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A HYDRODYNAMIC THEORY OF
HEAVY ION COLLISIONS

by

David Carl Mundinger

A THESIS

submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Physics

1979

ABSTRACT

A HYDRODYNAMIC THEORY OF
HEAVY ION COLLISIONS

by

David Carl Mundinger

A macroscopic description of heavy ion collisions
has been made based on the dynamics of an interacting
Fermi liquid. This framework is general enough to permit
the study of some of the bulk prOperties of the medium
such as the stress tensor and the equation of state.

A computation was done, first of all, for the one-
dimensional case using a Lagrangian technique. The
breakup threshold was calculated for the collision
between two semi-infinite slabs of nuclear matter. This
one-dimensional model contains many of the features
of heavy ion collisions. Based on these results a model
was made for the collision between two spherical nuclei.
The degrees of freedom that we emphasize are the centers
of the two nuclei, and the location of two boundaries
which separate the nuclear matter into a compressed
region and two uncompressed regions. Using the equation
of state, along with the conservation laws for mass and

momentum, we can generate equations of motion for each

of the dynamical variables. The momentum transport
between the two nuclei is taken from the theory of a
collisionless Fermi gas.

Angular distributions, energy loss, and fusion
cross-sections have been studied. Comparisons are made
with experiment and with other models that have been
suggested. Qualitatively the results are in accord
with the more computer-intensive time dependent Hartree-

Fock calculations.

TABLE OF CONTENTS

Page

LIST OF TABLES . . . . . . . . . . . . . . . . . . . iii
LIST OF FIGURES . . . . . . . . . . . . . . . . . . iv
I. INTRODUCTION . . . . . . . . . . . . . . . . 1

II. DERIVATION OF THE MACROSCOPIC EQUATIONS
OF MOTION FROM A MICROSCOPIC THEORY . . . . 6

III. BULK PROPERTIES AND MACROSCOPIC
PHENOMENA . . . . . . . . . . . . . . . . . . 19

IV. ONE-DIMENSIONAL MODEL . . . . . . . . . . . . 29

v. DESCRIPTION OF THE MODEL AND
COMPUTATIONAL TECHNIQUE . . . . . . . . . . . 39

VI. APPLICATION OF THE COMPRESSIBLE FLUID
MODEL TO ENERGY LOSS . . . . . . . . . . . . 49

VII. APPLICATION OF MODEL TO FUSION . . . . . . . 59
VIII. ANGULAR DISTRIBUTIONS . . . . . . . . . . . . 68
IX. EVALUATION OF RESULTS . . . . . . . . . . . . 78

X. A COMPARISON OF THE RESULTS WITH THE
OTHER MODELS O O O O O C O I O O O O O O O O 85

LIST OF REFERENCES . . . . . . . . . . . . . . . . . 94

ii

LIST OF TABLES

Table Page

1. The important quantities related to the
breakup threshold of nuclear matter.
The different values of O correspond to
different compressibilities . . . . . . . . 34

2. The critical velocity at which breakup
occurs for the different equations of
state. These are calculated using
Equation III-7 . . . . . . . . . . . . . . . 35

3. A summary of the various models which

have been used to describe heavy ion
collisions . . . . . . . . . . . . . . . . 93

iii

Figure

1.

LIST OF FIGURES

An energy density curve for a typical
fluid. The coordinate R is the average
separation between particles. At some
equilibrium spacing R0 the energy per
particle is a minimum . . . . . . . . . .

Slab collision at 1 MeV/A (c.m.). The
position of mass elements as a function
of time is Shown for half of a symmetric
collision of two slabs, each 9 fm. thick.
The slabs just make contact at t = O; the
equation of state is the fluid model with
O = 1/3. Maximum compression occurs at
40 fm/c and the system snaps at s 160 fm

The energy distribution of the quasi-Kr
products observed as a function of angle.
The reaction is 525 MeV 4Kr on a 208Pb
target . . . . . . . . . . . . . . . . .

The energy loss depends somewhat on the
equation of state. Here energy loss is
plotted as a function of angle. E/Einc
is the fraction of energy left in the
scattered nucleus.. . . . . . . . . . . .

The fraction of energy left as a function
of impact parameter at center of mass
energies of 2.5, 3.5, and 4.5 MeV/A . .

The fraction of energy left in the out-
going fragment versus the scattering
angle. The small numbers indicate the
impact parameter . . . . . . . . . . . .

Energy distribution of the outgoing
nuclei integrated over all angles. The
inset is experimental data taken from
Reference 46 . . . . . . . . . . . . . .

iv

Page

20

36

50

52

53

56

58

Figure Page

8. The energy-impact parameter area in which
fusion occurs. The extended curve repre—
sents a grazing trajectory. The compres-
sibility is K = 500 MeV . . . . . . . . . . 63

9. Energy dependence of the fusion cross
section for three different systems,
4OCa(4 OC a302r), etc. . . . . . . . . . . . 55

10. The 058? agsction for the reaction
40Ca( Zr) using two different equa-
tions ofa state. K = 200 MeV and K = 500
MeV represent the two extremes in
compressibility . . . . . . . . . . . . . . 65

11. The fusion regime for the reaction
4oCa(40Ca, 4OCa)40Ca. The textured area
represents a compressibility of K = 200
MeV. The enlarged area is for a com—
pressibility of K = 500 MeV . . . . . . . . 67

12. Angular distributions for the reaction
products of the collision between 620 MeV
6Kr ions and a 181Ta target . . . . . . . 70

13. Collision between two 118Pd nuclei at an
energy of 5 MeV/n (c.m.). The straight
line represents the boundary between the
two zones. The small arrows indicate its
direction of motion . . . . . . . . . . . . 71

14. Deflection function 6(b) for the collision
between two Calcium nuclei. The small
numbers indicate the center of mass energy
in MeV/n . . . . . . . . . . . . . . . . . 73

15. Deflection function for the reaction
4Kr(34Kr,34Kr)84Kr. The compressibility
is K = 200 MeV. The small numbers indi-
cate the center of mass energy in MeV/n . . 74

16. The dgglec4 ion function for the reaction
84 Kr( 48Kr) 4Kr using the K = 500
MeV equation of state. The small numbers
indicate the center of mass energy in
MeV/n . . . . . . . . . . . . . . . . . . . 75

Figure

17.

18.

19.

20.

21.

22.

Page

A comparison of the energy loss functions
using two differing forms for the stress
tensor 0 O O O O O I O O O O O O O O O O O 80

The effect of the relaxing stress tensor
on the fusion regime . . . . . . . . . . . 81

A comparison of the experimental results

of Reference 36 with the predicted

results of the simple model described in
Chapter VIII . . . . . . . . . . . . . . . 82

56na148tate Oenergies in the reaction
85Ca( 404Ca) The Coulomb barrier
in our model is at .85 MeV/n. The inci-
dent energy in the TDHF study is 1.738
MeV/n. To keep the same window we use
an energy of 2.1 MeV/n . . . . . . . . . . 37

The distance between centers of two Pb

nuclei as a function of time. The line

at the end indicates breakup. The solid

line is our calculation and the x'S are

from the TDHF calculations of Reference 4 33

Results of a calculation done using a

surface excitation model as described in
Reference 11 . . . . . . . . . . . . . . . 90

vi

CHAPTER I

INTRODUCTION

Machines with the capability of accelerating heavy
ions up to energies on the order of 5-50 MeV/nucleon
will soon be available. These energies represent a new
domain of nuclear physics. The velocities are high
enough to cause considerable interpenetration of the
two nuclei, yet they are not so high that the system
cannot respond collectively. It has been speculated
that perhaps new kinds of phenomena will be observed.1
It is important for us to gain theoretical understanding
of how nuclear matter behaves in this situation. We
recognize that a complete and detailed theory may be
unachievable at this stage. However, based on what
we do know about nuclear matter and based on experiments
that have been done, various ideas have emerged. A
number of attempts have been made at a theoretical
description of heavy ion collisions. These theories
are quite different from each other in the way they
approach the problem. Several of the models will be
described briefly here, and a more detailed account

of the results will be presented later.

The most fundamental calculations are based on
the time-dependent Hartree-Fock theory (TDHF). The
theory starts from basic principles. The nucleus is
visualized as a collection of quantum mechanical parti-
cles (nucleons) which interact via a two-body potential,
usually a Coulomb plus a parametrized nuclear force.
The nucleons are placed in states which are obtained
in a self-consistent way from the mean field. As the
system evolves, this procedure is repeated at each time
step. These calculations have the obvious advantage
of giving a detailed description of heavy ion collisions
and should be able to predict such things as fusion
cross—sections and other experimental observables. Much
discussion and work in this area has appeared.2’3’4’5’6
However, because the calculations are lengthy and con-
sume much computer time, a detailed picture has not
been presented. Results have been published for a few
systems, usually at just one energy. Another problem
with this theory is that it is unable to include two-
body collisions. This will not be a serious problem
at low energy but at higher energies two-body collisions
will become increasingly important.

The simplest model to have emerged thus far has
been that proposed by Gross and Kalinowski.7 This is

sometimes referred to as the classical friction model.

The number of degrees of freedom has been reduced dras-
tically. Instead of following the nucleon coordinates
as is done in the TDHF calculations, only the relative
coordinates of the two nuclei are used to describe the
system. A potential function is introduced which, along
with the frictional force, governs the time development
of the nuclear coordinates. This model offers a con-
venient picture of some of the more striking features
of heavy ion collisions including fusion and large
energy losses. It does not offer a very detailed de-
scription, and its usefulness is limited.

