penrose graphs

PENROSE GRAPHS

ASGHAR QADIR

Departmentof Mathematics,Universityof Islamabad,Pakistan

I

PHYSICSREPORTS(SectionC of PhysicsLetters)39 No. 2)1978)131-167.NORTH-HOLLAND PUBLISHING COMPANY

PENROSE GRAPHS

AsgharQADIRDepartmentof Mathematics,UniversityofIslamabad,Pakistan

ReceivedMay 1977

Contents:

1. Introduction 1332. Notation 34 6. Loop graphs ~5O3. The rules of Penrosegraphs 35 7. External lines with negativenumbers 1574. Treegraphs 137 8. Discussion 1645. Multi-line treegraphs 145 References 167

Abstract:

A graphicalrepresentationof scatteringprocesses,analogousin somewaysto Feynmangraphs,butavoidingthe infra-redandrenormalizationdivergences,is presented.Someof the methodsusedto calculatesuchgraphsare given.

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A. Qadir, Penrosegraphs 133

1. Introduction

In his Adam’s Prizeessay[1], R. Penrosesuggestedthata methodof trying to ‘quantizerelativity’would be to useadifferentmanifold______onein which the points were‘smeared-out’in effect. Helater achievedthis type of result with his twistors [2]. The twistorsalso appearednaturally whenheconsideredsolutionsof the zero rest-massfields equations[3, 4]. He thendevelopeda formalism toquantizecurvedspace-time[4]. To do this he constructeda Hamiltonian which was a function of atwistor Z~andits dualZa. This work led him to developadiagrammaticrepresentation[5], in somewaysanalogousto Feynmangraphs,to dealwith scatteringof particles.

So far the formalism can only be appliedin aconformally flat space-time,but thereis reasontohope [6] thatit canbeextendedto agenuinelycurvedspacetime,and deal with masses.The graphicalrepresentationappearedto work reasonablywell, but lacked a sound basis.This basis hasbeensupplied, to some extent, by Moore and Penrose[7], who have written a Hamiltonian for theelectromagneticpotentialwhich leadsto the identificationof a particular diagram(fig. 1(a)) with theFeynmanvertex (fig. 1(b)). It should be noticed that the diagramcan only be written for a virtualphoton,but not for a real photon.As the diagramscannot havean infra-reddivergence[8], therecanbe no renormalizationdivergence[8,9] either.

B A

0

1:’(a) (bl

Fig. 1.

ThoughI shall not go into the detailsof it, the Hamiltonianfor an electromagneticpotentialmightbe of interest.The Hamiltonian is a functionof two twistors H(Za, W~),such thatZ~and ~ areconjugatevariables.

H X’~w - £ F(U~,V~)V~X8L,,8U,~~WVP”~uv

WPV~U~V~UUX~where

9~UV=~UA~uJV

co~rr— ‘~~8TT,irr ,irr ,.irr3(E au~Au~Autiö

~ V = ~ V~dV~n dV~n dV~

andL,~,I~arethe infinity twistors.

134 A. Qadir. Penrosegraphs

F(U, V) is the field definedby

~AA’ = (~~~-)2~UA VA,F(ixCC V~,V~UD, —iy~’UD)~A~

where~AA’ is the electromagneticpotential,AB NB~A =UAVAdUBAdVBE �

UA and VB’ areexplainedin section8.

The condition ~DD = XDD ~ U~V” = 0, for which we require that F havesuitablesingularities.If the Hamiltonian is a function of a null twistor and its dual, H(Z”, Z~)with Z”Z,~.= 0, then

formally

H(Z”, 2~)=f pAA~ du

wheredu is definedin terms of the momentumpA4 by du = P~VAA..

The calculationspresentedhere will not be directly comparablewith the correspondingones forFeynmangraphs.Thesearedonein a twistor spacewhich correspondsto a compactified,complexifiedMinkowski space.For comparisonwith quantumelectrodynamicsin the high energylimit, the resultshave to be translatedinto momentum space,where they pick up a delta-function to expressmomentumconservation.For this reasonthe basiccalculationsareperformedin the twistor space.Atthe last stagetheycanbe translatedto momentumspace.The resultsdo in fact appearto correspondwith the resultsof quantumelectrodynamics.

I havefirst giventhe notation,thenthe rulesof Penrosegraphs,thenevaluatedvariousgraphs,andfinally discussedthem.

2. Notation

Twistor indiceswill be representedby lower casegreeklettersa,/3, y,..., spinorindicesby uppercaseunprimedandprimed latin lettersA,B,C,..., A’, B’, C’ Constanttwistors will be represen-ted by the earlier upper-caselatin lettersA”, B” C~,Do,. .. , and variable twistors by the laterupper-caselatin letters W”, X”,. . . , Y,~,,4,

I shallusePenrose’sabstractindexnotation [10],andwrite T~” asthe tensoron the manifold andas the componentsof the tensorin somecoordinateframe.Usingthis connectionI can formally

write~ J-’A...CA...C’

—

A II shall oftendispensewith indices [8, 11] andwrite for A~,B for B” etc.

for 2! A~”B~1,A B for 2! A1~B~1A B for 2! A(~B~),

IIA AB

A B for 2! At”B0~etc. I for A~B”, I for 2!A[G,B$)C”D~,

B CD


~i’~4’~for 2! A(~B~)C”D~etc. I I I for ~ I I I I for ~

ThusA B C D = �“~A~B~c7D8,A B C D =

AB CD AB CD I II I I =1 I I I

IIII=iiiI><IIlI,ABCD ABCD

ABCD ABCD tillI I =1 I I I XE PG Hetc.EFGH

GHAB AB ____ ____

I I I =—G H A B I I I A B I I I (I)EF CD I I I IEF LI I CDJ

If

The twistorswill be written in termsof their spinor componentsrelativeto someorigin by~ Na ~7A a

