Towardssolvingthehierarchyprobleminstringtheory

A"small"progressreportafter1st EastAsiaJointWSatHufei,2016

SatoshiIso (KEK,SOKENDAI)磯曉 이소사토시

withN.Kitazawa:北澤敬章 (TMU)키타자와 노리아키T.Suyama：須⼭孝夫(KEK) 수야마 타카오H.Ohta ：太⽥光(KEK) 오오타 히카루

2018Oct7@KIAS,Korea(Oct5-8)3rdEastAsiaJointWSonFieldsandStrings2018

Whichcamefirst,thechickenortheegg?

Genetictheorytellsusthattheeggcamefirstandmanypeoplemaythinkso....

Isthisalwaysthecase?ThisiswhatIwouldliketodiscusstoday.

1m10-4m 104m10-12m10-20m10-28m10-36m 1012m 1020m 1028m

Atom

Nucleus (Q

CD

)

GU

Tscale

Solar system

Our galaxy

Universe w

e can observe

Size of the earth

Plancklength

3Electro-w

eak (WS

)

DesertorIntermediate

scalesD

ark energy

Hierarchyofscalesinnature

livingcreatures

1029eV1021eV1013eV105eV10-3eV10-7 10-11eV10-35eV

Naturalnessproblem=Hierarchyofvariousscalesinnaturesatoshiiso

TodayIwillconcentrateonlyontheEWscale.

WhyEWscale100GeV ismuchlowerthanUVscalesGUTscale1016 GeV,ifexistsPlanckscale1018GeVStringscale1017 GeV?

V = −µ2|H|2 + λ(|H|2)2

1

Higgspotential

V = −µ2|H|2 + λ(|H|2)2

δV =1

2

!d4k

(2π)4Str log(k2 +M(φ2))

=Λ2

32π2StrM(φ2) + Str

M(φ2)4

64π2ln

"M2

Λ2− 1

2

#

1

quadraticdivergence

SUSY?

V = −µ2|H|2 + λ(|H|2)2

δV =1

2

!d4k

(2π)4Str log(k2 +M(φ2))

=Λ2

32π2StrM(φ2) + Str

M(φ2)4

64π2ln

"M2

Λ2− 1

2

#

StrM(φ2) = 0? ΛTC = MTC exp(−8π2

bg2 )r ≪ lstring ω ≪ mstringr

lstr∼ ω

mstr≪ 1

J = Mr2ω =V

glpstr

"r

lstr

#2 " ω

mstr

#(1)

v = rω ≪ 1

1

V

J2

2Mr2=

1

2mv2 =

mstr

2g

"r

lstr

#2 " ω

mstr

#2

=j2

2g

"lstrr

#2

(2)

M =mstr

g

V

lpstr(3)

=Λ2

32π2StrM(φ)2 + Str

M(φ)4

64π2ln

"M2

Λ2− 1

2

#(4)

1

V = −µ2|H|2 + λ(|H|2)2

δV =1

2

!d4k

(2π)4Str log(k2 +M(φ2))

=Λ2

32π2StrM(φ2) + Str

M(φ2)4

64π2ln

"M2

Λ2− 1

2

#

StrM(φ2) = 0? ΛTC = MTC exp(−8π2

bg2 )r ≪ lstring ω ≪ mstringr

lstr∼ ω

mstr≪ 1

J = Mr2ω =V

glpstr

"r

lstr

#2 " ω

mstr

#(1)

v = rω ≪ 1

1

V

J2

2Mr2=

1

2mv2 =

mstr

2g

"r

lstr

#2 " ω

mstr

#2

=j2

2g

"lstrr

#2

(2)

M =mstr

g

V

lpstr(3)

=Λ2

32π2StrM(φ)2 + Str

M(φ)4

64π2ln

"M2

Λ2− 1

2

#(4)

StrM(φ)2 = 0 (5)

1

Partiallistofsolutionstothehierarchyproblem:

(1)Supersymmetry:cancellationofquadraticdivergencesNoSUSYparticlesarefound,littlehierarchyproblem,...

(2)Technicolor:dynamicalgenerationofscaleslikeQCDLightHiggsdifficult,bigformfactor(composite),

Higgsisalightpseudo-NG?SO(4)/SO(3)etc.

(3)Multiverse/Anthropic?

(4)Classicalconformal:Coleman-Weinbergradiativebreaking

couplingtogravity->quantumscaleinvariance?

V = −µ2|H|2 + λ(|H|2)2

δV =1

2

!d4k

(2π)4Str log(k2 +M(φ2))

=Λ2

32π2StrM(φ2) + Str

M(φ2)4

64π2ln

"M2

Λ2− 1

2

#

StrM(φ2) = 0? ΛTC = MTC exp(−8π2

bg2 )

1

∆2R = V

24π2M4plϵ

ϵ ≪ |η| < 1, φ < cM2

MplV (φ) ∼ λ

4φ4!ln φ2

M2 − 12

"+ V0

1

Myfavoritemodel

PrimordialInflation

χSB inducesEWSB

QCDphasetransition

⇠ 100MeV

Super-coolinginEW<h>=0@T<100GeV(allparticlesmassless)

ReheatingatlowTReheating@highT

2ndInflation

DMproducedà diluteà "super-coolDM"(anappropriatenumberofe-folding)

ThermalinflationstartsatTeV scaleandendsat100MeV!

BigBangstarts

e-foldng <10

Super-cooleduniversewiththesecond(lowscale)inflationintheclassicallyconformalmodel

Serpico,Shimada,SI(2017)

gravitationalwave,PBH,Baryogenesis

1-pagesummaryofrecentprogressintheclassicalconformalpheno.

BBN~MeV

Anyway.......Therearemanyproposalstosolvethehierarchyproblem,butthereisonecommonbasicassumption

"CalculatetheHiggspotentialfirst!"

