User Note for Relativistic EOS Table(EOS2: 2010-version, with only nucleons)H. Shena1 , H. Tokib2 , K. Oyamatsuc3 , and K. Sumiyoshid4ab

Department of Physics, Nankai University, Tianjin 300071, China

Research Center for Nuclear Physics (RCNP), Osaka University, Ibaraki, Osaka 567-0047,

Japan
c

Department of Media Theories and Production, Aichi Shukutoku University, Nagakute-cho,

Aichi 480-1197, JapandNumazu College of Technology, Ooka 3600, Numazu, Shizuoka 410-8501, Japan

Abstract
This is a detailed note for users of the relativistic equation of state (EOS) of nuclear
matter.

Contents• 1 Introduction• 2 Relativistic mean-field theory• 3 Ideal-gas approximation• 4 Thomas-Fermi approximation• 5 Calculation procedure in detail• 6 Resulting EOS table• 7 Suggestions for using the EOS table1

e-mail: shennankai@gmail.com
e-mail: toki@rcnp.osaka-u.ac.jp
3e-mail: oyak@asu.aasa.ac.jp
4e-mail: sumi@numazu-ct.ac.jp
2

1

1

Introduction

We write this document as a guide for the use of the relativistic equation of state (EOS) table.
This EOS table is constructed in the following ranges:• temperature T [MeV]: −1.0 ≤ log10 (T ) ≤ 2.6; mesh of log10 (T ) = 0.04
• proton fraction Yp : 0.01 ≤ Yp ≤ 0.65; mesh of Yp = 0.01
• baryon mass density ρB [g/cm3 ]: 5.1 ≤ log10 (ρB ) ≤ 16.0; mesh of log10 (ρB ) = 0.1
We also add the results for the zero temperature case (T = 0) and the pure neutron mattercase (Yp = 0).
We have worked out consistent calculations for uniform matter and non-uniform matter inthe relativistic mean-field (RMF) framework [1, 2]. We use the Thomas-Fermi approximation
to describe inhomogeneous nuclear matter, which can be modeled as a mixture of free neutrons,
free protons, alpha-particles, and a single species of heavy nuclei. For extremely low density andfinite temperature, we approximate the nuclear matter as a classical ideal gas of free neutrons,
free protons, and alpha-particles. Antiparticles have some contribution when temperature isvery high. The thermodynamically favorable state is the one that minimizes the free energy
density in this model, and we determine the most favorable state of nuclear matter at eachtemperature, proton fraction, and baryon mass density.
The leptons can be considered as uniform non-interacting relativistic particles, which are
relatively easy to deal with. Hence, we provide the baryon EOS without the lepton contribution
in this table. Users are supposed to add the lepton contribution to the baryon EOS.
This document is arranged as follows. In Sec. 2, we describe the RMF theory. In Sec. 3,
we introduce the ideal-gas approximation to be used at low density. In Sec. 4, we describe theThomas-Fermi approximation, which is used for calculating non-uniform matter. In Sec. 5, we
make a detailed description of how to work out the EOS table. In Sec. 6, we list the definitionsof physical quantities in the EOS table. Sec. 7 is devoted to suggestions and discussions forusing the EOS table. Users who are not interested in the framework and calculating processes
can only read Sec. 6 and Sec. 7 for the purpose to use the EOS table correctly.

2

2

Relativistic mean-field theory

We adopt the RMF theory with nonlinear σ and ω terms to describe homogeneous nuclear
matter [3]. We start with the Lagrangian given by
LRMF = ψ̄ [iγµ ∂ µ − M − gσ σ − gω γµ ω µ − gρ γµ τa ρaµ ] ψ11
11+ ∂µ σ∂ µ σ − m2σ σ 2 − g2 σ 3 − g3 σ 4
2
2
3
4
1
1
1
− Wµν W µν + m2ω ωµ ω µ + c3 (ωµ ω µ )24
2
4
1 a aµν 1 2 a aµ
− Rµν R + mρ ρµ ρ .
4
2

(1)

Here, ψ denotes an SU(2) baryon field of mass M (proton and neutron), and σ, ω µ , and ρaµare σ, ω, and ρ meson fields with masses mσ , mω , and mρ , respectively. W µν and Raµν are theantisymmetric field tensors for ω µ and ρaµ , which can be written as
W µν = ∂ µ ω ν − ∂ ν ω µ ,(2)Raµν = ∂ µ ρaν − ∂ ν ρaµ + gρ ²abc ρbµ ρcν .

