User Note for Relativistic EOS Table(EOS4: 2020-version, with only nucleons)Hong Shen,1 Fan Ji,1 Jinniu Hu,1 and Kohsuke Sumiyoshi2
12School of Physics, Nankai University, Tianjin 300071, China
National Institute of Technology, Numazu College, Shizuoka 410-8501, Japan

Abstract
This is a detailed note for users of the relativistic equation of state (EOS) of nuclear matter.
The new EOS4 based on the extended TM1 (TM1e) model with L = 40 MeV is designed in thesame tabular form as the previous EOS2 based on the original TM1 model with L = 110.8 MeV.
The numerical procedure for constructing EOS table has been explained in previous note of EOS2.
For completeness and convenience in reading, we include all materials in the present document.

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 table

1

I.

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 several versions of the equation of state (EOS) within the relativistic
mean-field (RMF) framework [1–3]. 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 densityand finite temperature, we approximate the nuclear matter as a classical ideal gas of freeneutrons, free protons, and alpha particles. Antiparticles have some contribution whentemperature is very 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 each temperature, 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. II, we describe the RMF theory. In Sec. III,
we introduce the ideal-gas approximation to be used at low density. In Sec. IV, we describe
the Thomas-Fermi approximation, which is used for calculating non-uniform matter. In
Sec. V, we make a detailed description of how to work out the EOS table. In Sec. VI, we list
the definitions of physical quantities in the EOS table. Sec. VII is devoted to suggestions
and discussions for using the EOS table. Users who are not interested in the framework andcalculating processes can only read Sec. VI and Sec. VII for the purpose to use the EOS
table correctly.

II.

RELATIVISTIC MEAN-FIELD THEORYIn the present work, we employ the RMF model with an extended TM1 parametrization,
namely the TM1e model, to describe the nuclear system, where nucleons interact through theexchange of various mesons including the isoscalar-scalar meson σ, isoscalar-vector meson2ω, and isovector-vector meson ρ [4]. The nucleonic Lagrangian density reads
(
)]
∑ [
gρ
LRMF =
ψ̄i iγµ ∂ µ − (M + gσ σ) − γµ gω ω µ + τa ρaµ ψi
2
i=p,n
1
1
1
1
+ ∂µ σ∂ µ σ − m2σ σ 2 − g2 σ 3 − g3 σ 4
2
2
3
4
1
1
1 2
µ
µν
− Wµν W + mω ωµ ω + c3 (ωµ ω µ )2
4
2
4
)
)(
(
1 a aµν 1 2 a aµ
− Rµν R + mρ ρµ ρ + Λv gω2 ωµ ω µ gρ2 ρaµ ρaµ ,
4
2

(1)

where W µν and Raµν denote the antisymmetric field tensors for ω µ and ρaµ , respectively.
Under the mean-field approximation, the meson fields are treated as classical fields and thefield operators are replaced by their expectation values. In a static uniform system, thenonzero components are σ = ⟨σ⟩, ω = ⟨ω 0 ⟩, and ρ = ⟨ρ30 ⟩. We derive the equations of
motion for mesons and the Dirac equation for nucleons, which are coupled with each otherand could be solved self-consistently.
Compared with the original TM1 model adopted in Ref. [3], an additional ω-ρ coupling
term is introduced in the Lagrangian density (1), which plays a crucial role in determining
the density dependence of the symmetry energy. By adjusting the coupling constants, gρand Λv , it is possible to control the behavior of symmetry energy and its density dependence.
In the TM1e model, the slope parameter L = 40 MeV and the symmetry energy Esym =31.38 MeV at saturation density are obtained, which fall well within the constraints fromvarious observations. The corresponding values in the original TM1 model [5] are L = 110.8
MeV and Esym = 36.89 MeV, which are rather large and disfavored by recent astrophysicalobservations. In Table I, we present the coupling constants of the TM1e and TM1 models
for completeness. Note that the coupling constant gρ is different by a factor of 2 from theone in Ref. [3]. We follow here the convention of Ref. [4]. Compared the TM1e model withthe original TM1 model, only gρ and Λv related to isovector parts are different, while allother parameters remain the same. It is noteworthy that the TM1e model provides the same
isoscalar saturation properties and similar binding energies of finite nuclei as the original
TM1 model, whereas the density dependence of symmetry energy is very different.
For the Thomas–Fermi calculations of nonuniform matter, we need to input the energy
density and entropy density of uniform nuclear matter, which are given by√∑ 1 ∫ ∞)(
2
2 + M ∗2 f k + f k
dk
k
k
ϵ =
i−
i+
π2 0
i=p,n
1
1
1
+ m2σ σ 2 + g2 σ 3 + g3 σ 4
2
3
4
)
)(
(
1 2 2 31
4
+ mω ω + c3 ω + m2ρ ρ2 + 3Λv gω2 ω 2 gρ2 ρ2 ,
2
4
2

