What’s new - 2007: Open questions, expository articles, and lecture series from a mathematical blog

270 Pages·2008·0.73 MB·English

by  Terence Tao

190 CHAPTER2. EXPOSITORYARTICLES
butwewillnotneedtodosohere.
2.17.3 Dopplershift
Theformula(2.62)isalreadyenoughtorecovertherelativisticDopplershiftformula
(tosecondorderinv)forradiationmovingatspeedcwithsomewavenumberk.Math-
ematically, suchradiationmovingtotherightinaninertialreferenceframeF canbe
modeledbythefunction
Acos(k(x−ct)+θ)
forsomeamplitudeAandphaseshiftθ.Ifwemovetothecoordinates(t0,x0)=L (t,x)
v
providedbyaninertialreferenceframeF0,acomputationthenshowsthatthefunction
becomes
Acos(k (x0−ct0)+θ)
+
where k =(1−v/c+v2/2c2+O(v3))k. (actually, if the radiation is tensor-valued,
+
theamplitudeAmightalsotransforminsomemanner,butthistransformationwillnot
beofrelevancetous.) Similarly,radiationmovingatspeedctotheleftwilltransform
from
Acos(k(x+ct)+θ)
to
Acos(k (x+ct)+θ)
−
wherek =(1+v/c+v2/2c2+O(v3))k.Thisdescribeshowthewavenumberktrans-
−
formsunderchangesofreferenceframebysmallvelocitiesv. Thetemporalfrequency
ν islinearlyrelatedtothewavenumberkbytheformula
c
ν = k, (2.63)
2π
andsothisfrequencytransformsbythe(red-shift)formula
ν =(1−v/c+v2/2c2+O(v3))ν (2.64)
+
forright-wardmovingradiationandbythe(blue-shift)formula
ν =(1+v/c+v2/2c2+O(v3))ν (2.65)
−
forleft-wardmovingradiation. (Asbefore,onecangiveanexactformulahere,butthe
aboveasymptoticwillsufficeforus.)
2.17.4 Energyandmomentumofphotons
From the work of Planck, and of Einstein himself on the photoelectric effect, it was
knownthatlightcouldbeviewedbothasaformofradiation(movingatspeedc),and
also made up of particles (photons). From Planck’s law, each photon has an energy
of E =hν and (from de Broglie’s law) a momentum of p=±h¯k=± h k, where h
2π
2.17. EINSTEIN’SDERIVATIONOFE=MC2 191
is Planck’s constant, and the sign depends on whether one is moving rightward or
leftward. Inparticular,from(2.63)wehavethepleasantrelationship
E=|p|c (2.66)
forphotons.[Moregenerally,itturnsoutthatforarbitrarybodies,momentum,velocity,
and energy are related by the formula p= 1Ev, though we will not derive this fact
c2
here.] Applying (2.64), (2.65), we see that if we view a photon in a new reference
frameF,thentheobservedenergyE andmomentum pnowbecome
v
E =(1−v/c+v2/2c2+O(v3))E; p =(1−v/c+v2/2c2+O(v3))p (2.67)
+ +
forright-wardmovingphotons,and
E =(1+v/c+v2/2c2+O(v3))E; p =(1+v/c+v2/2c2+O(v3))p (2.68)
− −
forleft-wardmovingphotons.
Thesetwoformulae(2.67),(2.68)canbeunifiedusing(2.66)intoasingleformula
(E0/c2,p0)=L (E/c2,p)+O(v3) (2.69)
v
foranyphoton(movingeitherleftwardorrightward)withenergyE andmomentum p
asmeasuredinframeF,andenergyE0andmomentum p0asmeasuredinframeF.
v
Remark2.64. Actually,theerrortermO(v3)canbedeletedentirelybyworkingalittle
harder. From the linearity of L and the conservation of energy and momentum, it is
v
thennaturaltoconcludethat(2.69)shouldalsobevalidnotonlyforphotons, butfor
anyobjectthatcanexchangeenergyandmomentumwithphotons. Thiscanbeusedto
derivetheformulaE=mc2fairlyquickly,butletusinsteadgivetheoriginalargument
ofEinstein,whichisonlyslightlydifferent.