(3)

In the Lagrangian, the constants gσ , gω , and gρ are the coupling constants for the interactions
between mesons and nucleons, the coefficients g2 and g3 are the self-coupling constants forthe σ meson field, and c3 is the self-coupling constant for the ω meson field. It is known
that the inclusion of the nonlinear σ terms is essential to reproduce the properties of nuclei
quantitatively and provide a reasonable value for the incompressibility, while the nonlinear ωterm is added in order to reproduce the density dependence of the vector part of the nucleon
self-energy obtained in the RBHF theory. The Lagrangian contains the meson masses, thecoupling constants, and the self-coupling constants as parameters. We adopt the parameter
set TM1 listed in Table 1, which was determined in Ref. [3]. With the TM1 parameter set,
the symmetry energy is 36.9 MeV and the incompressibility is 281 MeV. The RMF theory
with the TM1 parameter set provides excellent properties of heavy nuclei including unstable
nuclei [3], and it is also shown to agree satisfactorily with experimental data in studies of nuclei
with deformed configurations [4]. The properties of the EOS at high density with the TM1parameter set were discussed in detail in Ref. [5], and the neutron star properties were studied
using this EOS in Refs. [2, 5].
3

Table 1: The parameter set TM1 for the RMF LagrangianParameter

TM1

M [MeV]938.0

mσ [MeV]

511.19777mω [MeV]783.0mρ [MeV]

770.0

gσ

10.02892gω

12.61394gρ

4.63219

g2 [fm−1 ]-7.23247g30.61833c3

71.30747

Starting with the Lagrangian, we derive a set of Euler-Lagrange equations. The Diracequation for the nucleon field is given by(iγµ ∂ µ − M − gσ σ − gω γµ ω µ − gρ γµ τa ρaµ ) ψ = 0,

(4)

and the Klein-Gordon equations for the meson fields are given by∂ν ∂ ν σ + m2σ σ = −gσ ψ̄ψ − g2 σ 2 − g3 σ 3 ,(5)∂ν W νµ + m2ω ω µ = gω ψ̄γ µ ψ − c3 (ων ω ν ) ω µ ,(6)∂ν Raνµ + m2ρ ρaµ = gρ ψ̄τa γ µ ψ + gρ ²abc ρbν Rcνµ .

(7)These equations are coupled nonlinear quantum field equations, which are very difficult to solve
exactly. We adopt the relativistic mean-field approximation as described in Ref. [6], in which
the meson fields are treated as classical fields, and the field operators σ, ω µ , and ρaµ are replacedby their expectation values hσi, hω µ i, and hρaµ i. In the present study, we first consider static
infinite matter so that we obtain simplified equations, where the derivative terms in the KleinGordon equations vanish automatically due to the translational invariance of infinite matter.
The spatial components of the vector meson fields vanish due to the rotational symmetry. For
4the isovector-vector meson field ρaµ , only the third isospin component has a non-vanishing value
because of the charge conservation. Hence, the equations for the meson fields are reduced to
´
gσ
1 ³ 2
3
,
h
ψ̄ψi
−
g
σ
+
g
σ
2
3
0
0
m2σ
m2σ
gω
1
≡ hω 0 i = 2 hψ̄γ 0 ψi − 2 c3 ω03 ,
mω
mω
g
ρ
≡ hρ30 i = 2 hψ̄τ3 γ 0 ψi.
mρ

σ0 ≡ hσi = −

(8)

ω0

(9)

ρ0

(10)

The Dirac equation then becomes³´−iαk ∇k + βM ∗ + gω ω0 + gρ τ3 ρ0 ψis = εis ψis ,(11)

where the index i denotes the isospin degree of freedom (proton and neutron), s denotes theindex of eigenstates of nucleon, M ∗ ≡ M + gσ σ0 is the effective nucleon mass, and εis is the
single-particle energy.
Nucleons occupy single-particle orbits with the occupation probability fis . At zero temperature, fis = 1 under the Fermi surface, while fis = 0 above the Fermi surface. For finite
temperature, the occupation probability is given by the Fermi-Dirac distribution,
1
1
·µq
¶
¸,
=
1 + exp [(εis − µi ) /T ]1 + exp
k 2 + M ∗ 2 − ν /T

(12)

1
1
·µq
¶
¸.
=
1 + exp [(−εīs + µi ) /T ]22∗
1 + exp
k + M + ν /T

(13)

fis =

i

fīs =i

Here the indices i and ī denote nucleons and antinucleons, and εis and εīs are the single-particle
energy of nucleons and antinucleons, respectively. The relation between the chemical potentialµi and the kinetic part of the chemical potential νi is give byµi = νi + gω ω0 + gρ τ3 ρ0 .

(14)

The chemical potential µi is related to the number density of nucleon ni as
γ Z∞
ni = 2
dk k 2 (fik − fīk ) ,
2π 0

(15)

where γ is the degeneracy factor for the spin degree of freedom (γ = 2). The quantum number
s is replaced by the momentum k when we do the integration in the momentum space instead5

of summing over the eigenstates. The equations of the meson fields can be written as´
gσ X γ Z ∞
M∗
1 ³ 2
2
3
q
dk
k
g
σ
(f
+
f
+
g
σ
)
−
ik
2 0
3 0 ,
īk
2
2
m2σ i 2π 2 0
m
2
∗
σ
k +M
gω
c3 3
=
(np + nn ) − 2 ω0 ,
m2ω
mω
gρ
=
(np − nn ) .
m2ρ

σ0 = −

(16)

ω0

(17)

ρ0

(18)