3

(2)

TABLE I: Parameters in the TM1e and TM1 models. The masses are given in MeV.
Model M

mσ

mω mρ

gσ

gω

gρ

g2 [fm−1 ]

g3

c3

Λv

TM1e 938 511.198 783 770 10.0289 12.6139 13.9714 −7.2325 0.6183 71.3075 0.0429
TM1 938 511.198 783 770 10.0289 12.6139 9.2644 −7.2325 0.6183 71.3075 0.0000

and
s = −

∑ 1 ∫ ∞
i=p,n

π2

[ k
(
) (
)
k
k
k
dk k 2 fi+
ln fi++ 1 − fi+ln 1 − fi+

0

(
) (
)]
k
k
k
k
+fi−
ln fi−+ 1 − fi−ln 1 − fi−
.

(3)

k
k
Here M ∗ = M + gσ σ is the effective nucleon mass. fi+
and fi−(i = p, n) denote theoccupation probabilities of nucleon and antinucleon at momentum k, which are given by the
Fermi-Dirac distribution,{[(√) ]}−1kfi±= 1 + exp
k 2 + M ∗ 2 ∓ νi /T,(4)with the kinetic part of the chemical potential νi related to the chemical potential µi as
νi = µi − gω ω −gρ
τ3i ρ.
2

The number density of protons (i = p) or neutrons (i = n) is calculated by
∫ ∞
( k
)
1
k
ni = 2
dk k 2 fi+
− fi−
.
π 0

(5)

(6)

Using the results of the TM1e model as input in the Thomas–Fermi calculation, we computethe average free energy density of nonuniform matter, and compare it with the one of
uniform matter. At a given temperature T , proton fraction Yp , and baryon mass densityρB , the thermodynamically stable state is the one having the lowest free energy density. We
determine the stable state and the phase transition between nonuniform matter and uniform
matter by minimizing the free energy density.

III.

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 andfinite temperature. We note that the connection between the RMF results and the ideal-gas
approximation is smooth.
4

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 by[]2
Zi = (nQ )ni /2 /(ni /2)! ,(7)( )3/2where we have used the abbreviation nQ = M2πT
. The factor (ni /2)! comes from avoiding
the double counting for ni /2 indistinguishable particles of spin up or spin down. We cancalculate the free energy density from the partition function asfi = −T ln Zi = −T ni [ln(2nQ /ni ) + 1] .

(8)

The entropy density is given by(si = −

∂fi
∂T

)

[

]
5
= ni ln(2nQ /ni ) +
,
2
ni

(9)

the internal energy density can be written as3
ϵ i = fi + T s i = T n i ,2the chemical potential is obtained through the relation()∂fi= −T ln(2nQ /ni ),
µi =
∂ni T

(10)

(11)

and the pressure is given by
[
pi =

]

∂ (fi /ni )= T ni .
n2i
∂ni
T

(12)

The partition function of alpha particles can be written asZα = [8nQ exp (Bα /T )]nα /nα !,(13)

where nα is the alpha-particle number density, and Bα = 28.3 MeV is the binding energyof an alpha particle taken from Ref. [6]. The thermodynamical quantities of alpha particles
are given by
fα = −T nα [ln(8nQ /nα ) + 1] − nα Bα ,[]5sα = nα ln(8nQ /nα ) +,23ϵα = T nα − nα Bα ,2µα = −T ln(8nQ /nα ) − Bα ,(14)

pα = T nα .

(18)

5

(15)
(16)
(17)

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 .

(19)

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
actually disappear. When we take into account the volume excluded by the alpha particles,
the free energy densities in Eq. (19) are given by
fn = −(1 − u) T ñn [ln(2nQ /ñn ) + 1] ,(20)

fp = −(1 − u) T ñp [ln(2nQ /ñp ) + 1] ,(21)

fα = −(1 − u) {T ñα [ln(8nQ /ñα ) + 1] − ñα Bα } ,(22)where u = nα vα is the fraction of space occupied by alpha particles with the effective volumeof an alpha particle vα = 24 fm−3 taken from Ref. [6]. We denote the effective number
densities of neutrons, 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 highdensity. Actually, we just use Eqs. (20) and (21) 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. (22) to calculate the free energy in the whole density range. It is considered as areasonable approximation because the number density of alpha particles is not so large eventhough 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 take some approximation in order to describe the range of low density and hightemperature. We take the nonrelativistic approximation of the RMF√ theory at low density,k2∗.
where the effective nucleon mass M ≈ M , < mesons >≈ 0, and k 2 + M ∗ 2 ≈ M + 2M
Then the occupation probability of the Fermi-Dirac distribution becomes()(23)
fis = ACi exp −k 2 /2M T ,()
A
fīs =
exp −k 2 /2M T ,(24)
Ciwhere 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 as
(
)(
)3/2
1
MT
ni = 2A Ci −
.
(25)
Ci
2π
6

