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arXiv:1010.0840v1 [nucl-th] 5 Oct 2010
CLUSTER EXPANSION OF COLD
ALPHA-MATTER ENERGY
F. Carstoiu1, S ¸. Mi¸ sicu1, V. B˘ al˘ anic˘ a1and M. Lassaut2,
1National Institute for Nuclear Physics and Engineering,
P.O.Box MG-6, RO-077125 Bucharest-Magurele, Romania
2Institut de Physique Nucl´ eaire
IN2P3-CNRS, Universit´ e Paris-Sud 11
F-91406 Orsay Cedex, France
(Received October 6, 2010)
Abstract
In the cluster expansion framework of Bose liquids we calculate ana-
lytical expressions of the two-body, three-body and four-body diagrams
contributing to the g.s. energy of an infinite system of neutral alpha-
particles at zero-temperature, interacting via the strong nuclear forces
exclusively. This is analytically tractable by assuming a density depen-
dent two-body correlation function of Gaussian type. For the α − α po-
tential we adopt the phenomenological Ali-Bodmer interaction and semi-
microscopic potentials obtained from the Gogny force parametrizations.
We show that under such assumptions we achieve a rapid convergence in
the cluster expansion, the four-body contributions to the energy being
smaller than the two-body and three-body contributions by at least an
order of magnitude.
Key words: Equation of state, nuclear matter, supernova explosion, clus-
ter expansion, quantum liquids.
1INTRODUCTION
Understanding the properties of α matter has retained a lot of attention
in recent years. This situation is mainly due to the believe that this type of
hadronic matter occurs in astrophysical environment in deconfined form. In
the debris of a supernova explosion, a substantial fraction of hot and dense
matter resides in α particles and therefore the equation of state of matter
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at subnuclear densities is essential in simulating the supernova collapse and
explosions and is also important for the formation of the supernova neutrino
signal [1].
The aim of the present work is to investigate the equation of state of α
matter from the standpoint of the cluster expansion method of Bose liquids.
We consider a cold (T=0) system of α-particles interacting only by means
of the strong nuclear force. Similarly to the case of ordinary nuclear matter
(composed of protons and neutrons) the Coulomb interaction is switched-off.
The internal structure of the α clusters is accounted only in the determination
of the α − α potential, the single particle structure being incorporated in the
cluster densities that are folded with the effective nucleon-nucleon (NN) inter-
action. In continuation to the previous assumption, no Pauli blocking effects
are included. Naturally, since the constituents of the α particles are fermions,
one should expect that the Pauli principle is manifest when two or more α
clusters start to overlap. As revealed by the work of R¨ opke and collab. [2] one
should expect from the action of this principle a dissolution of the α cluster
in protons and neutrons above the so-called Mott density, which presumably
lays between a fifth and a third of the nuclear matter saturation density. It
is well known [3], that two-body correlation functions (TBCFN) obtained by
minimization of the energy functional truncated at the lowest order may lead
to an unphysical deep minimum. We adopt the cluster expansion method of a
Bose liquid [4] and use the simple Jastrow ansatz involving state-independent
two-body correlation functions . These are taken in a Gaussian form, without
overshooting near the healing distance,
f(r) = 1 − e−β2r2. (1)
The parameter β is determined from the normalization condition for the cor-
relation function [5]
?∞
This condition ensures that the mean square deviation of the correlation func-
tion from unity is a small quantity and has an exponential healing. Con-
sequently the dependence β = β(ρ) , where ρ = Nα/V is the density of
α-particles, reads,
β =√πρ2 −
For the α − α interaction, we adopt two types of Gaussian-like potentials
containing a short range soft repulsive part and a long range shallow attractive
part. The first one is the S-state Ali-Bodmer (AB) potential [6]:
4πρ
0
drr2(f2(r) − 1) = −1 . (2)
??