The classical irrotational fluid drop model has
been the basis for several calculations concerning heavy

O I 0 8 9
Ion collISIons. '

The goal has been to isolate those
degrees of freedom which most completely describe the
whole system. A smaller number of parameters means
fewer equations of motion. Usually the surface is
parametrized by surfaces of revolution, and collective
variables such as the moments of the mass distribution
are chosen to represent the nucleus. The potential
energy can then be calculated from the configuration
and the equations of motion solved numerically. For
example, to study fusion one looks at the trajectory

of the system in this multi-dimensional space. If the

system enters a region where it gets trapped in a poten-

tial depression, one would say that the system is fused.

 

The fusion cross section as a function of energy has
been calculated for several systems by Nix and Sierk.8

An entirely different approach is to isolate those
collective quantum mechanical states which are most
important in absorbing energy from the relative motion.
This again reduces the number of equations that one
must solve. In the range of energies associated with
deeply inelastic collisions, surface excitations and
giant resonances may be the dominant transitions. Cal—
culations based on these ideas have been done for
several nuclei by Broglia and co-workers.10’11’12
Angular distributions, fusion cross sections, and deep
inelastic energy losses have been studied.

This thesis will develOp a hydrodynamic theory
of collisions, based on the idea that the dynamics of
nuclear matter can be represented by an equation of
state. To begin with, a set of classical equations
of motion will be derived from the microscopic ones.
Following this the bulk properties of a fluid will be
described. We also discuss some important macroscopic
phenomena which will be useful in discussing heavy ion
collisions within this theory. An equation of state
will then be postulated based on the observed properties

of nuclear matter. A model is constructed for the

one-dimensional slab collision, and the results are

presented. This model is then generalized to describe
a three-dimensional collision. The results of this cal-
culation are to be compared with experimental data and

the other models described above.

CHAPTER II
DERIVATION OF THE MACROSCOPIC EQUATIONS OF MOTION

FROM A MICROSCOPIC THEORY

In setting down a theory of heavy ion collisions
based on first principles, we will start with the
description of each nucleus as a collection of nucleons
which interact with each other according to the laws
of quantum mechanics. There is a many-body wave func—
tion W(ri,r2,...rn;t) which depends on the nucleon
coordinates and time, and obeys the Schrédinger

equation,

i g; W(r,t) = HW(r,t). (1)

This equation can be cast into a slightly different

form by multiplying it by W+(r't) on the left, then
subtracting from this the equation obtained by taking
the hermitian conjugate of (1), replacing r with r',

and multiplying on the right by W(r,t). After rearrang—
ing terms and using the chain rule for differentiation,

the result is

i g? (W+(r't)W(r,t)) = T+Hw — WHW+

If we say that H' Operates only on the primed coordinates
and H only on the unprimed, then this equation can be

written as

i g; N(r.r';t) = (H-H')N(r.r';t). (2)

This new quantity is called the N-body density matrix

and is defined as
+
N(r,r';t) = W (r',t)W(r,t).

Its physical meaning is somewhat abstract, but useful
quantities can be generated by integrating over the
appropriate variables, as will be seen in the following
discussion. The Hamiltonian H is an operator which

represents the total energy,

In configuration space H has the form

v(r.r.)

”IAK’.
M2
p

U

l<

U

N_V2 N
H — 2m + v(rirj)
i=1i<j
-V'2 N

 

1‘12

1 2 :
I _ I I
H - 2m + v(ri rj ).

I 1:1 i<j

By putting these expressions into Equation (2) we obtain

directly the equation of motion for the N-body density

matrix.37
' __ I. = —_ '.
I at N(r,r ,t) E 2m N(r,r ,t)
i=1
N
_ I I I.
+ E [V(ri.rj) v(ri,rj)]N(r,r ,t) (3)
i<j

Equation (3) is still very complicated and involves

all the degrees of freedom twice. At this point, to
make further progress, we adopt the following philosophy.
In order to obtain a simpler expression we will give up
the ability to describe the evolution of each degree of
freedom and try to find a small set of variables which

will describe the collective behavior of the system.38

It is toward achieving this end that we define the

n-body density matrix.

N(n)(r ,r ';t) = —-§i——-}fd3r W+(r$,r

n n (N-n)! p ;t)W(rn,rp;t)

P

(4)
The coordinates r1...rn are not integrated over. For

the coordinates r we set r = rp' and integrate.

n+1"'rN p

With this definition a set of N equations can be written

down. The nth equation is the equation of motion for

the n-body density matrix. To generate the nth equation

divide the coordinates into two groups;

rn = (r1,r2,...rn)
rp = (rn+1,...rN).
- _ . . _N_!_
In Equation (3) set rp — rp, multIply through by (N-n)! ,

and integrate over all the coordinates r ,
3 N! .8 . .
[d rp (N-n)! 1‘3? N(r,rn rp ,t)

, N (v?-v!2)
p (N-n)! IE: 2m ' II p’

i=1

N .
3 . u n .
+-./H rp IE: [v(rirj) v(rirj)]N(r,rnrp,t).

i<j

Evaluate this equation term by term. The left hand side is

clearly just

._ (n) .,
Iat N (rn,rn,t).

In the first term on the right hand side if i>n the
integral vanishes. What remains is

N 2,2
_ (Vi-Vi ’ N(n)

2m (rn'rn;t)°
1

i

10

In the second term on the right hand side if both i
and j are greater than n there is no contribution. If
both i and j are less than or equal to n, the sum comes

outside the integral giving

II

(n)

[v(r r.) — v(rir3)]N

I.
l j (rn,rn,t).

i<j

If i 5 n and j > n, the integral becomes

3
de rp[v(rirj) - v(ri r. 3)]N(r, r 'r. r ;t)

n j P

where rj is explicitly separated out of the set rp. It

is convenient to factor this integral as follows,

fdr[v(rr)-v(rr)]/drN(rrrpt)

There will be an identical contribution for each j in
the range n < j 5 N giving a factor of (N-n). Putting

the appropriate prefactor in gives

(n+11rnr. ,r' r. ;t).

3 .
-/U rj[v(rirj) - v(rirj)]N j n j'

Finally, putting these pieces together, we can write

the equation of motion for the n-body density matrix.

11

N (VZ-VZ)
i. (n) .. _. __i___i_ (n) ..
lat N (rn,rn,t) - 2 2m N (rn,rn,t)
i=1

n
+ z [v(rirj) - v(rirfinu‘n’

i<j

'.
(rn,rn.t)

n
3 , _ , (n+1) , .
+ 2 Id rj[v(rirj) v(rierN (rnrj,rnrj,t)
i=1
(5)
This equation expresses the time rate of change of the
n-body density matrix in terms of the space derivitives,

the two-body interaction, and the (n+1)-body density

matrix. If we are to obtain a closed set of equations

we must find an approximate expression for N(n+1) in

Na)

terms of N . This can be done if

(n), W“),

we take W(r,t) to be a Slater determinant wave function.

After separating out the time we can write

Mr) VéT Z(-1)p[¢p

(r1)¢p2(r2)...¢
P

1 p3(r3)]. (6)

¢p (ri) are normalized single particle states and the sum
1

over p represents a sum over all permutations.40 Putting

this expression back into Equation (4) we can generate a

(n+1)(r r.,r'r.) and the lower

relationship between N n j n 3

order density matrices. For the two-body case we get

12

N(2)(r1r2,riré) = N(1)(r1,r1)N(1)(r2,ré)

- N(1)(rl,ré)N(1)(r2.ri) (7)

From this point on the discussion will be limited to the
one-body equation and the superscripts will be omitted.
We can now derive a closed expression for N(r1,ri;t).

(The explicit time dependence has been omitted.)

_ 2_ ,2
(V1 V1 )

2m

 

L l _ I'
lat N(rl,r1) - N(r1,rl)

t/d3r2[v(r1r2) - v(rir2)] [N(rl,ri)N(r2r.2)
- N(r1r2)N(r2ri)] (8)

In this part of the discussion we will demonstrate
how several quantities can be formed which can be inter-

preted physically. First of all look at the one-body

density matrix.

*
N(rl'ri;t) =2 ¢a(rl’t)¢a(rl't)
C!

If r = r' this is the ordinary density p(r ,t). If
1 l (V -V' 1
1 1

the operator 2mi

acts on N(r1,ri;t) before setting

r1 = ri, the current density vector is obtained.

13

(v —v')
1 1 * .
TE ¢a‘r1't’¢a‘r1't’

d

_—L * - * —+

— 2mi‘EE:[¢d Vl¢d ¢dvl¢d1 — J(r1,t)
a

By a similar operation a tensor quantity can be formed

with units of a momentum flux density,

.— ' _ I
[‘Vl V1’1 (V1 V15
2i 2mi

 

N(r1,r1;t)]r1=r, = Tij(rl't)°

H

If ¢a(r) is interpreted as a probability amplitude then
it is consistent to identify these quantities with their
classical counterparts. Using these relations it is

straightforward to show that

_ |
”’1 V1

u, l. = .
[(v1+v1) 2mi N(rl,r1.t)1r = vr J(r.t).

D
1r1
and

(V -V') (V -V'
' I 1 1 0 1 1

 

jN(rllri;t)]r =r|

2mi 1 1

3
- EJ— Tij (rllt) o

The fact that N(r1,ri;t) is a function of two coordi-

nates makes it somewhat inconvenient to visualize in
physical terms. It would be more intuitive perhaps to

have a phase space function. This was introduced by

13

Wigner and is known as the Wigner function.

l4

hflx

3 . ,
f(r,p) = jr_é_§§ e 19 X N(r + g, r — )

(2")

Here we have defined relative and center of mass

coordinates
+ _ + _ +')
x — (r1 r1
+ + +
_ O
r — 35(r1 + r1)

v = av + v Vi = %V - V

_ l
Vr — V1 + V1 VX

ll
3X"
<
H
l
G
H‘

Evaluating the integrals of f(r,p)d3p will give addi-

tional physical insight.

ff (rrP)d3P
(VI-V

+ 3
Pf(r.p)d p
2mi

P-P-
[71—1 f(r.p)d3p =fi3xmr + x/2. r - x/2)><

3 .
.9L2§.pip e 1P X
(2n) 3

3 .
de3xN(r + x/2, r - x/2) —g—E—e-1p°x
(2w)3

0(r)

fi3xN(r + x/2, r - x/2)i‘7x5(x)

i)

= [m N(r1.ri)1r1=r. = m3(r)

l

(v -v').(v -v').
1 l 1 l l j . _
[ 21 2mi N(rl'rl)]r1=ri " Tij(r)

 

15

These integrals suggest that f(r,p) be interpreted as
a phase space distribution function.~

Equation (8) still contains the integral over a
product of one-body density matrices. The direct term
contains N(r2,r2) which is just p(r2). This can be

simplified wae define an equivalent one body potential.

