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f

This is to certify that the
dissertation entitled
FINAL STATE SCATTER I NG IN Z PRODUCT I ON
via THE QUARK-GLUON INTERACT I ON

presented by

THOMAS MI TCHEL EWING

has been accepted towards fulfillment
of the requirements for

EH) degree in PHYS T CS

imam“

Major professor

Date QA'AAQ— 2L /??0
/’ /

MSU is an Affirmative Action/Equal Opportunity Institution 0-12771

 

PLACE IN RETURN BOX to remove this checkout from your record.
TO AVOID FINES return on or before date due.

DATE DUE DATE DUE DATE DUE

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

MSU Is An Affirmative Action/Equal Opportunity Institution
c:\cIrc\ddeduaprn3—p.1

 

_
_
_
_
I

 

 

=‘inal State Scattering in Z Production via the
Quark - Gluon Interaction

BY

THOMAS MI TCHEL EWING

A DISSERTATION

Presented to
Michigan State Uniyercity
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY
DEPARTMENT OF PHYSICS

1989

 

 

ABSTRACT

Llytical calculation of photon scattering between final state quark and a final
e lepton in the process q 9 => qifq => 20 (1 => 1+ 1' (1. Result is found as
Ltio to the same process without final state scattering. Motivation for the
ulation is as a first step towards calculating photon scattering between final
e leptons and spectator quarks or debris particles, possibly predicting an
:rvable asymmetry in the 20 decay curve.

 

 

Dedication

thesis is dedicated to Jon Pumplin for the right suggestions.

‘0 L . 8 Community College, upecially Claude Watson and Melanie
, for providing their computer for Primins ‘hi' mm’cfiP‘

 

 

 

TABLE OF CONTENTS

Chapter 1 - Statement of the Problem

 

:tion 1.1 ...General Introduction 1
:tion 1.2 ...Introduction to this Thesis ...................................... 4
:tion 1.3 ...Division of the problem into

Allowable Helecity States ..................................... 7
:tion 1.4 ...Notation to be used ...... 9

 

(labelling of linear momenta,
metric used, gamma matrices,
the Epf function)

:tion 1.5 ...Preparing the Integrals 13
(choosing the frame of reference,
the J ACOBIAN and integrating out
the onoshell delta functions,

. divergence regulation)
:tion 1.6 Evaluating a Preliminary Diagram ......................... 16

 

Chapter 2 - The Calculation

:tion 2.1 Introduction 18
:tion 2.2 Calculating the BORN Term 19
(Spinor replacement,
removal of 2° vertex (.4 + 375)
and strong coupling constant
Polarisation vector replacement)
tion 2.3 Calculating the LOOP Term 21
(Spinor and Polarisation vector
replacements,
removal of 2° vertex (A + B75)
and strong coupling constant
Expansion of k”,
reduction to 19 irreducible integrals)
tion 2.4 Calculating the RATIO ......................................... 27
(FORTRAN Program,
Physically valid set of
input paramters)

 

 

 

 

 

 

 

 

Chapter 3 - The Result

Section 3.1 Introduction
Section 3.2 Choosing the Display Mode .........
Section 3.3 Presentation of Results .......
Section 3.4 Focus on M 2 Peak
Section 3.5 Edge Activity
Section 3.6 Connecting with the Center of Mass Reference System ........ 40
Section 3.7 Analysis of Results 44

 

 

 

 

 

 

 

Chapter 4 - Verification and Validation

 

 

 

 

Section 4.1 I ‘ J " 45
Section 4.2 Verification of Epf(P1, P3, P4, P5) 45
Section 4.3 Verification of Spinor replacement 45
Section 4.4 Verification of SCHOONSCHIP Born Term program .......... 45
Section 4.5 Verification of SCHOONSCHIP LOOP Term program

TRACE calculation ............................................ 46
Section 4.6 Verification of Divergence Regulation Coefficents ................ 46
Section 4.7 Verification of 19 Integrals, analytic solutions ...................... 47

Chapter 5 - Conclusion

 

 

 

 

Section 5.1 Introduction 49
Section 5.2 Conclusion 49
Section 5.3 Relation to other work 49
Section 5.4 Future Work 50

 

lllltstlI-I-Jl

 

e main text is written to be read quickly without detail and is thus not
15thy. The proofs and calculations are contained in a bulky set of Appen-
which are referenced in the text.

TAB LE OF APPENDICES

1dix A ..... Allowable Helecity Combinations

1dix B ..... Gamma matrices, metrics, and the Dirac equation

1dix C ..... Expanding Epf(P1, P3, P4, P5)

1dix D ..... Feynman Rules

idix E ..... Ignoring the 75 in A + .375.

1dix F ..... Evaluating the Jacobian

idix G ..... Absolving the Quark Singularity

1dix H ..... Spinor Replacement

1dix I ..... SCHOONSCHIP program to calculate and verify the BORN
term

1dix J ..... Polarization vector Replacement

[dix K ..... SCHOONSCHIP program for the LOOP term

)dix L ..... The Linear Expansion of Is”

dix M ..... Base Integral reduction

dix N ..... Nineteen Irreducible Integrals , Solved Analytically

dix O ..... FORTRAN program to calculate the result

dix P ..... Ascertaining the Physical Range of Parameters

dix Q ..... Verification of Spinor replacement

dix R ..... Verification of SCHOONSCHIP LOOP program,

TRACE calculation
5 ..... Verification of Divergence Regulation co—efiicent cancellation
T ..... Verification of 19 irreducible Integrals
U ..... FORTRAN Output - Sample Results
V ..... Selecting Representative Points for displayed output
W Evaluating a Preliminary Diagram

 

 

Secti

The I
uticl
four 1
date
The ‘
is ((
most
vii t

the i
Ion.
Wit]
cons
theo
tend
deie
FNI

Reg

Chapter One
Statement of the Problem

Section 1.1 ... General Introduction

The 1984 status of particle physics is summarised in the comprehensive review
article of Eichten, Hinchlifi'c, Lane, and Quigg (EHLQ)1. Of the traditional
four forces, the gravitational force is largely unincorportated (though since that
date more promise has evolved from the futher development of string theory).
The theory of strong interactions rests comfortably in quantum chromodynam-
ics (QCD), though the experiments are difficult and development is slow. The
most remarkable success is the union of the weak and electromagnetic theories
via the application of non-Abelian transformation invariance and spontaneous
symmetry breaking, introducing the Higgs sector. As described in that article,
the investigation of the Higgs sector will await the next generation of accelera-
tors.

With present machines the 1983 verification of the Z0 and W predicted masses’
constituted the first hallmark in the experimental resolution of the electro-wcak
theory. Following this, the second level of refinement will be experiments ex-
tending the precision of the theory, the most recent of these being the e+ e"
dctermlilnation of the Z0 mass and width at SLACa, and futher resolution at
FNAL .

Regarding the production of Z0, the seroth order (Drell-Yan) processes are:

¢1>_ _< q
0

a Z '7 a

q > < l—

where its interference with the photon process is noted.

The analytic calculation of these amplitudes presents no difficulty and is given
in EHLQI. Experimentally the measured invariant mass of the final state par-
ticles peaks at the z0 mass, (and for the case of a quark anti-quark pair would
appear as a two ._]_e_t phenonon).

 

 

 

Wit
crlc
The

mans—J

 

h the improvement in the experimental precision it is important to do the
ulations to first and higher order in perturbation theory.

first order diagrams are:
:lmm < —tnmnr

 

 

 

 

 

Ie calculations are summarised in the workshop of Berger, et al.4. The
esses, known as 2 particles in and 3 out, represent a degree upwards in the
:ulty of their analytic solutions. Futher, to forward the amplitudes to cross
ons, the kinematics of phase space of 2 in and 3 out processes including
soft gluon correction of the first two diagrams above is non-trival. Even so,
d in the precision of knowing the Z0 mass and width, the calculations have
carried out and occupy a standard place in the literature.4'5'6

articular current interest is great precision in the width of the of the 20, as
reflects the number of quark and lepton flavors that exist.
1 the EHLQ paper:

 

3 82
rzzw[1_2zw+ ZWJD

3x

e D is the number of kinematically acessible quark and lepton doublets....”

ave faith in first order experimental agreement, it must be held the pro-
5 measured are free from contamination by other processes that would be
tectable in the measuement. For example, the first order contribution of
l+l’q :

could bi
the final

This we
product
at SLA
The ca
telly a.
hilly I
t body
and V:
The cc
bill“
vtllm:

Even I
lttlng
Mic

 

 

be contaminated by the reseattering of one of the decay leptons with
al state quark.

 

 

'ould cause a change in the invariant mass plot from that inherently
ed by the Z0 decay for mesurements made at FNAL, though not those
.0.

lculational difficulty of diagrams containing a loop increases dramati-
I the number of vertices or ‘points’. Three and four point diagrams are
tandard. The above is a five point diagram. There must be developed
of calculational method making these diagrams ammenable. ’t Hooft
,tman have devoted a technique paper7 to procedures for these integrals.
nplexity and large number of terms involved necessitate the aid of alge-
anipulation programs, such as the SCHOONSCHIP program written by
15.

vrse than the reseattering that occurs above, (which is first order), reseat-
ould appear as a seroth order contamination via a spectator or debris

>————<

 

 

 

g the invariant mass plot to an even higher degree. This zeroth order
ination is very hard to estimate as it involves interaction with particles
olved in the Z0 production. An assumption can be made that the ratio
seroth order reseattering to the seroth order (Drell-Yan) diagram alone
oximately the same as the above loop diagram to the first order term
IO the effect of all reseattering can be estimated by calculating the latter

esis is the estimation of this effect via the calculation of the latter ratio.

striced to final state leptons, as the case for final state quarks is done
9

re.

1 1.2 ... Introduction to this Thesis

tittering, first order Zo production can take place according to the usual
hti—quark union decaying into a lepton pair:

Irate assessment of both the Z 0 runs and the width of the decay curve
a number of questions. To this end it is important to examine possible
tions to shifts of the peak and asymmetric distortions of this width.

ninetion of one possible source of contribution begins with the ques-

‘HAT IS THE PROBABILITY OF A LEPTON SCATTERING WITH
FATOR QUARK ON THE WAY OUT ?

>———-<

 

step, one can solve the reduced problem in which the spectator and

Ollt

: production quarks emerge from a common gluon source:

X"?

I is the calculation of the above five-point and equivalent diagrams
restriction of two on-shell lines (The introduction of the on—shell re-
is to make the LOOP integrations managable, as they reduce the
is to two dimensions. It is expected that if the diagrams here exam-
iee small effect under these restrictions, no great enhancement would
heir relaxation. This point is futher mentioned in Chapter 5 ‘Con-

 

ie two diagrams with the photon scattering switched to the other

 

 

t0“

e obtained, its interference RATIO to the process without scattering

:_<_+;—'\<+;[ “““ f+1 """ f +)-{ +)—<E|2

 

2

:_ """ < +r’\<

 

roduee an observable asymmetry in the 2° production curve.

 

asymmetry would experimentally appear in data gathered at FNAL
in data from SLAC, a comparison of the two would reveal an experi-
limit of resolution of this efi'eet.

{ginal motivation for the project was as a step in explaining some ex-
.tal results which later died away - excess photon production in the 20
3roeess not explainable by bremsstrahlung. The next step in that now
% motivation would have been:

 

 

 

 

7

it diagrams are hard to evaluate and the value of this work includes
opment of the skills and procedures for working them, independent of
.eular project. A follow up calculation would replace the scattering
ith a gluon and final state leptons with quarks, greatly enhancing the
:y in the ratio of the strong coupling constant to 5. This is essen-
same calculation except the interference term is zero due to the color
ess of the BORN and LOOP diagrams, the BORN term being a color
nd the LOOP a color octet. For this reason, the lepton anti-lepton
ion calculated here might still dominate, and is also easier to compare
ent experiments.

ve mentioned gluon scattering calculation has already been finished9
results will be compared with this photon scattering calculation in
i - ‘Conclusions’.)

the RATIO is immediately expected to be of order alpha (1/137)
e two added electromagnetic vertices, this could be considerbly im-
the relatively short ( 10‘24 sec. ) lifetime of the Z0 keeps the decay
L the locality of the exiting quark and available to the re-scattering

‘. is of the general type described by ’t Hooft and Veltman, though
the use of on-shell approximations, and uses the SCHOONSCHIP pro-
eltmau to calculate the TRACES and perform algebraic substitutions.

in 1.3 ... Division of the Problem into
Allowable Helicity Combinations

mass particles only plus or minus projections of the spin onto the
:nenta exist.

Ipton line :

 

 

 

 

 

 

itin

qua

lep‘

La]

1::
avarice-awa—

well known” and demonstrated in Appendix A that the final state lepton
anti-lepton must have opposing helicities, while the incoming and out-going
:ks must have the same helicity.

 

means for each helicity of the GLUON there are four allowable
m-quark combinations:

Initial Final

:1 Gluon Quark Quark Lept on Anti-lept on
(*1) (*1/2) (*1/2) (:1/2) (*1/2)
+ + + + -
+ + + - +
+ — - + -
+ - - - +
- + + + -
- + + - +
_ - _ + -
- — - — +

be various helicity states are orthogonal, as are all quantum mechanical
set states, interference with a Born helicity state will only be non-sero for
OP state of the same helicity.

eforc the answer is formed separately for each of the eight permitted helic-
.ates, labelled A-H in the above table.

 

Sul

Th1
use

Section 1.4 .... Notation
b-Section 1.4.1 ....Diagram Notation

e five point lepton scattering diagram will serve to illustrate the notation
d for all diagrams in the problem.

P3
:(A + B ‘75)?"
P4
11
7'6 1 111 P5

 

 

 

:linear momenta are labelled clockwise P1....P5 beginning with the GLUON.
is defined as the variable LOOP momentum and for convenience:

m=k—(P4+P5)

n=k—(P4+P5)+P1=m+P1

r vector dot products are indicated by the contiguous writing of two four
.ors, for example , P4P5 indicates the dot product of Pi with P5.

three part of a four vector is indicated by a bar; e.g. P4 is the space part
’4.

"section 1a402 ones The Metric

metric used is

9”” = 9n» =

COOL
COO-‘0
OHOO
HODO

 

 

and ti

The 6

wild

Sub-

The

to in

The
Ii: 3
equ

and the energy-momentum four vector-
Pp =EOIPZ)Py)P8
Pp = ”E01P21Pyst
The dot products in this metric are negative:

FTP” = —m2

This metric is consistent with the Minkowski metric P“ = Pu = (P,, P,, P“ iEo)
which is used by the SCHOONSCHIP program.

Sub-Section 1.4.3 .... Gamma Matrices
The original Dirac equation,
(a-p—E+,6~m)rb=0

to be consistent with the Klein-Gordan equation required that

ma, + (1,11, = 25,-,-

150.‘ + aifi = 0

and a and 5 must be Hermetian.

The Dirac equation may be written in terms of Gamma matrices for any met-
ric provided they and their dot products with p“ reproduce the original Dirac
equation, leading to

7H” + 7’7" = 29“”

and futher defining

‘75 = ‘31105261.?

 

mt

11

r the metric used here, the Gamma matrices, defined in terms of the Dirac
Ltrices, become:

7" = #30;

i change the Dirac equation to be:

(iv-p-mhb =0

becomes:

'75 = #717273

.' the Minkowski metric used by SCHOONSCHIP:

1" = ma.-

1‘ =—/3

h the same Dirac equation as above:

(w-p-m)¢=0

becomes:

n=¢f¢f

her detail and comparisons are in Appendix B.

>-Section 1.4.4 .... The Epf Emotion

varticular the Gamma matrix set used by SCHOONSCHIP introduces a
:tion Epf defined as

TRACE(75,4,B¢JD) = €450,975 £35070” = 4 Epf(A, B, c, D)

 

 

 

whe1

and
For

fun:
only

rel:
Ce:

Th1

:ere Cap-75 is the usual Levi-Cevita totally anti-symmetric four tensor.
the Minkowski metric is used by SCHOONSCHIP, then

P" = P, = (P,, P,, P,, we)

:1 the Epf function is always a pure imaginary number.

r this problem, as there are only four independent linear momenta, all Epf
tctions, following the usual properties of Levi-Cevita tensors, must reduce to
,y one Epf :

Epf(P1, P3, P4, P5)

is is a Lorentz invariant and may be calculated directly in any convenient
:rence frame and co-ordinate system, or by the usual evaluation of Levi-
vita tensor products via determinents, which of course gives the same result.

see are both done in Appendix C, and the result is

Epf(P1, P3, P4, P5) =

[4(P1P4)(P3P5)(P1Ps)(P3P4) — [(P1P4)(P3P5) + (P1P5)(P3P4) — (P1P3)(P4P5)]2] ”2

 

Section 1.5 Preparing the Integrals

Ib-Section 1.5.1 .... The Integral

1e Cutowsky rule for an on-shell condition gives:

1

WED?) =’ %(2”‘)25(’=2) 5([k — (P4 + P5)]’)

ith this, the amplitude of the sample diagram of the previous section is formed
the Feynman rules bf Appendix D:
t

M :2 S It I

.ere S collects all the multiplicative scale factors:

5 = (27d)2 e2

I 1
2 (271')4

i I is the LOOP integral of the above sample diagram:

6‘1: 50:”) 6w: - P4 — Psfl [7(P4)-ru (7-1:)(0 + 0'15 mum» W(P5)1“(7-M)(7-¢)(7-n)(A + Bur/MM»
[(k — P4P] [(k — P4 - P5 + P1P] [(k + Pa)2 + M? — iI‘M.)

b-Section 1.5.2 .... The Frame of Reference

e first delta function will dictate that m2 = hon.
e second, 6([k — (P4 + P5)]2), gives

k3—2k.(P4+P5)+(P4+P5)’=o

[with

k2=P42=P52=O
llts in

k . (P4+ P5): P4P5

ch assumes a particularly simple form only in the rest frame of P2 and R:

160(E4 + E5) = —P4P5

 

 

14

d, since also in this frame:

(P4 + 135)2 = —-(E4 + E5)2 = 2(P4P5)
ho = E4 = E5 = (—P4P5/2)1/2

ablishing that in this frame ho is CONSTANT.

1e enormous simplification gained when the magnitude of 7:- remains constant
:tates this rest frame of fi+fi to be the frame of reference for this diagram.
the diagram for anti-lepton scattering is evaluated by interchanging the la-
ls P3 and P4 (except for the lepton line trace), the same frame of reference
Ids throughout the problem.

)te that in this frame the energy, Ex, of any other particle, PX, of the problem
also simply given:

PX . (P4 + P5) = —2(ko)E:: = PXP4 + PXPs

_ PXP4 + PXPs

E: —2ko

b-Section 1.5.3 .... Removal of the Delta functions

: delta functions may be simply removed by integrating over d [It] and disc
vided the JACOBIAN between the set (llcl Jen) and the two arguments of
delta functions is included. This J acobian is calculated in the chosen frame
eference in Appendix F and evaluates to:

—1
J ' 4(P4P5)

Jacobian is now absorbed into the scale factor S:

1 . 1 —1
= E ‘2”): ‘2 [T275] 494%)

 

h ems)“ em a.“

15

Tthe remaining integration over lklz dflk,

|l¢|2 = :5? = constant

11 come outside the integral sign and be absorbed into the scale factor:
_1 .22 1 —1 —P4P5
5 ’ 2 (2“) e [(2104] 4(P4P5) l 2 l
_ i L
" 137 16:
hus the sample LOOP integral reduces to being over the directions of I:

= / 40. W(P4)7u(v-k)(0 +D'ul'vuv(P3)] [§(P5)‘7“('v-'n)('r-¢)(7-n)(A+ B'Vs)‘7"fl(fi)l
[(k — 194)!) [(1. - P4 — P5 + 191)?) [(h + P3)? + M: - sTM,]

ith the understanding that

I the frame of reference is E + R = 0
l W2 = kO’

i) lea = (—P4.P5/2)1/2

~) E: = W

is applies to all the various LOOP integrals in the problem.

b-Section 1.5.3 .... The Divergence Regulation

examining the above LOOP integral, the first denominator factor, arising
s the photon line promator, produces a singularity when the direction of
:comes co-linear with P4, resulting in a logrithmically infinite value for the
gral. This infinity does not occur in Nature - the process is finite - and that
:flected in the calculation by the coefficent of the logarithmically infinite

 

 

-gflfi

16

term becoming zero when the five point and the four point LOOP diagrams are
summed. To allow the doing of the diagrams separately, a small photon mass
is added to regularise the infinity. When the diagrams are added it is observed
that the cocfiicent of the log of this fictitious photon mass becomes identically
zero, producing no ill effects when the photon mass is then allowed to go to sero
and its log to infinity. The appearance of the sero coeflicent does in fact serve
as a check that the various parts of the calculation are fitting together correctly.

1-] 4m («mm m0 + Dvsmumn worm m)('1 6X1 n)(A + Bmthmn
[(k— P4)? + M2,] [(k— P4 P5 + P1)=) [(k + P3)? + M;- -.TM,]

Passing to the second denominator factor (k— (P4 + P5)-— P1)2 , it appears to
possess a similar infinity. It is shown in Appendix G that at the very direction
of k that this zero is produced, the value of the numerator also becomes sero,
rendering the singularity benign. The non-apearance of this infinity also serves
as a check that the integral calculation is proceeding correctly. A small quark
mass is carried to allow this check.

I = / an. [5(P4)1u(1~k)(0 + Dish-11103)] [5(P5)‘r”(7 m)(‘7 €)(‘Y n)(A+ Eva)1"u(P2)l
[(1. — P4)2 + M3] [(1. — P4 — P5 + P1)2 + M;w,] [(1. + P3)2 + M— :TM,]

Section 1.6 Evaluating a Preliminary Diagram

Prior to beginning the actual calculation it is expedient to calculate a prelimi-
nary diagram which contains all the essential features of the larger calculation,
yet is simplified sufiicently that it may be evaluated analytically (without the
aid of SCHOONSCHIP or projection techniques)

P3

, /
P3+k,M, / \k“ P4

1, M1 / P5, M5

 

 

 

and the companion anti-lepton scattering diagram.

All the external fermion particles carry Spin zero, which makes them unphysi-
cal, yet experience has shown that Spin does not greatly affect the magnitude
of the calculation.

The result of these diagrams should serve as an estimation for the larger calcu-
lation, and should that calculation differ significantly, the causes ascertained.

These diagrams are detailed and completely evaluated in Appendix W. The
results will be stated in Chapter 5 ‘Conelusions’, and compared with the
results of the larger calculation.

An advantage occurs from the simplicity of this diagram: it is possible to assign
a non-zero mass to the out-going quark, and to see the effect of varying this
mass on the result. The effect is also described in Chapter 5.

 

 

Chapter TWO
The Calculation

Section 2.1 .... Introduction

As the result will be the ratio of the LOOP plus BORN cross-section to the
BORN cross-section, two calculations are needed. The first, the Born ampli-
tude, is a tree structure and does not involve an integral. The calculation will
highlight a trick for projecting the spinor product of different linear momenta,
i.c. 17(P3)v(P4), onto a basis set more amenable to calculation, and will intro-
duce a projection of the gluon polarisation vector onto a chosen set of linear
momenta; both these projections will also be used in the more complicated
LOOP calculation. Other than these projections, the Born term calculation is
straight-forward.

The LOOP calculation , which of course involves an integration, will use the
same spinor replacement technique, which then allows the TRACE of the nu-
merator expressions to be taken. Then the same replacement of the polarisation
vector is made, resulting in an expression of several hundred terms, each of which
must be integrated.

Some of these terms contain a factor of Epf(k, A, B, C) in the numerator which
would make the integration over directions of I quite difficult. Improvement
in these terms is obtained by replacing k” in all Epf factors with a linear
expansion onto four momenta of the problem, after which all Epf’s become
Epf(P1, P3, P4, P5) which of course is the only indepedent Epf in the prob-
lem, and since it no longer depends on k”, can come outside the integral.

The approximately 50 remaining terms in the LOOP integral can be classi-
fied into integrals based on the number - from 0 to 3 - of PX - 1: factors in the
numerator and the number of similar denominator factors... three for the five
‘point diagrams, two for the four point. Each integral may then be reduced to a
1simplier integral through a process of adding and subtracting appropriate terms
1to allow some of these PX - 1: factors to cancel with denominator factors.

This reduction results in a set of 19 integrals which can not then be further
,reduced and must be solved analytically.

18

 

 

 

In this
for the
the Bo

square

This i1
licity s

Sectie

Sub-S

The B

point,

Sub.§

The fi
80m

son),

when

 

19

n this manner a single complex number is produced for the LOOP amplitude
'or the chosen helicity state and set of physical parameters. This is added to
he Born amplitude, an absolute square taken, and the RATIO to the absolute
quare of the Born term determined.

This is repeated for all physically allowable sets of input parameters and he-
icity states.

Section 2.2 Calculating the BORN Term

Sub-Section 2.2.1 ......Introduction

The Born term is the sum of two diagrams, called the five point and the four
mint, by analogy with their counterparts in the LOOP calculation.

+ 8107' (C + Duh. P3

:r-t P5

 

 

lub-Section 2.2.2 The BORN Amplitude

BORN = BORNfi" + BORNfour

he five point amplitude is:
om, , _ {wow + mumps» Wax. - 6X1 - P5 - 7 - P1)(A + B..)~,».(p2)]
[we ‘— —2(P1P5)MZP

= TRACE[s(P3)§(P4)(C + D75)7.,] Tmca(u(P2)s(Ps)(~, . s)(-7 - P5 .. 1-P1)(A + 8-15 )1']

DEN ;.
’ ‘ —2(P1P5)MzP

Jere M Z P is the 27° particle propagator:

MZP = 2(P3P4) + M3 - :TM,

 

 

and
301

80!!

Sub

may

Vie

the

5111

 

20

and the four point amplitude is:

_ iz(P4)(C + D75)7.,v(P3)] [5095)” + 37mm .101 + 7 -p2)(7 oe)u(P2)]

 

 

BORN’”' ' 2(P1P2)MZP
BORN _, TRACEI'(P3W(P4)(C + D75 )‘le TRACEI'(P2)7(P5)(A + 3‘75 )‘Yvh' ° P1 + ‘7 ° ”)0! ° ¢)]
’°" ' 2(P1P2)MZP

Sub-Section 2.2.3 ....Projecting the SPINOB. product

In Appendix B it is demonstrated that a spinor product of the form v(p)iZ(q)
may be replaced by (1 i 75 )7 . A where A is a four vector given by

A” = (p ° 09" + (9 ° ‘0?” " (P‘ 9)“: i EPfO‘: 9.10. 3)

\/ 16(p - 8)(q - 3)

where s p is an arbritary four vector of the problem, and the plus-minus signs
are determined by the combination of helecities.

This permits the substitutions:

v(P3)'ii(P4) = (1 i 75 )7 . B

 

 

 

 

and
u(P2)'1'1(P5) = (1 :t ‘75)7 - CU
here
B = (P3P5)P4u + (“NWWi 5‘me P3, P4, P5)
" \/16(P4P5)(P3P5)
nd
CU _ (P1P2)P5,. + (191175)qu - (P2P5)P1,. i Em“, P1, P2, P5)
" \/16(P1P2)(P1P5)
'hich enable the SCHOONSCHIP program of Appendix I to easily calculate

 

e TRACES.

b'section 2.204 oooooooRemoval Of 20 vertex (A + 375) and
strong coupling constant

 

 

The
pm

led

out
For
furw
Th

stn

Ne

21

The above introduction of the (1 :L- 115) factors into the TRACES nullifies the
presence of the 20 vertex coefficent (A + B75) as demonstrated in Appendix
E. The algebra of Gamma matrices causes the 75 in the coefficient to become
redundant and the remaining factor (A -— B)(A + B)(C' + D)(C — D) can come
outside the TRACE process where it simply cancels with the same factor that
will later come outside the LOOP integration before the final ratio is taken.
For this reason the Z0 vertex coefficient does not affect the ratio and will not
further be carried in the calculation of either the Born or the LOOP terms.
This cancellation is also true of the GLUON vertex coefficient, which will thus
also not be carried. This is quite desirable as it removes any presence of the
strong coupling constant with all its attendant uncertainties from the result.

Sub-Section 2.2.5 ... Projection of the Gluon Polarisation vector

Next the polarisation vector of the gluon is replaced by
e _ (P1P2)P5u + (P1PS)P2,1 i (Epflp, P1, P2, P5)
p _ ______________

,/(—P1P2)(P1P5)(P2P5)

where i is the GLUON-HELICITY. This replacement is derived in Appendix
J.

  
   
 
   
   
 
   

Sub-Section 2.2.6 ...... The BORN Term result

The Born term amplitude is now entirely evaluated in terms of the set of alge-
braic dot products that form the set of physical parameters for this problem.
The SCHOONSCEIP program of Appendix I outputs this expression in a FOR-
TRAN compatible form. It is then incorporated into a FORTRAN program
which will use it to determine a single complex number- the BORN amplitude-
for each physically permissible set of parameters for the chosen helicity state.

Section 2.3 ... Calculating the LOOP diagrams

Sub-Section 2.3.1 ...Introduction

 

 

22

The LOOP term is the sum of the four diagrams:

those for lepton scattering:

 

 

 

.1-

 

and those for anti-lepton scattering:

 

 

 

 

 

___%__<§_\'
\ <2?

 

ones a
manta
mate

[fin =

[1m 1

Sub-

In ea:
term
875):

113" :

Iii-r

{0110!
the I

23

Once the lepton scattering diagrams are solved, the anti-lepton scattering
nes are obtainable from that solution largely by re-labelling the linear mo-
ienta, ( except in the lepton TRACE itself) . Thus it is sufficent to demon-
trate the lepton solutions only:

__1_
2P1P2

Your =

Iloap = Ilia: + Ijour

,. . _ f 4‘“ TRACEIv<P3)r(P4)~/.(v-k)(c+ Duh.) TRACEIn<P2)v(P5)~r"(7-m)(-r-e)(~/-n)(A+ 87m")

[0: — P4)? + Mg] [(1. — P4 — P5 + P1)? + Mfufl] [(k + P3)? + M? — mm]

60» TRACE [0(P3)'¢7(P4)7u(‘7-k)(0 + D75 )‘7vl TRACEI'(P2)7(P5)7”(7-M)(A + 3707774)]

 

[(I: - P4)? + Mg] [(1: + P3)? + M? 41w.)

Bub-Section 2.3.2 ... Spinor and Polarisation Replacement,

Dropping of the 2° vertex coefiicent (A + 375)
and strong coupling constant

11 each integral the same spinor four vector replacements as used for the BORN
mm are made, with the concurrent dropping of the 20 vertex coefficent (A +
875), and the GLUON vertex strong coupling constant.

m.” I an. TRACEKIivaX'v-Bhubrkhyl TRACE!“tum-CU)7“(7-M)(7-<)('7-n)1"l

 
 
   
   
  
    
 
   

1

Jou- =

2P1P2

[(h — P4)2 + s13] [(k - P4 — P5 + Pi)2 + Mind] [(I: + Pa)2 + M? — il‘M.)

mu TRACEKI i 75)(7 - Blah-thy] TRACE [(1 i ‘75)(7 - CU)?”(‘7-m)7"(‘Y-€)l
{(k — P4)? + Mg] [(1. + P3)2 + M? — mu.)

 

RACES are taken by the SCHOONSCHIP program listed in Appendix K,
ollowed by substitution of the spinor and polarization four vectors, again as in
he Born terms.

ub-Section 2.3.3 ... Linear Replacement of k”

he result contains factors not present in the simplier Born calculation, due
the presence here of the LOOP momentum kn. These factors are of the form
pf(k,A,B, C), and an attempt to integrate with them as they stand would
e very unwieldy. Improvement is obtained by making a linear projection of
u in such factors onto a set of four chosen linear momenta of the problem -
1," S2", S3”, S4,, .

k" = (C1)51,, + (cz)sz,, + (cs)ss,, + (0954,,

 

Th1

 

 

24

.‘he 0 1, 02, 03, 04 can be found by forming the set of four dot products with
p:

51 .1: = (01)5151 + (02)5251 + (03)5351 + (04)5451
52 - k = (01)s152 + (02)5252 + (03)5352 + (005452
53 .1. = (01)5153 + (02)5253 + (03)5353 + (04)5453
54 . k = (01)5154 + (02)5254 + (035354 + (095454

This set of four equations can be inverted to yield 01, 02, 03, 04 in terms
4' the various dot products. This is done in Appendix L and yields:

01 =-.. 0011(31ok) + 0012(52 - k) + 0013(53 . k) + 0014(54-1.)
02 = 0021(51 - k) + 0022(52 . h) + 0023(53 - k) + 0024(54 - 1:)
03 = 0031(51 . k) + 0032(52 . k) + 0033(53 . k) + 0034(54 - I.)
04 = 0041(51 . k) + 0042(52 - k) + 0043(53 - k) + 0044(54 . k)

 
   
  
  
  

'th 0011,0012,0013, ......... 0043,0044 given in terms of the dot prod-
cts of external momenta, again in Appendix L

. . auxiliary arrangement of the same result is:

k, = (51 - lap-1,, + (52 ~ 1:)pr + (53 ~k)?3u + (54.1.)?4“
here
71,. .-= (0011)51,. + (0012)52,. + (0013)53,. + (0014)54,.
$2,. = (0021)51,. + (0022)52,. + (0023)53,, + (0024)54,,
$3,, = (0031)51,. + (0032)52,. + (0033)33,. + (0034054,,
$4,. = (0041)51,. + (0042)52,. + (0042053,, + (0044054,,

 

 

25

This, as well as being useful later, permits self consistency checks to be made
on the calculation of 0011, CC12,....CC44 as:

swam. = a:

which is obtained by dotting the above expression for k” with 51, 52, S3, S4.

51“k,, = (51#J=1,.)(s1.k)+(s1"f2,,)(52.k)+(51"$3,.)(53-Ie)+(s1#}'4,, )(54-1.)

and noting that 51 - 1:, 5'2 - k, 53 - k, 54 - I: are all independent.

For this problem:

51,, = P1,,
52,, = P3,,
53,, = P4”
54,, = P5,,

It is to be noted that in the Appendix L inversion, the condition for a valid in-
version - that the Wronskian determinant be non zero - is the same as will later
be required in the RESULTS Section for a set of parameters to be physically
acceptable and thus does not introduce a new restriction on the problem.

  
 
  
 
 
 
 
 
 
 
 
 
 
   

Sub-Section 2.3.4 Calculating the Remaining Set of Integrals

With the replacement in the Epf functions of k" by this linear projection onto
P1,P3,P4,P5 , all Epf’s reduce to Epf(P1,P3, P4,P5) times some number
of PX - 1: factors. The Epf(P1, P3, P4, P5) previously evaluated in Appendix
C comes through the integral sign and the terms remaining in the integration
number about 30. They can be classified according to the number of numerator
PX - 1: factors they contain.... from 0 to 3 for the five point integral and from

0 to 2 for the four point, reflecting the number of factors of k" in the original
numerator.

Each class, which is an integral, can be reduced to a simplier integral by the
process of adding and subtracting pieces which produce cancellation of a de-
nominator factor. This reduction is performed in Appendix M , and results in
a set of 19 integrals which can not then be further reduced and each must be
solved analytically.

 

 

The (
SCH(
of the
llbtlh

This
the B
Sub-

The I
insert

Sub-

‘At th

end t
sextet

26

The grouping and labelling of these 19 integrals is the last thing done by the
SCHOONSCHIP program , which then produces FORTRAN compatible output
of the calculation as it now stands, with each of the 19 integrals algebraically
labelled by a symbol.

This output is incorporated into the FORTRAN program already containing
the BORN Term result.
Sub-Section 2.3.5 .... Solving the 19 integrals

The 19 integrals are solved analytically in Appendix N and the solutions are
inserted into the above mentioned FORTRAN program as a called sub-routine.

Sub-Section 2.3.8 ....The LOOP term result

'At this point expressions solely in terms of dot products of external momenta
and the plus/minus sign of the Epf(P1, P3, P4, P5) parameter have been in-
serted into the FORTRAN program for the LOOP term. For each physically
allowable set of such parameters it will reduce to a single complex number.

 

El

 

27
Section 2.4 Calculating the RATIO

Sub-Section 2.4.1 Introduction

The FORTRAN program to generate the result is run separately for each of
the eight helicity states of the problem, identified by the letters A-H as per
the table in Section 1.2. The identifying letter is entered into the FORTRAN
program as a parameter before complilation time.

Due to their size (each runs over 1600 lines of FORTRAN code), the two outputs
of the SCHOONSCHIP LOOP code, one for LEPTON scattering, the other for
ANTI-LEPTON, (each combining internally a five and a four point diagram),
are complied separately as FORTRAN functions and joined to the above FOR-
TRAN program at LINK time.

The FORTRAN program contains permanently, (as functions), the eight output
amplitudes of the SCHOONSCHIP BORN TERM code, one for each helicity
state, and as a called sub-routine, the analytic solutions of the nineteen integrals.

Sub-Section 2.4.2 Physically Valid Set of Input Parameters
The program operates by examining in grid fashion all of phase space.
It is demonstated in Appendix P that all allowable values of the six dot prod-

ucts may be visualised as lying on triangular surfaces: ( remembering that dot
products in this metric are negative)

~P 1P3 -P3P4

-P1P5 -P4P5

-P1P4 -P3P5

2
Each corner of the triangles intersects the axis at £31.

l‘hese points are covered in grid fashion in the FORTRAN program by allowing:

E2
0 < —P1P4 < %

 

 

 

28

 

E2
0 < —P1P5 < .... —(—P1P4)

 

2
E2
—P1P3 = 3'" - (—P1P4) — (—P1P5)

and
E2
0 < —P3P4 < 7:3

2
o < —paps < §§”——(—P3P4)

2
—P4P5 = 5262 _(—p3p4)—(—paps)

The actual Em value of 120 GeV is, for ease of calculation, scaled down to
1.2 GeV, with the corresponding scaling of M, to .937 GeV and I‘, to .027 GeV.
As the result is a dimensionless RATIO, it is unchanged by sealing.

The further restriction, that only certain combinations of input-output four vec-
tors conserve separately all four components of energy—momenta (that is - only
certain directional cones are allowed), is represented here by the requirement
that P1,, be real, which is that-

4(P1P4)(P3P5)(P1P5)(P3P4) > [(PlP4)(P3PS)+(P1PS)(P3P4)—(P1P3)(I-"4P5)]2
again derived in Appendix P.

Ifthe selected set of six dot products passes this inequality test, the FORTRAN
program forwards it for processing. It processes both signs of Epf(P1, P3, P4, P5),
representing reflection of the output particles across the plane defined by the
two input particles. (In the FORTRAN program Epf(P1,P3,P4,P5) is la-
belled by the symbol EVL.)

Sub-Section 2.4.3 ... Processing the Point

The parameter labelling the helicity state is examined and the proper BORN
TERM function called to produce that amplitude. It contains both four and
five point diagram contributions, each multiplied by the appropriate propagator
of the other ready to be combined to a single number through the combining
denominator 4(P1P5)P1P2(MZP) , ( labelled PROPIV in the program )
The other multiplicative factors, those arising from the two Spinor replace-
ments and the gluon polarisation vector replacement, (labelled BOIV, CUOIV,
ind EOIV in the program), will also appear in the LOOP number and hence
iivide out when the ratio is taken. For this reason they are set equal to unity,
'or convenience.

 

29

To evaluate the LOOP amplitude, the nineteen integrals for this set of dot
products are evaluated, along with WIV, which is the determinant used in the
expansion of k,“ as per Appendix L. These results are fed into the linked LEP-
TON LOOP function which returns a single complex number.

As the AN TI-LEPTON diagram is the same as the LEPTON if one interchanges
P3 with P4 (except in the LEPTON line TRACE), the dot products involving
P3 or P4 are so interchanged, the sign of Epf(P 1, P3, P4, P5) switched to ac-
count for this interchange, the nineteen integrals re-evaluated along with WW,
and the results fed into the ANTI-LEPTON LOOP attached function, (which
has the LEPTON line TRACE in minus reverse form in preperation for the
switch), and the calculated number returned. (The minus sign is due to travers-
ing the variable LOOP momentum Ir." in the opposite sense when writing the
line TRACE for this diagram.)

 

Each of these numbers has already combined its four and five point contri-
butions with appropriate multipliers in preparation for combining into a single
number. The denominator for that combining is now folded in and the results
joined as a single number for the LOOP TERM.

That number is then sealed by the scale factors —(1/137) and 1/(161r). As

in the BORN TERM, the factors BOIV, CUOIV, and EOIV arising from the
Spinor and polarisation replacements are set to unity.

The LOOP number is added to the BORN number, squared, and divided by the

square of the BORN number to form a RATIO, which is written to an output

file along with the values of the six dot products that formed it.

The next point in the grid of phase space is examined and, if valid, processed.

This continues until all grid points have been examined.

 

 

Chapter Three
The Results

Section 3.1 ... Introduction

This chapter concerns itself with the presentation of the results of the FOR-
TRAN program which calculates the RATIO of the BORN + LOOP cross sec-
tion to the BORN cross section.

The results are presented separately for each of the eight allowed helicity states.
The display of the data is as generated, that is in grid fashion, one dimensionless
RATIO number for each physically allowable set of external dot products.
Rather than display the output from the small grid step sise which actually gen-
erates the results (this is too much data to comfortably display), the program
is rerun with a larger grid step convenient for displaying, the results of the two
runs carefully compared to ascertain nothing interesting was lost in going to the
larger grid size, and the larger sise displayed.

The smaller stepped data is enclosed in Appendix U in case it is desired to
examine it.

Section 3.2 ....Choosing the Display Mode

As indicated at the close of the previous chapter, those points of phase space
‘ which are energetically allowable lie on two plane triangular surfaces:

‘ -P1P3 -P3P4

-P1P5 -P4P5

-P1P4 -P3P5

1 Each corner of the triangles intersects the axis at EEK

The triangle on the right involves only outgoing dot products. For e_ach point on

. this triangle, ;a_ll__ points of the first (input) triangle are energetically allowable.
w The data is presented by selecting representative points on the output triangle
and for each of those points displaying the RATIOS for a representative set of
input points on the input triangle.

30

 

 

 

 

View 1
«mt

For ea
calculi
Pill pt

3'
v
A

\

 

31

View the output triangle straight on, and select 15 symmetrically spaced repre-
sentative points: (the alogarithim for selecting these is given in Appendix V)

-P3P4

 

 

For each output point generate an input triangle in which the RATIO has been
calculated for each of 15 representative points in conjunction with the fixed out-
put point:

.P1P3

 

 

 

 

The mi
points
that is

4(P1P
derived

Again,
the em

The (11
Mich
the set
changh

Sectio

The fol

 

32

The missing RATIOs are due to the fact that not all the energetically allowable
points also separately conserve each of four components of energy-momenta,
that is not all survive the inequality test:

1(P1P4)(P3P5)(P1P5)(P3P4) > [(P1P4)(P3P5)+(P1P5)(P3P4)—(P1P3)(P4P5)]7
ierived in Appendix P. ‘

Again, the display data has been carefully compared with that generated by
the smaller grid sise to assure it is indeed representative.

The dual values given for each RATIO represent the reflection of the output
particles across the plane of the input particles. This reflection does not change
the set of six dot products, but does alter the RATIO. It is accomplished by
:hanging the sign of Epf(P1, P3, P4, P5). ‘

Section 3.3 Results for the Eight Helicity States

The following eight pages contain the display of the eight helicity states.

 

 

to:
Z0

lll
III

41

Section 3.4 ... Focus on M, Peak '

Of particular interest is that area of phase space where the ’mass’ carried off
from the M, decay is equal to M, itself, that is: —2(P3P4) = M3 , and the
20 propagator becomes minimal:

MZP = 2(P3P4) + M} — iI‘M,
= —irM;

Therefore the values of P3P4 are slowly stepped through the M, peak in a sep-
arate running of the FORTRAN program. This is done for the helicity state A,
assuming it to be typical in this respect.

-I’3P4

 

 

.135, Q’s

.The results ( enclosed in Appendix U), show N_O particular distinction from
any other area of phase space in the range of RATIOs produced.

As one of the goals of this project is to look for asymmetries in the 20 de-
cay curve, the above data is plotted— RATIO vs. P3P4, as P3P4 is stepped
through the M, peak. In this graph, one point stands for the average value of
the RATIO on one of the lines in the above diagram, averaged over 1000 points.

 

 

RATIO

1.02

1.01

1.00

00

Illlllll'llllllllll
IIIII'II I'I'IIIII

 

42

 

 

0 ° 0

I. 0
.. o 0 ° 0
.. o 0 0 O O
.. o
I o
.. o
J: J l
' I I I I I I I I I J I I I I I I I

I I I I I I I i In I 7 I I I l I I I I
3300 4200 T 4600 pap,

U

M‘—

This plot shows & appreciable asymmetry, NOR shifting of the ZO peak.
Section 8.5 ... Edge Activity

All the above results work on the interior of the triangles, staying away from
the edges, that is away by 5 per cent.

 

 

3w
1'

ll 11
of t

of d
ThI

NI

-PxPs

 

-P1Pd ~P1P5
I’TI”T

-P8P(

 

-P:Ps/ £5

This is because as the edges are approached, the RATIO can become arbri—
tarily large due to the presence of dot products in denominator positions.

Section 3.6 Connecting with the Center of Mass System

It may be of interest to relate a set of dot products to outgoing scattering
angles in the laboratory (center of mass) system. To this end, the polar angles
of the three exiting particles in the center of mass system are derived in terms
of dot products in Appendix P.

They'are:
P1P3 ]

“(93)“ = 1 ’ 2 [P3P4 + P3P5
_P_1P_4_
P3P4 + P4P5
P1P5
”495%" = 1‘ 2 [m]

c040,)”, = 1 — 2[

Note that the new restrictions that seem to apply:

—(P1P3) < —(P3P4+ P3P5)

are air
Secti<

The p
in the
Ipece,
The 1:
The p
was II
For tl
angle:
in is
space
contr
l0 illI
The 1
WE
PIPE
press
forwI
parti
E

1+4

-(P1P4) < —(P3P4+P4P5)
—(P1P5) < —(P3P5+P4P5)

are already contained in the restriction that P1: be greater than zero.
Section 8.7 Analysis of the Results

The pattern of results is approximately the same for the eight helicity states,
in that almost all the RATIO’s are near unity except for one ‘corner’ of phase
space, where it might range from a value of 2 to near 50..

The largest ratio in the representative sets is 45.669, the smallest 0.0526.

The plot of RATIO vs. P3P4 showed _N_O_ noticable asymmetry as the M1 peak
was traversed, Egg shifting of the Z0 peak.

For the small grid size, the values grew arbritarily large as the edges of the tri-
angles were approached, as expected due to the presence of these dot products
in denominator positions. This increase is unimportant as the amount of phase
space near the edge goes to zero faster than the amplitude grows and does not
contribute to the integrated cross section. This is demonstrated in the solution
to the preliminary diagram in Appendix W.

The largest values of the RATIO appeared in the phase space ‘corner’ of large
P4P5 and small and equal P3P4 and P3P5. The ‘input’ triangle indicates small
P1P3 and P1P5 and large P1P4 . Comparison with the center of mass ex-
pressions of the preceeding section give particle 5 (the quark) staying in the
forward direction, particle 4 (the lepton) reversing direction to emerge at 180°,
particle 3 (the anti-lepton) emerging at 90°. Thus the largest effect of this pro-
cess produces a high transverse momentum product!

 

 

 

 

Secti
There
Seeti

The ‘
calm
In I
of em

Sect

Thin
Item
III A
and
aqua
(not
The

Illtll

iron
SCI

Sec

Chapter Four
Verification and Validation

Section 4.1 ... Introduction
There are six points at which verification of the calculation is enacted.
Section 4.2 Verifying Epf(P1,P3,P4,P5)

The value of Epf(P1,P3, P4,P5) in terms of dot products of the problem is
calculated two different ways in Appendix C - first by direct evaluation in a cho-
sen reference frame and co-ordinate system, and second via the usual method
of evaluating a determinent. Both methods give the same result.

Section 4.3 ... Verifying the Spinor replacement

This replacement, derived in Appendix H, for dis—similar momenta is non-
standard and is thus checked carefully.

In Appendix Q it is applied to two simplistic diagrams in which the amplitudes
and the spin averaged cross sections can be hand calculated. In each case, the
square of the amplitude, (using the replacement), is equal to the cross section,
(not using the replacement), verifying the correctness of it.

The second calculation is then slightly altered, to use the exact spinor replace-
ments for this problem, and a SCHOONSCHIP routine written to evaluate
and compare the amplitude squared and cross section. The successful compar-
ison validates the correctness of the exact replacements and the correct use of
SCHOONSCHIP syntax in evaluating the amplitude.

 

Section 4.4 ... Verifying the Born term

The eight helicity state amplitudes of the Born term are separately squared
and added. This sum is then subtracted from the spin averaged cross section.
If the amplitude calculations are correct, this will produce zero. It is checked
as the last calculation of the SCHOONSCHIP program in Appendix I which
generates the amplitudes. The sero is produced. This verifies that the least
standard piece of the calculation, the Spinor replacement for dis-similar mo-
menta, is working correctly. Less importantly, though of course also essential,
.t checks that the other factors of the Born amplitude are free of error.

As it operates on the amplitudes simultaneously as they are outputed into FOR-
I‘RAN compatible format, there is no further possibly of transcription errors in

45

 

 

theBor
Sectio:

it is m
puchd
tothe‘
Thh u
lfleIE
and pc
II per
precise
enacte
Thh c
eeted

produ
above
andti
Theh
the p
This 1
otter

i0 le

Sect

llth
mom
The
by t:
at u
coefl
me:
The

46

the Born amplitudes.
Section 4.5 Verifying the LOOP numerator

It is noted in Appendix G that the quark singularity is benign because for
precisely the value of Icy at which it occurs, the numerator also approaches sero
to the same order.

This zero is used as a check that the calculation is proceeding correctly.

After SCHOONSCHIP has taken the LOOP traces, substituted for the spinor
and polarisation vectors, and replaced I»:" in all Epf’s by its linear expansion (
as per Appendix L ), then the replacement in the resulting expression of In” by
precisely this value should still produce the zero, if all the substitutions were
enacted correctly. I

This check is performed by the SCHHONSHIP program of Appendix R, a trun-
cated copy of the main SCHOONSCHIP loop program ( Appendix K ) that
produces the FORTRAN compatible amplitudes. It is truncated after all the
above substitutions are made and before the base integral groups are recognised
and the nineteen integrals symbolically labelled.

The insertion of this expression for k” does produce the sero, indicating that at
this point all the above substitutions were made correctly.

This zero took a half hour of SUN time to produce, reflecting the large number
of terms present in the calculation at this point, all of which had to be correct
to produce the zero.

Section 4.6 ... Validation of Divergent parts of Nineteen Integrals

Although the calculation itself is finite, each separate diagram is infinite, and
most of the nineteen integrals possesses an infinite piece.
The infinite part of the diagram (and of the nineteen integrals) is regularized
by the use of small photon and quark masses which are allowed to go to sero
at the end of the calculation where the masses appear in log form, and their
coefficients must sum to zero over both diagrams to avoid divergence as the
mass goes to sero.
‘ The appearance of these zero co-efiicents verifys once again the correctness of
the substitutions already once checked by the replacement of k“ as per the above
section.
It goes further and verifies the base integral groupings of Appendix M and the
divergent parts of the analytic solutions of the nineteen integrals executed in
Appendix N.
This check is enacted by using the SCHOONSCHIP loop program of Appendix
K and instead of producing algebraic output, altering it to produce a numeric
result for a specific set of dot products.

At the beginning of the program the six symbolic dot products are replaced

 

with a
does n
sepsis
mine
the SC
of six
Titll
The m
ore e:
cards
The 0
reps
gehre
There
The I
pendi
prodI
It is

checl

Sect

 

with a set of six numbers that form a physically valid set. As SCHOONSCHIP
does not evaluate logs, any logs of the above six dot products are evaluated
separately and fed into the SCHOONSCHIP program. The same code that
evaluates the nineteen integrals in the FORTRAN program is then inserted in
the SCHOONSCHIP program, where it produces nineteen numbers from the set
of six dot products. (Alternatively, these could have been evaluated in FOR-
TRAN and the nineteen numbers fed in.)

The calculation is then performed numerically instead of algebraiclly, with the
one exception of the logs of the fictitious photon and quark masses, which are
carried as algebraic symbols.

The ouput of the SCHOONSCHIP program now becomes a complex number
representing the convergent part of the amplitude being calculated and two al-
gebraic symbols with coefficients for the logs of the photon and quark masses.

These coefficients must sum to zero over both diagrams.

The altered SCHOONSCHIP program to perform this check is listed in Ap-
pendix S. The valid set of six dot products were arbritarily chosen, and the
production of the zero co-efficents verified.

It is to be noted that the convergent parts of the nineteen integrals are not
checked by this result.

Section 4.7 Numeric semi-check of nineteen integrals

Finally the convergent and divergent parts of the nineteen integrals are semi-
checked by comparing (for a specific numeric set of dot products), the analytic
result with the result of an approximate numerical grid integration.

The result is only approximate as the numerical integration necessarily diverges.
This divergenge is examined by introducing small numeric photon and quark
masses and comparing runs with increasingly smaller masses, looking for an
extrapolation to zero mass.

The approach to zero is quite tame, as expected for a logarithmic divergence, and
even the presence of a minute mass, one-thousandth the dot product number,
keeps the numerical integration from turning upwards. As the one~thousandth
value is approached from the higher value of one-tenth, the integral does not
vary, indicating insensitivity to the presence of the mass.

The numeric integrations are thus performed with these one-thousandth masses
in place and the results compared with the analytic solutions.

The FORTRAN code for the numeric integrations and the results of the com-
, parisons with the analytic solutions are in Appendix T. In all cases the results
‘ are within 30 per cent for a large grid size and become better as the grid size is
reduced. In almost all cases the numeric results are above the analytic and by
about the same percentage, and in almost all cases change the same as the grid
size is reduced.

0 While not conclusive, this is indicative the analytic solutions are consistent as

 

 

a class.
This in
looking
terns.

cue tc
The Ie
of the

putt .
dents,
A! th
gent, i
wont
No re

48

a class.
This method was used to identify errors in the nineteen analytic solutions by
looking for those numeric comparisons that did not obey the systematic pat-
terns. It was successful in finding three errors which when corrected led in each
case to the integral falling into the pattern.

The search was instigated by the failure of the sero coeflicients of the log terms
of the above section to emerge. The three errors found were all in the divergent
parts of integrals and when corrected led to the emergence of the zero coeffi-
cients, indicating no more errors in the divergent parts.

As the convergent parts of the integrals are functionally similar to the diver—
gent, it is to be expected that any errors in them would produce similar pattern
anomilies in the analytic-numeric comparison as did the divergent.

No remaining anomilies are present in the comparisons.

 

Secti

Atth
analy
fions
wela
prfie

Sect

ltha
leted
It th
The
fide
reah
phh
Sev
ued
um
vec
an
den
“e
at
In
at
in

Chapter Five
Conclusion

Section 5.1 Introduction

At this point the calculation has been made, and the data summarised and
analysed. It is time to draw the conclusions of the project against the ques-
‘ tions raised in the Introduction, to see how it compares with estimations as
well as other actual calculations that have been performed and to see what next
projects might be undertaken.

Section 5.2 Conclusions

It has been ascertained that the influence of the re~scattering process here calcu-
lated is, except on the very edges of phase space and one ‘hot’ corner, minimal.
It thus offers little promise of producing an observable consequence.

The specter raised in the Introduction, that the measured width of the Z0 par-
ticle could be contaminated by the effects of rescattering is not at this point
realised. This must be qualified as there is definitely a ‘hot’ corner in the am-
plitude that might contribute on a detailed phase space evaluation.

Several key techniques for working with diagrams of this complexity were gath-
ered together: (a) the projection of a dis-similar momenta spinor product onto
a momentum basis set of the problem. (b) a similar projection of a polarisation
vector (c) the projection of the LOOP momentum k“ onto a momentum. basis
set of the problem, (d) the technique of reducing a loop integral containing three
denominator factors to a sum of integrals containing at most two (e) the various
steps for solving this still non-trival set of integrals (f) the set of checks enacted
at various points in the calculation to insure its correctness.

In addition it was determined once more that the Spin factors do not largely
affect the value of an amplitude, as the simplified preliminary diagram yielded
similar results to the full calculation.