1
2√2
??1/3
. (3)
v(r) = VRe−µ2
Rr2− VAe−µ2
Ar2.(4)
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In this expression, VA= 130 MeV, VR= 475 MeV, µA=0.475 fm−1, µR=0.7
fm−1. This potential obtained by a fit of the low energy α − α phase shifts,
can be considered as an approximation to the supersymmetric partner of the
deep potential of Buck et al. [7]. The second type of potential is a sum of
three Gaussian and is derived from two recent parametrizations of the Gogny
effective N − N force [8]. Two explicit forms, labeled (D1) and (D1N), are
given in [9]. We have checked that these potentials satisfying the integrability
condition
0r|v(r)|dr < ∞ are in agreement with the Levinson theorem in
the sense given in [10]. In the absence of the Coulomb interaction, the α − α
interactions (AB) and (D1N) provide a weakly bound g.s.
contrast to the aforementioned interactions, the α − α (D1) interaction is
characterized by the absence of bound states.
In section 2 we apply the cluster expansion method to our problem and
express all terms of the development in a compact form. Our results are
discussed in section 3.
?∞
Jπ= 0+. In
2CLUSTER EXPANSION METHOD
Let the energy of a system of strongly interacting N-bosons in the Jackson-
Feenberg form [11],
E =1
2ρN
?
drg(r)v∗(r) ,(5)
where g(r) is the radial distribution function and v∗is the effective Jackson-
Feenberg potential,
v∗(r) = v(r) −
¯ h2
2mα∇2lnf(r) .(6)
In the case of α-matter the energy Eq. (5) is measured relative to the rest
energy of a free α-particle. Using Eqs. (1) and (4) we obtain
2cβ2?
?1 − er2β2?2
where c = ¯ h2/2mα. The main trick allowing the cluster expansion of the above
expression consists in expanding the radial distribution function in powers of
the small parameter ω =?drh(r). It uses the fact that the function h = f2−1
is of short-range. Accordingly, the cluster expansion of g(r) reads,
v∗(r) =
3 − er2β2?3 − 2r2β2??
+ VRe−r2µ2
R− VAe−r2µ2
A, (7)
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g(r12)=f2(r12)
?
1
2h(r13)h(r23)h(r14)h(r24)
1
2h(r13)h(r14)h(r23)h(r24)h(r34)
?
1 + ρ
?
?
?
dr3h(r13)h(r23)
+ρ2
dr3
dr4
h(r13)h(r24)h(r34) + 2h(r13)h(r24)h(r34)h(r14)
+
+
?
+ O(ρ3)
?
. (8)
A diagrammatic expansion of the radial distribution function is depicted in
Fig. 1. For a given n-body diagram there are field points (open circles) and
dummy points (filled circles). For each dummy point there is an integration
ρ?d? ri. A bond between open points involves a factor f2in the integrand. Any
ground state energy per α-particle is then expanded in powers of the density :
other bond implies a factor h = f2−1 in the integrand. Mutatis mutandis, the
E = E2+ E3+ E4+ ...(9)
Above Enstands for the contribution to the energy per particle arising from
n-body diagrams. In what follows we calculate these first three terms of the
expansion.
g(r )=
12
+++ 2
+
2
1
1
2
+ .?.?.?.
+
Figure 1: Diagrammatic expansion of the radial distribution function. There
are 4 independent diagrams contributing to the 4th order term: ring, diagonal,
opened and connected.
Partial contributions to the total energy are displayed using the α − α
Ali-Bodmer interaction (Eq.(4)).
2.1 Two-body diagram E2
Let us split E2only into kinetic and potential components :
?
E2V=1
2Ω
where Ω is the integration volume. Consider the general 6-dimensional integral
?
E2K=1
2
ρ
Ω
dr1dr2f2(r12)
?
−¯ h2
2mα∇2lnf(r12)
?
, (10)
ρ
?
dr1dr2f2(r12)v(r12) ,(11)
I2=1
2
ρ
Ω
dr1dr2p(r12) , (12)
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where r12= r1−r2and p a generic function. With the unitary transformation
(unit Jacobian),
r1= R +1
2s ,(13)
r2= R −1
2s , (14)
the integral I2(Eq.(12)) reads:
I2=1
2
ρ
Ω
?
dRds p(s) = 2πρ
?∞
0
ds s2p(s) . (15)
Note that integration over the c.m. variable R gives the integration volume
Ω. Therefore we have
?