U(rl) = J{d3r2 v(rlr2)p(r2)

The direct interaction term becomes
l _ I
N(r1.r1)tud(rl) Ud(r1)1.

The exchange term is more involved since it involves

two density matrices.

3 .
fd r2 v(rlrz) N(rlr2)N(r2r1) (9)
This, as it stands, is not a local one-body potential.
An approximation can be made which will keep the equiva-

14

lent one-body potential local. Expand N(r2,ri) about

the p01nt r2 = r1.

N(r2,ri) = N(r1,ri) + N(r1,ri)-(r2 - r1)

(rz-r1)j+...

' .—
+ kViVjN(r1,r1) X (r2 r1)i
To the next lower approximation we can think of the
nucleus as an infinite_Fermi gas, in which case N(r1,r2)

is known explicitly.

16

r1+r2
31(kFlr1'r2')p( 2 ’

 

1
N (r ,r ) = _
PG 1 2 kFlr1 rzT

With these approximations expression (9) becomes

N(r r')./pd3r v(r r )N (r r )

1' 1 2 1 2 PG 1’ 2
+ V N(r r')./~d3r (r -r ) v(r r )N (r r )

j 1' 1 2 2 1 j 2 1 PG 2’ 1

3
' -

+ ViVjN(r1,rl)./~d r2(r2 r1)j v(r2,r1)NFG(r2,rl)

From the first term we can define an effective one-
body potential which can be combined with the direct
term to give a total effective one-body potential. The
second term vanishes in the infinite Fermi gas limit,
as do the cross terms in the third part. The third
term contributes a V2 which can be combined with the
kinetic energy term to give an effective mass. Experi—
mentally it is observed that the effective mass ratio
is not much different from 1,43 so, for purposes of
this theory, only the first part of the exchange term
will be used. It is combined with the direct term to

give
N(r1,ri)[U(r1) - U‘ri’l-

Expressed in the relative and center-of-mass coordinates

this becomes

N(r + x/2, r - x/2)[U(r + x/2) — U(r - x/2)].

17

If we now make the semi-classical approximation that
U(r) varies slowly over the nucleus, then the interaction

term becomes
N(r + x/Z, r - x/2)x-VrU(r).

The equation of motion (8) is now expressed in terms

ofaalocal one-body potential.

.3 (Vi'viz)
iSE-N(rl,ri) = ———§a——— N(rl,ri)
rl+ri
+ [(rl-ri)°(V1+Vi)U(--2-)]N(r1.ri) (10)

In the limit where r1 = ri the continuity equation is

recovered,

8

O

+ vr-3(r,t) = o (11)

Q)

t

I 0 ~ — . - '
If we operate first on the left With Prel — i(Vl V1)i

and then set r1 = ri we will get another convervation
law.
3 _-3_
5E (mJi(rlt)) - arj (Tij)
—--a— [x —§—-U(r)N(r + x/2 r - x/2)]
3x. j 3r. ’
l J
m§—-3(r t) + —§—-[T . + 6 .(pU-V)] = 0
3t ' 3r ij ij

av _
where 33 — U. (12)

18

This is the equation of momentum conservation. If we

define a stress tensor Pij as,

Tij + 6ij(pU-V) = Pij + mpuiuj

and combine Equations (12) and (11) we get Euler's
equation of motion,

+ +

u + +
—— + pu-Vu + V-P = 0.

pat

The Fourier transform of Equation (10) is the equation

of motion for the phase space distribution function.
Lin: p) +E-v f(r p) -vu-v f(r p) =0 (13)
8t ' m r ’ r p '

This is the Vlasov equation.15 It is just what one
would write down for a classical system of particles
if there were no collisions. Expressed in words, the
change in each coordinate is given by the velocity asso-
ciated with that coordinate, and the change in the
velocity is proportional to the force.

The validity of Equation (13) rests on the intro-
duction of a local one-body potential. The potential,
which will depend on the density, is a fundamental quan-

tity which must be chosen properly.

CHAPTER III

BULK PROPERTIES AND MACROSCOPIC PHENOMENA

In this section we will first discuss some of the
bulk prOperties of a general fluid. Then several exam-
ples will be presented which illustrate the different
kinds of macroscopic behavior. Consider an ordinary
fluid whose energy is known as a function of density.

A typical energy-density curve is shown in Figure 1.
Assuming that the fluid has an isotropic stress tensor,
which is the case for most normal fluids, we can define

a pressure which is also a function of the density.

_ _ 2g _ 2 3(E/n)
P _ 3V _ D So

At equilibrium density the energy function will have
a minimum. The pressure at this point will be zero.
Another important property of a fluid is its compres-
sibility. This is related to the second derivative

of the energy function.

3P
k = -v—| = —|
3V 90 p 00

A convention often used in nuclear physics is to define

the compressibility K,

19

20

”.1

 

 

 

Figure 1. An energy density curve for a typical fluid.
The coordinate R is the average separation between particles.
At some equilibrium spacing R0 the energy per particle is a
minimum.

21

Compressibility is not a well understood property of
nuclear matter. Depending on how one parametrizes the
interaction, the compressibility will vary. Empirically,
the best value for K is about 200 MeV.16 It will be a
goal of our model to show how the results depend on what
value is chosen for K. In physical terms if K is large
the energy will rise sharply as the density is increased.
Fluids with large K are relatively incompressible. A
softer equation of state, small K, corresponds to a more
compressible fluid. Three problems which will be of
particular interest in modeling nuclear collisions will
now be worked out.

The first question to consider is how the fluid
will respond when a portion of it is disturbed from
equilibrium by some outside force. The solution can

be easily worked out, if the disturbance is small, by

expanding the density and the velocity about their

equilibrium values.17 The equations of motion are:
8p + _
5? + V (on) - 0 (1)
+
an + +
p-a—E + (u°V)pu + VP = 0 (2)

In equilibrium, the fluid is at rest or in uniform

motion. No fluid element has any relative velocity.

22

By small disturbances, it is meant that fluctuations
in the density are small and the relative velocities

9.
remain small. Define the small function 6p(r,t) as
+
o(r,t) = 00 + 60(r,t).

Substituting this into the equations of motion and keep-
ing only first order terms, we find that small disturb-
ances propagate through the fluid. The function dp(;,t)
obeys a wave equation and the velocity of the disturb-
ance is closely related to the compressibility. The

continuity equation becomes, to first order

8 + +

SE 60(r,t) + oOV°u(r,t) — o. (3)
Euler's equation becomes

Bib} 2%" — 4
00 5E u(r,t) + cS v(r.t) - 0 ()

c: is defined as the derivative of the pressure with

respect to density, evaluated at equilibrium density.
Taking the time derivative of Equation (3) and subtract-

ing it from the divergence of Equation (4) gives

2

+ 32 +
V 5p(r,t) - —-§ {69(r,t)} = 0.

at

”I
(DNH

This is the equation for a wave whose velocity is

given as

23

_ 3P3:
Cs-(b—a)

In terms of the compressibility the sound velocity can

be written as

_ K 35
CS _ (9mpo)

If K = 200 then for nuclear matter the sound velocity

 

would be about .16 c. If large stresses are applied
to the system such that densities substantially lower
than 90 are reached then, as can be seen from Figure 1,
it is possible that %% will become negative. The point
in the equation of state where the sound velocity becomes
zero can be interpreted physically as that point where
the fluid becomes unstable and breaks up. This negative
maximum of the stress is a measure of the tensile
strength of the fluid.

The next example illustrates a fluid undergoing
a more massive disturbance, similar to the kind that
must be considered in a nuclear collision. Consider a
one dimensional slab of fluid traveling to the right
with velocity u0 and colliding with a stationary wall

at time t = 0.18

For the sake of simplicity it is
assumed that no energy is dissipated. Qualitatively,
the situation can be described as follows. Upon impact,

there will be an increase in the density in the region

24

close to the wall. This will cause a pressure which
slows down the incoming fluid elements. As more fluid

is forced to rest, the boundary between the region of
high density and normal density, hereafter called the
shock front, will move to the left. The important param-
eters to discover are: 1) how fast the shock front
propagates, and 2) what is the density of the fluid

in the compressed region. Again, the conservation laws
contain the solution. The following picture illustrates

the problem.

 

uncompressed compressed
region region
4..
00 Vs
u +
0

 

 

 

Consider what happens in a tube of unit area during

a time interval At. During this time, the shock front

will move a distance VSAt to the left. A fluid element
in the uncompressed region will move uOAt to the right.

In order to conserve mass, it must be true that,
pVSAt = po(uO + VS)At.

It is also clear that the rate at which momentum is

removed from the tube must be equal to the pressure in

the compressed part.