Section 5.3 ... Relation to Other Work

In the ‘Evaluation of a Preliminary Diagram’ in Appendix W, the restriction of
zero mass on the outgoing quark was lifted. It was found this did N21 produce
an enhancement, indeed that the diagram was maximum at the zero mass.
This can crudely be understood in that the region of interaction for re-scattering
of the outgoing particles is greatest if they all have like speeds; so if some have
the speed of light (sero mass), all should, for maximum affect.

Also from that diagram an estimation of about 1 per cent was made for the

49

 

result 0
A! to tl
diegrsn
erpeete
with th

The in!
quarts
(In GL
septum
otieet 1
counts

Seeth

The I:
the er
a the
acne
lead it
in thi:
effect
of the

50

result of the larger calculation. That estimation was fulfilled.

As to the restriction of the two internal lines being ON-SHELL, the preliminary
diagram has been elsewhere10 calculated with this restriction removed, and the
expectation (based on similar past calculations”) that the efi'ect is not enhanced
with the lifting of the restriction, fulfilled.

The larger calculation done with GLUON scattering and the M, decaying to
quarks has also been done elsewhere9 and results of the same low order found.
(In GLUON scattering, the interference term is of course missing due to COLOR
separateness between the BORN and LOOP digrams. This diminishment is
offset somewhat by the increased strength of the strong interaction coupling
constant over the electromagnetic one of the photon scattering.)

Section 5.4 ... Future Work

The most obvious next things to do would be to futher explore those corners of
the extended Dalits plot that generated the most promising effect. In particular,
as those corners were in conjunction with the production of a high transverse
momentum final state lepton, most acessible to experiment, see if these corners
lead to changes in the expected angular distribution of products. As the RATIO
in this corner reached a sizable value (near 50), a conclusion as to the negligible
effect of this re—scattering process can not be drawn until the weighting factors
of the phase space intergration are folded in.

 

1... Eic
2|” Be!
3'” hbl

4... Be:

5... Ch
5... Au
7_--- ’t
B... a.
9... J(
10 J,
11 A

I.‘ J

Footnotes

Eichen, et al. 1984 ‘Reviews of Modern Physics’ Vol 56, No 4, 579
Bergsma, et a1. 1983 ‘Physics Letters’ 1238, 269
Abrams, et a1. 1989 ‘Physical Review Letters’ Vol 63, No 7, 724

Berger, et a1. 1983 ‘Proceedings of the Drell-Yan Workshop’
Fermilab, Batavia, Illinois

Chaichain, et a1. 1979 ‘Physical Review D’ Vol 20, No 11, 2873
Aurenche and Lindfors 1981 ‘Nuclear Physics B’ Vol 185, 274

’t Hooft and Veltman 1979 ‘Nuclear Physics B’ Vol 153, 365

H. Strubbe 1974 ‘Computer Physics Communications’ Vol 8, 1

Jon Pumplin 1988 ‘Physical Review 0’ Vol 38, 1149

Jon Pumplin - personal communication

Abe, et a1. 1989 ‘Physical Review Letters' Vol 63, No 7, 720

Jon Pumplin, Wayne Repko, Jules Kovacs - personal communication

51

 

Bjorken and

Me: Kom

Bibliography

Bjorken and Drell - Relativistic Quantum Mechanics, McGraw-Hill, (1964)

Jules Kovacs - Lecture Notes - Relativistic Quantum Mechanics,
MSU, (1983-1984)

52

Consii

The pro

and sin

then

Then

giVes

And

the

Adi

Appendix A
Allowable Helicity Combinations

Consider a lepton line through one vertex:

 

 

 

7;!
The propagator numerator is:
Ed: 7,, v‘
and since
Hi 2 ii 1 4: 75
2
then 1
as 7””? = at < i 75) 7W?
2
Then using
75714 = ‘71175
gives
1 :1:
Hi 71:”? = id: 7# (775') 1’7
And since 1 i
< ‘75) vi ___ 0

 

2
(1 i715) v; =1}
2

the propagator is zero unless ii and 11 have opposite helecities.

Adding additional vertices to the electron line introduces additional gamma

53

matrii

The

lice:

How
line

54

matrices in pairs, not changing the result.

 

The same is true for any lepton line with internal line propagtors and ver—
tices that each contribute one gamma matrix.

However, if one particle is incoming, and the other outgoing, as in the quark
line for this problem:

 

 

arm

The propagator numerator is:
_ '7
ui 7,, u.‘

and by similar steps as above the helicites must be the same.

«...

‘l—V

 

The

to be<

and o

The 1
For a

In te

This
the ‘

Inc

am

Appendix B
Gamma Matrices, Metrics, and the
Dirac Equation

 

The original Dirac equation,

(am-E+mw=o

to be consistent with the Klein-Gordan equation required that
Orgaj + «1,0,- = 26,}
02 = :82 = 1
130; + 0m = 0
and a and [3 must be Hermetian.

The Dirac equation may be written in terms of Gamma matrices.
For any metric:
V’W” + 7"?” = 29“"

In terms of the original Dirac matrices. 75 is:

 

‘75 = i0'10'203

This evaluation for 75 into the original Dirac marices will be the same for all
the metrics here considered.

In one such metric, for example the one used in the text by Bjorken-Drell:

1 o o 0

,,,,_ _ 0 —1 0 o
9 ‘9""‘ 0 0 —1 0
0 o o —1

and the energy-momentum four vector-

Pu=E0+Pr+Py+P:

55

The dot

and the

change

75 beer

Anoth

and t

The

and

56

P“=E0—P1-_Py—P;

dot products in this metric are positive:

P“P,, = +m."z

the Gamma matrices, defined in terms of the Dirac matrices:

 

7° = 13
7". = Bar
ge the Dirac equation to:
(7 -p- "the: 0

_ ' 0 1 ,2 3

ther metric, the one used by ’t Hooft - Veltman is:

—1000
,,__0100
9‘9“”‘0010
0001

the energy-momentum four vector-

 

P“=Eo+Px+P,+P.-

P,,=——E0+P,,-i—Py—i-Pz
dot products in this metric are negative:
the Gamma matrices, defined in terms of the Dirac matrices:
7° = 17/3

7" = 1330';

change t

75 bec01

A varia

and it

The t

and 1

char

751

57

nge the Dirac equation to:
(7'7"19 - "IN = 0

)eCOflleS:
75 = i7°717"73

'ariant of the above metric is the Minkowski metric:

1000
0100
[JV—...—
9 "“9“"“6W‘ 0010
0001

l the energy-momentum four vector-

P“=Pp=P,+Py+Pz+iEo

e dot products in this metric are also negative:

.,
P“P = —m.‘

l the Gamma matrices, defined in terms of the Dirac matrices:
‘7" = i130?

7'4 = -l3

nge the Dirac equation to:
(hr-171W: 0

)ecomCS: q 4
75‘: 717‘737

 

 

conv
chos
(0'01
plan

P2-

P1

c0-

Th

"U“U“U'TU

Appendix C
Expanding Epf(P1, P3, P4, P5)

As Epf(P1,P3,P4.P-")) is a Lorentz i11variant,it may be evaluated in any
onveinent reference frame and co-ordinate oreintat.ion The reference system
hosen here is the usual one for this prolilem- the rest frame of P—4 and P J The
o—ordinate system is with P_4 taken as the plus z direction and P3 111 the x- 2
lane.

P3

F 1 p4

 

‘116 four vectors— P1,P3,P4.P5 - have their components evaluated for this
>—ordinate system in the solutions to integrals 15, I8, and 118. in Appendix N.
'he results are- using as usual

k0 = ,/—P4P5/2

11d working in the Minkowski metric—

 

 

4 = (0 0 +160, iko)
5 = (0 0. —ko 1.11:0)

_ W (P3P4— P3P5) _.(P3p4+P3P5)
3 —( H0 Zko )
1: (P1: Ply, (P1P4-kP1P5): ai-(P1P42-L-E’1P51)
here

P1_ (P1P4)(P3P5)+(P1P5)(P3P-1)-(P1P3)(P~1P5)>
’ _ i 2km/(P3P4)(P3P5)

58

 

P1P3 ap‘
products

As the E
tensor

we have.

= 11.141

Which r

This C:
'l

P1,}:

Whirl
catin
Posh.
the n

will 1

A110

59

4(P1P4)(P3P5)(P1P5)(P3P4) — [(P1P4)(P3P5) + (P1P5)(P3P4 — (P1P3)(‘P4P5)]=’
4ko'-’(P3P4)(P3P5)

appears in the above results. It is obtained in terms of the other dot
:ts by squaring P2 and noting the particles have zero mass-

P22 = 0 = P3P4 + P3P5 + P4P5 — P1P3 — P1P4 — P1P5

Epf function is defined in terms of the Levi-Cevita totally anti-symetric

Epft‘l, B, C, D) = fag751‘iaBpCv'7Dfi

'9. immediately:

Epf(Pl, P3, P4, P5) = 1(2k03)P3,P1,
1(P1P4)(P3P5)(P1P5)(P3P4)—[(P1P4)(P3P5)+(P1P5)(P3P4—(P1P3)(P4P5)]3)1/3

must be greater than 0i to make P lg real.

an be given a more synnnetric form by re-arranging the numerator of

4ab—(a+b—c)2:
b—[a2+b2+c2+2ab-2ac—2bc]:a2+b2+c2—2ab—2ac—2bc

Epf(Pl, P3, P4, P5) = ii([(PlP5)(P3P4)]2 + [(P1P4)(P3P5)]2
+[(P4P5)(P1P3)]3 — 2(P1P5)(P3P4)(P1P4)(P3P5)
(P1P5)(P3P4)(P4P5)(P1P3) — 2(P1P4)(P3P5)(P4P5)(P1P3))1/2

is transparently symmetric under the interchange of P3 with P4, indi-
that Epf(Pl, P3, P4, P5) will have the same absolute value for both the
>11 and electron Loop diagrams for this problem. and in order to differ by
nus Sign required by the interchange of two vectors in the Epf function.

ck up the oppositely signed square root of P13.
er possibly useful symmetric form could be generated by:

(a-b)"+(a—c13+(b—ci?-a=_12_c2

 

The result 1
nant expanr

[EpflPtf

giving the

Eu

-2(Pl

60

for Epf(Pl, P3, P4, P5) agrees with that obtained by the determi-

1sion of the four demensibnal Levi-Cevita tensor:

P12 P1P3 P1P4
mm 1022 P3P4
P1P4 P3P4 P33
P1P5 P3P5 P4P5

E’3. P4. P5)] 2: [Epf(P1, P3, P4, P5)] =

 

same result as above:
’(P1, P3, P4, P5) = ii([(P1P5)(P3P4)]2 + [(P1P4)(P3P5)]3

+[(P4P5)(P1P3)]‘~’ - 2(P1P5)(P3P4)(P1P4)(P3P5)
P5)(P3P4)(P4P5)(P1P3) — 2(P1P4)(P3P5)(P4P5)(P1P3))1’2

 

These are a
---Bjorlren and

 

Appendix D

Feynman Rules

a verbatim copy of Appendix B in RELATI\-'lSTIC QUANTUM MECHANICS
d Drell

 

61

“hunk-

1“

62

.AppmmflxJEg

Rules for
Feynman Graphs

lxprcssions for cross sections are divided into two parts: first the invariant ampli-
ide In, which is a Lorentz scalar and in which physics lies, and second, the phase
me and kinematics! factors. In terms of 1111, the expression for a difierential
F088 section do is, for apinleu particles and for photons only,
,-;(;)(m),mi . . .11;

)v, — v.) 201,, 24», 2mm)! 29.12:)! .

xmm(m+m- kOSmh

‘ 1

here u, - V 1pr + m! as usual and v1 and v, are velocities of the incident col-
lear particles. This expression is then integrated over all undetected moments

' - - k. of the final particles. The statistical factor S is obtained by including
factor l/ml if there are m identical particles in the final state:

s-flfi
0'

1r Dirac particles,x the factor 1/2u, is replaced by m/E” and the statistical factor
is again included; all other factors remain the same. .
A diflerential decay rate of a particle of man M is given in its rut frame by

I
- 1 ...i'. I_‘w“_..i cac(_zk)3

d“ ‘1 (¥) 2111””l 2.142,): 242.): (2’) 7 1-1 ‘

‘ If one adopts the convention that Dirac spinor-s be normalized to 2171 instead
to unity as in Eq. (11.2), Eq. (3.1) applies as well to fermions. The energy
ijection operators are then simply (in i i) in place of (A3).

35

 

Elects

He

in
/ 4‘
p, t

Fig. B“

action

 

6 3
Relativistic quantum mechanics

ctrodynann'cs of a Spin-zero Boson

Here there are three vertices. shown in Fig. 3.2, corresponding to the inter.

'
———— ----

'i‘o(p+P’)‘ /. z;\\

3.8
n Lagrangian density

131 - —g‘¢°-¢Y (i _ .9.) WA + c I-AI- .ptp- + “av”.

. 32; as. . p o . . . . . 0

rules for these vertices are:
l. A factor -ieo(p + p')., where p and p’ are the momenta in the charged line.
2. A factor +2ieo’g, for each “seagull” graph.
3. A factor iap’ for each mass counterterm.

l. A factor 5 for each closed loop containing only two photon lines, as shown
3. B3.

3. Renormalise the charge as in spinor electrodynamica.

 

 

Rulufor F073”

1; Melon-
There I

56: "

u illustnt

Mitt
ms com

61+

eynman graphs 289

Ieson-Nucleon Scattering
['here are four interaction terms in the charge-independent theory:

3c; - —.c; whim-w:- amzi‘rw: — Kau’:§°§:+ Mama-w:
nstrated in Figs. BA and B.5. The dotted line signifies that I - 0 only is

I
l
/
/
w A
X\
/ \
/ ““2 \g
/f \\

Fig. 3.4

mitted from the meson pair ij to the pair rs, as shown by rule 2 below. The
counterterms are treated as before and there is:

\ I
\\j I],
\ /
\\ /
Fig. 8.5 )0 so(-2i 5X GU68
/ \
/
/i \<
I/ \

l. A factor 9.7.1. at each meson-nucleon vertex giving relative coupling

gths of \/§ 9. for charged mesons and 11 for neutral ones to protons and

rons, respectively.
2. A factor —2iaxa.~,-a,. at each four-meson vertex in Fig. B.5. . .
B. A factor x for each closed loop containing two meson lines as in Fig. B.6.

Fig.3.6 ooo<: >000X§

 

 

 

t

0/

I

Electrodynl'

A vector
-g,,/k’ for n
«for photons

There on

p //-ie°[(p+l
l/'

' 1‘1.

/ ’iluz; ‘
r d
/

Fig. 8.7

interaction I

3' ' ‘lCoI

Th! rules to

1. A is
2. A is
3. A in
4. A f:
5. For
“Omnibus
thd C. N. ‘

11ml.

loop; m“

“d for M

65

Relativistic quantum mechanics

dynamics of Spin-one Boson

ector boson propagator is [—g," + k,k,/m'](lc' — 7M)“ in place of the
for massless photons, and the external line has a polarization factor c.
otons.

re are electrodynamics vertices shown in Fig. 13.7 corresponding to an

5.)“ u I

.__-_..-a _2———-——-5
[(fip’Ls.‘ 1” I‘o [291.5
p.’ gal-Pfl'u] ” / -“‘"‘

" -s_s,,]

(\ \\ x/
21.? \ x}
‘\ B / \
\ / \

\

>n Lagrangian density

..[(6L:) (Ar _ v)_(2)(‘47 O -A It)].
'0' 82,. ‘P" W . at. w, W I

+ co’iIA.A“w:¢' - A.¢“A'¢:l= + éu':¢:¢':
. for these vertices as illustrated are:

fector -ieo(r’ + puns + ieooapf. + iconoc-

flew? + ieo'l29um - anon. - 9-590]-

factor am... for each mass counterterm.

factor 34 for each closed loop containing only two photon lines.

>r the derivation of these rules from canonical theory, efiects of an
s magnetic moment term, and a regularization scheme see T. D. Lee
. Yang, Phys. Rem, 128, 885 (1962).

1 above examples matrices are arranged in “natural order." F or closed
means taking a trace. Isotopic indices are contracted with their mate
an end of a boson line. In taking polarization sums for photons

 

2 «mask,» - -a»-
A

actor mesons

2 manna,» .s -9” + 15%]?
A

Give

the A +

where i
in our

Then 1

Appendix E
Ignoring the 75 in A + B75

an an expression of the form

(liVs) 7-M 7-N(A+B7~5) 7-P

‘B75 can be commuted to the left:

(12!:75)(A=*:375) 7-M 7-N 7-P

Je second :L- depends on an even or odd number of commutations. though
ase it will always be even.

:ing 7575 :1
A :l: A75 i 375 :t B7§ = (A i B)(1:t 75)

66

 

Itisd

In gene:

and in

Where

P4+

Solv

So:

Appendix F
Evaluating the J ACOBIAN

lesired to evaluate a Jacobian such that

d(ko) d(|k|) = J d(lk|2 - 1.702) d(lc — (194+ 135))2

3.1, the Jacobian between (2:,y) and (fly) will be

61: Q1;
0f 69

QHQH
6f By

his case ., 9
f = as” - y‘
g = x2 — y? + 2y(E4 + E5) + 2194135
df = (210611 - (2y)dy
dg = df + 2(E4 + E5)dy

Ik .I and y = kc and use has been made of the rest frame property
0, and of:

g: (k —(P4+P5))2 = (134+135)2 —2(P4+P5)-k+k'—'

= 2P4P5 + 2ko( E4 + E5) + k?
= 2P4P5 + 2y(E4 + E5) + f

dg=df+2(E4+E5)dy
1
-...— ———————d
dy =2(E4+E5)df+2 (E4+E5)g

_1 y _i g —df d0 ]
d” ‘ fidf‘L Edy ‘ 23df+ x [2(E4+ E5) + 2(E4+ E5)

67

 

 

 

and using y

where use

and

68

_1-.__.‘/__ _1—
21 2::(E4+E5) 2(E4+E5)

 

J =
—1 1
2(E4+E5) 2(E4+E5)
ing y = 1: : lco gives
J _ 1

_ 4ko(E4 + E5)
J = ;

4(P4P5)

use has been made of

at?) => |k| = k0

6(k — (P4 + P5))2 => 1:2 + 2ko(E4 + E5) + 2P4P5 = 0
=> ko(E4 + E5) = -P4P5

“mun“—

 

 

 

Given

and BC

This e

or W}

Whic‘
(0 ~
and

Appendix G
obsolving the Quark Singularity

n the diagram:

 

 

P3
P2—— ------
P4
P I
1 P5
integral:

0): [F(P4h'u(')~k)(c+ Drum/11173)] [WPSHWT-mlh«lb-TINA+ Bo.=.)')"-U-(P'3)]
[(k — P4)2 + A13] [(k - P4 — P5 + Pl)2 + Mgmk] [(k + Pa.)2 + M? — mm]

ing with m2 = P12 = 0:

n..—_-m.+Pl 3 n2=2171-P1

lals 0 whenever m and P—l are co—linear:
2m - P1 = 2(m-fi— moEl)
= 2moE1 [0039 - 1]

ever

3—=(a--1)—1

leans, because of zero masses. the four vectors are also parallel. m =
3

n=m+P1=(u—1)P1+P1=0P1

69

 

 

 

‘V‘__

 

onnd
k2=0z

nng

Th5

Tm
Mm

me

70
.2 = 02m: = 0

0t free to assume any value. It is restricted by the other on-shell condition:

0.
m = (0' — 1)P1

m = k — (P4+ P5): (0 — 1)P1
k=(a-1)P1+P4+P5
k2 = 0 = (a — 1)(P1P4 + P1P5) + P4P5

; zero mass particles.

(0 _1)_ —P4P5
— P1P4+ P1P5
P4P5
o- = 1+

P1P4+ P1P5

means the value of k at the singularity is

P4P5
’“u = “(I + ”u ‘ (W) “(I

singularity uncovered is benign because the numerator of the above integral

becomes zero when m = (a — 1)P1 and n. = 0P1.
...(7 - m) (7' - c) (7' - 71).... = ....(7'-m) (7‘~c) (7-0131)...

: ...(o-—— 1)(7-P1) (7. c) (7 - 0P1)

using (7 - A) (7 - B) = — (7' - B) (7 -A) + A - B gives:

= ...(a -1)[-(7'-6) (7-P1)+ 6 - P1] (7' - 0P1)
:h, because 6 - P1 = 0 becomes

- ...(a - 1)cr(7' - 6‘) (7' - P1) (7‘ - P1)

= ...(0'-—1)o‘(7 -e) (P13)

:11 goes to zero as P12 . the same power as the denominator goes to zero.
0

:isasn2=o" Pl2

 

 

The fact tli

can be use
culates the
this value

71

‘he fact that the numerator goes to zero when
P4P5
lc = _ —
“ P4“+P5“ <P1P4+P1P5) P1"
an be used to check the correctness of the SCHOONSCHIP program that cal-

ulates the LOOP term. At various points in that program the substitution of
his value of k“ should produce zero. This check is made in Appendix R.

 

 

It is
replaced

where l
That tl
+t

XP

Since
nents

Simi'

Nor
My

Appendix H
Spinor Replacement

It is desired to show that a Spinor product of dis-similar momenta. may be
laced according to:

v‘ (p3)'*+)(p4)= —(1 “517-11"
ire W u is a four vector to be determined.

at this is a valid replacement may be seen by:

+(P3);(+-(p4)( i )(1 1 )< (1) (31):-“ 1>(7W_70W0)

1 —1 _ 1 —1I' a-W
( 1 _1 )x+(p3)1+ (p4)=—( 11) (_ 91— We >
1 —1 . 1—1
(1 _1>x+(p3)x+ (p4)=(1 :1)(i’-Io+a W )
\'+(p3)1'+‘(p4) =( We + 0.177 )
e(\' +(p3)\ +‘(p4)) and (lVo + a W) are 2 X 2 matrices , the four compo-

lets of ”I: may be found by comparison Q. E D

nilarly,
"('+)(P3)—(_)(P4l = -(1— 75 )7 - W

u‘+’(p3)fi<+’(p4)= «+7517- W
u‘ M123)? )(p41= -(---1—75)7 W

w that it is established that this replacement is valid ll“ max more sim-
be found by projecting onto a set of four linearh independent four \ect01 s-

W” = a.3(P3u)+a4(P4 )+a.5(P5 )_+a€,..jEpf(y P3 P4 P5)

72

 

 

and the co

Starting v1

multiply '

The con

These 1
R2“ ar

and u

glV‘e:

TRr

73

:l the co-efficeuts (a3, a4, as, aw!) determined as follows:

rrting with the defining statement for W“

(WWW = —((1 + 75h - Wm

iltiply both sides by (7“ )ga
(71“ )fia (11- T >05 2 TRACE( "—7y TV TV” + 75711711 VVV)

filmy“) = —4Wp

he conjugate of this yields

n-H-“um = 41-17;

hese two results may be dotted with different arbritary four vectors R1” and

i2“ and multiplied together to yield
amm- - R1)u(+)fi(+l(7 - raw“) = —16(W" -R.1)(W - R2)

TRACE{11’17(7-R1)ufi(7‘ -R2)} = —-16(VV‘ -R1)(ll7 - R2)

11d using projection operators:

v'(p3)'17‘(p3)' = (1:75) (7 - p3)

 

u+(p4)fi(p4)+ = (1 :75) (7 - p4)

 

gives

FRACE[(—1—i2£l(7-P3)(7-Rl)(1:75)(7-P4)(')'-R2)] = —16(II".R1)(1-I7.R2)

:1/2){TRACE[(-y-P3)(7-R1)(7-P4)(7-R2)]+TRAC-'E
= —16(W' -R1)(II=' - R2)

 

[75(1-P3it'1-R1)(7-P4)(1-R2)]}

 

 

 

Let R1 i
using

TRAC 1'

and

the ab:

0!

74

t R1 = P5. l/V ~ P5 can be found by temporarily letting R2 also equal P5:
ng
iACEIh-AXv-BMGM-Di] = 4 [(A - Bx-C - D) + (A - D1<B - C) — (A - C113 ' D1]

TRACE[75(7 ~ A)(7 - B)(7' - C)(7 - D)] = 4 Epf(A, B. C. D)
e above equation yields
1/2 _
[(P4P5)(P3P5)] : (W .135)“;
4
new W : IWIew, W" = IWIe'”.
ow let R2 once again be arbritary. The equation now assumes the form:

/2){TRACElt7-P3X7-P5)(7-P4)(7-R2)]+TRACE[75(7'P31(7-P5)(7-P4)(7-R2)]}

=16(W.R2)(/———(P4P5:(P3P0)e“9

nd taking the traces yields
![(P3P5)P4 . R2 + (P4P5)P3 ~ R2 — (P3P4)P5 . R2 -— Epf(P3, P5, P4. R2)]

=16e-‘9 ————(P4P5:(P3P°)(IV.R2)

[(P4P5)P3# + (P4P5)P4,, — (P3P4)P5,. — Epf(p, P3, P4. P5)]R.2“
= e'”x/16(P4P5)(P3P5)W,,R2“

rd since R2“ is totally arbritary,

i0
6 [(P4P-5)P3,,+(P4P5)P4u—( P3P4)P5,,—Epf(;r. P3. P4. P5)]

"‘ = \/16(P4P5)(P3P5)

 

SCE

 

Appendix I

SCHOONSHIP Program to Calculate
and Verify the BORN Term

75

 

M599; -..«

 

 

CCCCCCCCCCCC

76
START SCHOONSHIP PROGRAM

BORN TERM CALCULATION

this program produces the BORN terms for the various HELECITY
stares. As a check that these are taken correctly it addes
all eight of them together, takes absolute square, then
calulates the spin summed cross section, which does not

use the SPINOR replacement, and takes the difference in

the the two cross sections, which is zero if all is done
correctly.

it produces the output in FORTRAN compatible format

,K,CU,M,E,N4,N5,P1,P2,P3,P4,P5,PD
,FO,F1,F2,F3,F4,F5,F6,F7,F84,F85,F94,F95,F104,F105,F11
1,J2,M1,M2,K3,K2

13 = P34 + P35 + P45 - P14 - P15

SET SWITCHES

FOUR = 1
FIVE =

nstat

nlist

SET DOT PRODUCTS
P1P3 13
P1P4
P1P5
P3P4
P3P5
P4P5
tfix

C AA; A

P14
P15
P34
P35
P45

 

><><><><><><n'u'Unxxnnnn><Hfi1<nnnnnnnnnnnnnnnnnn
‘ognw

AAA;

IN THIS SECTION, TAKE THE Trace, REPLACE EACH K BY IT’S
LINEAR EXPANSION,
AND REDUCE TO A FINAL ANSWER.

+1
+1
+1

SCALE FACTOR FOR THE AMPLITUDE IS . .

xxxnnnnnnnnn

D
I
rn
F"

II II II

C
C
C
C TAKE THE Trace
C
Z

HA5=FIVEtFOtF1tF2¢F11tF5tF85tF95tF105

Z HA1

77
Z HA4=FOURtFOtF1tF2¢F11tF5¢F84tF94tF104
C
C SUBSTITUE ACCORDING TO HELICITIES AND DIAGRAM
C

Id,FO=O.5t(1+LHEL)¢G6(J1) + 0.5:(1-LHEL)*G7(J1)
Id,F1=(G(J1,B))

Id,F2=(G(J1,M1))

Id,F11=0.5t(1+QHEL)*GS(J2) + 0.5:(1-QHEL)¢G7(J2)
Id,F5=(G(J2,CU))

tyep

Id,F85=G(J2,E)

Id,F84=G(J2,M1)

Id,F95=G(J2,N5)

Id,F94=G(J2,N4)

Id,F105=G(J2,M1)

Id,F104=G(J2,E)

*yep
Id,Trick,Trace,J1

C

C Trick AND Trace

Id,Trick,Trace,J2
*yep
C

C START CONDENSATION SUBSTITUTIONS, SOME DIAGRAM DEPENDENT.

C

Id,2,Dotpr,N5(J‘) = FIVE:(P1(J)-P5(J))
Al,Funct,N5(K2’) = FIVEt(P1(K2)-P5(K2))
Id,2,Dotpr,N4(J') = - FOUR¢(P3(J)+P4(J)+P5(J))
Al,Funct,N4(K2‘) = - FOUR:(P3(K2)+P4(K2)+P5(K2))
Id,P4DP4=O

A|,PSDP5=0

A|,EDP1=0

C P output

*yep

C

C PUT IN THE ELSEWHERE EVALUATED SPINOR EXPRESSIONS.

C
Id,2,Dotpr,B(J')=B3tP3(J)+B4tP4(J)+BS*P5(J)+BEpf¢Epf(J,P3,P4,P5)
Al,Funct,B(K3')=B3aP3(K3)+B4tP4(K3)+BSsPS(K3)+BEpftEpf(K3,P3,P4,P5)
C P output

*yep .

Id,B3 BOIVtP4P5
Al,B4 BOIVtP3P5
AI,BS -BOIV:P3P4
Al,BEpf = + LHELtBOIV
Id,Trick
Id,PlDP1=0
A|,P3DP3=O
Al,P4DP4=0
AI,PSDP5=0
Id,PlDP3=P1P3
Al,PlDP4=P1P4
Al,PlDP5=P1P5
AI,PSDP4=P3P4
AI,P30P5=P3P5
Al,P4DP5=P4P5

C P output

 

 

’14

.._.._a- ))J>>

,yep 78
Id,2,Dotpr,CU(J')=
(CU1-CU2)tPl(J)+CU2tP3(J)+CU2tPD(J)+CU5:P5(J)

+CUEpftEpf(J,P1,P3,P5)+CUEpf¢Epf(J,P1,P4,P5)

A|,Funct,CU(K-)=

(CUl-CU2)tP1(K)+CU2¢P3(K)+CU2¢PD(K)+CU5¢P5(K)

+CUEpftEpf(K,P1,P3,P5)+CUEpf¢Epf(K,P1,P4,P5)

Id, CUl = — CUOIV¢(P3P5+P4P5-P1P5)

AI,CU2 = CUOIVtPlPS

A|,CU5 = CUOIV*(P3P4¢P3P5+P4P5)

A|,CUEpf = + QHELtCUOIV

*yep

Id,Trick

Id,EDP1=O

Id,2,Dotpr,PD(J')=P4(J)+P5(J)

Al,Funct,PD(K2')=P4(K2)+P5(K2)

Id,PlDP1=O

AI,P3DP3=0

Al,P4DP4=0

A|,P5DP5=0

Id,PlDP3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,P3DP4=P3P4

A|,P3DP5=P3P5

AI,P4DP5=P4P5

C P output

*yep

C NOW PUT IN THE POLARIZATION VECTOR EXPANSION.

C

Id,2,Dotpr,E(J')= E1tP1(J) e E3tP3(J) +
E44P4(J)+ES#P5(J)+EEpft(Epf(J,P1,P3,P5)+Epf(J,P1,P4,P5))

Al,Funct,E(K-)= E1:P1(K) + E3tP3(K) +
E4#P4(K)+E5#P5(K)+EEpf*(Epf(K,P1,P3,P5)+Epf(K,P1,P4,P5))
Id,E1 = EOIV*GHEL*(P3P5+P4P5-2#P1P5)

Al,E3 = EOIV:GHEL#P1P5

A|,E4 = EOIV¢GHELtP1P5

Al,E5 = -EOIVtGHEL#(P1P3+P1P4)

A|,EEpf= -EOIV

*yep

Id,Trick

Id,Epf(P1,P3,P4,P5) = EVL

Id,PlDP1=O

AI,P30P3=0

Al,P4DP4=O

AI,PSDP5=O

Id,PlDP3=P1P3

Al,PlDP4=P1P4

A|,PlDP5=P1P5

AI,P3DP4=P3P4

AI,P30P5=P3P5

Al,P4DP5=P4P5

C P output

ryep

Keep HA4,HA5
‘next

 

 

 

'lel
B l
Punc
Id,E

I
Keep
tnex
Z HA

Keep

(36‘
7-,

‘0
-x

nfihNNnnnmnxxxrsnnrsnfinn

.AAAI“- Hanan—4A

Z QA=-2#P1P51HA4 +2*(P3P4+P3P5+P4P5)tHA5
B EOIV, BOIV, CUOIV, EVL
Punch QA

Id,EVL=i*A

B PROIV,EOIV,BOIV,CUOIV,A,i
Keep QA

tnext

Z HA = QAaConjg(QA)

B PROIV,EOIV,BOIV,CUOIV,A,i
Keep HA

:next

 

IN THIS SECTION, TAKE-THE Trace, REPLACE EACH K BY IT’S
LINEAR EXPANSION,
AND REDUCE TO A FINAL ANSWER.

LHEL = +1
QHEL = +1
GHEL = -1

SCALE FACTOR FOR THE AMPLITUDE IS .
TAKE THE Trace

HBS=FIVE¢FO*F1*F2tF11tF5:F85*F95¢F105
HB4=FOURtFOtF1tF2tF11tF5tF84tF94tF104

SUBSTITUE ACCORDING TO HELICITIES AND DIAGRAM

nonNNnnnnnxxxnnnnnnnnr-sn

Id,FO=O.5¢(1+LHEL)tGG(J1) + O.5*(1—LHEL)¢G7(J1)
Id,F1=(G(J1,B))

Id ,F2=(G(J1, M11))

Id ,F11=0 5¢(1+QHEL)¢GG(J2) + o 5*(1- QHEL)¢G7(J2)
Id ,F5=(G(J2, CU))

 

yep
Id,F85=G(J2,E)
Id,F84=G(J2,M1)
Id,F95=G(J2,N5)
Id,F94=G(J2,N4)
Id,F105=G(J2,M1)
Id,F104=G(J2,E)
*yep
Id,Trick,Trace,J1
C

C Trick AND Trace
C

Id,Trick,Trace,J2

tyep

C START CONDENSATION SUBSTITUTIONS, SOME DIAGRAM DEPENDENT.
C

Id,2,Dotpr,N5(J’) = FIVE:(P1(J)-P5(J))
A|,Funct,N5(K2’) = FIVE¢(P1(K2)'P5(K2))

inbounw
' ll,Funct,N4l 2

1d,?lDP4:0

ll,P5l)P5=0

ll ,EDP1=0

t P will)“

W

t
t PUT IN THE

c 1
ld,2,l)otpr ,Bl:
ll ,Funct,B(K3
t P output

uev
Id,83 = BOIV-l
ALBA = BOIVt
ALBS = -BOIV
ll,BEpi = + L
Id,Trick
ld,PlDP1=0
ll ,PSDP3:0
ll,PtDP4=0
ll,P5DP5=0
ld,l’1l)P3=P1P
ll,P1llP4=P1P
ll ,PlDP5:P1F
ll ,P30P4:P3l
ll ,P3DP5:P3l
Al ,PtDP5=P4l
{P output
‘iep .
14.2,Dotpr,
(nu-cu:
*CUEpitEF
ll,Funct,Cl
(CUl-CUf
iCUEpitE
1d. (01 =
Al,CU2 : C
MUS : (
lWEpr =
tyep
Muck
ltEDP1=o
“121%“;
lrFUnct’
111mm:
MIP3DP3:
“MON:
“£5195:
llJ’lDPS
ltPnri
“this
“main
“$3st
ltPIDP!
(P on
.yep

8O

pr,N4(J') = — FOUR:(P3(J)+P4(J)+P5(J))
,N4(K2") = — FOUR:-(P3(K2)¢P4(K2)+P5(K2))
;o

0

put

IN THE ELSEWHERE EVALUATED SPINOR EXPRESSIONS.

ipr,B(J“)=83*P3(J)+84tP4<J1+es~Psm eases (J,P3,P4,P5)
;,B(K3')=83tP3(K3)+B4*P4(K3)+BS¢P5(K3)+BEpf:Epf(K3,P3,P4,P5)

2put

BOIV¢P4P5
BOIVtP3P5
-BOIV:P3P4
= + LHEL‘BOIV
<

[:0

3:0

4:0

5:0

3=P1P3
4=P1P4
5=P1P5
4=P3P4
S=P3P5
5=P4P5
tput

itpr,CU(J')=
-CU2)¥P1(J)+CU2#P3(J)+CU2*PD(J)+CU5$P5(J)
iT*Epf(J,P1,P3,P5)+CUEpf*Epf(J,P1,P4,P5)
It,CU K- =
.-CU2)¥PI(K)+CU2¢P3(K)+CU2$PD(K)+CU5#P5(K)
>TtEpf(K,P1,P3,P5)+CUEpftEpf(K,P1,P4,P5)

. = - CUOIVt(P3P5¢P4P5-P1P5)

= CUOIV‘PIPS

= CUOIV‘(P3P4+P3P5+P4P5)

3f = + QHEL'CUOIV

:k

i=0
otpr,PD(J')=P4(J)+P5(J)
:t,PD(K2')=P4(K2)+P5(K2)
P1=0

P3=0

P4:O

P5=0

P3=P1P3

P4=P1P4

P5=P1P5

P4=P3P4

P5=P3P5

'P5=P4P5

rutput

 

 

...—......- xn’n>>>>>

man

81

NOW PUT IN THE POLARIZATION VECTOR EXPANSION.

Id, 2 ,Dotpr, E(J )= E1*P1(J) + E3tP3(J) +
E4tP4(J)+E5tP5(J)+EEpf¢(Epf(J, P1 ,P3 ,P5)+Epf(J, P1, P4 ,P5))
Al ,Funct ,E(K )= E1#P1(K) + E3*P 3(K) +

E4:P4(K)+E5¢P5(K)+EEpft(Epf(K, P1 ,P3 ,P5)+Epf(K, P1, P4 ,P5))

Id,E1 = EOIV:GHEL¢(P3P5+P4P5- -2¢P1P5)

>
rn
A

II II II

EOIVtGHELtPlPS
EOIV*GHEL*P1P5
-EOIV*GHEL*(P1P3+P1P4)

AI ,EEpf: -EOIV
*yep

Id, Tri

Id ,Epi(P1, P3, P4 ,P5) = EVL
Id ,Pian

AI,P3DP3=0

A|,P4DP4=O

A|,PSDP5=O

Id,PlDP3=P1P3
Al,PlDP4=P1P4
A|,PlDP5=P1P5
AI,P30P4=P3P4
A1,P30P5=P3P5
A|,P4DP5=P4P5

C

P output

*yeP

Keep HA,HB4,HBS
enext

Z
B

QB = -2tP1P5*HB4 +2:(P3P4+P3P5+P4P5)¢HBS
EOIV,BOIV,CUOIV,EVL

Punch O8
Id,EVU=itA

B

PRDIV,EOIV,BOIV,CUOIV,A,i

Keep HA,QB
anext

Z
8

HB = QBtConjg(QB)
PROIV,EOIV,BOIV,CUOIV,A,i

Keep HB,HA

tnext

C

nnnnxxxnnnnnnnnn

IN THIS SECTION, TAKE THE Trace, REPLACE EACH K BY IT’S
LINEAR EXPANSION,
AND REDUCE TO A FINAL ANSWER.

LHEL = +1
QHEL = -1
GHEL = +1

SCALE FACTOR FOR THE AMPLITUDE IS . .

TAKE THE Trace

 

 

_“-HA—l)>)fi—4A. n>bfififl

82

HC5=FIVEmFOtF1:F2*F11:F5:F85wF95¢F105
HC4=FOURwFOPF1tF2tF11:F5tF84*F94:F104

hNNfi

C SUBSTITUE ACCORDING TO HELICITIES AND DIAGRAM

C

Id,F0=0.5#(1+LHEL)*G6(J1) + 0.5*(1—LHEL)*G7(J1)
Id,F1=(G(J1,B))

Id,F2=(G(J1,M1))

Id,F11=0.5t(1+QHEL)¢G6(J2) + O.5*(1-QHEL)*G7(J2)
Id,F5=(G(J2,CU))

eyep»
Id,F85=G(J2,E)
Id,F84=G(J2,M1)
Id,F95=G(J2,N5)
Id,F94=G(J2,N4)
Id,F105=G(J2,M1)
Id,F104=G(J2,E)

 

*yep
Id,Trick,Trace,J1

C

C Trick AND Trace

C
Id,Trick,Trace,J2
*yeP

C

C START CONDENSATION SUBSTITUTIONS, SOME DIAGRAM DEPENDENT.
C

Id,2,Dotpr,N5(J") = FIVE:(P1(J)-P5(J))
Al,Funct,N5(K2") = FIVE:(P1(K2)-P5(K2))
Id,2,Dotpr,N4(J') = - FOUR¢(P3(J)+P4(J)+P5(J))
Al,Funct,N4(K2”) = - FOUR:(P3(K2)+P4(K2)+P5(K2))
Id,P4DP4=O

Al,PSDP5=O

AI,EDP1=O

C P output

eyep

C

C PUT IN THE ELSEWHERE EVALUATED SPINOR EXPRESSIONS.

 

c
Id,2,Dotpr,B(J')=B3tP3(J)+B4tP4(J)+BS#P5(J)+BEpftEpf(J,P3,P4,P5)
Al,Funct,B(K3')=83¢P3(K3)+B4eP4(K3)+BS¢P5(K3)+BEpf¢Epf(K3,P3,P4,PS)

C P output
ryeP
Id,83 BOIVtP4P5

AI,B4 ; BOIVtPSPS
AI,BS = -BOIV¢P3P4
Al,BEpf = + LHELtBOIV

Id,P1nP1=o
A|,P3DP3=O
AI,P4DP4=O
AI,PSDP5=0
Id,PlDP3=P1P3
AI,P1DP4=P1P4
A|,PlDP5=P1P5
AI,P3DP4=P3P4
Al,PsDP5=P3P5

83
A|,P4DP5=P4P5
C P output

*yep
Id,2,Dotpr,CU(J~)=
(CU1-CU2)#Pl(J)+CU2*P3(J)+CU2:PD(J)+CU5*P5(J)

+CUEpftEpf(J,P1,P3,P5)+CUEpf#Epf(J,P1,P4,P5)

AI, Funct ,CU(K7)=

(CU1-CU2)#P1(K)+CU2tP3(K)+CU2tPD(K)+CU5¢P5(K)

+CUEpftEpf(K, P1, P3 ,P5)+CUEpf#Epf(K, P1, P4 ,P5)

Id, CU1= — CUOIV*(P3P5+P4P5- -P1P5)

Al,CU2 = CUOIVtPlPS

A|,CU5 = CUOIV*(P3P4+P3P5+P4P5)

Al,CUEpf = + QHELtCUOIV

wyep

Id,Trick

Id,EDP1=0

Id,2,Dotpr,PD(J~)=P4(J)+P5(J)

Al,Funct,PD(K2')=P4(K2)+P5(K2)

Id,P1DP1=O

A|,PBDP3=O

A|,P4DP4=O

Al,PSDP5=0

Id,PlDP3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,PSDP4=P3P4

Al,P30P5=P3P5

AI,P4DP5=P4P5

C P output

*yeP

C

C NOW PUT IN THE POLARIZATION VECTOR EXPANSION.

 

Id, 2 ,Dotpr, E(J )= E1*P1(J) + E3*P3(J)
E4tP4(J)+E5*P5(J)+EEpft(Epf(J, P1, P3 ,P5)+Epf(J, P1, P4 ,P5))

Al ,Func t ,E(K )= E1*P1(K) + E3:P3(K) +
E4¢P4(K)+E5¢P5(K)+EEpfa(Epf(K, P1, P3 ,P5)+Epf(K, P1, P4 ,P5))
Id,E1 = EOIV:GHEL:(P3P5+P4P5- -2tP1P5)

Al,E3 EOIV:GHEL¢P1P5

E01VtGHELtP1P5

-EOIV:GHEL¢(P1P3+P1P4)

Al ,EEpf: -EOIV

eyep

Id,Trick

Id,Epf(P1,P3,P4,P5) = EVL

Id,FlDP1=0

AI,P30P3=O

Al,P4DP4=O

Al,PSDP5=O

Id,PlDP3=P1P3

A|,PlDP4=P1P4

Al,PlDP5=P1P5

Al,PSDP4=P3P4

Al,P3DP5=P3P5

A|,P4DP5=P4P5

C P output

#yep

C

 

>
m
.5

II II II

84
Keep HA,HB,HC4,HC5
anext
Z QC = -2tP1P5tHC4 +2:(P3P4+P3P5+P4P5)4HC5
B EOIV,BOIV,CUOIV,EVL
Punch QC
Id,EVL=i#A
B PROIV,EOIV,BOIV,CUOIV,A,i
Keep HA,HB,QC
anext
Z HC = QCtConjg(QC)
B PROIV,EOIV,BOIV,CUOIV,A,i
Keep HC,HB,HA

 

tnext

C

C

C

C

C IN THIS SECTION, TAKE THE Trace, REPLACE EACH K BY IT’S
C LINEAR EXPANSION

C AND REDUCE TO A FINAL ANSWER.

C

C

C

X LHEL = +1

X GHEL = -1

X GHEL = -1

C

C SCALE FACTOR FOR THE AMPLITUDE IS ...
C

C TAKE THE Trace

C

Z HDS=FIVEtF0tF1tF2tF11tF5eF85tF954F105
Z HD4=FOUR¢F04F1*F2tF114F5tF84tF94aF104
C

SUBSTITUE ACCORDING TO HELICITIES AND DIAGRAM

no

Id,FO=O.5t(1+LHEL):GG(J1) 4 0.5:(1-LHEL)¢GY(J1)
Id,F1=(G(J1,B))
Id,F2=(G(J1,M1))
Id,F11=0.5t(1+QHEL):GG(J2) + 0.5:(1-QHEL)tG7(J2)
Id ,F5=(G(J2, CU))
ype
Id ,F85=G(J2, E)
Id, F84: G(J2, M1)
Id,F95=G(J2,N5)
Id,F94=G(J2,N4)
Id,F105=G(J2,M1)
Id,F104=G(J2,E)
'yeP

Id,Trick,Trace,J1

C

C Trick AND Trace
C

Id,Trick,Trace,J2
#yep
C

C START CONDENSATION SUBSTITUTIONS, SOME DIAGRAM DEPENDENT.
C

 

85

Id,2,Dotpr,N5(JT) = FIVE#(P1(J)-P5(J))
A|,Funct,N5(K2') = FIVE*(P1(K2)-P5 K2))
Id,2,Dotpr,N4(J~) = - FOUR*(P3(J)+P4(J)+P5(J))
A|,Funct,N4(K2-) = - FOUR:(P3(K2)+P4(K2)+P5(K2))
Id,P4DP4=O

A|,PSDP5=O

Al,EDP1=O

C P output

*yep

C

C PUT IN THE ELSEWHERE EVALUATED SPINOR EXPRESSIONS.
C

Id,2,Dotpr,B(J‘)=B3*P3(J)+B4*P4(J)+BS#P5(J)+BEpftEpf(J,P3,P4,P5)
Al,Funct,B(K3')=BStP3(K3)+B4*P4(K3)+B54P5(K3)+BEpf¢Epf(K3,P3,P4,P5)
C P output

*yeP

Id,83 = BOIVtP4P5

A|,B4 = BOIV4P3P5

A|,BS = -BOIV¢P3P4

Al,EEpf = + LHELtBOIV

Id,Trick

Id,P1DP1=O

Al,P3DP3=0

Al,P4DP4=0

Al,P50P5=0

Id,PlDP3=P1P3

Al,PlDP4=P1P4

Al,P1DP5=P1P5

Al,P3DP4=P3P4

Al,PSDP5=P3P5

Al,P4DP5=P4P5

C P output

*yep
Id,2,Dotpr,CU(J')=
(CU1-CU2)*P1(J)+CU2*P3(J)+CU2¢PD(J)+CU5:P5(J)

+CUEpf¢Epf(J,P1,P3,P5)+CUEpftEpf(J,P1,P4,P5)

Al,Funct,CU(K’)=

(CU1-CU2)#P1(K)+CU2¢P3(K)+CU2¢PD(K)+CU5¢P5(K)

oCUEpftEpf(K,P1,P3,P5)+CUEpftEpf(K,P1,P4,P5)

'Id, CU1 = - CUOIV:(P3P5+P4P5-P1P5)

Al,CU2 = CUOIV¢P1P5

Al,CU5 = CUOIV:(P3P4+P3P5+P4P5)

Al,CUEpf = + QHELtCUOIV

*yep

Id,Trick

Id,EDP1=0

Id,2,Dotpr,PD(J~)=P4(J)+P5(J)

Al,Funct,PD(K2')=P4(K2)+P5(K2)

Id,PlDP1=O

Al,P3DP3=0

A|,P4DP4=O

A|,PSDP5=O

Id,FlDP3=P1P3

Al,P1DP4=P1P4

A|,P1DP5=P1P5

Al,PSDP4=P3P4

A|,P3DP5=P3P5

AI,P4DP5=P4P5

 

 

86

C P output
*yeP

C NOW PUT IN THE POLARIZATION VECTOR EXPANSION.
C

Id,2,Dotpr,E(J")= E1*P1(J) + E3tP3(J) +
E4+P4(J)+E5+P5(J)+EEpft(Epf(J,P1,P3,P5)+Epf(J,P1,P4,P5))

Al,Funct,E(K')= E1+P1(K) + E3tP3(K +
E4+P4(K)+E5*P5(K)+EEpft(Epf(K,P1,P3,P5)+Epf(K,P1,P4,P5))
Id,E1 EOIV:GHEL*(P3P5+P4P5-2+P1P5)

EOIVtGHELtPlPS

EOIV4GHELtP1P5

-EOIVtGHEL*(P1P3+P1P4)

Al ,EEpf: -EOIV

3.
[TI
01;

II II II II

 

Id ,Epf(P1, P3, P4 P5):
Id ,P1DP1=O
Al,P3DP3=O
Al,P4DP4=0
Al,PSDP5=O
Id,FlDP3=P1P3
Al,PlDP4=P1P4
Al,PlDP5=P1P5
Al,P3DP4=P3P4
Al,P3DPS=P3P5
Al,P4DP5=P4P5
C P output
*yep

Keep HA,HB,HC,HD4,HDS

#next

Z QD = -2+P1P5+HD4 +2+(P3P4+P3P5+P4P5)+HDS
B EOIV,BOIV,CUOIV,EVL
Punch QD

Id,EVL=itA

B PROIV,EOIV,BOIV,CUOIV,A,i
Keep HA,HB,HC,QD

tnext

Z HO: QDtConj 9(OD)

B PROIV, EOIV, BOIV ,CUOIV,A,i
Keep HD, HC, HB, HA

 

 

anext

c 11111
C

C

C

C IN THIS SECTION, TAKE THE Trace, REPLACE EACH K BY IT’S
C LINEAR EXPANSION,

C AND REDUCE TO A FINAL ANSWER.

C

C

C

X LHEL = —1

X GHEL = +1

X GHEL = +1

C

C SCALE FACTOR FOR THE AMPLITUDE IS ...

.¥_.Arsnv r—cnnn

87

TAKE THE Trace

HES:FIVE*F0*F1*F2*F11tFStF85tF95tF105
HE4=FOUR¢F0tF1tF2tF11tFStF84tF94tF104

SUBSTITUE ACCORDING TO HELICITIES AND DIAGRAM

nnnNNnnn

Id,FO=0.5*(1+LHEL)*66(J1) + 0.5*(1-LHEL)*G7(J1)
Id,F1=(G(J1,B))
Id,F2=(G(J1,M1))
Id,F11=0.5¢(1+QHEL)¢GG(J2) + 0.5t(1-QHEL)¢G7(J2)
Id,F5=(G(J2,CU))

 

*yep
Id,F85=G(J2,E)
Id,F84=G(J2,M1)
Id,F95=G(J2,N5)
Id,F94=G(J2,N4)
Id,F105=G(J2,M1)
Id,F104=G(J2,E)
*yep
Id,Trick,Trace,J1
C

C Trick AND Trace

C
Id,Trick,Trace,J2
tyep

C

C START CONDENSATION SUBSTITUTIONS, SOME DIAGRAM DEPENDENT.

C

Id,2,Dotpr,N5(J”) = FIVE#(P1(J)-P5(J))
Al,Funct,N5(K2') = FIVE:(P1(K2)-P5(K2))
Id,2,Dotpr,N4(J“) = - FOUR:(P3(J)+P4(J)+P5(J))
Al,Funct,N4(K2”) = - FOUR:(P3(K2)+P4(K2)+P5(K2))
Id,P4DP4=0

Al,PSDP5=O

AI,EDP1=0

C P output

:yep

C

C PUT IN THE ELSEWHERE EVALUATED SPINOR EXPRESSIONS.
C

 

Id,2,Dotpr,B(J')=B3tP3(J)+B4¢P4(J)+BS¢P5(J)+BEpf¢Epf(J,P3,P4,P5)
AI,Funct,B(K3")=33.P3(K3)+B4.P4(K3)+35.95(K3)+8Epftepf(K3,?3,P4.P5)

C P output
*yep
Id,83 BOIVtPAPS

AI,B4 2 301v.P395
AI ,85 = -BOIV:P3P4
Al,BEpf = . LHELtBOIV

Id,FlDP1=0
Al,PSDP3=0
Al,P4DP4=0
Al,PSDP5=O
Id,FlDP3=P1P3
Al,PlDP4=P1P4
Al,PlDP5=P1P5

88

Al,P3DP4=P3P4
Al,P3DP5=P3P5
Al,P4DP5=P4P5
Cy Pp output

Ide 2 ,Dotpr, CU(J )=
(CU1—CU2)#Pl(J)+CU2¢P3(J)+CU2¢PD(J)+CU5¢P5(J)

+CUEpftEpf(J,P1,P3,P5)+CUEpf*Epf(J,P1,P4,P5)

A|,Funct,CU(K")=

(CU1-CU2)*Pl(K)+CU2*P3(K)+CU2¢PD(K)+CU5:P5(K)

+CUEpftEpf(K,P1,P3,P5)+CUEpf¢Epf(K,P1,P4,P5)

Id, CU1 = - CUOIV:(P3P5+P4P5-P1P5)

Al,CU2 = CUOIVtPlPS

Al,CU5 = CUOIV*(P3P4+P3P5+P4P5)

Al,CUEpf = + QHELtCUOIV

#yep

Id,Trick

Id,EDP1=O

Id,2,Dotpr,PD(J')=P4(J)+P5(J)

Al,Funct,PD(K2')=P4(K2)+P5(K2)

Id,PlDP1=0

Al,P3DP3=O

Al,P4DP4=O

Al,PSDP5=O

Id,FlDP3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,PSDP4=P3P4

Al,PSDP5=P3PS

AI,P4DP5=P4P5

C P output

*yep

C NOW PUT IN THE POLARIZATION VECTOR EXPANSION.
C

Id, 2 ,Dotpr, E(J )= E1tP1(J) + E3¢P3(J)
E4tP4(J)+E5mP5(J)+EEpft(Epf(J, P1 ,P3 ,P5)+Epf(J, P1, P4 ,P5))

Al, Funct, E(K )=E1tP1(K) + E3:P3(K) +
E4tP4(K)+E5#P5(K)+EEpf:(Epf(K,P1,P3,P5)+Epf(K,P1,P4,P5))
Id,E1 = EOIVtGHEL:(P3P5+P4P5-2¢P1P5)

Al,E3 = EOIVtGHELtPlPS

Al,E4 = EOIVtGHELePlPS

Al,ES = -EOIV#GHEL:(P1P3+P1P4)

Al,EEpf: -EOIV

*yep

Id,Trick

Id,Epf (P1,P3,P4,P5) =

Id,FlDP1=O

Al,P3DP3=0

Al,P4DP4=O

Al,P5DP5=O

Id,PlDP3=P1P3

Al,PlDP4=P1P4

Al,P1DP5=P1P5

Al,P30P4=P3P4

Al,P30P5=P3P5

Al,P4DP5=P4P5

C P output

 

 

89
*yep
c

Keep HA,HB,HC,HD,HE4,HE5
tnext

Z QE = ~2tP1P5tHE4 +2t(P3P4+P3P5+P4P5)*HE5
B EOIV,BOIV,CUOIV,EVL
Punch OE

Id,EVL=itA

B PROIV,EOIV,BOIV,CUOIV,A,i
Keep HA,HB,HC,HD,QE

:next

Z HE = QEtConjg(QE)

B PROIV,EOIV,BOIV,CUOIV,A,i
Keep HE,HD,HC,HB,HA

*next

IN THIS SECTION, TAKE THE Trace, REPLACE EACH K BY IT’S
LINEAR EXPANSION,
AND REDUCE TO A FINAL ANSWER.