L —‘~~A.e ~ eA

,~ A ‘~A’— ‘~~Ae~~1~L~

the e’s beingbasis spinstors[8], given by

— ~ _f1 0 0 0

(eA~)—(e ~ 0 0 0a — A _j’O 0 1 0

‘~ 0 0 1

3. The rulesof Penrosegraphs

I shall just state the rules, rather thanbother aboutwhere they comefrom. The reasoningissomewhatvague,and dependslargely on the fact that the methodworks. The original basiswas anattemptat a diagrammaticrepresentationwhich manifestlyconservesspin andhelicity. The detailsofthe reasoningaregivenelsewhere[5].

The rulesof writing Penrosegraphsareas follows:1. We havethe following typesof thingsto build the diagramsfrom:

a) Hollow vertices,0.b) Filled vertices,I.c) Lines carryingintegers,single a, doublea or triple a.

(36 A. Qadir, Penrosegraphs

2. Therearefour lines ata vertex carryingintegralnumberssuchthat their sum is zero(fig. 2(a) orfig. 2(b)) such that a + b + c + d = 0. The numberswith arrowsdenotehelicity, positive if the arrowpointstowardsa hollow vertex andnegativeif it points awayfrom the hollow vertex. The sum beingzero conserveshelicity.

3. A line canonly join oppositetypesof vertices.4. Externallinesmust carry positivenumbers.It shouldbe remarkedthat theserules neednot be heldto rigidly. An attemptis beingmadeto use

diagramswhich violatethe first part of rule 2. Also anothertype of line (carryinga number)is beingusedto join verticesof the sametype.The requirementin this caseis that the sumof all thenumbersat a vertex mustbe zero. However, in this paper,I shallstick strictly to the rulesstatedabove.

a c a

Ia) 1b)

Fig. 2.

The rules for correspondenceto masslessquantum electrodynamicsare as follows, at leasttentatively:

i) Two externallinesatavertex correspondto real particles.ii) The sum of the numberson the external lines is twice the modulusof the helicity of the

particles.iii) A virtual particleis representedby two linescoming from separatevertices,oneof eachtype.In particular a virtual photonattachedto an incoming and outgoing electron is given by fig. 3(a)

correspondingto fig. 3(b).The rulesfor writing the algebraicexpressionscorrespondingto the Penrosegraphsareas follows:I) Eachvertexcorrespondsto avariable twistor, the hollow vertex to a lower index andthe filled

vertexto anupperindex twistor.

(a) (b)

Fig. 3.


II) For each external line there correspondsa constanttwistor having an upper index if it is

attachedto ahollow vertexand a lower index if it is attachedto afilled vertex.III) A a B, A a!__B, A__a2__B correspondto (— 1)af(1 + a) I ) for a ~ 0, and to

\BI ) xlog (I )fora<0.

\B/IV) A contourintegral is performedat eachvertex aroundan appropriatelychosencontour. ‘the

result is divided by (2~.j)k_I_m, wherek is 3 times the numberof vertices,I is the numberof double

linesand m is twice the numberof triple lines.V) Any expressionto be integratedmustbehomogeneousof degree—4 in the variableover which

the integral is beingperformed.Sincelog (~~ is not homogeneous,for a vertex Wwe wouldwrite/p /Q\ \B/ IX IP X\

log (i / I )insteadoflo~(\ I ) ~andfortwoverticeslo~(I / I I )insteadoflo~ I\W W W W W Q W

4. Tree graphs

In this sectionI shall only considergraphswith no closed loops of double or triple lines. Suchgraphsarecalled treegraphs.I shall furtherrestrict myselfto graphswhoselinesonly carry zero,onthem. The contourintegral is ambiguouslydefinedasregardssign. This doesnot matter as the resultof the integrationis only an amplitude.The definition could be fixed so as to removethe ambiguity.(There is apossibilityof ambiguitydueto the existenceof other contours.)

1. Thesingle vertex (fig. 4(a))

B B

H ______________A ° ____ C A ° ~ C

0

D D

(a) Ib)

Fig. 4.

The algebraicexpressionfor this graphis (seeref. [5])

_l (_____

31 =1(2irl)thww W W ABCD

Ji I I IA B CD

138 A. Qadir, Penrosegraphs

Similarly the expressionfor fig. 4(b) is ~A B 1 C D

I I I I2. Two vertices

C 0

0 0

B ° _____________ EW 0 X

0 0

A F

Fig. 5.

The correspondingalgebraicexpressionis (seeref. [5])

11 91JWXY ____

W W W D E F! A B C

JAB CXXXX

3. Threevertices

B C

0 0

A 0 _____________ ° Dw 0 5

0 0

E

H 0 )‘ ° F

0

G

Fig. 6.

The algebraicexpressionfor this graph is

1 1 ~i,wXYW W C D E F G H W

I l I I I I I I I I IA BXXXXYYYY

Doing theW, X integrationsfrom the previousresults


if________ — —1‘~(2iri)~J~CD E PG HC D E I liP G H=C D F PG H

JABYYYY AB

usingthe identity (I).

4. Four vertices

Thereare two tree graphsof four vertices,one of which is much like the two and three vertexgraphs.However,the otherleadsto an interestingfive vertexgraph andwill be given here(fig. 7).

