146
174

 

_THS

PlON-NUCLEON SCATTERtNG AND
CHARGE SYMMETRY

The“: for ”to Degree 0* M. S.
MICHIGAN STATE UNIVERSITY

Justin Huang
1959

 

__ IrllmjltmflfllmiflfiWIm“

LIBR .4 R Y

Pliichigazlflt;1tc E
University E

’L______f

 

PION - NUCLEON SCATTERING AND CHARGE SYMMETRY

By .
Justin Huang

A THESIS
Submitted to the College of Science and Arts
Michigan State University of Agriculture and
Applied Science in partial fulfillment of

the requirements for the degree of

MASTER OF SCIENCE
Department of Physics

1959

ACKNOWLEDGEMENT

I am most grateful to Dr. J. Ballam for sponsoring

and suggesting the problem. His continued help and

.guidance is greatly appreciated. I am also deeply

indebted to John Scandrett for much help on the problem.
Thanks to Dr. W. Walker, Dr. J. Kovacs, and Dr. D.
Lichtenberg for checking parts of the manuscript and
making helpful suggestions. The help of Mrs. E. James
in scanning and preparing the graphs and David Balzarini
in scanning is also greatly appreciated.

PION - NUCLEON SCATTERING AND CHARGE SYMMETRY

By
Justin Huang

AN ABSTRACT

Submitted to the College of Science and Arts
Michigan State University of Agriculture and
Applied Science in partial fulfillment of

the requirements for the degree of

MASTER OF SCIENCE
Department of Physics

1959

 

ApprovedWfi/f ,3

ABSTRACT

An analysis of negative pion - neutron scattering
at 460.mev in a prOpane bubble chamber has been made in
order to verify the hypothesis of charge symmetry in
pion - nucleon interactions. A comparison of the ang-
ular distribution in ("7,n) and (3+,p) scattering at
equivalent energies was made for this purpose. Since
the (3+,p ),‘ datawere obtained in hydrogen, corrections
for nucleon motion, assuming a Fermi sphere of 191 mev/c,
as well as for diffraction from the carbon nucleus,
using a black sphere model, were carried out. Applying
a 12 test, the resulting (fl‘,n) angular distribution
fits a least squares fit of the (3+,p) data with a
probability of 0.054 for the entire curve and a prob-
ability of 0.20 for the part of the curve corresponding

to laboratory angles greater than 53 degrees.

TABLE OF CONTENTS
I. IntrOductionOOO0.0000000000000000000000000.000
II. Experimental Details.........................
III. Charge symmetryOOOOOOOOOOOOOOOOOOOOOOOOOOOOO
A. Correction for a Moving Target...........
BO Diffraction ScatterinSOOOOOOOOO0.00....00
Iv. conCIuSionOO000..00...OOOOOOOOOOOOOOOOOOOOOO.

Appendix.OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

BibliographyOOOOOOOOOOOOOOOOOOOOCOOOOOOOOOOO0.0...

12
12

25

26

27

55

I INTRODUCTION

An important particle in the mystery of nuclear
physics is the pi meson. According to Yukawa"s theory, the
pi meson plays the same role in a nuclear field as the
photon plays in the electromagnetic field} With the dis-
covery of the artificially produced pi meson; its mass,
spin, lifetime for weak decay and parity became well esta-
blished. The interaction of the pi meson with nucleons is,
however, not so well established. One method of studying
the interaction is to observe the scattering of pi mesons
from.nucleonse An experiment to measure pi meson proton
scattering is relatively simple to perform since there are
sources of free protons available. However, it is very
difficult to study pi meson neutron scattering, since
there are no stationary free neutrons available in nature.
All stable sources of neutrons are found in a bound state
(inside a nucleus). However, from.the principle of charge
symmetry, we can predict the interaction of a pi meson with
neutrons by studying the interaction of a pi meson with
protons.

To explain charge symmetry, it is convenient to
introduce the idea of isotopic spin? Isotopic Spin form-
alism makes use of the similarities between a proton and

a neutron. They have the same spin and mass, but a

different charge. Apart from electromagnetic effects,
they behave the same way in nuclear matter. They could
be thought of as just one particle (a nucleon) in two
different charge states. We introduce a new quantum
number "Mt" and give it a value of +1/2 for a proton and
-l/2 for a neutron. The state of a nucleon is now also
determined by this new quantum number. The Pauli ex-
clusion principle can now be generalized to exclude the
states of any two nucleons with all quantum numbers equal,
including Mt‘ The terminology "spin" was introduced
because of the analogous role that the charge states
play to the role of ordinary spin in atomic physica,if we
neglect spin-orbit interaction. Again in analogy to
ordinary spin, we could think of a three dimensional
"isotopic spin space", having the three Operators:
0/

x (II (D)
T .- 9—;

A0
(53)

The quantum number Mt would just be the projection of

B
I

F3
N
I

T on the z axis.

With this knowledge of isotopic spin space, we may
now proceed to define charge symmetry. The statement of
charge symmetry is that a 180 degree rotation of the
'isotopic spin vectors about any axis in the xy plane
does not change the interaction. The definition

5
effectively, Just changes the sign of Mt‘ We can now see
that from the previous assignment of Tz for a proton and
a neutron, (n,n) and (p,p) interactions should be ident-
ical. (n,n) equivalent to (p,p) is the definition of
charge symmetry that is commonly used. The advantage of
the first definition is that it can also be applied to
meson - nucleon systems. The pi meson is a charge
triplet3 with T2 - l,O,-l corresponding to a "f, '9,
and w"respectively. It now follows that by charge syms
metry, a proton can be interchanged with a neutron, a w+
meson can be interchanged with a ""meson, a WP meson can
be interchanged with a "9 meson, and vice versa in any
interaction. We can now clearly see that

a. (w+,P) and (3-,n)

b- (w-9P) and (n+,n)

c. <.~°.p) and <.°. n)-
should be identical interactions.