Andthenobtainthesolution =minimumofthepotential.onesolutiontooneHiggspotential

à Alsowearefacedwiththenaturalnessproblem

Whichcamefirst,thechickenortheegg?

EffectivePotentialSolution

Usuallycalculatethepotentialfirst,thenobtainasolution.

Isitpossibletoobtainasolutionfirst,thencalculatepotential?

AclassicalexampleoftheChickenfirstapproach

TherearemanyorbitsaroundtheSun:theEarth,theJupiter,Mercury..eachoftheorbitisatthebottomofthecorrespondingpotential.Buttheunderlyingdynamicsisthesame.

Solutiondeterminesthepotential

AsimilarmechanismtodynamicallygeneratetheEWscale.

Italkedaboutthebasicideaat1st EastAsiaJointWS@Huhei

IntheD-branemodelbuilding,thedistancebetweenbranes(=moduli)givesa vev ofthescalarfield.

Distance⇔ scaleofthegaugesymmetrybreakingN

M

U(N+M)à U(N)× U(M)

∆2R =

V

24π2M4plϵ

∝ λ

|η|3 → λ ∼ 10−14

ϵ ≪ |η| < 1, φ < cM2

MplV (φ) ∼ λ

4φ4!ln φ2

M2 − 12

"+ V0

χ+

#p2

a2+ (δm)2 +mχ(t)

2

$

χ = 0

mχ(t) = gφ(t) = gΦ sin(mefft) |βk|2 = e−πκ2κ2 = 1

|mχ|

!p2

a2 + (δm)2"

A =1

m2eff

#p2

a2+ (δm)2 + g2Φ2

$

q =g2Φ2

4m2eff

∂2zχ+ (A+ 2q cos z)χ = 0

χ+ 3Hχ+ g2φ(t)2χ = 0 φ+ 3Hφ+ λφ3 + g2χ2φ = 0

q =g2Φ2

m2eff

=g2Φ2

λΦ2=

g2

λ≫ 1

m2χ = g2(Φ2+ ⟨ϕ2⟩) m2

eff = λ(Φ2+ ⟨ϕ2⟩)+g2⟨χ2⟩ Φ ∼%⟨χ2⟩ +g2⟨χ2⟩φ2

V0 = λM4 > g2⟨χ2⟩⟨φ2⟩ = g2Φ4 Φ =

#λ

g2

$1/4

M

H2 =V0

3M2pl

∼ λM4

M2pl

m2eff = g2Φ2

Φ =

√λ

g

M2

MplL =

M

M2s

1

ThenaturalscaleofMshouldbethestringscale,nottheEWscale.

Hierarchyprobleminstringtheory=Difficultytogeneratethe EWscaleinstringtheory

satoshiiso 12

Ex：D3s+anti-D7onZ3 orbifold R4× (T2× T2× T2 )/Z3

Put4D3-branesona fixedpointofT6/Z3AssignmentofZ3 chargeforD3s

4

U(1)*U(1)

Quivergaugetheory

U(2)*U(1)*U(1)

vev

satoshiiso 13

AntiD7

D3

D3

D3

AttractiveforcebetweenD3sandanti-D7duetoopenstring1-loopamplitudes

RepulsivecentrifugalforcebyrevolutionofD3s

1-loopsuppressed

Highangularfrequency

∆2R =

V

24π2M4plϵ

∝ λ

|η|3 → λ ∼ 10−14

ϵ ≪ |η| < 1, φ < cM2

MplV (φ) ∼ λ

4φ4!ln φ2

M2 − 12

"+ V0

χ+

#p2

a2+ (δm)2 +mχ(t)

2

$

χ = 0

mχ(t) = gφ(t) = gΦ sin(mefft) |βk|2 = e−πκ2κ2 = 1

|mχ|

!p2

a2 + (δm)2"

A =1

m2eff

#p2

a2+ (δm)2 + g2Φ2

$

q =g2Φ2

4m2eff

∂2zχ+ (A+ 2q cos z)χ = 0

χ+ 3Hχ+ g2φ(t)2χ = 0 φ+ 3Hφ+ λφ3 + g2χ2φ = 0

q =g2Φ2

m2eff

=g2Φ2

λΦ2=

g2

λ≫ 1

m2χ = g2(Φ2+ ⟨ϕ2⟩) m2

eff = λ(Φ2+ ⟨ϕ2⟩)+g2⟨χ2⟩ Φ ∼%⟨χ2⟩ +g2⟨χ2⟩φ2

V0 = λM4 > g2⟨χ2⟩⟨φ2⟩ = g2Φ4 Φ =

#λ

g2

$1/4

M

H2 =V0

3M2pl

∼ λM4

M2pl

m2eff = g2Φ2

Φ =

√λ

g

M2

MplL =

M

M2s

M ≪ Ms (L ≪ Ls)

v0 =

&l

µω = µ ∼ g

4πMs

1

Lowvelocity

∆2R =

V

24π2M4plϵ

∝ λ

|η|3 → λ ∼ 10−14

ϵ ≪ |η| < 1, φ < cM2

MplV (φ) ∼ λ

4φ4!ln φ2

M2 − 12

"+ V0

χ+

#p2

a2+ (δm)2 +mχ(t)

2

$

χ = 0

mχ(t) = gφ(t) = gΦ sin(mefft) |βk|2 = e−πκ2κ2 = 1

|mχ|

!p2

a2 + (δm)2"