−1
2mα
E2K
=−1
2ρ
?
?
ds f2(s)
¯ h2
2mα∇2lnf(s)
?
?
=
2ρds f2(s)¯ h2
f(s)∇2f(s) − (∇f(s))2?
, (16)
E2V=1
2ρ
?
ds f2(s)v(s) .(17)
Summing up these two contributions we obtain,
E2=1
2ρ
?
drf2(r)v∗(r) ,(18)
or in analytical form
E2=1
4π3/2ρ
?
3√2c
β
+ 2VRFR− 2VAFA
?
. (19)
Defining the auxiliary function
χij(β,µ) =
1
(iβ2+ jµ2)
3
2
, (20)
we then have
Fi= −χ0,1(β,µi) + 2χ1,1(β,µi) − χ2,1(β,µi) ,
where (i = A,R). The dependence of E2 on density is displayed in Fig.
2. We observe that already the component E2 has a shallow minimum at
a density almost two times the saturation density of normal nuclear matter.
This is in contrast to the result of ref. [9], where it has been shown that
TBCFN’s obtained from Pandharipande-Bethe equation lead to a collapse of
(21)
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-25
-20
-15
-10
-5
0
5
0 0.010.02 0.030.04 0.05 0.060.070.08 0.09 0.1
ρα(fm-3)
E2 (MeV)
Figure 2: Density dependence of the E2component.
E2component. This effect arises entirely from the density dependence of our
particular functional form of TBCFN and its derivatives.
2.2Three-body diagram E3
The diagram corresponding to the three-body energy is given on the left
panel of Fig. 3. By definition,
?
Ω
?
Ω
E3K=1
2
ρ2
??
dr1dr2dr3f2(r12)h(r23)h(r31)
?
−¯ h2
2mα∇2
1lnf(r12)
?
, (22)
E3V =1
2
ρ2
??
dr1dr2dr3f2(r12)h(r23)h(r31)v(r12) .(23)
Like previously for the E2component we consider the following generic integral
?
Ω
I3
=
1
2
ρ2
??
dr1dr2dr3p1(r12)p2(r23)p3(r31)
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12
3
R
RR
1
23
Figure 3: E3diagram and δ-diagram
=
1
2
?
ρ2
Ω
??
3?
1
dri
3?
1
dRip1(R1)p2(R2)p3(R3)
×δ(R1− r1+ r2)δ(R2− r2+ r3)δ(R3− r3+ r1) .
Integration of rivariables is straightforward and we obtain,
?
Eq.(25) is useful for both numerical and analytical integration since the angu-
lar dependence is isolated in a δ function as shown in the right panel of Fig.
3. Introducing the double-folding δ-integral,
?
the integral I3is reduced to,
?
Applying this technique to the E3term, we have,
?
2mα
(24)
I3=1
2ρ2
dR1dR2dR3p1(R1)p2(R2)p3(R3)δ(R1+ R2+ R3) . (25)
Vδ
p,q(R1) =dR2dR3p(R2)q(R3)δ(R1+ R2+ R3) , (26)
I3=1
2ρ2
dR1p1(R1)Vδ
p2,p3(R1) . (27)
E3K=1
2ρ2
?
dR1f2(R1)−¯ h2
??
f(R1)∇2f(R1) − (∇f(R1))2?
Vδ
h,h(R1)
(28)
E3V=1
2ρ2
?
dR1f2(R1)v(R1)Vδ
h,h(R1) .(29)
Consequently the total three-body energy contribution reads,
E3= E3K+ E3V=1
2ρ2
?
dR1v∗(R1)f2(R1)Vδ
h,h(R1) ,(30)
where we have defined according to the prescription (26)
?
