdescribed

bythe

curvexi(ψ

),where

x1=x=rcos

ψ

x2=y=rsin

ψ

x3=z=

0.

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Now

introduceaprim

edcoordinate

systemthat

isrelated

tothe

originalsystemby

arotation

inthe

y-zplane

byan

angleα:

x=x ′

y=y ′cos

α−z ′sin

α

z=z ′cos

α+y ′sin

α.

The

rotatedequator,

which

weseek

todescribe,

isjust

thestandard

equatorin

theprim

edcoordinates:

x ′=rcos

ψ,

y ′=rsin

ψ,

z ′=0.

Using

therelation

between

thetw

ocoordinate

systemsgiven

above,

x=rcos

ψ

y=rsin

ψcos

α

z=rsin

ψsin

α.

Using

againthe

relationsbetw

eenpolar

andCartesian

coordinates,

cosθ=zr=

sinψsin

α

tanφ=

yx=

tanψcos

α.

(b)A

segment

ofthe

equatorcorresponding

toan

intervaldψ

haslength

adψ,so

theparam

eterψ

isproportional

tothe

arclength.

Expressed

interm

sof

themetric,

thisrelationship

becomes

ds2=gijdxi

dψ

dxj

dψdψ

2=a2dψ

2.

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Thus

thequantity

A≡gijdxi

dψ

dxj

dψ

isequal

toa2,

sothe

geodesicequation

(5.50)reduces

tothe

simpler

formof

Eq.(5.52).

(Note

thatweare

following

thenotation

ofLecture

Notes

5,exceptthat

thevariable

usedto

parameterize

thepath

iscalled

ψ,rather

thanλor

s.Although

Aisnot

equalto1as

weassum

edin

Lecture

Notes

5,itiseasily

seenthat

Eq.(5.52)

followsfrom

(5.50)provided

onlythat

A=con

stant.)

Thus,

ddψ

gijdxj

dψ

=12(∂i gk )dxk

dψ

dx

dψ

.

Forthis

problemthe

metric

hasonly

twononzero

components:

gθθ=a2,

gφφ=a2sin

2θ.

Taking

i=θin

thegeodesic

equation,

ddψ

gθθdθ

dψ

=12∂θ gφφdφ

dψ

dφ

dψ

=⇒

d2θ

dψ

2=

sinθcos

θ (dφ

dψ )

2

.

Taking

i=φ,

ddψ

a2sin

2θdφ

dψ

=0

=⇒

ddψ

sin2θdφ

dψ

=0.

(c)This

partis

mainly

algebra.Taking

thederivative

of

cosθ=

sinψsin

α

implies

−sin

θdθ=

cosψsin

αdψ.

Then,

usingthe

trigonometric

identitysin

θ=√1−

cos2θ,

onefinds

sinθ= √

1−sin

2ψsin

2α,

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sodθ

dψ

=−

cosψsin

α√

1−sin

2ψsin

2α.

Similarly

tanφ=

tanψcos

α=⇒

sec2φdφ=

sec2ψdψcos

α.

Then

sec2φ=

tan2φ+1=

tan2ψcos

2α+

1

=1

cos2ψ[sin

2ψcos

2α+

cos2ψ]

=sec

2ψ[sin

2ψ(1−

sin2α)+

cos2ψ]

=sec

2ψ[1−

sin2ψsin

2α],

Sodφ

dψ

=cos

α

1−sin

2ψsin

2α.

Toverify

thegeodesic

equationsof

part(b),

itis

easiestto

checkthe

secondone

first:

sin2θdφ

dψ

=(1−

sin2ψsin

2α)

cosα

1−sin

2ψsin

2α

=cos

α,

soclearly

ddψ

sin2θdφ

dψ

=ddψ(cos

α)=

0.

Toverify

thefirst

geodesicequation

frompart

(b),firstcalculate

theleft-hand

side,d2θ/dψ

2,using

ourresult

fordθ/dψ:

d2θ

dψ

2=

ddψ (

dθ

dψ )

=ddψ −

cosψsin

α√

1−sin

2ψsin

2α

.

After

somestraightforw

ardalgebra,one

finds

d2θ

dψ

2=

sinψsin

αcos

2α

[1−sin

2ψsin

2α ]

3/2.