25

p0u0(uO + VS) = P(O)

These two equations are enough to determine the density
of the compressed region and the velocity of the shock
front, provided we also know the equation of state.
This solution holds particular interest because it is
also the solution to the problem of the collision of
two slabs as seen in the center of mass frame. From
these results the evolution of the system can be
described quantitatively. At time t = O a shock-front
begins to move from the wall to the left. The velocity

of the front is given by

2
Vs = [P(o) - 00u01/pu0

The kinetic energy of motion goes into increasing the
density. This provides a pressure which brings the
moving matter to rest. The density in the compressed

region is given as the solution to the implicit equation
P(o) - Duo/(o - oo) = 0.

The shock front travels until it reaches the surface at
the left where it is reflected back to the right.
During this second stage fluid is ejected out of the
compressed zone to the left with velocity uo, and if
there are no cohesive forces between the fluid and the

wall it will leave having undergone an elastic bounce.

26

If the length of the slab is L, the time at which the
slab leaves the wall is t = 2L/Vs.

There is one more idealized problem that must be
solved before a complete description of the slab colli-
sion can be put together. This is the wave function of
suddenly released compressed matter. Imagine a semi-
infinite slab of matter in equilibrium and at rest.
Suddenly, the right hand side is pulled away with

velocity uo.

 

h*u

 

 

Because of the stress, a rarefied region will develop.19

How low the density drops and how the rest of the fluid
responds are the questions that must be answered. Because
there are no characteristic lengths, it can be argued

on dimensional grounds that the density and velocity

can be written as functions of a new variable C = x/t.

A different argument for this goes as follows. From

the theory of first order differential equations, it

is known that a boundary condition and a first order
differential equation are enough to uniquely specify

a solution for all values of the independent variable.

If the velocity and the density are known for some value

27

of g and it is also known how they change with respect

to a change in c, then a solution exists for all g.

Since the right hand edge is moving at v = uo, Q will

always be uO at this edge. The boundary condition is
v(; = no) = no.

At a point far enough away the density will be p0,

p({, = 00) 2 00'

Using the relation g; = -c%§ the continuity equation
can be rewritten
30 3V _
(v Q) 3C + pa; — O. (5)

Similarly, Euler's equation becomes

8

<2

+ c2 QR = O. (6)
s

o(v- C) C

Q)

C

These are two coupled first order equations for p(;)
and v(q). Thus 0 and v are specified for all values
of C. Dividing Equation (5) by Equation (6) gives a
relation between the sound velocity, the fluid velocity,

and C.

Putting this back into Equation (5) gives

28

 

Integrating across the rarefaction from v = 0 to v = u

tells how low the density will be at the edge.

cs(o)
u0 =f p dp (7)

 

Another important number that is more easily obtained
is how fast the edge can be pulled away before the fluid
breaks up. Setting the upper limit in the integral (7)

at p gives the upper limit on no. If the edge is pulled

c
away faster than this critical velocity the density
will become discontinuous.

With this result the slab collision can be under-
stood. After contact a compression wave propagates
in both directions outward from the point of contact.
Whentfluacompression wave reaches the outer edge it is
reflected back toward the center. When the two decom-
pression fronts meet back attfluacenter a rarefaction
will develop. If visualized in a frame moving with
one of the slabs it will appear as if the edge is sud-

dently pulled away. The solution given above tells

how low the density drops during the rarefaction.

CHAPTER IV

ONE-DIMENSIONAL MODEL

In order to describe the dynamics of nuclear matter,
it is crucial to have a realistic equation of state.
There are two properties of nuclei that must be reflected
in this equation of state. These are the equilibrium
density and the binding energy. Based on experiments
that measure the size and the mass of nuclei, these
values are taken as .16 nucleons/fm3 and 16 MeV/nucleon
respectively. Unfortunately there is no good theory

41 The

from which an equation of state can be derived.
best one can do at this stage is to write down a func-
tional form. If the nucleus were a non-interacting
Fermi gas then the energy per nucleon would be just

the average value of the single particle kinetic

20
energy,

2
_ .13.
E/A - 2m) .

If the particles also interact by a short range repul-
sive potential, then there is a contribution to the
energy proportional to the density, assuming that the

Fermi gas approximation is still valid. Finally, to

29

30

account for the saturation of nuclear matter we can
add a density or velocity-dependent potential, depending

21,22,23

on which model is used. This gives a contribu-

tion to the energy that varies as DO+1

, where 0 may
range between 1/3 and 5/3. This particular parametriza-
tion was suggested by Zamick24 and it is the one we

will use

2
B/A = <%)+ Ana/00) + B(o/oo)o+l.

Here 00 is the equilibrium density. For a given value
of o the two constants A and B are fixed by knowing

the binding energy and the saturation density. By con-
sidering different values of O we are, in effect, allow-
ing the compressibility to change. If we make the
Thomas-Fermi approximation and assume that the particles
always occupy a sphere of minimum radius in momentum
Space, then we can write the compressibility as a func-

tion of 0.24

2 2
P P

EB = binding energy

We must now specify the stress tensor. There are
two distinct limits that can be placed on the form of
Pij' In the simplest approximation we can say that Pij

is isotropic.

31

Pij = 6ijP(p)

This is the approximation that is usually made for
ordinary fluids. The stress tensor is also isotropic
in the Thomas-Fermi model.25 Physically this means
that collisions between the nuclei are frequent enough
so that a compression along one direction is followed
by an increase in pressure along all directions within
a time that is short compared to the time scale of the
nuclear collision. In this limit the single particle

kinetic energy can be written as

2

e is the Fermi energy at equilibrium density. The

F
pressure, which is now a scalar function of density,

is given as

p = 02 33 (E/A) = po [—52,- eF(o/oo)5/3 + Ana/co)

0+2

+ B(o + 1) (0/00) 1.

On the other hand, if the collision rate is low and
there is no mechanism by which momentum can be trans-
ferred from one direction to another, then the different
directions will remain uncoupled. This leads to the
kind of stress tensor that is used to describe an elas-

tic material. A compression along the z direction

32

causes a pressure along the z direction only. This
will change the dependence of the kinetic energy on

the density.

2 2 2

2 p P p
L - __X .1 _§ -3 l 2
<2m>‘<2m + 2m>+<2m>- 5 6F + 5 EF (“90)

We will now construct a model for the collision of
two slabs of nuclear matter. A Langrangian method is
used to numerically integrate the fluid dynamic equa-
tions.26 Individual fluid elements are tagged and may
be thought of as particles. Imagine a string of parti-
cles whose equilibrium spacing is r0. That is, if r > r0
there is an attractive force which pulls them together.
If r < r there is a repulsive force which pushes them

0
apart. At equilibrium the particle density is

If forces act which cause the particles to get closer
together the density will increase. If these forces
are conservative then the work done will cause an in-
crease in the internal potential energy. Similarly, a
decrease in density will also cause an increase in
energy. The force on each element is obtained from

the stress tensor.

33

r r r
(E—W‘ + Mfg) + B(o + 1) (;9)°+2}

'11
ll
HIH
o
A
uflw
m

3 (elastic)

Q
II

a = % (fluid)
An equation of motion is written for each fluid element,

mri = F(ri'ri+1) + F(ri,ri = F(ri).

-1)

From the configuration of the system at time t the force

is calculated. The new velocities are
Vi(t + At) = (F(xi,t)/m)At + vi(t)
The new positions are
xi(t + At) = xi(t) + vi(t)At.

One of the more interesting results of this model
is the energy at which the system breaks up. The most
important parameter in determining this breakup energy
is the tensile strength of nuclear matter. The critical
density, along with several other important characteris-
tics of the equation of state associated with the breakup
threshold, are presented in Table 1. The critical den-
sity remains between about .64 p0 - .68 p0 for all of
the cases considered. In the elastic model, however,

more energy is required to achieve critical density.

34

 

N

O
m
N
v
<r
\o

O

HIM

 

oflummam
m.m m.m mm. H
m.H v.~ «o. m
cfisam
~.N m.m mm. H
AmsM\>m2. oa suflmcmn u o mumum
A>mzv Aonm i Andvm whammmum ESEmeS HMUHuHHU mo cowumzvm

 

 

.mmfluflawnflmmmumaoo acmummmwc on ccommmuuoo o no mmsHm> uanmMMHc was
.Hmuums umoHosc mo caocmmunu moxwwun may op cmuwamu mmfluflucmsv ucmuuomefl one .H magma

35

 

 

m.H mmo. m
oaummHm
m.H moo. H
am. mac. m
anHm
H.H use. H
i>mzi mmumcm HmoHuHuo suHoon> HmUHuHuo u o mumum mo :oHumswm

 

 

mcoflumsqw ucmeMMflU

.hIHHH cowumsvm gnaw: cmumHsono mum mmmna .mumum Mo
on» How mnsooo stmoun nuanz um wufloon> HMUHDHHU was

.m mHnme

36

'5

fm

 

AA
5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

WW

 

 

 

 

 

 

 

4O 80 IZO ISO
(mk:

Figure 2. Slab collision at 1 MeV/A (c.m.). The posi-
tion of mass elements as a function of time is shown for half
of a symmetric collision of two slabs, each 9 fm. thick. The
slabs just make contact at t = 0; the equation of state is
the fluid model with o = 1/3. Maximum compression occurs at
40 fm/c and the system snaps at E 160 fm.

37

These numerical studies verify the behavior that was
described in Section 111.27 The numerical values for
the breakup energy and velocity are given in Table 2.
Figure 2 shows a graphic representation of a slab col-
lision. Each line represents the motion of a tagged
fluid element. In this calculation, we used a time
step At = 1 fm. The compression and rarefaction wave-
fronts are clearly visible. The first two stages of
the collision are completely described in the analytical
treatment given above. The third stage shows some
interesting features that require further discussion.
First of all, the energy required for breakup is some-
what higher than the predicted value.2.7 This happens
because the forces have a finite range. Even though
the density may have fallen below the critical value,
there are still forces between the fluid elements.
Furthermore, the development of an instability requires
a finite amount of time. The time required will depend
on the energy. If the energy is more than 50 percent
above threshold then the system will break up quickly.
On the other hand, if the energy is only slightly above
threshold, the rarefaction will have propagated a large
distance during the time it takes for the instability
to develop. During this stretching it may happen that

enough momentum is removed from the center of mass

38

motion so that the system does not come apart. Because
of the stretching, the energy associated with the rela-
tive motion of the slabs after separation can be con-
siderably less than the incident energy. This particu-
lar feature will come up again when we discuss energy

loss in deeply inelastic heavy ion collisions.