Vδ
h,h(R1) =dR2dR3h(R2)h(R3)δ(R1+ R2+ R3) . (31)
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For our particular selection of the TBCFN,
Vδ
h,h(R) =π3/2
β3
?
1
8e−R2β2−4√3
9
e
−2R2β2
3
+
√2e
−R2β2
2
?
, (32)
and
E3=π3ρ2
β3
?c
β
?
−
27
32√2−
29
24√3+
84
25√5
?
+ VRGR− VAGA
?
, (33)
where for (i = A,R),
Gi
=
1
16[χ1,1(β,µi) − 2χ2,1(β,µi) + χ3,1(β,µi) + 32(χ1,2(β,µi) − 2χ3,2(β,µi)
χ5,2(β,µi) − χ2,3(β,µi) + 2χ5,3(β,µi) − χ8,3(β,µi))].+
(34)
-2
0
2
4
6
8
10
12
14
16
18
0 0.01 0.020.030.040.05 0.060.07 0.080.09 0.1
ρα(fm-3)
E3 (MeV)
Figure 4: Density dependence of the E3component.
The density dependence of the three-body energy is given in Fig. 4. We
notice that contrary to E2which is attractive, E3is only weakly attractive at
low density and becomes strongly repulsive with increasing density.
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1
2
3
4
A
1
2
3
4
B
1
2
3
4
C
1
2
3
4
D
Figure 5: (A) Ring, (B) diagonal, (C) opened and (D) connected diagrams
contributing to the E4component.
2.3 The four-body diagram E4
The diagrams contributing to the E4component are depicted in Fig. 5.
They are dubbed as ring, diagonal, opened and connected diagrams. Explicit
expressions of these components are,
E4R=1
2
ρ3
Ω
?
4?
1
driv∗(r12)f2(r12)h(r23)h(r34)h(r41) ,(35)
E4D= 21
2
ρ3
Ω
?
?
4?
4?
1
driv∗(r12)f2(r12)h(r23)h(r34)h(r41)h(r24) ,(36)
E4O=1
2
1
2
ρ3
Ω
1
driv∗(r12)f2(r12)h(r23)h(r41)h(r24)h(r13) ,(37)
E4C=1
2
1
2
ρ3
Ω
?
4?
1
driv∗(r12)f2(r12)h(r23)h(r34)h(r41)h(r24)h(r13) .(38)
2.3.1 Calculation of E4Rring diagram
It is easy to see that the ring diagram contribution to the energy is given
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-12
-10
-8
-6
-4
-2
0
2
00.01 0.020.03 0.040.050.060.070.08 0.090.1
ρα(fm-3)
E4R (MeV)
Figure 6: Density dependence of the E4Rcomponent for the AB potential.
by,
E4R=1
2ρ3
?
4?
1
dRif2(R1)v∗(R1)h(R1)h(R2)h(R3)h(R4)δ(R1+R2+R3+R4)
(39)
or equivalently,
E4R
=
1
2ρ3
?
?
2?
3?
1
dRif2(R1)v∗(R1)h(R2)Vδ
h,h(R1+ R2)
=
1
2ρ3
1
dRif2(R1)v∗(R1)h(R2)Vδ
h,h(R3)δ(R3− R1− R2),(40)
in terms of the function Vδ
let us introduce,
h,hdefined in Eq.(31). In the hierarchy of δ kernel
Vδδ
p,q,r(R1) =
?
dR2dR3p(R2)Vδ
q,r(R3)δ(R3− R1− R2) .(41)
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The final integral is simply,
E4R=1
2ρ3
?
dR1v∗(R1)f2(R1)Vδδ
h,h,h(R1) . (42)
Using the expression for Vδ
Vδδ
h,h,h,
h,hgiven in Eq.(32) we obtain the expression of
Vδδ
h,h,h(R)=
π3
β6
?
1
24√3e−2
−2R2β2
5
3R2β2−
8
3√3e
3
8√2e−R2β2
−R2β2
3
2
+
12
5√5e
−
?