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The

right-handside

ofthe

firstgeodesic

equationcan

beevaluated

usingthe

expressionfound

abovefor

dφ/dψ,giving

sinθcos

θ (dφ

dψ )

2

= √1−

sin2ψsin

2αsin

ψsin

αcos

2α

[1−sin

2ψsin

2α ]

2

=sin

ψsin

αcos

2α

[1−sin

2ψsin

2α ]

3/2.

Sothe

left-and

right-handsides

areequal.

PR

OB

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15:G

EO

DESIC

SIN

AC

LO

SED

UN

IVER

SE

(a)(7

points)For

purelyradial

motion,

dθ=dφ=

0,so

theline

element

reducesdo

−c2dτ

2=−c2dt2+a2(t)

dr2

1−r2

.

Dividing

bydt2,

−c2 (

dτdt )

2

=−c2+

a2(t)

1−r2 (

drdt )

2

.

Rearranging,

dτdt= √

1−a2(t)

c2(1−

r2) (

drdt )

2

.

(b)(3

points)

dt

dτ=

1dτdt

=1

√1−

a2(t)

c2(1−

r2) (

drdt )

2.

(c)(10

points)During

anyinterval

ofclock

timedt,

theproper

timethat

would

bemeasured

byaclock

moving

with

theob

jectisgiven

bydτ,as

givenby

themetric.

Using

theansw

erfrom

part(a),

dτ=dτdtdt= √

1−a2(t)

c2(1−

r2p ) (

drp

dt )

2

dt.

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Integratingto

findthe

totalpropertim

e,

τ= ∫

t2

t1 √1−

a2(t)

c2(1−

r2p ) (

drp

dt )

2

dt.

(d)(10

points)The

physicaldistanced that

theob

jectmoves

duringagiven

time

intervalisrelated

tothe

coordinatedistance

drby

thespatialpart

ofthemetric:

d 2=ds2=a2(t)

dr2

1−r2

=⇒d =

a(t)√1−

r2dr.

Thus

vphys=d dt=

a(t)√1−

r2

drdt.

Discussion:

Acom

mon

mistake

was

toinclude

−c2dt2in

theexpression

ford 2.

Tounderstand

why

thisis

notcorrect,

weshould

thinkabout

howan

observerwould

measure

d ,the

distanceto

beused

incalculating

thevelocity

ofapassing

object.

The

observerwould

placeameter

stickalong

thepath

oftheob

ject,andshe

would

mark

offthe

positionof

theob

jectat

thebeginning

andend

ofatim

einterval

dtm

eas .

Then

shewould

readthe

distanceby

subtractingthe

tworeadings

onthe

meter

stick.This

subtractionis

equalto

thephysical

distancebetw

eenthe

twomarks,

measured

atthe

sam

etim

et.

Thus,

when

wecom

putethe

distancebetw

eenthe

twomarks,

weset

dt=

0.Tocom

putethe

speedshe

would

thendivide

thedistance

bydtm

eas ,which

isnonzero.

(e)(10

points)Westart

with

thestandard

formula

forageodesic,

aswritten

onthe

frontof

theexam

:ddτ

gµνdxν

dτ

=12(∂µgλσ )

dxλ

dτ

dxσ

dτ

.

This

formula

istrue

foreach

possiblevalue

ofµ,w

hilethe

Einstein

summation

conventionim

pliesthat

theindices

ν,λ,

andσare

summed.

Weare

tryingto

derivethe

equationfor

r,so

weset

µ=r.

Sincethe

metric

isdiagonal,

theonly

contributionon

theleft-hand

sidewill

beν=r.

Onthe

right-handside,

thediagonalnature

ofthemetric

implies

thatnonzero

contributionsarise

onlywhen

λ=

σ.The

termwill

vanishunless

dxλ/dτis

nonzero,so

λmust

be

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eitherror

t(i.e.,

thereis

nomotion

inthe

θor

φdirections).

How

ever,the

right-handside

isproportional

to

∂gλσ

∂r

.

Sincegtt

=−c2,

thederivative

with

respectto

rwill

vanish.Thus,

theonly

nonzerocontribution

onthe

right-handside

arisesfrom

λ=σ=r.