E

0

0 ° ___ F

C G

0 0 0

o 1w a 0 2 ____ H~I0 0

0 0 0

A L K

Fig. 7.

Following the typeof proceduregivenbeforeyields

14_l/ABC

5. Five vertices

There are four tree graphsof five vertices,threeof them beingsimilar to previousgraphs, thefourth is of moreinterest.It is

0

0° °F

a

C G

Ic o

B 0 1w v 0 Yj. H—I5

0 0 0

A K

N 0 Z ~ L

0

NFig. 8.

(40 A. Qadir, Penrosegraphs

It canbe seenthat

15= _________ 1

I I I I I I I I IABC DEF GHK LMN

We shallrepresentthis by fig. 9. It will be seenlater in this sectionthat this leadsto a new typeofdiagram.It alsogives someodd loop diagrams,oneof which reducesto the single vertex.

E0

D~4~E

A10 fo

B~f~<o>r4~H

01A K

0

M

Fig. 9.

6. Linescarrying non-zeronumbers

Considerthe graphgiven in fig. 10. The algebraicexpressioncorrespondingto this graph isf /W/WQ

tI/i I- 1 log\X/ P X 21(WX

‘6 (2iri)6 I W \~W W /D\2E P

I I I (I H I\A/B C\X,/X X

W W W W D E F QPut I = a, = /3, I = I = 1, I = I I = I = 1;

A B C P X X X X

c D

B ~ —1 E

Fig. 10.


,DEFQ2l~WXoodaAd$Ad7Ad8Ad�Ad~/I I I

ABCP

/DEFQ //DEFQ

__ lo~\J/ /1~~~J= I A B C P / dand/3AdyAd5Ad�Ad4~

6 ~ 2 2~L1Tl) a8yS$ D E P Q

f /D Q \I IXEF IDEPQIlogi I I I I I I

WBC ABCP

— i I \A P / daAdô— (2iri)

2 I a252 D E F QJ 1111ABCP

To do the a-integrationwe mustput a= 0 (in minusthe first derivativeof the residue).In this casewe have

Q/

I XEP /DEFQlogi I I I I I I I

WBC /ABCP

/DQEF DQEF\ /

~

XPBC XABC//ABCP

DQEF

1 XPBC d~

‘~‘ DQEF

XAPC ABCPInsteadof evaluatingat the secondorder poleat 5 = 0, we canevaluateat the pole dueto

QEF DEF\I~ I I I — I I I 1=0.ABC ABC!


This gives

f~ Q E F D E F(S I I I — I I I

l \PBC PBC dS¶2 ~2 ¶2 f ~‘(i~f 1/? f ~ABCP ABC\ ABC/ABC

QEFDEP DEFQEFIllIll—Ill—IIIP B C A B C PB C A B C

DEFQ/DEP\2I I I I f I I IABCP\ABC

Now Q D

QEFDEF DEFQEP EP EPII II

PB CA B C PB CAB C PB CAB C

B C A B C P’

EF

16=7 D E P\2

I I I I\ABC

This methodof calculationis very tedious.I shall now presenta restatementof a methodduetoPenrose[5] which simplifies the procedurefor evaluatingdiagramswhose lines have numbersonthem,enormously.It is basedon the fact that rule III gives

a a±I

AA B=A B.a (~)

C 0

C 0 c d~1

B~ —g E B II A F e E

Fig. II. Fig. (2.


Now, considerthe diagramgiven in fig. 11 with the conditionsthat a + b + c = g = d + e+ f. Theeffect of differentiatingwith respectto A is to changea to a + 1. This causesg to changeto g + I.

Thus one of d, e or f must increaseby 1. Lookingat the coefficientof in the expressionresulting

from the differentiationwith respectto ~, will give the diagram(shown in fig. 12). This is easilyseen

to be the sameasdifferentiatingthe initial diagramwith respectto ( . Carehasto be takenwhenusingthis method in loop diagrams. \ A

It is easyto checkthat = ‘6.

a(?)

We canusethis methodrepeatedlyto getfor the graph given in fig. 13(a),

BC

16(a)= IA\a =12 = (— 1~F(a+ 1)~A B C \“~

iI~ (I I I\DJ \DEF

Similarly we canevaluatethe diagramgiven in fig. 13(b),

ID E F G H\

b a b— a a — a B17 A b IA a’3 A b(1)I’(~1) C D E F G H~’a(~) By) 8(i)

DEFGH\”I GHCDE\b~ )L-+--LLLJ)

—‘ i~a+br~i ~ 1\ B \ B— ~ ij i ~a+ ii + i~ / C D E F G H\”~~5

(-HH--ELJJ\AB

B C B C

o a o a

A a~)~ —a ______ D A a.b -aw x w a o

0 0 -b 0

E E

____ yO ybF H F

0 0

G (a) . 0 (b)Fig. 13.

144 4. Qadir. Penrose graphs

This type of procedurecanbe carriedout for all the previoustreegraphs.A new type of graph is

producedby differentiating15 with respectto , , , and applying I I I I . The result is

non-zero,but cannot be expressedas the previoustypes of graphs.We shallwrite it as shownin fig.14.

Fig. (4.

a4

3A1aD1aG1aL1

= — 1 ~ I~

+2!(~ + ____ 9 y + ~9~yy

+ + 9 y 9~ + ___y 9

~ +~y9~yy9~9~

+ ~ + ___ ___ ___

+ 9~9y~~y + ___ ___ ___

where9=A~~,4~=D~i~,9GHK1, Y’’~~’1~,

~ 4=~,~

A. Qadir. Penrosegraphs 145

andL,-~...L-.J..,-Jagainrepresents�“~“~, the bumpsbeingusedto indicatethat thereareno contractionsat thosepoints.