CHAPTER V
DESCRIPTION OF THE MODEL AND

COMPUTATIONAL TECHNIQUE

We want to use the results of the one-dimensional
calculation to create a more general model of the col-
lision of two nuclei in three dimensions. The computa-
tional approach, however, will be quite different.
Recall the technique in the slab sollision was to tag
many fluid elements and then follow the motion of each
as the system evolved. This works if the system is
simple enough, but gets very complicated for a three
dimensional calculation. What can be done instead is
to parametrize the motion. This will reduce the number
of degrees of freedom, while still keeping a valid
description, provided the parametrization is chosen
properly. For example, knowing the form of the solution,
the one-dimensional slab collision can be described
with just one variable. This is the location of the

shock front.

 

Vo

 

IQ *
n.
o

39

40

The continuity equation becomes

L0 = po(£ — d) + od,

0

where L is the initial length of the slab, 2 is the
length at time t, and d is the location of the shock
front at time t. The fluid between £ and d moves to

the right with uniform velocity v Assuming the fluid

0.
to the right of the shock front in the compressed region

to be at rest, then Euler's equation becomes, in the

integral form,
oOv0(V0 + Vs) = P(o).

It is convenient to use the coordinate of the center

of mass x .
cm

E

_ _ l 2 l 2 _ 2
xcm - d/pxo(x)dx — [2 pd + 2 p0(2 d )]/Lp0
0

The velocity of the center of mass is

_ l -1 -l
Mvcm - pOULv0 + 2 RVs 2 LVS 2 dvo].

The acceleration of the center of mass is

Ma = M

cm vcm(d) = oovow0 + VS) = P(o). (1)

film

To solve the dynamics, a best guess is made for d. The

new configuration is specified and the acceleration of

41

the center of mass is calculated. A variation is then
done on d and again the acceleration is calculated.

The first derivative is defined as

Bacm _ acm(d0 + 6d) - acm(d0)

3d 6d °

 

On a small enough time scale, the force and the accelera—
tion will change slowly enough so that the force from

the previous time step can be used to calculate the
acceleration. Using a first order Taylor series expan-

sion the equation for 6d is

8a

_ cm
P(o) - acm(d0) + —3d 6d.

In the next step set d = d + 6d and again calculate
acm(d). If the quantity Macm(d) - P(o) has a magnitude
less than some small number, then this time step is
complete. If not, then d is put forward as the best
guess and the whole algorithm is reiterated.

For the three-dimensional model we will follow
this same general procedure. The system will be param-
etrized in such a way as to keep the number of degrees
of freedom small while still retaining the features
that are important to the theory. The features that
are to be included are the following. 1) The idea that

nuclear matter can be represented by an equation of

42

state and that during the collision a shock wave will
form. The velocity of the shock front and the density
that is maintained in the compressed region will depend
on this equation of state. 2) Because the particles
on the surface are less bound, there is an energy propor-
tional to the surface area. This causes a surface ten-
sion which tends to minimize the surface area. 3) The
charges on each nucleus give rise to a Coulomb repulsion.
The kinds of information to be supplied from the
model are the general features that can be observed
experimentally. The angular distribution, which in a
classical model comes from a deflection function, will
be studied. At lower energies fusion cross—sections
will be calculated. Energy loss and angular momentum
transfer are also topics that will be explored. All
of these results will depend on the equation of state.
The nature of this dependence will be of primary interest.
The collision of the two nuclei is parametrized in
the following way. The shape of each nucleus is assumed
to remain spherical. In fact, because of the compres-
sion there will probably be some distortion of the shape
in the region of higher density. This may be important
in dissipating energy. In this model we neglect this
distortion. Take the reaction plane to be the z = 0

plane. The coordinates (x,y) will specify the location

43

of the center of the sphere. These will be two of the
parameters used. The shock wave is taken to be a plane
wave which travels along the line connecting the centers
of the two spheres. The location of the shock front

will be the third parameter. The following figure illus-

trates a head-on collision at some instant in time.

>-'-)-<

 

 

Assuming the compressed region to be at uniform

density, the continuity equation becomes

3 3_ .
DO 3 “HR - fpdV+ foodV

The velocity field in the compressed region is taken
to be zero while the uncompressed region moves with
a uniform velocity v. Euler's equation will give the

shock velocity at the point of impact.
OOV(V + VS) = P(O)

VS is then assumed to remain constant during the col-

lision. For collisions at a non-zero impact parameter

44

the situation is similar if viewed from a rotating
frame.

As in the one-dimensional case the forces arising
from the pressure, Coulomb repulsion, and surface ten-
sion are taken to act on the center of mass coordinates.
The Coulomb repulsion is taken as the force between two
point charges located at the center of mass of each
nucleus. In calculating the force from the surface

tension we have used the following picture.

 

The surface energy is taken from the liquid drop model28

to be 1 MeV/fmz. If the separation between centers

changes then the surface area will change. The change
in area with respect to a change in separation is pro-
portional to the force. The area can be written as a

function of x,

x/R

Area = ZWRZ ‘Jf d(cos 6) = 217R2

X
(fi + l).

-1

45

The force is given as

Force = 0%; (Area) = 2wRo.

o is the surface energy and R is the radius of the
nucleus. Using this expression for the surface force
will underestimate its strength during the initial stage
of the collision since a neck region will be forming.
During the breakup stage when the neck is stretching

it will overestimate the surface force. However, since
the shape of the neck has not been parametrized this
seems to be the only reasonable assumption. The com-
pressed region will exert a pressure at the boundary
between the two nuclei. This has two components. There
is a radial force equal to the pressure times the area

of the boundary. The force is given by

Pr = P(o)n(R2 - x2).

where 2x is the distance between the centers of the
nuclei. This will be a repulsive force acting in the
same direction as the coulomb force. The surface force
is also radially directed but it is attractive.

The second component of the pressure is a tangen-
tial force which comes from the flux of particles, with
tangential momentum, across the boundary. This acts

like a friction force removing energy from the collective

46

motion. This energy goes into other degrees of freedom.
The momentum flux is calculated by integrating over

the phase space distribution function.

3
TH = prvLflrrmd P
This integral can be done explicitly using the Fermi

gas model. In momentum space the distribution can be

drawn as two displaced spheres.

Pll

F a
L

This leads to the following expression for T||

 

 

TII = 3per[1 - a2)(1 + a)3/3 + §a(1 + a)3
1 4
"Z(l+a) 1:
v v I
where a = ——, and v = ———. v is the Fermi velocity.
vF vF F

In the limit of small velocities this reduces to the

approximate expression

47

~ 1
TH = 3O€F\)(4).
The force itself is given by

_ 2_2
Ft - Tll (R X ).

Ft is taken to act on the center-of-mass coordinate of
each nucleus, changing the tangential component of its
momentum. The tangential force also causes some of

the orbital angular momentum to be transferred to intrin-
sic angular momentum of the nuclei.

The equations of motion for the center of mass

coordinates are

 

 

x a {xcm(x,y,d;t+At)-xcm(x,y,d;t)} Fx
acm(t) — 3? At - M—
y 3 ycm(X:Y,d;t+At) ' ycm(X:Y:d;t) FX
acm(t) — 5t { At }"fi—

These are both implicit equations for the parameters
x,y,d. The computational technique used is just a
generalization of the one dimensional case cited earlier.
Based on the previous history of the system, a best
choice is made for the new configuration. A variation
is then done on each parameter and the first derivatives
are calculated. The small changes are calculated from

a first order Taylor series expansion. This technique

48

seems to work very well. It converges rapidly onto

the correct solution, and the errors can be easily spec-
ified. It is also quite stable and allows a coarse

time step to be used. An interval of t = 1 fm/c was
used in the calculation. This provided sufficient
accuracy with a minimum of computer time. During the
final stage, when the system is breaking up, the lower
density causes a negative pressure which provides an
attractive force. If the velocity is larger than the
critical value obtained in the one-dimensional model,
then the system will break up. The amount of stretching
that takes place before separation is taken over

directly from the one-dimensional model.

CHAPTER VI
APPLICATION OF THE COMPRESSIBLE FLUID

MODEL TO ENERGY LOSS

In our model there are two mechanisms by which
energy is lost from the relative motion. For energies
well above the fusion barrier the primary cause of energy
loss is the flux of tangential momentum across the bound-
ary separating the two nuclei. This acts like a friction
force, removing momentum at a rate proportional to the
tangential velocity. The largest losses will occur
when the impact parameter is around .5 R. For collisions
at low energies, just above the fusion barrier, there
can be a considerable amount of energy lost from the
relative motion because of the finite distance the rare-
faction travels before the nuclei break apart. In this
case the energy goes into producing a rarefied region
of nuclear matter in which each nucleon is bound by
an energy less than 16 MeV. Both the frictional force
and stretching are one-body processes. In Section IX
we will also take into account the energy loss due to

nucleon-nucleon collisions.