Using

grr=

a2(t)

1−r2,

thegeodesic

equationbecom

es

ddτ

grrdr

dτ

=12(∂r grr )dr

dτ

dr

dτ,

orddτ

a2

1−r2

dr

dτ

=12 [

∂r (

a2

1−r2 )]

dr

dτ

dr

dτ,

orfinally

ddτ

a2

1−r2

dr

dτ

=a2

r

(1−r2)

2 (dr

dτ )

2

.

This

matches

theform

shownin

thequestion,

with

A=

a2

1−r2,and

C=a2

r

(1−r2)

2,

with

B=D

=E

=0.

(f)(5

pointsEXTRA

CREDIT

)The

algebrahere

canget

messy,but

itisnot

toobad

ifone

doesthe

calculationin

aneffi

cientway.

One

goodway

tostart

isto

simplify

theexpression

forp.

Using

theansw

erfrom

(d),

p=

mvphys

√1−

v2phys

c2

=m

a(t)

√1−

r2drdt

√1−

a2

c2(1−

r2) (

drdt )

2.

Using

theansw

erfrom

(b),thissim

plifiesto

p=m

a(t)√1−

r2

drdt

dt

dτ=m

a(t)√1−

r2

dr

dτ.

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Multiply

thegeodesic

equationby

m,and

thenuse

theabove

resultto

rewrite

itas

ddτ

ap

√1−

r2

=ma2

r

(1−r2)

2 (dr

dτ )

2

.

Expanding

theleft-hand

side,

LHS=

ddτ

ap

√1−

r2

=1

√1−

r2

ddτ

ap

+ap

r

(1−r2)

3/2

dr

dτ

=1

√1−

r2

ddτ

ap

+ma2

r

(1−r2)

2 (dr

dτ )

2

.

Insertingthis

expressionback

intoleft-hand

sideof

theoriginal

equation,one

seesthat

thesecond

termcancels

theexpression

onthe

right-handside,leaving

1√1−

r2

ddτ

ap

=0.

Multiplying

by √1−

r2,

onehas

thedesired

result:

ddτ a

p=

0=⇒

p∝1a(t)

.

PR

OB

LEM

16:A

TW

O-D

IMEN

SIO

NA

LC

URV

ED

SPA

CE

(40points)

(a)For

θ=con

stant,the

expressionfor

themetric

reducesto

ds2=

adu

2

4u(a−

u)=⇒

ds=

12 √a

u(a−u)

du.

To

findthe

lengthof

theradial

lineshow

n,one

must

integratethis

expressionfrom

thevalue

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ofuat

thecenter,

which

is0,

tothe

valueofuat

theouter

edge,which

isa.

So

R=

12 ∫a

0 √a

u(a−u)

du.

You

were

notexpected

todo

it,but

theintegral

canbe

carriedout,

givingR

=(π/2) √

a.

(b)For

u=con

stant,the

expressionfor

themetric

reducesto

ds2=udθ2

=⇒ds=√udθ.

Sinceθruns

from0to

2π,and

u=afor

thecircum

fer-ence

ofthe

space,

S= ∫

2π

0

√adθ=

2π √

a.

(c)Toevaluate

theansw

erto

firstorder

indumeans

toneglect

anyterm

sthat

would

beproportional

todu

2

orhigher

powers.

This

means

thatwecan

treatthe

annulusas

ifit

were

arbitrarilythin,

inwhich

casewecan

imagine

bendingit

intoarectangle

without

changingits

area.The

areais

thenequal

tothe

cir-cum

ferencetim

esthe

width.

Both

thecircum

ferenceand

thewidth

must

becalculated

byusing

themetric:

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dA

=circum

ference×width

=[2π √

u0]× [

12 √a

u0 (a−

u0 )

du ]

=π √

a

(a−u

0 )du.

(d)Wecan

findthe

totalarea

byim

aginingthat

itis

brokenup

intoannuluses,

where

asingle

annulusstarts

atradial

coordinateuand

extendsto

u+

du.

Asin

part(a),

thisexpression

must

beintegrated

fromthe

valueofuat

thecenter,

which

is0,to

thevalue

ofuat

theouter

edge,which

isa.