2.17.5 Einstein’sargument
WearenowreadytogiveEinstein’sargument. Considerabodyatrestinareference
frameFwithsomemassmandsomerestenergyE.(WedonotyetknowthatEisequal
tomc2.) NowletusviewthissamemassinsomenewreferenceframeF,wherevisa
v
smallvelocity. FromNewtonianmechanics,weknowthatabodyofmassmmovingat
velocityvacquiresakineticenergyof 1mv2. Thus,assumingthatNewtonianphysics
2
isvalidatlowvelocitiestotoporder,thenetenergyE0 ofthisbodyasviewedinthis
frameF shouldbe
v
1
E0=E+ mv2+O(v3). (2.70)
2
Remark2.65. Ifassumesthatthetransformationlaw(2.69)appliesforthisbody,one
can already deduce the formula E =mc2 for this body at rest from (2.70) (and the
assumptionthatbodiesatresthavezeromomentum),butletusinsteadgiveEinstein’s
originalargument.
192 CHAPTER2. EXPOSITORYARTICLES
WereturntoframeF,andassumethatourbodyemitstwophotonsofequalenergy
∆E/2,onemovingleft-wardandonemovingright-ward. By(2.66)andconservation
ofmomentum,weseethatthebodyremainsatrestafterthisemission.Byconservation
ofenergy,theremainingenergyinthebodyisE−∆E. Let’ssaythatthenewmassin
thebodyism−∆m. Ourtaskistoshowthat∆E=∆mc2.
To do this, we return to frame F. By (2.67), the rightward moving photon has
v
energy
∆E
(1−v/c+v2/2c2+O(v3)) ; (2.71)
2
inthisframe;similarly,theleftwardmovingphotonhasenergy
∆E
(1+v/c+v2/2c2+O(v3)) . (2.72)
2
Whataboutthebody? Byrepeatingthederivationof(2.69),itmusthaveenergy
1
(E−∆E)+ (m−∆m)v2+O(v3). (2.73)
2
Bytheprincipleofrelativity,thelawofconservationofenergyhastoholdintheframe
F aswellasintheframeF.Thus,theenergy(2.71)+(2.72)+(2.73)inframeF after
v v
theemissionmustequaltheenergyE0=(2.70)inframeF beforeemission. Adding
v
everythingtogetherandcomparingcoefficientsweobtainthedesiredrelationship∆E=
∆mc2.
Remark2.66. OnemightquibblethatEinstein’sargumentonlyappliestoemissionsof
energythatconsistofequalandoppositepairsofphotons.Butonecaneasilygeneralise
theargumenttohandlearbitraryphotonemissions,especiallyifonetakesadvantageof
(2.69); forinstance, anotherwell-known(andsomewhatsimpler)variantoftheargu-
mentworksbyconsideringaphotonemittedfromonesideofaboxandabsorbedon
theother. Moregenerally,anyotherenergyemissionwhichcouldpotentiallyinthefu-
turedecomposeentirelyintophotonswouldalsobehandledbythisargument,thanks
to conservation of energy. Now, it is possible that other conservation laws prevent
decomposition into photons; for instance, the law of conservation of charge prevents
anelectron(say)fromdecomposingentirelyintophotons, thusleavingopenthepos-
sibility of having to add a linearly charge-dependent correction term to the formula
E =mc2. But then one can renormalise away this term by redefining the energy to
subtract such a term; note that this does not affect conservation of energy, thanks to
conservationofcharge.
2.17.6 Notes
ThisarticlewasoriginallypostedonDec28,2007at
terrytao.wordpress.com/2007/12/28
LaurensGunnarsenpointedoutthatEinstein’sargumentrequiredtheuseofquan-
tum mechanics to derive the equation E = mc2, but that this equation can also be
derived within the framework of classical mechanics by relying more heavily on the
representationtheoryoftheLorentzgroup.
ThankstoBlakeStaceyforcorrections.