LHEL
GHEL
GHEL

-1
+1
-1

SCALE FACTOR FOR THE AMPLITUDE IS ...
TAKE THE Trace

HF5=FIVE+FO+F1+F2+F11tF5+F85¢F95+F105
HF4=FOUR+FotFltF2¢F11-F5:F84+F94+F104

SUBSTITUE ACCORDING TO HELICITIES AND DIAGRAM

nnnNNnnnnnxxxnnnnnnnnnn

Id,FO=O.5#(1+LHEL)tGS(J1) + 0.5#(1-LHEL)1G7(J1)
Id,F1=(G(J1,B))
Id,F2=(G(J1,M1))
Id,F11=0.5t(I+QHEL)tG6(J2) + O.5t(1-QHEL)$G7(J2)
Id,F5=(C(J2,CU))

*yep
Id,F85=G(J2,E)
Id,F84=G(J2,M1)
Id,F95=G(J2,N5)
Id,F94=G(J2,N4)
Id,F105=G(J2,M1)
Id,F104=G(J2,E)
‘yep
Id,Trick,Trace,J1
C

C Trick AND Trace

C
Id,Trick,Trace,J2
*yep

C

 

 

CCCTAAC $.1AAAIIAAWJJJ I I 1

90
C START CONDENSATION SUBSTITUTIONS, SOME DIAGRAM DEPENDENT.
C

Id,2,Dotpr,N5(J-) = FIVE:(P1(J)-P5(J))
Al,Funct,N5(K2~) = FIVE+(P1(K2)-P5(K2))
Id,2,Dotpr,N4(J-) = - FOUR:(P3(J)+P4(J)+P5(J))
A|,Funct,N4(K2') = - FOUR:(P3(K2)+P4(K2)+P5(K2))
Id,P4DP4=O

Al,PSDP5=O

Al,EDP1=O

C P output

tyep

C

C PUT IN THE ELSEWHERE EVALUATED SPINOR EXPRESSIONS.
C

Id,2,Dotpr,B(J')=B3+P3(J)+B4+P4(J)+BS*P5(J)+BEpftEpf(J,P3,P4,P5)
Al,Funct,B(K3‘)=B3*P3(K3)+B4*P4(K3)+85+P5(K3)+BEpf#Epf(K3,P3.P4.P5)
C P output

#yep

Id,B3 = BOIV+P4P5

A|,B4 = BOIV+P3P5

A|,BS = -BOIV+P3P4

A|,BEpf = + LHELtBOIV

Id,PlDP1=0

Al,P3DP3=O

Al,P4DP4=O

Al,PSDP5=O

Id,P10P3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,P3DP4=P3P4

Al,P3DP5=P3P5

Al,P4DP5=P4P5

C P output

t yep

Id, 2 ,Dotpr, CU(J )=

(CU1——CU2)¢P1(J)+CU2¢P3(J)+CU2+PD(J)+CU5+P5(J)

+CUEPf‘EPf(J,P1.P3,P5)*CUEPf‘EPf(J,P1,P4.P5)

A|,Funct,CU(K')=

(CU1-CU2)tP1(K)+CU2+P3(K)+CU2¢PD(K)+CU5¢P5(K)

+CUEpftEpf(K,P1,P3,P5)+CUEpftEpf(K,P1,P4,P5)

Id, CU1 = - CUOIV+(P3P5+P4P5-P1P5)

Al,CU2 = CUOIVtPlPS

Al,CUS = CUOIV+(P3P4+P3P5+P4P5)

Al,CUEpf = + QHEL#CUOIV

*yep

Id,Trick

Id,EDP1=O

Id,2,Dotpr, PD(J )=P4(J)+P5(J)

Al ,Funct ,PD(K2 )=P4(K2)+P5(K2)

Id ,PlDP1=0

AI ,P3DP3=O

Al,P4DP4=O

Al,PSDP5=O

Id,FlDP3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,P3DP4=P3P4

 

 

’r‘

... q-A-n1p>>>>

91
Al,P3DP5=P3P5
Al,P4DP5=P4P5
C P output
‘yep
C

C NOW PUT IN THE POLARIZATION VECTOR EXPANSION.
C

Id,2,Dotpr,E(J")= E1+P1(J) + E3tP3(J) +
E4¢P4(J)+E5tP5(J)+EEpf*(Epf(J,P1,P3,P5)+Epf(J,P1,P4,P5))

Al,Funct,E(K')= E1+P1(K) + E3+P3(K) +
E4:P4(K)+E5tP5(K)+EEpf¢(Epf(K,P1,P3,P5)+Epf(K,P1,P4,P5))
Id,E1 EOIV:GHEL+(P3P5+P4P5-2tP1P5)

Al,E3 EOIV#GHEL+P1P5

Al,E4 EOIV¢GHEL+P1P5

Al,ES -EOIV#GHEL¢(P1P3+P1P4)

Al,EEpf: -EOIV

*yep

Id,Trick

Id,Epf(P1,P3,P4,P5) = EVL

Id,PlDP1=O

Al,PsDP3=O

Al,P4DP4=0

Al,PSDP5=0

Id,PlDP3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,PBDP4=P3P4

Al,P30P5=P3P5

Al,P4DP5=P4P5

C P output

*yep

C

Keep HA,HB,HC,HD,HE,HF4,HF5
tnext

Z QF: -2+P1P5tHF4 +2:(P3P4+P3P5+P4P5)tHF5
B EOIV,BOIV,CUOIV,EVL
Punch QF

Id,EVL=itA

B PROIV,EOIV,BOIV,CUOIV,A,i
Keep HA,HB,HC,HD,HE,QF

tnext

Z HF = QFtConjg (OF)

B PROIV, EOIV, BOIV, CUOIV, A, i
Keep HF, HE, HD, HC, HB, HA

 

 

 

 

tnext

C I I
C

C

C

C IN THIS SECTION, TAKE THE Trace, REPLACE EACH K BY IT’S
C LINEAR EXPANSION,

C AND REDUCE TO A FINAL ANSWER.

C

C

C

X LHEL = —1

X QHEL = —1

X GHEL = +1

hfihNNhfifififi
w—T

.-AA‘HnnnH

92

SCALE FACTOR FOR THE AMPLITUDE IS ...
TAKE THE Trace

HG5=FIVEtFOtF1tF2tF11tF5tF85+F95tF105
HG4=FUUR¥FO¢F1tF2tF11¢F5¥F84tF94tF104

SUBSTITUE ACCORDING TO HELICITIES AND DIAGRAM

nnnNNnnnnn

Id,F0=0.5:(1+LHEL)¢66(J1) + 0.5:(1-LHEL)+G7(J1)
Id,F1=(G(J1,B))

Id,F2=(G(J1,M1))

Id,F11=0.5+(1+QHEL)+06(J2) + 0.5:(1-QHEL)*G7(J2)
Id,F5=(G(J2,CU))

*yep

Id,F85=G(J2,E)

Id,F84=G(J2,M1)

Id,F95=G(J2,N5)

Id,F94=G(J2,N4)

Id,F105=G(J2,M1)

Id,F104=G(J2,E)

*yep
Id,Trick,Trace,J1

C

C Trick AND Trace

Id,Trick,Trace,J2
tyep
C

C START CONDENSATION SUBSTITUTIONS, SOME DIAGRAM DEPENDENT.
C

Id,2,Dotpr,N5(J') = FIVEt(P1(J)-P5(J))
A|,Funct,N5(K2') = FIVE+(P1(K2)-P5(K2))
Id,2,Dotpr,N4(J') = — FOUR:(P3(J)+P4(J)+P5(J))
Al,Funct,N4(K2’) = - FOUR:(P3(K2)+P4(K2)+P5(K2))
Id,P4DP4=O

Al,PSDP5=0

Al,EDP1=O

C P output

tyep

C
C PUT IN THE ELSEWHERE EVALUATED SPINOR EXPRESSIONS.

C

Id,2,Dotpr,B(J')=83+P3(J)+B4#P4(J)+BS+P5(J)+BEpf+Epf(J,P3,P4,P5)
Al,Funct,B(K3”)=83:P3(K3)+B4-P4(K3)+BS+P5(K3)+BEpf¢Epf(K3,P3,P4,P5)
C P output

= BOIV+P4P5
AI,B4 = BOIV+P3P5
A|,BS = -BOIV¢P3P4
A|,BEpf = + LHEL¢BOIV
Id,Trick

Id,PlDP1=O

Al,P3DP3=0

Al,P4DP4=O

Al,PSDP5=O
Id,PlDP3=P1P3

 

 

.__“\-.H1>)>H>H

93

Al,PlDP4=P1P4
Al,PlDP5=P1P5
Al,P3DP4=P3P4
Al,P3DP5=P3P5
Al,P4DP5=P4P5
C P output
tyep
Id,2,Dotpr,CU(J')=
(CU1-CU2)+P1(J)+CU2¢P3(J)+CU2+PD(J)+CU5¢P5(J)
+CUEpftEpf(J,P1,P3,P5)+CUEpmepf(J,P1,P4,P5)
Al,Punct,CU(K')=
(CU1-CU2)#P1(K)+CU2¢P3(K)+CU2+PD(K)+CU5+P5(K)
+CUEpf+Epf(K,P1,P3,P5)+CUEpftEpf(K,P1,P4,P5)
Id, CU1 = - CUOIVt(P3P5+P4P5-P1P5)
Al,CU2 = CUOIVtPlPS

Al,CUS — CUOIV+(P3P4+P3P5+P4P5)
Al,CUEpf = + QHELtCUOIV

*yep

Id,Trick

Id , EDP1=0

Id,2,Dotpr,PD(J')=P4(J)+P5(J)
Al,Funct,PD(K2')=P4(K2)+P5(K2)
Id,PlDP1=0

Al,P3DP3=O

Al,P4DP4=O

Al,PSDP5=0

Id,PlDP3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,P3DP4=P3P4

Al,P3DP5=P3P5

Al,P4DP5=P4P5

C P output

#yep

C NOW PUT IN THE POLARIZATION VECTOR EXPANSION.
C

Id,2,Dotpr,E(J')= E1+P1(J) + E3*P3(J) +
E4+P4(J)+E5+P5(J)+EEpft(Epf(J,P1,P3,P5)+Epf(J,P1,P4,P5))

Al,Funct,E(K')= E1+P1(K) + E3+P3(K) +
E4tP4(K)+E5tP5(K)+EEpf+(Epf(K,P1,P3,P5)+Epf(K,P1,P4,P5))

Id,E1 = EOIVtGHEL¢(P3P5+P4P5—2mP1PS)

Al,E3 EOIV.GHEL:P1P5

Al,E4 EOIV+GHELtP1P5

Al,E5 -EOIV+GHEL:(P1P3+P1P4)

Al,EEpf: -EOIV

Id,Epf(P1,P3,P4,P5) = EVL
Id,PlDP1=0

Al,PSDP3=0

Al,P4DP4=O

Al,PSDP5=0

Id,PlDP3=P1P3
Al,PlDP4=P1P4
Al,PlDP5=P1P5
Al,P30P4=P3P4
Al,P30P5=P3P5

 

 

‘(‘(|l ....AIIW

 

94

Al,P4DP5=P4P5
C P output
*yep

C

Keep HA,HB,HC,HD,HE,HF,HG4,HG5
tnext

Z QG = -2#P1P5+HG4 +2¢(P3P4+P3P5+P4P5)¢HG5
B EOIV,BOIV,CUOIV,EVL

Punch QG

Id,EVL=i+A

B PROIV,EOIV,BOIV,CUOIV,A,i
Keep HA,HB,HC,HD,HE,HF,QG
tnext

Z HG = QGtConjg(QG)

B PROIV,EOIV,BOIV,CUOIV,A,i
Keep HG,HF,HE,HD,HC,HB,HA
#next

C 1.; I444 I I IIIIIII

 

IN THIS SECTION, TAKE THE Trace, REPLACE EACH K BY IT’S
LINEAR EXPANSION,
AND REDUCE TO A FINAL ANSWER.

LHEL
QHEL
GHEL

SCALE FACTOR FOR THE AMPLITUDE IS ...

-1
-1
—1

TAKE THE Trace

HH5=FIVEtFOtF1tF2¢F11+F5¢F85¢F95¢F105
HH4=FOUR+FO+F1tF2tF11tF5+F84tF94tF104

SUBSTITUE ACCORDING TO HELICITIES AND DIAGRAM

OOONNOOOOOXXXOOOHOGHOO

Id,FO=O.5+(1+LHEL)¢G6(J1) + 0.5:(1—LHEL):G7(J1)
Id,F1=(G(J1,B))

Id,F2=(G(J1,M1)

Id,F11=O.5t(1+QHEL)¢G6(J2) 4 O.5t(1-QHEL)+G7(J2)
Id,F5=(G(J2,CU))

‘yep

Id,F85=G(J2,E)

Id,F84=G(J2,M1)

Id,F95=G(J2,N5)

Id,F94=G(J2,N4)

Id,F105=G(J2,M1)

Id,F104=G(J2,E)

‘yep

Id,Trick,Trace,J1

C

C Trick AND Trace

C
Id,Trick,Trace,J2

Ic

MTMAAHC r.CCCIAC AIAAAIIAAAIWJJJJ‘ 1..

95

*yep
C START CONDENSATION SUBSTITUTIONS, SOME DIAGRAM DEPENDENT.
C

Id,2,Dotpr,N5(J') = FIVEt(P1(J)-P5(J))
Al,Funct,N5(K2') = FIVEt(P1(K2)-P5(K2))
Id,2,Dotpr,N4(J') = - FOUR+(P3(J)+P4(J)+P5(J))
Al,Funct,N4(K2') = — FOUR:(P3(K2)+P4(K2)+P5(K2))
Id,P4DP4=O

Al,PSDP5=O

Al,EDP1=0

C P output_

*yep

C

C PUT IN THE ELSEWHERE EVALUATED SPINOR EXPRESSIONS.
C

Id,2,Dotpr,B(J’)=B3*P3(J)+B4¢P4(J)+BS+P5(J)+BEpftEpf(J,P3,P4,P5)
Al,Funct,B(K3‘)=B3*P3(K3)+B4+P4(K3)+BS+P5(K3)+BEpf:Epf(K3,P3,P4,P5)

C P output

*yep

Id,83 = BOIVtP4P5
AI,B4 = BOIVtP3P5

AI,Bs = -BOIV*P3P4
‘Al,BEpf = + LHELtBOIV

Al,PBDP3=O

Al,P4DP4=O

Al,PSDP5=O

Id,PlDP3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,P3DP4=P3P4

Al,PSDP5=P3P5

Al,P4DP5=P4P5

C P output

’Yep

Id,2,Dotpr,CU(J')=

(CU1—CU2)+P1(J)+CU2+P3(J)+CU2+PD(J)+CU5+P5(J)

+CUEpftEpf(J,P1,P3,P5)+CUEpf-Epf(J,P1,P4,P5)

A|,Funct,CU(K')=

(CU1-CU2)¢P1(K)+CU2:P3(K)+CU2:PD(K)+CU5+P5(K)

+CUEpffiEpf(K,P1,P3,P5)+CUEpf¢Epf(K,P1,P4,P5)

Id, CU1 = - CUOIVt(P3P5+P4PS—P1P5)

Al,CU2 = CUOIVtPlPS

Al,CUS = CUOIV+(P3P4+P3P5+P4P5)

Al,CUEpf = + QHELtCUOIV

*YSP

Id,Trick

Id,EDP1=O

Id,2,Dotpr,PD(J')=P4(J)+PS(J)

AI,Funct,PD(K2")=P4(K2)+P5(K2)

Id,PlDP1=O

Al,P3DP3=0

Al,P4DP4=0

Al,PSDP5=0

Id,PlDP3=P1P3

Al,PlDP4=P1P4

 

 

96

Al,PlDP5=P1P5
Al,P3DP4=P3P4
Al,P3DP5=P3P5
Al,P4DP5=P4P5
C P output
:yep

C NOW PUT IN THE POLARIZATION VECTOR EXPANSION.
C

Id, 2 ,Dotpr, E(J )= EltP1(J) + E3+P3( J)
E4¢P4(J)+E5*P5(J)+EEpf¢(Epf(J, P1, P3 ,P5)+Epf(J, P1, P4 ,P5))

Al ,Funct ,E(K )= E1:P1(K) + E3:P3(K
E4:P4(K)+E5+P5(K)+EEpf¢(Epf(K, P1, P3 ,P5)+Epf(K, P1, P4 ,P5))
EOIV:GHEL+(P3P5+P4P5- -2*P1P5)
EOIV+GHELtP1P5

EOIVtGHELtPlPS

-EOIV+GHEL:(P1P3+P1P4)

Al NEE f: -EOIV

‘yep

Id,Trick

Id,Epf(P1,P3,P4,P5) =

Id,PlDP1=0

Al,PSDP3=O

Al,P4DP4=O

Al,PSDP5=O

Id,PlDP3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,PSDP4=P3P4

Al,P3DP5=P3P5

Al,P4DP5=P4P5

C P output

‘yep

Keep HA,HB,HC,HD,HE,HF,HG,HH4,HH5

inext

Z OH = —2¢P1P5¢HH4 +2:(P3P4+P3P5+P4P5)+HH5

B EOIV,BOIV,CUOIV,EVL

Punch QHA

Id ,EVL:

B PROIV, AEOIV, BOIV, CUOIV, A, i

Keep HA, HB, HC, HD, HE, HF, HG, OH

tnext

Z HH: QHtConjg(QH)

B PROIV, EOIV, BOIV, CUOIV, A, i

Keep HH, HG, HF, HE, HD, HC, HB, HA

tnext

Z HEL = PROIV¢t2+(HA+HB+HC+HD+HE+HF+HG+HH)

B PROIV, EOIV, BOIV, CUOIV, A

Id, A+t2= - (-4tP14+P15¢P34+P35+(P15tP34+P14¢P35-P13+P45)+#2)
Keep HEL

tnext

C
C
C THE EIGHT HELICITY STATE AMPLITUDES HAVE NOW BEEN
C CALCULATED.

>
mm
01-h
II II II II

C
C NOW FORM THE SPIN AVERAGED SUM CROSS SECTION

 

 

 

 

 

C AND C
C ABOVE
CDONEC

C IN '
C LIN!
C AND

VB,K,C
F m,
F124,F
I J1,J:

Z EXPS
Z EXPS
Z EXP4
Z EXP4
C St
Id,FO

Id,Fi:

97

AND COMPARE IT WITH THE SUM OF THE EIGHT
ABOVE SQUARED, AS A CHECK THAT THEY WERE
ONE CORRECTLY.

IN THIS SECTION, TAKE THE Trace, REPLACE EACH K BY IT’S
LINEAR EXPSANSION,
AND REDUCE TO A FINAL ANSWER.

,K,CU,M,E,E2,N4,N5,P1,P2,P3,P4,P5,PD
,F0,F1,F2,F3,F4,F5,F65,F74,F75,F85,F9,F105,F11,A1,A2,A3,AA,
24,F125,F135,F64,F84,F104,F134

1,J2,M1,M2,M3,K3,K2

SCALE FACTOR FOR THE AMPLITUDE IS ...
TAKE THE Trace

XP55=FIVEtFOtF1tF2¢F3¢F4¢F11tF5*F65*F75*F85#F9*F105tF125tF135
XPS4=FIVE¥FUUR*FO*F1#F2tF3tF4tF11#F5*F65*F75#F85*F9¥F104#F124*F134
XP45=FOUR¢FIVE¢FO¢F1mF2tF3tF4tF11tFStF64tF74tF84tF9tF105tF125tF135
XP44=FOUR¢FO¥F1¥F2*F3*F4*F11¢F5*F54¥F74tF84*F9*F104*F124#Fl34

SUBSTITUE ACCORDING TO HELICITIES AND DIAGRAM

F0 = 1

d,F0=0.5t0.5t(1+LHEL):G6(J1) + O.5¢O.5t(1-LHEL)tG7(J1)
F1=(G(J1,P4))

F2=(G(J1,M1))

F3=(G(J1,P3))

F4=(G(J1,M2))

F11 = 1

d, F11=0. 5:0. 5t(1+QHEL)#G6(J2) + O.5¢O.5t(1-QHEL):G7(J2)
F5=(G(J2, P$))

F65: (G(J2, M3))
F75=(G(J2, N5))
F85: (C(J2, M1))
F9=(G(J2, P2))
F10105= (C(JZ, M2))
F125=(G(J2, N5))
F135: (G(J2,M3))
F64: (G(J2,M1))
F74=(G(J2,N4))
F84: (G(J2,M3))
F104: (G(J2,M3))
F124=(G(J2,N4))
F134: (G(J2,M2))

p
Trick,Trace,J1
Trick AND Trace

Trick,Trace,J2
output

p
2,Dotpr,P2(K4') : P3(K4) . P4(K4) + P5(K4) - P1(K4)

 

 

Al,Pun:
CP 0|
tyep

C

C S'
C
Id,2,D
Id,2,D
Al,Pur
Id,2,[
Id,2,[
ALFUI
CP 1
Id,P4I
A|,P5‘
Al,Pl
Al,P3
Al,ED
ALE?
P 01.
W

D thnnnnmn

98

I,Funct,P2(K4') = P3(K4) + P4(K4) + P5(K4) - P1(K4)
P output
rep

START CONDENSATION SUBSTITUTIONS, SOME DIAGRAM DEPENDENT.

i,2,Dotpr,N5(J') FIVEt(P1(J)—P5(J))
i,2,Dotpr,N5(J') FIVE:(P1(J)-P5(J))
|,Funct,N5(K2") = FIVE:(P1(K2)—P5(K2))
i,2,Dotpr,N4(J') - FOUR*(P3(J)+P4(J)+P5(J))
i,2,Dotpr,N4(J~) — FOUR#(P3(J)+P4(J)+P5(J))
|,Funct,N4(K2") = - FOUR:(P3(K2)+P4(K2)+P5(K2))
P output
i,P4DP4=O
|,PSDP5=O
|,PlDP1 = O
|,PSDP3 = O
l,EDP1=0
|,E2DP1=0

output
rep

i,PlDP4=P1P4
|,PlDP3=P1P3
[,PlDP5=P1P5
I,P3DP4=P3P4
|,P30P5=P3P5
|,P4DP5=P4PS
P output
rep

Bep EXPSS,EXP54,EXP45,EXP44,HEL

1ext

EXP4 : Y:(Y1:EXP55 + Y23¢EXP54 + Y23:EXP45 + Y4¢EXP44)

i,Y1 = 4:(P3P4+P3P5+P4P5)t(P3P4+P3P5+P4PS)

I,Y23 = —2tP1P5:2:(P3P4+P3P5+P4P5)

I,Y4 : 4:P1P5:P1P5

i,Y=PROIVt#2tEOIV¢t2t(-4)tP15#(P13+P14+P15)t(P35+P45-P15)
EOIV,?ROIV

eep EXP4,HEL

1ext

SSEXP = EXP4:BOIV:n2¢CUOIVt42tBO#CUO

i,BO=16:P35:P45

i,CUO=16-(P34+P35+P45)tP15
PROIV,EOIV,BOIV,CUOIV

Bep HEL,SSEXP

1ext

NOW SUTRACT THE SPIN AVERAGED CROSS SECTION
FROM THE SQUARED SUM OF THE HELECITY
STATES. THE RESULT SHOULD BE ZERO.

DIFF : HEL - SSEXP
PROIV,EOIV,BOIV,CUOIV

3nd

 

 

 

 

The

Thi:
tick

wh

As

‘11th

Appendix J
Projecting the Gluon Polarization
Vector

The polarization vector of the GLUON particle may be expressed in Lorentz
invariant form by projecting it onto a set. of four vectors:

The task is to find the four coefficents, al, a4. (15, ac”
This can be done most simply in a co—ordinate system where the GLUON par-

ticle moves in the +2 direction. There the polaroization vector has the form:

cu = (1,520.0)

where i is the GLUON helecity state.

Assume the usual frame of reference, the. rest frame of 733 and m. then

._Pl= (0,0.E1,iE1)
P4 = (P4,.O,P4,,iE4)
P5 = (—P4,,0,—P4,,£E4)

E4 = k0
E1 = P1P4qikPlP5

= = P1P4+P1P5

P4 __ kolzflPlP-UU’lPS)
r — P1P4+P1P5

[co = —P4P5/2

99

 

 

With the
are, using

where {a
course w«

= (2314

and a:

then (

Whit

100

Nith these values of P1, P3,P4.P5, the non-zero terms of Epfm, P1. P4,P5)
Lre, using

Epf(A, B,C. D) = Cap-,5 AaBgC'ayDé

vherc fer/376 is the. usual Levi—Cevita totally anti—symetric four tensor and of
:ourse we are in the Minkowski metric.

E1)f([.l, P1, P4. P5) = €3314(2, P13, P41. P54)
+62413(2. P14. P41. P53) + €23-ut2a 1313.1344. P51.)
+62314(2,51,P4z~i34)+ €2431(21P14v-P431 P51)
+62413(2,iE1,P41-. -P4;) + (2341(2, E1. iE4, -P4¢-)
+62431(2iiE19P4:a _P41'-)

= €2314ltE1)(‘P4r)(iE4)]+62413[(iE4l(P4rl(—P4:)]+€2341l(51ltiE4)(-P4r)l+
€2431[(iE1)(P4;)(V—P417)]

= —2i(E1)(E4)(P4x) 6"",

and as P1, P4,P5 have no y component and {y 2 ii = (GHEL)1'

then “er! is immediately determined:

_ —GHEL
‘1‘" ’ 2(El)(P4,)(E4)

_ GH EL
- \/—2(P1P4)(P1P5)(P4P5)

As 6 has but three deminsions and the space is four deminsional:
eflPl“ = 0 = a4PlP4+ a5P1P5

which gives:
P1P5

“4 = rasp—rpz

 

 

and ‘

give

101

and using
cpP/i“ = P4, : a1P1P4+ a5P4P5

(”P5" = P5, = —P4: = u1P1P5 + a4P4P5

 

gives,
(11 : (PIP4 — P1P5) P4P5
(P1P4 + P1P5) \f—2(P1P4)(P1P5)(P4P5)
a4 = PIP-5
W
as = P1P4
—2(P1P4)(P1P5)(P4P5)

By gauge invariance any amount of P1” can be added to 6”. so a1 can be
set to zero and the result becomes

 

= 1 r r r _ ' . 1.P4.P-r
c,‘ \/_2(P1P4XP1P5KHP5)L(P1P0)P4u+(P1P4)P3,, (GHEL)Epf(p P 3)]

 

 

Appendix K

SCHOON SHIP Program for the
LOOP Term

 

 

102

XXXOOOZOO<MOOOOOOOOOOG

xnnnnnnnnnnxxxxnnnnxxxxxxnw‘uwonxxxxx

 

103

START SCHOONSHIP PROGRAM

Calculate the LOOP term, algebraically
with the 19 integrals labelled by a
symbol. Output is by set of helecity
states, and for either the two

lepton or the two anti-lepton
diagrams.

LNMP,LNMQ,PI,FPIIV
P1,P2,P3,P4,P5,Sl,S2,53,$4,U1,U2,U3,U4

FEED IN VALUES FOR THIS RUN
30,Ro

SET SWITCHES

FOUR = 1
FIVE =
ELECT = 1
POSIT = O
LHEL = +1
QHEL = +1
GHEL = +1
MPARIT = +1
utlimit,24000000
nstat
ninput
nlist

SET OUT PRODUCTS

 

P1P3 = ELECTnP13 + POSIT¢P14
P1P4 = ELECTtP14 + POSITtP13
PIPS = P15
P3P4 = P34
P3P5 = ELECTtP35 + POSITtP45
P4P5 = ELECTtP45 + POSITtPSS

FOR CONVEINENCE , DEFINE SOME EXPRESSIONS AND THEIR RECIPROCALS.
0145 = P1P4 + P1P5
01 = 0145
0345 : P3P4 + P3P5
MZP = 2tP3P4 + 03
NOW K CAN ALWAYS BE EXPANDED ON AN ARBRITARY BASIS SET $1,S2,S3,S4
K = C(1)tS1 + C(2)#S2 0 C(3):SS + C(4):S4
AND FOR THIS PROBLEM WE USE THE SET P1,P3,P4,P5
$0....

5152 = P1P3

$8333
XXXXXCCCCCCCCCX CCCVAVAvAvr1) iiiii

 

xnnnnnnnnnxxxxx

><><><><><><><nnn

nonNnnnnxnnnnHm<nnnnnnnnnns OXXX

...,

104
3153 = P1P4
5154 = P1P5
5253 = P3P4
3254 = P3P5
8354 = P4P5

TO THIS END, THE ORIGINAL EXPANSION OF K MAY BE OOTTED WITH EACH
OF THE BASIS TO FORM A 4 x 4 MATRIX, WHICH wHEN INVERTED
YIELDS ......
C(1) : CC(1,1)*SlDK + CC(1,2)*52DK + CC(1,3):S3DK + CC(1,4)¢S4DK
w 15 THE DETERMINENT FOR THIS INVERSION AND wrv IT’S RECIPROCAL
w : 5152:3334:(~SlS2#SSS4+SIS4:S2S3+5153#5284)
+ 5153.3254.( $152¢S3S4+SlS4:SZSS-Sl$3t$2$4)
+ $154t$2$3t( 5152:5354-5154.5253+S153.5254)

AND NOW, ALL THE CC’S USING THE NOTATION CC(4,3) = CC43

CC44 = WIV¢(-2)#(Sl$2#$1$3$$233)

CC43 = WIV*$1$2*(-$1$2t$3$4+$1$4152$3+SIS3ts2$4)
CC42 = WIVtSlSSt( $1$2¢S3S4+51$4t32$3-Sl$3#$2$4)
CC41 = WIVtS2S3t( 8132:8354-SlS4tS2$3+SlS3tSZS4)
CC33 = WIVt(-2)t($1$2t$1$4t$2$4)

CC32 = WIVtSIS4*( $1$2t$3$4-31$4tSZSS+SlSS$SZS4)
CC31 = WIV:SZS4#( SlS2tSB$4+SlS4tS2S3—SlS3tSZS4)

CC22 = WIVt(—2)¢($1$3*Sl$4t$3$4)
CC21 WIV#S3S4#(-Sl$2tS3S4+Sl$4nSZS3+SlS3tS2S4)
CC11 WIV:(-2)t(3253t$234¢33$4)

ix

 

IN THIS SECTION, TAKE THE TRACE, REPLACE EACH K BY IT’S .
LINEAR EXPANSION,
AND REDUCE TO A FINAL ANSWER. ‘

,CU,M,E
0,F1,F2
J2,M1,M

N,P1,P2,P3,P4,P5,PD
F3.F4,F5,F6,F7,F8,F9,F10,F11,A1,A2,A3,AA
2,K3,K2

I
I

SCALE FACTOR FOR THE AMPLITUDE IS ..

FSCAL = 2t(P34+P35+P45)

TAKE THE TRACE
EXP5=FSCALtFotF1tF2tF3*F4tF11tF5tF6eF7tF8tF9tF10
SUBSTITUE ACCORDING TO HELICITIES AND DIAGRAM

105

Id,Fo=0.5*((1+LHEL)*(GS(J1))+(1-LHEL)#(G7(J1)))
Id,F1=(G(J1,B))

Id,F2=ELECTt(G(J1,M2)) + POSIT#(G(J1,M1))
Id,F3=(G(J1,K))

Id,F4=ELECT¢(G(J1,M1)) + POSIT*(G(J1,M2))
Id,F11=O.5t((1+OHEL)*(G6(J2))+(1-QHEL)#(G7(J2)))
Id,F5=(G(J2,CU))

wyep

Id,F6=(G(J2,M2))

Id,F7=(G(J2,M))

Id,F8=G(J2,E)

Id,F9=(G(J2,N))

Id,F10=G(J2,M1)

*yep

Id,Trick,Trace,J1

C

C TRICK AND TRACE
C

Id,Trick,Trace,J2
*yep
C

C START CONDENSATION SUBSTITUTIONS, SOME DIAGRAM DEPENDENT.
C

Id,2,Dotpr,N(J") = M(J)+P1(J)
A|,Funct,N(K2") = M(K2)+P1(K2)

Id,Dotpr,M(J~)=K(J)-PD(J)
Al,Funct,M(J-)=K(J)-PD(J)
Id,KDPD=P4P5

Id,KDK=0
Id,2,Dotpr,PD(J')=P4(J)+P5(J)
Al,Funct,PD(J~)=P4(J)+P5(J)
Id,P4DP4=0

Al,PSDP5=0

Al,EDP1=0

C P OUTPUT

*yep

C

C PUT IN THE ELSEWHERE EVALUATED SPINOR EXPRESSIONS.
C

 

 

Id,2,Dotpr,B(J')=BS¢P3(J)eB4tP4(J)+BS¢P5(J)+BEPF¢Epf(J,P3,P4,P5)
Al,Funct,B(K3’)=BStP3(K3)+B4tP4(K3)+BS*P5(K3)+BEPFtEpf(K3,P3,P4,P5)
P OUTPUT

*yep
Id,Ba : BOIV:P4P5
AI,B4 = BOIVtP3P5

AI,Bs : —BOIV¢P3P4
Al,EEPF : + LHELABOIV¢(ELECT - POSIT)
Id,Trick

Id,KDK=0

Id,PlDP1=0

Al,PaoP3=o

Al,P4DP4:o

Al,P50P5=0

Id,PlDP3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,P3DP4=P3P4

Al,P30P5=P3P5 106

Al,P4DP5=P4P5
c P OUTPUT

tyep
Id,2,Dotpr,CU(J')=
(CU1—CU2)*P1(J)+CU2¢P3(J)+CU24PD(J)+CU5*P5(J)

+CUEPFtEpf(J,P1,P3,P5)+CUEPF*Epf(J,P1,P4,P5)

Al,Funct,CU(K' =

(CU1-CU2)*P1(K)+CU2tP3(K)+CU2*PD(K)+CU5:P5(K)

+CUEPFtEpf(K,P1,P3,P5)+CUEPF*Epf(K,P1,P4,P5)

Id, CU1 = - CUOIV:(P3P5+P4P5—P1P5)

Al,CU2 = CUOIVtPIPS

Al,CUs : CUOIv:(P3P4+P3P5+P4P5)

Al,CUEPF = + QHELtCUOIV

:yep

Id,Trick

Id,KDK=O

AI,KDPD=P4P5

Al,EDP1=O

Id,2,Dotpr,PD(J')=P4(J)+P5(J)

Al,Funct,PD(K2")=P4(K2)+P5(K2)

Id,PlDP1=O

Al,P3DP3=0

Al,P4DP4=O

Al,PSDP5=0

Id,PlDP3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,PSDP4=P3P4

Al,PBDP5=P3P5

Al,P4DP5=P4P5

C P OUTPUT

eyep

C NOW PUT IN THE POLARIZATION VECTOR EXPANSION.

 

C

Id,2,Dotpr,E(J')= E1¢P1(J) + E3tP3(J) +
E4tP4(J)+E5#P5(J)+EEPFt(Epf(J,P1,P3,P5)+Epf(J,P1,P4,P5))

Al,Punct,E(K')= E1:P1(K) + E3:P3(K) +
E4:P4(K)+E5¢P5(K)+EEPFt(Epf(K,P1,P3,P5)+Epf(K,P1,P4,P5))
Id,E1 = EOIVtGHELt(P3P5+P4P5-2tP1P5)

Al,E3 = EOIVtGHELtPlPS

Al,E4 = EOIV:GHEL:P1P5

Al,ES = -EOIV¢GHEL¢(P1P3+P1P4)

Al,EEPF = -EOIV

*yep

Id,Trick

Id,KDK=0

Id,Epf(P1,P3,P4,P5) = EVL

Id,PlDP1=O

Al,PSDP3=0

Al,P4DP4=0

Al,P5DP5=0

Id,PlDP3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,P3DP4=P3P4

Al,PSDP5=P3P5

 

 

Al,P4DP5=P4P5 107

C P OUTPUT
:yep
C

C REPLACE K BY IT’S LINEAR EXPANSION.
C

Id,2,Funct,K(J')=C(1)tP1(J)+C(2)#P3(J)+C(3)#P4(J)
+P5(J)t(1-C(3)-C(1)tD1¢P45IV—C(2)tD345¢P45IV)

Fyep

Id,Trick

Id,Epf(P1,P3,P4,P5) = EVL

C

C P OUTPUT

*yep

Id,C(1) = CC11¢PlDK+CC21¢P3DK+(CC31—CC41)*P4DK+CC41*P4P5

Al,C(2) : CC21tP1DK+CC22¢P3DK+(CC32-CC42)*P4DK+CC42¢P4P5

Al,C(3) : CC3ltPlDK+CC32¢P3DK+(CC33-CC43)*P4DK+CC43*P4P5
: CC41¢P10K+CC42tP3DK+(CC43-CC44)tP4DK+CC44tP4P5

THE AMPLITUDE IS NOW COMPLETELY
IN THE FORM OF DOT PRODUCTS
OF K WITH EXTERNAL MOMENTA.

THE DOT PRODUCTS OF K NEED TO
BE EVALUATED, FOR EXAMPLE...
(P3DK) . (P4DK) : (PSDK)

AND REMEMBER THIS FACTOR IS TO BE INTERGRATED OVER ALL K SPACE.

THIS DEFINES THE A3 MATRIX ...
A3(4,3,1) = (S4DK):(S3DK)¢(SIDK) INTEGRAL OF, OVER
ALL K SPACE.
AND SIMILARLY A2(J,K) AND A1(J) MATRICES.

 

THE A3, A2, AND A1 MATRICES ARE EVALUATED USING 19 SIMPLE
INTEGRALS, EVALUATED EARLIER .... II THRU I19....

REPLACE PDK ’8 BY A MATRICES
FOR THE FIVE POINT DIAGRAM, OR
SIMILAR B MATRICES FOR THE FOUR POINT.

nnnnnnnnnnnnnnnnonnnnnnnnnnnnn

Id,PlDK : AA(1)
Al,P3DK : AA(2)
Al,P4DK : AA(3
Al,PSDK : -AA(3) + P4P5

Id,AA(J'):AA(K")¢AA(L') = A3(J,K,L)
Id,AA(J-)tAA(K-) = A2(J,K)
Id,AA(J“) = A1(J)
Id,$ymme,A3,1,2,3,A2,1,2

_ AAA/\finnfifn

Keep EXPS 108
tnext

S A0

Z EXP51 = A0tEXP5

Id,A0*A3(3,3,3) : A3333
Al,A0*A3(2,3,3) = A3332
AI,A03A3(1,3,3) : A3331
Al,A0*A3(2,2,3) = A3322
AI,AOtA3(1,2,3) : A3321
Al,AO:A3(1,1,3) = A3311
AI,Ao:A3(2,2,2) : A3222
AI,AOtA3(1,2,2) = A3221
Al,A0tA3(1,1,2) : A3211
Al,A0:A3(1,1,1) : A3111
*yep

Id,A0tA2(3,3) A233
AI,A0¢A2(2,3) A232

Al,A0tA2(2,2)
AI,AO:A2(1,2)
AI,A0*A2(1,1)
Id,A0¢A1(3)
AI,A0*A1(2)
AI,AOxA1(1)
C P OUTPUT
*yeP

AI,Ao:A2(1,3) : A231

II II II
>>
H
M

AND IN TERMS OF THE 19 INTEGRALS,
THE ABOVE DEFINED D1 AND 03, THE
A3,A2, AND A1 MATRICES,
USING THE NOTATION ...
A3(1,3,3) = A3331

 

nnnnnnnnn

Id,A3333
AI,A3332
AI,A3331
AI,A3322
AI,A3321
AI,A3311
AI,A3222
AI,A3221
AI,A3211
AI,A3111
1d,A233
AI,A232
AI,A231
AI,A222
AI,A221
AI,A211
:yep
Id,A13 = -0.5:(16)

AI,A12 : 0.5.(14 - 03.110)
AI,A11 = 0.5t(15 + 2:01:110)
C

—o.5.(119)
-0.25:(113 - 03:19)
-0.25t(115 + 2:01:19)
-0.25#(114 - O.5¢D3¢I2 + 0.5:D3t03eIG)
-0.125t(4tPI + 2:01:12 - D3313 - 2*DltD3FIS)
-0.25*(116 . 01:13 + 2:01:01:Is)
0.5:(117 - Ds:A222)
(0.25:112 + 0.5:Dltl7 — 0.5:D3tA221)
(0.25¢(111 - 03:18) + DltA221)
0.5-(118 + 2tDltA211)
-0.5¢(19)
-O.25t(12 - D3t16)
-0.25¢(I3 + 2¢D1¢16)
0.5:(17 - D3¢A12)
(0.25#(11 + 2:01:14) - 0.5#03#A11)
0.5:(18 o 2tD1tA11)

 

C
C

 

109
-yep
Id,P13 : P34 + P35 + P45 - P14 - P15
C PUT IN Ao
C
Id,A0 = 110
c
C
C
C
C THE FIVE POINT ANSWER IS Now FORMED.

n

B WIV,D3,PI,FPIIV,CUOIV,BOIV,EOIV,EVL
Keep EXP51
*next

 

C NOW CALCULATE THE FOUR POINT AMPLITUDE

IN THIS SECTION, TAKE THE TRACE, REPLACE EACH K BY IT’S
LINEAR EXPANSION,
AND REDUCE TO A FINAL ANSWER.

BIKIwIMIE’NIPIIP2IP3'P4IP5IPD
C,Fo,F1,F2,F3,F4,F5,F6,F7,F8,F9,F10,F11,A1,A2,A3,AA
J1,J2,M1,M2,K3,K2

 

SCALE FACTOR FOR THE AMPLITUDE IS ...
FSCAL = 1.0

TAKE THE TRACE
EXP4=FSCALtFO¢F1tF2:F3tF4tF11tF5tF6tF7tF8tF9tF10
SUBSTITUE ACCORDING TO HELICITIES AND DIAGRAM

nnnNnnnnxnnnnHm<nnnnnnnnnn

Id,FO=O.5¢((1+LHEL)¢(G6(J1))+(1-LHEL)t(G7(J1)))
Id,F1=(G(J1,B))

Id,F2=ELECTt(G(J1,M2)) + POSIT¢(G(J1,M1))
Id,F3=(G(J1,K))

Id,F4=ELECTt(G(J1,M1)) + POSIT¢(G(J1,M2))
Id,F11=0.5¢((1+QHEL)w(66(J2))+(1-QHEL):(G7(J2)))
Id,F5=(G(J2,CU))

*yep

Id,F6=(G(J2,M2))

Id,F7=(G(J2,M))

Id,F8=G(J2,M1)

1d,F9=(G(J2,N))

Id,F10=G(J2,E)

‘yep

Id,Trick,Trace,J1

CL

__ A‘thnn r3)

110
TRICK AND TRACE

non

Id,Trick,Trace,J2
*yep
C

C START CONDENSATION SUBSTITUTIONS, SOME DIAGRAM DEPENDENT.

C

Id,2,Dotpr,N(J') = P3(J)+P4(J)+P5(J)
Al,Funct,N(K2') = P3(K2)+P4(K2)+P5(K2)
Id,MDM=0

Id,Botpr,M(J')=K(J)—PD(J)
Al,Funct,M(J')=K(J)-PD(J)
Id,KDPD=P4P5

Id,KDK=O '
Id,2,Dotpr,PD(J~)=P4(J)+P5(J)
Al,Funct,PD(J')=P4(J)+P5(J)
Id,P4DP4:o

Al,PSDPS=0

Al,EDP1=0

C P OUTPUT

‘yep

C PUT IN THE ELSEWHERE EVALUATED SPINOR EXPRESSIONS.
C

Id,2,Dotpr,B(J')=BS*P3(J)+B4*P4(J)+85*P5(J)+BEPFtEpf(J,P3,P4,P5)
Al,Funct,B(K3')=83:P3(K3)+B4¢P4(K3)+85:P5(K3)+BEPF¢Epf(K3,P3,P4,P5)
C P OUTPUT '

*yep

Id,B3 : Borv4P4Ps

AI,B4 : BOIVtPSPS

AI,BS = —BOIV¢P3P4

Al,EEPF : + LHEL:BOIV:(ELECT - POSIT)

Al,PlDP5=P1P5
Al,P30P4=P3P4
Al,P30P5=P3P5
Al,P4DP5=P4P5
c P OUTPUT

*yep
Id,2,Dotpr,CU(J')=
(CU1-CU2):P1(J)+CU2¢P3(J)+CU2GPD(J)+CU5:P5(J)
+CUEPF¢Epf(J,P1,P3,P5)+CUEPFtEpf(J,P1,P4,P5)
Al,Funct,CU(K')=
(CU1-CU2)#P1(K)+CU2*P3(K)#CU2*PD(K)¢CU5*P5(K)
+CUEPFtEpf(K,P1,P3,P5)+CUEPF¥Epf(K,P1,P4,P5)
Id, CU1 = - CUOIV*(P3P5+P4P5-P1P5)
Al,CU2 = CUOIVtPlPS
Al,CU5 = CUOIV¢(P3P4+P3P5+P4P5)
Al,CUEPF = + QHELtCUOIV
‘YGP

 

 

Id,Trick

Id,KDK=0

A|,KDPD=P4P5

Al,EDP1=0
Id,2,Dotpr,PD(J")=P4(J)+P5(J)
Al,Funct,PD(K2')=P4(K2)+P5(K2)
Id,PlDP1=0

Al,PBDP3=o

Al,P4DP4=0

Al,PSDP5=0

Id,PlDP3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,P3DP4=P3P4

Al,PSDP5=P3P5

Al,P4DP5=P4P5

C P OUTPUT

*yep

C NOW PUT IN THE POLARIZATION VECTOR EXPANSION.

C
Id,2,Dotpr,E(J')= E1*P1(J) + E3*P3(J) +
E4*P4(J)+E5tP5(J)+EEPFm(Epf(J,P1,P3,P5)+Epf(J,P1,P4,P5))
A|,Funct,E(K~)= E1*P1(K) + E3*P3(K) +
E4tP4(K)+E5¢P5(K)+EEPF*(Epf(K,P1,P3,P5)+Epf(K,P1,P4,P5))
Id,E1 = E01Vt0HELt(P3P5+P4P5-2$P1P5)
= EOIVtGHELePlPS
= EOIVGGHELtPlPS
Al,ES = —EOIV#GHEL¢(P1P3+P1P4)
F = —EOIV

Id,Trick
Id,KDK=0
Id,Epf(P1,P3,P4,P5) = EVL
Id,PlDP1=0

Al,P3DP3=O

Al,P40P4=O

Al,PSDP5=0

Id,PlDP3=P1P3
Al,PlDP4=P1P4
Al,PlDP5=P1P5
Al,P3DP4=P3P4
Al,P3DP5=P3PS
Al,P4DP5=P4P5

C P OUTPUT

ryeP

C REPLACE K BY IT’S LINEAR EXPANSION.
C

Id,2,Funct,K(J')=C(1):Pl(J)+C(2):P3(J)+C(3)*P4(J)
+P5(J)t(1-C(3)-C(1):DltP45IV—C(2)¢0345*P45IV)

tyep

Id,Trick
Id,Epf(P1,P3,P4,P5) : EVL
C P OUTPUT

#yep
1d,C(1) = CC11:P1DK+CC21:P3DK+(CC31-CC41)¢P4DK+CC41¢P4P5

 

 

AAAC ‘CC((I\I\l l l

112

Al,C(2) : CC21:P10K+CC22:PSDK+(CC32-CC42)tP4DK+CC42¢P4P5
Al,C(3) : CC31tPlDK+CC32¢PSDK+(CC33-CC43)MP4DK+CC43¢P4P5
Al,C(4) : CC41tPlDK+CC42tP3DK+(CC43-CC44)AP4DK+CC44¢P4P5
C P OUTPUT

tyep

THE AMPLITUDE IS NOW COMPLETELY
IN THE FORM OF DOT PRODUCTS
OF K WITH EXTERNAL MOMENTA.

THE DOT PRODUCTS OF K PNEED TO
BE EVALUATED, FOR Exm
(P3DK) (P4OK) (PSDK)

AND REMEMBER THIS FACTOR IS TO BE INTERGRATED OVER ALL K SPACE.

THIS DEFINES THE A3 MATRIX .
A3(4,3,1) = ($40K)*(SSDK)3(SlDK) INTEGRAL OF, OVER
ALL K SPACE.
AND SIMILARLY A2(J,K) AND A1(J) MATRICES.

THE A3, A2, AND A1 MATRICES ARE EVALUATED USING 19 SIMPLE
INTEGRALS, EVALUATED EARLIER .... 11 THRU 119.

REPLACE PDK ’5 BY A MATRICES
FOR THE FIVE POINT DIAGRAM, OR
SIMILAR B MATRICES FOR THE FOUR POINT.

nnnnnnnnnnnnnnnnnnnnnnnnnnnnnn

Id,PlDK : AA(1)
Al,PSDK : AA(2)
Al,P4DK : AA(3)

Al,PSDK = -AA(3
Id,AA(J'):AA(K')
Id,AA(J")tAA(K')
Id,AA(J') = A1(J)
Id,Symme,A3,1,2,3,A2,1,2
Keep EXP51,EXP4

.next

5 A0

Z EXP41= AO¢EXP4

+ P4P5
tAA(L-) = A3(J,K,L)
= A2(J,K)

Id ,Ao:A3(3, 3 ,3) = BsERR.
AI,AO*A3(2,3,3) : BSERR
Al A0:A3(1,3,3) : B3ERR
Al,A0tA3(2, 2,3) : B3ERR
AI,Ao.A3(1, 2,3) : B3ERR
Al,A0¢A3(1, 1,3) : B3ERR
AI,A04A3(2, 2,2) = B3ERR
Al ,AO:A3(1, 2,2) = BSERR
Al ,Ao.A3(1,1,2) : 83ERR
Al ,AOtA3(1, 1,1) = B3ERR

:yep

 

 

B233 113

8232
8231
8222

Id,Ao.A2(3,3)
AI,Ao.A2(2,3)
AI,AO:A2(1,3)
Al,AO*A2(2,2)
Al,AO:A2(1,2) B221
Al,Ao:A2(1,1) 3211
Id,AO*A1(3) = 813
Al,AO*A1(2) : B12
AI,A0*A1(1) =
C P OUTPUT
*yep

AND IN TERMS OF THE 19 INTEGRALS,

THE ABOVE DEFINED D1 AND 03, THE
A3,A2, AND A1 MATRICES,
USING THE NOTATION....
A3(1,3,3) : A3331

PUT IN "B" SERIES

finnnnnnnnnnnnn

-0.5*115
-0.25t(4*PI - 03:13)
—0.5#116

0.5:(112 - 03:812)
O.St(111 - 033811)
118

-0.5t13

0.5:(11 —03*15)
8

Id,8233
AI,B232
A|,8231
A|,B222
Al,3221
Al,8211
Id,Bl3
Al,812
AI,B11
C

 

C PUT IN A0
C

 

*yep

Id,P13 = P34 + P35 + P45 - P14 - P15
tyep

- Id,A0 = 15

C

C
C
C
C THE ANSWER IS NOW FORMED.

O

B WIV,03,PI,FPIIV,CUOIV,BOIV,EOIV,EVL
Keep EXP51,EXP41
tnext

C
NOW ADD THE FIVE AND FOUR POINT
DIAGRAMS TOGHETHER

AND PUNCH THE FORTRAN

C
C
C
C
C COMPATIBLE OUTPUT
C

114
Z EXP: EXP51 + EXP41
B WIV, 03, P1, FPIIV, CUOIV, BOIV, EOIV, EVL, 11, 12, 13, 14, 15, I6,
17,18,19,110,111,112,113,I14,115,116,117,118,11L
Punch EXP
tend

 

 

 

com]
inde
the ‘
co-ei

, and

As

tet

.... ., 9

Appendix L
The Linear Expansion of 13,,

An arbritary, though specific four-vector, k“, being composed of its four
components (kt, ky, 19,, k0), may always be projected upon four other linearly
independent four-vectors- 51,52,53,54. in that what we don‘t know about
the four components of k,‘ is traded for what we don‘t know about the four
co—eflicents of the expansion:

k“ = C1(51,,) + C2(52u) + C3(53,,) + C4(54,1)
and in general the co—efficents are functions of k

kg = 101(k)](Slu) + [02114)](521) + 103(k)l(33u) + [cr4(1—.))(54,,)

As the equation is linear. the co-efficents for any 117,, may be determined in
terms of the dot products:

51-k. 52-k. 53-k. 54-1:

or, to state it differently. what we'don‘t know about the four co—efiicents may
be futher traded for what we don‘t know about the four dot products.

This may be seen by forming the dot products:

51-k=C1(51-51)+C2(51-52)+C3(51-53)+C514( S4)
52-h:C1(52-51)+C2(52-52)+C'3(52-53)+C4(52 54)
53-11::C1(53-51)+C2(53-52)+C3(53-53)+C4(53 -54)
54-k=C1(54-51)+C2(54-52)+C3(54-5 3)+C4(54 54)
Which may be cast in matrix form
51-17 51-51 51-52 51-53 51-54 ('1
52-1: __ 52-51 52-52 52-53 52 54 ('2
53-1: — 53-51 53-52 53-53 53-54 C3
54-1: 54-51 54-52 54-53 54-54 C4

115

 

 

 

The. co

Let W

and t

Sin

De

116

,‘he co-eflicents C 1. C 2. C3. C4 are solved for in the usual Cramers rule method.
.et W be the determinant of the above matrix. It is solved to be:

w = —(5152)(5354){—(5152)(5354) + (5253)(5154) + (5254)(5153)}

-(5153)(5254){(5152)(5354) + (5253)(5154) + ~(5254)(5153)}
—(5154)(5253){(5152)(5354) — (5253)(5154) + (S254)(5153)}

ind the Cl numerator is the same with 51 replaced by k in the first column:
Clnumemmr = 2((51 -k)(5253)(5254)(5354)
—(5152)(5354){-(S2 ~k)(5354) + (5253)(54 -k) + (5254)(53 - 4)}
—(5153)(5254){(52 - k)(5354) + (5253)(54 - k) + -(5254)(53 -k)}

—(5154)(5253){(52 - k)(5354) — (5253)(54 Jr) + (5254)(5.s - 1.1)}

Clnumerator
C W

Similarly for the C2, C3, C4 numerators.

Define the elements of the inverse matrix to be:

5151 51-52 51-53 51-54 ’1 CC11 0012 CC13 CC14'

 

52-51 52-52 52-53 52-54 _ CC21 CC22 CC'23 CC24
53-51 53-52 53-53 53-54 — CC31 CC32 CC33 CC34
54-51 54-52 54-53 54-54 CC41 CC42 CC43 CC44
Then

CC11 CC12 CC13 CC14 51-]: C1

CC21 CC22 CC23 CC24 52-lc _ C2

CC31 C-‘C32 CC33 CC34 53-lc — C3

CC41 CC42 CC43 CC44 54-k C4

which gives four equations of the‘ form:

C1: CC11(51- k) + CC12(52 - 1:) + C'C13(53 - 1') + CC14(54 - 1')

Comparison with the Cramer‘s rule generated expressions for C'1.C-'2.C3.C-'4

 

yicl

117

yields the CC11 ............ CC44

0011 = W'12(S253)(5254)(53S4)
0012 = —W'1(5354){—(5152)(5354) + (5253)(5154) + (5254)(5153)}
0013 = —w-1(5254){(5152)(5354) + ($253)(5154) + —(5254)(5153)}
0014 = —w-1(5253){(5152)(S354) - (5253)(5154) + (5254)(5153)}

0021: 0012
0022 = w—12(5153)(5154)(5354)
0023 = -W‘1(5154){(5152)(S3S4) + (S253)(SIS4) + —(S2S4)(SlS3)}
0024 = —W'1(SIS3){(5152)(S3S4) —(S253)(5154)+(5254)(5153)}

0031 = 0013
0032 = 0023
0033 = w-12(SIS2)(SIS4)(S2S4)
0034 : —w-1(5152){—(5152)(5354) + (5253)(5154) + (5254)(5153)}

0041 = 0014
0042 = 0024
0043 = 0034
0044 = w-12(5152)(5153)(5253)

In the above the fact that the particles have zero mass. that is that 5,, 5‘‘ =
has been used.

 

 

118

Alternatively the terms may be rearranged to make explicit the dependence
on the dot products:

k” = (51 -k)f1,, + (52 - 1.1).72,‘ + (53 - k)f'3,. + (54 - k).7-'4,,
where

$1,, = (CC11)51,J + (CC12)52,, + (CC13)53,, + (CC14)54,,

$2,, = (CC21)51,, + (CC'22)52,, + (CC23)53,, + (C'C24)54,,

73,, = (CC31)51,, + (CC32)52,, + (C033)53,. + (CC34)54,,

$4,, = (C'C4l)51,. + (CC4‘2)52,, + (CC-'43)53,, + (CC44)54,,

As a. check that the matrix elements C C 11 ...... C C44 are correctly calculated.
they are compared with the results of the relation:

HOV-7W)“ = 55

which is obtained by dotting the above expression for k“ with 51.52. 53.54.