5. Multi-line treegraphs

Theseare graphswithout loops but with doubleor triple lines betweenvertices.I shall follow thepreviousprocedureof working out the simplestcasesin detail, andleavingthe morecomplicatedonesat the stagewherethe methodis obvious.

1. Double-linebetweentwo vertices

I~

Fig. 15.

The algebraicexpressionfor the abovegraphis

1 I________W W /W\2C D

J I I (I 11 IA B\X!XX

W W C D hR SPutting I = a, = /3, = ‘y, = 8, anddefining four twistorsP, Q~I and I suchthat

w . wI =1T=e’°cos~, I =~=sin4 (0~O<2ir)P QR SI =p=e’°cos4i, I =o=sinq~ (o~~cz!!)x x 2

so thatpir + o~= 1, and choosingthem suchthat

CDR CDSII I=I (=0 (B)ABQ ABP

we canevaluate~BWX,andthen19.The reasonfor this complicatedchoice is that the contourfor integration around the pole at

/W\2( I ) = 0 is not an S’ but an ~2• The point is that the doublepolecan be regardedin somesenseas\X/

(4.6 A. Qadir. Penrose graphs

two poles coming together.Two S”s would slip off the singularity.For simple poles the contoursessentiallyseperatethe two singularities.

Herewe will now have

~WX= +da Ad/3 AdyAdS A(cos2~. O—2isin3~cos~d9Ad~—2isin~cos3~d6Ad4~)CDRSI I I IABPQ

= —da A df3 A dy AdS A dO A (2i sin4 cos4)dqSCDRS

I I IABPQ

CDRS(I I I I

— A B P Q daAdf3Ad’yndSAdOA2isin4cos~d4’~ (2iri)5 — C D R ~ 2

\ABPQ/

CDRSI I I I

= A B P Q dO A 2i(sin 4i)(cos4)dçb2iri R S

(~CD

P Q

Using conditions(A) and (B) we get2,r ,r

CDRS~III

~_ A B P Q — ________________________

19 2iri IC D R C D S \2

I I I sin2 4~i+ I I I cos2~)\ABP ABQ

Changingvariablesto °° ‘~‘°

CDR CDSsin24 I I I +cos24 I I I

ABP ABQdxVi

2l(sln4)(cos4)d4 = C D R C D S

I I I — I I IABP ABQ


CDR CDSat4i=0,x= I I I =x1,at4=~,x= I I I =x2

ABP ABQ

CDR ~ X2 CDR S ICDR CDSI I I 1 f~~=—iI I I I I lxi I I

ABPQ(~2~1)x ABPQ/ ABP ABQ

Using condition(A) we get

R S

CDR CP S CD CD CD CDR SIIIXIII=Il II =11 hIllABP ABQ ABP ABQ AB ABPQ

lCD4=—l/ I I

lAB

2. Doublelines betweenthreevertices

B)W~X~Y:C

A ‘1~ ~R’ “s D

Fig. 16.

The algebraicexpressionfor the abovediagramis

1 1 ~WXY‘10 = (2iri)~JW W /W\

2 7Y\2Y ~

I l I (I III II IA B \XJ \X! C D

It can be shown(by usingthe previoustechniquetwice over) that this gives the sameresultasa singlew

vertex. The only point to note is that we can not use the previousresult by taking I integral as

I Xperformed,as the X integrationis over ~ now. The extensionsareobvious.

3. Doubleline with non-zeronumbers

Fig. 17.


The algebraicexpressionfor this can be got from the graph with lines carrying no numbersbydifferentiating,

a” (Dy‘9a = (— 1)°F(a+ 1)~c D a+I~By) 1 I I

A AB

4. Thetriple line graphL a

~ ~i ~

A ___________ B0 w 0 x 0

Fig. 18.

The algebraicexpressionfor this graphis

~1 1221WX~h11(2~7.j)4~W (W\3B

JA \X)X

W B IIIRS TPut I = a, I = /3 anddefinethe twistorsK, L , M, I , I and I suchthat

A XW RI =K=e’°cos4~cos~I; I =p=e°cost~cosi/1 (O~O<2ir)K xw . S

I = A = e~cosq5 sin f~ I = = e~lxcosçb sin fr (0 ~ x <2ir) (A)L XW TI =s=sinq5; I rsin4i (0~,~z~)M X

so that ,cp + Au + ~r = 1, andchoosethem suchthat

R S TI = I = I k(say)K L MR S T B S T B R T B R S

I = I = I = I = I = I = I = I = I = I = I = I =0 (B)A A A K K K L L L M MM

therebeing 15 conditionson the 6 twistorsto be chosen.As eachtwistor has4 components,thereissufficient freedom of choice, provided it gives the correct contours. The contour here is an

x S1 x S1 which is the correctonefor threepoles“pinching” the contourstogether.


WX_~~aAd$A(KdA- BRSR

I IAKLM

— da A d/3 A dO Ad~A (2i sin4, cos4, d4,)A (2i sin t/i coscli dcl’)— k3

— 1 da A d$ A dO A d~A 2i(sin çb)(cos4,) d4, A 2i(sin ~‘)(cos4ir) dcl’Ii1~(2iTi)4 B R S T ~

~)/(BPST)2

AKLM

k6(’I)[ . .