Here, np and nn are the proton and neutron number densities as defined in Eq. (15), and wedenote nB = np + nn as the baryon number density. We solve these equations self-consistently.
The thermodynamical quantities are described in Ref. [6], and we simply write the expressions
here. The energy density of nuclear matter is given by² =X γ Z ∞q1
1
1
k 2 + M ∗ 2 (fik + fīk ) + m2σ σ02 + g2 σ03 + g3 σ04
2
3
4
0
i
1
1
+gω ω0 (np + nn ) − m2ω ω02 − c3 ω04
2
4
1 2 2
+gρ ρ0 (np − nn ) − mρ ρ0 ,
2
2π 2

dk k

2

(19)

the pressure of nuclear matter is given byp =X γ Z ∞
i

6π 2

0

k2

1
1
1
(fik + fīk ) − m2σ σ02 − g2 σ03 − g3 σ04
2
3
4
k2 + M ∗2

dk k 2 q

1
1
1+ m2ω ω02 + c3 ω04 + m2ρ ρ20 ,
2
4
2

(20)

and the entropy density is calculated bys =X γ Z ∞i

2π 2

0dk k 2 [−fik ln fik − (1 − fik ) ln (1 − fik )−fīk ln fīk − (1 − fīk ) ln (1 − fīk )] .

3

(21)

Ideal-gas approximation

The interaction between nucleons is negligible at low density, so that we can treat nucleons
as non-interacting Boltzmann particles at low density. We use the ideal-gas approximation
to describe the mixed gas of neutrons, protons, and alpha-particles at low density and finite
temperature. We note that the connection between the RMF results and the ideal-gas approximation is smooth.
6

For Boltzmann particles with spin 1/2, mass M , and number density ni (i is n or p), we
start with the partition function given byhi2

Zi = (nQ )ni /2 /(ni /2)! ,where we have used the abbreviation nQ =

³

MT
2π

´3/2

(22)

. The factor (ni /2)! comes from avoiding the

double counting for ni /2 indistinguishable particles of spin up or spin down. We can calculate
the free energy density from the partition function asfi = −T ln Zi = −T ni [ln(2nQ /ni ) + 1] .

(23)

The entropy density is given byÃ∂fi
si = −∂T!

·¸= ni ln(2nQ /ni ) +
ni

5
,
2

(24)

the internal energy density can be written as
3²i = fi + T si = T ni ,2(25)

the chemical potential is obtained through the relationÃµi =and the pressure is given by

∂fi
∂ni
"

pi =

!

= −T ln(2nQ /ni ),

(26)

T

#

∂ (fi /ni )= T ni .
n2i
∂ni
T

(27)

The partition function of alpha-particles can be written asZα = [8nQ exp (Bα /T )]nα /nα !,(28)

where nα is the alpha-particle number density, and Bα = 28.3 MeV is the binding energy ofan alpha-particle taken from Ref. [7]. The thermodynamical quantities of alpha-particles are
given byfα = −T nα [ln(8nQ /nα ) + 1] − nα Bα ,

(29)

¸

·

sα = nα ln(8nQ /nα ) +

5
,
2

3
T nα − nα Bα ,2
= −T ln(8nQ /nα ) − Bα ,

(30)

²α =

(31)

µα

(32)

pα = T n α .

(33)7For a mixed gas with the neutron number density nn , the proton number density np , andthe alpha-particle number density nα , the free energy density is given by
f = fn + fp + fα .

(34)

We have to take into account the volume of the alpha-particle, otherwise the alpha-particle
fraction would be some big number at high densities, where the alpha-particles should actuallydisappear. When we take into account the volume excluded by the alpha-particles, the freeenergy densities in Eq. (34) are given by
fn = −(1 − u) T ñn [ln(2nQ /ñn ) + 1] ,(35)

fp = −(1 − u) T ñp [ln(2nQ /ñp ) + 1] ,(36)

fα = −(1 − u) {T ñα [ln(8nQ /ñα ) + 1] − ñα Bα } ,(37)where u = nα vα is the fraction of space occupied by alpha-particles with the effective volume ofan alpha-particle vα = 24 fm−3 taken from Ref. [7]. We denote the effective number densities ofneutrons, protons, and alpha-particles as ñn = nn /(1−u), ñp = np /(1−u), and ñα = nα /(1−u).
The inclusion of the volume excluded by the alpha-particles has negligible effect in the low
density region, while it is necessary for the calculation at high density. Actually, we just use
Eqs. (35) and (36) for the case of ñn + ñp < 10−5 fm−3 , while the RMF theory is used to describe
nucleons when ñn + ñp > 10−5 fm−3 . For alpha-particles, we use Eq. (37) to calculate the freeenergy in the whole density range. It is considered as a reasonable approximation because thenumber density of alpha-particles is not so large even though the total baryon density is high,and the alpha-particle fraction tends to vanish in the high density limit.
For very high temperature, we need to consider the contribution from antiparticles. The
alpha-particle fraction is very small at high temperature, so we can treat the matter as uniform
matter of nucleons and antinucleons. We have included the freedom of antiparticles in the RMF
theory, but the RMF code meets some difficulty at extremely low density, so we need to takesome approximation in order to describe the range of low density and high temperature. We
take the nonrelativistic approximation of the RMF theory at low density, where the effectiveq∗nucleon mass M ≈ M , < mesons >≈ 0, and82k. Then the occupation
k 2 + M ∗ 2 ≈ M + 2M

probability of the Fermi-Dirac distribution becomes
³

´

³

´

fis = ACi exp −k 2 /2M T ,fīs =(38)

A
exp −k 2 /2M T ,
Ci

(39)

where A = exp (−M/T ), and Ci = exp (µi /T ) with µi being the chemical potential of protons(i = p) or neutrons (i = n), which is related to the number density ni asµ1
ni = 2A Ci −
Ci

¶µ

MT
2π

¶3/2

.