A =1

m2eff

#p2

a2+ (δm)2 + g2Φ2

$

q =g2Φ2

4m2eff

∂2zχ+ (A+ 2q cos z)χ = 0

χ+ 3Hχ+ g2φ(t)2χ = 0 φ+ 3Hφ+ λφ3 + g2χ2φ = 0

q =g2Φ2

m2eff

=g2Φ2

λΦ2=

g2

λ≫ 1

m2χ = g2(Φ2+ ⟨ϕ2⟩) m2

eff = λ(Φ2+ ⟨ϕ2⟩)+g2⟨χ2⟩ Φ ∼%⟨χ2⟩ +g2⟨χ2⟩φ2

V0 = λM4 > g2⟨χ2⟩⟨φ2⟩ = g2Φ4 Φ =

#λ

g2

$1/4

M

H2 =V0

3M2pl

∼ λM4

M2pl

m2eff = g2Φ2

Φ =

√λ

g

M2

MplL =

M

M2s

M ≪ Ms (L ≪ Ls)

v0 =

&l

µω = µ ∼ g

4πMs v = ωd ∼ v0

Ms≪ 1 a ∼ v0

1

∆2R =

V

24π2M4plϵ

∝ λ

|η|3 → λ ∼ 10−14

ϵ ≪ |η| < 1, φ < cM2

MplV (φ) ∼ λ

4φ4!ln φ2

M2 − 12

"+ V0

χ+

#p2

a2+ (δm)2 +mχ(t)

2

$

χ = 0

mχ(t) = gφ(t) = gΦ sin(mefft) |βk|2 = e−πκ2κ2 = 1

|mχ|

!p2

a2 + (δm)2"

A =1

m2eff

#p2

a2+ (δm)2 + g2Φ2

$

q =g2Φ2

4m2eff

∂2zχ+ (A+ 2q cos z)χ = 0

χ+ 3Hχ+ g2φ(t)2χ = 0 φ+ 3Hφ+ λφ3 + g2χ2φ = 0

q =g2Φ2

m2eff

=g2Φ2

λΦ2=

g2

λ≫ 1

m2χ = g2(Φ2+ ⟨ϕ2⟩) m2

eff = λ(Φ2+ ⟨ϕ2⟩)+g2⟨χ2⟩ Φ ∼%⟨χ2⟩ +g2⟨χ2⟩φ2

V0 = λM4 > g2⟨χ2⟩⟨φ2⟩ = g2Φ4 Φ =

#λ

g2

$1/4

M

H2 =V0

3M2pl

∼ λM4

M2pl

m2eff = g2Φ2

Φ =

√λ

g

M2

MplL =

M

M2s

M ≪ Ms L ≪ Ls

1

Solution:HierarchyofEWscale

Merry-go-roundscenario(namedbyKimyeong )

N.Kitazawa SIPTEP,2015

Itispossibletomakeaclassicallystablestatewithashortdistance r<<lstring .

Butthelargeangularfrequencyà twoproblems・DispersionrelationofHiggs

violatesLorentzsymmetry(Coriolisforce)

・closedstringemissionà unstable

∆2R =

V

24π2M4plϵ

∝ λ

|η|3 → λ ∼ 10−14

ϵ ≪ |η| < 1, φ < cM2

MplV (φ) ∼ λ

4φ4!ln φ2

M2 − 12

"+ V0

χ+

#p2

a2+ (δm)2 +mχ(t)

2

$

χ = 0

mχ(t) = gφ(t) = gΦ sin(mefft) |βk|2 = e−πκ2κ2 = 1

|mχ|

!p2

a2 + (δm)2"

A =1

m2eff

#p2

a2+ (δm)2 + g2Φ2

$

q =g2Φ2

4m2eff

∂2zχ+ (A+ 2q cos z)χ = 0

χ+ 3Hχ+ g2φ(t)2χ = 0 φ+ 3Hφ+ λφ3 + g2χ2φ = 0

q =g2Φ2

m2eff

=g2Φ2

λΦ2=

g2

λ≫ 1

m2χ = g2(Φ2+ ⟨ϕ2⟩) m2

eff = λ(Φ2+ ⟨ϕ2⟩)+g2⟨χ2⟩ Φ ∼%⟨χ2⟩ +g2⟨χ2⟩φ2

V0 = λM4 > g2⟨χ2⟩⟨φ2⟩ = g2Φ4 Φ =

#λ

g2

$1/4

M

H2 =V0

3M2pl

∼ λM4

M2pl

m2eff = g2Φ2

Φ =

√λ

g

M2

MplL =

M

M2s

M ≪ Ms (L ≪ Ls)

v0 =

&l

µω = µ ∼ g

4πMs

1

Toavoidtheseproblems,itisnecessarytomakeabound(orresonant)statewithω <<mstr .à Weneedweakattractiveforce:V<<(mstring)3 r2Asimplewaytoavoidω=mstr istoconsider

FlatmodulisuchasDp - Dp ,

BPS=nointeractionatrestbut

v--dependentattractiveforceisgeneratedwhentheyaremovingwithaconstantvelocity

r

V = −µ2|H|2 + λ(|H|2)2

δV =1

2

!d4k

(2π)4Str log(k2 +M(φ2))

=Λ2

32π2StrM(φ2) + Str

M(φ2)4

64π2ln

"M2

Λ2− 1

2

#

StrM(φ2) = 0? ΛTC = MTC exp(−8π2

bg2 )r ≪ lstring ω ≪ mstringr

lstr∼ ω

mstr≪ 1

J = Mr2ω =V

glpstr

"r

lstr

#2 " ω

mstr

#(1)

v = rω ≪ 1

1

V

J2

2Mr2=

1

2mv2 =

mstr

2g

"r

lstr

#2 " ω

mstr

#2

=j2

2g

"lstrr

#2

(2)

M =mstr

g

V

lpstr(3)

=Λ2

32π2StrM(φ)2 + Str

M(φ)4

64π2ln

"M2

Λ2− 1

2

#(4)

StrM(φ)2 = 0 (5)

−Mv4

r7−p+

J2

Mr2(6)

1

V = −µ2|H|2 + λ(|H|2)2

δV =1

2

!d4k

(2π)4Str log(k2 +M(φ2))