A=π ∫

a

0 √a

(a−u)

du.

You

didnot

needto

carryout

thisintegration,

butthe

answer

would

beA

=2πa.

(e)From

thelist

atthe

frontof

theexam

,the

generalform

ulafor

ageodesic

iswritten

asdds [

gijdxj

ds ]

=12∂gk

∂xi

dxk

ds

dx

ds.

The

metric

components

gij

arerelated

tods2by

ds2=gij d

xidxj,

where

theEinstein

summation

convention(sum

overrepeated

indices)is

as-sum

ed.In

thiscase

g11 ≡

guu=

a

4u(a−

u)

g22 ≡

gθθ=u

g12=g21=

0,

where

Ihave

chosenx

1=uand

x2=θ.

The

equationwith

du/dson

theleft-

handside

isfound

bylooking

atthe

geodesicequations

fori=

1.Ofcourse

j,k,

and must

allbe

summed,

butthe

onlynonzero

contributionsarise

when

j=

1,and

kand

are

eitherboth

equalto

1or

bothequal

to2:

dds [

guudu

ds ]

=12∂guu

∂u (

du

ds )

2

+12∂gθθ

∂u (

dθ

ds )

2

.

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dds [

a

4u(a−

u)du

ds ]

=12 [

ddu (

a

4u(a−

u) )](du

ds )

2

+12 [

ddu(u) ](

dθ

ds )

2

=12 [

a

4u(a−

u)2 −

a

4u

2(a−u) ](

du

ds )

2

+12 (

dθ

ds )

2

=18a(2

u−a)

u2(a−

u)2 (

du

ds )

2

+12 (

dθ

ds )

2

.

(f)This

partissolved

bythe

samemethod,but

itissim

pler.Here

weconsider

thegeodesic

equationwith

i=

2.The

onlyterm

thatcontributes

onthe

left-handside

isj=

2.Onthe

right-handside

onefinds

nontrivialexpressions

when

kand

are

eitherboth

equalto

1or

bothequal

to2.

How

ever,the

termson

theright-hand

sideboth

involvethe

derivativeof

themetric

with

respectto

x2=θ,and

thesederivatives

allvanish.

So

dds [

gθθ dθ

ds ]

=12∂guu

∂θ (

du

ds )

2

+12∂gθθ

∂θ (

dθ

ds )

2

,

which

reducesto

dds [

udθ

ds ]

=0.

PR

OB

LEM

17:R

OTA

TIN

GFR

AM

ES

OF

REFER

EN

CE

(35points)

(a)The

metric

was

givenas

−c2dτ

2=−c2dt2+ [d

r2+r2(dφ+ωdt)

2+

dz2 ]

,

andthe

metric

coefficients

arethen

justread

offfrom

thisexpression:

g11 ≡

grr=

1

g00 ≡

gtt=

coefficient

ofdt2=−c2+r2ω

2

g20 ≡

g02 ≡

gφt ≡

gtφ

=12 ×

coefficient

ofdφdt=r2ω

2

g22 ≡

gφφ=

coefficient

ofdφ

2=r2

g33 ≡

gzz=

coefficient

ofdz2=

1.

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Note

thatthe

off-diagonalterm

gφtmust

bemultiplied

by1/2,

becausethe

expression3∑µ=

0

3∑ν=

0

gµνdxµdxν

includesthe

twoequal

termsg20 dφdt+

g02 dtdφ,

where

g20 ≡

g02 .

(b)Starting

with

thegeneral

expression

ddτ

gµνdxν

dτ

=12(∂µgλσ )

dxλ

dτ

dxσ

dτ

,

weset

µ=r:

ddτ

grνdxν

dτ

=12(∂r gλσ )

dxλ

dτ

dxσ

dτ

.

When

wesum

overνon

theleft-hand

side,the

onlyvalue

forwhich

grν =

0is

ν=

1≡r.

Thus,

theleft-hand

sideis

simply

LHS=

ddτ (

grr dx

1

dτ )

=ddτ (

dr

dτ )

=d

2r

dτ

2.