Chapter 3
Lectures
193
194 CHAPTER3. LECTURES
3.1 Simons Lecture Series: Structure and randomness
OnApr5-7, 2007, IgaveoneoftheSimonsLectureSeriesatMIT(theotherlecture
serieswasgivenbyDavidDonoho). Igavethreelectures, eachexpoundingonsome
aspectsofthetheme“thedichotomybetweenstructureandrandomness”(seealsomy
ICM talk [Ta2006], [Ta2006a] on this topic). This theme seems to pervade many of
theareasofmathematicsthatIworkin,andmylecturesaimtoexplorehowthistheme
manifestsitselfinseveralofthese.Inthefirstlecture,Idescribethedichotomyasitap-
pearsinFourieranalysisandinnumbertheory. Inthesecond,Idiscussthedichotomy
inergodictheoryandgraphtheory,whileinthethird,IdiscussPDE.)
3.1.1 Structure and randomness in Fourier analysis and number
theory
The“dichotomybetweenstructureandrandomness”seemstoapplyincircumstances
inwhichoneisconsideringa“high-dimensional”classofobjects(e.g. setsofintegers,
functionsonaspace,dynamicalsystems,graphs,solutionstoPDE,etc.). Forsakeof
concreteness, let us focus today on sets of integers (later lectures will focus on other
classesofobjects). Therearemanydifferenttypesofobjectsintheseclasses,however
onecanbroadlyclassifythemintothreecategories:
• Structured objects - objects with a high degree of predictability and algebraic
structure. AtypicalexamplearetheoddintegersA:={...,−3,−1,1,3,5,...}.
NotethatifsomelargenumbernisknowntolieinA,thisrevealsalotofinfor-
mationaboutwhethern+1,n+2,etc. willalsolieinA.Structuredobjectsare
beststudiedusingthetoolsofalgebraandgeometry.
• Pseudorandomobjects-theoppositeofstructured;thesearehighlyunpredictable
andtotallylackanyalgebraicstructure. Agoodexampleisarandomlychosen
setBofintegers,inwhicheachelementnliesinBwithanindependentproba-
bilityof1/2. (Onecanimagineflippingacoinforeachintegern,anddefining
Btobethesetofnforwhichthecoinflipresultedinheads.) Notethatifsome
integernisknowntolieinB,thisconveysnoinformationwhatsoeveraboutthe
relationship of n+1, n+2, etc. with respect to B. Pseudorandom objects are
beststudiedusingthetoolsofanalysisandprobability.
• Hybridsets-setswhichexhibitsomefeaturesofstructureandsomefeaturesof
pseudorandomness. AgoodexampleistheprimesP:=2,3,5,7,.... Theprimes
have some obvious structure in them: for instance, the prime numbers are all
positive,theyareallodd(withoneexception),theyarealladjacenttoamultiple
ofsix(withtwoexceptions),andtheirlastdigitisalways1,3,7,or9(withtwo
exceptions). Ontheotherhand,thereisevidencethattheprimes,despitebeing
adeterministicset,behaveinavery“pseudorandom”or“uniformlydistributed”
manner.Forinstance,fromtheprimenumbertheoreminarithmeticprogressions
weknowthatthelastdigitsoflargeprimenumbersareuniformlydistributedin
theset{1,3,7,9};thus,ifN isalargeinteger,thenumberofprimeslessthanN
ending in (say) 3, divided by the total number of primes less than N, is known
3.1. SIMONSLECTURESERIES:STRUCTUREANDRANDOMNESS 195
to converge to 1/4 in the limit as N goes to infinity. In order to study hybrid
objects, oneneedsalargevarietyoftools: oneneedstoolssuchasalgebraand
geometrytounderstandthestructuredcomponent,oneneedstoolssuchasanal-
ysisandprobabilitytounderstandthepseudorandomcomponent,andoneneeds
tools such as decompositions, algorithms, and evolution equations to separate
thestructurefromthepseudorandomness.
A recurring question in many areas of analysis is the following: given a specific
object (such as the prime numbers), can one determine precisely what the structured
componentsarewithintheobject,andhowpseudorandomtheremainingcomponents
oftheobjectare?Onereasonforaskingthisquestionisthatitoftenhelpsonecompute
various statistics (averages, sums, integrals, correlations, norms, etc.) of the object
beingstudied. Forinstance,onecanaskforhowmanytwinpairs{n,n+2},withnbe-
tween1andN,onecanfindwithinagivenset. InthestructuredsetAgivenabove,the
answerisroughlyN/2. FortherandomsetBgivenabove,theanswerisroughlyN/4;
thus one sees that while A and B have exactly the same density (namely, 1/2), their
statistics are rather different due to the fact that one is structured and one is random.