Experiments done at energies just above the Coulomb

barrier show a large cross section for strongly damped

49

50

 

 

 

 

 

 

 

 

 

 

1 I j j
9-69.79’
9.49-59'
)2
1 ' -
5.
'% d'
‘2
o . 4
‘2’ , L
w 9.39.49-
U
5 4 4
\
“b J , 9.59.69-
‘0
I
V' 9-29-39. 1 2
' ’ $49.29* "
J l 1 7 j
zoo 300 400 200 300

Kinetic enemy (MeV)

Figure 3. The energy distribution of the quasi-Kr
products observed as a function of angle. The reaction
is 525 MeV 84Kr on a 203Pb target.

d'ovda ac mb MeV“'.sr-'

products observed as a function of angle.

C)

50

 

 

 

 

 

 

 

 

 

 

Kinehc energy (MeV)

j t 6-69.79'
9.49.59°
)—
«I
)-
9.39.49-
4 A
J ._ 9-59.69'
r
9-29.39” -4 2
* 6.19.29' "
1 l l ,
zoo 300 400 200 300

Figure 3. The energy distribution of the quasi-Kr

is 525 MeV 84Kr on a 208Pb target.

The reaction

51

collisions. Looking at the energy loss as a function
of angle one finds, aside from the elastic peak, a broad
energy distribution centered near the Coulomb energy.
Figure 3 shows a typical distribution as seen in the
lab frame.29 Energies below the Coulomb barrier are
commonly observed which indicates that there is some
deformation of the system before separation.

The following figures illustrate the results of
the compressible fluid model as it applies to energy
damping. The two fragments are assumed to follow Coulomb
trajectories after scission. The kinetic energies
referred to include this Coulomb energy. Figure 4 shows
how the energy loss depends on the compressibility.
The ratio of the scattered particle's kinetic energy
to the incident energy is plotted as a function of angle
for two different equations of state. The two cases
considered correspond to compressibilities of 200 MeV
and 500 MeV. These represent the extreme values of
K. The nuclear system used in this example is 84Kr-I- 84Kr
at an incident energy of 4 MeV/nucleon in the center of
mass frame. The energy of the outgoing fragment ranges
from Einc' for a nearly grazing collision to a value
slightly below the Coulomb energy of two touching
Spheres. The energy loss is less for the softer equation

of state (K = 200). To understand this result recall

52

 

J

K=2oo ,

’ K=SOO

, a4Kr 49‘)“
E ==4|MeY/h

 

8 (degree!)
ISO

 

 

i J
50 IOO
Figure 4. The energy loss depends somewhat on the
equation of state. Here energy loss is plotted as a function
of angle. E/Einc is the fraction of energy left in the scat-

tered nucleus.

 

3.5 45....

2.5

 

”Kr+°4Kr
K: 200 MeV

impact poromflor

 

 

.5

 

Figure 5. The fraction of energy left as a function
of impact parameter at center of mass energies of 2.5,
3.5, and 4.5 MeV/A.

54

from Table 1 that the critical velocity, which is a
measure of the strength of a fluid, is lower for a softer
equation of state. This lower critical velocity means
that the system will not stretch as much before snapping,
leaving more energy in the relative degree of freedom.
Figure 5 shows the fraction of energy left as a
function of impact parameter. For an incident angular
momentum greater than the grazing value there will be
no energy loss in this model. In fact, there may be
some energy loss due to Coulomb excitations. This
mechanism has not been considered here. As the angular
momentum decreases to slightly below the grazing value
there is still very little overlap between the colliding
nuclei. The length of time during which contact is
maintained is also short. Based on this one would
expect the energy loss to be small. As the impact
parameter becomes even smaller the energy loss increases
sharply until it reaches a maximum value at about b = .5.
In this range the contact between the two nuclei is
solid and there is substantial energy loss due to the
flux of tangential momentum. These results up to this
point show very little energy dependence other than
the obvious dependence of L9 on Einc' For small impact
parameters the energy loss does depend somewhat on the

incident energy. High energy collisions are quick and

55

the nuclei have a tendency to bounce off from each other
with very little dissipation of energy. As the energy
decreases the nuclei begin to appear sticky to each
other. The analogy to silly putty, which has been used
to describe heavy ion collisions, is apprOpriate here.
In Figure 6 we present the results in the form of a
Wilczynski plot.45 This is the kind of information

one would obtain from experiment. At the more forward
angles there may be two or three distinct energy peaks.
This happens because two or three very different trajec-
tories may finally scatter into the same angle. The
largest energy losses are observed at quite forward
angles. The cause of this is the onset of nuclear
orbiting which will be discussed further at a later
point. Again at the large angles which correspond to
small impact parameters there is some energy dependence.
For low energy collisions just above the Coulomb barrier
there will, of course, be no back scattering because

of fusion. The small numbers on each curve give the
impact parameter that corresponds to the angle. This
provides a relative measure of the cross sections. This
particular plot is for 40Ca + 40Ca at energies over

the range of 2 to 5 MeV/nucleon. Using the results

of energy loss as a function of impact parameter it

is possible to derive another kind of cross section

56

 

l l '

“Co + “Ca
K=ZOO MeV

 

 

9 (degrees)

 

l

 

 

l l
50 I00 I50

Figure 6. The fraction of energy left in the out-
going fragment versus the scattering angle. The small
numbers indicate the impact parameter.

57

Physically this corresponds to plotting the number of
counts in the energy range dE as a function of energy,
integrated over all angles. Figure 7 shows do

3%?
E for the reaction 84Kr + 84Kr at 5 MeV/nucleon. The

versus

fairly sharp peak comes from the peak in energy loss

at b E .6 R.

58

 

T 7

a "Kr +"Kr

é K=2oo MeV
15me

3‘.

\

Z

(D
lO-S
5-

ENERGY (MeV/n)

 

 

 

 

3 4

Figure 7. Energy distribution of the outgoinc
nuclei integrated over all angles.

CHAPTER VII

APPLICATION OF MODEL TO FUSION

When two heavy ions collide with energy in the
range of 1-6 MeV/nucleon, and if the angular momentum
is not too large, then there is a good possibility that
the two nuclei will fuse together, forming a compound
nuclear system. Cross sections for fusion have been
measured on a variety of systems at energies up to about
5 MeV/nucleon. (Energies quoted will always refer to
the center of mass coordinate system.) A detailed
account of many fusion experiments has been compiled

by Birkelund and co-workers.30

From these experiments
the general features of the fusion cross section can

be described. At energies below the Coulomb barrier
fusion is improbable. As the energy increases above

the Colomb barrier, the cross section rises sharply.

At about 4 MeV/nucleon it reaches a maximum. As the
energy is increased further the cross section starts

to decline. Above 5 or 6 MeV/nucleon it is expected

to continue this decline, but there is no data available

to verify this. The maximum cross section is typically

one barn.

59

60

Before presenting the results of our calculation
it will be useful to discuss briefly how fusion occurs
in this compressible fluid model. For energies below
the Coulomb barrier no contact is made and the fusion
cross section is predicted to vanish. At these energies
the nuclei follow purely Coulomb trajectories. As the
energy is increased the range of impact parameters over
which contact occurs becomes larger. When contact is
made the surface tension and the cohesiveness of nuclear
matter favor fusion. As the energy is increased further
contact is made with relative velocity. In this case
a region of higher density is formed. Some of the rela-
tive kinetic energy will go into increasing the density
and the rest will be dissipated by the frictional force,
exciting internal degrees of freedom. The compound
system will go through compression, decompression, and
rarefaction. The relative velocity of the two nuclei
along with the impact parameter will determine the inten-
sity of the oscillations, which in turn will determine
whether the system remains fused or breaks apart during
the rarefaction. Depending somewhat on the equation
of state, nuclear matter has enough cohesiveness to
hold the system together if the rarefaction is not too
strong. At low impact parameters a larger portion of

the kinetic energy goes into the compression. This

61

energy, stored in the first stage of the collision,

is then available in the second stage to accelerate

the two nuclei apart. This phenomena suggests that

the two nuclei are bouncing off from each other. The
critical energy for breakup is lower for the smaller
impact parameter. The critical energy will increase

as the impact parameter increases because of the increas-
ingly larger amounts of energy dissipated by the tangen-
tial friction. This has the effect of damping the oscil-
lations and reducing the intensity of the rarefaction.

As the collisions become more grazing the overlap be-
comes less and the centrifugal force will set the upper
limit on the range of impact parameters for which fusion
occurs at a given energy. All these arguments can be
incorporated in a very simple model if friction is
neglected. The nuclei follow Coulomb trajectories up

to the point of contact. At this point, if the radial
velocity is less than some critical value, which will
depend on the equation of state, and if the centrifugal
force plus the Coulomb repulsion is less than the surface
tension, then the two nuclei will remain fused. If

these conditions are not satisfied the system will break
up and a scattering event will take place. This simple
model exhibits the features of the more detailed model

regarding the fusion cross section. It allows one to

62

calculate directly the fusion cross section for any
nuclear system. The effect of including friction will
be to increase the cross section at the higher energies.
The results of the full calculation have been
tabulated for a wide range of nuclear systems, from

4OCa to 134X

e. Figure 8 shows the range of impact
parameters for which fusion occurs as a function of
energy. One can clearly see the bouncing effect at

low impact parameters. At the highest energies, fusion
takes place only at b E .5 R. It is in this range that
the largest amount of energy is lost. The top smooth
curve represents a grazing collision. For impact
parameters above this line, the scattering is pure
Coulomb. Note also that at higher energies the maximum
impact parameter decreases as the energy increases.

This indicates the increasing importance of the centrif-
ugal force. From Figure 8 it is clear that larger
systems are more likely to break apart after having

made contact. This is due largely to the Coulomb repul-
sion. Figure 9 shows the total cross section as a func-
tion of energy. The peak occurs at between 2 and 3 MeV/
nucleon. The fraction of the total cross section
available for fusion decreases as the number of nucleons
increases. Figure 10 shows the dependence of the fusion

cross section on the equation of state. The reaction

63

 

Jawumiod JOde!