In this paper, we are interested in an experimental
verification of this principle of charge symmetry in
pion - nucleon scattering at pion energies of 460 mev.
The method used is to compare the angular distributions
of negative pions scattered from neutrons and of positive
pions scattered from protons in the laboratory system.
Previous work similar to this has been performed by
J. Ashkin4 et al. The method used by Ashkin was to

compare the total cross sections of positive and negative

pions scattered from deuterium.

Since the protons in this experiment are free and
the neutrons are bound in carbon nuclei, both experiment-
ally obtained distributbns must be transformed before the
comparison can be made. This is described in detail in
Section III.

The main proton data‘werc obtained by W. D. 97111135
who did (37",p) scattering at 500 mev using a hydrogen
bubble chamber. As will be described in Section III,
data from (3+,p) scattering at 850 mev in a hydrogen

chamber by Erwin and Kopp6

and at 1.1 bev in a prOpane
chamber by Glaser et al7 will be used in the analysis.

The neutron data were obtained from Walker et al8
in a propane chamber exposed to #60 mev w- mesons at

the cosmotron.

II EXPERIMENTAL DETAILS

The pion-neutron data was abstracted from 20,000
pictures taken of a prOpane bubble chamber built by
Professor W. D. Walker of the University Of Wisconsin
which was exposed to a 460 mev negative pion beam at the
cosmotron of the Brookhaven National Laboratory. There
was no magnetic field. The chamber was photographed by
two cameras with their Optical axes at right angles to
each other. The use of the maximum possible stereo angle
allowed for the accurate measurement of the most important
quantity in the experiment -the space angles between the
various tracks. A sketch of the Optical system is shown
in figure 1.

In the scanning of the pictures, every event in-
volving an incoming pi meson was recorded. These events

were broken down as follows:

Hydrogen
elastic ""+ p . w*'+ P
charge exchange fl"+.p . "o e n
"fpproduction w? +.p - sf + ”T'+ n
“9 production "7 + p . “9 + "7'+ p
Carbon
quasi elastic "7'+ p . W7'+ p
0

charge exchange ”7 + p - n + n

 

 

 

 

 

 

 

BOTTOM
/ CAKRA

BEAM DORECTION
FIGURE /

 

. "4- production a." + P . 1r+ + W- + n
"9 Production w-d+ P _ W_O + w7'+ P
deflection

diffraction off carbon nucleus
s9 prod» on neutron Ff +»n- _ “o + w? + n
stars _

The number of events for each catagory is listed in Table l.

 

 

A-

description number
elastic — — — _298-
quasi elastic 569
w? prod. (hydrogen & carbon) 182
17° prod. " " " 105
charge exchange " " 464
deflections > 6° 905
deflections < 69 219
WP prod. on neutron 20
stars with 0 "7 442
stars with l “T 545
stars with 2 w- 72
Total 5621

 

 

 

 

Table l .

7
The eventsof most interest to us were the (1(a)) elastic
scatterings from the hydrogen in the chamber and the
("7,nfi)elastic scatterings from.the neutrons bound in
the carbon of the chamber. In the former case, the kine—
matics of the event is completely determined from the
angles that the scattered pion and the proton make with
the initial direction of the incoming pion. An equally
good set of variables is the angle of the scattered pion
and the range of the proton. The elastic events are sep-
arated from the quasi-elastics by a measurement of the
cOplanarity of the event. With these parameters, one can
measure the angle of emission of the scattered pion in
the center of mass system. Or from the Lorentz trans-

formation, one has (symbols defined on page 15)

(11:); Pose = X ( ”C035" * Mk)

(:2) Pswe = ”My
Aeset of curves giving 9% vs GT with 9* and Pkfincoming)
as parameters has been plotted so that once the space
angles have been measured, the incoming momentum and the
corresponding center of mass angle can be quickly deter-
mined (See figure 15).

In order to calculate the space angles, the following,

measurements are made on each elastic scattering (these

are all made with respect to a set of fixed axes in the

chamber and in each of the two views).

 

Fig. 2 Deflections.

a. Incoming beam angle.

b. Angle Of the scattered pion.

c. Angle of the scattered proton.

d. Co-ordinates of the point of collision.
e. Range of the proton (when available).

f. Tangents of the cOplanarity angles, where
available.

These data must then be combined to find the space angle.
In this process, the two most important corrections are
for the conical projection and the index of refraction
of the prOpane (1.25). The details of these calculations
are shown in the Appendix. A program for these calcu-
lations was made for the digital computer Mistic of the
computer laboratory, Michigan State University.

We are concerned in this paper mainly with the
deflections. The small angle deflections are considered
to be mainly diffraction scatterings from carbon, while
the large angle ones are mainly pion-neutron scatterings.
A typical deflection is shown in figure 2. It is char-
acterized.by a sharp break in a fast pi beam accompanied
by no visible recoil tracks. A total of 1090 of these
deflections were measured and calculated for their space
angles. The following quantities were measured from the

conical plane (film).