=Λ2

32π2StrM(φ2) + Str

M(φ2)4

64π2ln

"M2

Λ2− 1

2

#

StrM(φ2) = 0? ΛTC = MTC exp(−8π2

bg2 )r ≪ lstring ω ≪ mstringr

lstr∼ ω

mstr≪ 1

J = Mr2ω =V

glpstr

"r

lstr

#2 " ω

mstr

#(1)

v = rω ≪ 1

1

V

J2

2Mr2=

1

2mv2 =

mstr

2g

"r

lstr

#2 " ω

mstr

#2

=j2

2g

"lstrr

#2

(2)

M =mstr

g

V

lpstr(3)

=Λ2

32π2StrM(φ)2 + Str

M(φ)4

64π2ln

"M2

Λ2− 1

2

#(4)

r ≫ lstrStrM(φ)2 = 0 (5)

−Mv4

r7−p+

J2

Mr2(6)

1

Intheclosedstringregion,

Nominimumexists.Potentialbarrieratr<lstr

closedstringpicture

Intheclosed-stringdominatedregion(r>>lstr ),therepulsiveforcesurpassestheattractiveforce.

However,"BeautyisAttractive" Kofman LindeLiuMalonyMaCllister Silverstein2004

Masschangesrapidly.à non-adiabaticprocessOpenstringparticlesareproduced(=preheating).

à loseenergyandthetrajectoryshrinks.à andwhenthetrajectorybecomescircular,

nomoreparticleproductionoccurs.

Isitpossibletomakea"boundstate"withr<<lstrbetweenrevolvingD-branes?

Dp Dp

ManyworksonD0-branes(N=16SU(N)supersymmetricQM)Wittenindex(Piljin Yi)Tr(-1)F =1---- thresholdboundstate

masslessgravitonQuantumboundstate(Kabat Pouliot)---- resonantstateE>0

Thesestatesarenearthegroundstate.Whatwewantishighlyexcited,butalmoststableresonantstates.

(likeasolarsystem,notlikeahydrogenatom)

2 4 6 8 10

1

2

3

4

5

?openstringregion

Poorman'sCalculationofattractivepotentialbetweenrevolvingDp,inparticularp=0.

D0 D0

Itisstraightforward,butnotsotrivialbecauseofthetime-dependentboundaryconditions:

Intherotationalframe,theboundaryconditionbecomessimple,butthesystemisinteracting:

T.Suyama,H.Ohta,SItoappearsoon

Furthermore,bytakingvariationwithrespecttotheTfields,

ThusT– fieldmustsatisfyàattheboundaries:

Theseconditionscanbesimplifiedbyintroducinganewvariable

σ＝０

σ＝π

Withthese(andafewmore)changesofworldsheetfields,theactionofopenstringsstretchedbetweenrevolvingD0sbecomes

OpenstringworldsheetactionbetweenrevolvingD0s

boundaryconditions

One-loopamplitudeofopenstringbetweenrevolvingD0sperturbationwithrespecttotheangularvelocityωuptoω2

V = −µ2|H|2 + λ(|H|2)2

δV =1

2

!d4k

(2π)4Str log(k2 +M(φ2))

=Λ2

32π2StrM(φ2) + Str

M(φ2)4

64π2ln

"M2

Λ2− 1

2

#

StrM(φ2) = 0? ΛTC = MTC exp(−8π2

bg2 )r ≪ lstring ω ≪ mstring

1

Theassumptionsofourcalculationare

lowlyingopenstringstatesdominate

perturbationisvalid

V = −µ2|H|2 + λ(|H|2)2

δV =1

2

!d4k

(2π)4Str log(k2 +M(φ2))

=Λ2

32π2StrM(φ2) + Str

M(φ2)4

64π2ln

"M2

Λ2− 1

2

#

StrM(φ2) = 0? ΛTC = MTC exp(−8π2

bg2 )r ≪ lstring ω ≪ mstringr

lstr∼ ω

mstr≪ 1

J = mstrV

lpstr

"r

lstr

#2" ω

mstr

#(1)

v = rω ≪ 1

1

Effectivepotentialinducedbymassivestates:Doubleexpansionwithrespecttow^2andr^2.

The term e−2r2t/πα′is nothing but the effect of stretched open strings with distance

2r, and the additional factor!1− v2

3

"−1/2comes from the v2 correction to the

momentum integration.To determine V2(r,ω), we need to evaluate the integrals of the form

Ik(r) :=

# ∞

0

dt

2ttkf(t, r)e−2πt. (4.6)

The integral with k ≤ 12 is defined by the analytic continuation for k. We expand

the integrand in r2 and integrate term by term. As a result, we obtain

Ik(r) =$16πα′%− 1

2 (2π)−kΓ

&k − 1

2

'

×(1−

&k − 1

2

'r2

π2α′ +1

2

&k2 − 1

4

'&r2

π2α′

'2

+O(r6)

). (4.7)

Using this expression, we obtain

V2(r,ω) =1296 + 106α′ω2

8√α′

+1296 + (−730 + 432π2 + 1296π2ϵ0)α′ω2

16√α′

&r2

π2α′

'

−1296 + (−2430− 864π2 + 2592π2ϵ0)α′ω2

64√α′

&r2

π2α′

'2

+O(ω4, r6).

(4.8)

Note that since our calculation is performed in the Wick rotated metric, the po-tential in the Lorentzian metric is obtained by replacing ω2 with −ω2. The relevantpart of the potential is written as a positive power series of (r/lstr)2 and (ω/mstr)2;

V2 ∼ mstr

*c1 + c2

&w

mstr

'2

+ · · ·+&

r

lstr

'2

+ mstr

*c3 + c4

&w

mstr

'2

+ · · ·+&

r

lstr

'4

+ · · · (4.9)

where we defined (2πα′)1/2 = lstr = 1/mstr and c1 > 0. As expected, the leadingorder potential is proportional to m3

stringr2 and strongly attractive. Angular-

frequency corrections to the potential are given by mstringω2r2. Whether it is at-tractive or repulsive depends on the regularization constant ϵ0 and we cannot sayeither here. In the superstring case, the potential must vanish in the ω → 0 limitand this ω-dependent term give the leading order (massive state) contribution tothe potential.