— A1dOA d~A 2l(sln 4i)(cos4,) d4, A 2i(srn4i)(cos ~c) difr

(21ri)2 T (B\3 6 3

il Jk(Kp+Au+11Lr)J \A/

havingdonethe a and /3 integralsandusingcondition(B). Using condition(A) anddoingthe 0 andxintegrals

,r1

2

= f f 2(sin 4,)(cos4,) d4, x 2(sin *)(cos 4’) d4’ = = A 0 B.

jtI 1’Thus onecanget the resultgenerally A ~ A B

Fig. 19.

/B \~“~

a B=lim(—l)”F(l+a+�)( I\A

The triple vertexcanbe usedfor “gluing” appropriatefree-ends(i.e. of oppositetypeswith the samenumberson them) together.We shall considerthis type of uselater, in the loop graphs.

5. Thefour line graph K~\ ,/ N /.

Fig. 20.

The algebraicexpressionfor this is

1 f.~WX

‘12 = (21~i)~~(W\

4~

J~x)


III IPRS TDefine the twistorsK, L,M,N, I , I , I ,and I suchthat

w PI = K = e’°cos4, cos4’ cos o = = e’°cos4, cos4’ cosK X

W RI =A=e’°cos4,cos4’sinw~ I =p=e’°cos4,cos4’sinwL X

W . S . W TI =~=e’~cosØsin4’; I =u=e~cos4,sin4’: I =~=sin4,; I =r=sin4,

M X N X

so that icir + Ap + pu+ yr = 1, and choosethem suchthat

R S TP ST PR T PR S

I=I=I=I=I=I=I=I=I=I=I=I=oKKKL L LMMMNNN

P R S TI = I = I = I =k (say)

K L M N

thus giving a contourover an S4. There are 16 conditionson the 8 twistors, thus giving sufficient

freedomof choice,

~WX=(KdA Ad~LAdV+AdpAdVAdK+pdPAdKAdA+PdKAdAAdp)A

A(lTdpAduAdr+pduAdrAdIT+udrAdIrAdp+rdITAdpAdu)

PRS T

KLMN

— 1 ~ dO A do Ad~A 2i(sin 4,)(cos4,) d4, A 2i(sin 4’)(cos 4’) d4’ A 2i(sin w)(cos~) dai‘12 3~’ 4 4 12 16

(21rl) j k (KIT + Ap + pu + vr) k 1k

=J sin2~d4’Jsin24’d4’J sin2wdw = 1.

6. Loop graphs

Unlike the Feynmangraphs,the loop-graphsheredo not diverge.Unfortunatelytheyarenot mucheasierto calculatethan the Feynmangraphs.They do not, in generalcorrespondto the sametype ofscatteringprocesses.They do not necessarilyincluderenormalizationdiagrams,in fact, the electron—electron scatteringgraph appearsto be a single loop, and the electron—photonscatteringgraph adoubleloop. Beforegoing on to considerscatteringgraphsI shallusethe triple line to gluetogethertwo free endsof the graphof two vertices.

A. Qadir, Penrosegraphs (51

1. Thetwo vertexclosedloop

B C /~E~

jo 0 Ii~0 ,[ _________________ 0 / ) ~ a -a a

A Oglue L - — ~1w -a x / / Z

a a / ////

/ _z

Fig. 21(a).

B C

A ° ‘~‘ ° x~/o

a ay~

Fig. 21(b).

Thisgives the diagramwhich, expressedalgebraically,is

1 1 ~YYZF(1+a)C D y a±1 ~ y 4-a (if a<3)

f/y\a3 /y y Q\

— 1 1~k)log ~ j .~,)~YZF(1+a) •fC D (i a~3).

J ABZ!

IIM NDefinetwistorsK, L, I and I suchthat

MN NM CDN CDMI = I =1; I = I =0; I I = I I I =0.KL KL ABK ABL

Put

Y Y M NI=K, l=A, =p, I=i’K L Z Z

suchthat

iqi + Ày = 1.


Choosean origin, anda frameof reference,with respectto which

Ya = (2e~’cosc, 2 sinc, e’1 cosh, ~ sin h)ze = (2e~’~cosc,2e~~’sin c, e’1 cosh, e’~sinh)

K~= (0,0, 1,0), L~=(0,0,0,1)

Ma=(0,O,0,1,0), Na=(0,0,0,l)

0<b,f,g,k~2ir,

Thus we have

CDY CD CDYI I I I I /Kp+Al1\ I I IABZ AB(MN) AB~

\K LI

Proceedingas beforeonegets

~YZ = 16e2” dg A db A df A dk A (sin2c) dc A (sin2h)dh.

Thus for a <3, we get

1’(a+l) l6dbAdfA(sin2c)dcA(sinh)dhAe2”dkAdg‘13 (2iri)4 C D

I (1 + 4ehI~)S_~2

= 16f(a+1)f sin2cdcA (sin2h)dh =_f2F(~~J(l+Z

4~s~a

havingput e = z .~.(1/i) e1~~dk = dz.Clearly a ~ 0, as we could not havethe negativenumberson the externallines E and F. Also for

a = 0, we seethat~ = 0, as

I zdzt’(l+4z)~

0~Thus

a=1

AB

D a=2.

AB

The abovecontourdoesnot work for a ~ 3, as onegetson to the wrong Riemannsheet.This mayI Q YQ

be thought of as the problem of not having the freedom to chooseP and I so that I = I = 1.P Z


However,generalizingthe previousresultsuggeststhatonecanput

114(1)a C D

AB

or ~ 111a

Fig. 22.

Onecandifferentiatethe abovewith respectto A to obtain

•

Fig. 23.

4. ThesimplestscatteringgraphB C

0 0

A wJ. 0 0 a~ ~ m’

‘I ii’

0 0

/ a p \

H ~ 0 ‘~‘Y ° E

0 olF

Fig. 24.