The

RHSincludes

everycom

binationofλand

σfor

which

gλσdepends

onr,

sothat

∂rgλσ =

0.This

means

gtt ,

gφφ ,

andgφt .

So,

RHS=

12∂r (−

c2+r2ω

2) (dt

dτ )

2

+12∂r (r

2) (dφ

dτ )

2

+∂r (r

2ω)dφ

dτ

dt

dτ

=rω

2 (dt

dτ )

2

+r (

dφ

dτ )

2

+2rω

dφ

dτ

dt

dτ

=r (

dφ

dτ+ω

dt

dτ )

2

.

Note

thatthe

finalterm

inthe

firstline

isreally

thesum

ofthe

contributionsfrom

gφtand

gtφ ,

where

thetw

oterm

swere

combined

tocancel

thefactor

of1/2

inthe

generalexpression.

Finally,

d2r

dτ

2=r (

dφ

dτ+ω

dt

dτ )

2

.

Ifone

expandsthe

RHSas

d2r

dτ

2=r (

dφ

dτ )

2

+rω

2 (dt

dτ )

2

+2rω

dφ

dτ

dt

dτ,

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thenone

canidentify

theterm

proportionaltoω

2as

thecentrifugal

force,andthe

termproportional

toωas

theCoriolis

force.

(c)Substituting

µ=φ,

ddτ

gφνdxν

dτ

=12(∂φgλσ )

dxλ

dτ

dxσ

dτ

.

But

noneof

themetric

coefficients

dependon

φ,sothe

right-handside

iszero.

The

left-handside

receivescontributions

fromν=φand

ν=t:

ddτ (

gφφdφ

dτ+gφtdt

dτ )

=ddτ (

r2dφ

dτ+r2ω

dt

dτ )

=0,

so

ddτ (

r2dφ

dτ+r2ω

dt

dτ )

=0.

Note

thatone

cannot“factor

out”r2,since

rcan

dependon

τ.Ifthis

equationisexpanded

togive

anequation

ford

2φ/dτ

2,theterm

proportionaltoωwould

beidentified

asthe

Coriolis

force.There

isno

termproportional

toω

2,since

thecentrifugal

forcehas

nocom

ponentin

theφdirection.

(d)If

Eq.(P

17.1)of

theproblem

isdivided

byc2dt2,

oneobtains

(dτdt )

2

=1−

1c2 [(

drdt )

2

+r2 (

dφdt+ω )

2

+ (dzdt )

2 ].

Then

usingdt

dτ=

1(

dτ

dt )

,

onehas

dt

dτ=

1√√√√

1−1c2 [(

drdt )

2

+r2 (

dφdt+ω )

2

+ (dzdt )

2 ].

Note

thatthis

equationis

reallyjust

dt

dτ=

1√

1−v2/c2,

adaptedto

therotating

cylindricalcoordinatesystem

.

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HE

STA

BIL

ITY

OF

SC

HW

AR

ZSC

HIL

DO

RB

ITS ∗

(30points)

Fromthe

metric:

ds2=−c2dτ

2=−h(r)

c2dt2+h(r) −

1dr2+r2dθ2+r2sin

2θdφ

2,

(S18.1)

andthe

conventionds2=gµνdxµdxνweread

thenonvanishing

metric

components:

gtt=−h(r)c

2,grr=

1h(r)

,gθθ=r2,gφφ=r2sin

2θ.

(S18.2)

Weare

toldthat

theorbit

hasθ=π/2,

soon

theorbit

dθ=

0and

therelevant

metric

andmetric

components

are:

ds2=−c2dτ

2=−h(r)

c2dt2+h(r) −

1dr2+r2dφ

2,

(S18.3)

gtt=−h(r)c

2,grr=

1h(r)

,gφφ=r2.

(S18.4)

Wealso

knowthat

h(r)=

1−RSr.

(S18.5)

(a)The

geodesicequationdd

τ [gµνdxν

dτ ]

=12∂gλσ

∂xµ

dxλ

dτ

dxσ

dτ,

(S18.6)

forthe

indexvalue

µ=rtakes

theform

ddτ [

grrdr

dτ ]

=12∂gλσ

∂r

dxλ

dτ

dxσ

dτ

.