(40)

The thermodynamical quantities can be approximated as
µ

²i
pi
si

¶µ

¶

µ

¶

1
M T 3/2 3
T ,= 2A Ci +
Ci
2π
2
¶
µ
¶µ
1
M T 3/2
T,= 2A Ci +
Ci
2π
· µ
¶
¶
µ ¶¸ µ5
1
1
M T 3/2
A= 2A
Ci +− Ci ln (ACi ) −
.
ln
2
Ci
Ci
Ci
2π

(41)
(42)
(43)

The contribution from antiparticles is negligible when Ci À 1/Ci , where the above expressions
agree with the ideal-gas approximation.

4

Thomas-Fermi approximation

In the range, T < 14 MeV and ρB < 1014.2 g/cm3 , where heavy nuclei may be formed in orderto lower the free energy, we perform the Thomas-Fermi calculation based on the work doneby Oyamatsu [8]. In this case, the non-uniform matter can be modeled as a mixture of freeneutrons, free protons, alpha-particles, and a single species of heavy nuclei, while the leptons
can be treated as uniform non-interacting particles separately. For the system with fixed proton
fraction, the leptons play no role in the minimization of the free energy. Hence we mainly pay
attention to baryon contribution in this calculation. Hereafter, we will not mention the leptons
frequently, while we should keep in mind that there exists a uniform lepton gas everywhere.
We assume that each heavy spherical nucleus is located in the center of a charge-neutral
cell consisting of a vapor of neutrons, protons, and alpha-particles. The nuclei form a bodycentered-cubic (BCC) lattice to minimize the Coulomb lattice energy. It is useful to introduce
9the Wigner-Seitz cell to simplify the energy of a unit cell. The Wigner-Seitz cell is a sphere
whose volume is the same as the unit cell in the BCC lattice. We define the lattice constant aas the cube root of the cell volume. Then, we have
Vcell = a3 = NB /nB ,(44)

where NB and nB are the baryon number per cell and the average baryon number density,
respectively. We define the baryon mass density as ρB = mu nB with mu being the atomic mass
unit. We calculate the Coulomb energy using this Wigner-Seitz approximation and add anenergy correction for the BCC lattice [8]. This energy correction is negligible unless the nuclear
size is comparable to the cell size.
We assume the nucleon distribution function ni (r) (i = n for neutron, i = p for proton) in
the Wigner-Seitz cell asni (r) = ³´·³ ´ti ¸3


out
 nin
1− r
+ nouti − nii ,0 ≤ r ≤ Ri , nout ,Ri ≤ r ≤ Rcell ,

Ri

i

(45)

where r represents the distance from the center of the nucleus and Rcell is the radius of the
Wigner-Seitz cell defined by the relation
Vcell ≡

4π 3R .
3 cell(46)

out
are the densities at r = 0 and r ≥ Ri , respectively. The
The density parameters nin
i and niparameters Ri and ti determine the boundary and the relative surface thickness of the heavy
nucleus.
For the distribution function of alpha-particle nα (r), which should decrease as r approaches
the center of the heavy nucleus, we assumenα (r) =·³ ´tp ¸3
 −nout+ nout1− r
α ,α0 ≤ r ≤ Rp , nout ,Rp ≤ r ≤ Rcell ,

Rp

α

(47)

which could give nα (r = 0) = 0 and nα (r > Rp ) = nout
α . Here we use the same parameters Rp
and tp for both proton and alpha-particle distribution functions in order to avoid the presenceof too many parameters in the minimization procedure. The parameters Rn and tn may besomewhat different from Rp and tp due to the additional neutrons forming a neutron skin in the10

surface region. For a system with fixed temperature T , proton fraction Yp , and baryon massoutdensity ρB , there are eight independent parameters among the ten variables, a, nin
n , n n , Rn ,outout
tn , ninp , np , Rp , tp , and nα . The thermodynamically favorable state is the one that minimizes

the free energy density with respect to these eight independent parameters.
In this model the free energy density contributed from baryons is given by
f = Fcell / a3 = ( Ecell − T Scell ) / a3 ,(48)

where the free energy per cell Fcell can be written as
Fcell = (Ebulk + Es + EC ) − T Scell = Fbulk + Es + EC .

(49)

The bulk energy Ebulk , entropy Scell , and bulk free energy Fbulk are calculated byZ

Ebulk =

Zcell

Scell =

Zcell

Fbulk =cell

² ( nn (r) , np (r) , nα (r) ) d3 r,(50)s ( nn (r) , np (r) , nα (r) ) d3 r,(51)

f ( nn (r) , np (r) , nα (r) ) d3 r.