- 1 1 ~wXYz“~(2iri)’~,LW W W C D Y Y Y Y G H W

J I I I I I I I I I I IABXXXXEFZZZZ

The simplemethodof connectingtwo of the free endsof the graph(given in fig. 25) by the triple line0

u -_~!_-c~.—~_-v cannot beusedas it wouldgive zero. Onecouldtry adjustingit by changing


C

0 0

A 0 0 0

0 0

H 0 0 LK 0 ° E

0 0

F

Fig. 25.

1 —lthe numbersso that u —~-—— v couldbe used,but the previousdiagramwould not givethis one by differentiation as a closed loop is involved. We thereforehave to evaluateit by firstprinciples.

IIMNIIR SDefining the twistorsK, L, I I P, Q, I and I suchthat

W N Y SI = I = I = 1=1L X Q Z

andputting

W W W C D M Y YI=a,I /3,IK,II&Ip,I�,I4,A B K X X X F F

Y G H RI IT, I ~y, I =~, I =p;P Z Z Z

~,wxyz—~ Ad/3AdKAd~Ad8AdpAd�Ad4,Ad1TAd7Ad~AdP- CDMN GHRS

I I I I I I I IABKL CDPQ

CDMN GHRSI I I I I I I IABKLEFPQ

‘15 — (2iri)12

fda Ad/3AdKAd~Ad8AdpAdEAd4,Ad1rAdyAd~AdpICDMN G HR SCDMNGHRS

1w w y yI I I ——~

J ABKLABKLEPPQEFPQ


CDMNGHRSI 11111

ABKLEFPQ— (2iri)4

dK A d~A dir A dp

M NR SM NR SI I I (I

< XCD ZGH XCDIZGHI I I I I —I— I I

WAB WAB YEF YEF4 I I IK LK LP QP Q

CD GH CD GNPut I I = c.~, I I = 4, I I I ~ I I I = ~~,anddo theK, IT. p andp integrations

AB AB EF EFin thatorder.Now

MN MN MN RS RS RS

~1 __________

CDMNGHRSI I I I I I I I

JABKLEFPQ dp.Adp15 (2

1ri)2 M N S R M N S R

I I I I

I I IXKLZ XPQZ

Now

MNSR NSR MSR MNSR NSR MSRI I I I I I I I I I I

=~ o• - ; AV =p~V -AVXIIZ II II lIz lIZ lIZ

KL KLZ KLZ XPQ PQ PQ


CDMN GHRSI I I I I I

1ABKL EFPQ115

LIT1

f dpI SMNS SMNR RMNS RMNR

___ ~ ___ A V +~ ~ 1L ~!7”\+~ ~ ___

J ~)(~2 \LKPQ ~)HH2)~/5 R\2

CDMN GHRS I MN

\ I I I 1 - 2

LKPQ —Il

RMNRSMNS

41 0 A ___ ___

I I I I ILKPQLKPQ

CDGH 1GHCD 1CDGHPutting a = I I I I , b = I I I I ‘ c = I I I I , and4ABEF ABEF ABEFusing the identity (I), we get

~ = —(2ab+ 2bc + 2ca = a2— b2 — c2)~2.

Penrosehasshown[5] thatusingmomentumstateswe canget

B C

0 0

0 0A —0

I~~(l)= 1 I

H ___ - ___ E0 0

0 0

6 F

Fig. 26.


B C

I 0

A 0 -t L_0

2 2— — 8 3 ~ a—cI~4(1,1)— 0 ° — 8D~aF”8B

83H8 ~l +~ I~4.

H -1 E

0 II

6 F

Fig. 27.

This hasbeenevaluated[8], and gives avery complicatedresult which I shall not presenthere. Itappearsto correspondto electron—electronscattering.

7. Externallines with negativenumbers

Althoughexternallinesarenot allowedto havenegativenumbers,to be interpretedas “particles”,theymayperhapsbe interpretedas virtual “particles”. It is, therefore,interestingto look at them asparts of other graphs,or for physical interpretationas virtual particles.I shall follow the previousprocedureof startingwith the simplestgraphsandbuilding them up.

1. Thesinglevertexand a minusoneC

P

0

B w-I

0~0

A

Fig. 28.


hI6(21ri)3~w ~ log (i/i).

PutW W W W

I a, I f3’ I I =1.A B C Q

158 A. Qadir, Penrose graphs

Then

daAd/3Ady

I I I IABCQ

I (~TT ~__ 1Iv~____I = 1 log A B C Q/ /daAd/3Ady________ 216 (2iri)~ I I af3yABCQ

~ [1111111- 1 log(/3 A P C Q-P A B C)-logA B C Qd/3

2iri I I I-— ABCQ

~~lIIAPCQ

I I I I I IABC PAB CQ

2. The singlevertexand a minus two

CP

0

I,

B 2 ________ Djw-2

A

Fig. 29.


I~7= (2iri)3 w w ~ ~ log ~I / I ). Put I = a, I = /3’ I = I = 1.1 f ~liWF(3) W 1w lw W w ~‘ ~i(1)1D D P A B C D

r

___ L (J’ I I 1)11111 _______Thenwe have /~= 1 2 log A B C D —logA B C DdaAd/3Ady117 (2iri)3 I I I a/3~y

ABCD


I I I I I I I I1 2 log(/3A P C D-P A B C)-logA B C D d/3

21ri I I IABCD

/1 I I I\2APCD~ 1

I I I / I IABCP/ABCD

It is obviousthat for higher orderswe just haveto differentiatethe log termsextratimes. Thus

CP

0 / a‘I I I I’

(a) a / IA P C D\I B 0 = ______ ______w—a

0 ABCP ABCD

A

Fig. 30.