(HZ/Q)

ENERGY (MeV/n)

l J l l l

l 2 3 4 5

 

 

 

 

 

 

Figure 8. The energy-impact parameter area in
which fusion occurs. The extended curve represents a
grazing trajectory. The compressibility is K = 500 MeV.

64

is 40Ca + 40Ca at K = 200 MeV. The effect is not large

and seems to have no effect on the low energy collisions.
As the energy increases the cross section falls off

more rapidly for small K. This happens because the
critical velocity increases as the compressibility
increases. In effect the softer fluid is less sticky

and unable to hold together under large tear away forces.

65

 

 

 

 

] I I T’ I
”Ca
5 o - “Kr ‘
.9
’\
2% n4
Xe
25 '- ..
1 I 1 4 1
I 2 3 4 5

ENERGY (MeV/n)

Figure 9. Energy dependence of the fxsion8 ross
section for three different systems, 40Ca( Zr),
etc.

 

66

 

”Co +‘°Co
K=SOO MeV
9
3
SOO-cr H
K =ZOO

 

 

ENERGY (MeV/n)
l

l J 1

l 2 3 4 5

 

Figurs 10. The cross section for the reaction
40Ca(40Ca, oZr) using two different equations of state.
K - 200 MeV and K = 500 MeV represent the two extremes

in compressibility.

67

 

0.80 P

b/2R
0.60 -

 

 

 

O.4O -

020' '3 ' ff? H

 

 

 

 

 

l l l
| .00 2.00 3.00 4.00 5.00
E /n (M eV)

Figure 11. The fusion regime for the reaction
4oCa(40Ca,40Ca)40Ca. The textured area represents a compres-

sibility of K = 200 MeV. The enlarged area is for a compres-
sibility of K = 500 MeV.

CHAPTER VIII

ANGULAR DISTRIBUTIONS

Many experiments have been done which measure the
angular distribution of the products of heavy ion col-
lisions. In describing any scattering event, the wave
number k and the physical size of the system are the
important parameters in determining whether the behavior
is wave-like or particle-like. If kR < 1 a wave descrip-
tion is necessary. On the other hand, if kR >> 1, then
it is valid to describe the scattering with a classical
trajectory. For typical heavy ion collisions, at ener-

1 and R is

gies above the Coulomb barrier, k > 10 fm-
the order of 5-10 fm. For these collisions, it is valid
to define an impact parameter and to describe the scat-
tering in terms of a deflection function 6(b). The

deflection function itself is not experimentally observ-

able, but the cross section can be obtained as

 

An mentioned before, most of the experiments have
been done at energies near the Coulomb barrier. Also,
the systems tend to be asymmetric with a large target

and a small projectile. Typical beams are 160, 20Ne,

68

69

and 46Ar. For targets, 208Pb, 107Ag, and 197Au are

commonly used. The main feature of the measured cross
sections is a side peak at angles near the Coulomb
grazing angle. Figure 12 shows some angular distribu-

tions for the reaction products of 84Kr + 181Ta at

3.3 MeV/nucleon.31 For angles greater than 8 ,tfluacross
section drops off rapidly. As the energy increases the
peak moves forward and becomes sharper. This side
peaking tends to support a theory which gives short
interaction times, t < 400 fm/c. If the interaction

time were long, the two nuclei would lose their indi-

vidual identity. The cross section would be isotropic.

 

This would result in a angular distribution. Some—

sine
times instead of a side peak the cross section decreases
monotonically with increasing angle. This kind of

behavior is usually associated with orbiting, or nega-
tive angle scattering.33 There is in fact experimental

evidence which indicates that negative angle scattering

oes occur.

Figure 13 shows several time slices of a typical

118? 1mm. The

scattering event. The reaction is d +
energy, in the center of mass frame, is 5 MeV/nucleon.
The impact parameter is .4 R, which gives an initial
orbital angular momentum of L = 130 h. After impact

the compound system goes through compression, decompres-

sion, and then a rarefaction, during which a neck region

70

$510 + 620Mev 33K:

 

   
 
   

(07 , . . . , j . T T : T r I n I
) W .[ q
.) ( r. .. _
q» XIO q» j ‘ ).. ..
v- -' ah I)- -l
4 . ( ) 0 044
-2)‘
'0’. F2) .1» 5

I 63)-

 

(-7)
2 -b
(~8)

 

 

 

A

k6)

 

(2)
#4 -
33 (~3)
. . LR :
39 , t4) «4)
._4\ _
‘ -5)
;;A\\(5)
)

 

 

'3
_)- '
fine
1:43 d- -
(9)
I66 I 1 I 1 I ' 1 ‘ ' l '4 l 1 I l i 9 1 l
30 90 I50 30 90 ISO 30 90 ISO 30 90 ISO
9cm. (deg)

Figure 12. Angular distributions for
products of the collision between 620 MeV

a 181Ta target.

he reaction
Kr ions and

71

 

 

 

 

 

 

1=0 t=20fm/c f=40fm/c

t= 60 fm/c t= 80fm/c = |00
f=|20 t=l40 t=l60
Figure 13. Collision between two Pd nuclei at an

energy of 5 MeV/n (c.m.) . The straight line represents the
boundary between the two zones.
its direction of motion.

The small arrows indicate

72

is formed. The rarefied region causes a negative pres-
sure which slows down the outward motion. At t= 140 fm/c
the neck snaps. The energy that had gone into creating
the rarefied region now goes into large collective vibra-
tions in the outgoing nuclei. The energy lost to fric-
tion is about 1 MeV/nucleon, and the energy in the vibra—
tion is also about 1 MeV/nucleon. The relative kinetic
energy has thus been reduced to about 3 MeV/nucleon.

The angle at which scission occurs is ¢ = 104 degrees.
From this point the scattered particle follows a Coulomb
trajectory with an assymptotic angle at 8 = 46 degrees.
Because of the tangential friction the orbital angular
momentum in the outgoing channel has been reduced to

94 h.

Figures 14, 15, and 16 show families of deflection
functions for energies in the range from just above the
Coulomb barrier to 5 MeV/nucleon. Most of the features
of these curves have been discussed in previous sections.
The large impact parameters b > bg correspond to elastic
Coulomb scattering. As b decreases below the grazing
value the deflection function is pulled to more forward
angles by the nuclear interaction. The peak in the
deflection function at b = bg is sometimes called the
Coulomb rainbow. This is associated with a peak in the

cross section, for the nearly elastic scattering events,

73

 

 

 

 

I I I
CD
8:
o
(a
(00 ~ 3 i
m
2’.
. ’ J
O 5 § ’ . 44
40 40
C0 + Co
-IOC>- .
K = 2 00 MeV
impact parameter
1 l 1

 

 

.25 .50 75

Figure 14. Deflection function 6(b) for the collision
between two Calcium nuclei. The small numbers indicate the

center of mass energy in MeV/n.

74

 

 

 

 

 

I I I I I I I
GD \\\\\\\
E
m .
(Q
ICXD- 3 .
m \
‘m’ ‘\
O A ' if
84 04
K r + K r
' K = 2 00 MeV ‘
-I00 - -
impact parameter
1 1 1 1 l 1 1

 

 

.25 .50 .75

Figur8415. Deflection function for the reaction

84Kr(84Kr, Kr)84Kr. The compressibility is K = 200 MeV.

The small numbers indicate the center of mass energy in
MeV/n.

75

 

 

 

 

 

 

I ' I 1 r
- 1
GD
5:
m
(00- 3 J
m
m 25
O 2 c - c
04 04
Kr + Kr
- K: 500 MeV ‘
'I00 "' 1
~ w
impact parameter
_1 l J l l J
.25 .50 .75
Figure 16. The deflection function for the reaction

84Kr(84Kr,84Kr)34Kr using the K = 500 MeV equation of
state. The small numbers indicate the center of mass

energy in MeV/n.

76

at the Coulomb grazing angle. In our model there is

no Coulomb rainbow because of the discontinuous nature

of the surface tension. The deflection function flattens
out again at more forward angles. This gives a peak

in the cross section for the deeply inelastic events.
Figure 14 shows a family of curves for the nuclear system
40Ca + 40Ca. In this case the deep inelastic peak occurs
at a negative angle. In Figure 15 the reaction

84Kr + 84Kr is shown. Comparing these two cases shows
how the angular distribution changes as the size of the
nucleus increases. The most noticeable effect is that
the strongly damped peak moves toward more positive
angles. This result can be understood by considering

how the forces change with nuclear size. The surface

tension goes like A1/3. The cohesive force increases

2/3

as A , as does the centrifugal force. The Coulomb

repulsion varies like A4/3. As the nucleon number

increases the Coulomb plus centrifugal force increases

faster than the attractive forces. Because of this,

the breakup will occur more quickly for the larger

system, and there will be less tendency to orbit and

scatter to negative angles. In Figure 16 are the results
84 84

for the same system, Kr + Kr, using a different

equation of state. The compressibility for this case

77

is 500 MeV. The most noticeable difference is an
increase in the fusion regime. The nuclear rainbow does

not appear to depend much on the equation of state.

CHAPTER IX

EVALUATION OF RESULTS

In comparing the predictions of this model with
the results of experiment we find fairly good agreement
in some areas and poor agreement in others. Regarding
energy loss, one would expect from the results of experi-
ment to find large energy losses for any collision in
which the colliding nuclei interpenetrate deeply. The
results of our model agree with this up to the low impact
parameters, but the energy loss suffered during head-on
collisions is not as great as would be expected. Further
development of this model must include a mechanism by
which energy can be dissipated from the radial motion.
Taking into account the relaxation of the stress tensor
due to nucleon-nucleon collisions will account for some
energy loss. Also, by allowing a more general velocity
field in the compressed region, we would expect the col-
lective motion to become more randomized. Both of these
improvements would increase the energy loss at low impact
parameters. To gain some insight as to the size of these
effects a calculation was done in which the stress tensor
was allowed to relax from the elastic form to the fluid

form in the following way.