(52)Here ² ( nn (r) , np (r) , nα (r) ), s ( nn (r) , np (r) , nα (r) ), and f ( nn (r) , np (r) , nα (r) ) are thelocal energy density, entropy density, and free energy density at the radius r, where the system
can be considered as a mixed uniform matter of neutrons, protons, and alpha-particles. Notethat these local densities at each radius are calculated by treating neutrons and protons in the
RMF theory as described in Sec. 2 for the case of nn (r) + np (r) > 10−5 fm−3 , while we can use
the method described in Sec. 3 for low density case of nn (r) + np (r) < 10−5 fm−3 .
As for the surface energy term Es due to the inhomogeneity of the nucleon distribution, we
take the simple form

Z

Es =

cell

F0 | ∇ ( nn (r) + np (r) ) |2 d3 r.

(53)

The parameter F0 = 70 MeV · fm5 is determined by performing the Thomas-Fermi calculations
for finite nuclei so as to reproduce the gross properties of the nuclear mass, charge radii, andthe beta stability line as described in the Appendix of Ref. [8].
The Coulomb energy per cell EC is calculated using the Wigner-Seitz approximation withan added correction term for the BCC lattice [8]EC =1Z
e [np (r) + 2nα (r) − ne ] φ(r)d3 r + 4EC ,2 cell11(54)

where φ(r) represents the electrostatic potential calculated in the Wigner-Seitz approximation,
ne is the electron number density of a uniform electron gas (ne = Yp nB ), and 4EC is thecorrection term for the BCC lattice, which can be approximated as
4EC = CBCC

(Znon e)2
.
a

(55)

Here a is the lattice constant as defined in Eq. (44), the coefficient CBCC = 0.006562 is takenfrom Ref. [8], and Znon is the non-uniform part of the charge number per cell given by
Znon =

Z Rp
0



Ã

r
out 
out
(nin
p − np − 2nα ) 1 −Rp!tp 3 4πr 2 dr.

(56)

Because of the long-range nature of the Coulomb interaction, the Coulomb energy will dependon the lattice type. This dependence was discussed in more detail in Ref. [8]. The systemprefers the BCC lattice because the BCC lattice gives the lowest Coulomb energy.

5Calculation procedure in detail

We calculate the EOS table in the range mentioned in Sec. 1. For each T , Yp , and ρB , weperform the minimization of the free energy for non-uniform matter using the Thomas-Fermi
method described in Sec. 4. We also do the minimization for uniform matter with respect
to converting two protons and two neutrons into an alpha-particle, in which there is only one
independent parameter. Here, the phase of heavy nuclei formed together with free nucleons andalpha-particles is referred to as non-uniform matter, while the phase of nucleons mixed withalpha-particles without heavy nuclei is referred to as uniform matter. By comparing the freeenergy of non-uniform matter with that of uniform matter, we determine the most favorable
state of nuclear matter at this T , Yp , and ρB . The density of the phase transition between
uniform matter and non-uniform matter depends on both T and Yp . The non-uniform matter
phase can exist only in the low temperature range (T < 14 MeV).
We construct the EOS table in three dimensions (T , Yp , and ρB ). We first fix T , and thenfix Yp and ρB . We finish the following steps in order to work out an EOS table with fixed T
where the non-uniform matter phase exists.
• 1 calculate input table in the RMF theory for the use of minimization12

• 2 do minimization for uniform matter at low density• 3 do minimization for non-uniform matter at medium density• 4 calculate the RMF result for uniform matter without alpha-particles at high density• 5 determine the phase transition and connect the free energy• 6 calculate the pressure and chemical potentials from the free energy• 7 construct the finial EOS table at this T
Now, we explain each step in more detail.
• 1 The calculation of the minimization procedure needs the free energy per baryon as afunction of nn and np as input table, and this input table should be big enough in order to getgood numerical accuracy in doing linear approximation. We calculate this input table in the
RMF theory which can describe the homogeneous matter of protons and neutrons. This input
table covers the following rangep
Yp = nnn+npfrom 0.0

to 0.5mesh = 0.0005

point = 1001

nB = nn + np (fm−3 ) from 0.000010 to 0.000100 mesh = 0.000001 point = 91
from 0.000110 to 0.001000 mesh = 0.000010 point = 90
from 0.001100 to 0.060000 mesh = 0.000100 point = 590
from 0.061000 to 0.160000 mesh = 0.001000 point = 100
The density meshes are determined by the requirement to get good accuracy for the minimization results. In this input table, we list the quantities of the energy per baryon E and theentropy per baryon S. The free energy per baryon can be obtained by F = E − T S. We notethat the data at nn + np > 0.16 fm−3 is not necessary for the minimization procedure, while theideal-gas approximation is used to describe neutrons and protons when nn + np < 10−5 fm−3 .
• 2 At low density and finite temperature, the thermodynamically favorable state might bea uniform matter which is a mixed gas of neutrons, protons, and alpha-particles. For each Yp
and ρB , we do minimization in order to find out the alpha-particle fraction having the minimum
free energy. Since nB = nn + np + 4nα and Yp = (np + 2nα )/nB are fixed in this case, there is13