Other contributions VN (r,ω) with N > 2 can be obtained similarly, but thecalculations become tedious as N increases.

14

By using (??) for Tr0, (??), (??) and (??) for Tr1, and the results in Appendix?? for Tr2, we can obtain the explicit forms of the partial traces up to O(v2) as

Tr0!e−2πt(H0(v)+1)

"= F (t, r)

#1− 2πv2

$− α′

8r2+ ϵ0

%t

&,

Tr1!e−2πtH0(v)

"= 3F (t, r)

#1− 2πv2

3

$25α′

8r2− 4

π2+ 3ϵ0

%t+

16

3v2t2

&,

Tr2!e−2πt(H0(v)−1)

"= 9F (t, r)

#1− 2πv2

9

$151α′

8r2− 18

π2+ 9ϵ0

%t+

32

3v2t2

&,

(4.4)

where

F (t, r) := f(t, r)e−2πt

$1− v2

3

%−1/2

, f(t, r) ='8π2α′t

(− 12 e−

2r2

πα′ t. (4.5)

The term e−2r2t/πα′is nothing but the effect of stretched open strings with distance

2r, and the additional factor)1− v2

3

*−1/2comes from the v2 correction to the

momentum integration.To determine V2(r,ω), we need to evaluate the integrals of the form

Ik(r) :=

+ ∞

0

dt

2ttkf(t, r)e−2πt. (4.6)

The integral with k ≤ 12 is defined by the analytic continuation for k. We expand

the integrand in r2 and integrate term by term. As a result, we obtain

Ik(r) ='16πα′(− 1

2 (2π)−kΓ

$k − 1

2

%

×,1−

$k − 1

2

%r2

π2α′ +1

2

$k2 − 1

4

%$r2

π2α′

%2

+O(r6)

-. (4.7)

Using this expression, we obtain

V2(r,ω) =1296 + 106α′ω2

8√α′

+1296 + (−730 + 432π2 + 1296π2ϵ0)α′ω2

16√α′

$r2

π2α′

%

−1296 + (−2430− 864π2 + 2592π2ϵ0)α′ω2

64√α′

$r2

π2α′

%2

+O(ω4, r6).

(4.8)

Note that since our calculation is performed in the Wick rotated metric, the po-tential in the Lorentzian metric is obtained by replacing ω2 with −ω2. The relevantpart of the potential is written as a positive power series of (r/lstr)2 and (ω/mstr)2;

V2 ∼ mstr

.c1 + c2

$w

mstr

%2

+ · · ·/$

r

lstr

%2

+ mstr

.c3 + c4

$w

mstr

%2

+ · · ·/$

r

lstr

%4

+ · · · (4.9)

14

Bosoniccase:thefirstexcitedmassivestate

Insuperstring

M = mp+1str V (set the string coupling g = 1 for simplicity) rotate with an angular

momentum L = Mr2ω. Then the centrifugal repulsive force is given by Mrω2 andunder the angular momentum conservation, its centrifugal potential becomes

VL =L2

2Mr2. (5.2)

As we saw in the case of D0-particles in the bosonic string, strong attractive potentialis generated by the one-loop open string amplitude: c1m3

strr2 for r ≪ lstr. In the Dp-

brane case, an integration over the position of the end point of open strings gives thevolume factor V ∝ M and the most dominant part of the effective potential will begiven by cM(r/lstr)2 for r ≪ lstr. Then we have a classical solution with ω ∼ mstr.However, such a classical bound state with large angular velocity ω is unstableagainst emitting closed string emission. Furthermore, if we lived on such a D3-brane,the Lorentz symmetry is strongly violated; thus a construction of the standard modelbased on this kind of D-brane configurations is not phenomenologically viable.

Alternatively we may consider a pair of D-branes revolving around each otherin superstring theory. When ω = 0, such configurations of D-branes is BPS, andtherefore, there is no force between them. If they are revolving around each other,it would generate attractive potential as in (??). But in superstring cases, manyterms cancel due to supersymmetry; especially ω-independent terms are completelycancelled. Thus in the typical form of the effective potential generated by massivestates (??), some of the coefficients must vanish: e.g. c1 = c3 = 0. It is furtherknown that when two branes are moving with a constant relative velocity, the v2

terms also cancel. It is, however, not clear whether such cancellations occur in therevolving configuration. For revolving Dp-branes (for odd p), massless open stringstates induce the following Coleman-Weinberg type effective potential;

V!

i

(−1)Fini

"#r

lstr

$2

+ Ci

#ω

mstr

$2% p+1

2

log

"#r

lstr

$2

+ Ci

#ω

mstr

$2%

(5.3)

where ni is the number of degrees of freedom of the field i. Supersymmetry imposesthat

&i(−1)Fini = 0. Under the angular momentum conservation, ω2 is replace by

(J2/Mr2)2. If (r/lstr)2 terms and (ω/mstr)2 terms can be balanced, together withthe centrifugal potential L2/2Mr2, there mimght exist bound states of revolvingD-branes. And, if it is the case, the rotation of the D-branes must be extremelyslow;

ω

mstr∼ r

lstr≪ 1. (5.4)

In applications to phenomenological and cosmological applications, this propertywill save the problems of strong Lorentz violation as well as the rapid closed stringemissions. We hope to report a further investigation on this possibility in future.