51%,, = (Sl“f1p)(Sl-k)+(51“.72p)(52-A:)+(Sl”}‘3u)(53-1-)+(Sl“f411)(5‘4-k)

and noting that 51 - k, 52 - k, 53 - k. 54 - k are all independent.

For this problem:

51,, = P1,,
52,, = P3,,
53,, = P4,,
54,, = P5,.

L454... 54—.- a.

A.

Appendix M
Base Integral Reductions

After the TRACE is taken and the four linear substitutions invoked, the
)OP calculation has a numerator containing terms of three, two, one, or none
:tors of dot products with k“.

.ch term, here called a ’base integral’, can be integrated separately, and is
tentially reducable to a simplier integral via cancellation of the dot products
1:“ with similar dot products in the denominator.

1r example, the ’base integral’

)—/ d94- (P3-k) (P1-k)(P1./c)
_ (-2P4-k+Jl/[.;-’)(2P1 °kfi21,P1P4+P1P5)+1\/[3

quark

3(2,1,1

 

)(2P3 - 1: +313 - Trim)

rith some co-efficent), from the five point diagram;

1d
dQL- (P4 - 15)

31(3) = (—2P4 - k + 111$)(2P3-k + M? _ ‘TM-z)

 

*ith some co-efiicent), from the four point diagram, which only has two de-
-minator factors to start with.

l establish some notation:

lose ’base integrals’ of the five point diagram. (with three denominator fac-
rs), will be called ’A’ type integrals and will be sub-typed A1. A2, or A3 to
licate the number of dot products of It:,, present. The first integral above is
type A3.

nilarly, the four point diagram integrals (wit-11 two denominator factors) will
typed ’B’, and sub-typed B1 or B2 to indicate the number of dot products
Icy present in the numerator. The second integral above is of type Bl.

ther the designation will be indexed to show specifically. which dot prod-
ts of k“ are present. This index corresponds to the vectors position in the
,1ansion of k” effected in Appendix L:

k, =(01)P1,. + (02)P3,, + (03)P4,. + (C4)P5,,

119

 

120

ning the index 1 to P1, the index 2 to P3, the index 3 to P4, and the index
P5. The index 4 is removable by the on—shell condition:

P5-k:—P4-k+P4P5

; the first example integral above is A3(2, 1,1) and the second is B1(3)

reduction of all base integrals by the cancellation of factors is now enacted.
ich case reduction is carried out until no futher cancellation is possble. The
integral is then eXpressed in terms of nineteen irreducable integrals which
t be solved analytically. These nineteen integrals are listed and solved in

endix N.

1e reduction, whenever a term proportonal to M7 or Mquark is generated it
“opped, in recogniton that both these masses will ultimately go to zero.

do). (P44) (P44) (P44)

 

 

 

 

 

 

 

 

3‘3’3) = / (—2P4 . k + Mg)(2P14 - 2(P1P4+ P1P5) + 11- jua,,)(2P3 - I.- +413 413141,.)
_/dn,, (P44) (P44) (P44)
‘ (...)(...)(...)
__ do). ('—2(P4‘)4) (P44) (P4-k)]
" ”2 U (...)(...)(...)
_ d0), (—2(P4)4+M.';’) (P44) (P44) _ 111’ 4:2,.
”2 [/ (...)(...)(...) 1 7 (...)(...)(...)]
do, (—2(P4)4+M:,-’) (P44) (P44) 412,
_ —0
”2 [/ (...)(...)(...) /(...)(...)(...)l
d9), (P44) (M4)]
= —l 2
/ l] -(...)<...)
= —1/2(119)

(In, (P44) (P44) (P34)
3’3’2) "‘ / (—2P4 4 +4-13)(2P14 - 2(P1P4 + P1P5) + 11I5,,,,,.,.)(2P3 - k + M? - mu.)

 

121

 

=_1/2[/dm (P44) (P3-k)]

 

 

.(...)(...)
, do). (P44) (2P34+(A-1;-’—iI‘M,)) . . ' dn (P4-k))]
. 12 . _ .2_ .,
l/ (/ -(...)(...) “”~' mm] 4.14...)

= —1/2 [1 /2 (I13 — (4.13 — KM, )19)]

 

1)_/ do; (P4 k) (P4 k) (P1 4)
" (-2P4-lc+M'-’)(2P1 4— 2(P1P4+P1P5)+M S;-’,,a,.,)(2P3 k+M§— mu.)

 

 

 

(112),, (P4 1:) (P1-k)]
:3 —l 2
.4 [f (19); (P4-k) (21:(1-)k(—)2(P1P4+P1P5)) +2(P1P4+P1P5)/d9:( ()f-‘likq

= ..1/4 [115 + 2(P1P4 + P1P5)19]

2 2 _/ (152;. (P44) (P34) (P3 1:)

:_,/2 U419). (P34) (P3-k)]

 

 

 

 

 

 

4 i/ 412,. (P3-Ic')(2P3-lc+(M;-i1‘411:)‘)__(M;_iFM;)/d9k (ID-3'0]
_ .(...)(...) . ° '(---)("°)
... . . . 2 1:2— 1: o . da'
4 I14—1/2(Mf—iI‘A-I;)(fdm ”Trill; ’“1*))—(.n; —21‘M:)/.(m:(m))]

—1/4 [114 — 1/2(Mf — 27PM, )12 + 1 /2(.-1-[f — mu. NM} — mu, )16]

 

122

,2 1)=/ (112,, (P4 4) (P3 k) (P1 4)
' ’ (—2P44+M,=’ )(2P1 k—2(P1P4+P1P5)+Mj,,a,,)(2P3 k+M:— inn.)

-—1/2[/W]

 

d0), (P1-k)(2P3-k+(Mf—i1‘1t[;)) 4 ., fdm (P1-4.)]
. 4 _ ,~_, .. __
ll/ .(...)(...) (M: AM) .(...)(...)

 

 

an (Pl-k) .m 2P14-2 P1P4 P1P .40
I _"— - (M3 .— irM,)1/: " ( ( + 5)) + :(P1P4+ PIPS) "
c(...)o o(...)(...) o(...)(...)

do (2191 4 — 2(P1P4 P1P5 m
I[1/2 (/ 1. + )) +:,P,p,+ p1p3)/ ( E) >- 1/24Mf - lrfll;) (I3+2(P1P~1+ P1P5)16)]
. ... .

 

 

o(...)l

[/4 [1/2 (471' + 2(P1P4 + P1P5)12)—1/2(1\If—i1‘1)-I,)(I3 + 2(P1P4 + P1P5)16)]
[/8 [4n + 2(P1P4+ P1P5)I2 — (ME — iriuzns — 2(P1P4+ P1P5)(Mf—if1\-I;)IG]

31 1)_/ (10;. (P44) (P14) (P14)
’ ’ " (—2P44+M,'-’)(2P1-k—2(P1P4+P1P5)+A-Ijm,)(2P34+Mg—irM,)

=_1/2 [/ko (Ill-k) (Pl-k)

 

[/4 [/4112], (Pl-k) (2f(l-;c(—)2(P1P4+P1P5))+2(P1P4+P1P5)/d9:()fl)dc)]

 

 

an 2P14—2P1P4 PIPS) do
'4[116+(P1P4+P1P5)(/ " l 0‘ ): ) + )+2(P1P4+P1P5)/" H" )>]

./4 [116 + (P1P4-l- P1P5)I3+ (l/2)2(P1P-l+ PlP5)2(PlP-l+ P1P5)I6)

 

123

2 2):/ 019:: ('P3'k) (P3-k) (P3-k)
, . (-2P4.Ic +2143)(2P1.k — 2031194.). P1P5)+1\Ifa,.k)(2P3-k +1113 — iI‘AL)

U

 

 

(191: (P3 - H2]

tU ko (P.1c) (P3-k) (2P3-k+(M;—i1‘111;))_(MZ:»_ITM:)/ ( )( ).

(...)(...)(...)
= 1/2 [I17 — (1113 — il‘Mz)A2(2, 2)]

/ alm- (P3-k) (PB-k) (P1-k)
(—2P4 - k +1113)('2P1-k — 2(P1P4 + P1P5) + 11flft‘a,.k)(2P3 . k + M? — iI‘AL)

)U do,c (Pm/c) (P1-k) (2P3-k+(M}—ier)) _W?_im[:)/dm (P3-k)(P1-k)]
(...)(...)(...) ~ (...)(...)(...)

an F: - k 2P1»k — 2 PIP-l PxPs 417 P3 2 k .,
1/2 (/ " ( ) ( ( + ”+2(P1P4+P1P5)/—" ( )) - (M; — IFAI:)A212‘~H]
(...)(...)o (...)(...).

= 1/2 [1/2 (112 + 2(P1P4 + P1P5)I7) — (ME — iF1)[;)A2(2,1)]

¢,2.1)=

 

/ (1m. (PB-k) (P1-k) (P1-k)
(—2P4 . k + M;~’)(2P1-Ic — 2(P1P4 + P1P5) + MZWM2P3 - k + M} - iI‘AL.)

(I
,[ ko (P1-k) (P1-k) (2P3-k+ (ME — iI‘M;
' (...)(...)(...)

40 P ~k 2P 41—2 PIP-l PlPs d0 Pl~k .,
1/2 (/ " ( 1 ) ( l ( + )) +2”?qu + P1PS)/ ———" )( )) - (M; - '1'M:)A:I(1.l)]
(...)(...)a _ (... (...).

m 2P1-k—2lPIP4 P1P!) 40. ,
1/2(111 +(P1P4 +P1Ps) [/ k ( + n + :(mm + P1P5)/ ‘ J) - (M; — :I‘1\l;)A2(l.l)]
(..4)(...). (...)(...)-

:[1/2 (111 + (P1P4 + PIP-5) [11 + 2(P1P4 + P1P5)I4]) — (.113 — ir.\1_.)A2(2. 1)]

‘,1,1)=

) —(1113—ir111;)A2(1,1)]

 

 

 

 

 

124

The previous may be done another way.

 

 

A3(2 1 1)_/ (1521. (P3-k) (P1-k) (Pl-k)
’ ' ’ (—2P4-k+111;-’)(2P1.Ic—2(P1P4+P1P5)+Mgwk)(2P3-Ic+M§—mm)
=1/2[ (1121. (P31) (PL?){(2P)1('l;_2(P1P4+P1PJ))+2(P1P4+P1P5)A2(2,1)]

 

-L- '2 - 2 —' 1.
=1/2 [1/2 (j an. (P1 ’(‘ fa(.k;"“"= II1‘")-(M,2—x'l"M:)18>+ 2(P1P4+P1P5)A2(2.1)]
.

= 1/2 [1/2 (111 ~ (ME — iI‘M;)18) + 2(P1P4 + P1P5)A2(2,1)]

dm. (Pl-k) (P1-k) (Pl-k)
/(—2P4 ~ k + A!.',-’)(2Pl - k — 2(P1P4 + P1P5) + 1115“”): )(2P3 . k +1113 — 11111:)
[/ d9), (P1-k) (P1-k) (2P1'k—2(P1P4+P1P5))
=1/2
(...)(...)(...)
=1/2[118 + 2(P1P4+ P1P5)A2(1.1)]

 

A3(1,1,1) =

+ 2(P1P4+ P1P5)A2(1. 1)

 

 

 

125

dQL. (P4- k) (P4- k)

 

 

 

 

A ' =
2(3’3) j(—2P4-k+M_3)(2P1-lc—2(P1P4+ P1P5)+M§ua,k)(2P3-Ic+1113 41111;)
:(/d_______Qk (P4 k) (P4 k)
...... P)(...)
d121(— k) (P4-In]
—1 2
/ [ 2(...)( ...)(...)
d9); (—2( (P4)-Ic+M-) (P4-k) 9 do)
——1 2 7 — -- ——
/ U: (...)(...)(...) [UV/(...)(...)(...)]
(112,: (—2(P4)-k+M?) (P4-1c) dm
——1 2 7 —0 —
/ U (...)(...)(...) /(...)(...)(...)J
__ 412k (P4-k)
_ W U -<...>(...) 1
= —1/2 19

“(3 2)_/ d9): P(4 1) (P3 k)
’ ‘ —2P4 k+M°)(2P1 k—2(P1P4+P1P5)+M;u;m)(_2P3-k+M3—i1‘1\I:)

__ 4121 (P3-k)
WU -(...)(...) 1

=_ (1521 (2P3-k+(M§—iI‘M;))_ /2_, ‘ (19,; )]
1/2[1/2(/ .(...)(...) (M; ‘FM‘)/.(...)(...)

= -1/2 [1/2 (12 — (ME — i1“M.-)16)]

 

 

 

 

A2(3 1)—/ 491. (P4-k) (P1-k)
, — ("2P4'kfluvgmpl"'—2(P1P4+PIP-5)+.\/f.1.1,.1.)(2P3-I.-+M;—;r,\/:)

_1/2 U d9). (P1-1)]
.(...)(...)

126

412 2P1-lc—2P1P4 .r
k ( -(...)(...) +P1P)))+2(P1P4+P1P5)/

= -1/4 [13 + 2(P1P4 + P1P5)16]

 

—_1/4[ 110). ]

.(...)(...)

A2(2 2):/ (101 (P3- 1:) (P3 k)
, ( —.2P4 H1113 )(2P1 1c—2(P1P4+P1P5)+quX21231“);_,TM)

11m (P3 1:) (2P3 k+(ME —1r11-1)) 4 , 1112). (P3-k)
=1 2 / 1: — . -~ /———-]
/ [ (...)(...)(...) (M ’FM') (...)(...)(...)
= 1/2 [17 — (1113 — 1111.. 141(2)]

ko (P3 K) (P1-k)

 

 

A2 2(’2 1)=/(— 2P4 Ic+M;’)(2P1 k— 2(P1P4+ P1P5)+1\1 umx2P3 (Hug—11111)
d9): (P1-k) (2P3-k +(11-13— 1111.)) , /dm (P1~1.)]
=12 — M:—.I‘\I. ——
/ [ (...)(...)(...) ( - ’ ’ ‘) (...)(...)(...)

 

 

an 2 .11.: r
:1]: [1/2 (/ 1. 1 P1 (P1P4+P1Po))+2(P1P4+P1P5)/( )1 ’
...... C

1 )1 1. )-(Mf-ir‘)\l;)A1(1)]

=1/2[1/2(11 + 2(P1P4 + P1P5)I4) — (ME — 11‘111: )A1(1)]

/ 1191- (P1 1') (P1 In)
(— 2P4 k+M=)(2P1 k— 2(P1P4+P1P5)+M;uark)(2P3 (+1113- 1111:)
[/ (IQ)g (P1-k) (2P1 k-2(P1P4+P1P5))
=1/2
(...)(...)(...)
= 1/2[18 +2(P1P4+ P1P5)A1(1)]

A2(1 1)=

 

+ 2(P1P4 + P1P5)A1(1)]

 

 

 

 

 

127
111(3):] 1112). (P4-k)
(—2P4 - k + 1113)(2P1-k — 2(P1P4 + P1P5) + Mfwm2P3 - k + M? - iI‘MJ
z/dm (P4-k)
, (...)(...)(...)
.1121 (—2(P4)-k)]
=—1 2 ——
/ U (...)(...)(...)
1112). (-2(P4)-k+1112) 4 (1:21-
=—1 2 7 —M-/————
/ [/ (...)(...)(...) 7 (...)(...)(...)]

 

=_]/2 /d§2,. (—2(P4)- :+M;) _0/ 1112),
(...)(...)(...) (...)(...)(...)

= -1/2 [/ .(_d,s)2(l)]

= -—1/2(I6)

 

_/ (19). (P3 - k)
" (—2P4 - k + 1113)(2P1-k — 2(P1P4 + P1P5) + 1115””)(2P3 - k + M3 — 1111:)

l

A1(2)

 

 

_ 1112,. (2P3 - k+ (ME — 11M,» _
’1/2U (...)(...)(...)
= 1/2 [14 — (111,2 — 1111.)11o]

., . d9);
’1- : — '- ——
(11. 11“NJ/(...)(...)(...)]

A1 1 _ c112,, (P1-k)
( )_ / (-2P4 - k + 11$)(2P1-k — 2(P1P4 + P1P5) + 111;”,].)(2P3 - k + 1113—1111.)
_1/2]/d§21 (2P1 - k - 2(P1P4+ PIP-5))
' (...)(...)(...)
=1/2[15 + 2(P1P4 + P1P5)110]

 

 

+ 2(P1P4 + PIP-SKID]

m1

 

 

128

 

B1(3):/ (1121 (P41)
. (—2(P4-k)+M$)(2P3-k+1\f} 41111.)

_/ko (P4-k)
‘ (....)(....)

= —1/2 U110). (—2(P4).1-.)]
(....)(.....)

c112,. (—2(P4) - k+ 1113) ., (112).
: — 2 _ , -
1/ U (....)(....) M7/(....)(....)]
(1521, '(—2(P4)-k+M'-’) 412).
= —1 2 W - /
/ U (....)(....) 0 (...)(...)J

= ~12 V 37.31

= —1/2(13)

 

 

 

 

 

ko (P3 ~ k)

B1(2) = (4124.1 + 1113)(2P3 - k + 1113—1111..)

 

 

_ 452,. (2P3-k+(ME—iI‘AL))_ 2_, , / dm ]
“I/QU (....)(...) (M; 71711:) (----)(.-.)

= 1/2 [11 — (113 — 11111.. )15]

 

131 1 _ d91- (Pl-k)
( )—/(—2P4-k+M.3)(2P3-k+M§—ir1\1’;)

=18

_ (1m. (P4-k) (P4-1“)
B2(3,3)_/(41%],+A13)(2p3.1+1113—11111.)

 

 

129

_/df21 (P4-k) (P4-k)
_ (....)(...)

11—— .1 1

= —l/2 I15

32 3 2 _ 1112). (P4-k) (P31)
( ’ )‘ (—2P4-lc+M.;-’)(2P3-k+M§’—iI‘M=)

= _1/2 U «1521. (P3 - 1)]
d9;

= —l/2 [1/2 (/df21-, (2P3 - k+()M; — iI‘M:)) _ (M? _ iI‘M;)/ .(m:))]

o(....

 

 

 

= —1/2[1/2(4:r — (ME — iI‘M; )13)]

_ (112,. (P4-k) (P1-k)
132(311) - (42124.1; + 113)(2P3 ~ k + M? — 11111..)

419- Pl-k)
= —1/2[_/ Lo(.(...) ]

= —1/2(116)

(19). (P31) (P31)
B20” = / (—2P4 - (v + 1113)(2P.3 -1-+ 1113—1111:)

 

130

 

= 1/2 [112 — (ME — 11M. )B1(2)]

c191 (P3-k) (P1-k)

B2 2,1 = _
( ) (—2P4-k+fif$)(2P3-k+1\13—zI‘11[;)

 

= ”2 [j (191: (P1-k) (22193)] k)+(M; —iI‘J\'1'.~))_(ME_I.I,M;)-/d§2)E ()EPl).L-.):]

= 1/2 [111 — (ME — iI‘M: )18]

13211 —/ d911P1.1~.)(p1.).)
(, )— (—2P4-k+M.';’)(2P3.k+ME_irM:)

 

= I18

 

Appendix N
Analytic Solutions of Nineteen
Irreducable Integrals

The calculation involves nineteen integrals which may not be futher reduced
and must be done analytically.

They are here listed, and in the remainder of this Appendix‘ analytically eval-
uated.

The notation to delineate each of the separate nineteen integrals is based on
the following:
I—/ (101. (P4-k) (P1-k) (P3-k)
'— (—2P4 - k + 111$)(2P1-k - 2(P1P4 + P1P5)+ Afft‘arl.)(2P3 - k + M? -— 7T1U;)
The numerator contains three factors as does the denominator. As illustrated
in Appendix M, some of these factors may be cancelled by the judicous adding
and subtracting of balancing terms. Those which remain form the nineteen ir-

reducable integrals. They are listed below in a notation where o ‘s replace those
factors of the notation integral above that have been removed by cancellation.

For example, in this noatation the integral

(1:2,. (P3-k)
(2P1 ~12 — 2(P1P4+ P1P5) + M2 m)

=/dm ..(....)(:) w

The nineteen irreducable integrals are:

 

 

_ d9]. 0 a -
11 ‘/ (...) ' ~

/ ({Ql' a o A
12 =

 

131

132

 

I4:

 

IG:

 

18:

I9:

/
/
/
/
/
/
/

I10 =

 

 

133

112 =

 

 

 

 

115 =

 

116 =

/
/
f3. . 1...)
/
/

 

 

117 = [d91- O O (..)

 

118 : jdflk o (...)' o

 

119 = der. (...)' o o

The integrals may be grouped into 5 classes for solution- 11 — I3. I4 -IG.
17-19. Ill —116, and 117 - 119. which are now solved for. The integral
110 is '. through on-shell conditions. expressable in terms of I1 — 19.

The Class of Integrals II - I3

These integrals are simple and may be done directly.

They are of the form:

dflk
I = —
/ (...)

In the co—ordinate system where the only three momentum in the integral is
taken to be the z direction they assume the form:

 

 

d(cos6')dd> 1 d: 21r a +b
—— = 27r —-———— = — log ——
4,,(acos(9+b) _1aa:+b a —a.+b
11 _ / d9}; _ / ko
— —2(P4 - k) + A13 — —2ko'-’ c080 + 21:03 + A172
__ 271' A13
‘ —2k02 °g «aka?

 

11- —2« IO <—2(P4P5))
’ P4P5 g M;

12_/ (ml.
‘ 2(P1-lc)

— 2(P1P4 + P1P5) + A'quuark
and using 2(E1)ko = —(P1P4+ P1P5)

-_ J/ d§2k

2(E1)ko c089 + 2(El)ko + Affuark

271’

 

4(El)ko
= —— o .,
2(E1)ko g lug-Md.

M

2
quark

—27r 1 —2(P1P-1+P1P5)
:————o
12 P1P4+P1P5 g

135

13— / ————_d9“
' 2(P3 - k) + ME — ITM:

and using 2(E3)ko = —(P3P4 + P3P5)

 

’ 2(E3)ko c036 — 2(133 )k0 + M? —i1‘M:

271' ME — il‘M;

 

 

= 2(E3)ko l°g —4(E3)ko + N; — if)”;

—2
I3

ME — iI‘M:

 

7r
‘ P3P4 + P3P5 l°g (2(P3P4)

+ 2(P3P5) + M? — 1T1”:

)

 

136

The I4 - 16 Class of Integrals

These integrals are of the form:
I = (19;.
tux...)

 

In a co-ordiuate system where. the left denominator defines the z direction and
the right denominator is in the x-z plane, they assume the general form:
I _ / c191.
_ (c c050 + d) (e c039 + f sine cosqS + g)

 

then using

 

/“ dd, — 27 a > b > c
o a+bcos¢$ — (a1’—b'-’)1/2

 

1 d1 ,
[=2 ' '7 ‘I '1 '1 '7
"/1(cx+d)((e-'+f-):c-+2gew+g-—f-)1/-

Then transforming a: = C: + d

6-” dx
c+d 1' ("T-1" + m: + p) -

where
m = c: + f2
n. = 2(gec — dm)

0,

p : dam — dedc+ c2(g2 — f’)

which is solved using

/ da: __—_110 2x/p.\'+nz+2p
xXl/2 - J13 g 1?

X=m32+nz+p p > 0

In general these solutions are intracable. however in the special cases involved
in these integrals- d = it + A13, the solutions become:

first for d = —c + M2

2% [log ((9+\/€2+f2)2)+log(!I-\/€2+f2>_10g(’2(')]

=c‘(g+€) (9+9)? g+ €'-’+f2 M2

 

 

 

 

 

137

and for the other case (1 = +c + M?

2" (9-6)2 g+\’/e2+f2 (1‘12)
I:—-— l —— l —————— _
C(g—e) [0g ((g+‘/ez+fz)2) + 0g (9- fe3+f2) log 2c J

 

Now to use this general solution to find I4, 15 and IS.
I4 _ / - ko
’ (-2P4 - k + A13) (2131-1.- — 2(P1P4 + P1P5)+ Mfum)

_ in.
— / (-2lco'-’cos€ + 21:02 + A13)(2P1;(ko)cos6' + 2P1,(ko)sin.6 cosq‘) + 2(lco)El + Affuarl.)

 

then using P1; = P—gfz = %E3 + E1 = Pl—Péfmmi from 2(ko)E1 =
-—(P1P4+P1P5)

and P1, =(E1— P13)"2 = WNW
and noting that .,
c = -2ko'
d = 21w2 + M: = —-c+ A13
6 = P1P4 — P1P5
f = 2 (P1P4)(P1P5)
g = —(P1P4+ P1P5) + Mfwk
e2 + f2 = (2(ko)E1)2 = (P1P4 + 10119.5)2
9 + e = -2(P1P5) + Mia“

.gives

 

—2« (P1P4 + 13mm?) Mimi. < —2(P4P5))
= ———-—— —— 1 —— — lo —q—
I4 2(P4P5)P1P5 l°g( 1311352 + °g (—2(P1P4+P1P5) g M;

 

—1 P1P4+ P1P5 , ]
: _..___ __—_. __ P1P4 P1P? 12 0.? P4135 11
(P4P5)P1P5 [rl0g( PIP-5 )+0)( + )) + )( l

 

138

aim.
(—2P4 - k + 111.?) (2P3 - k + 1113 — mm.)

15:

_ ' d9;-
- / (i—2k03c059 + 21:03 + Mf)(2P3Ako)cos€ + 2P3£(ko)sin.0 cosqS - 2(ko)E'3 + ME - EAL.)

fin

 

 

then using P3; = if? = 323% + E3 = (”Rafa”) from 2(ko)E3 =
-(P3P4 + P3P5)

 

and P3, = (E33 - 1033)”2 = \/(P3P4)(P3P5)/ko
and noting that
c. = —2k0"2
d = 21:02 + Mj" = -c + M?
e = P3P4 — P3P5
f = 2\/(P3P4)(P3P5)
g = (P3P4 + P3P5) + M} — 271w:
e3 + 1" = (2(ko)E3)3 = (P3P4 + P3P5)2
g + c = 2(P3P4) + M} — iI‘M;

 

gives, using the symbol M Z P for the Z particle proprogator

MZP = 2(P3P4) + ME — {I‘ll-I:
r 2 2 _ f . _¢ ‘. r
21r [I ((2(P3P-))+MZP) )+lo < M: :I‘M, >_ (108 2(P1P)))]

 

 

 

 

 

 

I5 :
(P4P5)}UZP IUZP2 2(P3P5l-i- JUZP AL?
r
: (P4P51)A'IZP [27f log (2(P3133)Z+PAIZP) — (P3P4 + P3P5)I3 + (P4P5)Il]
16 _ (19;.
—/ (2P1- k - 2(P1P4 + P1P5l+ .l/qguarkM'Zpli - l’ + .lf? - ITAL)

(IQ;-
- ./ (2(ko)Elcosfl + 2(ko)El + fiffuark )(2P3:(ko)c059 + 2P3,(ko)sinfi coso - 2(ko)£‘3 + M? — ITAL)

 

139

their using P3: = P—ZIP—T : 12,-’13?- + E3 \
t
w' 1 P3P4 + P3P5 P1P4 + P1P5
E3 = —* 171 = —_
2k0 2kg

to give
-2k0(P1P3+ (P1P4+P1PR(OI:3P4+P3P5))

P3” _ P1P4+ P1P5
with P3; = (E33 — P3§)1/'~’

and noting that
c : —(P1P4 + P1P5)

d = c + MEWL.
2(P1P3)(P4P5) — (P1P4+ P1P5)(P3P4+ P3P5)
e: (PlP4+P1P5)
f = 2(lco)P3._t
g = —(P3P4 + P3P5)+Mf—1TM;
e? + f3 = (2(1-10)1§3)2 = (P3P4 + P3P5)2

2(P1P3)(P4P5) R
—. — l' I _ =
g e ' 203%”sz (P1P4+P1P5) (P1P4+P1P5)

where

 

R = (2(P3P5) + MZP)(P1P4+ P1P5)— 2(P1P3)(P4P5)

using as usual,the symbol AIZP for the Z particle proprogator

MZP = 2(P3P4) + ME — 11w;

 

 

16 —27r R2 > 1 ( ME — I'm/1,. ) log Mimi
: —— _ O ——_____ _ .—
R l°g ((P1P4 + P1P5)2(2(P3P5) + MZP)2 g 2(P3P5) + MZP 2c
—1 R
= — , + P3P4 + P3P5)I3 — (P1P4 + P1P5)12]
R [4T log <(P1P4 + P1P5)(2(P3P5) + 112m) (

140

The 17 - I9 Class of Integrals

These integrals are of the form:

I7 — 19 = / __dQ“(A ‘ "’
(...)(...)

and in each case the solution will involve already solved integrals of the denom-

inator only:
d9 .

d9.
Iright:IR=/.( 1)

dQ.
. Iboth=IB=/(H)(L)

 

 

 

The integrals of this class assume the general form:

I __ / ko (AA/coking cosd) + Ay(ko)sin0 sinq’i + .-l:(ko)cos€ — .-lo(ko))
— (c cosl9+d)(€ c059+fsin€ cos¢+g)

using the previously defined co-ordinate system of class 14 -IG.

Each term of the numerator may be solved separately. In an obvious notation;

 

Iy equals zero by d> symmetry. and by inspection:

10 = —Ao(k0)IB

I, is easily reduced:

A,lco f sinfi c0345 + 6 c050 + g — e c050 — g
f (ccosfl+d)(ecos€+fs1'n€ coso+{/)

 

 

1,:

I fir/co I I 6 (c (0304-4—11) ]
I — f L -g B — r/ (r 00594-100 (050+fsinficmo+g)
Arko

ed
: f [IL—(g—Tlla-EIR]

 

 

 

141

and the final term

 

 

 

I _A_.ko/ ccosfl+d—d
.- - f (c cosfi + d)(e 6039 + f 51711.9 cosd) + g)
:Lko
: :f [In —dIB]

Combining the four terms:

k0 .4,er

(IR —dIB)+

 

 

A. ed 6
I=—Ao(k0)IB-i- h [IL—(g—T)IB';IR]

C

Now to use this general solution to find 17,18 and 19.
I7 _ / (19;. P3 - k
‘ (—2P4 1+ 11,2)(2P1-k — 2(P1P4 + P1P5)+ Min“)

ko(P3,.(ko)sin9 cos¢ + P3y(ko)sin9 sinzb + P3:(ko)c050 - E3(ko))

I7 = (—2Ico'-’cos€ + 2k02 + A[$)(2P1.._(ko)cos€ + 2P1,(ko)sin€ (03¢ + 2(ko)El +1lf‘fuar‘.)

. _ fi-fi _ P1P4 _ P1P4-P1P5
then usmg P1; — E4 — £4 + £1 — 21-0

from 2(ko)E1= -(P1P4+ PIP-5)
and P1, = (1311’ — P13)”2 = ‘/(P1P4)(P1P5)/ko

and noting that
C = ~2k02

d = 21.03 + M; = —c + 113
e = P1P4 — P1P5
f = 2 (P1P4V)(P1P5)
g = —(P1P4 + P1P5) + Min“.
e? + f3 = (2(ko)E1)3 = (P1P4 + P1P5)2
g + e = —2(P1P.5) + 1‘15"”;-

and adding

—-‘.— : .P4—P3Pr
_P.5 P4_P3P4 + £3 =P3 )

P3“?- 134 210

 

142

and to get P3,

- 3 = P1P3 + E1E3 = Per35 + PlyP3y + P1,;P3;

El

with P1,, = 0
P3,; = 1/P1;[P1P3 + E1E3 - Pl;P3:]

V r F 1 _ 1' — . 3'
= ko/ (P1P4)(P1P5) [P1P3+ (P1P4+ P1P;::I:3P4+ P3130) _ (PIP—l PlP;/:E)I:3P4 P3PJ)

 

S
I 2(k0) (P1P4)(P1P5_)

where

S = P1P4(P3P5) + P1P5(P3P4) — P1P3(P4P5)

and using the general form gives:

P3P4 — P3P5 £3

— dI
21cc c (IR B)

17 = —ko(E3)IB +

(P1P4 -— PIP-'3)
—2/c02

(5)060)

____.___—.—.‘

1
+ ________———
2kot/(PIP4)(P1P5) 2 (PlP4)(PlP-5)

(P3P4 —- P3P5)
2(P4P5)

S (P1P4 — P1P5)
___———— 11 2 P1Pr I4 — ——————
+4(P1P4)P1P5[ + ( °) P4P5

[IL+2(PIP5)IB— In

I7 = (P3P4)I4+

12]

 

 

18 _ ko Pl -k
“ / (—2P4 - k +111$)(‘2P3 -1- +113 arm)

 

ko(Pl,(ko)sin6 c0565 + P1,,(Irolsinfisino + FLU-(1)0190 — EMA-0))
— / (—2I€02(‘O$6 + 21602 +1lI.;-’)('2P3_~(I\'o)c039 + 2P3;(k0)sin(9 (0.51;) — 2(kolE3 + .lf? -— il‘dL)

 

. "‘."-' , . P3P4-P3P?
then usnig P3; = BEE—5:! = P2? + E3 = 21:0 )

 

 

143

from 2(ko)E3 = —(P3P4 + P3P5)

and P3, = (1332 — P33)”2 = ix/(P3P4)(P3P5)/l:o
P1; = (P1P4 — P1P5)/2ko

Ira S

x/(P3P4)(P3P5)m§

1
P1, = F[P1P3 + E1E3 — P1:P3:] = i

and noting that
c = -2ko2

d = 210? + 113 = —c+ 113
e = P3P4 - P3P5
f = aha/W
g = (P3P4+ P3P5) + ME — iI‘M:
e2 + f3 = (2(ko)133)=’ = (P3P4 + P3P5)2
g + e = 2(P3P4) + 113 — iI‘M:

gives, using the symbol M Z P for the Z particle proprogator
MZP = 2(P3P4) + M? — iI‘M;

(P1P4 — PIP-5) [ca
2120 —2/co'-’

5' k0
+———————(IL-(
2k0y/(P3P4)(P3P5) 2V (P3P4)(P3P5)

18 = —ko('E1)IB + (IR — 210313)

 

(P3P4;P3P5)

MZP)IB— 21:02

13)

(Note the :t arbritariness became self-cancelling due to AffL—° form.)

_ r S
(P1P4 P1P) I“ [11 — (MZP)15 -

- K VT
18 ‘ (P1P4)I"+ 2(P4P5) 4(P3P4)P3P5

 

(P3P4 — P3P5)13
, P4P5

19 d§2k 134 'k
‘/ (2P1-k — 2(P1P4 + P1P5) + 1113

quar

“(217.3% + M} - ITAL)

 

144

ko(P4,(ko)sin0 cosd) + P4 (ko)s-in€ sinq’) + P4. (ko)cost9 — E4(ko))
(2(ko)E1cos€ + 2(ko)E1 + M2 k)(2P3 (ko)c099 + 2P3 (k0)sin.0 cos¢>- 2(k0)E3 +1lf?— iI‘Ms)

quar

 

 

 

their using P3; = 5%]? = % + E3
wit1
P3P4 + P3P5 P1P4 + P1P5
E3 = —— = __
21cc E1 2ko
to give
(P1P4+P1P5NP3P4+P3P5
P3: _ —2Ico(PlP3+ “.0, ’
“ P1P4+ P1P5
and
P31 = (1:32 — P33)”2
P1P
_ —[_ (——3) — 2(P1P3)(E1/E3)]1/-
11 1/2
= (:2)
where
A = P1P4 + P1P5
and

T = 2(P1P3)[PlP3(P4P5)—-P1P4(P3P4)—PIP4(P3P5)—P1P5(P3P4)-—P1P5(P3P5)]

P4 _P4 P1_P_1_P4+k0_ k0(P1P4—P1P5)
5 ’ —El “ E1 ‘ P1P4+P1P5)
P4, = 1/P3,[P4P3 + E4E3 — P4.P3 ]
P P
— _1/P3,[P3P4+ ko(E3) — (ID—2,1133 + E3)(—1—— + k0 )]

= (t) (:33).

U = P3P4(P1P5)2 + P3P4(P1P4)P1P5+ P4P5(PlP3)PlP-l

where

—(P1P4)2(P3P5) — P1P4(PIP5)P3P5 — PlP5(PlP3)P~lP5

and noting that
-(PIP4 + PlPS) : —-.~l

 

 

145

d = C + Affirm-k
ii"

6:

tel

where

W = 2(P1P3)(P4P5) — (P1P4 + P1P5)(P3P4 + P3P5)
1/2
f = 2(ko)P3, = 2ko (%)
g = (P3P4 + P3P5) + ME — :TM:
e2 + 12 = (2(ko)E3)2 = (P3P4 + P3P5)"’

2(P1P3)(P4P5) _ R

_ =2PP MZ — — '
g e (3 5)+ P (P1P4+P1P5_) (P1P4+P1P5)

 

where

R = (2(P3P5) + MZP)(P1P4 + P1P5) — 2(P1P3)(P4P5)

using as usual,the symbol )1»! Z P for the Z particle proprogator
MZP = 2(P3P4) + ME — 1PM,

—k P1P4— PIP? —l.‘.
Lil—”TOUR _ #1113)

A2 ”2 [so A2 ”2 U _ .2
+(?) (72%) (7F) (fi)(IL—(R/-~1)IB+(u/A)IR)

P4P5(PJP4—P1P5)
7T”

19 = --ko2IB +

19 = (P4P5/2)16 —

P4P5 (P1P4 — P1P5)
———_A—

U , 2
2 16+ fiuz— (R/.4)16+ (ii/A )13)

 

 

 

146

The 110 Class Of Integral

It is desired to solve 110.

I10 —/ m"
' (—2P4 .1: + 11$)(2P1-k — 2(P1P4 + P1P5) + 111; ”)(2P3 - k + 1113-11‘114)

u

 

This is effected by noting from Appendix L:

ku = (51 Mm“ + (S2 - k)f2,, + (53 1173,, + ($4 - 1V4“

where
$1,, = (CC11)Sl,J + (CC12)S2u + (CC13)S3p + (CC14)S4u
$2,, = (CC21)51,. +(CC22)52“+(CC23)S3,1+(CC'24)S4;,
73,, = (CC31)51,‘ + (CC32)52,‘ + (C'C'33).S'3p + (C-‘C'34)S4u
f4“ = (CC-‘41)Sl,, + (CC-‘42)S2u + (CC43)53,‘ + (C'C44)S4,‘

where CC 11 ....... CC44 are given in that Appendix as assemblages of dot prod-
ucts.

In this problem-

Sl = P1
52 = P3
53 = P4
54 = P5
so...
k” = (P1 - k).7:1,, + (P3 - Hf?” + (P4 . k).'F'3,i + (P5 ~ k)f4,,
where

flu = (CC11)P1,, + (C7C12)P3,. + ((‘C'13)P4,. + mamas”
T211 = (CC21)P1,. + (002-2)qu + (('('-23)P4,, + (("('2.1)Pr,,,
$3,, = (CC-'31)P1,. + (C'C'32)P3,, + (51-'33,}:4“ + ((1734) P5"
$4,, =(CC41)P1u + (CC-'42)P3,, + (c.'(743)p4u + (CC44)P5“

 

 

 

147

F uther, in this problem. the on-shell condition

6[(P4+ P5 — k)]2 = 0

implies
P5-k: P4P5—P4-k

giving
k,J =(P1~kg)f1u + (P3 - k).’F'2,, + (P4 - k)(.7-'3,, — fi1#)+(P4P5).7-"4,,

Use of this is made by noting the other on-shell condition of this problem-
6(k2) = 0
0 = H‘ku =
f1,,fl“(P1'k)(P1-k)+ 72”.???”(P3 - k)(P3 - k)
+(1‘3u — $4..)(f31‘ — f4“)(P4 - 1:)(P4- k)+ f4..1—”4"(P4P5)2
+2?1,,.7-‘2“(P1 - k)(P3 - k) + 2fl,1(.7-'3“ — f4”)(Pl - k)(P4 - k)
+2.7:lpf4“(Pl - k)(P4P5) + 2f2“(.’F3” — f4")(P3 - 1:)(P4 - 1:.)
+2?2,,.7-‘4“(P3 - Ic)(P4P-5) + 2(J-"3p — .774“).7-'4“(P4P5)(P4 ~ 1:)

As 1:3 = 0. any integral of 1:2 will also equal 0. In particular the integral of k2
over the above 110 will be zero.
0 _ / dm we“
‘ (—2P4 - k + 113)(2P1-Ic — 2(P1P4 + P1P5) + 1115“,. )(2P3 - k + 113 — inn.)

When the above expression for k.“ k” is inserted into this integral. it produces a
sum of integrals which still equals 0.
For example. one such term would be:
1191 flufl“(P1 1k)(Pl -1~)
/(—2P4 - k + M.;-,’)(2Pl - k — 2(P1P4 + P1P5l+ 1115M”. )(2P3 - k + 1113—1'I‘11L)

This integral. except for 71,..7-‘1". is defined in Appendix M to have the notation
A2(1.1). The other integrals appearing in this sum haw.- a similar notation
defined in that Appendix. Using this notation. the expression for 0 becomes:

0 = f1,,f1"A2(1.1)+ f2,.f2"A2(2. 2) + (73,, - 74,. )(f3" — f4“)A2(3. 3)

 

 

148

+f4uf4"(P4P5)2110 + vignmzm, 1) + 2f1,.(.7:3” — .7-‘4“)A2(3. 1)
+2?1,..7~'4“A1(_1)(P4P5) + 22:2),(1314 — f4”)A2(3. 2)
-+2f2,,$4"A1(2)(P4P5) + 2(13). — $4,.)f4"(P4P5)A1(3)

In Appendix M. it is found that
A2(3,3) = —1/2 19
A2(3,2) = —1/2 [1/2 (12 — (ME —1rM.)16)]
A2(3, 1) = ~1/4 [13 + 2(P1P4 + P1P5)IG]
A2(2. 2) = 1/2 [17 — (111.2 — irM.)A1(2)]
A2(2, 1) = 1/2 [1/2 (11 + 2(P1P4 + P1P5)I4)—(1\13—irM,.)A1(1)]
A2(1. 1) = [I8 + 2(P1P4 + P1P5)A1(1)]
141(3) = -1/216
A1(2) = 1/2 [14 - (ME - if)”: )110]
A1(1) = 1/2 {15 + 2(P1P4 + P1P5)110]

After these are substituted in the expression for 0. that expression becomes
an algebraic expression [or 110, which may be solved to yield:

_ Iloupper

110 =
Ilolower

where

110m," = fl‘,fl“(l/2(IS) + 1/2(P1P4 + P1P5)15
+f2,,f2#(1/2(17) — 1/4(Mf — irM. )14)

—(f3,. — 74,.)(131‘ - f4")(1/2)19
+f'1,,f2“(1/2(11)+ (P1P4 + P1P5)I4) — 1/8(Mf — if)”; )15)
—f1,.(J-'3” - f4”)(1/2(I3) + (P1P4 + P1P5)16) + f1,,.7~'4“(P4P5)15
—J-'2,,(}'3“ — f4“)(l /2(12) — (1113 — irM. )16)
+f2,,f4“(P4P5)I4 — (73,. — P4,, )f4”(P4P5 )16

and
11mm, = .7-‘1,,.7-'1“((P1P4 + P1P5)3) + f2,,f2“(1/4)(M§’ — iI‘AI; )1’
+f4uf4”(P4P5)2 + .7-‘1,.f2“(—1/4(P1P4 + P1P5)(.-\!.'-’—i1‘.-U:))
+f1,.}'4“(P4P-5)(2(P1P4 + P1P5))+ f2,,}'4" )1—(1/3 — 112)]. ))

As 11......19 are now known. 110 is now also known.

 

149

The Ill - 116 Class of Integrals

These integrals are of the form:

111 —116 = / ———d9”((‘4)' 1“)

and in each case the solution is reducable to integrals already solved of the

denominator only:
dflk

I = —-
D (...)
The integrals of this class assume the general form:
I: /d§2k (Az (ko)sin0 cosd>+ A y(ko)sin0 sin¢+ A: (ko)cos€ — Ao(l'o))
(c c050 + (1)

using the previously defined co-ordinate system of class 14 -16.

Each term of the numerator may be solved separately. In an obvious notation;

I, and Iy equal zero by (1) symmetry. and by inspection:

10 = —Ao(k0)ID

I: is easily reduced:
I _A:ko cc059+d-—d

‘ c (cc059+d)

 

so.

k0 [47f — dID}

 

A
I = -Ao(k0)ID +

New to use this general solution to find 111 - I16.

11 de11
1 ‘/(‘1P4l+\!)

er(Pl, (ko)sin9 c056") + P1 y-(lo)sin() sino + P1 (Lo)cosfl - 151(k))
111:] (— —2ko- cosfi+2ko~+ AL? )

 

 

       

then using P1; = FE? = P115? = BAP—é'gf—“fi
from 2(ko)E1= --(P1P4 + P1P5)
and noting that

c = —2k02

d = 2102 + M3 = —c + 1113

P1P4+P1P5 P1P4-P1P5 P1P4—P1P5
111: ————— ____—— _—
( 2 )H ( 2(P4P5) )4”+( 2 )“

P1P4— P1P5

P4P5 )+ (P1P4)Il

Ill:21r(

and I 12 is the indentical with P1 replaced by P3

(1521. P3-k
112 ’ / (—2(P4 - k) + 1113.)

P3P4 - P3P5

P4P5 )+(P3P4)Il

112 = 27r<

113 (19,. P4 . k
=/(2(P1 k)— 2)(P1P4+P1P5)+quua,k)

113 _ (191 P4 - k
’ (2(ko)E1c059 + 2(ko)El + Mia“)
c = —(P1P4 + P1P5)
d=C+AIg2uark :C+O=C

’74' ”PT _PlP4+ko__Lo(P1P-1—P1P5)
E1 ’ (PIP-1+P1P5)

 

P4;=

2 (P1P4 P1P5))ko( —4, _ 12)
11:1 = 'k" 12 + (—ko(PlP4 + P1P5) (P1P4 + PIP-‘3)

 

 

151

= —21r(P4P5‘ (P1P4 — PIP-'3) ((P4P5)(P1P5)>
(

’(P1P4+P1P.5)'-' P1P4+P1P5)

 

 

 

114 z] (19;, P3 - k
. (2(P1-l-.)—2(P1P4+P1P5)+nguark)
I14 = /’ d011‘P3 - k .,
(2(ko)Elcos€ + 2(ko)E1+ Mq'uark)

c: —(P1P4+ P1P5)
d = C + ALiner-k. = C
P1P4 P1P5 P3P4 P3P5
P3_ = —2ko(P1P3+‘ + “1‘0, + ’
‘ . P1P4+P1P5
_ - [—2(P1P3)(P4P5) + (P1P4 + P1P5)(P3P4 + P3P5)]
‘ 2ko(PlP4+ P1P.5)

,
114 = (w) 12

+ —2(P1P3)(P4P5)+(P1P4+P1P5)(P3P4+P3P5) 411- +12
2(P1P4+P1P5) P1P4+P1P5

 

 

 

 

 

 

115 _/ (1121. P4 .1.
‘ (2(P3 - k) + 1113 — 11M.)

d9;c P4-k

115 = (2E3(ko)cos€ — 2(101133 + 1113 —1tr111..)

 

and noting that
c = -(P3P4 + P3P5)
d = (P3P4 + P3P5) + M} — mu.
E4 = 1:0
flfi P1P4 (P1P4—PJP5)

P4: = 1:1 = 731‘ + k” = "k" PIP-1+ PIP-5)

(P3P4 - P3P5)
(P3P4 + P3P5)?

 

115 = —Ico"’I3+ (4.0 ) (—lco) (41 — [P3P4 + P3P5 +1113— 1111.] 13)

152

_ (P4P5)
‘ 2(P1P4 + P1P5)'-’

 

(19;; P1 - k
116 =
/(2(P3-k)+1lff—1T1U;)

I16 _/ (19;; P1 - Ic
_ (2E3(k0)cos€ — 2(ko)E3 + 1M} — iI‘AL.)

 

and noting that
c = _(P3P4 + P3P5)
d = (P3P4 + P3P5)+1’t[3—i1‘1\=[;

_ , r
E1: (P1P4+ P1P1)

ko

P_3-fl
PL: 1
. E3 +E

(P1P4+P1P5)(P3P4+P3P5)
_ —2k0(P1P3 + 1112
P3P4 + P3P5

_ — [—2(P1P3)(P4P5) + (P1P4 + P1P5)(P3P4 + P3P5)]
‘ 2ko(P3P4 + P3P5)

 

 

 

 

gives

I3

P1P4 PlPF
(<__;__2)

 

__ [2(P1P3)(P4P5) — (P1P4 + P1P5)(P3P4 + P3P5)]
2(P1P4+ PIP-5)

) (—47r(P3P4 - P3P5) + I3 [4(P3P4)(P3P5) + P3P4 — P3P5)] MZP)

 

> (4,1 — [P3P4 + P3P5 + M} - 171311;] 13)

153

The I17 - I19 Class of Integrals

These integrals are of the form:

1191, (A ~ 1‘.)2

117 — 119 =

Most of the solution is expressable in terms of previously used I L,I R. and I B
and integrals also already solved involving just 911; of the numerator factors and
one or both of the denominators:

d9 .4 1k
ITopLejt = ITL =/_:(—):‘l

d9: A - k
ITOPRight = 171-? = / —-:—%—)——)

11:21. (.4-1-.)

ITopBO‘h = [TB =/ (...)(---)

The solution is effected by expandingjust one of the numerator factors into four
terms:

I _ d9; (A,(ko)s1'.n0 cosd> + Ay(ko)s1'110 Sin-d5 + A:(Ico)cosfl - ADM-0)) (A - k)
_/ (c c056 + d)(e c050 + f sine cosd: + g)

using the previously defined co-ordinate system of class I4 - I6. producing the
four terms, in an obvious notation:

First solve Io, by inspection:

Io = —Ao(k0)ITB

and then I ,

 

 

.4-k
I: = if 0 [1111 - dITBl
and 1: fill-1‘0 ((1 (
Ir: f ITL-(fl—TNTB";]TR

These were done in the usual way of adding and subtracting the terms required

 

 

154

to produce cancellation of denominator factors.

1,, is harder. it is approached by also expanding the other numerator factor:

I _ / d0); (Ay(lco)sin0 sindi) (.4,(ko)sin0 cosd: + .4y(ko)si11.9 51nd: + A:(ko)cosfl - 140(130) )
y — (c c056 + d)(e c039 + f sine c0542 + g)

Three of the terms are zero by symmetry. The remaining term:

ko (sin-'39 sinqu)
(c c056 + d)(e c050 + f sine cosq') + g)

 

n=Aflf

Aglw2 dflk (f2 — fzcoszfi — fzsinzfl 51.11.2113)
f2 (c c059 + d)(e c059 + f sin0 cosgb + g)

 

19:

which three terms may be called:

By inspection:
Iyl = 113110313

 

and
I -A3k03 ko (czcosgfi + 2cd c050 + d2 — 2cd 0050 — (I?)
y: — c3 / (c c050 + d)(e c059 + f sing cosqS + g)
—A§ko2 ._, 3 d IC'
Iyz = c2 {—2113 + 2d 13 — d 13 + .111 + c( 1]
_Azk 2 9
I92 = _zv—° [dIR — d'IB — c(IC-')]
where

IC _ d9]; C059
_ / (e c050 + f sing cos¢ + g)

and remains to be evaluated.

Lastly, ., - ., ’
Ailco2 (IS-2;. (—fzsin'0.s1n-o)

I93 = f“-3 (c (030 + d)(c c050 + f sing cnso + .11)

Then. using

I2 sinzflsi112¢ = [(6 cost? + g) + (f sin.9coso)]2—2(e cosfl+g)(f sinflcosq))—(e cosfl+g)2

155

= [(e c039 + g) + (f sin9c05¢)]2—2(e cosfi+g) [(6 c050 + g) + (f si11.0cos¢)]+(c c056+g)2
: [(e c050 + g) + (f sin10c05¢)][—(e c050 + g) + (f sinflcosm] + (e c030 + g)2

Aiko" / ko(e c030 +g) _ / dfli. (f sinflcosd) ) _ (191. (ezcosgfi + 296 c050 +92)
f2 (c c089 + d) (c c050 + d) (c c050 + d)(e c056 + f 51119 cos¢ + g)

 

Iya =

The second term is zero.

The remaining become:

 

ko (c0529) ]

—e ed 2ge 3 2
[74“ + 71L — gIL + 7”}? _ dIB) + g 18 + 6 (c c059 + d)(e c056 + f sine cosqS + 9)

Then. using the results of [ya on the cos? integral:

 

-A§k02 —e ed
I93 =

f3 —47T-(9-—)IL +—(I11-d13)+o IB-—(d11=z dq'IB—CUC'))

and the complete integral for this class of integrls:

I=Io+Iz+Ir+Iy1+Iy2+Iy3

 

14.1:
117 — I19=—Ao(ko)I1-3+ °[Im — dIral+

 

.1-4 L d
0 [In - (9 — 'e—lITB - SITE]
f c c

 

11312
+ 431.0 13 + 02[d1,1 — d- 13 — cum]

 

‘ ; [_—477-(9-—)1L+—(IR-dfs)+9213-:—:(dIR- dqls-CUC') )]

It remains to evaluate IC.
(19;. cost?
IC = / .
(e c050 + f 51119 c0547 + 9)
Using
21 dd) _ 271'
o a + bcosaS — ((13 — 1,2)1/2

2 1 1rd:
1C.— 7/;((53+f2)13+29€r+92—f2l1/'

a>b>c

 

 

 

156

1
IC = 21r/ +dx—
_1(m1:-+n.1:+p)1/'-’

where
.,

1n=62+f'
112296
19:112—1‘2

which is solved using

1 -
1. .1_(1)(;_m).,..1

 

1 X”? W 2112
Noting
X1+1> = (9+ «:12
-\'(-1) = (y - 6)."
and

(g+€)\/€"’ +f3+:2(c2+f'-’)+2(.<1c)= (9+ \/e?+f'-')(e+\/e?+f?)
(g-e)\/63+f'~’—2(62+F)+2(yc) (g—¢e2+12)(e+\/e?+13)
gives
Ic=L 2e__z:__,og 2+__ x/e'+f
(12+P) W g—W

The log is the same as appears in the II - 13 class integrals and will be express-
able in terms of them.

This class of integrals thus has solution:

 

”,4 k0 6d
I17-119: —Ao(k0)IrB+ACko [Ira - d1r8l+ —f_ 0[ITL - (a - —)ITB - 4711]

 

4‘43“? —e(f'-’+e'-’) ) _fi (21_.d_'_,_£‘.’2>1 N]
-r f2 [( C(fz-l-62) +C/C 47l'+(.<7 C)IL+ fl (:1) (.l C). 3+

where

: N(d —2(f‘ +e e2)- 2g__e>IR+ (j—_3 +(-') 371' __g_-( fglog(_r/+1/c-+f')

c (f—-+_e'-’)./2 g_‘/,.'.'_*_f2

157

 

 

 

AJ: A,k d
117 - 119 = -Ao(k0)Ira+ c 0 [Ira - d1r13]+ f o [In — (y - 517”” - EITR]
Azkog
+ ”1
f-

ed cd 2 d 2 2 27rgt 9+ “(2+)”
[(5] — :) IL — (g — 7) I3 +1/c ([;(f + e )— 2ge] In + ——\/€2+—f2 log (g———_ W))l

158

Now to use this general solution to find 117, 118 and 119.

117—] 1112111131):
“~ (—2P4 . k + Mjf)(2P3 . 11+ 1113 — ier)

dQL.(P3,(ko)sin9 cos¢ + P3y(ko)si11.9 5111.115 + P3;(ko)cos(7 - E3(Ico))(P3 - k)

 

117 =

 

(—2k02c050 + 211:02 + Mf)(2Pl;(ko)cosfi + 2P1,(ko)sin.0 cosd: + 2(ko)E1 + Mimk)
IL=II 13:12 IB=I4 ITL=IIZ [73:11.4 ITB=I7

The parameters are the same as for the I7 solution.