only one independent parameter, the alpha-particle fraction Xα = 4nα /nB , in this minimization
procedure.
• 3 The non-uniform matter phase exists in some middle density range where the heavy
nuclei can be formed in order to lower the free energy. The starting density of non-uniform
matter phase depends on the temperature strongly, while the ending density is nearly independent of the temperature. Both of them have a weak dependence on Yp . We first try to findout the starting density at T , and then calculate the non-uniform matter results in this range
by using the minimization codes with respect to some independent parameters as described inSec. 4. Actually, we have to use a few codes to do this calculation because of the numericaldifficulty in the minimization procedure. As the density ρB increases with fixed T and Yp , theheavy nucleus fraction increases, while the free nucleon fractions and alpha-particle fraction
decrease. When one of them decreases to a small value of about 10−5 , it brings difficulty inthe minimization code, and lose good accuracy in the results. In this case, we use anothercode which takes this composition out by deleting one independent parameter. It may run into
difficulty again as the density increases, and then we delete one more parameter. We list allcodes used in this procedure in Table 2. We determine which code should be used by comparing the free energies, and choose the one with the lowest free energy. Generally speaking,the heavy nuclei use up free protons (Yp < 0.45) or free neutrons (Yp > 0.45) quickly after the
non-uniform matter appears. The alpha-particles will disappear as the density increases, andthen free neutrons (Yp < 0.45) or free protons (Yp > 0.45) disappear. Finally, only heavy nuclei
are formed without free particles outside (this pure nucleus phase does not occur at Yp < 0.3).
When T > 4 MeV, the calculation becomes relatively easy because the favorable state is always
the one including all compositions. As the temperature increases, the heavy nucleus fraction
decreases, and it disappears completely when T > 14 MeV.
• 4 As the density ρB increases beyond 1014.2 g/cm3 , the non-uniform matter phase disappears, and it becomes a homogeneous matter of neutrons and protons with less alpha-particles.
The alpha-particle fraction Xα decreases as the density ρB increases. We switch off the alphaparticle degree of freedom when Xα < 10−4 , and use the RMF code to calculate the results of
uniform matter without alpha-particles in the high density range.

14

Table 2: The minimization codes for the non-uniform matter phase
code compositions
nameof non-uniform

matternpah

free protons,

alpha particles,

heavy nuclei

8

free protons,

alpha particles,

heavy nuclei

7alpha particles,

heavy nuclei

7

heavy nuclei

7

heavy nuclei

6

heavy nuclei

6

heavy nuclei

6

heavy nuclei

5

free neutrons,

pah
nah

free neutrons,

nph

free neutrons,nhfree neutrons,phparameter
numberfree protons,

free protons,ahalpha particles,h• 5 We determine the phase transitions in the whole density range at each Yp with fixed
T by comparing the free energies calculated with the uniform matter codes and non-uniform
matter codes listed in Table 2. All the coming results in steps 2, 3, and 4 are the inputs forthe judging code, and we get two important output files from it, EOS.FE and EOS.TAB. In
EOS.FE, the free energy per particle F is written as a function of Yp and ρB with fixed T ,which can be used for calculating pressure and chemical potentials. In EOS.TAB, all quantities
that are needed for the finial EOS table are included.
• 6 We calculate pressure and chemical potentials through the following thermodynamical
relationsp =hn2B (∂F/∂nB )

i
T,Yp

,

(57)

µn = [ ∂(nB F )/∂nn ]T,np ,(58)µp = [ ∂(nB F )/∂np ]T,nn ,(59)

where F is the free energy per baryon written in EOS.FE obtained in step 5, while nB is thebaryon number density related to the baryon mass density as ρB = mu nB with mu = 931.494
MeV being the atomic mass unit [9]. nn = (1 − Yp )nB and np = Yp nB are the neutron andproton number densities, respectively. The numerical derivatives are calculated by a five-point
15

differentiation method. In the high density range where alpha-particles have been switched off,
we take the exact results calculated in the RMF theory instead of the numerical differentiations.
The output files of this step are PRE.TAB for pressure and CHE.TAB for chemical potentials.
• 7 We combine the files EOS.TAB, PRE.TAB, and CHE.TAB in order to get the finial
EOS table at this T . Due to the use of many codes listed in Table 2 for the non-uniform matter
phase, some fractions might be equal to zero at some densities in the file EOS.TAB. This isbecause these compositions have been switched off for getting good accuracy. However, weknow that the fraction should be a small value at finite temperature if the chemical potential
is finite. In this case, we calculate the fraction Xi by the chemical potential µi and write it in
the finial EOS table,out
Xi = (nout)/(nB Vcell ),
i V

(60)

3
where Vcell is the cell volume as defined in Eq. (44), while V out = Vcell − 4π
is the volume
RA3

outside the heavy nucleus with RA being the maximum of Rp and Rn considered as the boundaryis the free neutron number density (i = n) or free proton numberof the heavy nucleus. nouti
density (i = p). For alpha-particles (i = α), noutshould be four times of the alpha-particleinumber density outside the heavy nucleus, because there are two protons and two neutrons in
an alpha-particle. The number densities outside the heavy nucleus can be obtained throughthe relations,
µ

nout
n
nout
p
nout
α

¶

M T 3/2µn= 2exp( ),
2π
T
¶3/2
µ
MT
µp
= 2
exp( ),
2πT¶3/2µµα + Bα
MT
exp(
),
= 8
2π
T