3. Thesinglevertexand two minus ones

P

B -l

0‘S

A

Fig. 31.


:8= (2~i)~~ w~~ylo~( ~‘/ ~‘)log (‘~‘,/~‘)F(3).

Put I =a, I =/3~I =v I =1,A B P R

(60 A. Qadir, Penrose graphs

__ [(~~ ~~/“ I I”) ~ ~/‘I Ii)]- 1 I’(3)jog R A B P R A B P -logy, log R A B P R A B P

‘8(2iri)2 I I IRABP

da A df3 n dyX a$3

__£ [(~~ III I][(r DII IjI J J’(3) log R B P — log y — log R A B P log R B P — log R A B P

(21T1)2J I I I IRABP

df3 Ad’)’X ~

— 1 1 R(/3,y) d/3Ady“(2

1Ti)2+l I I I /33

JRABP

We haveto put /3 = 0 in the secondderivativeof R(f3, y) with respectto /3 and divide theresultby 2!,

I I IF I I I I I I I 1 I I I I I ID(Q \_2! Q A R PUog(S A B P-yS A B R—/3S A R P)-log(R A B P)1%1~jJ~7/ — ______________ ______________ ______________

(Q A B P-yQ A B R-f3Q A R P)

I I I II I I I I I I I I I I I I I-2! S A R PLlog(Q A B P--yQ A B R-j3Q A R P)-logy-logR A B P

1111 1111 1111S A B P—7S A B R-/3S A R P

______ p)2 [log(~~ i~ti ~1~’-y~ )~~1~)______

(QABPQABR)2 III I-logR A B P

1111 I 1111 1111

— I I ~ 7 I I [log(Q A B P—-yQ A B R)—logy—logR A B P(S A B P-S A B R)

2

I I I I I I+2 ____ QARPSARP ____

1111 IIIIIIII(QABP—

7QABR)(SABP-7SABR)


Put

‘18 ~1 +J2+2J3.

Let the singularitybe due to the pole

I I I I/I I I IyyiQ A B PIQ A B R.

Thento evaluateJ2 onehasto do acontourintegralround a cut, as the contouris not allowedto passbetweeny = 0 and ‘y = y~,andnot surroundingthe pole (seefig. 32)

Fig. 32.

I I/I I I I772S A B P/S A B R.

The cut maythen be evaluatedas the sumof the contourintegralsround ‘V = y~and y -~ ~. Now, as

I I I I I I I I I Ilog(Q A B P/y-Q A B R)—*logR A B Q.

Thus thereare no poles as y -.~ ~. Hencewe may evaluate.12 at the pole y = 72 and J1 andJ~at thepole y = y~.Sincethe directionsof thecontourssurroundingthe poles 7 = y~and y = 72 are opposite,the resultof the contour integralaroundy = 72 mustbe multiplied by (— 1)

/1 I I 1\2 I I I I I I I I I IA R P) log(S A B P-yS A B R)-logR A B

I I I; 2~’ ~2I I I I\QARB/ iTl~.YYilRABP

Performingthe integralandsimplifying gives

Il I I I\2 I I I I~~IQARPl SABR~ kI I I I/I I I I I I I

\RABP/RAB QQAB S

162 A. Qadir. Penrose graphs

/IIII\2 1111 1111 [~III~jS A R ~\ log(Q .A B P/y—Q A B R)-logR A B2’~ [ I I I I 2iri(y—

72)2 I I I I ‘V

\S A B P/ R A B R

which gives

/T I I I\2 I I I IJjSARP~ QABP

2~IIlIJIIlI[III\RABP/QABSSABP

f[III rI II

_____ QARP_SARP d

I I I I I I I I I I I ‘V

R A B P Q A B P (S A B P-yS A B R)2iri(y-y1)

I I III I I- QARPSARP

— — F I I I I I I(RABP)

2QAB S

adding2.13 to J1 + J2, and simplifying we get

I I I~ I I I I I I I I I I I I- Q A R P Q A R S S A B P-i-S A R PS A Q P Q A B R~III[IIIIIIIIIII

QABSSABPPABRRABQ

One can similarly evaluatethe vertex with threeminus ones,but the answeris very complicatedand theredoesnot seemto be any application.

4. Thedoublevertexand two minusones

A

-a

B 0 ~ -1

“P

--S

C 0 -I

0

0

Fig. 33.


This graph is of interestas it seemsto leadto the analogueof the Feynmanvertex. The algebraicexpressionfor it is

h19(2~i)6fw ~ (W)2 C ~

Doing the contourintegralof W as before,we get

‘19 (2~i)3JI I II ~I I ~I C ~lo~ (~/~).ABPXABQX~

C D R SPut I = I = I = ~‘ I = l.

x x x xCDRS(I I I I

A B P Q logpd~AdSAdp‘19 (21ri)

3 C D R S C D R SI I I I I I IABPX ABQX

CDRS

I IA B P Q logpdp

2iii CDRSCDRS-~---L-~t -~-4---~tABPXABQX

We cannot evaluatethis atp = 0, as thatis a logarithmicsingularity.Hencewe evaluateit attheotherpole(in the denominator)

CDRS /cDR\ /CDR

ABPQ lOI~~~\_lO(14BQ‘9CDRSCD ~ 8%CDS

\2~iH3,I ~

/CDS CDR /CDR CDSIII III,,,,,,.”III III

1ogABP ABQ ABP ABQCD

jt~

(64 A. Qadir. Penrose graphs

One can thusget the graph

A

-Q

B w -1

‘20 C _____ -1 (Cr9

Fig. 34.