—4 _ P1P4 __ P1P4-P1P5
- £4 + E1 " 21:0

 

using P1; =
from 2(Ico)E1= —(PlP4 + P1135)
and P1, =(E1— P13)”2 = :t‘/(P1P4)(P1P5)/ko

and noting that ,,
c = —2ko'
d = 2ko2 + 111.3 2 —C+ AL?
6 = P1P4 — P1P.5
f = 12W
g = —(P1P4 + P1P5) + 1115.111
62 + f2 = (2(10)E1)2 = (P1P4 + P1P.5)2
g + e = —2(P1P5) + 111‘wa
ge = - [P1P42 — P1P52]

 

and adding
2
_2” .-1 L_ W _ 2" 10g 11.1.1.1 = 12
./—ez + f2 °5 9 _ C2 + f2 ” (P1P4 + P1P5) -2(P1P4+P1P5)
fl - PE P3P4 P3P4 — P3P5
- = _— = _— E3 = ———.—.——
P3‘ E4 E4 + 21-0

and to get P3,-
_1-_3 = P1P3 + EIE3 = P1,P3, + P1yP3y + P1:P3:

 

159

which, since P1,, = 0
P3, = 1/P1x[PlP3+ E1E3 - P1,,P33]

(P1P4 + P1P5)(P3P4 + P3P5) (P1P4 - P1P5)(P3P4 - P3P5)
411202 W 41902

 

 

= iko/\/(P1P4)(P1P5) [P1P3 +

___ 1:5
2(ko)\/(P1P4)(P1P5)

 

 

where

S : P1P4(P3P5) + P1P5(P3P4) — P1P3(P4P5)

Now in addition, is needed P33:
q

P33 ..—. E32 - P3; - P3;

_ 4(P3P4)P3P5(P1P4)P1P5 — S2
’ 4k02(P1P4)P1P5

and using the general form gives:

 

 

 

 

 

_ P3P4 — P3P5 _A P3P4 — P3P5 _
117 _ —ko(E3)I1 + 14—21-02 (114) + 2 (11)
(5)0”) 1 [ — P1P4 - {1})”(114) + 2(P1P5)17]
2ko\/(P1P4)(P1P5) 2\/(P1P4)(P1P5) (—2ko~

4(P3P4)P3P5(P1P4)P1P5 — 52]
4(P1P4)(P1P5)4(P1P4)P1P5

 

{—2(P1P5)Il — 4(P1P5)2I4 + 21mg (—(P1P4 + P1P5)'-’ + P1P42 — P1P52) 12]

where the :1: arbritariness has cancelled out.

 

-. P3P4 - P3P5 (.S') P1P4 — P1Pr1 _]
= . 112— I14+2 P1P? I
1” (PBP‘HH‘L 2(P4P5 (114)+4(P1P4)(P1P5) [ (P4P5 ( ’l ‘

 

~ F —:'~’ ., P1 4 P1Pr
[4(P3P4lp3p5(P,lP4)Plf’ S J [—2(P1P5)Il — 4(P1P5)-I4— (( P + r ’l) (2P1P5)12]
16(P1P4)-(P1P5)- P4P.1

 

 

160

d9): (Pl-k)",

118 =
(—2P4 - k + 1H,?)(2P3 . k + AIS—17111;)

 

_/ dQL.(Pl,(Ico)sin0 cosd) + Plg(ko)sin€ sinqS + Pl;(ko)cosfl — E1(ko)) (P1 ~13)
— (—2k03cos(9 + 21m2 + A!$)(2P3;(ko)cosfl + 2P3a.(ko)sin9 cosc’: — 2(lco)E3 + 1U} —iI‘1\/;)

IL=Il IR=I3 13:15 ITL=111 ITR=116 [73:18

The parameters are the same as [or the 18 solution.

- _ fi-P—4 _ P3P4 _ P3P4—P3P5
Usmg P3; — E4 — —E4 + E3 — ————._,ko

 

from 2(ko)E3 = —(P3P4 + P3P5)

and P3, = (E33 — P33)”2 = i\/(P3P4)(P3P:3)/ko
P1; = (P1P4 — P1P5)/2ko

 

iko S
P1, :1 P3, P1P3 E1E3 — P1..P3.. = —— q
/ [ + ] x/(P3P4)(P3P5) 2ko-

and noting that 0
c : —2ko’

d = 21:02 + M: = —c+ M:
e = P3P4 — P3P5
f = am

= (P3P4 + P3P5) + M3 — iI‘M:

= —(P3P4 — P3P5) + MZP

e2 + f3 = (2(ko)133)=' = (P3P4 + P3P5)"’
g + c = 2(P3P4)+J!f’—i1'.\l:
g6 = (P3P4 — P3P5) [—-(P3P4 + P3P5) + MZP]

 

161
using. as usual, the symbol M Z P for the Z particle proprogator
MZP = 2(P3P4) + M} — 171‘111z

and -now adding

271' 0g (9 + x/e- + j”) _ 27r )log (2(P3P4)+ MZP) = 13

 

 

_1 _
,F—ez+ f2 g_,/—e2+ f3 (P3P4+P3P5 M3 41w:
fl - P—4 P1P4 P1P4 — P1P5
P1” E4 - ‘54—”1 "—2?—

Now in addition, is needed P13:
P1; = E12 — P13r — P13

_ 4(P3P4)P3P5(P1P4)P1P5— S?
‘ 4k03(P3P4)P3P5

and using the general form gives:
P1P4— P1P5 P1P4—P1P5

 

 

 

118 = —ko(E1)18 + 2(_2k02 (116)+ —-——2—(18)
—————(S)(k°) —————l [111— ————-——P'5P4 _ P,,3P'J(116)— (P3P4 —- P3P5)I8]
2ko‘/(P3P4)(P3P5) 2 (P3P4)(P3P5) (—2ko-

4(P3P4)P3P5(P1P4)P1Pr — 52]
16(P3P4)=(P3P5)=
[(MZP)II — (MZP)2I5 — 22:2 (—(P3P4 + P3P5): — (P3P4 — P3P5) [—(P3P4 — P3P5) + MZPD

where the d: arbritariness has cancelled out.

P1P4 — P1P$ r (5) [ P3P4 — P3P5 J
= 1 0+ 1 — ——nc — MZP 18
118 (P “D “18* 2(P4P5) 1 4(P3P4)(P3P5) (P1P5) ( )

4(P3P4)P3P-5(PlP-1)PlP-r — 52]
16(P3P4)'-’(P3P-5)'-’

162

l
P4P5

 

[(Msz — (MZP)"’15 — {(P3P4 + P3P5)2(P3P5) + (P3P4 — P3P5)A~IZP}IS]

 

119=/( ko(P4-k)3

2P1 - k — 2(P1P4 + P1P5)+1L[:uark)(2P3-k + M; 41311:)

_/ th(P4x(k0)sin6 cosqS + P4y(k0)sin0 sings + P4;(ko)c059 - E4(ko)) (P4 - k)
' (2(ko)E1cos6 + 21101131 + M; M )(2P3;(ko)c050 + 2P3,(1o)sma cos¢ — 2(ko)E3 +1113 — 1mm

U

ILZIZ [12:13 IB=IG ITLZII3 ITRZII5 ITB=19

The parameters are the same as for the 19 solution.

 

 

 

 

 

 

Using P3z = 7;? = % + E3
with
E3 _ P3P4+ P3P5 E1 _ P1P4+ P1P5)
_ 21cc _ 21:0
to give
P1P4+P1P5 P3P4+P3P5)
P3- = —2ko(PlP3 + t 4503
“ P1P4+ P1P5
and
P31 = (E32 — P313)”2
1/2
P1P3 2
= _ — — 2(P1P3)(E1/E3)
E1
- :1: l 1/'_’
_ A2
where
A = P1P4 + P1P5
and

T = 2(P1P3)[P1P3(P4P5)—Pl P4(P3P4)—P1P4(P3P5)—PlP-5(P3P-l)—P1P5( [33135)]

163

 

Eiffi P1P4 (P1P4— P1P5)
P4. = = ——- 1 = — . —-—-—————
* E1 E1 “0 [‘0 P1P4+P1P5)
P4, = 1/P3,[P4P3 + E4E3 - P4CP3...]
= 1/P3:[P3P4 + ko(E3) — (5%? + E3)(%fi + k0)]

A" 1’3 U
“*(E) (:13)

U = I-"3P4(P1P.’5)2 + P3P4(P1P4)P1P5 + P4P5(P1P3)P1P4

where

—(P1P4)2(P3P5) — P1P4(P1P5)P3PF — P1P5(P1P3)P4P5

and noting that '
c = —(P1P4 + PIP-5) = A

d=c+Mfuark =c+0=c

..q
<4

6:

i=~|

where
W = 2(P1P3)(P4P5)—(1P1P4 + P1P5)(P3P4 + P3P5)(P1P4 + P1P5)
T 1/2
f = 2(lco)P3, = $21.30 (71—2)
9 = (P3P4 + P3P5) + MS — mu:
e2 + f2 = (2(Ico)E3)2 = (P3P4 + P3P5)2
ge = W/A {—(P3P4 + P3P5) + ME — mm]

2(P1P3)(P4P5) _ R-
g — e = 2(P3P5)+ MZP - (P1P4+ 121105) ‘ (P1P4+ P1P5)

where
R. = (2(P3P5) + MZP)(P1P4 + P1P5)— 2(P1P3)(P4P5)
usingras usual. the symbol M Z P for the Z particle proprogator

MZP = 2(P3P4) + M? — iI‘M:

 

and adding

21: 0 9+ M; +72 _ 2,1 10 (2(P3P4)+MZP) _13
,/’—_e2 + f2 g g _ F2 + f2 ’ (P3P4+P3P5) ‘5 1113 — 11111; ‘
and now in addition, is needed Ply:

P43 = E42 — P43 — P43

2 _ k02(P1P4— PIP-‘3)"z _ U:

= k
0 A2 A31“
_ 2(kog)(P1P4)(P1P5) _ U2
’ A? .421

and using the general form gives:

k03(P1P4 — PIPS)
A3

<—> <1) <1) <> (43+

 

I15+

. 2 _ .r
119: (—ko=)19+( (————————L" (P1P: P1P”) 19

TV

”2) 11.5 + (1—3) 19 — (P3P5 —- P3P4 + 1112mm)

, 2 '1 IV I
[(S) 12 — (E) 16 —14 ([P3P4+ P3P-Sl' - L4- l—(P3P4 - P3P5) + NZ”) 13]

where the :t arbritariness has cancelled out.

 

119 = ((P4P5)(P1P5))19 _ ((P4P5)(P1P:1— P1P5))115
A 2A-
+-U— (113 + (3)115 + (33) 19 — (P3P5 — P3P4 + 112mm)
2T A2 A
—P4P5(P1P4)P1P5 (73
+l 2T — T179]

2 " ‘1. n r
[(11) 12 _ (E) 16 — l4 ([P3P4+ P3P5]' — 714121134 — 12,1231.» 112191) 13]

 

 

Appendix 0

FORTRAN Program to Calculate the
Result

165

 

166

PROGRAM thesis
IMPLICIT NONE
INTEGER LHEL,QHEL,GHEL
REAL ECM,PI,ALPHA
COMPLEX MZ
PARAMETER (LHEL = +1, QHEL = +1, GHEL = +1, ECM = 5.80)
PARAMETER (ALPHA = 1.0/137,PI = 3.14157,MZ = (93.5,2.7))
INTEGER EPF_PARITY
REAL P1P3,P1P4,P1P5,P3P4,P3P5,P4P5,LIMIT,P1Y_$QD,P1Y
REAL PP1P3,PP1P4,PP3P5,PP4P5
REAL WIV,PPIIV
REAL PROPIV,E0,BO,CUO,EOIV,BOIV,CUOIV,RATIO
COMPLEX Il,IZ,I3,I4,I5,IG,I7,18,I9,I10,Ill,I12,Il3,114,115
COMPLEX 116,117,118,119
COMPLEX EVL,BORN_TERM,BORN,LOOP_TERM,MZP,MZPIV
COMPLEX NUMERATOR ,ELECTRON_LOOP_TERM,POSITRON_LOOP_TERM
COMPLEX E_ LOOP, P _LOOP
CHARACTER—SET_HEL_CODE#1,HEL CODEtl
OPEN ( UNIT = -10, FILE = ’TOUTPUT’, STATUS = ’NEW’ )
LIMIT = -2#ECMI¥2
HEL_ CODE = SET_ HEL _CODE(LHEL, QHEL, GHEL)
PRINTt, ’HEL CODE = ’ ,HEL_ CODE
DO 1000 P3P4 = 0 ,LIMIT, -10
DO 1000 P3P5 = 0,LIMIT-P3P4,-10
P4P5 = LIMIT - P3P5 - P3P4
DO 1000 P1P4 = 0,LIMIT,-10
DO 1000 P1P5 = O,LIMIT-P1P4,-10
P1P3 = LIMIT - P1P5 - P1P4
P1Y_SQD = ( 4tP1P4tP1P5tP3P4tP3P5)
- (P1P5tP3P4+P1P4*P3P5-P4P5#P1P3)#12
PRINTt ’P1Y_SQD= ’ ,P1Y 5 D
IF ( P1Y _SQD .GE. 0 ) THEN

PRINTI, ’P1P3 = ,P1P3
PRINT: ’P1P4 = ’,P1P4
PRINT:,’P1P5 = ’,P1P5
PRINT:,’P3P4 = ’,P3P4
PRINT:,’P3P5 = ’,P3P5
PRINT: ’P4P5 = ’,P4P5

P1Y= EPF_ PARITYIP1Y _SQD:#(O. 5)
PRINT: ’PlY = ’ ,P1Y
EVL: CMPLX(O. 0 ,P1Y)
PRINT‘, ’EVL = ’,EVL
BORN_TERM= BORN(HEL_ CODE, P1P3, P1P4, P1P5,
P3P4, P3P5, P4P5 ,EVL)
MZP = 2¢P3P4 + MZ
MZPIV = 1.0/MZP
PROPIV = 1. O/(- -2tP1P5#2t(P3P4+P3P5+P4P5))
BORN_ TERM = PROPIV:MZPIV:BORN_ TERM
E0 = -4tP1P5¢(P3P5+P4P5- -P1P5)t (P3P4+P3P5+P4P5)

E0IV==11.0/E0t¢(0.5)

80 = 1

80 = 0.125:(P3P4tP3P5)tt(0.5)
BOIV = 1.0/BO:¢(0.5)

cuo = 0.125:(P3P4+P3P5+P4P5)tP1P5

CUO = 1
CUOIV = 1.0/CUO:¢(O.5)

 

 

++++

+++++

0941+

§

167

FPIIV = 0.251(1/PI)
BORN_TERM = E01V¢BOIV¢CUOIV*BDRN_TERM
PRINT:,’BURN TERM = ’,BORN_TERM

H
03
II II II II II II II II II II
A
~O
O
V

 

(010)

H
H
01
II II II II II II II II II II II
A
O
0
V

w 0

CALL INTEGRALS (P1P3,P1P4,P1P5,P3P4,P3P5,P4P5,
I1,I2,13,I4,I5,16,I7,18,I9,110,

111,112,113,114,115,116,117,118,119,
w1v,FP11v,

MZ)
ELECTRON_LO0P_TERM = E_LO0P(HEL_CODE,P1P3,P1P4,P1P5,
P3P4,P3P5,P4P5,
11,12,13,14,15,16,17,18,19,110,
I11,112,113,Il4,I15,116,I17,118,119,
WIV,FPIIV,BOIV,CUOIV,EOIV,EVL,

MZ)
ELECTRON_LO0P_TERM = ELECTRON_LDOP_TERM/(2¢P4P5)

PP1P3 = P1P4

PP1P4 = P1P3

PP3P5 = P4P5

PP4P5 = P3P5

EVL = -EVL
E0 = -4tP1P5¢(PP3P5+PP4P5-P1P5)#(P3P4+PP3P5+PP4P5)
E0 =

1
EOIV = 1.0/E0t¢(0.5)

80 = 1
80 = O.125¢(P3P4¢PP3P5):¢(0.5)
BOIV = 1.0/BO:¢(0.5)
CUO = 1
CUO = 0.125s(P3P4+PP3P5+PP4PS)*P1PS
CUOIV = 1.0/CUOtt(0.5)
CALL INTEGRALS (PP1P3,PP1P4,P1P5,P3P4,PP3P5,PP4P5,
I1,I2,I3,I4,15,16,I7,18,19,110,
Ill,112,113,Il4,I15,116,117,118,119,
WIV

, IIV,

POSITRON_LO0P_TERM = P_LUOP(HEL_CDDE,PP1P3,PP1P4,P1P5,
P3P4,PP3P5,PP4P5,
11,12,13,14,15,16,17,18,19,11o,

168

4.

I11, 112, I13, I14, 115, 116, I17, 118, I19,
+ WIV, FPIIV, BOIV, CUOIV, EOIV, EVL,

MZ
POSITRON_ LOOP TERM: POSITRON LOOP _TERM/(2tPP4P5)
LOOP _TERM = ELECTRON LOOP TERM - PUSITRON LOOP_ TERM
LOOP _TERM: LUOP TERMtBOIVACUOIVtE6IV
LOOP TERM LOOP _TERM/(2.(P3P4+P3P5+P4P5))
PRINTP, ’LOOP TERM = ’ ,LOOP TE RM
LOOP TERM: LOOP _TERMLALPHA/(BMPIMP4P5)
NUMERATOR = LOOP TERM 4 BURN_ TERM
RATIO = NUMERATD§1CUNJG(NUMERATUR)
RATIO = RATIO/(BORN_ TERMICONJO(BORN_ TERM))
PRINTI, ’RATIO = ’ ,RATIO
WRITE ( 10, 200 ) HEL CODE, P1P3, P1P4, P1P5 P3P4, P3P5,
+ P4P5 RATIO
200 FORMAT (A1,6F7. 2, F12. 7)
ENDDO

ENDIF
1000 CONTINUE
END

CHARACTER FUNCTION $ET_ HEL _CUDE(LHEL, OHEL, GHEL)
INTEGER LHEL, QHEL Om
IF ((LHEL Eq. .1) .AND. (OHEL .Eq. 1)
4 .AND. (GHEL .EO 41))m
PRINTt,’A’
SET_ HEL_ CODE = ’A’

RETURN
ELSEIF -((LHEL .Eq. +1) .ANDTH (OHEL .eq. 1)
+ (G GHEL EQ -1))TH
PRINT#, ’8’
SET HEL_ CODE = ’B’

RETURN
ELSEIFA ((LHEL .Eq. .1) .AND. TH(OHEL .Eq -1)
4 . (c GHEL Eq 41))m
NDPRINT. C
$ET_HEL_CODE = ’C’

RETURN
ELSEIF ((LHEL .Eq. 41) .AND. (OHEL .EQ. -1)
,4 .AND. (GHEL .EQ. —1)) THEN
PRINT-, ’0’
SET_ HEL_ CODE: ’0’

RETURN
ELSEIF ((LHELH .Eq. E.1) AND. (QHEL .Eq 1)
+ (GH +1))W
PRINT: 'E'ELI
PRINT:,’EVL = ,EVL
$ET_ HEL_ CODE = E

RETURN
ELSEIFA((LHELH.EQ.E-1) .ANDm(QHEL .EQ. 1)
. + (CH --1))TH
APRINTt, ’F’
SET_ HEL_ CODE = ’F’

RETURN
ELSEIFA((LHELH.EQ.E-1) '??DW(QHEL .EQ. -1)
4r (CH +1
APRINT: ’G

 

 

5ET_

 

169
HEL_CODE = ’6’
RN

R
ELSEIF A((LHEL .EO. -1) .AND TH(OHEL .EO -1)

(c GHEL .EO -1))m
PRINT4, ’H’
SET_HEL_CUDE : ’H’
RETURN
ENDIF
RETURN
END

COMPLEX FUNCTION
REAL P13, P14, P15,

BORN(HEL_ CODE,P13,P14,P15,P34,P35,P45,EVL)
P34, P35 P45

COMPLEX EVL, OA, OB, OC, OD, OE, OF, OG, OH
CHARACTER HEL_ CODEtl

IF (HEL_ CODE .EO
BORN : OA(P13,

IF (HEL_ CODE EO.
BORN: OB(P13,
RETURN

ENDIF

IF (HEL_CODE .EO
BORN : OC(P13,
RETURN

ENDIF

IF (HEL_ CODE EO.
BORN: OD(P13,
RETURN

ENDIF

IF (HEL_ CODE EO
BORN: OE(P13,
RETURN

ENDIF

IF (HEL_ CODE .EO.
BORN: OF(P13
RETURN

ENDIF

IF (HEL_ CODE EO.
BORN : OG(P13,
RETURN

ENDIF

IF (HEL_ CODE ..EO
BORN: OH(P13,
RETURN

ENDIF

’A’) THEN
P14, P15, P34, P35 P45, EVL)

’8’) THEN
P14, P15, P34 P35, P45 ,EVL)

’C’ THEN
P14, P15, P34, P35, P45, EVL)

D) THEN
P14, P15, P34, P35, P45 ,EVL)

’E’) THEN
P14,P15,P34,P35,P45,EVL)

’F’) THEN
P14, P15, P34, P35, P45, EVL)

'c') THEN
P14,P15,P34,P35,P45,EVL)

’H’) THEN
P14, P15, P3O P35, P45 ,EVL)

 

170

RETURN
END

SUBRDUTINE INTEGRALS (P1P3,P1P4,P1P5,P3P4,P3P5,P4P5,
Il,I2,13,I4,IS,IB,I7,18,IQ,110,
111,112,113,114,I15,116,117,118,119,
WIV,FPIIV,

D3)

nnnn

COMPLEX D3,MZP,R,LN3,LN5,LN6,MZP1,ZIlOD,RUPPER,

c MZPUP,MZPIV,RIV,ZIIOU,UPPER,
c I1,12,13,14,15,16,17,18,19,110,196,IR,
c 111,112,113,114,115,116,117,118,119

REAL LN1,LN2,LN4,RLOWER,MZPLW,IQU,19L,193,IW,IS,
c LOWER,WIV

REAL P1P3,P1P4,P1P5,P3P4,P3P5,P4P5,API,FPIIV,K02,K02IV

DOUBLE PRECISION CC44,CC43,CC42,CC41,CC33,CC32,CC31,
c CC22,CC21,CC11

API = 3.142857143

P13IV = 1/P1P3

P14IV = 1/P1P4

P151V = 1/P1P5

P34IV = 1/P3P4

P3SIV = 1/P3P5

P4SIV = 1/P4P5

MZP = 2tP3P4 + 03

5152 = P1P3

5133 = P1P4

5154 = P1P5

3253 = P3P4

5234 = P3P5

5354 = P4P5

W = 5152:5384:(~51S2¢$3$4+8154tSZS3+$1$3*$254)
C + $153#$254t( $1$2tS3$4+$1$4t$2$3-S153:5254)
C + SlS4tSZ$3t( 5152:5354-3154tSZSB+Sl$3tSZ$4)

WIV = 1.0/W

CC44 = WIVt(-2)t(SlS2tSISBtSZS3)
CC43 = WIV*51$2*(-51$2t53$4+5154t5253+3153152$4)
CC42 = WIVtSlS3t( 5152:533495154#SZS3-SlSStSZS4)
CC41 = WIV#8253#( S152:S3S4~$1$4¢$2$3+$153t$2$4)
CC33 = WIV:(-2)t(51$2t$1$4¢$2$4)
CC32 = WIVtSlS4t( 5152:5354-5154t32$3+$133t$2$4)
CC31 = WIV*$2$4¢( 5152*535443154:5253-Sls3t5254)
CC22 = WIV:(~2)¢(Sl$3*$1$4t$3$4)
CC21 = WIV#S3S4*(-$1$2*$3S4+$1$4*5233+$1$3t5254)

CC11 = w1v¢(-2).(5253‘3234.sas4)
R = -2¢P4P5:P1P3+2:P3P5¢(P1P4+P1P5) + MZP¢(P1P4+P1P5)

LN1 = Lee (-2¢P4P5 )
LN2 = LOG(-2¢(P1P4‘P1P5))

LN3 = LOG((2¢P3P5+MZP)/03)

LN4 = LOG((P1P4+P1P5)/P1P5)

LNS = LOG(MZP/(2.P3P5+MZP))

LN6 = LOG((2¢P3P5+MZP)t(P1P4+P1P5)/R)
K02 = 0.5a(-P4P5)

Kosz = 1/K02
0145 = P1P4 + P1P5

110
120
130
140
150

 

L

 

 

171

01 = 0145

D145IV = 1/0145
0345 = P3P4 + P4P5
03451v = 1/0345

MZP = 2*P3P4 + D3

11 = ~2tAPItP4SIV*(LN1)

I2 = -2tAPI*Dl451V*(LN2)

I3 = +2:API¢D345IV¢LN3

R = —2tP4P5tP1P3+2*P3P5*01+MZP¢01
MZP1 = MZP

RUPPER = CONJG(R)

RLOWER = RtCONJG(R)

MZPUP = CUNJG(MZP1)

MZPLW = MZP1¢CONJG(MZP1)

RIV = RUPPER/(RLUWER)

MZPIV = MZPUP/(MZPLWO

I4 = - apitP451VtP15IVtLN4 -
0.5:01*P45IV#P151V¢12 - 0.5*P151V*Il

I5 = P45IV¢MZPIV#(—4*api*LN5 - 0345:13 +P4P5*I1)
I6 = RIV#(+4*api*LN6 - 0345*I3 + 01:12)
I7 = 0.25:(P3P5:P15IV+P3P4tP14IV-P1P3tP4P5tP14IVtP15IV)t

C (11 - (P1P4-P1P5)*P45IV*12 + 2tP1P5*I4)
C —0.5t(P3P5—P3P4)#P45IV#I2 + P3P4tI4
18 = 0.25*(P1P4#P34IV+P1P5#P3SIv-P1P3¢P4P5$P34IV¥P351V)t
C (II - (P3P4—P3P5):P4SIV#I3 - MZPtIS)
C -0.5¢(P1P5-P1P4)‘P451V*I3 +P1P4tIS
IQU = P3P4tP1P4wP1P5+P3P4#PlP5*P1P5+P4P5tP1P3#P1P4
C — P4P5tP1P3tP1P5-P1P4tP1P4tP3P5-P1P5tP1P4tpaps
I9L = 2¢P1P3t
C (P1P3tP4P5—P1P44P3P4-P1P4tP3P5-P1P5#P3P4-P1P5*P3P5)
= 2tP4P5¢P1P3t0145IV¢0145IV-(P3P4+P3P5)*01451V
196 = -2¢P4P5¢P1P3#01451V + 2¢P3P5 + MZP
19 = O.5¢I9Ut(1/IQL):(I2+193*I3—196¢16)
c + 0.5*P4P5:(P1P5-P1P4)*D145IV¢(D14SIV:I3+I6)
C + 0.5*P4P5*15

111 = 2.API.P451V*(P1P4-P1P5) + P1P4tI1
112 = 2‘API‘P4SIV$(P3P4-P3P5) + P3P4$Il
113 = -2¢API#P4P5t01451Vt01451V:(P1P4 — P1P5) +

C P4P5¢P1P5t01451V312
114 = 0.5:(P3P4 + P3P5)tI2 +

C 0.5:(0145:0345-2tP1P3tP4P5)#014SIVt( 4tAPI#DI45IV + I2)

115 = -2:API¢P4P5tD345IV¢DS45IVa(P3P4 - P3P5) +

C 0.5tP4P5t(1+034SIV10345IV¢(P3P4-P3P5):(0345+03))tI3
I16 = 0.5!0145tI3 4

C 0.5#(0145#0345 - 2tP1P3tp4P5)m

C D3451Vt034SIV*( 4#API - I3t(0345 + 03))

117 = P3P4t17 + 0.51P451Vt(P3P4-P3P5)¢Il4 +
0.25tIS#P14IV#P151V#(112 - (P1P4-P1P5):P451th14 + 2tP1P5tI7) +
o.0625#P14IVtP14IV#P15IV‘P1SIVt(4tP3P4tP3P5$P1P4tP1P5-IS*IS)t
(-2tPlP5:I1 - 4¢P1P51P1P5t14 - P451Vt0145¢2tP1P5¢I2)

118 = P1P4#18 + 0.5:P451Vt(P1P4-P1P5)t116 +
0.25nIStP34IV:P351V:(I11-(P3P4—P3P5)tP451VtI16-(2tP3P4+03):18)+
0.0625tP34IVtP34IVtP351V¢P35IVt(4tP3P4tP3P5¢P1P4tP1P5-I$¥IS)t

(
(MZP)#Il-MZPtMZP#15
-P45IV~(0345¢2¢P3P5+(P3P4-P3P5)¢MZP)¢I3

)
IW = 2¢P1P31P4P5 - 0145:0345

1090
1100
1110
1120
1130
1140
1150
1160
1200
1210
1220
1230
1240
1250
1260
1270
1290
1300

1360
1370
1380
1390
1400
1410
1420
1430
1440
1450
1460
1470
1480
1490
1500

1560

1580
1590
1

1700
1620
1630
1640
1720
1730
1740
1750
1760
1770
1780
1790
1800
1810
1820
1830

 

 

 

172
IR = -2#P4P5#P1P3 + 01*(2*P3P5 + MZP)
119 = P4P5tP1P5tDl4SIV419 -
0.5*0145IVt014SIVtP4P5t(P1P4—P1P5):I15 +
0.5:I9U¢(1/19L)#
(113 9 01451Vt01451VtIWt115+(01451VtIW-P3P4-P3P5-03)#19) +

—0.5*P4P5*P1P4tP1P5*(1/I9L)
-0.25*IQU*I9U*(1/IQL)t(1/I9L)
)t
(
C (0145IV*IR)*I2
C —01451V¢01451V1IR1IR416
C -01451Vt(D345:DB45+DI45IV#IW*(P3P4-P3P5-MZP))*I3
C

hfinfihfififl

) ,
U10U1=+2.*CC41:CC31*P4P5+2.tCC41tCC21*P3P5+2.*CC41:CC11*P1P5+2.*
CCC31*CC21*P3P4+2.tCC31tCC11tP1P4+2.tCC21tCC11¢P1P3
U10U2=+CC42tCC31tP4P5+CC42¢CC21tP3P5+CC42¢CC11tP1P5+CC41*CC32#P4
CP5+CC41*CC22tP3P5+CC4laCC21*P1P5+CC32*CC21tP3P4+CC32tCC11tP1P4+C
CC31tCC22tP3P4+CC314CC21:P1P4+CC22:CC111P1P3+CC21#:21P1P3
UlDU3=+CC43tCC31:P4P5+CC43*CC21tP3P5+CC43aCC11*P1P5+CC41*CC33xP4
CP5+CC411CC32tP3P5+CC41#CC31tP1P5+CC33#CC21tP3P4+CC33tCC11¢P1P4+C
CC32tCC31tP3P4+CC32tCC11tP1P3+CC31*CC21*P1P3+CC31**2*P1P4
U10U4=+CC44¥CC31#P4P5+CC44*CC21#P3P5+CC44#CC114P1P5+CC43ICC414P4
CP5+CC43tCC21*P3P4+CC433CC11tP1P4+CC42tCC41*P3P5+CC42:CC31*P3P4+C
CC42:CC11*P1P3+CC41¢CC31tP1P4+CC41tCC21tP1P3+CC41**2*P1P5
U20U2=+2.tCC42tCC32¢P4P5+2.*CC42*CC22#P3P5+2.tCC42tCC21tP1P5+2.i
CCC32¢CC22¢P3P4+2.:CC32tCC21*P1P4+2.tCC22tCC21$P1P3
U2DU3=+CC43*CC32:P4P5+CC43tCC22¢P3P5+CC43¢CC21*P1P5+CC42tCC33tP4
CP5+CC42tCC32tP3P5+CC42tCC31*PlP54CC33:CC22:P3P4+CC33#CC21tP1P4+C
CC32*CC31#P1P4+CC32#CC211P1P3+CC32**2*P3P4+CC31*CC22*P1P3
U2DU4=oCC44tCC32tP4P5+CC44tCC22tP3P5+CC44tCC21tP1P5+CC43tCC42tP4
CP5+CC43*CC22*P3P4+CC43¢CC214P1P4+CC42*CC41*P1P5+CC42tCC32*P3P4+C
CC42tCC21tP1P3+CC42tt2tP3P5+CC41tCC32tP1P4+CC41tCC22#P1P3 _
U30U3=+2.tCC43tCC33tP4P5+2.tCC43tCC32tP3P5+2.tCC43tCC31tP1P5+2.t
CCC33¢CC32tP3P4+2.tCC33tCC31tP1P4+2.tCC32tCC31tP1P3
U3DU4=+CC44tCC33tP4P5+CC44:CC32:P3P5+CC44:CC31tPlPS+CC43¢CC42tP3
CP5+CC43tCC411P1P5+CC43#CC32#P3P4#CC43#CC31tP1P4+CC43t¢2tP4P5+CC4
c2‘cc33‘P3P4*CC42$CC31*P1P3+CC41fiCC33*P1P4*CC41tCC32¥P1P3
U4DU4=+2.tCC44tCC43:P4P5+2.tCC44tCC42tP3P5+2.aCC44tCC41tP1P5+2.t
CCC43tCC42tP3P4+2.tCC43fiCC41tP1P4+2.tCC42tCC41tP1P3+O.
' ZIIOU = U10U1t(0.5t18+0.5*01#IS)+U20U2#(0.5tI7-O.25t03*14)
C - (U30U3-2:U30U4+U40U4):0.5¢19
C + U10U2t(0.5#I1+01#I4-0.5*03tI5)
C - (U10U3-UIDU4)*(O.5:13+D1t16)
C + U10U4tP4P5tI5-(U20U3-U20U4)t0.5t(I2-D3tI6)+U20U4tP4P5tI4
C - (U30U4-U40U4):P4P5¢I6
Z1100 = U10U1#01t01+U20U2*0.25t03tD3+U4DU4tP4P5*P4P5
C -1.0tU10U2t01t03+2¢U10U4tP4P5tDl-U20U4tP4P5tD3
UPPER = -ZI10UtCONJG(ZIlOD)

LOWER = ZIIoO.c0NJc(ZIIOO)
110 = UPPER/(LOWER)

RETURN

END

COMPLEX FUNCTION E_LOOP(HEL_COOE,P13,P14,P15,P34,P35,P4s,
+ 11,12,13,I4,Is,Ie,I7,Ie,19,I1o,

1840

1850
1860
1870
1880
1890
1900
1910
1920
1930
1940
1950
1960
2000
2010
2020
2030
2040
2050
2060
2070
2080
2090

2110
2120
2130
2140
2150
2160
2170
2180
2190
2200
2210
2220
2230
2240
2250
2260
2270
2280
2290
2300
2310
2320
2330
2340
2350
2360

5010

 

 

173

+ 111,I12,113,114,115,116,117,118,119
+ WIV,FPIIV,BOIV,CUOIV,EOIV,EVL,
+ 03)

REAL P13,P14,P15,P34,P35,P45
REAL WIV,FPIIV,BOIV,CUOIV,EOIV

COMPLEX EVL,ELA,ELB,ELC,ELD,ELE,ELF,ELG,ELH

COMPLEX Il,I2,13,I4,I5,16,I7,I8,19,110,111,I12,I13,I14,I15,
+ 116,I17,I18,119

COMPLEX 03

CHARACTER HEL_CODEtl

 

IF (HEL_CDDE .EO. ’A’) THEN
E_LOOP = ELA(P13,P14,P15,P34,P35,P45,
Il,12,13,14,15,16,I7,I8,I9,110,
111,I12,Il3,I14,115,I16,117,118,119,
WIV,FPIIV,BOIV,CUOIV,EOIV,EVL
03)

¢+++

R
ENDIF

IF (HEL_ CODE ..EQ ’8’) THEN
L=DOP ELB(P13 P14 ,P1,5 P34, P35, P45 ,EVL)
ETURN

R
ENDIF

IF (HEL_CODE .EQ. ’C’) THEN
E_LOOP = ELC(P13,P14,P15,P34,P35,P45,EVL)
RETURN

ENDIF

IF (HEL_CODE .EO. ’0’) THEN
E_LOOP = ELD(P13,P14,P15,P34,P35,P45,EVL)
RETURN

ENDIF

IF (HEL_ CODE .EQ. EN
EL OOP =ELE(P13, EP14 P15, P34 ,P35, P45 ,EVL)

 

 

ENDIF

IF (HEL6 CODE EO. 'F’) THEN
=ELF(P13, P14 P15, P34 ,P35, P45 ,EVL)

ENDIF

IF (HEL_CODE .EO. ’0’) THEN
E_LOOP = ELG(P13,P14,P15,P34,P35,P45,EVL)
RETURN

ENDIF

IF (HEL_ CODE .EQ ’H’) THEN
E_LOOP: ELH(P13, P14, P15, P34 ,P35 P45 ,EVL)
RETURN

ENDIF

RETURN

174
END

COMPLEX FUNCTION P_LO0P(HEL_CODE,P13,P14,P15,P34,P35,P45,
11,12,I3,I4,IS,IS,I7,I8,I9,IIO,
111,112,113,114,115,116,Il7,118,119
WIV,FPIIV,BOIV,CUOIV,EOIV,EVL,

O3)

++++

REAL P13,P14,P15,P34,P35,P45

REAL WIV,FPIIV,BOIV,CUOIV,EOIV

COMPLEX EVL,PLA,PLB,PLC,PLD,PLE,PLF,PLG,PLH

COMPLEX Il,I2,13,14,15,16,I7,18,19,110,111,I12,113,114,I15,
+ 116,117,I18,119

COMPLEX D3

CHARACTER HEL_CODE¢1

IF (HEL_CODE .EQ. ’A’) THEN
P_LOOP = PLA(P13,P14,P15,P34,P35,P45,

+ Il,12,I3,I4,I5,I6,I7,18,19,I10,
+ I11,112,113,Il4,115,116,117,118,I19,
+ WIV,FPIIV,BOIV,CUOIV,EOIV,EVL
+ 03)
RETURN
ENDIF

IF (HEL_ CODE .EQ. ’8’) THEN
PLOOP: PLB(P13, P14 ,P15, P34, P35 ,P45 ,EVL)
ETURN

ENDIF

IF (HEL_CODE .Eq. ’C’) THEN
P_LOOP = PLC(P13,P14,P15,P34,P35,P45,EVL)
RETURN

ENDIF

IF (HEL_CODE .EO. ’0’) THEN
P_LOOP = PLD(P13,P14,P15,P34,P35,P45,EVL)
RETURN

ENDIF

IF (HEL_ CODE .EQ
PLOOP: PLE(P13, EP14, P15, P34, P35, P45 ,EVL)
RETURN

ENDIF

IF (HEL_ CODE .EQ. ’F’) THEN
“LOOP PLF(P13, P14, P15, P34, P35, P45 ,EVL)
RETURN

ENDIF

IF (HEL_ CODE .EQ ’6’) THEN
PLOOP = PLG(P13, P14, P15, P34 ,P35, P45 ,EVL)
RETURN

ENDIF
IF (HEL_CODE .EQ. ’H’) THEN

 

175
P_LOOP = PLH(P13,P14,P15,P34,P35,P45,EVL)
RETURN

ENDIF

RETURN

END

C Expression QA
COMPLEX FUNCTION QA(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45
COMPLEX EVL
COMPLEX TLPART,HDPART,XTOTAL

XTOTAL = 0
HDPART = 1
TLPART=256tP34tP35tP45tP14tPlSt*2-256tP34tP35tP45tP15##3+512*P3
4aP35*P45tt2tP14*P15+768#P34*P35*P45##2tP15tt2-512tP34tP35tP45¢
*3:P15+256¢P34tP35**2*P45—P14#P15+256tP34tP35##2:P45*P15t#2-256
tP34tP35t#2tP45#*2tP15+512tP34tP35*t2tP14tP15*t2—256*P34tP35t*2
*P15*t3-256*P34*P45**2*P14*P15**2-256*P34*P45$*2*P15#*3+256*P34
tP45t:3tP14tP15+512tP34tP45tt3*P15tt2-256tP34tP45tt4tP15+256tP3
4tt2*P35tP45tP14tP15+1024tP34t12*P35tP45tP15t#2-512tP34##2#P35#
P45:#21P15+256:P34¢t2tP351P14*P15*t2-512tP34tt2¢P35tP15#*3-256#
1'P34tt2tP45tP14tPlstt2
TLPART=TLPART-512#P34**2tP451P15**3+256#P34t*2*P45¢t2*P14#P15+1
1 O24tP34t¢2tP45t*2tPlStt2—512tP34tt2tP45tt3tP15+512tP34tt3tP45tP
1 15*t2-256*P34¢t3#P45##2tP15-256RP34:#3tP15t#3+256tP35tP45t*2tP1
1 4tP15t#2+256tP35tP45tt2tP15tt3-256tP35:P45tt3tP15#t2+512tP35t*2
1 #P45*P14#P15*t2+256tP35tt2tP45tP15nt3-512tP35t#2tP45*t2tP15*t2—
1 256RP35tt3tP45tP15t*2+256:P35t#3tP14:P15tt2
XTOTAL=XTOTAL+HDPARTtTLPART
HDPART=EVL -
TLPART=-256tP34tP35tP45tP15+512tP34wP35tP15tt2o256tP34tP45tP15t
1 42-256tP34tP45*‘2*P15-256*P34#t2tP45#P15+256tP34t#2tP15tt2+256t
1 P35tP45¢P15¢t2+256¢P35*¢2RP15¢*2
XTOTAL=XTOTAL+HDPART¢TLPART
QA=XTOTAL

HHHHHI—‘H

END
tend

C Expression QB
COMPLEX FUNCTION OB(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45
COMPLEX EVL
COMPLEX TLPART,HDPART,XTOTAL

XTOTAL=O.

HDPART=1

TLPART:512:P34#P35¥P45**2aP14tP15+256tP34tP35tP45¢*2IP151t2-102
4tP34tP35tP4Stt3tP15+256tP34tP35t*2RP45*P14tP15—256tP34tP35tt2t
P45sP153#2-512tP34tP35tt2tP45tt2¢P15+256tP34:P45t#3tP14tP15+512
tP34tP45tt3tP15t#2-512PP341P45tt4tP15~2$6tP34t¢2tP35tP45¢P15tt2
-256tP34:t2tP35tP45tt2tP15+256¢P34tR2tP45tt2tP150t2-256tP34t‘2‘
P45tt3tP15+768tP35tP45tt3!P14tP15+512tP35¢P45tt3tP15tt2-768OP35
tP45t:4:P15+768tP35t:2PP45t~2tP14¢P15+256uP35tt2:P45#t2tP15:t2-
768tP35:t2tP45tM3sP15+256tP35tt3tP45:P14tP15-256tP35tt3tP45tt2:

HHHHHHH

 

176
1 P15+256tP451t4tP14tP15
TLPART=TLPART+256tP451t4tP15tt2-256tP45tt51P15
XTOTAL=XTOTAL+HDPARTtTLPART
HDPART=EVL .
TLPART=-256*P34tP35tP451P15-256:P34*P45tt2*P15-512PP354P45**2#P
1 15-256tP35t*2tP45*P15-256*P45**3#P15
XTOTAL=XTOTAL+HDPARTnTLPART
QB=XTOTAL
RETURN
END

tend

C Expression QC
COMPLEX FUNCTION QC(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45
COMPLEX EVL
COMPLEX TLPART,HDPART,XTOTAL

XTOTAL=O.
HDPART=1
TLPART=256tP34tP35tP45t*2tP14¢P15+512tP34tP35tP45tt2¢P15t*2-512
*P34*P35*P45**3#P15+512*P34*P35#*2*P45‘P14*P15+256*P34*P35**2*P
45tP15**2-1024tP34tP35**2tP45:*2tP15-512tP34tP35tt3tP45tP15+256
¢P34tP35~t3tP14*Pl5—256*P34tP35tt3¢P15t12+2561P34#*2*P35*P45*P1
5tt2-256tP34t*2tP35tP45t*2:P15-256¢P34¢t2tP35tt2tP45tP15-256PP3
4tt2tP35*#2cP15t#2+256#P35tP45#*3tP14#P15+256tP35tP45tt3tP15*#2
~2561P35tP45tt4wP15+768tP35**2*P45**2*P14*P15+512tP35tt2tP45tt2
#P15tt2-768tP35tt2#P45¢*StPl5+768tP35tt3¢P45¢P14#P15+256#P35##3
tP45:P15#*2
TLPART=TLPART—768tP35tt3¢P45*t2tP15-256tP35t#4tP45tP15+256*P35t
1 t4¢P14tP15
XTOTAL=XTOTALCHDPARTtTLPART
HDPART=EVL -
TLPART=256tP34RP35tP45tP15+256¢P34tP35-t2NP15+256tP35¢P45tt2wP
1 5+512tP35t#2:P45*P15+256¢P35t¢3tP15
XTOTAL=XTOTAL+HDPARTtTLPART
QC=XTOTAL
RETURN
END

HHHHHHHH

send

C Expression QD
COMPLEX FUNCTION QD(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45
COMPLEX EVL
COMPLEX TLPART,HDPART,XTOTAL

XTUTALzo .

HDPART=1
TLPART=256tP34¢P35¢P45¢P14¢P15¢t2+512#P34tP35tP45tP15t33+256tP3
1 4tP35tP45*:2tPl4tP15-1280tP34tP35tP45tt2tP15t#2-256iP34tP35tP45
1 tt3tP15+512tP34tP35tt2OP45tP14tP15-512tP34tP35*t2tP45tP15n#2-51
'1 2¢P34¢P35tt2tP45tt2tP15-256tP34¢P35tt2tP14tP15tt2-256tP34tP35#t
1 3*P45tP15+256tP34tP35t:3tP14tP15+512¢P34tP45t#2tP14tP15tt2+768¢
1 P34tP45tt2tP15tt3-768#P34tP45tt3¢P15##2+256:P34tt2tP351P45tP14t
1 P15-1024tP34tt2cP35tP45tP15tt2-512tP34tt2tP35tP45tt2tP15-256tP3
1 4t:2PP35tP14tP15t¢2+256~P34tt2tP35¢P15¢t3~512tP34tt2tP35#t2tP45

1 #P15

 

 

 

tend

TLPART=TLPART+256*P34tiézp3st*2*P14*P15-256*P34¥*2#P35‘*2*P15“
2+256tP34tt2¢P45tP14*P15~t2+768tP34t*2tP45*P15tt3—768tP34tt2*P4
5*t2#P15#t2-256tP34t*3tP35tP45tP15-2561P34tn3tP35tP15tt2—2561P3
4*t3¢P45tP15*t2+256tP34*t3tP15**3+512tP35*P45¢t2tP14tP15*t2¥256
*P35*P45t*21P151*3-512:P35*P45**3#P15**2+256#P35#*2tP45¢P14MP15
tt2-256tP35tR2tP45tt2tP15tt2+256tP45t#3*P14*P15t*2+256tP45*t3tP
15tm3-256tP45*t4*P15tt2

XTOTAL=XTOTAL+HDPARTtTLPART

HDPART=EVL

TLPART=2561P34tP35$P45tP15-256tP34tP35:P15#t2+256tP34¢P35tt2tP1

1 5-512tP34tP45tP15:t2+256tP34tt2tP35tP15-256*P34tt2tP15t32-256tP

1 35*P451P15tt2-256tP45t*2#P15¢#2

XTOTAL=XTOTAL+HDPARTATLPART
QD=XTOTAL

RETURN

END

HHHHHH

C Expression OE

tend

HHHHHHHH

COMPLEX FUNCTION OE(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45

COMPLEX EVL

COMPLEX TLPART,HDPART,XTOTAL

XTOTAL=0.
HDPART:1
TLPART=—256*P34tPBStP45tP14tP15*t2—512tP34tP35*P45tP15¢t3—256tP
34tP35¢P45tt2tP14tP15+1280*P34#P35#P45t*2tP15#*2+255*P34¢P35*P4
5*t3-P15-512tP34tP35tt2tP45tP14*P15+512tP34tP35tt2tP45tP15t*2+5
12*P34tP35tt2tP45tt2tP15+256*P34tP35wt2tP14tP15*t2+256tP34*P35¢
#3tP45tP15-256tP34tP35t#3:P14*P15-512*P34#P45t12¢P14tP15t*2-768
tP34tP45¢#2tP15tt3+768tP34tP45tt3tP15t*2—256tP34tt2tP35tP45tP14
1P15+1024tP34tt2tP35#P45:P15*:2+512*P34tt2tP35tP45tR2tP15+256tP
34*t2tP35*P14:Pls:t2-256xP34tt2tP35*P15tt3+512#P34tt2*P35t*2*P4
5tP15
TLPART=TLPART-256*P34¢t2tP35t#2tP14tP15+2561P34t*2tP35tt21P151t
2-256tP34-*2*P45*P14:P15¢t2-768:P34tt2tP45tP15tt3+768¢P34tt2tP4
5tt2tP15#t2+256tP34*t3tP35tP45¥P15+256tP34#t3#P35#P15t*2+256#P3
4tt3tP45tP15tt2-256#P34tt3tP15t:3-512tP35tP45t:2tP14tP15*t2-256
#P35:P4Stt2tP15tt3+512RP35tP45#t3tP15*t2-256tP35¢¢2tP45tP14*P15
tt2+256tP35¢*2tP45tt2tP15tt2—256fiP45tt3tP14¢P15¢t2~256tP45#~3PP
15tt3+256tP45tt4tP15t¢2
XTOTAL=XTOTAL+HDPARTtTLPART
HDPART=EVL
TLPART=256*P34#P35tp45#P15-255tp34tp35fipl5t#2+256*P34#P35*#2*P1
1 5-512:P34:P45¢P15:t2+256tP34tt2tP35aP15-2564P34tt2tP15tt2—256tP
1 35tP4StP15tt2-256tP45tt2tP15tt2
XTOTAL=XTOTAL+HDPARTtTLPART
QE=XTOTAL
RETURN
END

HHHHI—‘H

C Expression OF

COMPLEX FUNCTION QF(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45

COMPLEX EVL

COMPLEX TLPART,HDPART,XTOTAL

 

 

178

XTOTAL=O.
HDPART=1
TLPART=-256*P34*P35tP45t*2#P14*P15-512tP34tP35#P45#*2#P15**2+51
2tP34tP35tP45¢t3tP15-512tP34tPSStt2#P45#P14tP15-256tP34tP35**2*
P45tP15tt2+1024tP34tP35tt2*P45*t2tP15+512tP34tP35tt3tP45tP15—25
6*P34¢P35#*3:P14*P15+256*P34*P35**3tP15*#2—256tP34tt2tP35tP45#P
15*t2+256*P34¢#2:P35*P45t#2PP15+256#P34#*2:P35*¢2¢P45¢P15+256¢P
34**2*P35¢*2tP15tt2-256tP35tP454:34P14tP15-256:P35¢P45##3*P15¢t
2+256tP35tP45t*4tP15-768tP35tt2tP45t#2tPl4tP15-512tP35tt2tP4SRt
2*P15t12+768*P35**2*P45t¢3*P15-7681P35**3*P45*P14*P15-256*P35**
3¢P45*P15t*2
TLPART=TLPART+768*P35*t3tP45**2*P15+256tP35tt4tP45tP15-256tP35t
1 t4tP14tP15
XTOTAL=XTOTAL+HDPARTtTLPART
HDPART=EVL
TLPART=256tP34tP35tP45tP15+256tP34*P35#*2:P15+256#P35#P45*t2tP1
1 5+512tP35tt2¢P45tP15+256¢P35tt3¢P15
XTOTAL=XTOTAL+HDPARTtTLPART
QF=XTOTAL
RETURN
END

tend

HHHI—‘t—‘D—‘HH

C Expression QG
COMPLEX FUNCTION QG(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45
COMPLEX EVL
COMPLEX TLPART,HDPART,XTOTAL

XTOTAL=O.
HDPART=1 .
TLPART=-512¢P34*P35tP45¢*2tP14tP15—256tP34tP35tP45tt2tP15#t2+10
24*P34PP35tP45t*3tP15-256tP34tP35tt2#P45tP14tPlS+256tP34¢P35tt2
tP45tP15tt2+512¢P34tP35ta2tP45t*2tP15-256tP34tP45:#3#P14:P15—51
2tP34tP45¢t3tP15tt2+512tP34tP4St*4tP15+256tP34tt2tP35tP45tP15tt
2+256tP34tn2tP35tP45t*QtPl5-256tP34tt2tP45tt2tP15tt2+256tP34t¥2
tP45tt3tP15-768tP35tP45t*3tP14tP15-512tP35tP45Rt3tP15**2+768tP3
StP4Stt4tP15-768tP35*t2nP45tt2tP14tP15—256tP35tt2tP45tt2tP15tt2
¢768tP35tt2tP45tt3tP15-256tP35tt3tP45tP14tP15+256tP35¢t3¢P45tt2
#P15-256:P45t:4tP14tP15
TLPART=TLPART-256tP45#¢4tP15tt2+256tP45ttStP15
XTOTAL=XTOTAL+HDPART¢TLPART
HDPART=EVL
TLPART=-256¢P34tP35tP45tP15-256tP34:P45tt2tP15-5124P35¢P45*#2#P
1 15-256tP35tt2*P45tP15-256tP4SCR3tP15
XTOTAL=XTOTAL+HDPARTtTLPART
QG=XTOTAL
RETURN
END

HHHHHHHH

tend

C Expression QH
COMPLEX FUNCTION QH(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45
COMPLEX EVL
COMPLEX TLPART,HDPART,XTOTAL

 

 

179

XTOTAL=O.
HDPART=1 ~
TLPART=-256*P34*P35tP45:P14*P15*t2+256tP34¢P35*P45*P15t¢3-512:P
34*P35tP45w#21P14tP15-768*P34*P35#P45t*2:P15*t2+512¢P34tP35tP45
tt3*P15-256tP34tP35t#2tP45*P14tP15-256tP34*P35tt2tP45tP15fi#2+25
6*P34tP35t:2tP45tt2tP15-512tP34tP35t*2tP14tP15t#2+256tP34tP35tt
2*P15*t3+256tP34tP45tt2tP14tP15*t2+256¢P34¢P45tt2tP15¢*3—256tP3
4tP45t#3tP14tP15-512uP34tP451*3*P15*#2+256¢P34¢P45tt4tP15-256:P
34**2‘P35*P45*P14*P15-1024*P34tt2&P351P45tpl5t#2+512‘P34*‘2‘P35
*P45*:2tP15-256tP34t*2tP35:P14tP15*#2+512tP34t#2¢P35tP15*¢3+256
tP34tt2tP45*P14*P15#*2
TLPART=TLPART+512tP34¢*2tP45tP15tt3-256tP34t#2tP45tt2tP14tP15~1
1 024*P34tt2tP45#t2tP15*¢2+512¢P34t#2tP45:t3tP15—512#P34#*3¢P45*P
1 15*t2+256¢P34t*3#P45t12tP15+256tP34¢t3tP15t*3-256tP35tP45tt2tP1
1 4*P15**2-256tP35tp45**2*P15**3+256*P35*P45*13*P15**2-512#P35t*2
1 *P45tP14xP15tt2~256tP35tt2tP45tP15*#3+512*P35tt2tP45tt2tP15tt2+
1 256~P35*t3tP45tP15#t2-256tP35t#34P14#P15*t2
XTOTAL=XTOTAL+HDPARTtTLPART
HDPART=EVL
TLPART=-256tP344P35:P45¥P15+512rP34*P35¢P15**2+256*P34*P45*P15#
1 *2-256tP34*P45tt2tP15-256tP34:t2tP45tP15+256xP34tt2tP15**2+256*
1 P35:P45*P15tt2+256tP35¢t2tP15tt2
XTOTAL=XTOTAL+HDPARTgTLPART
QH=XTOTAL
RETURN
END

HHHD—‘HHHH

C Expression ELC

tend

COMPLEX FUNCTION ELC(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45

COMPLEX EVL

COMPLEX TLPART,HDPART,XTDTAL

XTOTAL=O.
HDPART=1
TLPART=256*P34*P35tP45tt2tP14tP15+512tP34‘P35tP45tt2tP15ttQ—512
tP34tP35#P45#t3#P15+512tP34tP35tt2tP45tP14tP15+256¢P34tP35tt2tP
45*P15"2‘1024‘P34fip35#*2*P45fi*2‘P15-512*P34*P35**3*P45¥P15+256
tP34#P35¢:3tP14tP15-256nP34sP351t3tP15t#2+256tP34t#2tP35tP45tP1
5t:2-256tP34tt2tP35tP45tx2tP15-256tP34tt2tP35:#2tP45#P15-256tP3
4tt2tP35tt2tP15-t2o256tP35tP45tt3tP14tP156256tP35tP45tt3tP15t‘2
—256*P35tP45#t4tP15+768tP35#t2tP45Nt2tP14tP15+512tP35¢#2tP45##2
tPIStt2-768tP35tt2tP45tt3aP15+768tPastt3tP45:P14#P15+256#P35#:3
nP45tP15t¢2
TLPART=TLPART-758*P35"3‘P45**2#P15—255¥P35¢*4‘P45‘P150256tP35‘
1 t4tP14tP15

XTOTAL=XTOTAL+HDPART¢TLPART

HDPART=EVL
TLPART=256tP34tP35tP451P15+256tP34tP35tt2¢P15+256tP35tP45tt2tP1
1 5+512tP35tt2tP45tP15+256tP35tt3tP15

XTOTAL=XTOTAL+HDPARTtTLPART

ELC=XTOTAL

RETURN

END

HHHHHHHH

C Expression ELD

COMPLEX FUNCTION ELD(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45

 

 

hIF‘F‘h‘P‘P‘F‘PJ

h‘k‘h‘h‘h‘h‘

1
1

*end

C Expr

hlh‘h‘h‘hlh‘h‘h‘

180

COMPLEX EVL
COMPLEX TLPART,HDPART,XTOTAL

XTOTAL=O.
HDPART=1
TLPART=256#P34tP35tp45*P14*P15tt2+512tP34tP35*P45*P15##3+256*P3
4*P35tP45tt2tP14tP15-1280*P34tP35tP45tt2tP15tt2-256tP34tP35tP45
tt3tP15+512tP34¢P35tt2tP45*P14*P15-512*P34tP35tt2tP45tP15tt2—51
2tP34¢P35t#2tP45t#2—P15-256tP34tP354t2tP14¢P15**2—256fiP344P35tt
3*P45mP15+256tP34tP35t#3tP14tP15+512*P34tP45*t2tP14tP15t12+768t
P34tP45t*2tP15tt3-768tP34tP45:t3tP15*t2+256tP34¢#2#P35*P45*P14¢
P15-1024tP34tt2tP35tP45tP15tt2-512mP34*t21P35*P45tA2tP15-256tP3
4t*2tP35tP14tP15tt2+256tP34tt2tP35tP15tt3-512tP34tt2tP35tt2tP45
tP15
TLPART=TLPART+2561P34t*2tP35‘*2‘P14*P15-256*P34**2*P35**2*P15*$
2+256tP34t#2tP4SPP14tP15tt2+768¢P34tt2tP45tP15t*3-768tP34tt2tP4
5t*2tP15**2-256tP34t*3tP35tP45tP15-256mP34tt3tP35tP15tt2—256tP3
4**3tP45tP15*#2+256tP34tt3tP15**3+512*P35tP45*t2tP14tP15:*2+256
tP35¢P45t*2tP15**3-512tP35tP45**3:P15*t2+256tP35tt2tP45tP14tP15
ti2—256tP35tt2tP45:*2tPlS::2+256*P45**3*P14*P15tt2+256tP45tt3tP
15*:3-256*P45tt4tP15tt2
XTOTAL=XTOTAL+HDPARTtTLPART
HDPARTzEVL
TLPART=255*P34tP35¥P45tpl5-256tp34*P35*P15**2+256#P34#P35#*2#P1
5—5121P34tP45tP15tt2+256wP34t:2*P35:P15-256tP34tt2*P15t#2-256#P
35tP45*P15t*2—256tP45tt2tP15tt2
XTOTAL=XTOTAL+HDPART:TLPART
ELD=XTOTAL
N

END

ession ELE

COMPLEX FUNCTION ELE(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45

COMPLEX EVL

COMPLEX TLPART,HDPART,XTOTAL

XTUTAL=O.