Expanding

out

ddτ [

1h

dr

dτ ]

=12∂gtt

∂r (

dt

dτ )

2

+12∂grr

∂r (

dr

dτ )

2

+12∂gφφ

∂r (

dφ

dτ )

2

.

Using

thevalues

in(S18.4)

toevaluate

theright-hand

sideand

takingthe

derivativeson

theleft-hand

side:

−h ′

h2 (

dr

dτ )

2

+1h

d2r

dτ

2=−12c2h ′ (

dt

dτ )

2−12h ′

h2 (

dr

dτ )

2

+r (

dφ

dτ )

2

.

*Solution

byBarton

Zwiebach.

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Here

h ′≡dhdrand

wehave

supressedthe

arguments

ofhand

h ′to

avoidclutter.

Collecting

theunderlined

termsto

theright

andmultiplying

byh,

wefind

d2r

dτ

2=−12h ′h

c2 (

dt

dτ )

2

+12h ′h (

dr

dτ )

2

+rh (

dφ

dτ )

2

.(S18.7)

(b)Dividing

theexpression

(S18.3)for

themetric

bydτ

2wereadily

find

−c2=−hc2 (

dt

dτ )

2

+1h (

dr

dτ )

2

+r2 (

dφ

dτ )

2

,

andrearranging,

hc2 (

dt

dτ )

2

=c2+

1h (dr

dτ )

2

+r2 (

dφ

dτ )

2

.(S18.8)

This

isthe

most

usefulform

ofthe

answer.

Ofcourse,

wealso

have

(dt

dτ )

2

=1h+

1h

2c2 (

dr

dτ )

2

+r2

hc2 (

dφ

dτ )

2

.(S18.9)

Weuse

now(S18.8)

tosim

plify(S18.7):

d2r

dτ

2=−12h ′ (

c2+

1h (dr

dτ )

2

+r2 (

dφ

dτ )

2 )+

12h ′h (

dr

dτ )

2

+rh (

dφ

dτ )

2

.

Expanding

out,the

termswith

(drdτ )

2cancel

andwefind

d2r

dτ

2=−12h ′c

2+ (

rh−12h ′r

2 )(dφ

dτ )

2

.(S18.10)

This

isan

acceptableansw

er.One

cansim

plify(S18.10)

furtherby

notingthat

h ′=RS/r2and

rh=r−

RS :

d2r

dτ

2=−12RSc2

r2

+ (r−

32RS )(

dφ

dτ )

2

.(S18.11)

Inthe

notationof

theproblem

statement,

wehave

f0 (r)

=−12RSc2

r2

,f1 (r)

=r−

32RS.

(S18.12)

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(c)The

geodesicequation

(S18.6)for

µ=φgives

ddτ [

gφφdφ

dτ ]

=12∂gλσ

∂φ

dxλ

dτ

dxσ

dτ.

Sinceno

metric

component

dependson

φ,the

right-handside

vanishesand

weget:

ddτ [

r2dφ

dτ ]

=0→

ddτL=

0,

where

L≡r2dφ

dτ.

(S18.13)

The

quantityL

isaconstant

ofthe

motion,nam

ely,it

isanum

berindependent

ofτ.(d)

Using

(S18.13)the

second-orderdifferentialequation

(S18.11)for

r(τ)takes

theform

statedin

theproblem

:d2r

dτ

2=f0 (r)+

f1 (r)r4

L2≡

H(r)

,(S18.14)

where

wehave

introducedthe

functionH(r)

(recallthat

Lis

aconstant!).

The

differentialequation

thentakes

theform

d2r

dτ

2=H(r)

.(S18.15)

Sinceweare

toldthat

acircular

orbitwith

radiusr0exists,

thefunction

r(τ)=r0

must

solvethis

equation.Being

theconstant

function,the

left-handside

vanishesand,

consequently,the

right-handside

must

alsovanish:

H(r

0 )=f0 (r

0 )+f1 (r

0 )r40

L2=

0.

(S18.16)

Toinvestigate

stabilityweconsider

asm

allperturbationδr(τ)

ofthe

orbit:

r(τ)=r0+δr(τ)

,with

δr(τ)r0at

someinitial

τ.