/DS DR DS DR

-l 1 i31’ ___ ___ ___120C D I C D SC D RC D S~C D R

II III III III IIIAB ABP ABP ABQ ABQ

DI CDR CDSICDS CDR

Ill/Ill(I I~ ABP ABQ/ABP ABQ

\AB!

8. Discussion

We have seenthe methodsof evaluatingthe Penrosegraphs,and the difficulty of computationinvolved. It might be wonderedwhy oneshould botherwith computingthem.The answeris that theyare not meantto replace Feynmangraphs,but to provide a formalism which avoids the divergenceproblemsof quantumelectrodynamics.

It shouldbe notedthat the formalism presentedhereis conformallyinvariant.This invariancecanbe brokenby introducingthe infinity twistor into the expressionto be integrated.This is representedgraphically by brokenlines (carryingsomenumber)joining verticesof the sametype.

The correspondencebetweenthe Feynmanvertex (fig. 35a) and the Penrosegraph (fig. 35b) isbasedon the following argumentof Moore and Penrose[7]. The Hamiltonian given in section 1expressesthe graph(given in fig. 36). Now operatingon the electro-magneticpotentialgiven in section1 by the D’Alembertianoperator,it can be shownthat [8, 12]

L14,AA = (2~i)2~ XAWA(Ua/aVj)(1~a/aUJ)F(U, V)~UV


Iv

X 01 -

0~-t

(a)(bI

Fig. 35,0, ~

0

Fig. 36. Fig. 37.

whereL.....J representsI”~jl represents‘“8’ and B/BVC) representsa/3V”, 8/BUJ represents8I8U~.Usingsomenaturalassumptionswe get

1w ip\ lu u\ ___

i J1og~/~)lo~(~/~) (Y3b8v1)(~aIaUJ)F(U, V)~UV= 2 uH(X, W) (2iri) I

xwhich correspondsto the graph (given in fig. 37), ElF being the source.If the sourceis an electroncoming in andgoing out, the graphbecomesas shownin fig. 38. This is to correspondto the Feynmangraph (fig. 35a).

0~ -,“0_ 0 0 - 01

‘aI /‘ ~-

-I 1 - 0

‘S5 0

“P

Fig. 38. Fig. 39.

166 A. Qadir. Penrose graphs

Fig. 40.

Using the correspondencegiven above it is found that fig. 39 correspondsto fig. 40. Whencalculatedin momentumstatesthe resultis found to agreewith the high energylimit of the quantumelectrodynamicpredictions.The Penrosegraph for y—e scatteringis given in fig. 41(a) correspondingto fig. 41(b) in Feynmangraphs.Again the result correspondsto the high energylimit of the quantumelectrodynamicpredictions.Since the results are basedon a conformally invariant formalism, theycan be expectedto correspondto the high energylimit.

~Io-I~~

a) )b)

Fig. 41.

The really hopeful feature of the Penrosegraphs is that they may not admit the infra-reddivergencegraph,while retainingthe Feynmanvertex for a virtual photon.The renormalizationgraph

(a) Ib)

Fig.42.

fig. 42(a) is found to be proportional[9] to fig. 42(b), i.e. in somesensethat—‘c7—— a(~Fig. 43.

)cooi >=~=<(a) )b)

Fig. 44.

Similarly fig. 44(a) is proportionalto fig. 44(b), i.e. in somesenseFig. 45.


Theresultseemsobvious,but the actualcalculationof thegraphandthe constantof proportionalityisvery complicated.

Notice that the Penrosegraphsare so definedas not to give renormalizationdivergence.It maythereforebe hopedthat the formalism may lead to a theory free of divergenceproblemsdue torenormalization.Howeverit should be mentionedthat Sparlinghasshownthat the topologiesof thecontoursusedto integrateoverare not unique[13].

Acknowledgements

I should like to expressmy gratitudeto J. MooreandG. Sparlingfor theirwork on Penrosegraphs,andespeciallyto R. Penrose,the inventorof thegraphs.I mustalso thankG.S. for going throughthemanuscriptandoffering manyusefulsuggestions.

References

[1] R. Penrose,An analysisof theStructureof Space—Time,AdamsPrizeEssay(Princeton,1967).[2] R. Penrose,Twistor Algebra,J. Math.Phys.8 (1967) 345—366.[3] R. Penrose,Solutionsof theZero Rest-MassField Equations, I. Math. Phys. 10 (1969)39—39.

[4] R. Penrose,Twistors andthe Quantizationof Curved Space—Time,Intern. J.Theoret.Phys.1(1968)61—99.[51R. Penroseand M.A.H. MacCallum, Twistor Theory: An approachto the Quantizationof CurvedSpace—Time,Phys. Reports6C No.4(1973)

243—315.[61R. Penrose,private communication.[7] R. Penrose,privatecommunication.[8] A. Qadir, Twistor Fields, Ph.D. thesis, Birkbeck College,London (1971).

[9] G. Sparling, private communcation.[10] R. Penrose,Structureof Space—Time,in: BattelleRencontres,ed.C.M. DeWitt (Benjamin,1968).[11] R. Penrose,Applicationsof NegativeDimensionalTensors,in: CombinatorialMathematics,ed.J.Welsh (London, 1971).[12j R. PenroseandA. Qadir,Field Equationsin Twistors, under preparation.(131 C. Isham, R.PenroseandD.W. Sciama,editors, Quantum Gravity (Oxford University Press,1975).

penrose graphs

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