HDPART=1

TLPART=-256tP34#P35#P45¢P14tP151t2-512tP34$P35tP45:P15##3-256#P
34:P35*P45¢t2¢P14tP15+1280*P34tP35tP45:t2tP15tt2+256tP34tP35¢P4
5tt3tP15—512tP34tP35tt2nP45tP14tP15+512tP34tP35*t2tP45tP15tt2+5
12*P34PP35tt2tP45-t2¢P15+256tP34tP35t¢2:P14tP15tt2+256tP34tP35*
t3tP45tP15—256#P34tP35t*3tP14tP15-512tP34tP4stt2tP14tP15tt2-768
tP34tP45¢#2tP15tt3+768¢P344P45t*3tP15#*2-256#P34tt2tP35tP45tP14
-P15+1024¢P34t¢2tP35tP45tP15¢t2+512RP34¢*2tP35tP45t*2tP15+256#P
34:t2RP35tP14tP15:t2-256tP34tt2tP35tP15tt3+512tP34t¢2tP35tn2tP4
5tP15

TLPART=TLPART-256tP34tt2tP35¢t2tP14tP15+256tP34t#2tP35tt2tP15—t

P‘h‘h‘h‘h‘h‘

2-256tP34t¢2tP45tP14tP15tt2-768¥P34##23P45#P15##3+768¢P34$¥2tp4
5t#2tP15*t2+256¢P34tt3tP35tP45tP15+256tP34t¢3tP35tP15¢t2+256tP3
4tt3¢P45tP15¢t2-256tP34t:3:P15tt3-512tP35tP45*t2tP14tP15tn2-256
¢P35tP451¢2tP15tt3+512¢P35¢P45tt3tP15tt2-256tP35tt2tP45tP14tP15
tt2+256tP35¢¢2tP45tt2¢P15¢t2-256tP45tt3tP14¢P15#:2~256tP45:t3¢P
15*t3+256tP45tt4tP15tt2

XTOTAL=XTOTAL+HDPARTtTLPART

HDPART=EVL

 

 

181

TLPART=256tP34tP35tP45tP15-256tP34*P35tP15t#2+256tP34tP35#*2*P1
1 5-512tP34tP45tP15tt2+256tP34t*2tP35tP15-256tP34tt2tP15tt2—2564P
1 35*P45tP15tt2-256tP45tt2tP15tt2
XTOTAL=XTOTAL+HDPARTtTLPART
ELE=XTOTAL
RETURN
END
tend

C Expression ELF
COMPLEX FUNCTION ELF(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45
COMPLEX EVL
COMPLEX TLPART,HDPART,XTOTAL

 

XTOTAL=O.
HDPART=1
TLPART=—256*P34tP35*P45t*2*P14*P15-512tP34tP35tP45tt2tP15*t2+51
2*P34tP35tP45#*3tP15-512*P34*P35tt2:P45tP14tP15—256tP34tP35*12t
P45tP15*t2+1024tP34tP35t¢2tP45t*2:P15+512tP34tP35*t34P45tP15-25
GXP34*P35t#3tP14tP15+256tP34tP35tt3tP15t*2~256*P34**2tP35tP45tP
15:t2+256*P34*t2tP35tP45tt2tP15+256tP34tt2tP35t*2tP45tP15+256¢P
34*#2tP35t*2:P15**2-256tP35tP45t*3PP14tP15-256tP351P45*#3:P15¢#
2+256tP35tP45t*4tP15-768tP35tt2tP45##QtP14tP15—512tP35:t2tP45*t‘
2*?15‘*2+768*P35**2‘?45**3‘P15-768#P35**3tP45tP14‘P15-256tp35fit
3tP45tP15tt2
TLPART=TLPART+768tP35**3*P45*#2#P15+255*P35**4*P45*P15-256*P35*
1 ¢4tP14rP15
XTOTAL=XTOTAL+HDPART¢TLPART
HDPART=EVL
TLPART=256tP34~P35tP45tP15+256¢P34*P35*tZtPlS+256tP35tP45tt2tP1
1 5+512tP35tt2tP45tP15+256tP35¢t3tP15
XTOTAL=XTOTAL+HDPARTtTLPART
ELF=XTOTAL
RETURN
END

HHHHHHHH

 

tend

C Expression ELC
COMPLEX FUNCTION ELG(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45
COMPLEX EVL
COMPLEX TLPART,HDPART,XTOTAL

XTOTAL=O.
HDPART=1
TLPART=-512*P34*P35tP45*#2¢P14#P15-256tP34tP35*P45tt2tP15##2+10
24tP34tP35tP45:#3tP15—256tP34tP35:t2tP45tP14tP15+256tP34tP35tt2
aP45¢P15tt2+512tP34tP35¢¢2¢P45:t2¢P15-256tP34tP45tt3tP14tP15-51
2tP34tP45tt3tP15¢t2+512tP34tP45tt4tP15+256tP34t¢2¢P35tP45tP15#t
2+256tP34tt2tP35¢P45tt2tP15-256tP34t42tP45:42¢P15tt2+256¢P34t¢2
tP45t*3tPls-768tP35tP45t:3tP14tP15—512tP35tP45tt3tP15tt2+768¢P3
5-P45tt4tP15-768tP35tt2tP45t:2*P14tP15-256-P35tt2tP45¢t2tP15tt2
+768tP35¢t2tP4Stt3tP15-256tP35*$3:P45#P14:P15+256tP35tt3¢P451:2
tP15-256tP45tt4tP14tP15
TLPART=TLPART-256tP45#¢4tP151¢2+256tP45tt5¢P15
XTOTAL=XTOTAL+HDPARTMTLPART
HDPART=EVL

HHHHHHHH

182

TLPART=-256*P34tP35tP45tP15-256tP34tP45tt2:P15—512*P35*P45t*2tP
1 15-256‘P35**2*P45$P15-256ip45*#3tpl5
XTOTAL=XTOTAL+HDPARTtTLPART
ELG=XTOTAL
RETURN
END

tend

C Expression ELH
COMPLEX FUNCTION ELH(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45
COMPLEX EVL
COMPLEX TLPART,HDPART,XTOTAL

 

XTOTAL=O.
HDPART=1
TLPART=-256*P34tP35tP45tP14tP15t#2+256*P34tP35*P45¢P15t*3-512tP
34tP35tP45t:2tP14tP15-768tP34tP35¢P45t*2tP153#2+512*P34*P35*P45
##3#P15-256tP34tP35:t2tP45tP14tP15—256*P34tP35t#2tP45tP15tt2+25
6*P34tP35t*2#P45t:2tP15-512tP34tP35tx2tP14tP15tt2+256¢P34tP35tt
2*P15t*3+255*P34*P45t42*P14fiP15*42+255tp34tp45#*2*P15**3-255’P3
4*P45tt3tP14tP15-512tP34tP45t*3tP15tt2+256tP34tP45tt4tP15-256wP
34*t2*P35tP45tP14*P15-1024tP34t#2:P35*P45¢P15¢t2+512tP34tt2tP35
¢P45tt2tP15-256*P34tt2:P35tP14tPlStn2+512tP34t¢2tP35tP15**3+256
tP34tt2tP45tP14tP15tt2
TLPART:TLPART+512$P34t*23P45*P15**3*256*P34#*2*P45**2*P14*P15-1
1 024tP34tt2¢P45*t2tP15t*2+512tP34tt2tP45tt3tP15-512tP34tt3tP45tP
1 15**2+256*P34#*3*P45**21P15+256*P34#*3‘P15**3-256*P35#P45$#2#P1
1 4tP15tt2-256tP35tP45tt2tP15nt3+256tP35eP45-t3tP15::2—512tP35tt2
1 tP45tP14tP15t#2-256tP35tt2tP45tP15xt3+512tP35t*2tP45tt2tP15#*2+
1 256tP35tt3tP45tP15tt2-256:P35**3tP14tP15t#2
XTOTAL=XTOTAL+HDPART¢TLPART
HDPART=EVL
TLPART=-256tP34tP35tP45tP15+512tP34tP35tP15:t2+256tP34tP45tP15t
1 *2-256xP34tP45tt2tP15-256tP34tt2tP45tP15+256tP34t*2tP15tt2+256t
1 P35tP45tP15¢t2+2561P35tt2tP15tR2
XTOTAL=XTOTAL+HDPARTOTLPART
ELH=XTOTAL
RETURN
END

HHHHHHHH

 

C Expression ELB

COMPLEX FUNCTION ELB(P13,P14,P15,P34,P35,P45,
11,I2,I3,I4,15,16,I7,IB,IQ,IIO,
I11,112,113,I14,I15,116,I17,118,119,
WIV,FPIIV,BOIV,CUOIV,EOIV,EVL '
D3) V

§+C+

REAL P13,P14,P15,P34,P35,P45
REAL WIV,FPIIV,BOIV,CUOIV,EOIV

COMPLEX EVL

COMPLEX I1,I2,I3,I4,I5,16,I7,I8,I9,IIO,I11,I12,I13,Il4,I15,
+ I16,I17,I18,119

COMPLEX 03

COMPLEX TLPART,HOPART,XTOTAL

XTOTAL = o i
HDPART = 1 T

 

*end

183
TLPART=256tP34tP351P45tP14tP15t*2-2561P34#P35tP4SRP15#*3+512*P3
4tP35tP454¢2¥P14*P15+768#P34#P35*P45*#2tP15*t2-512¢P34*P35#P45*
¢3*P15+256*P34#P35*t2tP45tP14*Pl5+256*P34tP35tt2tP4SMP15t:2-256
tP34tP35:t2tP45¢t2¢P15+512tP34tP35t¢2tP14tP15t*2-256:P34*P35:t2
:P15t*3-256tP34tP45t*2tP14*P15:t2-256tP34tP45tt21P15**3+256*P34
*P45t*3*P14#P15+512¢P34*P45*t3$P15tt2-256*P34tP45t¢4:P15+256*P3
4**2MP35*P45#P14*P15+1024tP34tt2:P35#P45#P15#*2-512:P34t*2*P35t
P45t*2!P15+256#P34*t2tP35tP14tP15tt2-512tP34tt2tP35tP15tt3-256t
P34##2#P45¢P14¢P15*t2
TLPART=TLPART-512*P34tt2tp45*P15**3+256#P34#t2tp45**2*P14*P15+1
1 024*P34tt2tP45*t2#P15tt2-512tP34tt2tP45t#3tP15+512*P34**3¢P45*P
1 15**2-256tP34t#3tP45r#2¥P15-256*P34tt3¢P15tt3+256¢P35tP45tt2tP1
1 4*P15tt2+256tP35¢P45tt2tP15tt3-256tP35tP45t*3:P15¢*2+512*P35tt2
1 *P45*P14*P15#t2+256¢P35¢*2tP45*P15*t3—512*P35t#2tP45tm2tP15**2-
1 256#P35**3*P45*P15*#242564P35*#3#P14*P15**2
XTOTAL=XTOTAL+HDPARTwTLPART
HDPART=EVL
TLPART=-256*P34tP35tP45tP15+512tP34tP35tP15tt2+256tP34tP45tP15t
1 t2-256tP34tP45tt2tP15-256tP34tt2tP45fiP15+2561P34t*2tP15tt2+256t
1 P35:P45*P15*t2+256tP35tt2¢P15tt2
XTOTAL=XTOTAL+HDPARTtTLPART
PLA=XTOTAL
RETURN
END

HHHHHHHH

C Expression PLB

tend

COMPLEX FUNCTION PLB(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45

COMPLEX EVL

COMPLEX TLPART,HDPART,XTUTAL

XTOTAL=0.
HDPART=1
TLPART=512tP34#P35tP45t‘2tP14tP15+256tP34tP35tP45‘*2*P15t*2-102
4tP34tP35tP45tt3tP15+256tP34tP35t*2tP45tP14tP15-256*P34:P35*t2:
P45¢P15¢t2-512tP34tP35t*2:P45¢t2tP15+256¢P34tP45¢~3¢P14tP15+512
tP34tP45ta3tP15tt2—512tP34tP45t¢4¢P15-256tP34#t2tP35¢P45tP15tt2
—256tP34t¢2¢P35tP45tt2tP15+256tP34tt2tP45t:2xP15tt2-256tP34ct2t
P45:t3tP15+768tP35cP45tt3tP14tP15+512xP35RP45nt3tP15tOZ-7684P35
tP4Stt4tP15+768tP35Rt2tP45t#2tP14tP15+256¢P35¢t2tP451t2tP15t:2-
768tP35at2tP45tt3¢P15+256tP35¢¢3#P45tP14vP15-256tP35tt3wP45tt2t
P15+256¢P45fit4¥P14tP15
TLPART=TLPART+256¢P45tt4¢P15¢t2—256tP45t¢5*P15
XTOTAL=XTOTAL+HDPARTtTLPART
HDPART=EVL
TLPART=—256#P34tP35tP45*P15-256tP34tP45t#2tP15-512tP35tP4SRt2tP
1 15-256:P35*t2tP45tP15-256tP45tt3wP15
XTOTAL=XTOTAL+HDPARTtTLPART
PLB=XTOTAL
RETURN
END

HHHHHI—‘HD—l

C Expression PLC

COMPLEX FUNCTION PLC(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45
COMPLEX EVL

 

 

184
COMPLEX TLPART,HDPART,XTOTAL
XTOTAL=O.
HDPART=1

TLPART=256*P34*P35*P45**2fiP141P15+512*P34*P35*P45#*2*P15*#2-512
tP34tP35tP45**3tP15+512tP34*P35t*2tP45:P14tP15+256*P34*P35tt2#P
45*P15t*2-1024*P34¥P35**2tP45**2*P15-512tP34*P35**3*P45*P15+256
tP34tP35*t3*P14*P15-256tP34tP35tt3tP15#*2+256*P34¢#21P35tP45tP1
5t*2-256tP34t*2tP35tP45t*2tP15-256tP34tt2tP35*t2tP45tP15-256tP3
4¢t2tP35#t2tP15tt2+256tP35tP45¢t3tP14tP15+256tP35¢P45t*3:P15tt2
-256*P35tP45*t4*P15+768*P35:t2¢P45t#2:P14:P15+512:P35tt2tP45¢A2
tP15*¢2—768*P35#t2*P45**3*P15+768*P35t*3tP45tP14tP15+256*P35¢t3
tP45tP15tt2

TLPART=TLPART-768tP35##3tP45t¢2*P15~256#P35**4tP45tP15+256*P35t

1 t4~P14xP15 '

XTOTAL=XTOTAL+HDPART:TLPART
HDPART=EVL
TLPART=256*P34tP35tP45tP15+256tP34tP35¢t2¢P15+256¢P35¢P45tt2tP1
1 5+512tP35tt2tP45*P15+256tP35tt3RP15
XTOTAL=XTOTAL+HDPARTtTLPART
PLC=XTOTAL
RETURN
END

HHHHHHHH

tend

C Expression PLD
COMPLEX FUNCTION PLD(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45
COMPLEX EVL
COMPLEX TLPART,HDPART,XTOTAL

XTOTAL=O.
HDPART=1
TLPART=256*P34tP35nP45#P14tPl5**2+512mP34tP35#P45*P15¢R3+256tP3
4¢P35tP45¢‘2tP14*Pl5-1280tP34tP35tP45¢t2tP15*t2-256tP34tP35*P45
tt3tP15+512tP34tP35tt2tP45tP14tP15-512tP34tP35mt2tP45tP15t#2-51
2tP34tP35tt2tP45tt2tP15-256tP34tP35tx2tP14tP15tt2-256tP34tP35t:
3-P45:P15+256:P34#P35¢#31P14tP154512tP34tP45t¢2tP14¢P15t:2+768¢
P34tP45tt2tP15tt3—768tP34tP45t#3tP15*¢2+256tP34:#2tP35*P45tP14t
P15-1024tP34t¢2IP35tP45fiP15t*2-512tP34fit2tP35tP45tt2tP15-256tP3
4¢t2tP35tP14tP15tt2+256¢P34tn2tP35xP15tt3-512tP34tt2tP35tt2tP45
tPlS
TLPART=TLPART+256tP34¢#2tP35tt2tP14tP15-256tP34t¢2tP35¢t2tP15¢t
2+256tP34st2xP45tP14tP15:t2+768tP34:t2tP45tP15tt3—768tP34tt2xP4
5tt2tP15tt2-256tP34tt3¢P35#P45¢P15-256:P34tt3tP35tP15RR2-256eP3
4tt3tP4SXP15tt2+256tP34tt3tP15tt3+512tP35tP45tt2tP14tP15t¢2+256
tP35tP45tt2¢P15¢t3~512tP35¢P45tt3tP15¢#2+256*P35:m2¢P45¢P14¢P15
tt2-256tP35:t2tP45tt2tP15tt2+256tP45tt3tP14¢P15*#2+256tP45tt3tP
15tt3—256tP45tt4tP15tt2
XTOTAL=XTOTAL+HDPART¢TLPART
HDPART=EVL
TLPART=256#P34tP35tP45*P15—256tP34¢P35¢P15t*2+256tP34¢P35**2:P1
1 5-512¢P34tP45tP15tR2+256tP34tt2tP35tP15-256tP34tt2tP154t2-256tP
1 35tP45tP15tt2—256tP45tt2xP15#:2
XTOTAL=XTOTAL+HDPART¢TLPART
PLD=XTOTAL
RETURN
END

HHHHHHHH

HHHHHH

 

 

 

:end 185

C Expression PLE
COMPLEX FUNCTION PLE(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45
COMPLEX EVL
COMPLEX TLPART,HDPART,XTOTAL

XTOTAL=O.
HDPART=1
TLPART=-256*P34tP35tP45tP14tP15t*2-512tP34tP35tP45tP15tt3—256tP
34:P35tP45**2tP14tP15+1280tP34tP35*P45¢¢2tP15t*2+256tP34tP35tP4
5tt3¢P15-512tP34*P35tt2tP45tP14tP15+512*P34tP35tt2tP45tP15tt2+5
12tP34tP35:*2tP45:*2tP15+256tP34#P35t*2tP14tP15*#2+256#P344P35t
t3tP45tP15-256:P34*P35*t3*P14tPI5-512tP34tP45t:2tP14*P15**2—768
#P34xP45tt2tP15tt3+768tP344P45tt3tP15tt2-256tP34t#2tP35tP45*P14
*P15+1024#P34*t2tP35tP45¢P15t*2+512#P34*t2tP35*P45**2*P15+256¢P
34tt2tP35tP14tP15¢*2-256tP34tt2tP35tP154t3+512tP34tt2¢P35**2tP4
5*P15
TLPART=TLPART-256*P34¢t2tP35t:2tP14tP15+256tP34:*2:P35**2*P15t¢
2—256tP34t*2tP45tP14*P15t#2-768tP34t*2*P45*P15**3+768tP34**2xP4
5tt2tP15t*2+256tP34t*3*P35tP45*P15+256*P34**3:P35*P15*t2+256¢P3
4t*3tP45tP15t#2-256¢P34#*3tP15**3-512:P35*P45#t2tP14tP15tt2-256
tP35tP45tt2tP15*t3+512*P35tP45tt3tP151t2-256tP35t#2tP451P14tP15
tt2+256tP35t*2tP45tt2tP15tt2-256tP45*t3tP14tP15tt2-256tP45tt3tP
15t¢3+256tP45et4tP15tt2
XTOTAL=XTOTAL+HDPARTtTLPART
HDPART=EVL
TLPART=256tP34tP35tP45tP15-256tP34tP35tP15tt2+256*P34*P35*#2#P1
1 5—512tP34tP45tP15tt2+256tP34tt2tP35tP15-256tP34tt2tP15*t2—256*P
1 35¢P45tP15t*2-256tP45t*2tP15t*2
XTOTAL=XTOTAL+HDPARTRTLPART
PLE=XTOTAL
RETURN
END

HHHHHHHH

HHHHHH

#end

C Expression PLF
COMPLEX FUNCTION PLF(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45
COMPLEX EVL
COMPLEX TLPART,HDPART,XTOTAL

XTOTAL=O.
HDPART=1
TLPART=-256tP34tP35tP45tt2#P14tP15-512tP34tP35tP45tt2tP15tt2+51
2tP34¢P35tP45¢#3tP15—512tP34tP35t#2tP45tP14tP15-256tP34#P35t#2:
P45¢P15*t2+1024tP34tP35tR2¢P45#t2tP15+512sP34tP35*t3tP454P15-25
6tP34RP35tt3tP14tP15+256tP34tP35t¢3tP15tt2-256tP34tt2tP35tP45tP
15¢t2+256tP34t#2tP35tP45t#2:P15+256*P34**2¢P35:a2tP454P15+256tP
34¢:2tP35t*2tP15tt2-256tP35‘P45tt3—P14tP15—256#P35¢P45tt3tP15tt
2+256tP35tP45:t4tP15—768:P35*t2tP45tt2tP14tP15—512tP35tt21P45tt
2tP15*:2+768tP35#t2tP45tt3tP15-768:P35¢n3tP45tPl4:P15-256:P35t:
3tP45tP15tt2
TLPART=TLPART+768tP35¢t3tP45t:2tP15+256tP35tt4tP45tP15-256tP35t
1 t4tP14¢P15
XTOTAL=XTOTAL+HDPARTtTLPART
HDPART=EVL

HHHHHHHH

 

 

186
TLPART=256tP34tP35tP451P15+256tP34tP35:*2:P15+256¢P35¢P45**2*P1
1 5+512¢P35tt2tP45tP15+256tP35tt3tP15
XTOTAL=XTOTAL+HDPART*TLPART
PLF=XTOTAL
RETURN
END

tend

C Expression PLG
COMPLEX FUNCTION PLG(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45
COMPLEX EVL
COMPLEX TLPART,HDPART,XTOTAL

XTOTAL=O.
HDPART=1
TLPART=-512¢P34tP35tP45tt21P14*P15-2561P34*P35¢P45*t2¢P15tt2+10
24#P34*P35*P45t*3tP15-256tP34tP351*2:P45tP14tP15+256¢P34tP35**2
tP45tP15t42+512tP34iP35tt2*P45*t2tP15-256tP34tP45tt3tP14tP15-51
2*P34tP45tt3tP15tt2+512¢P34tP45t*4*Pl5+256*P34**2*P35*P45*P15nx
2+256tP34tt2tP35tP45**2tP15-256tP34t*2tP45t*2:P15t*2+256tP34*t2
*P45t:3*P15-768tP35wP45*t3*P14*P15—SthP35tP45**3¢P15tt2+768tP3
5#P45t*4tP15—768tP35tt2tP45t42:P14tP15-256#P35$t2:P45*t2tP15¢*2
+768tP35tt2tP45tt3tP15-256¢P35t*3:P45*P14#P15+256#P35¢*3tP45*t2
tPlS—256tP45tt4tP14tP15
TLPART=TLPART—256¢P45tt4tP15t:2+256¢P45*#5tP15
XTOTAL=XTOTAL+HDPARTRTLPART
HDPART=EVL
TLPART=~256tP34tPBStP45tP15-256:P34*P45**2tP15-512tP35tP45*t2*P
1 15—256*P35##2tP45¢P15-256¢P45¢#3tP15
XTOTAL=XTOTAL+HDPARTxTLPART
PLG=XTOTAL
RETURN
END

HHHHHHHH

tend

C Expression PLH
COMPLEX FUNCTION PLH(P13,P14,P15,P34,P35,P45,EVL)
REAL P13,P14,P15,P34,P35,P45
COMPLEX EVL
COMPLEX TLPART,HDPART,XTOTAL

 

XTOTAL=0.
HDPART=1
TLPART=-256tP34tP35tP45tP14tP15:t2+2561P34~P35¢P45tP15tt3-512tP
34*P35tP45tt2¢P14tP15-768tP34:P35:P45#¢2*P15xt2+512tP34tP35tP45
##3#P15-256tP34tP35tt2*P45tP14tP15-256tP34tP35tR2:P45¢P15tt2+25
6tP34tP35t#2#P45tt2tP15-512#P34tP35*#2tP14tPlSttZ+256tP34tP35tt
2tP15tt3+256tP34tP45tt2¢P14tP15tt2+256tP34tP45#t2tP15tt3-256tP3
4tP45t*3MP14tP15-512tP34tP45tt3tP15*t2+256tP34tP45¢¢4¢P15—256¢P
34#*2‘P35*P45¥P14tP15-1024*P34**2iP35tP45*P15##20512tp34‘*2‘P35
tP45tt2tP15-256tP34t#2tP35tP14:P15:t2+512tP34cR2tP35tP15tt3+256
fiP34tt2‘P45tP14tP15‘*2
TLPART=TLPART¢512tP34tt2tP45:P15*t3-256tP34tt2tP45tt2tP14tP15-1
1 024tP34tt2¢P45¢t2tP15:t2+512¢P34tt2*P45:t3tP15-512tP34tt3tP45tP
1 15*R2+256tP34nx3xP45tt2tP15+256tP34tt3tP15¢t3-256tP35tP45tt2tP1
1 4tP15tt2-256tP35tP45tt2tP15tt3+256tP35tP45$t3tP15tt2-5121P35tt2
1 tP45tP14tP15tt2—256tP35:t2:P45tP15t¢3#512tP35tt2tP45tt2tP15tt2¢

HHHHHHHH

 

 

187

1 256tP35t#3#P451P15¥t2-256fiP35tt3tP14tP15tt2
XTOTAL=XTOTAL+HDPART¢TLPART
HDPART=EVL _
TLPART=—256tP34tP35tP45tP15+512*P34fiP35tP15tt2+256tP34tP45tPISR
1 t2—256tP34tP45tt2tP15-256tP34t#2tP45*P15+256#P34tfi2¢P15¥¢2+256¢
1 P35tP45tP15tt2+256tP35tt2¢P15112
XTOTAL=XTOTAL+HDPARTtTLPART
PLH=XTOTAL

END

 

 

 

 

 

Appendix P
Ascertaining the Physical Range of
Parameters

All interacting systems of two particles in and three particles out possess six
degrees of freedom- five continuous ones and a plus-minus degree which repre—
sents the reflection of the three output particles across the plane defined by the
two input particles.

P3

P2————- ——————
P4

 

 

l ' P5

It is common to work in the center of mass system and represent the degrees
of freedom by the energies and directions of the particles. This representation
is desribed here though not used. For this calculation it is more appropiate to
work in the rest frame of P3 and E. with the degrees of freedom representated
by ranges of dot products.

Section 1 Center of Mass System
Sub-Section 1.1 ...Fixing the Parmneters

For the input particles, one parameter EC," fixes their values:

E E

u_ cm cm
P1 —(O,0.—-—2 .——2)
.E.", E",

u- ...._'- _-_
P2 —(0.0. 2 . 2 )

(The assignment of the minus sign to one of the input paticle's momentum is
the fixing of the plus-minus degree of freedom.)
For the output particles. three parameters fix the «:liro-cticms:

Pick particle 4 to be in the x-z plane:

P4“ = [E4sin(94). O. E4cos(fi4). E4]

188

 

 

 

189

and 04 becomes the second parameter.

Then 65 and #5 become. the third and fourth parameters. fixing the direction of
P5".

P5” : [E5sin(65)cos(¢5), E5si11(95_)sin(d>5) E5 cos( (05) E]
and P3 becomes fixed by energy-momentum conservation:

P3” = {—(P43+P5r), —(P4y+P5y), —(P4:+P52)! (Erin —E4-E5l]

Now select E3 as the fifth and last parameter.

Then E4 and E5 are determined by the simultaneous solution of:
E4 + E5 = Ecn) _ E3

and the zero mass of P3
1932 = 0 =>

E3': [E4 s111((94)+E5 sin(05)c os(a’>5]' +[E5 sin((95)sin(q35)]' +[E4 cos(94)+E5 cos(65 )]2

Sub-Section 1.2 The Center of Mass Outgoing Polar Angles

To find the polar angles made by the three outgoing particles in the center
of mass frame of reference system. first note. taking particle 3 as an example.

that:
(P1+P2)-P3=—- --(E1+E2)E3=— 2(E1)E3

but also:
‘ (P! + P2) - P3 = (P3+ P4 + P5) - P3 = P3P4+ P3P5
giving:
4 P3Pr
(E1)E3 __P3P :2 0

Then to find cos 63 :

_-1 P3: lP1||P3| ”0‘93——(El)53 mwg

but also:
' 1~ 3=P1P3+(E1)E3

 

 

 

190

giving:
P1P3
6 = 1 —-—-
“’5 3 + (E1 1133
then using the above for (E1)E3 :
2(PlP3)

cos 93 = l — —*
P3P4 + P3P5

and similarly for cos 04 and cos 65:

2(P1P4)
P3P4 + P4P5
_ 2(P1P5)

P3P5 + P4P5

cos ()4 = 1—

cos ()5 =

Section 2 Another System
For this problem the chosen set is:

the five continuous-
P 1P4
P 1 P5
P3P4
P3P5
P4P5

 

and the plus-minus degree-
the sign of Epf(Pl.P3. P4,P5)

As these are all Lorentz invariants, the combinations of values that are physically
allowable may be ascertained in any convenient reference frame and co-ordinate
system. The reference frami- chosen here is the usual one for this problem- the
rest frame of W and P5 . The co—ordinate system is with W taken as the plus
2 direction and P73 in the x-z plane.

The four vector P2 is removable by energy-momentum conservation

P2=P3+P4+PF—Pl

The remaining four vectors- P1.P3.P4. P5 - have their components evaluated

 

191

for this co—ordinate system in the solutions to Integrals 15. 18. and 118. in Ap-
pendix N.
The results are» using as usual [:0 = \/-P4P5/2 :

P4 = (0, 0. +110, k0)
P5 = (0. 0, -ko. k0)

\/(P3P4)(P3P5) 0 (P3P4— P3P5) _(P3P4+ P3P5)
k0 ' ' 21:0 ‘ 21m

(P1P4 - P1P5) _ (P1P4 + P1P5)]

 

 

 

 

”=1

 

 

P1: [Pl”P1”' 2ko 21:0

where

 

P1 _i (P1P4)(P3P5)+(P1P5)(P3P4)-(P1P3)(P4P5)
"' — 2ko\/(P3P4)(.P3P5)

P 2
13’ 4k03(P3P4)(P3P5)

P1P3 appears in the above results. It is obtained from the above expression for
P2 by squaring and noting the particles have zero mass-

P22 = 0 = P3P4 + P3P5 + P4P5 — P1P3 — P1P4 — PIP-5
Selecting a value for the Center of Mass energy imposes a restriction on the set
of dot products

(P1+ 192)2 = (P3 + P4 + 1913)2 = -E;-’,,,

and removing P2 as usual—

P1P3 + P1P4 + P1P5 .—_ P3P4 + P3P5 + P4P5 = —E,f:’,,, /2

The allowable values Inay be visualized as lying rm triangular snrfm-ns

_ 4(P1P4)(P3P5)(P1P5)(P3P4) — [(P1P4)(P3P5) + (PlP5)(P3P4 — (P1P3)(P4P5)]

1)

192

-PlP3 -P3P4

-P1P-5 -P4P-5

 

\

-P1P4 -P3P5

o o ' E2
where each corner of the triangles mtersects the axis at —.,m .

Thase points are generateable in the output FORTRAN program by allowing

0 < -P1P4 < 5%
EB",
0 < —P1P5 < ——2'———(—P1P4)
E2

—P1P3 : —-§1'- — (—P1P4) — (—P1P5’)

and q
Ezm

2

 

O < -P3P4 <

E?
0 < -—P3P5 < —§"l—(—P3P4)
Efm

2 — ('—P3P4) - (—P3P5)

 

—P4P5 :

These possibilities are then restricted by the requirement that P1y be real.
which is that-

4(P1P4)(P3P5)(P1P5)(P3P4) > [(P1P4)(P3P5)+(P1P5)(P3P4)—(P1P3)(P4P5)]1’

A set of six clot products which is on the two triangle-s and also obeys this
restriction may be regarded as a physically realizable point and [.‘rrcv'essed.

The remaining degree of freedom. the plus-minus sign appears as that Sign

 

193

which is selected for MP1?” (with P23, picking up the opposing sign ). As

Epf(P1.P3.P4,P5), evaluated in Appendix B in this same reference frame
and co-ordinate system is equal to i(2k02)(P3.,)(Ply). the sign of Ply enters
the calculation as the sign of Epf(P1. P3, P4..P5).

 

Appendix Q
Verification of Spinor Replacement

Section 1 Introduction

In Appendix B it is demonstrated that a spinor product of the form v(p)i?(q)
may be replaced by (1 i 75h - A where A is a. four vector. and is given by

:1 = (77 ° Sign + (q ' SlPu - ('17 ' q)s,, i Epf(p.. q.p.s)
" \/16(P - sxq . s')

where s" is an arbritary four vector of the problem. and the sign of 75 is deter-
mined by the combination of helecities.

This permits the substsitutions:

«Pawn-1) = (1:1: 75 )7 . B

 

 

 

and
v(P2)TI(P5) = (1 :1; 75 )7 - CU
where
B _ (P3P5)P4,, + (P4P5)P3u —- (P3P4)P5,, i Epf(11.. P3. P4. P5)
“ ' ./16(P4P‘5)(P3P5)
and

_ (P1P2)P5.. + (P1P5)P2,, - (P2P5)P1,. i Ep m1. P1. P2. P5)

. f _
CI “ 16(P1P2)(P1P5)

which enable the SCHOONSCHIP program of Appendix I to easily calculate
the TRACES.

Here this result is checked in two examples by squaring the amplitude it leads
to and comparing it to the cross section obtained by the standard method spm

sum replacement of: 11(1)) 17(1)) = ‘1 '17

194

195

Section 2 Example 1

Consider the following simplistic diagram in which the polarization of the inci-
dent gluon and the exiting Z O are simply set equal to their linear momenta:

 

 

 

P2 _ _ _ _ _ _fzf P3
'7 :2 P3
n
P! m; P5
(1 = P1

Overall energy momenta conservation dictates:
P1+ P2 : P3 + P5
which gives immediately. using zero mass particles:
P 1P2 = P3P5

P2P3 = P1P5
P1P3 = P2P5
P2P5 = P3P3 - P1P5

and the amplitude is proportional to:

E(PS') 7"61 7'" 7"62 NP?)

Then using

(7.5,)(741) = (7-P1)7-(P1-P5) =—[(7-P1)(7495)]:[(7-P5)(7-P1)-P1P5l

gives:
E(PS) [(7 - P5) (7 PI) — P1P5] (7- «2) v(P2)
and because of the Dirac equation i7(P5) 7 - P5 = 0 becomes:
—(P1P5) TRACE [r(P2)U(P5)1 .62]
and using the above definition of C 'I ’:

—(P1P5) TRACE [(1 +7s)(7 -CC') (1 42)]

 

M6

= —4(P1P5)[CU . P3]

Using the expansion of CU:

= —4(P1P5)[(P1P2)P3P5 + (P1P5)P2P3 — (P2P5)P1P3 :1: Em P3. P1. P2. P5)]
\/16(P1P2)(P1P5)

 

noting P3 = P1 + P2 — P5 makes the Ep f zero

2 ‘4lP1P5lw- (WP-5)"
16(P1P2)(P1P5)

 

 

_ —4(P1P5)[—2(P3P5)(P1P5)]
' 16(P1P2)(P1P5)

 

and squaring to produce the cross section

Amplitude? = 4(P1P5)2(P3P5)P2P3

This is to be checked with the spin summed cross section:

E(PS) 7-61 7-12 7-63 v(P2)ii(P2) '7 -€2 7-17 7-61 u(P5)

= —(PIPS)2 TRACE [ll—E11) 7:135 7-P3 7 -P2 7-193]

_—_ 4(P1P5)3(P3P5)P2P3

giving agreement.
Section 3 ..... Example 2

Consider a second example :

(it-#87517" (CT-+015)» P3
_2_\ ..... .\.<
P4
11

Pl 7-! P5

 

 

197

First evaluate the spin summed cross section.

The quark line. yields:
E(P5) 7-6 7‘-n 7‘” 11(P2)T1(P2) 711 7.12 7-6‘ u(P5)

= TRu‘iCET U(P5)E(P5) .7. . 6 7, . 1l- 7.11 (_£_-;_’)_5_) (7 . 132).“! 7. '11- 7 . (a ]

then using the Example 1 relationship between 6. n. and the Dirac equation:

= 4.___—(P12P5l' TRACE[(1+75) 7~P5 7'” 7 '1’? 7'" l

r 2
= 412211 [4( 125,192“ + P5,.P2. — (P2P5lgpu) — 4 Epf(P5.u. P2. ”)1

and from the lepton line:

T11194) 7‘uv(P3W(P3) Vim-(P4)

TR.4CE[ (1+75) ‘1"P4 7., 7'-P3 7]

l
2
(4)[4 (P4"P3“ + P4"P3" — (P3P4)g‘“’) — 4 Epf(P4.11., P3. V) ]

_ l
— 2
and the product:
= 16(P1P5_)2{(P4P5)P2P3+P2P4)P3P?—P3P4(P2P5)-EPf(P4. P2. P3. P5)
(P3P5)P2P4 + (P2P5)P4P-r — P3P4(P2P5) -— Epf(P4. P4. P3. P2)
-(P3P4)P2P5 — (P3P4)P2P5 + 4(P3P4)P2P5 + 0
-Epf(P5.P3, P2.P4) — Epf(P5.P4. P2.P3) + 0

+[Epf(P5. 11, P2. v1] ... [Epf(P4.11. P3. um

” P4P5 P3P5
=16(P1P5)~[ 2(P4P5)P2P3+2(P2P4)P3P5+2l P2194 p2p3 l l

= 16(P1P5)2 4 (P4P5)P2P3

Now calculate and compare the amplitude squared.
if(P4) ‘1 '6 7" 1‘(P3)‘17(P5) ’1 '6' ”.- "1 7,, F(PL’)
= TRACE [v(P3)‘1Z(P4) 7 -€ 7" ] TR.4C'E[I'(P2)TI(P5) 7 -c' 7 ~11 7‘, ]

198

= TRACE [(1+75) 7-8 7" ]2(P1P5) TRACE[(1+7'5) 7 -CU 7,. ]
= 4 (13") (P1P5) 4 (CU..)

 

Then using
B _(P3P5)P4 +(P4P5)P3 — (P3P4)P5,.i 13pm P3 P4 P5)
”= ./16(P4P.5)(P3P5)

and a slightly different CU
(P2P3)P5u+(P3P5)P2u — (P2P5)P3 — Epf“: P3 P2 P5)

CU”:
./16(P1P3)(P3P5)
with
B - CU = o + P4P5(P3P5)P2P3 + P4P5(P2P3)P3P5 + o
—P3P5(P2P5)P3P4 + P3P52(P2P4) + P2P3(P3P5)P4P5
—P3P5 [ Epf(P4, P3, P2.P5)] — P3P5 [ 5mm. P3. P4. P5)]
; [Epf(p,P3.P4,P5)]* [Epf(p.P2,P3,P5)]

= 3(P4P5)P2P3(P3P5) — P3P5(P2P5)P3P4

P2P3 P2P4 P2P5
P32 P3P4 P3P5

+P3P53(P2P4) —
P3P5 P4P5 P52

 

all over appropiate denominator,
4(P4P5)P2P3(P3P5)

= ______________———
\/16(P3P4)(P3P5)./16(P1P3)(P3P5)

.—_ 4(16)(P1P5)2(P4P5)(P2P3)

 

giving agreement.

Now do again using the regular projection of ('1':

(P1P2)P5u+(P1P5)P2u-(P2P5)Plu wiEpflp Pl. P2. P5)

CUu=16(P1P2)(P1P5)

199

This will give a result that SCHOONSCHIP wiil be asked to calculate the final
step of.

B . CU =
-P4P5(P2P5)P1P3 + P4P5(P1P5)P2P3
+P4P5(P1P2)P3P5 - (P4P5) Epf(P3. P1. P2. P5)
-P3P5(P2P5)P1P4 + P3P5(P1P5)P2P4
+P3P5(P1P2)P4P5 - (P3P5)Epf(P4. P1. P2. P5,)
+P3P4(P2P5)P1P5 - P3P4(P1P5)P2'P5 + (P2P5)Epf(P1. P3, P4. P5)
-(P1P5)Epf(P3, P3, P4. P5)+ [Epf(p. P3. P4. P5)] 4 [Epf(p. P1. P2. P5)]

= —P4P5(P2P5)P1P3 + P4P5(P1P5)P2P3 - P3P5(P1P5)P2P3
-P3P5(P2P5)P1P4 + P3P5(P1P5)P2P4 + 2(P4P5)P1P2(P3P5)
+(P2P5 + P1P5 - P3P5 + P4P5) Epf(P1. P3. P4. P5)

P1P3 P1P4 P1P5
+ P2P3 P2P4 P2P5
P3P5 P4P5 P53

= -2(P4P5)P2P5(P1P3 + 2(P4P5)P1P5(P2P3)
+2(P4P5)P1P2(P3P5) + 2(P4P5) Epf(P1. P3. P4. P5)

_ 2(P4P5) [P1P2(P3P5) + P1P5(P2P3) - P2P5(P1P3) + EPflPl. P3. P4. P5)]

 

 

\/16(P1P2)P1P5(16)P3P5(P4P5

This is evaluated by the attached SCHOONSCHIP program to be:
Amplitude? = 64(P1P5)"-(P4P5)P2P3

giving agreement.

Appendix R

Verification of SCHOON SHIP LOOP
Program, TRACE Calculation

200

13nx><><><><><><><nnnznn<>mnnnnnnnnnnnnn

P

201

START SCHOONSHIP PROGRAM
ALGEBRAIC TEST OF LOOP - TRACE PROGRAM

this program tests the spinor, polarization replacements
and the expansion of Ku , by after these replacements
have all been made, replacing Ku by an expression that
should force a zero result , as per this appendix,

if all these replacements have been made correctly.

LNMP,LNMQ,PI,FPIIV
P14,P13,P15,P34,P35,P45
P1,P2,P3,P4,P5,Sl,$2,$3,$4,U1,U2,U3,U4

FEED IN VALUES FOR THIS RUN
20

SET SWITCHES

FOUR = 0
FIVE = 1
ELECT = 1
POSIT = O
LHEL = ‘1
QHEL = +1
GHEL = +1
MPARIT = +1

ninput

Outlimit,12000000

OOOOOOXXXXOGOODXXXXXXOV

C

nlist
SET DOT PRODUCTS

ELECTtP35 + POSITeP45
ELECTeP45 o POSITePSS

P3P5
P4P5

P1P3 = ELECT#P13 + POSITeP14
P1P4 = ELECTeP14 + PUSITePl3
P1P5 = P15
P3P4 = P34

FOR CONVEINENCE , DEFINE SOME EXPRESSIONS AND THEIR RECIPROCALS.
0145 = P1P4 o P1P5
01 = 0145

0345 = P3P4 o P4P5
MZP = 2eP3P4 + 03

now K CAN ALWAYS as EXPANDED ON AN ARBRITARY BASIS SET 51,sz,ss,s4
K = C(1):Sl . C(2)a52 4 C(3):$3 4 C(4):S4
AND FOR THIS PROBLEM we use THE ssr P1,P3,P4,P5

202

C
C SO.
C
X 5152 = P1P3
X 3153 = P1P4
X 5154 = P1P5
X 5253 = P3P4
X 5234 = P3P5
X 5354 = P4P5
C
C TO THIS END, THE ORIGINAL EXPANSION OF K MAY BE DOTTED WITH EACH
C OF THE BASIS TO FORM A 4 X 4 MATRIX, WHICH WHEN INVERTED
C YIELOS ......
C
C C(l) = CC(1,1)t$1DK + CC(1,2)#SZDK + CC(1,3)e$30K + CC(1,4)eS4DK
C
E W IS THE DETERMINENT FOR THIS INVERSION AND WIV IT’S RECIPROCAL
X W = 5132:5354e(-Sl$2¢S3S4+Sl$4eSZSS+SIS3eSZS4)
+ 5153.52544( 515245354451$445253-515345254)
+ 5154:5253e( $152eSSS4-S154eSZS3+SIS3eSZS4)
C
C AND NOW, ALL THE CC’S USING THE NOTATION CC(4,3) = CC43
C
X CC44 = WIVt(-2)t($1$2t5133#3253)
X CC43 = WIVtSIS2e(-Sl$2¢S3S4+SIS4#52$345153.5284)
X CC42 = WIVe51$3e( 5152:5334+SIS4¢5253-5133t5234)
X CC41 = WIVt5253¢( $152¢S3$4-SIS4tSZSS+Sl$3e$2$4)
X CC33 = WIVe(—2)e(8152eSIS4eSZS4)
X CC32 WIVtSIS4t( 5152:3354-51S4eSZSS+5153eSZS4)

w1v45254.( 5152453544515445233-515345254)
WIV:(-2)¢(Sl$3:$1$4¢$3$4)
w1v45354-(-$1$2.$354+$1$4452534515345254)
WIV:(-2)4(S253¢SZS44$3S4)

CC31
CC22
CC21
CC11

fix

fitfi“¢$tttfifi¢fitfifitfitttttttt‘fittittt...fitttttfitttfitttttttfitttttt‘

IN THIS SECTION, TAKE THE TRACE, REPLACE EACH K BY IT’S

LINEAR EXPANSION,
AND REDUCE TO A FINAL ANSWER.

,P1,P2,P3,P4,P5,PO
3,F4,F5,F6,F7,F8,F9,F10,F11,A1,A2,A3,AA
K3,K2

SCALE FACTOR FOR THE AMPLITUDE IS ...
FSCAL = 0.5:FPIIVO(1/137):(FIVE:(P34+P35+P4S) + FOUR)

TAKE THE TRACE

nonnxnnnnHm<nnnnnnnnnne fiXXXX

 

 

 

203

g EXP=FSCALtFOtF1tF2¢F3tF4¢F11tFSeFGtF7eF8¢F9¢F10
E SUBSTITUE ACCORDING TO HELICITIES AND DIAGRAM
Id,F0=O.5¢((1+LHEL)¢(66(J1))+(1-LHEL)¢(G7(J1)))
Id,F1=(G(J1,B))

Id,F2=ELECTt(G(J1,M2)) + POSIT¢(G(J1,M1))
Id,F3=(G(J1,K))

Id,F4=ELECT¢(G(J1,M1)) + POSIT:(G(J1,M2))
Id,F11=O.5e((1+QHEL)¢(GB(J2))+(l-QHEL)¢(G7(J2)))
Id,F5=(G(J2,CU)) .

ryep

Id,F6=(G(J2,M2))

Id,F7=(G(J2,M))

Id,F8=FIVE¢(G(J2,E)) + FOUR¢(G(J2,M1))
Id,F9=(G(J2,N))

Id,F10=FIVE¢(G(J2,M1)) + FOUR¢(G(J2,E))

ryep
Id,Trick,Trace,J1

C

C TRICK AND TRACE
C

Id,Trick,Trace,J2

*yep
c

C START CONDENSATION SUBSTITUTIONS, SOME DIAGRAM DEPENDENT.
C

Id,2,Dotpr,N(J') = FIVEt(M(J)+P1(J)) - FOUR¢(P3(J)+P4(J)+P5(J))
Al,Punct,N(K2') = FIVE:(M(K2)+P1(K2)) - FOUR.(P3(K2)+P4(K2)+PS(K2))
Id,MDM=O

Id,00tpr,M(J”)=K(J)-P0(J)

Al,Funct,M(J')=K(J)-P0(J)

Id,KDPD=P4P5

Id,KOK=O

Id,2,00tpr,P0(J')=P4(J)+P5(J)

Al,Funct,PD(J')=P4(J)+P5(J)

Id,P4DP4=O

Al,P50P5=O

Al,EOP1=O

C P OUTPUT

' C Id,2,Dotpr,K(J") = P1(J)

C Id,2,Funct,K(J') = P1(J)

crop

C

C PUT IN THE ELSEWHERE EVALUATED SPINOR EXPRESSIONS.

c
Id,2,Dotpr,B(J’)=83eP3(J)+B4¢P4(J)+BSePS(J)+BEPFtEpf(J,P3,P4,P5)
AI,Funct,B(K3')=83:P3(K3)+84tP4(K3)+BS¢P5(K3)+BEPFtEpf(K3,P3,P4,P5)

C P OUTPUT

Al,s4 ; BOIV¢P3P5
A|.Bs = -BOIV:P3P4
Al,EEPF = . LHEL¢BOIV:(ELECT - POSIT)

Id,P10P1=O
Al,P3DP3=O

204
Al,P4DP4=O

Al,PSDP5=O

Id,PlDP3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,P3DP4=P3P4

Al,P3DP5=P3P5

Al,P4DPS=P4P5

c P OUTPUT

ryep
Id,2,Dotpr,CU(J')=
(CU1-CU2):P1(J)+CU2¢P3(J)+CU2:PD(J)+CU5:P5(J)

+CUEPFtEpf(J,P1,P3,P5)+CUEPFtEpf(J,P1,P4,P5)

Al,Funct,CU(K')=

(CU1-CU2)tPl(K)+CU2¢P3(K)4CU2:P0(K)+CU5¢P5(K)

¢CUEPFrEpf(K,P1,P3,P5)+CUEPFtEpf(K,P1,P4,P5)

Id, CU1 = - CUOIVO(P3P5+P4P5-P1P5)

Al,CU2 = CUOIVePlPS

Al,CUS = CUOIVt(P3P4+P3P5+P4P5)

Al,CUEPF = + QHELtCUOIV

*yep

Id,Trick

Id,KOK=O

AI,KOPD=P4P5

Al,EDP1=0

Id,2,00tpr,PD(J")=P4(J)+P5(J)

Al,Funct,P0(K2')=P4(K2)+P5(K2)

Id,P10P1=O

Al,P30P3=0

Al,P40P4=O

Al,P50P5=0

Id,PlDP3=P1P3

Al,P10P4=P1P4

Al,PlDP5=P1P5

Al,P3DP4=P3P4

Al,P30P5=P3P5

Al,P40P5=P4P5

C P OUTPUT

eyep

C NOW PUT IN THE POLARIZATION VECTOR EXPANSION.

C

Id,2,00tpr,E(J')= E11P1(J) + E3-P3(J) +
E4¢P4(J)¢E5tP5(J)¢EEPFo(Epf(J,P1,P3,PS)+Epf(J,P1,P4,P5))

Al,Punct,E(K')= EltP1(K) + E3:P3(K) o
E4:P4(K)+E5AP5(K).EEPFt(Epf(K,P1,P3,P5)4Epf(K,P1,P4,P5))

Id,E1 EOIVtGHELt(P3P5oP4P5-2tP1P5)

Al,E3 EOIVOGHELOPIPS

EOIVeGHELePlPS

-EOIVeGHELe(P1P3+P1P4)

Al,EEPF = -EOIV

, DK=0 .
Id,Epf(P1,P3,P4,P5) = EVL
Id,PlDP1=0
Al,P3DP3=0
Al,P4DP4=O

205

Al,PSDP5=0
Id,PlDP3=P1P3
Al,PIOP4=P1P4
Al,PlDP5=P1P5
AI,P30P4=P3P4
Al,P30P5=P3P5
AI,P4OP5=P4P5
C P OUTPUT
tyep

c REPLACE K BY IT’S LINEAR EXPANSION.
Id,PSDK = -P4OK + P45

Id,2,Funct,K(J')=C(1)tP1(J)+C(2)eP3(J)+C(3)AP4(J)
+C(4)¢P5(J)

‘Yep

Id,Trick

Id,Epf(P1,P3,P4,P5) = EVL

C

C P OUTPUT

ryep

1d,C(1) = CC11tP10K¢CC21eP3DK+(CC31-CC41)tP4DKoCC41eP4P5
Al,C(2) = CC21eP10K+CC22tP3DK+(CC32-CC42)tP40K+CC42¢P4P5
Al,C(3) = CC31eP10K+CC32¢P30K+(CC33-CC43)tP4DK+CC43eP4PS
Al,C(4) = CC41¢P10K+CC42¢P3DK+(CC43-CC44)¢P4DK¢CC44eP4P5
C P OUTPUT

‘YOP

Id,Epf(P1,P3,P4,P5) = EVL

B WIV,CUOIV,PI,FPIIV,BOIV,EOIV,EVL
Keep EXP

tnext

C now begin the replacement of Ku by that which should force a zero.