78

79

.. = S€IUld + e-t/T [8?Ia5tlc — s€¥Uld]
1] 1] 1] 13
_ 1000

The relaxation time was taken to be fm/c.

_ —E;—
This is based on a calculation of the collision rate

in an infinite Fermi system having a deformed Fermi
surface, in the limit of small deformations.42 (Nuaresults
are shown in Figure 17 along with the standard results
discussed earlier. It is clear from the figure that

the scattered products have less energy. This is because
for a given density the pressure decreases as Sij relaxes
toward the fluid model. Now there is less pressure
during the decompression stage than during the compres-
sion stage. Using this form for Sij will also change

the fusion cross section. The results are presented

for comparison in Figure 18. The effect is not large,
but at low impact parameters the energy range for fusion
is increased by about,10%. The fusion cross sections
calculated using this compressible fluid model are quite
consistent with the results of experiments. Many more
experiments must be done, using larger nuclei and higher
energies, before a complete picture of fusion can be
presented. Figure 19 shows the data from an experiment
done with 16O + 26Mg. Cross sections were measured at

various energies up to about 4 MeV/nucleon.36 The solid

curve is from a calculation made using the simplified

 

(U/AaW) AOBBNB

e 4
2 a“Kr + “Kr
E: 4MeV/n
K=200 MeV
IF 1

impact parameter

 

 

1 L L

.25 .5 .75

 

Figure 17. A comparison of the energy loss functions
using two different forms for the stress tensor.

81

 

-'a)

#6)

mod)»;

9
Jaiaummd

   

.i»

”Ca (”Ca . EmZr)

 

 

 

.o 20 so
ENERGY (MeV/n)

Figure 18. The effect of the relaxing stress tensor
on the fusion regime.

82

 

ISO + 26Mg

 

 

 

 

q
)0» g; .
S /r—Jb
2.
r- q
.5“ 1

 

 

 

ENERGY (MeV/h)
l l l
| 2 3 4

 

Figure 19. A comparison of the experimental results
of Reference 36 with the predicted results of the simple
model described in Chapter VIII.

83

model for fusion in which friction was neglected. The
agreement with experiment is quite good. The low cross
sections predicted at the higher energies will increase
when the tangential friction is included. Our model,

of course, predicts no cross section for fusion at ener-
gies below the Coulomb barrier. Experiments do show

a small but finite cross section for energies slightly
below Vc' Experiments done with 10B + 14N have shown
measurable cross sections for fusion at energies down
to .25 MeV/nucleon below the Coulomb barrier.35 This
corresponds to an interaction radius of about twice the
mass radius. For these events the model we have de-
scribed is not adequate. Several other experiments have
also been analyzed, and agreement is quite good at the
low energies.30 At the higher energies the predicted
fusion cross sections are somewhat smaller than the mea-
sured values. This would seem to indicate that there
are other mechnisms by which energy is dissipated.

It is difficult to compare the angular distributions
predicted by our model directly with experimental data
but some general comparison can be made. Typical data
shows either a monotonically decreasing cross section
as the angle increases, or a side peak at angles near

to, or somewhat forward of, the grazing angle. From

the results presented in Figures 14, 15, and 16 one can

84

clearly see the side peak in the deflection function.

However, it is at an angle considerably forward of the

Coulomb grazing angle. For 40Ca + 40Ca the peak is at

about 10 degrees. For the reaction 84Kr + 84Kr the side

peak is even more forward at about 5 degrees.

CHAPTER X
A COMPARISON OF THE RESULTS

WITH THE OTHER MODELS

In comparing the results of the compressible fluid
drop model with the other calculations that have been
done, it should first be noted that there are some fun-
damental differences. One example of this is the exist-
ence of an angular momentum window in the fusion cross
section. This is a characteristic predicted by some
of the models and not by others. Another example of
a basic difference is the kind of scattering rainbow
predicted by the model. In the following discussion
we will consider each aspect of heavy ion collisions
and briefly describe what the various models have to
say about it.

First of all consider the energy loss. In the clas-
sical friction model of Gross and Kalinowski7 the col-
liding nuclei undergo a rapid energy loss as a result
of a large radial friction. Similarly, in the TDHF cal-
culations and in the surface excitation model, the energy
of the scattered particle decreases as the impact param-
eter decreases. Other than an understandable dip near

the fusion regime, this decrease is monotonic. In

85

86

Figure 20 a comparison is shown between our results and
a TDHF calculation done by another group.4 The energy
was chosen to give the same low 2 window size. The
results are comparable, however, the TDHF calculation
does not exhibit any elastic bouncing at low impact
parameters as is characteristic of our results.

The next aspect of heavy ion reactions we will
discuss is fusion. In this area the TDHF calculations
are in good agreement with our results. Using a TDHF

118Pd + 118Pd at an energy of

method, the reaction
1.35 MeV/nucleon has been studied by Cusson and co-
workers.3 The equations of motion were solved for
several values of the impact parameter. In this calcu-
lation it was found that for b less than .27 the system
remained fused. In our calculation we found fusion for
an impact parameter of less than .33. This number is
critically dependent on what value is taken for the
interaction radius and so we consider this to be in good
agreement. An example of a very large system is the
reaction 208Pb + 208Pb. The energy here is also only
slightly above the Coulomb barrier. This result illus-
trates the effect of the Coulomb repulsion on fusion.
Fusion does not occur even at zero impact parameter.

Figure 21 shows the relative separation of the two nuclei

as a function of time in a head-on collision. One curve

87

 

 

 

 

2 t 1 ‘
“E
§ X
0 x
. 0‘1 .
(3%) (avIID
>.
s " r
g ' ’ FUSION "
LU I) x ’0‘"
G
\V‘
a?“
I I I I I I I I I J I

 

Figure 20. Final state energies in the reaction
40Ca(40Ca,40Ca)40Ca. The Coulomb barrier in our model is
at .85 MeV/n. The incident energy in the TDHF study is
1.738 MeV/n. To keep the same window we use an energy of
2.1 MeV/n.

88

 

 

SEPARATION Um)

5 ' b=0
. E= l.9| MeWn

 

 

I I I I I I L

O ICC 200 300 400 500 600 700
TIME (tm/c)

 

Figure 21. The distance between centers of two Pb
nuclei as a function of time. The line at the end indicates
breakup. The solid line is our calculation and the x's are
from the TDHF calculations of Reference 4.

89

represents the TDHF result, the other is from the com—
pressible fluid drop model. At somewhat higher energies
another TDHF calculation was done for the reaction

40Ca + 40Ca. In this case the energy was 1.738 MeV/
nucleon, or about .82 MeV/nucleon above the Coulomb
barrier. These workers found that there was an angular
momentum window in which fusion occurred. This feature
is also present in our model at slightly higher energies.
In terms of the impact parameter they found that the
system remained fused if b was in the range from .17 to
.46. For impact parameters outside this window, a scat-
tering event takes place. This fusion window is also
present in the surface excitation model.

The results of a calculation done by Broglia and
others are shown in Figure 22. The cases presented in
this study represent energies in the range from 3 to
6 MeV/nucleon above the Coulomb barrier. The fusion
window is quite small and sits around b/2R = .5. Looking
more closely at the results for the reaction Ar + Ag we
find that at 5.24 MeV/nucleon the window is between
b/ZR = .43 and b/2R = .59. At the highest energy
6.98 MeV/nucleon the window lies between b/2R = .52 and
b/ZR = .59. This behavior is consistent with the

expected decrease in fusion cross section as the energy

 

 

 

 

 

 

 

 

 

90, A! 9 A9 9C9. Kr 9 BI
60 ~ 150 —
120 ‘ “\
0' ..... - \‘
20 __ \\. r’,—\.,~ 50 - “-.
.. \. ' ' . “~
‘60 iii“? ‘0 - (7M-
-20 .. A _H .-br.—._.-._.l-_._._.
1--..-. -1..- ”1.. ..__,. p 200 «00 600 (
EU, (MeV) E (MeV) “725.!“
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,-"‘"“‘
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Figure 22. Results of a calculation done using a
surface excitation model as described in Reference 11.

91

increases. For any impact parameter one can calculate
the distance of closest approach on the Coulomb trajec-
tory. If we do this for the largest impact parameter
in the fusion regime, we find that the distance of
closest approach ranges between .8 and .9 times the mass
radius. The penetration depth required increases as
the energy increases. It is also interesting to cal-
culate the radial velocity at the point of impact using
the lower impact parameter. In all cases this velocity
is between .065 c and .075 c. This suggests that the
radial velocity at the point of impact is an important
parameter in determining whether or not the system fuses.
In the friction model this fusion window is not present.
All trajectories with angular momentum less than some
critical value Lcr will lead to fusion. This is also
the case for the incompressible fluid model. This model
has been used by Nix and Sierk to calculate fusion cross
sections.8 Their results show a smaller cross section
than ours with a peak at about .5 MeV/nucleon above the
Coulomb barrier. It should be noted that in this calcu-
lation there was no dissipation.

The final topic we will consider is the angular
distribution. In our model this is dominated by a
nuclear rainbow which appears at an angle more forward

than the Coulomb grazing angle. This peak is made up

92

of reaction products that have suffered a considerable
energy loss. Both the friction model and the surface
excitation model exhibit a Coulomb rainbow. This
appears near the grazing angle and consists of nuclei
that have been scattered with very little loss of
energy. The TDHF calculations should be able to predict
angular distributions. However, because of the length
of the calculation, there is not enough information to
construct a deflection function. In Table 3 we present
a summary of the results of the various models that have

been considered.

93

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