As for the prime numbers, nobody knows for certain what the answer is (although
the Hardy-Littlewood prime tuples conjecture [HaLi1923] predicts the answer to be
roughly1.32 N ),becausewedonotknowenoughyetaboutthepseudorandomness
log2N
oftheprimes.Ontheotherhand,theparitystructureoftheprimenumbersisenoughto
showthatthenumberofadjacentpairs{n,n+1}intheprimesisexactlyone: {2,3}.
Theproblemofdeterminingexactlywhatthestructuredandpseudorandomcompo-
nentsareofanygivenobjectisstilllargelyintractable. However,whatwehavelearnt
inmanycasesisthatwecanatleastshowthatanarbitraryobjectcanbedecomposed
into some structured component and some pseudorandom component. Also there is
oftenanorthogonalityproperty(ordichotomy):ifanobjectisorthogonal(orhassmall
correlationwith)allstructuredobjects, then it isnecessarilypseudorandom, andvice
versa. Finally,wearesometimesluckyenoughtobeabletoclassifyallthestructured
objectswhicharerelevantforanygivenproblem(e.g.computingaparticularstatistic).
Insuchcases,onemerelyneeds(inprinciple)tocomputehowthegivenobjectcorre-
lates with each member in one’s list of structured objects in order to determine what
thedesiredstatisticis. Thisisoftensimpler(thoughstillnon-trivial)thancomputing
thestatisticdirectly.
To illustrate these general principles, let us focus now on a specific area in an-
alytic number theory, namely that of finding additive patterns in the prime numbers
{2,3,5,7,...}. Despite centuries of progress on these problems, many questions are
stillunsolved,forinstance:
• (Twin prime conjecture) There are infinitely many positive integers n such that
n,n+2arebothprime.
• (SophieGermainprimeconjecture)Thereareinfinitelymanypositiveintegersn
suchthatn,2n+1arebothprime.
• (Even Goldbach conjecture) For every even number N ≥4, there is a natural
numbernsuchthatn,N−narebothprime.
196 CHAPTER3. LECTURES
Ontheotherhand,wedohavesomepositiveresults:
• (Vinogradov’stheorem)[Vi1937]ForeverysufficientlylargeoddnumberN,there
arepositiveintegersn,n0 suchthatn,n0,N−n−n0 areallprime. (Thebestex-
plicitboundcurrentlyknownfor“sufficientlylarge”isN≥101346 [LiWa2002];
theresulthasalsobeenverified¡for7≤N≤1020[Sa1998].
• (vanderCorput’stheorem)[vdC1939]Thereareinfinitelymanypositiveintegers
n,n0suchthatn,n+n0,n+2n0areallprime.
• (Green-Tao theorem)[GrTa2008] For any positive integer k, there are infinitely
manypositiveintegersn,n0suchthatn,n+n0,...,n+(k−1)n0areallprime.
• (Apolynomialgeneralisation)Foranyinteger-valuedpolynomialsP(n),...,P(n)
1 k
withP(0)=...=P(0)=0,thereareinfinitelymanypositiveintegersn,n0such
1 k
thatn+P(n0),...,n+P(n0)areallprime.
1 k
Asageneralrule,itappearsthatitisfeasible(afternon-trivialeffort)tofindpatterns
intheprimesinvolvingtwoormoredegreesoffreedom(asdescribedbytheparameters
n, n0 in above examples), but we still do not have the proper technology for finding
patterns in the primes involving only one degree of freedom n. (This is of course
an oversimplification; for instance, the pattern n, n+2, n0, n0+2 has two degrees of
freedom,butfindinginfinitelymanyofthesepatternsintheprimesisequivalenttothe
twin prime conjecture, and thus presumably beyond current technology. If however
onemakesanon-degeneracyassumption,onecanmaketheaboveclaimmoreprecise;
see[GrTa2008b].)