Substitutingthis

into(S18.15)

weget,to

firstnontrivialapproxim

ation

d2δr

dτ

2=H(r

0+δr)

H(r

0 )+δrH

′(r0 )

=δrH

′(r0 ),

where

H′(r)

=dH

(r)

dr

andweused

H(r

0 )=

0from

(S18.16).The

resultingequation

d2δr(τ)dτ

2=H

′(r0 )δr(τ)

,(S18.17)

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isfam

iliarbecause

H′(r

0 )is

justanum

ber.The

conditionof

stabilityis

thatthis

number

isnegative:

H′(r

0 )<

0.Indeed,

inthis

case(S18.17)

isthe

harmonic

oscillatorequation

d2x

dt2=−ω

2x,

with

replacements

x↔δr,

t↔τ,−ω

2↔H

′(r0 ),

andthe

solutiondescribes

boundedoscillations.

Sostability

requires:

StabilityCondition:

H′(r

0 )=

ddr [

f0 (r)

+f1 (r)r4

L2 ]

r=r0

<0.

(S18.18)

This

isthe

answer

topart

(d).−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−

Forstudents

interestedin

gettingthe

famous

resultthat

orbitsare

stablefor

r>

3RS

wecom

pletethis

partof

theanalysis

below.

First

weevaluate

H′(r

0 )in

(S18.18)using

thevalues

off0and

f1in

(S18.12):

H′(r

0 )=

ddr [−

12RSc2

r2

+ (1r3 −

3RS

2r4 )

L2 ]

r=r0

=RSc2

r30

−3L

2

r50

(r0 −

2RS ).

The

inequalityin

(S18.18)then

givesus

RSc2−

3L

2

r20

(r0 −

2RS )

<0,

(S18.19)

where

wemultiplied

byr30>

0.Tocom

pletethe

calculationweneed

thevalue

ofL

2for

theorbit

with

radiusr0 .

This

valueisdeterm

inedby

thevanishing

ofH(r

0 ):

−12RSc2

r20

+(r

0 −32RS )L

2

r40

=0→

L2

r20

=12

RSc2

(r0 −

32RS )

.

Note,

incidentally,that

theequality

tothe

rightdem

andsthat

foracircular

orbitr0>

32RS .

Substitutingthe

abovevalue

ofL

2/r20in

(S18.19)weget:

RSc2−

32RSc2

(r0 −

32RS ) (r

0 −2RS )

<0.

Cancelling

thecom

mon

factorsofRSc2wefind

1−32(r

0 −2RS )

(r0 −

32RS )

<0,

which

isequivalent

to32(r

0 −2RS )

(r0 −

32RS )

>1.

Forr0>

32RS ,

weget

3(r0 −

2RS )

>2(r

0 −32RS )

→r0>

3RS.

(S18.20)

This

isthe

desiredcondition

forstable

orbitsin

theSchw

arzschildgeom

etry.

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19:P

RESSU

RE

AN

DEN

ER

GY

DEN

SIT

YO

FM

YST

E-

RIO

US

ST

UFF

(a)Ifu∝

1/ √

V,then

onecan

write

u(V+∆V)=u

0 √V

V+

∆V

.

(The

aboveexpression

isproportional

to1/ √

V+

∆V,and

reducesto

u=u

0

when

∆V

=0.)

Expanding

tofirst

orderin

∆V,

u=

u0

√1+

∆VV

=u

0

1+

12∆VV

=u

0 (1−

12∆VV )

.

The

totalenergyis

theenergy

densitytim

esthe

volume,

so

U=u(V

+∆V)=u

0 (1−

12∆VV )

V (1+

∆VV )

=U

0 (1+

12∆VV )

,

where

U0=u

0 V.Then

∆U

=12∆VVU

0.

(b)The

work

doneby

theagent

must

bethe

negativeofthe

work

doneby

thegas,

which

isp∆V.So

∆W

=−p∆V

.

(c)The

agentmust

supplythe

fullchange

inenergy,

so

∆W

=∆U

=12∆VVU

0.

Com

biningthis

with

theexpression

for∆W

frompart

(b),onesees

immediately

that

p=−12U

0

V=

−12u

0.