Z expo = DOAEXP
Id,Ainbe,DOtWIV = 1

Id,00 — W

B 0145IV,CUOIV,PI,FPIIV,BOIV,EOIV,EVL
Keep expo

tnext

C

Z expOO = expo
Id,2,00tpr,K(J') = (P4(J) 4 P5(J)) - 014SIVtP4P5-P1(J)
C Id,2,Funct,K(J')= (P4(J) o P5(J)) - DI451V¥P4P5tP1(J)
Id,Epf(P1,P3,P4,P5) = EVL

O

Id,P1DP1 =
Al,P1DP3 = P1P3
Al,PlDP4 = P1P4
Al,PlDPS = P1P5
Al,P3DP4 = P3P4
Al,PSDPS = P3P5
Al,P4OP5 = P4P5
Al,PlDPl = o
Al,PaoPs = o
Al,P4DP4 = o
Al,PSDPS =

P

O
Id,P13 = 34 4 P35 4 P45 -P14 - P15

206

Id,Epf(P1,P3,P4,P5) = EVL

B 014SIV,CUOIV,PI,FPIIV,BOIV,EOIV,EVL
Keep expOO

enext

Z expl = DOtexpOO
Id,Ainbe,DO¢Dl45IV‘2 = 1

Id,00 = P14tP14 + 2tP14tP15 +P15PP15
B 014SIV,CUOIV,PI,FPIIV,BOIV,EOIV,EVL
Keep expl

#next

Z exp2 = DOtexpl

Id,Ainbe,DOtDl4SIV = 1

Id,00 = P14 4P15

B 014SIV,CUOIV,?I,FPIIV,BOIV,EOIV,EVL
tend

C
C the test is now finished.

C the remainder of the program is not

C subject to this test as all Ku’s will

C now be replaced by A and B expresssions.
C
C

Appendix S

Verification of Divergence Regulation
Co—efficent Cancellation

207

THIS
PROGR

IT CA
SET 0

IT LO
LOG P
OF TH
THE C
INDIC
AND V

INTEG

LNMP,L
P1,P2,
D3 = 9

30,R0
P13 =

P14
P15
P34
P35
P45
X P13

FOUR =

XO‘U‘V‘OOXXXXXXXXGOOOXXXXX xznnx<mn‘nnnnnnnnnnnnnnnnnnnnnnnnn

P1P3 =
X P1P4 =

208

START SCHOONSHIP PROGRAM

CHECK DIVERGENCE CO-EFFICENT
CANCELLATION

PROGRAM CHECKS THE SCHOONSHIP LOOP
AM BY COMPUTING A NUMERICAL RESULT

FOR AN ARBRITARILY CHOSEN SET OF PARAMETERS.

LCULATES AND USES THE NUMERICAL

VALUE OF THE 19 INTEGRALS FOR THIS

F NUMERICAL INPUT.

OKS FOR THE COEFFICENT OF THE
HOTON MASS AND THE COEFFICENT

E LOG QUARK MASS TO COME THROUGH
OMPLICATED CALCULATION AS ZERO,
ATING THAT THE TRACE CALCULATION
ARIOUS SUBSTITUTIONS, AND AT

LEAST THE DIVERGENT PART OF THE 19

RALS HAVE BEEN CODED CORRECTLY.

NMQ,PI,FPIIV
P3,P4,P5,Sl,52,S3,S4,U1,U2,U3,U4
3.7:93.7 - 2.7:93.7¢i

FEED IN VALUES FOR THIS RUN

-0.5¢66.0605782596549785908571076
:66.0605782596549785908571076

-O.5t70¢70

-0.5¢15t15

-O.5t92¢92

-O.S¢20:20

-O.5t25425

= P34 4 P35 + P45 - P14 - P15

SET SWITCHES

SET DOT PRODUCTS

ELECTeP13 + POSITeP14
ELECT¢P14 + POSIT4P13

209

X P1P5 = P15
X P3P4 = P34
X P3P5 = ELECT¢P35 + POSITeP45
X P4P5 = ELECTeP45 + POSITtPSS

RECIPROCALS

C
C
C
C
X P13IV = ((P1P3)tt-1)
X P14IV - ((P1P4)e¢-1)
X PISIV ((P1P5)¥¢-1)
X P34IV ((P3P4)ee-1)
X P3SIV ((P3P5)te-1)
X P45IV ((P4P5)eA-1)

C
C FOR CONVEINENCE , DEFINE SOME EXPRESSIONS AND THEIR RECIPROCALS.

C
x K02 = 0.5¢(-P4P5)

X K02IV = ((K02)ee-1)
X 0145 = P1P4 + P1P5
X DI = 0145
X 0145IV = ((0145)¢e-1)
X 0345 = P3P4 + P4PS
X 034SIV = ((0345)ee-1)
X MZP = 2eP3P4 + 03
C
C
C NOW K CAN ALWAYS BE EXPANDED ON AN ARBRITARY BASIS SET 51,32,53,S4
C
C K = C(1)eSI + C(2)e32 + C(3)tS3 + C(4)tS4
C
C AND FOR THIS PROBLEM WE USE THE SET P1,P3,P4,P5
C
C SO.
C
X 5152 = P1P3
X 5153 = P1P4
X 5154 = P1P5
X 5253 = P3P4
X 5254 = P3P5
X 5354 = P4P5
C
C TO THIS END, THE ORIGINAL EXPANSION OF K MAY BE DOTTED WITH EACH
C OF THE BASIS TO FORM A 4 X 4 MATRIX, WHICH WHEN INVERTED
C YIELOS ......
C
C C(l) = CC(1,1)¢SIDK + CC(1,2)¢SZDK + CC(1,3)#530K + CC(1,4)eS4DK
C
C W IS THE DETERMINENT FOR THIS INVERSION AND WIV IT’S RECIPROCAL
C
X W = SlSZeSSS4e(-SlS2eSSS4+SIS4eSZSS+SIS3eSZS4)
4 SlSSeS2S4e( SIS2¢SSS4+SIS4e52S3-515345254)
+ SIS4¢SZS3t( 5152:5354-5154$5253+5153e5284)
m = (Orin-1)

AND NOW, ALL THE CC’S USING THE NOTATION CC(4,3) = CC43

xnnnx

CC44 = WIV:(-2)e(SIS2¢5153tSZS3)

 

 

 

210
X CC43 = WIVtSISZt(~SlS2tSSS44SlS4tS253+SI$348254)
X CC42 = WIVtSlS3t( S152:S3$4+SIS4:3253-5153t8234)
x CC41 = WIV:$2$34( $152:$354-$154.5233+51$3.szs4)
X CC33 = WIV:(-2)t($1$2t$1$4e$2$4)
X CC32 = WIVtSIS4:( $1521S354—SlS4#S2S3+$1$345254)
X CC31 = WIV:SZS4*( SlSZtS3S4+SlS4tS2S3-SlSStS2S4)
X CC22 = WIV‘(-2)t($1$3t$1$4t$3$4)
X CC21 = WIV#SSS4t(-SIS2#S3S4+SIS4*SZSS+SISS*SZS4)
é CC11 = WIVt(-2)e(3253:5254t3334)
tfix
C
C THE NINETEEN INTEGRALS WILL NOW BE EVALUATED.
E THEY WILL MAKE USE OF SIX LOGS WHICH ARE ...
C LN1 = LN(-2tP4P5)
C LN2 = LN(—2t(P1P4+P1P5))
C LN3 = LN((2¢P3P5+MZP)/03)
C LN4 = LN((P1P4+P1P5)/P1P5))
C LN5 = LN(MZP/(2tP3P5+MZP))
C LN6 = LN((2:P3P5+MZP)¢(P1P4+P1P5)/R)
E WHERE R = -2tP4P5eP1P3+2tP3PSt(P1P4oP1P5)+MZPt(P1P4+P1P5)
C
C
C SET LOG VALUES
C
Z LN1 = ELECTe(6.43775164973640149840303733)
+ POSIT:(5.99146454710798198687044715)
Z LN2 = ELECT‘(8.54188580400660892766904137)
o POSIT:(8.43141741439483350344920756)
Z LN3 = ELECTt(-3.494604862965 - 1.863668333657ti)
4 POSITt(-3.090158418852 - 2.427214215569ei)
Z LN4 = ELECT.(3.12578540180218879567703223)
4 POSITe(3.01531701219041337145719842)
Z LNS = ELECTO( .4167792613628 + 1.216889362411ei)
+ POSITt(.01233281724963 o 1.780435244322wi)
Z LNG = ELECTt(-.6569902501244 - 1.378299363866ei)
t POSITt( .3589882796597 - 1.320996180611ni)
C .
Keep LN1,LN2,LN3,LN4,LNS,LN6
nnext
C
Z I1 = -2tPIoP45IVt(LN1 - LNMP)
Z I2 = -2¢PI40145IV:(LN2 - LNMQ)
Z IS = +2tPIt0345IVtLN3
C
Keep LN4,LN5,LN6,11,12,I3
enext

Z R = -2nP4P54P1P342tP3P5301AMZPaDl
Z MZP1 = MZP

Keep LN4,LN5,LN6,II,12,I3,R,MZP1,PPD
enext

Z RUPPER = Conjg(R)

Z RLOWER = ReConjg(R)

Z MZPUP = Conjg(MZP1)

Z MZPLW = MZP1.Conjg(MZP1)

 

 

 

211
Keep LN4,LNS,LN6,Il,12,I3,RUPPER,RLOWER,MZPUP,MZPLW

tnext

Z RIV = RUPPER/(RLOWER)

Z MZPIV = MZPUP/(MZPLW)

Keep LN4,LN5,LN6,Il,IZ,I3,RIV,MZPIV

#next

Z I4 = - PI¢P4SIVtPISIVtLN4 -
0.5tDltP45IVtPISIVt12 — 0.5tPISIV¢I1

Z IS = -P4SIVtMZPIVt(4¢PI¢LN5 +D345tI3 - P4P5¢Il)
Z 16 = RIV¢(4tPItLN6 - 0345:13 + 01:12)

C

Keep 11,12,13,I4,15,16,PPDIV,RIV

tnext

C

Z 17 = 0.251(P3P5tP151V+P3P4¢P14IV-P1P3tP4P5¢P14IV¢P151V)e
(II - (P1P4-P1P5)¢P451Vt12 + 2tP1P5#I4)
- O.5¢(P3P5-P3P4)¢P4SIV¢12 + P3P4.14
Z 18 = 0.25:(P1P41P34IV+P1P5tP3SIV-P1P3tP4P5¢P34IVtP3SIV)t
(Il - (P3P4-P3P5)¢P4SIV¢I3 - MZPtIS)
- 0.5:(P1P5-P1P4)tP4SIV¢I3 +P1P4¢IS
Z IQU = P3P4tP1P4tP1P5+P3P4IPIPStP1P5+P4P5¢P1P3tPIP4
- P4P5tP1P3tP1P5-P1P4#P1P4tP3P5-P1P5#PIP4#P3P5
Z 19L = 2tP1P3t
(P1P3tP4P5-P1P4aP3P4-P1P4tPsPS-PlP5:P3P4-P1P5¢P3P5)

Keep 11,12,13,14,15,16,17,Is,190,19L,R1v

tnext

C

X 193 = 2¢P4P5tP1P3tD14SIVtDl4SIV-(P3P4+P3P5)tDl451V
X 196 = ~2¢P4P5¢P1P3¢DI4SIV + 2tP3P5 + MZP

X 19F = O.5¢19U¢((IQL)¢$-1)¢(12+193tIS-196wIG)

Z I9 = 19F o O.5¢P4P5¢(P1P5-P1P4)tDl451Vt(Dl4SIV¢I3+16)
+ 0.5:P4P5t16 T
Keep Il,I2,I3,I4,IS,16,17,18,I9,I9U,19L,RIV

‘next

Z I11 = 2tPItP4SIV¢(P1P4-P1P5) + P1P4¢II

Z 112 = 2¢PItP4SIVt(P3P4-P3PS) + P3P4¢II

Z 113 = -2¢PI¢P4P5¢Dl45IV¢Ol45IVt(P1P4 - P1P5) +
P4P5tP1P5¢Dl4SIV¢12

Z 114 = O.5t(P3P4 o P3P5)¢I2 +

0.5:(Dl45t0345-2tP1P3cP4P5)$014SIV¢( 4¢PI¢DI45IV o 12)
Keep I1,I2,13,14,I6,IS,I7,I8,IQ,111,II2,Il3,Il4,IQU,IQL,RIV

tnext

Z 115 = -2tPI¢P4PStD345IV¢D34SIVt(P3P4 - P3P5) +

O.S¢P4P5.(10034SIV.D345IV¢(P3P4-P3P5)t(D$45+D3))¢13
Z 116 = 0.5.0145t13 o

0.5#(0145$D345 - 2tP1P3¢P4P5).
D345IVtD34SIVt( 4¢PI - I3#(D345 + D3))
C
Keep Il,12,13,14,IS,IG,I7,IB,19,I11,112,Il3,114,115,116,I9U,IQL,
RIV

tnext

Z I17 = P3P4eI7 + 0.50P4SIV¢(P3P4-P3P5)#114 +
O.25tIS¢P14IVtP15IV:(112 - (P1P4-P1P5):P4SIV¢II4 + 2tP1PS¢I7) +
0.0625tP14IV¢P14IV¢P1SIV¢P151Vt(4tP3P4¢P3P5tP1P4tP1P5-IS¢IS)t
(-2tP1P5tIl - 4tP1P5tP1P5¢I4 - P451VtDl45t2tP1P5tIZ)
Z 118 = P1P4tI8 + 0.5tP4SIVt(P1P4-PIPS)tIIG +
O.25:IS:P34IV:P351V:(111 - (P3P4-P395).P451v.115-(2¢P3P4+03).18).

 

 

212

0.0625tP34IV¢P34IVtP351VtP351V¢(4tP3P4tP3P5tP1P4tPlPS-IStIS)t
((MZP)t11-MZP:MZP¢IS-P451Vt(0345t2tP3P5+(P3P4-P3P5):MZP)¢I3)
éd,IS = P1P4tP3P5 + P1P5:P3P4 - P1P3¢P4P5
Keep 11,12,13,14,15,16,17,18,19,111,112,113,114,115,116,19U,19L,
117,118
tnext
C
Z I19=P4P5tP1P5¢Dl451Vt19-0.StDl4SIVtOI451VtP4P5t(P1P4-P1P5)3115 +
O.5¢19Ut((19L)tt-1)t
(113 + 01451Vt01451Vt1Wt115 - Dl451VtIR)¢19) +
(-0.5tP4P$¢P1P4tP1P5t((IQL)tt-l)-O.25¢19U¢19U¢((19L)t.-1)¢((19L)¢t-1))¢

(DI4SIVtIR)t12 .
-Dl451V#Dl451V¢IRtIRt16
-DI451Vt(D345tD345+Dl451V¢IW¢(P3P4—P3P5-MZP))tI3

C

C

Id,Iw = 2.P1P3.P4P5 - 0145.0345

Id,IR = -2¢P4P5¢P1P3 + Dlt(2¢P3P5 + MZP)

C

C

Keep 11,12,13,I4,15,16,17,I8,19,111,112,113,114,115,116,
117,118,119

tnext ,

C .

Z ZIIOU = UlDU1¢(0.5¢18+0.5*DI#15)+U20U2¢(O.5¢17-0.25t03t14)
- (U30U3-2tU3DU4+U4DU4)t0.5¢19
+ UlDU2t(O.5fiIl+Dl¢I4-O.5103t15)
- (U10U3-U10U4)¢(O.5t13+01t16)

+ UlDU4tP4P5¢15-(U2DU3-U2DU4)$0.5:(IZ-O3t16)+U20U4¢P4P5¢I4

- (U3DU4-U4DU4)¢P4P5¢16

Z 21100 = U10U1§OlfiDl+UZDU2tO.25tDS¢O3+U4DU4tP4PStP4P5
-1.0tUlDUQtDlt03+2tU10U4tP4P5¢01-UQDUAtP4P5¢D3

Id,2,Dotpr,U1(J')=CC11tP1(J)+CC21¢P3(J)+CC31:P4(J)+CC41¢PS(J)

Al,2,Dobpr,U2(J')=CC21tP1(J)+CC22¢P3(J)+CC32¢P4(J)+CC42:P5(J)

Al,2,Dotpr,U3(J')=CC31tP1(J)+CC32¢P3(J)4CC33¢P4(J)+CC43¢P5(J)

Al,2,Dotpr,U4(J')=CC41tP1(J)+CC42¢P3(J)+CC43¢P4(J)+CC44¢P5(J)

Id,Numer,P10P1,0,P3DP3,0,P4DP4,0,PSOP5,0

Al,PlDP3=P1P3

Al,PlDP4=PlP4

Al,PlDP5=P1P5

Al,PaDP4=P3P4

Al,P3DP5=P3P5

Al,P4OPS=P4PS

Keep 11,12,13,I4,15,16,17,18,19,111,112,113,114,115,116,
117,118,119,Z110U,21100

tnext

Z UPPER : -ZIlOUtConjg(ZIlOO)

Z LOWER = ZIlODtConjg(211OD)

Keep 11,12,13,I4,15,16,17,18,19,111,112,113,Il4,115,116,
117,118,119,UPPER,LOWER

tnext

Z 110 = UPPER/(LOWER)
Keep Il,I2,I3,I4,IS,16,17,18,I9,111,112,113,114,115,116,

117,118,119,110
tnext

 

 

 

213
ALL THE NINETEEN INTEGRALS ARE NOW CALCULATED.

tttttttifittttitlttttttitfifitttttttfitfifittfitttfittttttttttttttttfifiit

IN THIS SECTION, TAKE THE TRACE, REPLACE EACH K BY IT’S

LINEAR EXPANSION,
AND REDUCE TO A FINAL ANSWER.

B,K,CU,M,E,N,P1,P2,P3,P4,P5,PD
C,FO,F1,F2,F3,F4,F5,F6,F7,F8,F9,F10,F11,A1,A2,A3,AA
J1,J2,M1,M2,K3,K2

SCALE FACTOR FOR THE AMPLITUDE IS ...

FSCAL = 0.5.FPIIV.(1/137).(P34.P35+P45)

TAKE THE TRACE
EXP5=FSCALtFO¢F1¢F2tF3tF4tF11tFStFGtF7tF8tF9¢F10
SUBSTITUE ACCORDING TO HELICITIES AND DIAGRAM

nnnNnnnnxnnnnHm<nnnnnnnnnnnn

Id,FO=O.5t((1+LHEL):(66(J1))+(1-LHEL).(G7(J1)))
Id,F1=(G(J1,B))

Id,F2=ELECT¢(G(J1,M2)) + POSITA(G(J1,M1))
Id,F3=(G(J1,K))

Id,F4=ELECT¢(G(J1,M1)) + POSIT:(G(J1,M2))
Id,F11=O.5¢((1+QHEL):(66(J2))+(1-QHEL)t(G7(J2)))
Id,FS=(G(J2,CU))

:yep
Id,F6=(c(J2,M2))
Id,F7=(c(J2,M))
Id,F8=G(J2,E)
Id,F9=(G(J2,N))
Id,F10=G(J2,M1)

' *yep

Id,Trick,Trece,J1

C

C TRICK AND TRACE
C

Id,Trick,Trace,J2

Ayep
C
C START CONDENSATION SUBSTITUTIONS, SOME DIAGRAM DEPENDENT.

c

Id,2,Dotpr,N(J") = M(J)+P1(J)
Al,Funct,N(K2") = M(K2)+P1(K2)
Id,MDM=O
Id,Dotpr,M(J')=K(J)-PD(J)
Al,Funct,M(J')=K(J)-PD(J)
Id,KDPD=P4P5

Id,KDK=O

 

 

 

214
Id,2,Dotpr,PD(J')=P4(J)+P5(J)
Al,Funct,PD(J")=P4(J)+P5(J)
Id,P4DP4=O
Al,PSOP5=0
Al,EDP1=O
C P OUTPUT
Ayep
C

E PUT IN THE ELSEWHERE EVALUATED SPINOR EXPRESSIONS.
Id,2,Dotpr,B(J')=B3tP3(J)+B4tP4(J)+BS¢P5(J)+BEPFtEpf(J,P3,P4,P5)
Al,Funct,B(K3')=BS¢P3(K3)+B4¢P4(K3)+85¢P5(K3)+BEPF¢Epf(K3,P3,P4,P5)
C P OUTPUT

*yep '
Id,B3 BOIVtP4P5

AI,B4 BOIVtPSPS

A|,BS -BOIV¢P3P4

Al,EEPF = + LHEL:BOIV¢(ELECT - POSIT)
Id,Trick

1d,KDK=O

Id,PlDP1=O

Al,P3DP3=0

Al,P4DP4=O

Al,PSDPS=O

Id,PlDP3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,P30P4zP3P4

Al,PBDPS=P3P5

Al,P4DP5:P4P5

C P OUTPUT

#yep
Id,2,Dotpr,CU(J")=
(CU1-CU2):P1(J)+CU2¢P3(J)+CU2¢PD(J)+CU5¢PS(J)

+CUEPFtEpf(J,P1,P3,PS)+CUEPFtEpf(J,P1,P4,P5)

Al,Eunct,CU(K')=

(CU1-CU2):P1(K)oCU2-P3(K)+CU2tPD(K)4CU5tP5(K)

+CUEPFtEpf(K,P1,P3,P5)+CUEPFtEpf(K,P1,P4,P5)

Id, CU1 = - CUOIV¢(P3P5+P4PS-P1PS)

-A|,CU2 = CUOIVtPlPS

Al,CUS = CUOIVt(P3P4+P3P5+P4PS)

Al,CUEPF = + QHELtCUOIV ’

*yep

Id,Trick

1d,KDK=0

AI,KDPD=P4P5

Al,EDP1=O '

1d,2,Dotpr,PD(J')=P4(J)+P5(J)

Al,Funct,PD(K2')=P4(K2)+P5(K2)

Id,PlDP1=O

Al,P3DP3=O

Al,P4DP4=O

Al,PSDPS=O

Id,PlDP3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,P3DP4=P3P4 '

Al,P3DP5=P3P5

 

 

215
Al,P4DP5=P4P5
C P OUTPUT

‘YOP
C Now PUT IN THE POLARIZATION VECTOR EXPANSION.

C
Id,2,Dotpr,E(J')= EltP1(J) + E3tP3(J) +
E4-P4(J)+E5¢P5(J)+EEPF¢(Epf(J,P1,P3,P5)+Epf(J,P1,P4,P5))
Al,Eunct,E(K')= E1:P1(K) + E3:P3(K) +
E4:P4(K)+E5tP5(K)+EEPFO(Epf(K,P1,P3,P5)+Epf(K,P1,P4,P5))
Id,E1 = EOIVtGHELfi(P3P5+P4P5-2#P1P5)
Al,E3 = EOIVNGHELDPIPS
= EOIVtGHELtPlPS
= -E01V-GHEL:(P1P3oP1P4)
Al,EEPF = -EOIV

Id,KDK=0

Id,Epf(P1,P3,P4,P5) = EVL

Id,PlDP1=O

AI,P3DP3=O

Al,P4DP4=o

Al,PSDPS=O

Id,PlDP3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,P30P4=P3P4

Al,P30P5=P3P5

Al,P4DP5=P4P5

C P OUTPUT

‘YGP

C REPLACE K BY IT’S LINEAR EXPANSION.

C

Id,2,Funct,K(J')=C(1)tP1(J)+C(2)tP3(J)+C(3)tP4(J)
.P5(J).(1-C(3)-C(1).D1.P451v-C(2).D345.PASIV)

tyep

Id,Trick
1d,Epf(P1,P3,P4,PS) = EVL
C

C P OUTPUT

= CC11-P1DK.CC21.PSDK+(CC31-CC41).PADK+CC41.P4Ps
A|,C(2) = CC21tPIDKoCC22-P30Ko(CC32-CC42)-PADK.CC42.P4P5

Al,C(3) CC31*P10K+CC32tP3DK+(CC33-CC43)#P4DKOCC43UP4P5
Al,C(4) CC41-PIDK9CC42IP3DKo(CC43-CC44)tP4DK¢CC44tP4P5
C P OUTPUT

‘YOP

THE AMPLITUDE IS NOW COMPLETELY
IN THE FORM OF DOT PRODUCTS
OF K WITH EXTERNAL MOMENTA.

THE DOT PRODUCTS 0F K NEED TO
BE EVALUATED, FOR EXAMPLE...
(PSDK).(P4DK).(PSDK)

nnnnnnnnnn

 

 

216
AND REMEMBER THIS FACTOR IS TO BE INTERGRATED OVER ALL K SPACE.

THIS DEFINES THE A3 MATRIx.. .
A3(4,3,1) = (S4DK)¢(530K):(SlDK) INTEGRAL 0F, DvER
ALL K SPACE.
AND SIMILARLY A2(J,K) AND A1(J) MATRICES.

THE A3, A2, AND A1 MATRICES ARE EVALUATED USING 19 SIMPLE
INTEGRALS, EVALUATED EARLIER. .11 THRU 119..

REPLACE PDK ’3 BY A MATRICES
FOR THE FIVE POINT DIAGRAM, OR
SIMILAR B MATRICES FOR THE FOUR POINT.

nnnnnnnnnnnnnnnnnnnn

Id,PlDK
Al,P3DK
AI,P4DK
Al,PSDK -AA(3) + 4P
Id,AA(J' )tAA(K ) AA L =
Id,AA(J )tAA(K' ) = A2(J ,K)
1d ,AA(J ) = A1(J )

Id ,Symme, A3, 1, 2, 3, A21

Keep 11,12,13,14,15,16,17,18,19,111,112,113,114,115,116,

117, 118, 119, 110, EXPS

III II II II
>
>
(.0

P 5
( ) A3(J,K.L)

S
Z EXP51 = AOtEXPS

Id,Ao.A3(3,3,3) = A3333
AI,Ao.A3(2,3,3) = A3332
AI,Ao.A3(1,3,3) = A3331
AI,Ao.A3(2,2,3) = A3322
Al,Ao-A3(1,2,3) = A3321
AI,Ao.A3(1,1,3) = A3311
AI,Ao.A3(2,2,2) = A3222
AI,AOaA3(1,2,2) = A3221
AI,AO.A3(1,1,2) = A3211
Al ,AO.A3(1,1,1) = A3111
”0
Id ,Ao.A2(3, 3) = A233
AI ,AO.A2(2, 3) = A232
AI,Ao.A2(I,3) = A231
Al,Ao-A2(2,2) = A222
Al,A0tA2(1,2) = A221
AI,Ao.A2(1,1) = A211

Id,AO.A1(3) = A13
AI,AO.A1(2) = A12
AI,AO.A1(1) = A11
C P OUTPUT

eyep

C

C
C AND IN TERMS OF THE 19 INTEGRALS,

 

217
C THE ABOVE DEFINED 01 AND 03, THE
C A3,A2, AND A1 MATRICES,
C USING THE NOTATION. . ..
C A3(1,3,3) = A3331
C
C

Id,A3333 = -O.5¢(119)

AI,A3332 = -0.25¢(113 - 03:19)

AI,A3331 = -o.2s.(IIS + 2.01.19)

AI,A3322 = -0.25t(114 - 0.5303312 + 0.5103RD3t16)
AI,A3321 = -0.125:(4OPI + 2:01:12 - 03‘13 - 2301:03316)
AI,A3311 = -O.25¢(116 + 01:13 + 2:01:01316)

AI,A3222 = o.5¢(117 - D3¢A222)

AI,A3221 = (0.25.112 o 0.5:DI¢I7 - 0.5:D3¢A221)
AI,A3211 = (O.25¢(Ill - 03:18) + 01¢A221)

AI,A3111 0.5:(118 + 2tDl¢A211)

Id,A233 = -0.5t(19)

AI,A232 = -0.25¢(12 - 03.15)

A|,A231 = -0.25‘(I3 9 2‘01‘16)

AI,A222 = 0.5¢(17 - 033A12)

AI,A221 = (0.25t(11 o 2‘01t14) - 0.5tD3tA11)
A',A211 = 0.5¢(18 0 2*01'A11)

‘YQP

1d,A13 = -0.5¢(16)

AI,A12 = 0.5:(14 - 03.110)
AI,A11 = 0.5¢(IS 4 2¢Dlt110)
C

C

C
*yeP
C

C PUT IN A0
C
1d,AO = 110

THE FIVE POINT ANSWER IS NOW FORMED.

a LNMP,LNMO,PI,FPIIV,CUOIV,BOIV,EOIV,EVL
Keep 11,12,IS,I4,IS,IS,IT,18,19,111,112,113,II4,115,116,
117,113,119,IIO,EXP51

(5(3F5firifi

.
3
O
X
C?

ti.it00.000.00.000000000ittfifitfittttttttltttttttttittttdtfittttttt

NOW CALCULATE THE FOUR POINT AMPLITUDE

IN THIS SECTION, TAKE THE TRACE, REPLACE EACH K BY IT’S

LINEAR EXPANSION,
AND REDUCE TO A FINAL ANSWER.

'<(5(5(5(3r5(3(5(5(5(5(3(3

B,K,CU,M,E,N,P1,P2,P3,P4,PS,PD

 

 

218
FO,F1,F2,F3,F4,F5,F6,F7,F8,F9,F10,F11,A1,A2,A3,AA
,J2,M1,M2,K3,K2

1‘...“
HR.

SCALE FACTOR FOR THE AMPLITUDE Is ...
FSCAL = O.5¢FPIIV¢(1/137)

TAKE THE TRACE
EXP4=FSCAL¢F0¢F1¢F2¢F3tF4tF11tFStFGtF7¢F8tF93F10
SUBSTITUE ACCORDING TO HELICITIES AND DIAGRAM

Id,FO=O.5¢((1+LHEL)¢(06(J1))+(1-LHEL)¢(G7(J1)))
Id,F1=(G(J1,B))

Id,F2=ELECT¢(G(J1,M2)) . POSIT¢(G(J1,M1))
Id,F3=(G(J1,K)) -
Id,F4=ELECT.(c(J1,M1)) . POSIT¢(G(J1,M2))
Id,F11=o.5.((1+OHEL).(O5(J2))+(1-OHEL)1(c7(J2)))
Id,F5=(G(J2,CU))

nnnNnnnnxnnnnH'n

‘YOP
Id,F6=(G(J2,M2))
Id,F7=(G(J2,M))
Id,F8=G(J2,M1)
Id,F9=(G(J2,N))
Id,F10=G(J2,E)
*yep
Id,Trick,Trace,J1
C

C TRICK AND TRACE
C
Id,Trick,Trace,J2

*yeP
C

C START CONDENSATION SUBSTITUTIONS, SOME DIAGRAM DEPENDENT.

C

Id,2,Dotpr,N(J') = P3(J)+P4(J)+P5(J)
A|,Funct,N(K2”) = P3(K2)+P4(K2)+P5(K2)
Id MDM=0

Id,Dotpr,M(J')=K(J)-PD(J)
Al,Funct,M(J')=K(J)-PD(J)
Id,KDPD=P4P5

1d,KDK=0
Id,2,Dotpr,PD(J')=P4(J)+P5(J)
Al,Funct,PD(J')=P4(J)+PS(J)
Id,P4DP4=0

Al,PSDP5=0

Al,EDP1=0

C P OUTPUT

Prep

C

C PUT IN THE ELSEWHERE EVALUATED SPINOR EXPRESSIONS.

C
Id,2,Dot r,B J“ =83:P3(J)+B4tP4(J)+85¢P5(J)+BEPF‘Epf(J,P3,P4,P5)
Al,Funct?B(K§f);BanP3(K3)+B4SP4(K3)+BS¢P5(K3)+BEPF:Epf(K3,P3,P4,PS)

C P OUTPUT

 

 

 

219

*yeP

Id,B3 BOIVCPAPS

A|,B4 BOIV¢P3P5

A|,85 - -BOIV#P3P4

Al,EEPF = + LHEL:BOIV:(ELECT - POSIT)

Id,Trick

Id,KDK=0

Id,PlDP1=0

Al,P30P3=0

Al,P4DP4=0

Al,PSDPS=O

Id,PlDP3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,P3DP4=P3P4

Al,P3DP5=P3P5

Al,P4OP5=P4P5

C P OUTPUT

Prep

Id,2,Dotpr,CU(J')=

(CU1-CU?)3P1(J)+CU2OP3(J)+CU2:PD(J)+CU5¢P5(J)

+CUEPFtEpf(J,P1,P3,P5)+CUEPFtEpf(J,P1,P4,P5)

A|,Funct,CU(K')=

(CU1-CU2)¢P1(K)+CU2¢P3(K)+CU2OPD(K)+CU5:P5(K)

+CUEPFtEpf(K,P1,P3,P5)+CUEPFtEpf(K,P1,P4,P5)

1d, CU1 = - CUOIVO(P3P5+P4P5-P1P5)

Al,CU2 = CUOIVtPlPS

Al,CUS = CUOIV¢(P3P4+P3P5+P4P5)

Al,CUEPF = + QHELFCUOIV

Myep

Id,Trick

1d,KDK=0

AI,KDPD=P4PS

Al,EDP1=0

1d,2,Dotpr,PD(J")=P4(J)+P5(J)

Al,Funct,PD(K2')=P4(K2)¢P5(K2)

Id,PlDP1=0

Al,P30P3=0

A|,P4DP4=0

Al,PSDPS=0

Id,PlDP3=P1P3

Al,PlDP4=P1P4

Al,PlDP5=P1P5

Al,P3DP4=P3P4

Al,P30PS=P3P5

Al,P40P5=P4P5

C P OUTPUT

*yep
C
C NOW PUT IN THE POLARIZATION VECTOR EXPANSION.

C

Id,2,Dot r,E J’ = EltP1(J) + E3tP3(J) +
E4tP4(J§+ESSP51J)+EEPFt(Epf(J,P1,P3,PS)4Epf(J,P1,P4,P5))

A|,Funct,E(K')= E1tP1(K) + E3¢P3(K) +
E4:P4(K)+E5tP5(K)+EEPFt(Epf(K,P1,P3,P5)+Epf(K,P1,P4,P5))
Id,E1 = EOIVDGHELO(P3P5+P4P5-2tP1P5)

Al,E3 = E01VtGHELtP1PS

Al,E4 = EOIVOGHELSPIPS

 

 

 

220
Al,ES = -E01VtGHEL#(P1P3+P1P4)
Al,EEPF = -E01V
‘YOP
Id,Trick
Id,KDK=O
1d,Epf(P1,P3,P4,P5) = EVL
Id,PlDP1=O
Al,P30P3=O
Al,P4DP4=0
Al,P50P5=0
Id,PlDP3=P1P3
Al,PlDP4=P1P4
Al,PlDP5=P1P5
Al,P30P4=P3P4
Al,P3DP5=P3P5
Al,P4DP5=P4P5

 

C P OUTPUT

*yeP

C

C REPLACE K BY IT’S LINEAR EXPANSION.

C

Id,2,Funct,K(J')=C(I).P1(J).C(2).P3(J)+C(3).P4(J)
.P5(J).(1-C(3)-C(1).DI.PASIv-C(2).0345.P4SIV)

*yep

Id,Trick

Id,Epf(P1,P3,P4,P5) = EVL

C

C P OUTPUT

.yep

1d,C(1) = CC11¢PIDK+CC21tP3DK+(CC31-CC41)MP4DK+CC4IMP4P5

Al,C(2) = CC21¢P10K+CC22¢P30K+(CC32-CC42)tP4DK+CC42tP4P5

Al,C(3) = CC31.PIDK+CC32.P30K.(CC33-CC43).PADK+CC43.P4P5

Al,C(4) = CC41.PIDK+CC42.PSDK+(CC43-CC44).P4DK+CC44.P4P5

C P OUTPUT

¢yep

THE AMPLITUDE IS NOW COMPLETELY
IN THE FORM OF DOT PRODUCTS
OF K WITH EXTERNAL MOMENTA.

 

 

THE DOT PRODUCTS OF K NEED TO
BE EVALUATED, FOR EXAMPLE...
(PSDK).(P4DK).(PSDK)

AND REMEMBER THIS FACTOR IS TO BE INTERGRATED OVER ALL K SPACE.

THIS DEFINES THE A3 MATRIX. . ..
A3(4,3,1) = (SADK).(SSDK).(SIDK) INTEGRAL OF, OVER

ALL K SPACE.
AND SIMILARLY A2(J,K) AND A1(J) MATRICES.

nnnnnnnnnnnnnnnnnnnnnnn

THE A3, A2, AND A1 MATRICES ARE EVALUATED USING 19 SIMPLE

 

221

C INTEGRALS, EVALUATED EARLIER .... I1 THRU 119....
C

C

C REPLACE PDK ’5 BY A MATRICES

C FOR THE FIVE POINT DIAGRAM, OR

C SIMILAR B MATRICES FOR THE FOUR POINT.

C .

Id,PIDK = AA(1)

Al,P3DK = AA(2)

Al,P4DK = AA(3)

Al,PSDK = -AA(3) + P4P5

Id,AA(J"):AA(K')¢AA(L') = A3(J,K,L)

Id,AA(J')tAA(K') = A2(J,K)

1d,AA(J") = A1(J)

1d,Symme,A3,1,2,3,A2,1,2

Keep 11,12,13,14,15,16,I7,18,19,111,112,113,114,115,116,
117,118,119,110,EXP51,EXP4

.next

S A0

Z EXP41 = A0tEXP4

Id,A0#A3(3,3,3) = B3ERR

Al,A0¢A3(2,3,3) - B3ERR

AI,AOtA3(1,3,3) = B3ERR
AI,A0*A3(2,2,3) = B3ERR -
Al,A0#A3(1,2,3) = B3ERR
A1,A0¢A3(1,1,3) =.B3ERR
AI,A0*A3(2,2,2) = B3ERR
AI,A0¢A3(1,2,2) = B3ERR
AI,AOtA3(1,1,2) = B3ERR
AI,AOtA3(1,1,1) = B3ERR
*yeP

Id,A0¢A2(3,3) = 8233
AI,A0tA2(2,3) = 8232
AI,AOtA2(1,3) = 3231
A|,A0tA2(2,2) = 3222
Al,A0¢A2(1,2) = 3221
AI,AODA2(1,1) = 8211

Id,Ao.A1(3) = 313
AI,AO.A1(2) = 312
AI,AOMA1(1) = 811
C P OUTPUT

'yoP

AND IN TERMS OF THE 19 INTEGRALS,
THE ABOVE DEFINED DI AND 03, THE
A3,A2, AND A1 MATRICES,
USING THE NOTATION....
A3(1,3,3) = A3331

PUT IN '8' SERIES

nnnnnnnnnnnnn

Id,8233 = -0.5¢115
A|,B232 = -0.25t(4tP1 - 03:13)
Al,8231 = -0.53116

 

 

222
, = 0.5.(112 - 033812)
A|,8221 = 0.5.(111 - 03.311)

A|,3211 I18

Id,BIS = -0.5tI3

Al,312 = 0.5!(11 '03115)
AI,311 = I8

C

C PUT IN A0

C

Id,A0 = 15

C

C

C

C

C THE FOUR POINT ANSWER IS NOW FORMED.
C

B LNMP,LNMQ,PI,FPIIV,CU01V,BOIV,E01V,EVL

Keep EXP51,EXP41

tnext

C

C

C NOW ADD THE FIVE POINT AND FOUR POINT RESULTS
C

C

2 EXP = EXP51 + EXP41

B LNMP,LNMQ,PI,FPIIV,CU01V,BOIV,E01V,EVL

tend

 

 

Appendix T

Verification of 19 Irreducible Integrals

223

 

367
370

377
380

224
PROGRAM check 19 integrals
COMPLEX 03,MZP,R,LN3,LN5,LN6,MZP1,ZIIOD,RUPPER,

C MZPUP,MZPIV,RIV,ZIlOU,UPPER,I,

C 11,12,13,I4,15,16,17,I8,I9,110,196,IR,

C 111,112,113,114,115,116,117,118,119,

C g,p,xx,i$u,i5l,z,111,n
REAL LN1,LN2,LN4,RLOWER,MZPLW,19U,19L,193,IW,IS,

C LOWER,S,MP,MQ,x,c,d,e,f,m,za,zO,21,NN,NX,NZ,N0,AX,NY,NY1
REAL P1P3,P1P4,P1PS,P3P4,P3P5,P4P5,API,FPIIV,K02,K021V,NX1,NZl
DOUBLE PRECISION WIV,CC44,CC43,CC42,CC41,CC33,CC32,CC31,

C CC22,CC21,CC11

CAM : 2.7

“P = 1

Mu : 1

M2 = 93.7

03 = ( 8779.69,-252.99)

API = 3.142357143
FPIIv = 0.25-(1/API)

PRINT¢,API

PRINT-t, ’03 = ’,03
PRINTc,’P3P4 = 92’
P3P4 = 92
PRINT:,’P3P5 = 20’
P3P5 = 20
PRINT:,’P4PS = S’
P4P5 : 5

P3P4 = -O.StP3P4:t2
P3P5 = -O.5tP3P5¢t2
P4P5 = -O.5¢P4P5tt2
P14MAX (P3P4+P4P5)GP4P5¢(P3P4+P3PS+P4PS)

P14MAX = P14MAX/((P3P4+P4P5)n(P3P5+P4P5) - P3P4tP3P5)
PRINTt, ’P14MAX = ’,-((2t(-P14MAX))¢:(O.S))
PR1NT¢,’P1P4 = 70’

P1P4 = 70

P1P4 = -0.5¢P1P4¢#2

R = P3P4 + P4P5

S = P1P4t(P3P5+P4PS) - P4P5t(P3P4+P3P5+P4P5)
T = P1P4¢P3P4¢P3P5

DISC = (16¢Tt(T-Rt$))tt(0.5)

PISUP = (-(2thS - T) o DISC)/(2:ROR)

IF (PISUP) 370,367,367

P15UP = 0

PISDN = (~(2thS - T) - DISC)/(2¥ROR)

IF (PISDN) 380,377,377

P150" = 0

PRINTt, ’PISUP = ’,-((2t(-P15UP))tt(0.5))
PRINTt,’P150N = ’,-((2t(-P150N))tt(0.5))
PR1NT¢,’P1P5 = 15’

P1P5 = 15
P1P5 = -O.SaP1P5t32
P1P3 = P3P43P3P5wP4PS-PIP4-P1PS

PRINT¢,'P1P3 = ’,P1P3
PRINT‘,’P1P3 = ’,(-2#P1P3)#¢(0.5)
P13IV = 1/P1P3

P14IV = 1/P1P4
P1SIv = 1/P1Ps
P34IV = I/P3P4
P35IV = 1/P3P5

090
110
120
130

150

170

180

190

210
220
230
240
250
260
270
280
290

310
320
330
340
350

370
380
390
400
410
420
430
440
450
460
470
480
490
500

520
530
540
550
560
5708
580
590
600
610

 

 

225

P45IV = 1/P4P5 520
MZP = 23P3P4 3.03 630
8152 = P1P3 549
3133 = P1P4 650 '
5154 = P1P5 660'
3253 = P3P4 670
5254 = P3P5 680
S334 = P4P5 690
W = 5152353543(-SIS23$354+SIS4352533515338254) 700
C + SlS33SZS43( $1523S3$4+Sl$43SZS3-SIS33SZS4) 710
C + 5134352533( $1S23$354-$134332533313333234) 720
WIV = 1.0/W 7.
CC44 = WIV3(-2)3($152351333SZS3) 740
CC43 = WIV3$1$23(-S18235354+SIS43SZSS+SIS33SZS4) 750
CC42 = WIV351533( SlS23S3S4+SI$438253-513335254) 760
CC41 = WIV3S2533( $1523S3S4-SIS4352833315335254) 770
CC33 = WIV3(-2)3(51323515435234) 780
CC32 = WIV331S43( $1523S354-SIS43$253351533SZS4) 790
CC31 = WIV3SZS43( $152353543515438253-513335254) 800
CC22 = WIV3(-2)3(SIS33SIS43S3$4) 810
CC21 : WIV3S3SA3(-SIS23S354+SI54382533513335254) 820
CC11 = WIV3(-2)t(52533525435354) . 830

 

PRINT3,'WIV = ',va

 

 

PRINTt,’CC44 = ’,CC44

PRINTt,’CC43 = ’,CC43

PRINT3,’CC42 = ’ CC42

PRINTt,’CC41 = ’,CC41

PRINTt,’CC33 = ’,CC33

PRINTt,’CC32 = ’,CC32

PRINTt,’CC31 = ’,CC31

PRINTt,’CC22 = ’,CC22

PRINTt,’CC21 = ’,CC21

PRINTt,’CC11 = ’,CC11

R = -23P4P53P1P3+23P3P53(P1P4+P1P5) + MZP3(P1P4+P1P5) 340
LN1 = LOG (-23P4P5 3 MP332)

LN2 = LOG(-23(P1P4+P1P5) + M0332)

LN3 = LOG((23P3PS+MZP)/D3) 870
LN4 = LOG((P1P4+P1P5)/P1PS) 880
LNS = LOG(MZP/(23P3PS+MZP)) 890
LN6 = LOG((23P3P5+MZP)3(P1P43P1PS)/R) 900
80 = (83P3P53P4P5)33(-O.5) 910
CUO = (83P1P53(P3P43P3PS3P4PS))33(-0.5) 920
E0 = (-43P1P53(P3P5+P4P5-P1P5)3(P3P43P3P53P4PS))33(-0.5) 930
EVL1=(P1P53P3P43P1P43P3P5-P4P53P1P3)332 940
EVL = ( 43P1P43P1P53P3P43P3P5 - EVL1)33(0.5) 950
PRINT3,’80 = ’,80 - 960
PRINTt,’CUO = ’,CUO 970
PRINTt,’E0 = ’,E0 980
PRINTt,’EVL = ’,EVL 990
PRINTt,’LN1 = ’,LN1 1000
PRINTt,’LN2 = ’,LN2 1°10
PRINTt,’LN3 = ’,LN3 1020
PRINT‘,’LN4 = ’,LN4 1030
PRINTt,’LNS = ’,LN5 1°40
PRINT‘,’LN6 = ’,LN6 105°
K02 = O.$3(-P4PS) i833
K021V — 1/K02 1080

0145 = P1P4 + P1P5

1135

1137

1145

1147

1155

1157

1163
1165

1167

226

DI = 0145

DI4SIV = 1/0145

D345: P3P4 3 P4P5
D3451V= 1/0345

MZP = 23P3P4 3 D3

I1 = -23AP13P45IV3(LN1)

I = 0

S =.00001

Print3,’s= ’,s

X = -1

I = I 3 S/(P4P53(X-1) 3 MP332)
X = X35

1F ( X-1) 1135,1137,1137
PR1NT3, ’CHECK 11:1323AP1
120 = -23API301451V3(LN2)

S =. 00001

pr rin at; ,’3= ’,3

X = -1

I = I 3 S/(- (P1P43P1P5)3 (X31) 3 M0332)
X = X 3 S

1F ( X—l 1) 1145,1147,1147
PRINT3, ’CHECK 12: I323AP1
I3 = 323API3D3451V3LN3

I =

S =.00001

9

IF ( X-l) 1155,1157,1157

PRINT3, ’CHECK 13: ,1323API
PRINT3, ’11 = ’,11
PRINT3, ’12 = ’,12

PRINT3, ’13 = ’ ,13

R = -23P4P53P1P3+23P3P53Dl3MZP3DI
MZP1: MZP

RUPPER = CONJG(R)

RLOWER = R3CONJG(R)
MZPUP=CONJGGMZP1)

MZPLW = MZP13CONJG(MZP1)
RIV = RUPPER/(RLOWER)
MZPIV= MZPUP/(MZPLW)
PRINT3, ’MZPIV = ’ ,MZPIV

ZO s 23((P1P43P1PS)33(0. 5))

I = 0

S =.01

print3,’s- ’,s

X = -1

Y = 0

ll =ZO3 sin(acos(X))

ZA = 213cos(Y)

111 = (P1P4-P1P5)3X32A-P1P4-P1P53MQ332
I = I 3 S3S/((P4P53(X-1) 3 MP332)3111)
Y = Y 3 S

1F (Y - 23API) 1165,1167,1167

X = X 3 S

IF ( X-l) 1163,1169,1169

1090
1100
1110
1120
1130

' 1140

1150

1160

1170
1180
1190
1200
1210
1220
1230
1240
1250
1260
1270
1280

 

 

1169 .PRINT:,’CHECK 14 = .' I 227

I4 = - api3P45IV3P151V3LN4 -

C 0}53DI3P451V3P151V312 - 0.53PISIV3II 1300
print3,’I4 2 ’ ,14
c = p4p5
d = -c 3 mp332
e = p1p4-p1p5
f = 23((plp439195)33(0 5))
9 = -(plp4+9195) + mq332
m = (plp4+p195)*32
n = 23(g3o3c - d3m)
p = d3d3m - 239393c3d 3 c3c3(g3g - ftf)
x = c 3 6
xx = m3x3x 3 n3x 3 p
z = (p3xx)33(0.5)
u

 

'4u = log((23z 3 n3x 3 23p)/x)
x = -c 3 d
xx = moxtx 3 ntx 3 p
z = (p3xx)33(0.5)
i4| = log((23z 3 n3x 3 23p)/x)
i4 = 323api3(p33(-O.5)3(i4u-i4l))
print3, ’i4detl = ’,i4
20 = 23((P3P43P3P5)33(0.5))
I = 0
S =.01
print3,’s= ’,s
X = -1
1183 Y = 0
21 =ZO3 sin(acos(X))
1185 ZA = 213cos(Y)
III = (P3P4-P3P5)3X32A3P3P43P3P53d3
I = I 3 S3S/((P4P53(X-1) 3 MP332)3111)
Y = Y 3 S
IF (Y - 23API) 1185,1187,1187
1187 X = X 3 3'
IF ( X-l) 1183,1189,1189
1189 PRINT3,’CHECK IS = ’, I
15 = P451V3MZPIV3(-43api3LN5 - 0345313 3P4P5311)

 

 

prinbt,’15 = ' ’15

p4p5

-c 3 np332

p3p4-p3p5

23((p3p43p3p5)3¢(0.5))
+(939439395) + 63

(p3p43p3p5)332

23(g3o3c - 63m)

dtd3m - 2393.3c3d 3 c3c3(gcg - ftf)
c 3 d

x = n3x3x 3 nox 3 p

z = (p3xx)33(0.5)

iSu = Iog((23z 3 n3x 3 23p)/x)

x = -c 3 d

xx = m3x3x 3 03x 3 p

z = (p3xx)33(0.5)

35' = 109((232 3 n3x 3 23p)/x)

i5 = 23api3(p33(-O.S)3(35u-35l))
print3, ’iSdetl = ’,35

o = (23919339495-(plp4+plpS)3(9394+p395))/(91p4+9195)

X X‘U 3 3‘0 43. 0.0

 

1193

1195

1197
1199

2163

2165

2167
2169

-<k4k4h4h4-<)<‘O(ntqru

228

0 = ((p3p43p3p5)332 - o3o)33(0.5)

= 0

:.01

RINT3,’S = ’,S

= -1

= 0

1 =ZO3 sin(acos(X))

A = 213cos(Y)

11 = o3X3ZA3P3P43P3P53d3 .
= I 3 S3S/((-(P1P43P1P5)3(X31) 3 M9332)3III)
= Y 3 5

IF (Y - 23API) 1195,1197,1197

X = X 3 S

IF ( x-1).1193,1199,1199

PRINTt,’CHECK 16 = ’, I

16 = RIV3(343api3LN6 - 0345313 3 01312)
PRINT3,’16 = ’,16

-p1p4-p1p5

3c 3 mq332 -
(2*plp33p495-(9194+plp5)3(p3p43p395))/(plp4+9195)
3(p3p43p3p5) 3 d3

20

(P3P4*P395)“2

23(g3e3c - d3n)

d3d3n - 23g3otc3d 3 c3c3(gtg - ftf)
c 3 d

xx = m3x3x 3 n3x 3 p

z = (p3xx)33(0.5)

i6u = log((23z 3 n3x 3 23p)/x)

x = -c 3 d

XX = M’X.X Q n.x 4 p

z = (p3xx)33(0.5)

i6l = Iog((23z 3 n3x 3 23p)/x)
36 = 23api3(p33(-O.5)3(i60-i6l))
print3, ’i6dotl = ’,36

Z0 = 23((P1P43P1P5)33(0.5))

N0 = 0.53(P3P43P3P5)
AX = .53((P1P43P1P5)33(~0.5))3(P1P43P3P53P1P53P3P4-P1P33P4PS)

xuaa-«moan
II II II II II II II II ll

0
O
.01
r'nt3,’s= ’,s
1

I
s
p
X
Y
N2 = O.S3(P3P4-P3P5)3X

NX = AX3sin(acos(X))

21 = ZO33in(acos(X))

ZA = 213cos(Y) .