One useful tool for establishing some (but not all) of the above positive results is
Fourier analysis (which in this context is also known as the Hardy-Littlewood circle
method). Ratherthangivethetextbookpresentationofthatmethodhere,letustryto
motivatewhyFourieranalysisisanessentialfeatureofmanyoftheseproblemsfrom
theperspectiveofthedichotomybetweenstructureandrandomness,andinparticular
viewing structure as an obstruction to computing statistics which needs to be under-
stoodbeforethestatisticcanbeaccuratelycomputed.
Totreatmanyoftheabovequestionsconcerningtheprimesinaunifiedmanner,let
usconsiderthefollowinggeneralsetting.Weconsiderkaffine-linearformsψ ,...,ψ :
1 k
Zr→Zonrintegerunknowns,andask
Question3.1. Doestherethereexistinfinitelymanyr-tuples~n=(n ,...,n )∈Zr of
1 r +
positiveintegerssuchthatψ (~n),...,ψ (~n)aresimulatenouslyprime?
1 k
For instance, the twin prime conjecture is the case when k=2, r =1, ψ (n)=
1
n, and ψ (n)=n+2; van der Corput’s theorem is the case when k=3, r=2, and
2
ψ (n,n0)=n+(j−1)n0for j=0,1,2;andsoforth.
j
Becauseofthe“obvious”structuresintheprimes,theanswertotheabovequestion
can be “no”. For instance, since all but one of the primes are odd, we know that
there are not infinitely many patterns of the form n,n+1 in the primes, because it
is not possible for n,n+1 to both be odd. More generally, given any prime q, we
know that all but one of the primes is coprime to q. Hence, if it is not possible for
3.1. SIMONSLECTURESERIES:STRUCTUREANDRANDOMNESS 197
ψ (~n),...,ψ (~n) to all be coprime to q, the answer to the above question is basically
1 k
no(modulosometechnicalitieswhichIwishtoglossover)andwesaythatthereisan
obstructionatq. Forinstance,thepatternn,n+1hasanobstructionat2. Thepattern
n,n+2,n+4 has no obstruction at 2, but has an obstruction at 3, because it is not
possibleforn,n+2,n+4toallbecoprimeto3. Andsoforth.
Anotherobstructioncomesfromthetrivialobservationthattheprimesareallpos-
itive. Hence, if it is not possible for ψ (~n),...,ψ (~n) to all be positive for infinitely
1 k
manyvaluesof~n,thenwesaythatthereisanobstructionatinfinity,andtheanswerto
thequestionisagain“no”inthiscase.Forinstance,foranyfixedN,thepatternn,N−n
canonlyoccurfinitelyoftenintheprimes,becausethereareonlyfinitelymanynfor
whichn,N−narebothpositive.
It is conjectured that these “local” obstructions are the only obstructions to solv-
abilityoftheabovequestion. Moreprecisely,wehave
Conjecture 3.2. (Dickson’s conjecture)[Di1904] If there are no obstructions at any
primeq,andtherearenoobstructionsatinfinity,thentheanswertotheabovequestion
is“yes”.
This conjecture would imply the twin prime and Sophie Germain conjectures, as
well as the Green-Tao theorem; it also implies the Hardy-Littlewood prime tuples
conjecture[HaLi1923] as a special case. There is a quantitative version of this con-
jecture which predicts a more precise count as to how many solutions there are in a
givenrange,andwhichwouldthenalsoimplyVinogradov’stheorem,aswellasGold-
bach’sconjecture(forsufficientlylargeN);see[GrTa2008b]forfurtherdiscussion. As
onecanimagine,thisconjectureisstilllargelyunsolved,howevertherearemanyim-
portantspecialcasesthathavenowbeenestablished-severalofwhichwereachieved
viatheHardy-Littlewoodcirclemethod.
OnecanviewDickson’sconjectureasanimpossibilitystatement: thatitisimpos-
sibletofindanyotherobstructionstosolvabilityforlinearpatternsintheprimesthan
theobviouslocalobstructionsatprimesqandatinfinity.(Itisalsoagoodexampleofa
local-to-globalprinciple,thatlocalsolvabilityimpliesglobalsolvability.) Impossibil-
itystatementshavealwaysbeenverydifficulttoprove-onehastolocateallpossible
obstructions to solvability, and eliminate each one of them in turn. In particular, one
has to exclude various exotic “conspiracies” between the primes to behave in an un-
usuallystructuredmannerthatsomehowmanagestoalwaysavoidallthepatternsthat
oneisseekingwithintheprimes. Howcanonedisproveaconspiracy?