(61)
(62)
(63)

where µα = 2µn + 2µp is based on the equilibrium condition. The finial EOS table at this T is
the output file of this step.
We have to go through these steps for each T where the non-uniform matter phase exists. We
also add the results for zero temperature case (T = 0). There are two main differences between
T = 0 and T 6= 0 cases. (1) At T = 0, there are no alpha-particles and free protons (or freeneutrons) outside the heavy nucleus, so we need not calculate so many codes listed in Table 2.
The starting density of the non-uniform matter phase is below 105 g/cm3 at T < 0.4 MeV, so
16we can skip step 2 in this case. (2) The ideal-gas approximation is not available at T = 0, so weuse a function Ckf2 based on the Fermi-gas model to express the nonrelativistic kinetic energy
at low density. Note that the free energy at T = 0 is equal to the internal energy, because theentropy is equal to zero in this case.
When the non-uniform matter phase disappears at high temperature, the calculation becomes relatively easy. We can skip step 3 which is the most complicated part in this calculation.
As the temperature increases, the alpha-particle contribution gets less and less. We completely
neglect the alpha-particle contribution at T > 100 MeV. On the other hand, the alpha-particle
contribution is also dependent on the density. As the density ρB increases, the alpha-particle
fraction Xα increases, but it drops rapidly at high density due to the finite volume effect ofalpha-particles. In the high temperature range 14 MeV < T < 100 MeV, we take into accountthe alpha-particle contribution from ρB = 1010 g/cm3 to the density where Xα becomes smaller
than 10−4 . If Xα is smaller than 10−4 in the whole density range, we neglect the alpha-particle
contribution and calculate Xα by the chemical potential µα . The difference between the idealgas approximation and the RMF results is mainly due to the relativistic effect, and it increases
as the temperature increases. We use the RMF theory to perform the calculation in the wholedensity range at T > 30 MeV.
At the end, we calculate the pure neutron matter results (Yp = 0). There is only one
composition in this case, so we can use the RMF theory to perform the calculation in the wholedensity range, and get the final EOS table for Yp = 0.

6

Resulting EOS table

In this section, we explain the resulting EOS table and list the definitions of the physicalquantities tabulated. We provide the resulting EOS by three tables, which are named as(1) eos2.tab (main EOS table, size: 139.81 MB)• temperature T [MeV]: −1.0 ≤ log10 (T ) ≤ 2.6; mesh of log10 (T ) = 0.04
• proton fraction Yp : 0.01 ≤ Yp ≤ 0.65; mesh of Yp = 0.01
• baryon mass density ρB [g/cm3 ]: 5.1 ≤ log10 (ρB ) ≤ 16.0; mesh of log10 (ρB ) = 0.1
17

(2) eos2.t00 (EOS at T = 0, the same range of Yp and ρB , size: 1.54 MB)(3) eos2.yp0 (EOS at Yp = 0, the same range of T and ρB , size: 2.18 MB)
One can download them from the following websites:
http://physics.nankai.edu.cn/grzy/shenhong/EOS/index.html
http://www.rcnp.osaka-u.ac.jp/∼ shen/In the model used in this calculation, we consider four compositions which are free neutrons,
free protons, alpha-particles, and a single species of heavy nuclei. Generally speaking, there are
two phases existing in the range covered by this EOS table. The phase where heavy nuclei areformed is referred to as non-uniform matter, while the phase without heavy nuclei is referred
to as uniform matter. The phase transition can be found by the variation of the heavy nucleus
fraction between XA = 0 and XA 6= 0.
We write the EOS table in the following order: first fix T which is written at the beginningof each block in the table, second fix Yp , third fix ρB . The blocks with different T are divided
by the string of characters ‘cccccccccccc’. For each T , Yp , and ρB , we write all quantities in one
line by the following order:• (1) logarithm of baryon mass density: log10 (ρB ) [g/cm3 ]• (2) baryon number density: nB [fm−3 ]The baryon number density is related to the baryon mass density as ρB = mu nB with
mu = 931.494 MeV being the atomic mass unit taken from Ref. [9].
• (3) proton fraction: Yp
The proton fraction Yp of uniform matter is defined by
Yp =np + 2nα
np + 2nα=,nBnn + np + 4nα(64)where np is the proton number density, nn is the neutron number density, nα is thealpha-particle number density, and nB is the baryon number density.
For non-uniform matter case, Yp is the average proton fraction defined by
Yp =

18

Np
,
NB

(65)

where Np is the proton number per cell, and NB is the baryon number per cell given by
Z

Np =
NB =

Zcell
cell

[ np (r) + 2nα (r) ] d3 r,(66)[ nn (r) + np (r) + 4nα (r) ] d3 r.