. (43)
The corresponding expression for E4Ris:
E4R=π9/2ρ3
β6
?c
240
49√7
β
?3
?
4+
3
128√2+
7
4√3+
−
12
25√5−
+ HRVR− HAVA
?
.(44)
In the equation (44) the Hi’s are defined by
Hi=
1
16(−6χ1,2(β,µi) + 12χ3,2(β,µi) − 6χ5,2(β,µi) − 64χ1,3(β,µi)+
χ2,3(β,µi) + 128χ4,3(β,µi) − 2χ5,3(β,µi) − 64χ7,3(β,µi)
+χ8,3(β,µi) + 96χ2,5(β,µi) − 192χ7,5(β,µi) + 96χ12,5(β,µi)) ,
where (i = A,R).
From the inspection of Fig.6 we infer that the ring contribution to the
four-body energy has the same behavior as E3save for a factor -1.
2.3.2 Calculation of E4Ddiagonal diagram
In order to isolate the angular variables, we proceed as follows,
E4D
=21
2
ρ3
Ω
?
?
4?
5?
1
driv∗(r12)f2(r12)h(r23)h(r34)h(r41)h(r24)
E4D
= 21
2
ρ3
Ω
1
dRi
4?
1
driv∗(R1)f2(R1)h(R2)h(R3)h(R4)h(R5)
×δ(R1− r1+ r2)δ(R2− r2+ r3)δ(R3− r3+ r4)
×δ(R4− r4+ r1)δ(R5− r2+ r4) . (45)
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Performing the same manipulations as above we obtain,
E4D= 2 ·1
2ρ3
?
5?
1
dRif2(R1)v∗(R1)h(R2)h(R3)h(R4)h(R5)
δ(R5+ R1+ R4)δ(R5− R2− R3) . (46)
We observe that variables Ri,i = 1,2,3,4 are decoupled. It can be seen
easily that:
?
E4D= ρ3
dR1dR4dR5f2(R1)v∗(R1)h(R4)h(R5)Vδ
h,h(R5)δ(R5+ R1+ R4) .
(47)
Finally we have,
E4D=π9/2ρ3
288β6
?
c
β
?
27
8−26624√3 − 7344
121√11+580608
49√7
+2432√6 − 10368
25√5
?
−
?
945
2√2+ 74√3 −
15552
169√13−442368
361√19
+ VRDR− VADA
,(48)
and the Di’s are given by,
Di
=−9[χ1,1(β,µi) + χ3,1(β,µi) − 2χ2,1(β,µi) − 4(χ6,5(β,µi) + χ16,5(β,µi))
+8(χ3,4(β,µi) + χ11,4(β,µi) + χ11,5(β,µi))
−16(χ2,3(β,µi) + χ8,3(β,µi) + χ7,4(β,µi))
+25(χ5,3(β,µi) +√2(χ22,7(β,µi) +√2χ8,7(β,µi)))
−26√2χ15,7(β,µi)
−27(χ10,9(β,µi) + χ28,9(β,µi))
+28(χ6,7(β,µi) + χ5,7(β,µi) + χ19,9(β,µi) + χ19,7(β,µi) + χ20,7(β,µi))
−28(χ8,11(β,µi) + χ10,11(β,µi) + χ30,11(β,µi) + χ32,11(β,µi))
+29(χ5,8(β,µi) + χ21,8(β,µi) + χ19,11(β,µi) + χ21,11(β,µi))
−29(χ3,5(β,µi) + χ13,5(β,µi) + χ12,7(β,µi) + χ13,7(β,µi))
+210(χ8,5(β,µi) − χ13,8(β,µi))
where (i = A,R).
?
, (49)
2.3.3Calculation of E4Oopen diagram
Performing the usual manipulations on the E4Oterm we obtain,
E4O
=
1
4ρ3
?
5
?
1
dRif2(R1)v∗(R1)h(R2)h(R3)h(R4)h(R5)
12
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-2
0
2
4
6
8
10
00.010.020.030.040.050.060.070.080.090.1
ρα(fm-3)
E4D (MeV)
Figure 7: Density dependence of the E4Dcomponent.