POSITION 0F BEAM m CHAMBER IN ”V2

+2

-IO

300

F IG_ 3 P05/ T/ON w CHAMBER vs HOMEM'UH
FOR 52' FROM 0°. /°

0
O

O

400 soo 600 700 soo soo
MOMENTw m MEV/c

Bottom view:

a. beam angle.

b. deflected angle.

c. position of beam in chamber.
Side view:

a. beam angle.

b. deflected angle.
The purpose of the third measurement in the bottom view
was to separate the events into different momentum groups.
However, a preliminary study made of the sensitivity of
the measurement to momentum was negative. About 300
(W7,p)}elastic events were analyzed for the dependence
of beam angle and position in chamber to momentum. The
momentum of the (u',p)3e1astics were found through other
criteria' already mentioned. An example of the results
is shown in figure 5. The idea of dividing our events
into momentum groups was thus quickly abandoned and an
average value of 590mev/c was adepted for all incident
pions. The space angles were calculated using the anal-
ysis scheme derived in the Appendix and by the digital
computer Mistic. The approximation that all events occur
at the center of the chamber was made to simplify the
computations.(an error of at most 1° in the lab. system).

Using the values <£137,H) - 50 mb and <§(fl',c)
equal to 316 mb determined from experiments by Cool et 8?;10

the value for V(deflection) can be determined from

lO

 

(3) _ _§Kdeflection) . number=gg_ge§;gc§10ns
8 ¢E(W;H)+5CE(W:C) total number of events

The value for 4(def1ection) came out to be 106 mb. The
differential cross section dV/dR for deflections can now
be calculated using

(4) SE; a EQ;_9§_§Eterva13X--B€s-$§-l§§§£¥§l-x((defl,)
dJZ 4w' total no. of defl.

The angular distribution dV/dR.vs cos 9 for all deflect-
ions is given in Table 2. The distribution is also
illustrated in figures 17 and 18 (the ”I curve). The
total cross sections obtained from Cool et al were for
anti-coincidence angles up to 6°. This means that all
events with angles less than 60 are counted as going
straight through. Therefore, in our calculatiOns, we
must neglect all deflections with angles less than 6°.
The events classified as deflections have actually
a third possible origin besides the two already listed.
This other possibility is a 137, 2n star with a fast pi
coming off. As a rough check on how many of these events
we could have included as deflections, a survey was made
in our data for 13', 1p, ?n stars with a fast pi and a
backward scattered proton. It is assumed that in these
events, a neutron would also have been given off by the
carbon nucleus. It is then assumed that the number of

lw', 2n stars would not exceed the number of ln', lp, 1n

11
stars. The number Of INT, 1p, 1n stars found was roughly
28, negligible compared to the total of 1124 deflections
found. We shall thus consider the deflections to be just

composed of (n-}n) and diffraction scatterings.

Table 2. Angular Distribution for Deflections.

 

 

 

-9-1ab. (deg;)_ No. dT}dR mb/str Error mb/str
6.8 69 ' 508 37
8-10 74 257 . 50
10-12 84 240 26.2
12-14.l 82 188 20.8
14.1-16.3 80 151 16.9
16.5-18.2 69 150 15.7
18.2-20 47 88.9 13.0
20-25.1 48 45.8 6.6
23.1—25.9 42 40.0 6.2
25.9-28.3 53 31.4 5.5
28.5-32.9 50 25.8 5.4
52.9-56.9 31 14.8 2.7
36.9-45.6 50 9.53 1.54
45.6-66.4 59 3.76 0.49
66.4-90.0 26. 1.72 0.54
90.0_120 20 0.76 0.17
120-180 18 0.69 0.16 J

 

 

 

III CHARGE SYMMETRY

To test the principle of charge symmetry, we will
compare the angular distributions related to the (w*,p)
and (w',n) scatterings. There are two major difficulties
with the (3-,n) data. As mentioned previously, we can
not in general distinguish between a (fi',n) and a dif-
fraction scattering, especially small angle scatterings.
Secondly, the target neutron is bound inside a carbon
nucleus; and according to the Fermi gas model, it has
a momentum associated with it. We can reduce the second
problem by considering a model prOposed by Serberll’lz.
In this model, if the target nucleus is small and the
incoming momentum is high, we can consider the problem
as just the collision of the pion with one nucleon inside

the carbon nucleus. The problem is now similar to an

ordinary scattering except that the target is in motion.

A. Correction for a Moving Target.

let us first correct for the moving target. We have
a choice of two approaches. One is to smear out the (3+,p)
angular distribution, that is to give the stationary tar-
get nucleon various momenta that it would have if it were
bound inside a carbon nucleus. A further explanation of

this method will be given later. The other approach is to

15
unsmear the (fl',n) angular distribution, that is to find
out the most likely distribution it would have if the tar-
get nucleon were stationary. The former is the much
simpler approach and will be adOpted here.

The first question that comes to mind is what para-
meter determines a scattering. In the center of mass co-
ordinate system, this parameter is just the momentum P‘
of the particle. However, the (n‘,n) scattering data is
available only in the laboratory system. Therefore, we
must compare our distribution in the laboratory system.
This means that a Lorentz transformation would have to
be performed from the center of mass to the laboratory
system. In this transformation, another parameter, the
velocity of the center of mass "B" is introduced. Thus,
in the laboratory system, our scattering is completely
determined.by two parameters, P‘ and B.