NN = NX3cos(Y) 3 NZ 3 NO -
II (P1P4-P1P5)3X3ZA-P1P4-P1P53MQ332

3 $3$3NN/((P4P53(X-1) 3 MP332)3111)

S
- 23API) 2165,2167,2167
3 S
X-1) 2163,2169,2169
3,’CHECK I7 = ’, I
3((P3P43P3P5)33(0.5))

2
0.5 P1P4 P1P5
0.5:E(P3P43P3P5)33(~0.5))3(P1P43P3P53P1P53P3P4-P1P33P4P5)

IN‘UF4><F4-<h4
m
u
A
)(c2-<34ll

3303“
II II II a

 

 

 

2263

2265

2267
2269

2363

2365

2367
2369

229

0
.01
'nt3,’s=

1

9

r ,s

I
S
p
x -
Y 0
NZ
NX

Zl

0.53(P1P4-P1P5)3X

AX3sin(acos(X))

ZO3sin(acos(X))
ZA 213cos(Y)
NN NX3cos(Y) 3 NZ 3 N0

III = (P3P4-P3P5)3X3ZA3P3P43P3P53D3

= I 3 §3S3NN/((P4P53(X-1) 3 MP332)3111)

3 §

F (Y - 23API) 2265,2267,2267

= X 3 S

F ( X-1) 2263,2269,2269
RINT3,’CHECK I8 = ’, I
= (2*plp3tp495-(9194+plpS)3(p3p4+p3p5))/(plp439195)
0 = ((P3P43P3P5)332 - e3o)33(0.5)
N0 = 0.53(P4P5)

u = p3p43p1p53p1p5 3 p3p43p1p43p1p5 3 p4p53p1p33p1p4
C - p1p43p1p43p3p5 - p1p43p1p53p3p5 - p1p53p1p33p4p5

NO VHXH-(H

t = 23p1p33(p1p33p4p5-p1p43p3p4-p1p43p3p5-p1p53p3p4-p1p53p3p5)

:X = ((-p4p5/2)33(O.5))3u/((p1p43p1p5)3t33(0.5))
= 0

S =.01

print3,’s= ’,s

X = -1

Y = 0

NZ = 0.53P4P53(P1P4-P1P5)3X/(P1P43P1PS)

NX = AX3sin(acos(X))

21 = ZO3sin(acos(X))

ZA = Zl3cos(Y)

"N = NX3cos(Y) 3 NZ 3 NO

111 = o3X3ZA3P3P43P3P53D3

I = I 3 S3S3NN/((-(P1P43P1PS)3(X31) 3 M9332)3111)

Y = Y 3 5

IF (Y - 23AP1) 2365,2367,2367

X = X 3 5

IF ( X-l) 2363,2369,2369

PRINT3,’CHECK I9 = ’, I
I7 = 0.253(P3P53P15IV3P3P43P14IV-P1P33P4P53P14IV3P151V)3

C (11 - (P1P4-P1P5)3P451V3I2 3 23P1P53I4)
C- 0.53(P3P5-P3P4)3P451V312 3 P3P43I4

18 = 0.253(P1P43P34IV3P1P53P351V-P1P33P4P53P34IV3P35IV)3
C (11 - (P3P4-P3PS)3P451V313 - MZP315)
C - 0.53(P1P5-P1P4)3P451V3I3 3P1P4315

IQU = P3P43PIP43P1P53P3P43P1P53P1PS3P4P53P1P33P1P4
C - P4P53P1P33P1P5-P1P43P1P43P3P5-P1P53P1P43P3P5
I9L = 23P1P33
C(P1P33P4P5-P1P43P3P4-P1P43P3P5-P1P53P3P4-P1P53P3P5)
193 = 23P4P53P1P3301451V3DI4SIV—(P3P43P3P5)3014SIV
196 = -23P4P53P1P33014SIV 3 23P3P5 3 MZP

I9 = 0.53I9U3(1/19L)3(123193313-196316)
C 3 0.53P4PS3(P1P5-P1P4)3Dl451V3(D1451V3I3316)
C 3 0.53P4P5316

PRINT3,’I7 = ’,17

PRINT3,’18 = ’,I8

1360
1370
1380
1390
1400
1410
1420
1430
1440
1450
1460
1470
1480
1490
1500
1510
1520

 

 

 

230

PRINT3,’I9 = ’,19

PRINT3,’I9U = ’,I9U

PRINT3,’I9L = ’,19L

PRINT3, ’I93 = ’ ,193

PRINT3,’196 = ’ :196 1530
III: 23API3P4SIV3(P1P4- -P1P5) 3 P1P4311

PRINT3, ’111 = ’ ,Ill 1540
NX = 23((P1P43P1P5)33(0. 5))

9

3
('00
DH

r ,’s= ,s

I
H

2763

2-<><‘U (DH
X
HI

0
=NX 3 sin(acos(X))
NX13cos(Y)
(P1P4-P1P5)3X3NX23P1P43P1P5
I 3 $3330. 53NN/(P4P53(X-1) 3 MP332)
.Y 3S
IF (Y - 23API) 2765,2767,2767
2767 X = X 3 S
IF ( X-l) 2763, 2769, 2769
2769 PRINT3, ’CHECK 11: I
. 112 = 23API3P4SIV3(P3P4-P3P5) + P3P43I1
PRINT3, ’112= ’ ,I12
NX = 23((P3P43P3P5)33(0. 5))

2765 NX2

z
2

II II

-< H II II

 

0

.0055

“nt3 ,5
-1

2963

z-<><‘o (DH
X

=NX 3 sin(acos(X))
NX13cos(Y)
(P3P4-P3P5)3X3NX23P3P43P3P5
3 $3530. 53NN/(P4P53(X- -1) 3 MP332)
3S
IF (Y - 23API) 2965,2967,2967
=X 3 S
IF ( X-1 ) 2963, 2969, 2969
2969 PRINT3, ’CHECK 12: I
113: —23API3P4P53Dl451V3Dl4SIV3(P1P4 - P1P5) 3 1560
C P4P53P1P5301451V3I2
PRINT3,’II3 = ’ ,113
N0 = 0 53(P4P 5)
N2 = 0. 53P4P53(P1P4-P1P5)/(P1P43P1P5)
= ( NO332 - NZ332)33(0. 5)

NH

II II
-< H II II

2965

X

2967

 

1570

0
.0

...;

I
,8

3063
NZ3X

NX3sin(acos(X))
=NX13cos(Y) 3 N21 3 N0

3 S3S3NN/(-(P1P43P1P5)3(X31) 3 M9332)

nt,
0

|§22<X: MH§
XN
HHII II -° II II

3065

«H.

s
F (Y - 2am) 3065 3067,3067

=X 3 5
IF ( X- I) 3063, 3069, 3069

XH-(H

3067

 

231
3069' PRINT3,’CHECK 113 = ’, I
114 = 0.53(P3P4 + P3P5)312 3 1580
C 0.53(014530345-23P1P33P4P5)3Dl451V3( 43AP1301451V 3 I2) 1590 _
PRINT3,'114 = ’,114 _3
NZ = (23919339435-(9194+0195)3(p3p4+p395))/(Plp4+plps)
NX = ( (p3p43p3p5)332 - NZ332 )33(0.5)

I = 0
S =.01
PRINT3,’S = ’,S
X = -1
3193 Y = 0
N21 = NZ3X
NX1 = NX3 sin(acos(X))
3195 NX2 = NX13cos(Y)
NN = N213NX23P3P43P3P5
I = I 3 $3530.53NN/(-(P1P43P1P5)3(X31) 3 N9332)
Y = Y 3 5
IF (Y - 23API) 3195,3197,3197
3197 X = X 3 5

IF ( x-1) 3193,3199,3199

3199 PRINT3,’CHECK 114 = ’, I
115 = -23API3P4P53D3451V303451V3(P3P4 - P3P5) 3 1600
c O.53P4P53(1303451V3D34SIV3(P3P4-P3P5)3(0345303))313 1700

PRINTt,’I15 8 ',I15
N0 = 0.53(P4P5)
NZ ‘ 0.53P4P53(P3P4-P3P5)/(P3P43P3P5)

2
X
III

 

( N0332 - NZ332)33(0.5)
I = 0
5 =.01
print3,’s= ’,s
X = -1
3363 Y = 0
N21 = NZ3X
NX1 = NX3sin(acos(X))
3365 NN = NX13cos(Y) 3 N21 3 NO
I = I 3 $3S3NN/(-(P3P43P3P5)3(X-1) 3 D3)
Y = Y 3 5

IF (Y - 23API) 3355,3367,3367
3367 X = X 3 5

IF ( X-I) 3363,3369,3369
3369 'PRINTO,’CHECK 115 = ’, I

116 :_O.53Dl45313 3 1620
c 0.5.(0145.034s - 23P1P33P4P5)3 1630
c 03451v303451v.( 43API - 133(0345 + 03)) 1640

PRINT3,’I16 = ’,116
NZ - (23919330495-(91p439195)3(p3943p395))/(plp43p195)

2
X
III

((939439395)332 - "1332)33(0-5)
I = 0 ,
S =.01
PRINT3,’S = ’,S
X = -1
3593 Y = 0
NX1 = NX 3 sin(acos(X))
3595 NX2 = NX13cos(Y)
NN = NZ3X3NX23P3P43P3P5
I = I 3 S3530.53NN/(-(P3P43P3P5)3(X-l) 3 03)
Y = Y 3 3
IF (Y - 23API) 3595,3597,3597

 

3597

3599

4163

4165

4167

4169

4263

4265

4267

4269

232

X = X 3 3
IF ( x-1) 3593,3599,3599
PRINTt,’CHECK 116 = ’,I

20 = 23((P1P43P1P5)33(0.5))

N0 = O.53(P3P43P3P5)

NX = 0.53((P1P43P1P5)33(~0.5))3(P1P43P3P53P1P53P3P4-P1P33P4P5)
NZ = O.53(P3P4-P3P5)

NY = (N0332 - N2332 - NX332)33(O.5)

I = 0

s = .01

pruntt,’s= ’,s

X = -1

Y = 0

N21 = NZ3X

NX1 = NX3sin(acos(X))

NYl = NYtsin(acos(X))

Zl = ZO3sin(acos(X))

ZA = Z13c0s(Y)

NN = NX13cos(Y) 3 NYltsin(Y) 3 N21 3 N0

III = (P1P4-P1P5)tX3ZA-P1P4-P1P53MQ332

I = I 3 $353NN3NN/((P4P53(X-1) 3 MP332)#III)
Y = Y 3 5

IF (Y - 23API) 4165,4167,4167

X = X 3 3

IF ( x-1) 4163,4169,4169
PRINTt,’CHECK 117 = ’, I

ZO = 23((P3P43P3P5)33(0.5))

N0 = O.53(P1P43P1P5)

NX = O.53((P3P43P3P5)33(-0.5))3(P1P43P3P53P1P53P3P4-P1P33P4P5)
NZ = 0.53(P1P4—P1P5)

NY = (N0332 - NZ332 - NX332)33(0.5)

I = O

S =.01

pruntt,’s= ’,s

X = -1

Y = O

NZ1 = NZtX

NX1 = NX3sin(acos(X))

NYl = NYtsin(acos(X))

21 = 203sin(acos(X))

ZA = Z13cos(Y)

NN = NX13cos(Y) 3 NYltsin(Y) 3 NZ1 3 NO

(P3P4-P3P5)3X3ZA3P3P43P3P53D3
3 S3S3NN3NN/((P4P53(X—1) 3 MP332)3III)

II III-3

3 S
- 23API) 4265,4267,4267

3 S
X-l) 4263,4269,4269

'0F4><h3-<h4h3
11
A
><-<-<H

F
RINT3,’CHECK 118 = ’, I
e = (239193*p495-(plp4+plps)*(p3p43p395))/(plp439195)
20 = ((P3P43P3P5)332 - ete)33(0.5)
N0 = 0.53(P4P5)
u = p3p43p1p53p1p5 3 p3p43p1p43p1p5 3 p4p53p1p33p1p4
C - p1p43p1p43p3p5 — p1p43p1p53p3p5 - p1p53p1p33p4p5
t = 2391933(9193*p4p5-plp4*9394-91p4*9395-919539394-PlpS*p395)
NZ = 0.53P4P53(P1P4-P1P5)/(P1P43P1P5)
NX = ((-p4p5/2)33(0.5))tu/((-(p1p43p1p5))3t33(0.5))
NY = (N0332 - NZ332 - NX332)33(O.5)

 

 

4363

4365

4367

4369

5163

I = 0

S =.01
printt,’s= ’,s
X = -1

Y = O

NZ1 - N23X

NX1 = NX3sin(acos(X))

NY1 NY3sin(acos(X))
Z1 = ZO3sin(acos(X))
ZA = ZI*COS(Y)

NN =

NX13cos(Y) 3 NYltsin(Y) 3 NZI 3 NO
III = 83X3ZA3P3P43P3P5303

I = I 3 $3$3NN3NN/((-(PIP43P1P5)3(X3I) 3 MQ:*2)3III)
Y = Y 3 5

IF (Y — 23API) 4365,4367,4367

X = X 3 5

IF ( x- -1) 4363,4369, 4369

PRINT3 ’CHECK I19: ’ I

IS: PIP4¢P3P5 3 P1P53P3P4 - P1P33P4P5

117: P3P4317 3 O. 53P4SIV3(P3P4-P3P5)3II4 3
C 0. 253153P14IV3P15IV3(112 - (P1P4- -P1P5)3P4SIV3II4 3 2*P1P53I7) 3
C O. 06253?14IV3P14IV3PISIV3PISIV3(43P3P43P3P53P1P43P1P5— -I$3IS)3
C (~23P1P53II - 43P1P53P1P5314 - P45IV3DI45323P1P5312)

118: P1P4318 + 0.53P451V3(P1P4-P1P5)3116 +
C 0.2 253133P34IV3P351V3(111-(P3P4-P3P5)3P4SIV3116-(23P3P43D3)318)3
C 0. 06253P34IV3P34IV3P3SIV3P351V3(43P3P43P3P53P1P43P1P5-IS3IS)3

(
(MZP)3Il-MZP3MZP3IS
-P4SIV3(0345323P3P53(P3P4-P3P5)3MZP)313

)
IW = 23P1P33P4P5 - 014530345
IR = -23P4P53P1P3 3 013(23P3P5 3 MZP)
119: P4P53P1P53014SIV3I9 -

nnnn

C 0.530145IV3DI45IV3P4P53(P1P4—P1P5)3I15 3

C 0.53I9U3(1/19L)3

C (I13 3 DI4SIV3DI45IV3IW3IIS3(DI4SIV31W-P3P4-P3P5-03)319) 3
C

C -0.53P4P53P1P43P1P53(1/I9L)

C -0.253I9U319U3(1/19L)3(IIIQL)

C )3

C(

C (DI4SIV3IR)312
C -Dl4SIV3Dl451V3IR3IR316
C ~014SIV3(DS453D3453DI451V3IW3(P3P4-P3P5-MZP))3I3
C)

PRINTt,’I17 = ’,Il7

PRINTt, ’118 = ’ ,118

PRINT3, ’119— ’ ,119

ZO- = 23((P1P43P1PS)33(O. 5))

NO 0. 53(P3P43P3P5)

NX O. 53((P1P43P1P5)33(-0. 5))3 (P1P43P3P53P1P53P3P4- -P1P33P4P5)
NZ 0. 53(P3P4- P3P5)

(NO332 - NX332 - N2332)33(0.5)
0

Z
-<
I
H

4X2 0! H
II II -' II II
a
d

#4!
V»
II
m

0 I

 

 

1710
1720
1730
1740
1750
1760
1770
1780
1790
1800
1810
1820
1830

1840

1850
1860
1870
1880
1890
1900
1910
1920
1930
1940
1950
1960
1970
1980
1990

 

5165

5167

5169

 

234

= NZiX
NX1 = NX3sin(acos(X))
= NY3sin(acos(X))

21 = 203sin(acos(X))
ZA = 213cos(Y)
NN = NX13cos(Y) 3 NYltsin(Y) 3 N21 3 N0

III = (P1P43P1P5)3X3ZA-P1P4-P1P53M0332

I = I 3 StS/((23NN3D3)3(P4P53(X-1) 3 MP332)3III)

Y = Y 3 3

IF (Y - 23API) 5165,5167,5167

X = X 3 5

IF ( X-l) 5163,5169,5169

PRINTt,’CHECK I10 = ’, I
U1DU1=+2.3CC413CC313P4P532.3CC413CC213P3P532.3CC413CC113P1P532.3
CCC313CC213P3P432.3CC313CC113P1P432.3CC213CC113P1P3
U10U2=3CC423CC313P4P53CC423CC213P3P53CC423CC113P1P53CC413CC323P4
CP53CC4I3CC223P3P53CC413CC213P1P53CC323CC213P3P43CC323CC113P1P43C
CC313CC223P3P43CC313CC213P1P4+CC223CC113P1P33CC213323P1P3
UIDU3=3CC433CC313P4P53CC433CC213P3P53CC433CC113P1P53CC413CC333P4
CP5+CC41*CC323P3P53CC4I3CC313P1P53CC333CC213P3P43CC333CC11*P1P43C
CC323CC313P3P43CC323CC113P1P3+CC313CC213P1P33CC313323P1P4
UlDU4=3CC443CC313P4P53CC443CC213P3P53CC443CC113P1P53CC433CC413P4
CP53CC433CC213P3P43CC433CC113P1P43CC423CC413P3P53CC423CC313P3P43C
CC423CC113P1P33CC413CC313PIP43CC413CC213P1P33CC413323P1P5
U2DU2=32.3CC423CC323P4P532.3CC423CC223P3P532.3CC423CC213P1P532.3
CCC323CC223P3P432.3CC323CC213P1P432.3CC223CC213P1P3
U20U3=3CC433CC323P4P53CC433CC223P3P53CC433CC213P1P53CC423CC333P4
CP53CC423CC323P3P53CC423cC313P1P53CC333CC223P3P43CC333CC213P1P43C
CC323CC313P1P43CC323CC213P1P33CC323323P3P43CC313CC223P1P3
U20U4=3CC443CC323P4P53CC443CC223P3P53CC443CC213P1P53CC433CC423P4
CP53CC433CC223P3P43CC433cC213P1P43CC423CC413P1P53CC423CC323P3P43C
CC423CC213P1P33CC423323P3P53CC413CC323P1P43CC413CC223P1P3
U30U3=32.3CC433CC333P4P532.3CC433CC323P3P532.3CC433CC313P1P532.3
CCC333CC323P3P432.3CC333CC313P1P432.3CC323CC313P1P3
U3DU4=3CC443CC333P4P53CC443CC323P3P53CC443CC313P1P53CC433CC423P3
CP53CC433CC413P1P53CC433CC323P3P43CC433CC313P1P43CC433323P4P53CC4
C23CC333P3P43CC423CC313P1P33CC413CC333P1P43CC413CC323P1P3
U4DU4=32.3CC443CC433P4P532.3CC443CC423P3P532.3CC443CC413P1P532.3
CCC433CC423P3P432.3CC433CC413P1P432.3CC423CC413P1P330.

ZIIOU = U10U13(O.53IB3O.5301315)3U20U23(0.5317-0.25303314)
C - (U3DU3-23U30U43U40U4)30.5319
C 3 UlDU23(0.5311301314-0.53033IS)
C - (UIDU3-U10U4)3(O.5313301316)
C 3 UIDU43P4P53IS-(U20U3-U2DU4)30.53(12—03316)3U20U43P4P5314
C - (U3DU4-U4DU4)3P4P53I6

ZIlOD = U10U13013013U20U230.253033033U4DU43P4PS3P4P5
C -1.03U10U2301303323UIDU43P4P53DI~U20U43P4P53D3

UPPER ~ZIIOU3CDNJG(ZIIOD)

LOWER ZIIOD3C0NJG(ZIIOD)

110 = UPPER/(LOWER)

PRINT3,’110 = ’,110

END

 

2000
2010
2020
2030
2040
2050
2060
2070
2080
2090
2100
2110
2120
2130
2140
2150
2160
2170
2180
2190
2200
2210
2220
2230
2240
2250
2260
2270
2280
2290
2300
2310
2320
2330
2340
2350
2360
2370
5010

 

235

_HSCOOO‘DUAS:[EH1NCJCK.LDG$8

3 IF F$HODE() .EGS. "BATCH" THEN 6 EXIT
3 RUN CHECK
3.142857
D3 I (8779.690.-252.9900)
P3P4 =
P3P5 I
P4P5 I
P14HAX = 394.14882
P1P4 I
P15UP = 0.0000000E300
PISDN I “23.00508
P1P5 I
P1P3 I “2182.000
PIPS I 66.06058
80 I 1.4142136E-03
CUO I 4.8393107E-04
E0 I 3.4219094E-05

EVL I 899726.6

LN1 I 5.993961

LN2 I 8.542081

LN3 I (~3.090159.-2.427214)
LN4 I 3.125785

LN5 I (1.2334378E-02.1.780435)
LN6 I (0.449509la-1.008301)

CHECK 11 I (0.188166130.0000000E300)
CHECK 12 I (2.0959381E-02o0.0000000E+00)
CHECK 13 I (4.2695566E-03na.3561315E-03)
11 I (0.1883816.0.0000000E+00)

12 I (2.0953396E-0200.0000000E300)

13 I (4.3826397E-03o3.4424122E-03)
HZPIV I (1.9288890E-03.1.5457852E-03)
CHECK 14 I (1.280062OE-03oo.0000000E300)
14 I (7.5656187E-04o0.0000000E+00)
i4d¢t1 I (1.0:32990E-03.o.00000005300)
CHECK 15 I (2.2208611E-04.2.6291382E-04)
15 I (1.2245527E-04.2.1099522E~041
i5det1 I (1.4313067E-04o2.0318378E-04)

S I 5.0000002E-04

CHECK 16 I (8.5144438E-06a4.8571172E-05)
16 I (9.2996797E-06.4.2994034E-05)
ibdctl I (5.9150848E-Obo4.8653204E-05)
CHECK 17 I (-5.436626.0.0000000E+00)
CHECK 18 I (-O.5183365.-0.6392409)
CHECK 19 I (~4.7151849E-04.-2.4306704E~03)
17 I (~4.466591.0.0000000E+00>

18 I (-0.3241602.-0.4922062)

19 I (-2.1155991E-04.-5.3360919E-04)

111 (-388.0707.0.0000000E300)

112 - (-¢74.o4ee.o.00000005300)
113 - (-0.6314969.0.0000000E300)
113 - (~80.43964.0.000000083001
115 : (-O.2913233.-6.77223438-02)
116 . (-7.577301.-a.xoeaea>

117 - (10072.33.0.0000000£+00)
119 = (469.3852.892.2405)

119 = (~103.7464.-261.4523)

‘ut'

r

. qu : c1 nQDQnIOt-OA Q 07183995-063

 

 

 

 

236

I ‘0 1883816-0 3000000E+00)

I (2.0953396E-02.0.0000000E+00>

OJ I (4.3826397E-03.3.44241225-03)

HZPIV I (1.9288890E-03a1.5457852E-03)
CHECK 14 I (1 2800620E-03.0 0000000E+00)
14 I (7.5656187E-04o0.0000000E+00>
i4det1 I (1.ossze¢os—os.o.00000005300)
CHECK 15 I (2.2208611E-O4.2 6291382E-04)
’5 I (1.2245527E-04.2.1099522E-04)
'Sdetl I (1 4313067E~04.2.03183785-04)

= 9 9999998E’03

CHECK 16 I (3.6448710E-05.2.6391298E-04)
16 I (9.2996797E-Obo4.2994034E-05)
zodetl I (5.9150848EIO6a4.8653204E-05)

"HOO
if)...

 

0

n»

d

 

CHECK 17 I {-5.436626.0.0000000E+00)
CHECK 18 I {-0.5183365.-0.6392409)
CHECK 19 I 1-4.7151849E-04.—9.4306704E-03)

17 I (~4.466591o0.0000000E+00)
18 I (-0.3241602o-0.4922062)
19 I (-2.1155991E-04o-5.3360919E-04)

_HscboosDUAe;IENINCJCK.LOG.1

 

111 I (-388.0707,0 0000000E+001
112 I (-674.0468.0 0000000E300)
113 I (-0.6314969.C 0000000E300)
114 I (-80.43964.0 0000000E*00)
115 I (-0.2915253.-6.7722343E-02)
116 I (-7.577801.-8.108868)

117 I (10072.33.0 0000000E+00)
118 I (469.3852o892.2405)

119 I (-103.7464.-261.4523)

s I 1.00000005-03
CHECK 110 I (7.6385157E-O7.2.8371851E-06)
130 I (5.7549897E-O7.4.04321748-06)
xFOR-F-ENDDURREA. end-o3-File during read

unit -4 File SYS‘INPUT:.5

user PC 000037E2
—RH$-E-EOF. end of file detected
ZTRACE-F-TRACEBACK. symbolic stack dump 3ollows

module name routine name lane 3.} pc

oooocooa
ooooarea
ooooacao

CHECK191NTEGRAL CHECK191NTECRALS 477 gg°°§716
EHING Job termxnated at 1-JUL-19Ba 19:56'00.85 00;FE2

Accountxng Information'

BUIIEFQG I/O count: 68 PQIR martin S

Direct.1/0 covnt 58 p2.. p.9. 621.9:i:1392 529
Page faults. 539 Haunted valum.,. ' 7B7
Charged CPU time- 0 00-10.31.50 Elapsed txmet '

0
0 OO:45:41_.

237

_HSCOOOIDUAS'EENINGJCK.LDG;1

CHECK I7 I (—5.436626.0.0000000E+00)

SI 9 9999998E-03

CHECK 18 I (-O.5183365.-O.6392409)

s: 9.9999998E-03

CHECK I9 I (~4.7151849E-O4.-2.4306704E-03)
27 I (~4.466591.0 00000008300)

15 I t-O.3241602,-O.4922062)

1° I (-2.1155991E-04.-5.3360919E-O4)

SI 5 0000002E-04

CHECK 11 I <-387.2682.0.0000000E+00)
EI 5 0000002E-04 '
CHECK 12 I (-674.6837.0.0000000E+00)
sI 5 OOOOOOZE-O4

CHECK 113 I {-0.3193493o0.0000000E+00)

E I 5.0000002E-04

CHECK 114 I {-70.59887.0.0000000E300)
sI 5 OOOOOOZE-04

CHECK 115 I (-0.5000000.-0.5654656)

S I 5.0000002E-04

116 I (-8.238111.-12.45711)
(-388.0707.0.0000000E300)

n
I
m
n
x

111

112 I (-674.0468.0.00000005300)
113 I (-O.6314969.0.0000000E300)
114 I (-80.43964.0.0000000E300) ,
115 I .(-0 2915253.-6.77Z234BE-02)
116 I (-7.577801:-8.108868)

SI 9.9999998E-03
CHECK 117 I (23191.27.O.OOOOOOOE3OO)
SI 9.9999998E-03
CHECK 118 I (1225.885.1556.307)
sI 9.9999998E-03
CHECK 119 I (1.5155446E-OZ-2.4578951E-02)
117 I (10072.33.0.0000000E300>
118 I (469.3852oe92.2405)
119 I (-103.7464.3261.4523)
s I 9.9999998E-03
CHECK 110 I (1.0828019E-06.3.0718329E-06)
110 I (5.7549897E-O7;4.0432174E-06)
xFDR-F-ENDDURREA. end-of—File during read
unit -4 file SYS!1NPUT:.5
user PC 00004474
~PHS-E-EOF. end 03 File detected
kTRACE-F-TRACEBACK. symbolic stack dump 3ollowg

module name routine name 11“. Tel PC
OOOOCCDé
OOOOCBEB
00009860
CHECK191NTEGRAL CHECK191NTEGRALS 646 30009313
EWING Job terminated at 4-JUL-19BB 11:08:05.74 00°3A74

Accounting information: - a - _.

Buffered 1/0 count' 72 Peak workin -

Direct 1/0 count; 69 Peak Page 1:1:.:1:1?.: 535
Page faults: 539 Haunted VOIUm.‘- I. 787
Charged CPU time: 0 01:28:07.84 ‘

Elapsed time: 0 01'39‘22

 

 

 

 

238

CHECK 17 I (-5 436626a0.0000000E+OO)
SI 9.9999998E-03

CHECK 18 I (-0.5183365.-O.6392409)

SI 9.9999998E-03

CHECK 19 I (-4.6512054E-04.-2.4429935E-03)
17 I (-4.466591.0.0000000E+00)

18 I (-0.3241602:-0.4922062)

19 I (-2.1155991E-04.-5.3360919E-04)
111 I (-388.0707.0.0000000E300)

SI 9.9999998E-03

CHECK 11 I (~487.2787.0.0000000E300)
112 I (-674.046810.0000000E300)

SI 4.9999999E-03

CHECK 12 I (~753.0253.0.0000000E300)
113 I (-0.6314969.0.0000000E+00I

S: 9.9999998EI03

CHECK 113 I (-1.124600.0.0000000E*00)
114 I (-80.43964.0.0000000E300)

S I 9.9999998E-03

CHECK 114 I (-327.2178.0.0000000E300)
115 I {-0.29152531-6.7722343E‘02)

:I 9.9999998E-03

CHECK 115 I (-0.2977193.-6.8340585E-02)
116 I (-7.577801o-8.108868)

S I 9.9999998E-03

CHECK 116 I (-12.40340o-13.92181)

SI 4.9999999E-03

CHECK 117 I (21520.46.0.0000000E+00)
S. 4.9999999E-03

CHECK 118 I (967.8455o1364.273)

s- 3.99999995-03
CHECK 119 - <1.6154986E-02.1.6571850E-02>

117 I (19075.81.0.00000005*00)
118 I (749.0870.1194.346)
119 I (1.3806876E-02.8.2‘38448E-03)

s - 9.9999998E-03
CHECK 110 - (1.0650439E-06.3.07624605-06)

110 I (5.7549897E-07.4.0432174E-06)
ZPOR-F-ENDDURREA. end-of-file during read
unit -4 Qile SYS‘INPUT:.5
user PC 00004548
-RHS-E-EOF. end of file detected

 

 

xTRACE-F-TRACEBACK. symbolic stack dump follows 1 PC
module name routine name line re
OOOOCEDé
OOOOCDEB
00009A60
00009516
CHECK19INTEGRAL CHECK191NTEGRALS . 657 00 00003848
suxuo Job terminated at 5-NDV-i988 16.14.22. .
Ac t‘ information: '
Bu::::e:n1/0 count: 67 Peak working set size. 536
Direct I/o count: 54 Peak page_311e Size: 837
Page faults- 539 Hounted golumes: 0 .
' 0 00:01:38 80 Elapsed time: 0 00 01:49.

Charged CPU time:

 

239

CHECK 17 I (-5.436626.0 0000000§+00)
SI 9.9999998E-03
CHECK IS I (-0.5183365.-O.6392409)
SI 9.9999998E-03
CHECK 19 I (-4.6512054E-04.-2.4429935E-03)
17 I (-4.406591.0.0000000E+OO>
18 I (-O.3241502.-O.4922002)
19 I (~2.1155991E-O4.-5.3300919E-04)
111 I (—388.0707.0.0000000E+001
SI 9.99?999BE-03
CHECK 11 I {-487.2787.0.0000000E+00)
112 I (-674.0468.0.0000000EIOO)
SI 4.9999999E-03
CHECK 12 I (-753.0253.0.0000000E+OO)
113 I (-0.0314969.0.0000000E+OO)
SI 9.9999998E-03
CHECK 113 I (-1.124600.0.0000000E+OO)
114 I (-80.439b4.0.0000000E+OO>
S I 9.9999998E-03
CHECK 114 I (-327.2178.0.0000000E+OO)
115 I {-0.2915253.-6.7722343E-02)
SI 9.9999898E-03
CHECK 115 = (-O.2977193.-6.8340585E-02)
116 I (-7.577801.-8.108868)
s - . 9.99999995—03
CHECK 116 I {-12.40340.-13.92181)
SI 4.9999999E-03
CHECK 117 I (21520.46.0.0000000E+OO>
SI 4.99999?9E-03
CHECK 118 I (967.8455.1364.273)
s- 4.99999995-03
CHECK I19 I (1.6154986E-02.1.6571850E-02)
117 I (19075.81.0.0000000E+OO)
118 I (749.0870.1194.3401
119 I (1.3806B76E-02.B.243844BE-03)
S I 9.9999998E-03
CHECK 110 I (1.0650439E-06.3.0762460E-06)
110 I (5.7549897E-07.4.0432174E-06)
ZFOR-F-ENDDURREA. end-of—file during read
unit -4 file SYSSINPUT:.;
user PC 00004548
-RHS-E-EOF. end of file detected
ZTRACE-F-TRACEBACK. Sumbolic Stack dump Follows

module name routine name line

657

O CHECK191NTECRALS
CHECK1.INTEGRAL s-NDV’lqea 15:14:22-00

EHINC Job terminated at

Accounting information:

Buffered 1/0 count: 67
Direct 1/0 count: 54
Page taultS: 53?
Charged CPU time: 0 00:01:38.80

 

rel PC

0000CE06
OOOOCDEB
00009A60
00009516
00003348

 

 

 

240
,ISCOOCSDUAa-tsuruoicx LOG:2

.IECV :7 I (-5 430020 0. OOOOOOOEIOC)

SI 9 °°999§8E~03

CHECP 18 I {-0.5183305.-0.6392409)

s: 8 89999985-03

CHECK 1° I {-4 71518495-04.-2u¢300704E-03)

1‘ I --4.400591.O.0000000EIOO)

I 1-0.3241002,-0.4922062)
i-2.11558915-0Ir-5u3360919E-04)

I (~388 0707.0.0000000E+00>

9 99998885-03

Ct. '1t- I - (P48772792: 0 0000000900,-
I 1-674 0468,0.0000000E+OO)
4.9998999E-03

CHECK 12 I 1—753m6212.0-00690605+88+—

I13 I (-O.6314969.0.00000005+OO)

SI 9 9899998E- 03

“HECV 113- --‘.-t 1434596. W9

.14 = (-80 43964 O OOOOOOOEIOO)

5 I 9.9999998E-03

.: 'CHECK 114 I <-327v+87ir8r8800008€+009-

.. I: I (-0.2915253.-b.7722343E-02)

L. sa 9 9999998E-03

m-b-HECK’ '1 :5 I “-0. 29861-89.~77-602097¢E-629 -

18 I '-7 577801.-8.108868)

30 I 9 a999998E-03
3:‘“€HE€K—ilo'I“"1I12.403*2r-13r927819'

;. SI 5.0000002E-04

.. CHECK 117 I (19989.04.0.0000000E¢00)
—-SI"-9.-°99999BE-08—"III ..

CHECK 118 I (1225 885.1556.307)

SI 9 9999998E'03
-—eHECK-+i9-I-—+i:—5r&5§‘65—6292~45789fiiE-621
3: 117 I (18432.83 0.0000000EIOO)
as 118 I (697.3430.1197.104)

q-——++9-—m+-e~909+952£—02v+~4¢eaterE—ee+—

z, a I 9 99999985-03
.. CHECK 110 I (1.0828019E-Ob.3.0718329E-06)
~-1i—O -.-—(.5 #W‘FPE-OH.0432-17IE-06+ -

ZFDR-F-ENDDURREA. end-oG-file during read
- unit -4 file SYS‘INPUT:.3
'I"—~vser-PC'6990I45t'I'II .....

~9HS-E-EOF. end of file detected

27RACE-E-TRACEBACK. symbolic stack dump Follows

..a n g"
I

-- lec'em rod-co
..a

" a; .‘E "

O
f
O

U. 0.0

 

-_- .-. .-—-

 

(-
L.’

 

 

.? "module”name - ~ rootine name - ~- -- v-~ - itne rel PC
:3 ooooccoe
.. - --« --- --- -I-—--——-———-~—- --I--«8000C859
00009860
,_ 00009316
;. CHECK191NTEGRAL CHECK191N¥EORALS----~ ---~ - -----6‘¥-~--- 00003fi£1
' EHINC Job terminated at 8-JUL-1988 20:08:20.36
- Accountrng rn+ormatt0n
Bu‘fered 1/0 count 68 Peak working Set Size: 535
Direct 1/0 count; :9 Peat page File Size: 787
§‘---the Tovitet ‘59¢—————-Hoon+ed-vo+vmet< <%
Charged CPU time. 0 00:22:52.56 ElapSed tine: 0 00:32:51.

0

 

 

 

 

 

 

 

Appendix U

 

 

 

 

 

FORTRAN Output Sample Results

 

241

 

 

 

e‘_ _‘h‘ ' -" .

 

 

47
47
4
4
-u
-n
4
4
47
-n
4
4
47
-u
4
4
4
4
47
-n
4

~27
~27
~17
~17
~7
~7

-u
47

~17
~17

~7
~27
~27
~17

~7

~7
~37
~37
~27
~27
~17

~7
~7
~27
~27
~17
~17
~7
~7

.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
.28
~27.
~27.

28

.28
.28
~7.
.28
.28
.28
-7.
.28
.28
.28
.28
~17.
.28
.28
.28
.28
.28
.28
.28
~17.

28

28

28

.28
.28
.28
.28
.28
.28
.28
.28

~10.
~10.
~10.
~10.
~20.
~20.
~20.
~20.
~30.
~30.
~30.
~30.
~40.
~40.
~40.
~40.
~50.
~50.
~10.
~10.
~10.
~10.
~20.
~20.
~20.
~20.
~20.
~20.
~30.
~30.
~30.
~30.
~30.
~30.
~40.
~40.
~40.
~40.
~10.
~10.
~10.
~10.
~10.
~10.
~20.
~20.
~20.
~20.
~20.
~20.
~20.
~20.
~30.
~30.
~30.
~30.
~30.
~30.

88888888888888888888888888888888888888888888888888888

 

~40.00
~40.00
~50.00
~50.00
~30.00
~30.00
~40.00
~40.00
~20.00
~20.00
~30.00
~30.00
~10.00
~10.00
~20.00
~20.00
~10.00
~10.00
~40.00
~40.00
-50.00
~50.00
~20.00
~20.00
~30.00
~30.00
~40.00
~40.00
~10.00
~10.00
~20.00
~20.00
~30.00
~30.00
~10.00
~10.00
~20.00
~20.00
~30.00
~30.00
~40.00
~40.00
~50.00
~50.00
~10.00
~10.00
~20.00
~20.00
~30.00
~30.00
~40.00
~40.00
~10.00
~10.00
~20.00
~20.00
~30.00
~30.00

24

~10.00
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.00
~10.00
~10.00
~10.00
~10.00

8888888888888888888888888888888888888888888888888888

 

2

~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~20.
~20.
~20.
~20.
~20.
~20.
~20.
~20.
~20.
~20.
~20.
~20.
~20.
~20.
~20.
~20.
~20.
~20.
~20.
~20.
~30.
~30.
~30.
~30.
~30.
~30.
~30.
~30.
~30.
~30.
~30.
~30.
~30.
~30.
~30.
~30.
~30.
~30.00
~30.00
~30.00

8888888888888888888888888888888888888888888888888888888

47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47
47

HHHHHHHHHHHHHHHHHHHOHHHHHHHHHHHHHHHHHHOHHHHHHOHHHHHHHHOHCH

.0001823
.9995154
.0004421
.9997410
.0012978
.0009155
.0004318
.0001115
.0014023
.0016459
.0006382
.0005274
.9998592
.0003670
.0004268
.0004545
.0004847
.0004878
.0000499
.9999858
.0001203
.0000110
.0001280
.0001181
.0000650
.0000538
.0000736
.0000571
.0002115
.0002140
.0000709
.0000695
.0000588
.0000566
.0000805
.0000807
.0000527
.0000530
.9999997
.0000006
.0000060
.0000060
.0000273
.0000241
.0000234
.0000242
.0000116
.0000124
.0000155
.0000159
.0000300
.0000296
.0000151
.0000153
.0000153
.0000157
.0000250
.0000250

 

 

 

~47.
~47.
~37.
~37.
~27.
~27.
~17.
~17.

~7.

47
47
47
47
47
47

4.

4.
47
47
47
47
47
47
47
47

4.

-7.
47
47
47
47
47
47
47
47

~7.
~17.
~17.
-7.
~7.
~17.
~17.
~7.
~7.
~7.
~7.
~37.
~37.
~27.
~27.
~17.
~17.
~37.
~37.
~27.
~27.

~10.

~10

~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~20.
~20.
~20.
~20.
~20.
~20.
~20.
~20.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~10.
~20.
~20.
~20.
~20.
~20.
~20.
~30.
~30.
~30.
~30.
~40.
~40.
~40.
~40.
~50.
~50.
~10.
~10.
~10.
~10.
~10.
~10.
~20.
~20.
~20.
~20.

8888888888888888888888888888888888888888888888888888888888

I
Hi.
00

8888888888888888888888888888888888888888888888888888888888

~40

-w

~30.
~40.
~40.
~50.
~50.
~10.
~10.
~20.
~20.
~30.
~30.
~40.

~10.
~10.
~20.
~20.
~30.
~30.
~40.
~40.
~50.
~50.
~30.
~30.

~40.
~20.
~20.
~30.
~30.
~40.
~40.
~20.
~20.
~30.
~30.
~10.
~10.
~20.
~20.
~10.
~10.
~20.
~20.
~30.
~30.
~40.
~40.
~10.
~10.
~20.
~20.

243
~10.00 ~40.
~10.00 ~40.
~10.00 ~40.
~10.00 ~40.
~10.00 ~40.
~10.00 ~40.
~10.00 ~40.
~10.00 ~40.
~10.00 ~40.
~10.00 ~40.
~10.00 ~40.
~10.00 ~40.
~10.00 ~40.
~10.00 ~40.
~10.00 ~40.
~10.00 ~40.
~10.00 ~40.
~10.00 ~40.
~10.00 ~50.
~10.00 ~50.
~10.00 ~50.
~10.00 ~50.
~10.00 ~50.
~10.00 ~50.
~10.00 ~50.
~10.00 ~50.
~10.00 ~50.
~10.00 ~50.
~20.00 ~10.
~20.00 ~10.
~20.00 ~10.
~20.00 ~10.
~20.00 ~10.
~20.00 ~10.
~20.00 ~10.
~20.00 ~10.
~20.00 ~10.
~20.00 ~10.
~20.00 ~10.
~20.00 ~10.
~20.00 ~10.
~20.00 ~10.
~20.00 ~10.
~20.00 ~10.
~20.00 ~10.
~20.00 ~10.
~20.00 ~10.
~20.00 ~10.
~20.00 ~20.
~20.00 ~20.
~20.00 ~20.
~20.00 ~20.
~20.00 ~20.
~20.00 ~20.
~20.00 ~20.
~20.00 ~20.
~20.00 ~20.

00

~20.

~20.

 

 

8888888888888888888888888888888888888888888888888888888888

~17.
~17.
~17.
~17.
~17.
~17.
~17.
~17.
~17.
~17.
~17.
~17.
~17.
~17.
~17.
~17.
~17.
~17.

~7.

~7.

~7.

~7.

~7.

~7.

~7.

~7.

~7.

~7.
~37.
~37.
~37.
~37.
~37.
~37.
~37.
~37.
~37.
~37.
~37.
~37.
~37.
~37.
~37.
~37.
~37.
~37.
~37.
~37.
~27.
~27.
~27.
~27.
~27.
~27.
~27.
~27.
~27.
~27.

 

HHHHOHHHHHHHHHHHHHHHOHHHHHOHHHOHOHOHOHHHHHHHHHOOOHOHOHHOHO

.9999987
.0000116
.9999945
.0000080
.0000180
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.9999896
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.0005606
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.0000486
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.0000381
.0000353

 

 

 

 

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HHOHHHOHOHOHHHOHOHOHOHHHHHOCOHHHHHHOOHOHHHOOHHHHHHHHHHHHHH

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II
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245

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Appendix V
Choosing the Output Data Display
Points.

Given that the allowable output data. points lie on a triangular surface:
~P3P4

-P4P5

—P3P5

2
where each corner of the triangles intersects the. ans at 3.,

Each edge of the triangle is obviously of length (fl) (5%)

To find the angle between the plane of the triangle and any of the three co-
ordinate planes. for example the x-y plane:

The normal to the triangle is N1 =: i + j + k

The normal to the x-y plane is N2 = k

N1 'N2 =1=IN1HN2|c030 = (1)(\/8)cosfi

cosfl—-1—- sinfl-[g
'75 75

So if p measures the distance ‘up‘ the face of the triangle from the base line in
the x-y plane, p is related to the 'z' co—ordinate. that is the value. of —P3P4.
by:

\f2-
—P3P4 :: ——
P J;

247

 

 

 

 

 

 

 

 

248

433134

-P4P5

 

~P3P5

Viewing the triangle straight on and selecting 15 symmetrically spaced rep-
resentative points.

Allow the internal distance between any two points to be one unit, with an
additional 3 unit to the corner. as shown.

-P3P4

 

 

4 units

 

 

 

 

 

 

 

 

249

Then 5 1/2 units equals the edge length of (72) (5%“) or

lunitz (121.) (,5) (57372;)

The width of each border is i 78 units, and the vertical distance between rows
is % 78 units, so if N = 0, 1, 2, 3, 4 indicates the horizontal row number, then
the distance p from the lower edge to that row is given by:

p = [G 73) + N G- 7§)] [length of one unit]

2

_(_)(E;.,,(,..,

and since \/_
2

—P3P4= —

p 4

then 2 E
gm 1 r
(a) (——. >) (2 +1)

and the values of —P3P4 to run the FORTRAN program at to generate these
display points are given by allowing N to range through its values 0.1.2.3.4.
In a similar fashion the values of the other axis, -P3P5 and —P4P5 are gen-

... _2_ E3. .1.
’P3P”’(11)(2’)(2+L)
2 E3", 1

and the set N,L,M locates the display point.
The set of three values for the other triangle, —P1P3, —P1P4. —P1P5 are gen-

erated in an identical fashion.

_ l .1. 3 E_gm
—[(4\/§)+N(27§)] [1172’ ]

 

erated,

 

 

 

 

Appendix W
Evaluation of a Preliminary Diagram

Section 1 Introduction

Consider the diagram of a single, massive, incoming particle of mass Ml > M“
which decays to M; and an outgoing ’quark‘. Assign Spin zero to all external
particles. Allow the outgoing ‘quark’ to have mass, MS.

Section 2 The Diagrams

 

This four point lepton scattering diagram will serve to illustrate the notation
used for both diagrams in the problem.

 

 

P3
/

P3 + k, M. / \k“ P4

/
/
P1, Ml __f / m P5. MS
7 / }
The linear momenta are labelled clockwise P1....P5 beginning with the GLUON.

Icy is defined as the variable LOOP momenta and for conveinence:

m:k-(P4+P5)

Four vector dot products are indicated by the contiguous writing of two four
vectors, for example, P4P5 indicates the dot product of P:1_with P5.
The three part of a four vector is indicated by a bar: e.g. P4 is the space part

of P4.

250

 

251

The anti-lepton scattering diagram is:

 

 

 

 

P3
/ .
P3 + k, M. / \k~
/ P4
/
P1, Ml 7-6 / / l 1]] P5, M5
1
Section 3 Preparing the Calculation
The Cutowslry2 rule for an on—shell condition gives:
1 1 n o o .,
. ,, ~2.T~6k'6k—P4 P. - M'
(k3)[(k-(P4+P5))3+Mg] : 2( T?) ( ) ([ ( + 5)] + 5)

With this, the amplitude of the sample diagram of the previous section is formed
by the Feynman rules of Appendix D:

Let
A! = 5* I

where S collects all the multiplicative scale factors:

1
(270"

 

S = g (27:7)? e2

 

and I is the LOOP integral of the above diagram:

I _ d4]: 6(k2) 5w: — P4 — P5)2 + Mg] (1: + P4). 9"” (P4 + P5 — k + P5).
"/ [(1. — P4P] [(k + P3)? + M; — mu.)

Sub-Section 3.1 The flame of Reference

2

The first delta function will dictate that m2 = k0 .
The second, 6([k — (P4 + P-5)]2 + 1115'). gives

1:?—2k-(P4+P5)+(P4+P5)'-’+M§=0

252

and with

’92 =3 P42 :: O
and

P52 = 4115?
results in

k - (P4+ P5) = P4P5

which assumes a. particularly simple form only in the rest frame of P3 and F5:

ko(E4 + E5) = -P4P5

And, since also in this frame:

(P4 + P5)2 = —(E4 + 13.5)? = 2(P4P5) — M:

from the above result

(P4P5)2

———:—- = 2(P4P5) — 11153
ko~
then 9
v) (P4P5)'
ko" -

 

_ ‘7 Mg — 2(P4P5)

establishing that in this frame k0 is CONSTANT.

The enormous simplification gained when the magnitude of 8 remains constant
dictates this rest frame of P71 + P5 to be the frame of reference for the lepton

diagram (and similarly P_3 + P5 for the anti-lepton).

Note that in this frame the energy, Ex. of any other particle, PX. of the problem
is also simply given: .
P4P5

= PXPd + P.\'P5
k0

 

PX ~ (P4 + P5) = -E,(E4 + E5) = E,

or .,
k0" . .
E? = —— .\ P4 PA P5

 

 

253

Sub-Section 3.2 The SPIN Factor

The SPIN factor in the above integral can be evaluated to:
(k + P4)” 9“” (P4 + P5 — 15+ P5).

2(P5 ~ 1:) + 2(P4P5)

: 2P5 - ‘ — 2(ko)E5 + 2(P4P5)

then using the above equivalent for (k0)E5

=2P5-F—2

k0?

_ .3 1'
P4P5(P4P5 A15)+2(P4P-))

 

Sub-Section 3.3 Removal of the Delta functions

Using
(141: = d [kl dko k02 ko

The delta functions may be simply removed by integrating over (1' [kl and dko
provided the J ACOBlAN between the set ( [1:]. kc) and the two arguments of the
delta functions is included. This Jacobian is the same as for the larger problem.
It is calculated in the chosen frame of reference in Appendix F and evaluates
to:

—1

J:———

4(P4P5)

The Jacobian is now absorbed into the scale factor S:

_ 1 ,2 2 1 —1
S" 5 (27”) 6 [(2.44] 4(P4P5)

1

5=;_____

1
137 (2a) 4(P4P5)

Thus the sample LOOP integral reduces to being over Mgr/Qt.
k0? may be absorbed into the scale factor:
1 k0?

, 1
>=—————-

137(27)4(P~1P5)

 

254

leaving the integral over the directions of 1::

___ I191 (k + P4)” g“" (P4 + P5 -— k + P5)..
[Ur ~ P4)'-’] [(k + P3)? + 113 - {I‘ll-1:]

with the understanding that

#

(i) the frame of reference is P3 + 5 == 0

(ii) )1)? = 1.02

(P4195)?

one I)
("1) L0 " big-2(P4P5)

 

(iv) tow.) = P3; (PXP4 + PXP5)

This applies to both LOOP integrals in the problem.

Sub-Section 3.4 The Divergence Regulation

In examining the above LOOP integral. the first denominator factor. arising
)agator. produces a singularity when the direction of

from the photon line pro;
It becomes co-linear with P4, resulting in a. logritlnnically infinite value for the

integral. This infinity does not occur in Nature - the process is finite — and
that is reflected in the calculation by the coefiicent of the logritlnnically infinite
term becoming zero when the LEPTON scattering and the ANTI-LEPTON.
scattering LOOP diagrams are summed. To allow the doing of the diagrams
separately, a small photon mass is added to regularize the infinity. When the
diagrams are added it is observed that the coefficent of the log of this ficticous

photon mass becomes identically zero. producing no ill effects when the photon

mass is then allowed to go to zero and its log to infinity. The appearance of
the zero coefficent does in fact serve as a Check that the various parts of the

calculation are fitting together correctly.

I— fink (k-i-P4lug'w (P4+P5—k+P5)"/‘
'- [(1.2 .— P4)2 + 113] [(i' + P312 +‘l13’ir‘m

 

 

255
Section 4 The Calculation
I /' th [2(P5 - k) + 2(P4P5)]
(-—2P4 - k + M?) (2P3 - k + M? — iI‘M;)

In a co-ordinate system with P4___ in the +2 direction and using the above ex-
pansion of the SPIN factor with P5: ——P4:

 

_ / dn.[— —2|P5|kocosa—2 ,;;,:5,(P4P5— Mg)+2(P4P5)]
" (—2ko-cos€ + 2ko- + Mg.) (2P3 - k + M} —i1‘M;)

ko ( a. 0050 + b.)
(cc056+ d.) (2P3-k+ Mg"— {I‘ll/1;)

 

where __
a. = —2 [PSI kc
and since |_5| = (7| : E4 = 1:0
a : —2ko2
1‘02
(2 = —-2 P4P5— M 2 P4P
(P40P5)( 52)_,_ ( 5)

('2 —2k02 = a

d=2k02+Mz.=—c+M$

I—' 119;.(“7 [cc059]+ +/ (19,,(b—L d)
_ ) 2P3 k+M.-—i.I‘M. )

(ccos9+d)(2P3 k+1\1-—1FAI- (ccosfi+d)(

I=Ia+Ib

First solve Ia:
ko( c cost? +d) =/ Q;-
(cc059+ d) (2P3- L+M-—11‘I\I ) (2P3-k+M;—’—i1‘.\[:)

noting 5%: 1.

As this is integrated over all angles, the co—ordinate system may be switched to

256

P3 in the +2 direction.

I =/+1 qu d(cos€)
° _1 2033ch c059 — 2(E3)ko + M} — irM;

 

_ —21r ,0 (-4(E::ko+ Alf—11PM:
‘ 2(E3)ko g M2 — iI‘M
I _ _ mp4? ,0 -—4 ,;(P3P4+ (P3P5)+ ’\I;-1'I‘M..
° ko'—’(2P3P4 + 2P3P5) M- — iI‘M-

Now to solve Ib. It remains in the original co—ordinate system of P4 in the
+2 direction and taking P3 to be in the x-z plane.

             

ad d9.
I = b— ——
b ( c )j (~2kozcosfi + 21:02 + 1113) (2P3=(ko)co:9 + 2P3,(ko)sin€ cow — 2(ko)E$+ ME “'1‘sz
. a _
Usmg -; _ 1
(b d) (112,.
— i (c c056 + d) (e c059 + f 51110 cosq’: + g)

 

noting P3- Péfi— —-P',33% + E3 = 3%}? + E3
then 0
C : -—2k0'
d: 21:02 +M2— - —c+ M2
e- — 2(ko)P3-- __ 2(P3P4) + 2(lco)E3—— = 2(P3P4) + ——-—-(P3P4 + P3P5)

g = TEE—“0:11:31?“ P3P5) + M:—1'I‘M

e? + f2 = (2(1201E313 = [32%— ———2—5(P3P4 + 193135)]

then using

~t__i——:.-q’2:r7"q a >b>c
0 a+bcosd> (a- —b-)1/-

257

1 (11‘

11> =(b—dl27111(.cx+d) ((63+f2)$2+2962:+g3—f2)1/3

Then transforming a: = or + d

 

I : (b -— (0217/ . 9 o
-c+d :1: (mx- + 11.1: +p)1/-

where .
m. = 62 + f2
n = 2(gec — dm.)
p = dim — 2gedc + c2(g2 — f2)

which is solved using

/ dz :1, 2m+nx+2p

 

 

o
zXl/z \/1-) g .1:

X=mx3+nr+p p > 0

__(b-dm (gm/£5472)" , g—W)_, (35)]
1“ C(9+6) [log( (9+6)2 )+Og(g+\/e=’+f2 09 M;

Now noting ., .
g + e = 2(P3P4) + M; - zI‘M:

= AI Z P
using the symbol A! Z P for the Z particle proprogator

MZP .—. 2(P3P4) + M3 — iI‘M:

g + Me? + f2 = M3 - iI‘Mz

 

 

 

41202 o .
_ 2 2 -_-.- - P3P4 + P3P5 + M; - If)”:
g e + f P4P5( )
b - d —2ko-?(P4P.5 — Mg) + 211941195)2 +1
6 = 412021104195 )
_ -2(P4P5)

[.102

258

gives 2

—4, P4P .3—'I-3 —4k°’ ' 13—' / ‘~’
,6: kw; 5) 10g ((111, . my.) )+,Og P4P5(P3P4+P3I’5)+M; er; _ 10g 4ko
(, o )MZP MZP~ I113 — MI; 1113

 

Combining Ia and lb, folding in the scale factor S = —’k———— and

02
137(27r)4(P4P5) ‘
~ 2 _ 1941952
replacmg k0 — m

. 1
AI = S I = S L, I M : ——— ' .
* *i + bl 4,137) tunes

—2 10 (ME—mm)?
MZP g MZP?

_( 2 + 1 1 -[,\,542(f:,};ilp5,,(P3P4+P3P5)+Mf—iI‘AL
MZP 2(P3P4)+2(P3P5)) 0g IlIE—iFM:

if? )l-1°(/‘PW ll
MZP g [11.152-21P4P5HWH

The anti-lepton scattering diagram produces the indentical result with P4 and
P3 interchanged and an overall minus Sign to account for traversing the variable
LOOP momentum k“ in the opposite direction when writing the lepton line
TRACE.

When the two diagrams are combined. the leading term, which is symmetrical
in P3 and P4, vanishes, and the co-elficent of the log(1l--I.';’) term becomes zero.
leaving:

 

 

l .
It/Ibou, = m tunes

2 1 1 ~5§L¥§£fi<P3P4 + P3P5) + M? — 1m,
_ ’__ / -
(MZP + 2(P3P4) + 2(P3P5)> 0g Mg — 1.er

—W%(P3P4 + P4P5) + 11.13 — $115)]

2 1
... / 1 q .
+ (MZP + 2(P3P4) + 2(P4P5)> [0g ( M; ~1.1111:
2 _ lo <(P4P5)3[A[53 — 2(P3P5)]>]
MZP g (10313.5)?[1152 — 2(P4P5)]

First to examine the effects of Ill-'32 . the mass of the outgoing ‘quark‘. Re-
membering that dot products in this metric are negative. it is clear from its

259

position in the factors that the matrix element obtains its greatest value when
MS is zero ! W ith that replacement we have:

1
A10 =-———_— t.
b th 4(13” 117265

_( 2 + 1 ,0 1+ 2(P3P4+ P3P5)
MZP 2(P3P4 + P3P5) g M} — iI‘M:
2 1 2(P3P4 + P4P5)
+ (MZP + 2(P3P4+ P4P5l> i1°g<1+ M} —1'.1‘M: >i
_ 2 _ ,0 <P4P5
MZP g P3P5
To examine the effect of setting either (P3P4 + P3P5) or (P3P4+ P4P5) to

zero and thereby gaining a large co—efficent. noting that log(l+;1:) :> r. as 1' — 0
shows that these terms will at most become unity.

It is the final term that dominates:

M _ 1 2 >10 (P4P5>]
”°""4(137) MZP g P3P5

Notice this term, (as well as the terms above) goes to zero when P3P5 = P4115,
as it should, for in this case there is no distinguishing the lepton scattering di-
agram from the anti-lepton one, and the two are subtracted.
It becomes a. maximum when either P3P5 or P4P5 is zero. indicating one of the
leptons is co—linear (assuming MS = 0) with the outgoing ‘quark‘ .
Conservation of energy-momemtum dictates that:
11112 = (P3 + P4 + 135)2
: 2(P3P4 + P3P5 + P4P5)

so that P3P5 and P4P5 must add to a constant:
IlIl2
P3P5+ P4P5 : T — P3P4

P3P5 = [112—1: - P3P4] -— P4P5

_M—

l (RIP-5) log .1113—21P3P4)

09 ' r = 21P-1P5)
P3P‘) 1' A113—21P3P-1)

260

Thus lo ”I” ra hsa I L . .° ._ 2(P-1P5) ,
9p3p5 g P 5 09(,_,,) vs. .1. “It-112. _ M1,_2,P3P4,,

 

 

 

 

 

 

To realize the divergence does not contribute (due to the disappearance of phase
space where it occurs), find the average between 0 and 1/2:

L
Average = 1/ f(;z:)d:1:
L o

1 1,2, x d
Average : 1—/—2— f0 0g(l———:r) :1

1/2 1/2
= 2 (/ log(;1:) dz —/ 109(1— 2:) d3)
0 o

/10g(;r) dz = :1: 109(3) — 1'

and using:

and l‘Hopital‘s rule:

261

gives

Average = 2 (log (3) - 1)

Average = —4

so approximately,

so an ‘average’ value of the matrix element is

1 1
4(137) MZP

 

Mboth = (4)

and since the matrix element for the same diagram without scattering (the

BORN term) is just:
1

AIZP

3
then the RATIO of @7323?” becomes:

AIB om :

 

'1

4 4 '
RATIO = 1+ 2 (4(137)) + (4037))

so the scattering diagrams can expect to contribute on the order of 1.0 per cent.

 

 

UNIV. LIBRQRIES f

“Tittijfliul

@78i5ii785