(67)Here, np (r) and nn (r) are the proton and neutron density distribution function given by
Eq. (45), and nα (r) is the alpha-particle density distribution function given by Eq. (47).
• (4) free energy per baryon: F [MeV]The free energy per baryon is defined as relative to the free nucleon mass M = 938 MeV
in the TM1 parameter set as
F =

f
− M.
nB

(68)

• (5) internal energy per baryon: Eint [MeV]
The internal energy per baryon is defined as relative to the atomic mass unit mu = 931.494
MeV as²− mu .
nB

Eint =(69)• (6) entropy per baryon: S [kB ]The entropy per baryon is related to the entropy density via
S=

s
.
nB

(70)

• (7) mass number of the heavy nucleus: A
The mass number of the heavy nucleus is defined by
A=

Z RA
0

[ nn (r) + np (r) ] 4πr2 dr,(71)

where nn (r) and np (r) are the density distribution functions given by Eq. (45), and RA is
the maximum of Rp and Rn , which is considered as the boundary of the heavy nucleus.
• (8) charge number of the heavy nucleus: Z
The charge number of the heavy nucleus is defined by
Z=

Z RA
0

np (r) 4πr2 dr.

19

(72)

• (9) effective nucleon mass: M ∗ [MeV]The effective nucleon mass is obtained in the RMF theory for uniform matter. In the
non-uniform matter phase, the effective nucleon mass is a function of space due to inhomogeneity of the nucleon distribution, so it is meaningless to list the effective nucleon
mass for non-uniform matter. We replace the effective nucleon mass M ∗ by the freenucleon mass M in the non-uniform matter phase.
• (10) free neutron fraction: Xn
The free neutron fraction is given byout
Xn = (nout)/(nB Vcell ),
n V

(73)

3
where Vcell is the cell volume as defined in Eq. (44), V out = Vcell − 4π
RAis the volume
3

outside the heavy nucleus, noutis the free neutron number density outside the heavynnucleus, and nB is the average baryon number density.
• (11) free proton fraction: Xp
The free proton fraction is given byout
Xp = (nout)/(nB Vcell ),
p V

(74)

where noutis the free proton number density outside the heavy nucleus.
p• (12) alpha-particle fraction: XαThe alpha-particle fraction is defined by
Xα = 4Nα /(nB Vcell ),(75)

where Nα is the alpha-particle number per cell obtained by
Z

Nα =

cell

nα (r) d3 r,(76)

and nα (r) is the alpha-particle density distribution function given by Eq. (47).
• (13) heavy nucleus fraction: XA
The heavy nucleus fraction is defined by
XA = A/(nB Vcell ),
20

(77)

where A is the mass number of the heavy nucleus as defined in Eq. (71).
• (14) pressure: p [MeV/fm3 ]The pressure is calculated through the thermodynamical relationhp = n2B (∂F/∂nB )

i
T,Yp

.

(78)

• (15) chemical potential of the neutron: µn [MeV]The chemical potential of the neutron relative to the free nucleon mass M is calculated
through the thermodynamical relationµn = [ ∂(nB F )/∂nn ]T,np .

(79)Here nn = (1 − Yp ) nB .
• (16) chemical potential of the proton: µp [MeV]The chemical potential of the proton relative to the free nucleon mass M is calculated
through the thermodynamical relationµp = [ ∂(nB F )/∂np ]T,nn .

(80)Here np = Yp nB .
We have done the following check for the EOS table:(1) the consistency of the fractions,
Xn + Xp + Xα + XA = 1.

(81)(2) the consistency of the relation between F , Eint , and S,
F = Eint − T S + mu − M.

(82)(3) the consistency of the thermodynamical quantities,
F = µn (1 − Yp ) + µp Yp −

p
.
nB

(83)

In general, these consistency relations can be satisfied within a few thousandths. Physicalconstants to convert units are taken from Ref. [9].
21

7Suggestions for using the EOS table

This relativistic EOS table is designed for use in supernova simulations, so we perform thecalculation at each T , Yp , and ρB . Comparing with EOS1 done in 1998, we have improved thenumerical codes to take regular logarithmic meshes for log10 (T ) and log10 (ρB ), and use a linear
mesh for Yp in EOS2.
In order to perform the minimization of the free energy for non-uniform matter, we haveto parameterize the nucleon distributions, so that some quantities in the EOS table like A,Z, Xn , Xp , Xα , and XA are dependent on this parameterization method. One should keep inmind their definitions when these quantities are used. We suggest users pay more attention
to the thermodynamical quantities like F , Eint , S, p, µn , and µp , which are supposed not tobe sensitive to the parameterization method. The thermodynamical quantities are found to be
more smooth in the resulting EOS, while the use of many codes in Table 2 may bring somefluctuations in the fractions, especially in the temperature range 1 MeV < T < 4 MeV, where
the connection is very complicated.
We hope this EOS table can be used in your calculation successfully. We emphasize againthat this EOS table includes only baryon contributions, so you have to add the lepton contributions when you use it.

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[3] Y. Sugahara and H .Toki, Nucl. Phys. A579, 557 (1994).
[4] D. Hirata, K. Sumiyoshi, B. V. Carlson, H. Toki, and I. Tanihata, Nucl. Phys. A609,
131 (1996).
[5] K. Sumiyoshi, H. Kuwabara, and H. Toki, Nucl. Phys. A581, 725 (1995).
[6] B. D. Serot and J. D. Walecka, Adv. Nucl. Phys. 16, 1 (1986).
22

[7] J. M. Lattimer and F. D. Swesty, Nucl. Phys. A535, 331 (1991).
[8] K. Oyamatsu, Nucl. Phys. A561, 431 (1993).
[9] C. Amsler et al. (Particle Data Group), Phys. Lett. B667, 1 (2008).

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