δ(R1+ R3+ R4)δ(R5− R1− R2) .(50)
The latter equation is simply,
E4O=1
4ρ3
?
dR1f2(R1)v∗(R1)[Vδ
h,h(R1)]2, (51)
and after substitution of eq.(1),
E4O=π9/2ρ3
β6
?
c
β
?
−
48
19√19+
1024+191√3
336
169√13+63 − 64√3
?
294√7
+
20√6 − 27
150√5
−
3
22√11−
177
256√2+
3
576
+ PRVR− PAVA
?
(52)
where,
Pi
=
1
2χ1,1(β,µi) −255
1
256χ4,1(β,µi) +1
256χ2,1(β,µi) +63
4χ3,2(β,µi) −1
128χ3,1(β,µi)
+
2χ5,2(β,µi)
13
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+1
4χ7,2(β,µi) +
8
3√3χ7,3(β,µi) +1
−1
−8χ19,6(β,µi)
4
3√3χ4,3(β,µi) −1
4χ5,3(β,µi)
4
3√3χ10,3(β,µi)
−
2χ8,3(β,µi) +
4χ11,3(β,µi) − 8χ7,6(β,µi) + 16χ13,6(β,µi)
-0.5
0
0.5
1
1.5
2
2.5
0 0.010.020.030.04 0.05 0.060.07 0.08 0.090.1
ρα(fm-3)
E4O (MeV)
Figure 8: Density dependence of the E4Ocomponent.
The density dependence of the E4Ocomponent is displayed in Fig. 8.
2.3.4Calculation of E4C connected diagram
In this subsection we estimate,
?
1
E4C=1
4
ρ3
Ω
4?
driv∗(r12)f2(r12)h(r23)h(r34)h(r41)h(r24)h(r13) .(53)
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Using the same technique as above we introduce δ kernels,
E4C=1
4ρ3
?
6?
1
dRiv∗(R1)f2(R1)h(R2)h(R3)h(R4)h(R5)h(R6)
δ(R6+ R3+ R4)δ(R5− R2− R3)δ(R6− R1− R2) . (54)
The identities
(2π)3δ (R6+ R3+ R4) =
?
?
?
dq1exp(q1.(R6+ R3+ R4)) , (55)
(2π)3δ (R5− R2− R3) =dq2exp(q2.(R5− R2− R3)), (56)
(2π)3δ (R6− R1− R2) =
are introduced in Eq.(54). The latter equation is then expressed in terms of
the Fourier transform of h, labelled?h and that of v∗f2, labelled G. More
1
2048π9ρ3
i=1
For the sake of simplicity use is made of the scaling qj?→ 2√2βqjin Eq.(58)
in order to eliminate the β dependence in?h. Introducing,
h∗(q) = −2e−2q2+
such that?h(2√2βq) ≡ (√π/b)3h∗(q), the equation (58) becomes,
E4C=
β6√π3ρ3
i=1
dq3exp(q3.(R6− R1− R2)) ,(57)
precisely,
E4C=
?
3?
dqi?h(q1− q2)?h(q1+ q3)?h(q2+ q3)?h(q1)?h(q2)G(q3) .
(58)
1
2√2e−q2,(59)
4√2
?
3
?
dqih∗(q1− q2)h∗(q1+ q3)h∗(q2+ q3)
h∗(q1)h∗(q2)G(2√2βq3) . (60)
The auxiliary integral
J(q3) =
?
2?
i=1
dqih∗(q1− q2)h∗(q1+ q3)h∗(q2+ q3)h∗(q1)h∗(q2) , (61)
entering the definition of E4c, has been performed by using the Cartesian
coordinates. We obtain after calculations J(q3) = π3Q(q3) with,
−9√2 + 2√3
72
Q(q)=
exp(−2 q2) +
8
13√13
exp
?
−22
13q2
?
15
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