Let us now look into the actual Fermi momentum Of
a nucleon inside a nucleus. Several momentum distri-
butions have been proposed which have given reasonable
fits to experimental datala. One is the Gaussian
distribution, 2

3h. ‘bP

where the average kinetic energy of a nucleon is 19.5mev
which corresponds to a momentum of 191 mev/c. We shall

use the Gaussian distributioETto adopt the value of

14
191 mev/c for the momentum of a nucleon inside a nucleus.
Assuming that the nucleons are isotrOpic in direction, the
momenta can be represented by'a sphere of 191 mev/c where
the density of scattering events is uniform over the sur-
face of the Sphere. A three dimensional momentum diagram
of the initial state of a scattering event is illustrated
in figure 4.

The smeared out angular distribution must be the net
effect of every possible scattering on the Fermi sphere
of 191 mev/c described above. Thus, in smearing out a
distribution, we must consider all scatterings on the
Fermi sphere. This would involve an integration over the
whole surface of the sphere, a messy calculation to per-
form. To simplify the task, we shall instead slice up
the sphere into 52 parts such that each of the 52 dissected
parts represents an equal surface area on the original
sphere. Taking a representative point on the outside
surface of each of the 52 parts, we shall calculate an
angular distribution for that point. The net effect of
the whole Fermi sphere will then be the average of the
52 angular distributions.

To obtain an angular distribution for a particular
point on the sphere, we must remember that the laboratory
angular distribution depends only on P* and B. Available
to us are the experimental angular distributions of

(n+,p) scattering with stationary targets at energies of

twee $0 225%? it OK

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17 Joe. A ..

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. e0-
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1610.... ue® moo

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do
bu

15
.5 6 7

500:me , 850 mev , and 1.1 bev as shown in figures 5,
6, and 7. The corresponding momenta available in the
center of mass system are 405 mev/c, 549 mev/c, and
641 mev/c respectively. This is calculated from con-
sidering the Lorentz transformation of the momentum vector
from the center of mass co-ordinate system to the labor-

atory co-ordinate system.
(6) p“ r. 303' 5E)

where V
(7) 5 = (Br " 3») "7' (is f ”P)
(8) 3’ 2 </ .— 52) “V2

From figures 5, 6, and 7, we can make a cross plot of
dvyaw vs P* with cos 6"as the parameter. The plots are
shown in figures 8 and 9. From figures 8 and 9, we can
determine the center of mass distribution of any (w+,p)
scattering, moving or stationary target once P‘ is deter-
mined. Thus, for a particular point on the Fermi sphere,
we can obtain a center of mass angular distribution by
solving for P‘ of that particular point.

Inorder to Obtain P‘ for a particular point on the
Fermi sphere, we need only specify the angle q’(see fig.4)
that is related to the point. This is evident from
equation 6, 8, and the following,

(7a) fl )/
A P : (Fe:+: P‘ eéficosq)“
T

(9)

16,

 

 

 

Figure:10‘ Dissection of Fermi Sphere.

For our choice of qfls, let us first slice the Fermi sphere
into eight sections as illustrated in figure 10. Again
referring to figure 4, this is done by taking eight dif-
ferent values OfCX. The «'s were chosen such that each.
center of mass angular distribution corresponding to the
particular'q willLbe weighted equally (equal solid angle
involved); Table 5 lists the one chosen and the corres-e
ponding cos eve, B's, P*"s, are and P;Fs. The equations
used to compute the variables are listed right below the
'Table. The variable I::will be explained later.

For each of the eight values of a listed in Table 5,
we can determine a center of mass angular distribution by
using figures 8 and 9. we must next consider transforming
these;eight angular distributions into the laboratory sys-

tem. To transform to the laboratory system, let us again

1?

 

 

 

 

 

 

=‘=" ‘ “‘ ’ ‘ 1
PW - 590 mev/c and IF:p - 191 mev/c
_ _ _—=_— — —_— __ — — 4
0((deg.) costx B P” mev/c l4)(deg.) P; mev/c
-— _ — ——— — — — —--"T
54.4 0.875 0.471 546. 9.21 855
51.5 0.625 0.446 558 12.58 771
68 . 0 0. 575 0.420 570 15 ~69 700
82.8 0.125 0.592 582 17.95 621
97;. 2 -0. 125 0. 561 598 19 . 58 545
112.0 -0.57.S 0.528 414 20.23 477,.-
_l45:6_ _-0.8'Z_5_ _ 0.247 _ _455 17.78 525 _ _ __J
(6) [3* z X (P‘ 55)
2 2.
<7.) a = 8%21‘0 + PM»
-1/
(s) X = Q '- 32) a
./
2 /2
(9) PT ' (PM? 7‘ 255059
(10) 9ou ‘ ’52 “W “L 'D

, -5712 1.
(11) £3: 76‘ -

H

 

18

'Ol

 

 

flamlL

consider the Lorentz transformation of the energy-momentum
four vector.

, s x
(12) longitudinal [3(sz -‘-'- b’ ( P2959 +55 )

I __ i" It
(15 )3 transverse P swe - ID 5W9

where 0"and 9* are illustrated in figure 11. Dividing
equation 12 by 15, we get

it 9(- ‘K g
(04);: 72w 9’ : ewe —:- (a’wse + .65 #9)

We can now transform our center of mass angular distri-

butions to the laboratory system by using equation 14.
However, by referring to figure 11, we notice that

the angle 0' is measured with respect to the direction of

8. With a stationary target, this presents no difficulties

since B is along the same direction as the incident part-

icle in the laboratory system. But with a moving target,

19
B is in a different direction with the incident particle
in the laboratory system. The desired angle "9" is the
angle between the direction of the incident particle and
the scattered particle in the laboratory system.