Acknowledgements

19

C2 =C4 =0?àWearecheckinghowitdeviatesfromconstantvelocitycase

Effectivepotentialinducedbymasslessstatesstretchedbet.D0sThe typical form of the effective potential induced by the massless state (4.27)

is nothing but the Coleman-Weinberg potential for quantum particles in (1 + 0)-dimensions. Finally in the section, we briefly comment on the relation. For com-parison and future generalizations to Dp-branes, we will consider general cases in(1 + p)-dimensions. In 1 + p-dimensional field theory, one-loop integral of a scalarfield with mass m is given by

−Tr log(∆+m2)−1/2 =

∫ ∞

ϵ

dt

2t

∫dp+1k

(2π)p+1e−(k2+m2)t

∼∫ ∞

ϵ

dt

2tt−

p+12 e−m2t ∼

{(m2)

p+12 p = odd

(m2)p+12 logm2 p = even.

(4.28)

If mass is given by vev of some scalar field φ, it gives the well-known Coleman-Weinberg potential. In our case, mass is generated by the distance r betweenD0-particles and the angular velocity ω. Thus the effective potential induced byopen string massless states is given by the typical form (4.27), namely p = 0 casein (4.28), and the instability is associated with a possibility that the mass squaredm2 becomes negative at large ω.

5 Conclusions and discussions

We have investigated an open string stretched between two D0-branes which revolvearound each other. To quantize the corresponding worldsheet theory, we introducednew fields which satisfy simple boundary conditions. This makes the worldsheet ac-tion quite complicated. We analyze the resulting system perturbatively, using theformalism developed in [IOS]. One of the advantages of using the method of the im-proved perturbation is the fact that H0(v), containing all perturbative informationon the energy spectrum, commutes with H0 by construction. This enables us toseparate the massless contributions from the massive ones, and also to expand thetrace as in (3.6).

As an application, we obtained the one-loop partition function, and extract thebehavior of the effective potential between the pair of D0-branes induced by theopen string when the separation of the D0-branes is small. Here we summarize ourresults for the effective potential. It is given by various contributions from masslessand massive open string states (after the replacement of ω2 by −ω2) as

V ∼ mstr

⎡

⎣3∑

i=0

√(r

lstr

)2

+ Ci

(ω

mstr

)2

+

(c1 + c2

(ω

mstr

)2)(

r

lstr

)2⎤

⎦ (5.1)

where C0 = 1/32, C1 = −11/32, C2 = C3 = −13/32 and c1 = 162√

2/π. Theresult shows that there is an attractive force between the rotating D0-branes. Acontribution from open string tachyons is neglected in this expression.

Finally we briefly discuss if there is a possibility that Dp-branes with volume Vcan make a bound state by revolving each other. Suppose that Dp-brane with mass

18

The typical form of the effective potential induced by the massless state (4.27)is nothing but the Coleman-Weinberg potential for quantum particles in (1 + 0)-dimensions. Finally in the section, we briefly comment on the relation. For com-parison and future generalizations to Dp-branes, we will consider general cases in(1 + p)-dimensions. In 1 + p-dimensional field theory, one-loop integral of a scalarfield with mass m is given by

−Tr log(∆+m2)−1/2 =

∫ ∞

ϵ

dt

2t

∫dp+1k

(2π)p+1e−(k2+m2)t

∼∫ ∞

ϵ

dt

2tt−

p+12 e−m2t ∼

{(m2)

p+12 p = odd

(m2)p+12 logm2 p = even.

(4.28)

If mass is given by vev of some scalar field φ, it gives the well-known Coleman-Weinberg potential. In our case, mass is generated by the distance r betweenD0-particles and the angular velocity ω. Thus the effective potential induced byopen string massless states is given by the typical form (4.27), namely p = 0 casein (4.28), and the instability is associated with a possibility that the mass squaredm2 becomes negative at large ω.

5 Conclusions and discussions

We have investigated an open string stretched between two D0-branes which revolvearound each other. To quantize the corresponding worldsheet theory, we introducednew fields which satisfy simple boundary conditions. This makes the worldsheet ac-tion quite complicated. We analyze the resulting system perturbatively, using theformalism developed in [IOS]. One of the advantages of using the method of the im-proved perturbation is the fact that H0(v), containing all perturbative informationon the energy spectrum, commutes with H0 by construction. This enables us toseparate the massless contributions from the massive ones, and also to expand thetrace as in (3.6).

As an application, we obtained the one-loop partition function, and extract thebehavior of the effective potential between the pair of D0-branes induced by theopen string when the separation of the D0-branes is small. Here we summarize ourresults for the effective potential. It is given by various contributions from masslessand massive open string states (after the replacement of ω2 by −ω2) as

V ∼ mstr

⎡

⎣3∑

i=0

√(r

lstr

)2

+ Ci

(ω

mstr

)2

+

(c1 + c2

(ω

mstr

)2)(

r

lstr

)2⎤

⎦ (5.1)

where C0 = 1/32, C1 = −11/32, C2 = C3 = −13/32 and c1 = 162√

2/π. Theresult shows that there is an attractive force between the rotating D0-branes. Acontribution from open string tachyons is neglected in this expression.

Finally we briefly discuss if there is a possibility that Dp-branes with volume Vcan make a bound state by revolving each other. Suppose that Dp-brane with mass

18

mass2 acquiredbytheeffectofrevolution

NotethatC1 isnegativeforvectors(positiveforKKscalar)whichindicatesthatthesystemisunstableforlargeω .

c.f.Fieldtheorycalculation

grals can be performed easily. We obtain

V10 = −! ∞

0

dt

2t

!dk

2πe−2πtE0(k,r,ω) =

"#####$

−α′ω2 + 8%1 + ϵ0π

2α′ω2& r2

π2α′

32α′'1− 1

3ω2r2

(

(4.22)

V11 = −! ∞

0

dt

2t

!dk

2πe−2πt(E1(k,r,ω)−1) =

"#####$

11α′ω2 + 8%1 + ϵ0π

2α′ω2& r2

π2α′

32α′'1− 1

3ω2r2

( .