Togiveanexampleofwhatsucha“conspiracy”mightlooklike,considerthetwin
prime conjecture, that of finding infinitely many pairs n,n+2 which are both prime.
This pattern encounters no obstructions at primes q or at infinity and so Dickson’s
conjecture predicts that there should be infinitely many such patterns. In particular,
there are no obstructions at 3 because prime numbers can equal 1 or 2 mod 3, and
it is possible to find pairs n, n+2 which also have this property. But suppose that
it transpired that all but finitely many of the primes ended up being 2 mod 3. From
lookingattablesofprimesthisseemstobeunlikely,butitisnotimmediatelyobvious
how to disprove it; it could well be that once one reaches, say, 10100, there are no
more primes equal to 1 mod 3. If this unlikely ”conspiracy” in the primes was true,
thentherewouldbeonlyfinitelymanytwinprimes. Fortunately, wehaveDirichlet’s
198 CHAPTER3. LECTURES
theorem,whichguaranteesinfinitelymanyprimesequaltoamodqwhenevera,qare
coprime,andsowecanruleoutthisparticulartypeofconspiracy. (Thisdoesstrongly
suggest,though,thatknowledgeofDirichlet’stheoremisanecessarybutnotsufficient
condition in order to solve the twin prime conjecture.) But perhaps there are other
conspiraciesthatoneneedstoruleoutalso?
To look for other conspiracies that one needs to eliminate, let us rewrite the con-
spiracy“allbutfinitelymanyoftheprimesare2mod3”inthemoreconvolutedformat
1
0.6<{ p}<0.7forallbutfinitelymanyprimes p
3
where{x}isthefractionalpartofx. Thistypeofconspiracycannowbegeneralised;
forinstanceconsiderthestatement
√
0<{ 2p}<0.01forallbutfinitelymanyprimes p (3.1)
Again,suchaconspiracyseemsveryunlikely-onewouldexpectthesefractional
partstobeuniformlydistributedbetween0and1,ratherthanconcentrateallinthein-
terval[0,0.01]-butitishardtorulethisconspiracyoutapriori.Andifthisconspiracy
(3.1)wasinfacttrue,thenthetwinprimeconjecturewouldbefalse,ascanbequickly
seenbyconsideringtheidentity
√ √ √
{ 2(n+2)}−{ 2n}=2 2mod1,
which forbids the two fractional parts on the left-hand side to simultaneously fall in
theinterval[0,0.01]. Thus,inordertosolvethetwinprimeconjecture,onemustrule
out (3.1). Fortunately, it has been known since the work of Vinogradov[Vi1937] that
√
{ 2p} is in fact uniformly distributed in the interval [0,1], and more generally that
{αp} is uniformly distributed in [0,1] whenever α is irrational. Indeed, by Weyl’s
famousequidistributiontheorem(seee.g. [KuNe1974]),thisuniformdistribution,this
isequivalenttotheexponentialsumestimate
∑ e2πiαp=o(∑ 1),
p<N p<N
andwenowseetheappearanceofFourieranalysisinthissubject.
One can rather easily concoct an endless stream of further conspiracies, each of
which could contradict the twin prime conjecture; this is one of the reasons why this
conjecture is considered so difficult. Let us thus leave this conjecture for now and
considersometwo-parameterproblems. Considerforinstancetheproblemoffinding
infinitelymanypatternsoftheformn,n+n0,n+2n0+2(i.e. arithmeticprogressions
oflength3,butwiththelastelementshiftedby2). Onceagain,theconspiracy(3.1),if
true,wouldobstructsolvabilityforthispattern,duetotheeasilyverifiedidentity
√ √ √ √
{ 2n}−2{ 2(n+n0)}+{ 2(n+2n0+2)}=2 2mod1
√
which is related to the fact that the function 2n has a vanishing second derivative.
(Notehoweverthatthesameconspiracydoesnotobstructsolvabilityofanunmodified
arithmeticprogressionn,n+n0,n+2n0.Thishighlightsaspecialpropertyofarithmetic