To correct for this shift in angles, consider the
diagram illustrated in figure 12. The primed co-ordinate
system refers to the scattering system, the system which
gives the angle 9"with reSpect to the direction of B.
The unprimed co-ordinate system refers to the reference
system, the system which gives the angle 9 with resPect
to the incident pion beam in the laboratory system. We

can write for the unit vectors 51 and $2,

(15) E : S/NLO Z + €552.03. + OE

(16) "r; : ewe; 31/49 I + cos an, .5 + 3.107050 A

The actual scattering angle ew_can now be obtained from

the following expression,

__ __ I x /
(17) ",1; : C5569“, 1' 3000501909qu 7“ (3052,0605?-

We shall now complete the dissection of the Fermi
sphere. Cutting along the angle 0', we shall split each
of the eight sections shown in figure 10 into four equal
parts. For simplicity, let us choose for the angle 0';
the values 0°, 90°, 180°, and 270°. Equation 17 now
simplifies to

SKEWED SCATTERING 1‘

 

 

 

 

 

 

 

 

 

 

 

z!
\ ¢l
"1 e
y. REFERENCE VECTOR
A:
" y
REFERENCE SYSTEM
1" —
I" REFERENCE VECTOR
r2 I»
77'
SCATTERED
VECTOR 1r
xi
/
A
..' /$h9'1r
'/
z /

 

FIGURE /.':3

F761 13 K/NE/‘m TIC (M? W? S

i

1r (c of m)

o'
\9

 

 

h®

Lab. (deg)

9:»

20

(18) 0' - 0°, 180° ms 977 = Cos was 0,;
(19) o' .. 90° Cos e,r : COS<9,,/'w)
(20) s: a 270° C05 O,T = cos(6,,'+@

For each of the dV/dfl vs cos 9;, we now have in its place
four dV/dQ vs cos 9" angular distributions.

Let us now return to equation 14 which transforms
the angular distributions from the center of mass to the
laboratory co—ordinate system. Instead of doing all the
calculations using this equation, we shall make use of
the already available kinematic curves shown in figure 15.
These curves were calculated using equation 14, but for
a stationary targetl4. The curves give a relationship
between 9;, 95, 9:, and P", However, we can just as well
think of the curves as a relationship between 9%, 95, 9;,
and B. This would make the curves adaptable to moving
target scatterings. To utilize the kinematic curves, let
us think of the collision with a moving target as one of
an incident pion beam with momentum P; and a stationary
target. Pé‘of course, must be chosen such that B is

preserved.

(21) 5: f7),

a._______w_ , 2 .2 A???)
”:~ s//r?,'?”+ 772,,‘ M?) I T I.fl+v’fl I R
x, I

Values for P;:are listed in Table 5. By using Egg we are
really using B in disguise. NOw we may adopt figure 15

for our computations.

_- so- so- so- No

Sec. 0 we

¢.0

 

 

 

0&6... Mme. use. wok OREERSm 071..

.2 OR

0.0

0.0

0..

 

21
In detail, here is how the computations were done.
a. From figures 8 and 9, determine angular distri-
butions for each Of the P*'s listed in Table 5.
b. For each of the PT's, transfer over to the lab-
oratory system (9; with respect to B) using figure 15.
First take a particular value of cos 9: and note the cor-

responding value of dVVdR. From figure 15, find the cos 9;

which corresponds to the value of cos 9: and I;_under
consideration. This value of cos 9; will now correspond
to the above value of dV/dfl.

c. For each cos 9;, we have four corresponding values
for cos 9w (0' - 0°, 90°, 180°, &.270°). These are cal-
culated from equations 18, 19, and 20. Values of’a)may
be obtained from Table 5. An example of the angular dist-
ributions with respect to e for 0( . 145.60 and 0' - 0°,
90°, 180°, and 2709 are shown in figure 14.

d. We now have 52 angular distributions from the
eight values of‘d and the four values of 0'. The net
effect is just the average of them all. The next step is
to add up the 52 distributions and divide by 52.

The resultant curve is shown as the solid curve in
figure 15. The dotted curve represents the angular dist-
ribution for a stationary target at a pion laboratory
momentum of 590 mev/c. The stationary target curve was
Obtained by

a. Finding the center Of mass angular distribution

 

 

FIG. /5 71”“— P scams/Ne

\ STATIONARY
- TARCET '

  
 

MOVING
\ ./ TARGET

 

I.O 0.6 0.2 02 ~06 -l.0
C O S O L AB

FIG. /6 77’1P SCATTERING

f'r--“

LELASTIC
f .

 

 

 

 

|.0 0.6 0.2 :02 -0.6 To
008 0 LAB

22
from figures 8 and 9 (P* - 587 mev/c).

b. Transferring to the laboratory system using the
kinematic curves in figure 15. (9 . 0' since u)- 0 for a
stationary target)

As can be seen from figure 15, there is very little differ-
ence between the two distributions.

As an experimental check on the results, we made use
of the elastic (stationary target) and quasi-elastic
(moving target) data mentioned in section II. Angular
distributions were calculated for the two and are shown
in figure 16. The curves are in good agreement with the
predicted results of stationary and moving target angular

distributions.