(4.23)

To evaluate the integrals for the remaining two eigenvalues, we make the follow-ing approximation )

1

π2+

(α′)2

r2k2 → α′

r|k| (4.24)

for small r. Then, the integrals can be performed, and we obtain

V12 + V13 = −! ∞

0

dt

2t

!dk

2πe−2πt(E2(k,r,ω)−1) −

! ∞

0

dt

2t

!dk

2πe−2πt(E3(k,r,ω)−1)

=

"#####$

13α′ω2 + 8*1 +

%ϵ0π

2 − 2&α′ω2

+ r2

π2α′

8α′'1− 1

3ω2r2

( − ω2

'1− 1

3ω2r2

(2 .

(4.25)

The effective potential induced by the massless modes becomes a sum of (4.22),(4.23) and (4.25),

V1 =3,

i=0

V1i (4.26)

and in the Lorentzian metric, ω2 is replaced with −ω2. In the limit, r ≪ lstr andω ≪ mstr, each term is written in the form of

mstr

-'r

lstr

(2

+ C1

'ω

mstr

(2

+ C2

'r

lstr

(2' ω

mstr

(2

(4.27)

At ω = 0, the potential becomes r/l2s , which is proportional to the length of thestretched string. In the Lorentitzian metric, the coefficient C is positive for a trans-verse massless state, namely a contribution from αi

−1|k⟩ in (4.22). On the contrary,

contributions from αT,X,Y−1 |k⟩ in (4.23) and (4.25), C1 is negative. This indicates

that these vector fields destabilize the configuration of revolving D0-branes whenthe angular velocity becomes faster; ω/mstr !

.|C1|r/lstr.

17

Closedstringemissionwithspinsissuppressedforω <<mstring

angularmomentumisconservedà randω arerelated:

V = −µ2|H|2 + λ(|H|2)2

δV =1

2

!d4k

(2π)4Str log(k2 +M(φ2))

=Λ2

32π2StrM(φ2) + Str

M(φ2)4

64π2ln

"M2

Λ2− 1

2

#

StrM(φ2) = 0? ΛTC = MTC exp(−8π2

bg2 )r ≪ lstring ω ≪ mstringr

lstr∼ ω

mstr≪ 1

J =V

glpstr

"r

lstr

#2" ω

mstr

#(1)

v = rω ≪ 1

1

V

J2

2Mr2=

1

2mv2 =

mstr

2g

"r

lstr

#2 " ω

mstr

#2

=j2

2g

"lstrr

#2

(2)

1

+thefollowingattractivepotential

ThetotalpotentialperunitvolumeforrevolvingDp-braneis

V = −µ2|H|2 + λ(|H|2)2

δV =1

2

!d4k

(2π)4Str log(k2 +M(φ2))

=Λ2

32π2StrM(φ2) + Str

M(φ2)4

64π2ln

"M2

Λ2− 1

2

#

StrM(φ2) = 0? ΛTC = MTC exp(−8π2

bg2 )r ≪ lstring ω ≪ mstringr

lstr∼ ω

mstr≪ 1

J = Mr2ω =V

glpstr

"r

lstr

#2 " ω

mstr

#(1)

v = rω ≪ 1

1

V

J2

2Mr2=

1

2mv2 =

mstr

2g

"r

lstr

#2 " ω

mstr

#2

=j2

2g

"lstrr

#2

(2)

M =mstr

g

V

lpstr(3)

1

M = mp+1str V (set the string coupling g = 1 for simplicity) rotate with an angular

momentum L = Mr2ω. Then the centrifugal repulsive force is given by Mrω2 andunder the angular momentum conservation, its centrifugal potential becomes

VL =L2

2Mr2. (5.2)

As we saw in the case of D0-particles in the bosonic string, strong attractive potentialis generated by the one-loop open string amplitude: c1m3

strr2 for r ≪ lstr. In the Dp-

brane case, an integration over the position of the end point of open strings gives thevolume factor V ∝ M and the most dominant part of the effective potential will begiven by cM(r/lstr)2 for r ≪ lstr. Then we have a classical solution with ω ∼ mstr.However, such a classical bound state with large angular velocity ω is unstableagainst emitting closed string emission. Furthermore, if we lived on such a D3-brane,the Lorentz symmetry is strongly violated; thus a construction of the standard modelbased on this kind of D-brane configurations is not phenomenologically viable.

Alternatively we may consider a pair of D-branes revolving around each otherin superstring theory. When ω = 0, such configurations of D-branes is BPS, andtherefore, there is no force between them. If they are revolving around each other,it would generate attractive potential as in (5.1). But in superstring cases, manyterms cancel due to supersymmetry; especially ω-independent terms are completelycancelled. Thus in the typical form of the effective potential generated by massivestates (4.9), some of the coefficients must vanish: e.g. c1 = c3 = 0. It is furtherknown that when two branes are moving with a constant relative velocity, the v2

terms also cancel. It is, however, not clear whether such cancellations occur in therevolving configuration. For revolving Dp-branes (for odd p), massless open stringstates induce the following Coleman-Weinberg type effective potential;

V!

i

(−1)Fini

"#r

lstr

$2

+ Ci

#ω

mstr

$2% p+1

2

log

"#r

lstr

$2

+ Ci

#ω

mstr

$2%

(5.3)

where ni is the number of degrees of freedom of the field i. Supersymmetry imposesthat

&i(−1)Fini = 0. Under the angular momentum conservation, ω2 is replace by

(J2/Mr2)2. If (r/lstr)2 terms and (ω/mstr)2 terms can be balanced, together withthe centrifugal potential L2/2Mr2, there mimght exist bound states of revolvingD-branes. And, if it is the case, the rotation of the D-branes must be extremelyslow;

ω

mstr∼ r

lstr≪ 1. (5.4)

In applications to phenomenological and cosmological applications, this propertywill save the problems of strong Lorentz violation as well as the rapid closed stringemissions. We hope to report a further investigation on this possibility in future.