25
B. Diffraction Scattering.

New that we have corrected for the moving target,
the only thing left is the correction for the diffraction
scattering from the carbon nucleus. Assuming that the
incident meson wave sees the carbon nucleus as a black

Sphere, we have for the angular distributioE,5

O/V’(d1ff)9/T__(dlff) [ J: (A3”? Silk/J2

 

(22 ) a}; = 3m 6*

where k* - h/P‘ and R - rko are the wave number and radius
of the nucleus respectively. The deflections, as mentioned
before, consist of both the (wf.n) and the diffraction
scatterings. Therefore, before we can compare the (w',n)
and (nI,p) interactions, we must add the diffraction
effect to the transformed ("5,p) scattering or subtract
the diffraction effect from the deflections. We shall
do the former.

At first one might expect that since there are Six
neutrons and six protons inside a carbon nucleus, we can

simply write
(25 ) dI(d£v[/'chxml> : if (d'I/‘flg’TC 7‘20») + 6 _0/:1' (77‘,— 7L)
(215?. EU? 63?.

However, there is the prOblem that some of the nucleons
are shielding the others which will reduce the factor six
being used above. To obtain the correct factor, we turn
to the (n',p) data listed in Table 1. We are reasonably
sure in this case that we are able to distinguish between

24
an elastic event (free proton) and a quasi-event (proton
bound inside a carbon nucleus). In prOpane (C5H8)’ if
there were no shielding effects, we would expect a 8/18
ratio of elastics to quasi-elastics. The actual ratio
was 298/569, indicating some shielding effects. Taking
this into cOnsideration, the relation

(24) 8/5m - 298/56‘9
should hold where "m" is the effective number of un-
eshielded protons (or neutrons) inside the carbon nucleus.
We can now write after solving for "m" in equation 24,

iii/(def!) J gym/Io + 3.3 £0579

(25 ) de " c/JZ

Or if charge symmetry is to hold, we must have
Egon-71L 427W”)- éffi/IVO: 33 5550579
(26) 3.32/12 “Cm d5? ° 002
In equation 22, there are still two unknown con-
stants,«¢(diff.) and the radius parameter ro to be deter-
mined. By trial and error methods, HVf.- 25 mb/str and
r0 - 1.1 x 10'13 cm. were found to give the best results.

Equation 22 can now be written as

915(de9_ 25 aim-meme“) 2
(27) 0152 _ Ski/6*

 

 

Values Of the d¢(diff.)/dflicurve were only taken to the
first minimum (around 55°).

Rewriting equation 26, we have A)
c/o’MEf/Q d<r(77*}>) ck (diff
:7 . J + __ .-

320 ,
!,

 

24o . ' ,.,
d
—— 200
I60

\n

l20

 

g
00

80
40

0
l0 0.95 0.90 0.85 0.80

008 0 LAB

STR

 

7L.\\f I

‘

IS
I4 ‘1

 

 

 

 

 

 

 

LI:
:2 A
dO'
—— :0
d0 E
s
I A
6 \‘VL+
MB. 4 NE
STR  1\
2 M 3
\1\1 I
.0 \I’

0.6 0.2 02 -O.6
008 0 LAB

'|.O

25
as the condition for the validity of charge symmetry.
The observed wf(dcfl.)/dn and the calculated value of
5.3 dV(‘1r.+,p)/d52 + «(aura/cm are shown in figures 17
and 18 as the dashed curve and histogram respectively.
A fiftest was applied to the curves, the result being
6(probability) - 0.054 for the entire curve and a €(prob-
ability) - 0.20 for the part of the curve corresponding
to laboratory angles greater than 55°. A 742 test with,
a €(pr0bability) of greater than 0.01 is sufficient for
one not to reject a hypothesis. The results to our
test thus gives us close enough agreement for us to say
that charge symmetry has not been violated. As expected,
the fit for laboratory angles greater than 55° is better
than the fit at smaller angles where we had to apply the
black sphere model to correct for the diffraction effect.
It is also interesting to note that the value of 1.1 x 10'13
cm. for the radius parameter is in good agreement with the

Stanford16 determination of ro (1.07 x 10'15 cm.).

IV CONCLUSION

Two immediate remarks can be made. One is that a
moving target does not change the angular distribution
appreciably. This is illustrated in figures 15 and 16.
The other is that charge symmetry is not violated in the
(fi',n) and (3+,p) scatterings. This is illustrated in
figures 17 and 18. A closer look at figure 17 would also
suggest that a better fit could be made in the region
from cos 9 equal 0.82 to 0.92. A smaller value for the

radius parameter r accomplishes this. However, a smaller

0
r0 also swings the curve off in the region around cos 0
equal 0.98. It was decided that the black sphere approx—
imation made earlier for the nucleus was responsible for
this. To obtain more precise results, a model of the

nucleus with a diffuse edge must be considered.

APPENDIX
ANALYSIS SCHEME FOR THE MEASUREMENT OF SPACE ANGLES

After obtaining the data, the problem.is how to com-
puts the space angle from the films taken. As mentioned
before, the pictures were taken by two mutually perpend-
icular cameras in order to obtain a stereoscopic view of
the events. The main difficulty is that the cameras must
be placed at a finite distance away from the chamber. As
a consequence, the events seen on the films represent the
Aconical projections of the true epace events. In this
section, we shall set up an analysis to measure the Space
angle between two intersecting vectors in space knowing in
the mutually orthogonal side and bottom.views,-the conic-
ally projected angles, the co-ordinates of the vertices
of the conically projected angles, and the co-ordinates
of the points of projection. Correction factors for the
change in index of refraction in traveling from propane
to air are taken into consideration. A ceplanarity test
for events with three prongs (three intersecting vectors
in space) is also considered. The analysis scheme has
been set up in particular to solve for space angles in a
(w’,p) scattering experiment, but may be applied to a

(n-,n) scattering and other types of problems too.