Acknowledgements

19

The term e−2r2t/πα′is nothing but the effect of stretched open strings with distance

2r, and the additional factor!1− v2

3

"−1/2comes from the v2 correction to the

momentum integration.To determine V2(r,ω), we need to evaluate the integrals of the form

Ik(r) :=

# ∞

0

dt

2ttkf(t, r)e−2πt. (4.6)

The integral with k ≤ 12 is defined by the analytic continuation for k. We expand

the integrand in r2 and integrate term by term. As a result, we obtain

Ik(r) =$16πα′%− 1

2 (2π)−kΓ

&k − 1

2

'

×(1−

&k − 1

2

'r2

π2α′ +1

2

&k2 − 1

4

'&r2

π2α′

'2

+O(r6)

). (4.7)

Using this expression, we obtain

V2(r,ω) =1296 + 106α′ω2

8√α′

+1296 + (−730 + 432π2 + 1296π2ϵ0)α′ω2

16√α′

&r2

π2α′

'

−1296 + (−2430− 864π2 + 2592π2ϵ0)α′ω2

64√α′

&r2

π2α′

'2

+O(ω4, r6).

(4.8)

Note that since our calculation is performed in the Wick rotated metric, the po-tential in the Lorentzian metric is obtained by replacing ω2 with −ω2. The relevantpart of the potential is written as a positive power series of (r/lstr)2 and (ω/mstr)2;

V2 ∼ mstr

*c1 + c2

&w

mstr

'2

+ · · ·+&

r

lstr

'2

+ mstr

*c3 + c4

&w

mstr

'2

+ · · ·+&

r

lstr

'4

+ · · · (4.9)

where we defined (2πα′)1/2 = lstr = 1/mstr and c1 > 0. As expected, the leadingorder potential is proportional to m3

stringr2 and strongly attractive. Angular-

frequency corrections to the potential are given by mstringω2r2. Whether it is at-tractive or repulsive depends on the regularization constant ϵ0 and we cannot sayeither here. In the superstring case, the potential must vanish in the ω → 0 limitand this ω-dependent term give the leading order (massive state) contribution tothe potential.

Other contributions VN (r,ω) with N > 2 can be obtained similarly, but thecalculations become tedious as N increases.

14

Dp-Dp potential(p=odd)

Thepotentialhasaminimumaround

V = −µ2|H|2 + λ(|H|2)2

δV =1

2

!d4k

(2π)4Str log(k2 +M(φ2))

=Λ2

32π2StrM(φ2) + Str

M(φ2)4

64π2ln

"M2

Λ2− 1

2

#

StrM(φ2) = 0? ΛTC = MTC exp(−8π2

bg2 )r ≪ lstring ω ≪ mstringr

lstr∼ ω

mstr≪ 1

1

Totalangularmomentum

Forp=0,J<<1à onlys-wave(groundstate)isallowed.

Thusonlythresholdboundstateexists.Forp>0Jcanbelargerthan1.AresonantstatewithhigherJmayexist.

consistentwiththeassumptionsofourcalculations

V = −µ2|H|2 + λ(|H|2)2

δV =1

2

!d4k

(2π)4Str log(k2 +M(φ2))

=Λ2

32π2StrM(φ2) + Str

M(φ2)4

64π2ln

"M2

Λ2− 1

2

#

StrM(φ2) = 0? ΛTC = MTC exp(−8π2

bg2 )r ≪ lstring ω ≪ mstringr

lstr∼ ω

mstr≪ 1

J =V

glpstr

"r

lstr

#2" ω

mstr

#(1)

v = rω ≪ 1

1

Summary

Ifresonantstatesexist,itmustbeahierarchicalsolution

whichmeans E EW <<mstring .

Wehavecalculatedtheeffectivepotentialcorrespondingtothesolution.

Forp>0,theremayexista"classical"boundstatesatisfying

Lorentzviolation,instabilitybyradiationà OK

"Thechicken-firstapproach"tohierarchyproblem

Futureissues:(1) SmartercalculationbyD-braneEFTint-depbackground(2)constructphenomenologicallyrealisticmodels(3)SUSYbreaking: TeV SUSY?solutiontolittlehierarchy?

V = −µ2|H|2 + λ(|H|2)2

δV =1

2

!d4k

(2π)4Str log(k2 +M(φ2))

=Λ2

32π2StrM(φ2) + Str

M(φ2)4

64π2ln

"M2

Λ2− 1

2

#

StrM(φ2) = 0? ΛTC = MTC exp(−8π2

bg2 )r ≪ lstring ω ≪ mstringr

lstr∼ ω

mstr≪ 1

1

StationarysolutionsofD-branes

V = −µ2|H|2 + λ(|H|2)2

δV =1

2

!d4k

(2π)4Str log(k2 +M(φ2))

=Λ2

32π2StrM(φ2) + Str

M(φ2)4

64π2ln

"M2

Λ2− 1

2

#

StrM(φ2) = 0? ΛTC = MTC exp(−8π2

bg2 )r ≪ lstring ω ≪ mstringr

lstr∼ ω

mstr≪ 1

J = Mr2ω =V

glpstr

"r

lstr

#2 " ω

mstr

#(1)

v = rω ≪ 1

1

V

J2

2Mr2=

1

2mv2 =

mstr

2g

"r

lstr

#2 " ω

mstr

#2

=j2

2g

"lstrr

#2

(2)

M =mstr

g

V

lpstr(3)

=Λ2

32π2StrM(φ)2 + Str

M(φ)4

64π2ln

"M2

Λ2− 1

2

#(4)

r ≫ lstrStrM(φ)2 = 0 (5)

−Mv4

r7−p+

J2

Mr2(6)

1

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