(x wkbmvfix

20:33,: no E:
6:33 .. . («~23

 

 

 

 

 

 

I!

 

 

 

 

 

28
In figure 1A is shown one view of the cameras. In

practice what is usually done is to mark the surface of
the chamber with identification markers a specified dist-
ance apart. These markers will show up in the films taken
and can be used to calibrate the actual size of the events.
The film will then be projected back onto a screen to
about twice the size between the original markers. This
is illustrated in figure 1A where the back side of the
chamber now represents the conically projected plane. Now
we are ready to proceed to derive an expression for the
space angle 0 in terms of measurements made on the conic-
ally projected plane (the screen) and the co-ordinates of
the points of projection.

A. Derivation of Equations for the Orthogonally Projected
Angles.

From figure 1A, the following sets Of triangles are

similar.

(a) AADO and. ACHO

(b) ADFO and AHGO
Let us take as our space point (xi, Ye: z;) and as our
conically projected point (X50 Ys'.’ ZSL)' The distances
are then as follows: I

(O) AD .. Z3;

(d) CH - z;

29
(e) GH 3 x;
(f) FD . X5;
(3) D0 -' D5 + ds/np
(11-) H0 ' D5 + (ds/2 - :YL )/np
whereen, represents the index of refraction for prOpane.
The last two distances D0 and H0 are corrected for the
change in Optical path due to a change in the index of
refraction in traveling from air to prOpane. Since cor-
responding sides are proportional in similar triangles, we
have from the relations (a) and (b), I
AD/HC‘n CHVHO
(1A)
FD/DO - GH/HO

or rewriting as

ZS.) /<Ds + dS/nf’)

H

3t/[Q+t(%-279]

 

(21!) I ‘-
XsL/(Dfids/“O = e/[Ds +7171 92* All
Next define a
(5A) {a 1' 7,";( f " ‘19
We then have
. (D5 + G’s/71F)
Zei- " 3" D. + {.1
(4A)
_ : 34L D + dséfl)
x... (“t—”JD. + Y“

let us now take two arbitrary Space points (xl,yl,z,)
and (x2,y2,22) and their corresponding conical projection
(X91,YSZ,Z$2) and (XS.,Y$‘,Z5,). These two points form a vector
512 in three space, and their corresponding conical pro-

jections form a vector in plane "s". The vector being

50
(st-XSI, 0, Zsz'ZST)’ The tangent of the angle 9; that this

vector makes with the x axis in plane "a" is
I ‘ _ - 1 .1. _ '
(5A) 75” es ‘ (Z52 “‘31) ’ (x5?- X5”)

Substituting equation 4A into 5A and simplifying, we have

’: (Ba-39+ 32(W-/
S nf+’32

<¢2-]¢')+ 12(st‘czé/nfeé’-)
Simplifying the term in brackets, ££+dkhf+zg
DS+ Og/flP+ Vs:

(6A) 7A1 9.

___

 

 

 

 

 

 

(7A) . , / 3 Ti?“
05 ., 45/7? +3; 791% 'fsia‘t’z
Next define -;
(8A) A ‘ {791% T CW2 I 9°

We may now rewrite tan 9; as

(32‘3r)+(€' 74> + M (WM
/ + €H(Vz’7;)+(7‘2'2€)
But (22-z,)/(x2-x,) is just the tangent of the orthogonally

 

(9A) 754.; 65’ ='

 

projected angle eg in plane "8*"

and (y2 -y )/x2-x,) is
just the tangent of the orthogonally projected angle 9" in
plane "b*" (XY plane). The star "n" denotes the ortho-
gonally projected plane as contrasted with the conical

plane without the star. We may now write tan 9; as
(100 75w 93' = (25mg + AAA/ab) +(/+ 2’24 Axe)

Next to simplify the expression 22A and xZA using equa-

tion 2A. -/
12 /st 3 [ACE-DNC’Q]

(11A) fizz/ESL = [AWQWQJ

51

Therefore,

242/4 = x52 / ($195+ C15)
(12A)
32A : Zsz/ (771° De + C15)

Taking equation 10A and rewriting in terms of tan 9"
II I ’I /
72A! 68 3 7AM! 65 + é/Veg (ZZAZZKWGS ‘- 32 A)

 

 

(15A) , I] X52 I Z
I; 6 - $2
7/1"N 93 : 72M, (95 7‘ ZEWQA 0; 1.300572” 5 ds+ EDS]
Let
I ’ _

(14A) A’ z ”PQMS ( X52 ifs/65 232
Therefore, ” I I 0

(15A) 75'” es ‘ 72” 8. + A 75’” ‘96

Let us now turn to the XY plane "b" where our other
camera is located. The situation in plane "b" is exactly
analogous to that in plane "s" and all our equations would
be correct if we would change our subscripts "s" to "b" or
"b" to "s" and our Z's to Y's or Y's to Z's. Equations

14A and 15A could now be written as
’ — ——J——- )< JEN63’-ZEI)
(16A) 8 ' ”PDe*db( ‘2 6 b2
II / I ”
(17A) 72., 0,5 = Awe. + B Awe.

Equations 15A and 17A can now be solved explicitly for e;

and 9% (two equations in two unknowns). The result being
7%.. a,” (m a; + A’Awa’) —.'- (x- B’AQ
(It/w 66’ + 874N659 “—2- 0- BA)

Il

(18A) #N 9e”

52
We now have explicit expressions for our orthogonally
projected angles in terms of measurable quantities - 9;,
and D

9b., X32, 232’ X132, Zb2 and. the conStantS DP, D5, be

H. Correction for Beam Angle.

It should be noted that the angles we have referred
to so far are all measured with respect to the x axis. In
a usual scattering experiment what we are interested in is
the angle the scattered particle makes with the incident
beam of bombarding particles. This beam is usually making
a small angle with the x axis and must be corrected for.
From the well known trigonometric formula for the sum of
tangents of angles,

m a, (a, a: - t4~(9,9+( / + mews” ”Waves?

(19A) 7L/M/ @ (fan 95” “ fl”! fig) 7" ( / + f/wfé/ 7:14! 9.4” + [94/ 25/79/4659
5

where ' is the beam angle (angle it makes with the x axis).
Making the assumption that the beam angle is always very

small, we have

7AM 95 1 (72M 55’; 525) + ( H 13., 75/459: * Jé’fiVA/Qb)

(20A) 7295 : (2%,Ne;’-fe;)+<I+mmeg’+fl;79~99

as our true orthogonally projected angle with respect to
the beam.

The tangent of the Space angle can now be written as

(21A) 7‘4N6 = (flu/296 + 5363/2

i\\
>x

\\

‘ ‘\\\\ n

 

3; l\\\
\\\\\\ in?!

3/?)

52%

FIGURE 3A

 

 

 

0
OPTIC AXIS INTO PAPER

 

*5

VIEW "s"
+2

DIRECTION OF BEAM
Q:

 
  
 

6‘s,n's< O ‘ 9'9““; >0
X
93$?» 0 v V 6'.,fl',< O

 

 

 

G
OPTIC AXIS INTO PAPER

33
C. COplanarity Test.

Inorder for an event to be classified as elastic, it
must all lie in the same plane. Referring to figure 2A,
the angle ¢ is the angle between the event plane and the
chosen reference plane. If an event is to be classified
as cOplanor, each track composing the event must make the

same angle fl with respect to the reference plane.
22 7‘ Sb y: - 77 (91:— 113/65” 1.9 _ m e.
A ' 3 “fir'ir z ~+3 .-.:-—* - ~+f—f’

( ) km 52 ' (32 ‘ 3. ”(120' to) 7"” 8°

 

&
where the superscript 0 refers to the fact that these

quantities are with respect to the beam track.

D. Sign Convention.

To satisfy our equations, we must restrict ourselves
to certain sign conventions in the conically projected
plane (screen). By referring to equation 5A and others,
we see that the convention used in figure 3A is a suitable

one o

E. Summary.

We have now all the equations necessary for solving
9 and fl'by'measuring in the conical plane,.R%, 96, Kb, Yb,
' 9's, Xs’ Zs’ The eight equations are listed below.

8,
V
firm 6 = (7%in + 7%”265) 2

764w 9b : f4” 91: / 7L4” 85

(21A)

(22A)

54

W ab = (m 66'”fi£)+(/+Jzzm~e”+ngme)

(20A) I
m =(mes-xz;)-=-</+w4~@'+wa
me,” = (m e’w’ms’) H ”4’59

(18A) 7%; e: = (7%” GSI+r4754~QQ '3‘ (/ v43)

(14A) A’ :_ (XS five; - 22>//?5

(16A) 5' : (X4, 7;?qu ij/(vs

The description of each of the quantities is listed in
Table IA

 

 

 

 

 

 

view symbol description 7
"b" view .Rb angle beam makes with x axis
9% angle particle makes with x axis
Xb
co-ordinates of vertex
Yb
"3" view 9; angle particle makes with x axis
‘Qé angle beam makes with x axis
Xs
co-ordinates of vertex
Zs
Table IA.

A program for the calculation of 9 and s using the
eight equations listed above has been set up for the com-

puter Mistic located at Michigan State University.

10.
11.
12.
15.
14.
15.
16.

17.

BIBLIOGRAPHY

H. Yukawa, Proc. Phys.-Math. Soc. Japan, 17, 48 (1955).

R. G. Sachs, "Nuclear Theory," Addison Wesley 1955,
p a 156-161 0

H. A. Bethe & F. de Hoffmann, "Mesons," Row, Peterson,
1955, sect. 51d.

J. Ashkin, Phys. Rev. 96, 1104 (1954).
W. J. Willis, Ph.D. Thesis, Yale Univ. (1958).
Private Communication with Dr. Erwin.

D. Glaser & Rollig, "Proceedings of the 1958 Conference
on High Energy Physics at CERN."

Private Communication from Dr. Ballam.

R. L. Cool, D. Clark, 0. Piccioni, Phys. Rev. 105,
1082 (1956). "'

Private Communication with Dr. Cool.

R. Serber, Phys. Rev. 2%, 1114 (1947).

J. Cladis, Thesis, Univ. of California, (1952).
Marshak, "Meson Physics," McGraw-Hill, 1952, p. 85-88.
We thank Dr. L. Leipuner for the results.

marshak, log. git, p. 266.

B. Hahn, D. Ravenhall, & R. Hofstadter, Phys. Rev.
101, 1151 (1956).

The statistics in the problem does not warrant a more
detailed treatment of the momentum.

|||l|||||||1111111111 ||l||||||||||l|||||1||||||||||||

 

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