The thermodynamical quantities can be approximated as()3/2 ( ))(13MTϵi = 2A Ci +
T ,
Ci
2π
2
(
)3/2
)(
1
MT
pi = 2A Ci +T,Ci2π( )] (
)
)3/2
[ (
1
1
A
MT
5
Ci +
− Ci ln (ACi ) −ln
si = 2A
.
2
Ci
Ci
Ci
2π

(26)
(27)
(28)

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

THOMAS-FERMI APPROXIMATIONIn the range, T < 14 MeV and ρB < 1014.2 g/cm3 , where heavy nuclei may be formed inorder to lower the free energy, we perform the Thomas-Fermi calculation based on the work
done by Oyamatsu [7]. In this case, the non-uniform matter can be modeled as a mixtureof free neutrons, free protons, alpha particles, and a single species of heavy nuclei, while theleptons 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 abody-centered-cubic (BCC) lattice to minimize the Coulomb lattice energy. It is useful tointroduce the Wigner-Seitz cell to simplify the energy of a unit cell. The Wigner-Seitz cell isa sphere whose volume is the same as the unit cell in the BCC lattice. We define the lattice
constant a as the cube root of the cell volume. Then, we have
Vcell = a3 = NB /nB ,(29)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 andadd an energy correction for the BCC lattice [7]. This energy correction is negligible unlessthe 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 as[( ) t i ]3)
 ( inout+ nout1 − Rrini − nii , 0 ≤ r ≤ Ri ,(30)
ni (r) = out
Ri ≤ r ≤ Rcell ,ni ,7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(31)

out
The density parameters nin
are the densities at r = 0 and r ≥ Ri , respectively.
i and niThe parameters Ri and ti determine the boundary and the relative surface thickness of theheavy nucleus.
For the distribution function of alpha particle nα (r), which should decrease as r approaches the center of the heavy nucleus, we assume
[( ) t p ]3out
−nα 1 − Rrp+ noutα , 0 ≤ r ≤ Rp ,nα (r) =
(32)
 out
nα ,
Rp ≤ r ≤ Rcell ,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 thepresence of too many parameters in the minimization procedure. The parameters Rn and tn
may be somewhat different from Rp and tp due to the additional neutrons forming a neutron
skin in the surface region. For a system with fixed temperature T , proton fraction Yp , andbaryon mass density ρB , there are eight independent parameters among the ten variables, a,
out
in
out
out
nin
n , nn , Rn , tn , np , np , Rp , tp , and nα . The thermodynamically favorable state is the onethat 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 ,(33)

where the free energy per cell Fcell can be written as
Fcell = (Ebulk + Es + EC ) − T Scell = Fbulk + Es + EC .
The bulk energy Ebulk , entropy Scell , and bulk free energy Fbulk are calculated by∫
Ebulk =ϵ ( nn (r) , np (r) , nα (r) ) d3 r,∫cell
Scell =s ( nn (r) , np (r) , nα (r) ) d3 r,
cell∫
Fbulk =f ( nn (r) , np (r) , nα (r) ) d3 r.

(34)

(35)
(36)
(37)

cell

Here ϵ ( nn (r) , np (r) , nα (r) ), s ( nn (r) , np (r) , nα (r) ), and f ( nn (r) , np (r) , nα (r) ) are
the local 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 alphaparticles. Note that these local densities at each radius are calculated by treating neutrons8and protons in the RMF theory as described in Sec. II for the case of nn (r) + np (r) >10−5 fm−3 , while we can use the method described in Sec. III 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∫
Es =F0 | ∇ ( nn (r) + np (r) ) |2 d3 r.
(38)
cell

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, and the beta stability line as described in the Appendix of Ref. [7].
The Coulomb energy per cell EC is calculated using the Wigner-Seitz approximation withan added correction term for the BCC lattice [7]∫1EC =e [np (r) + 2nα (r) − ne ] ϕ(r)d3 r + △EC ,(39)
2 cellwhere 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 △EC isthe correction term for the BCC lattice, which can be approximated as△EC = CBCC

(Znon e)2
.
a

(40)

Here a is the lattice constant as defined in Eq. (29), the coefficient CBCC = 0.006562 is takenfrom Ref. [7], and Znon is the non-uniform part of the charge number per cell given by∫ Rp
Znon =

[
out
out
(nin
p − np − 2nα ) 1 −

0

(

r
Rp

)tp ]34πr2 dr.

(41)

Because of the long-range nature of the Coulomb interaction, the Coulomb energy will
depend on the lattice type. This dependence was discussed in more detail in Ref. [7]. The
system prefers the BCC lattice because the BCC lattice gives the lowest Coulomb energy.

V.

CALCULATION PROCEDURE IN DETAIL

We calculate the EOS table in the range mentioned in Sec. I. 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. IV. 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
and alpha particles is referred to as non-uniform matter, while the phase of nucleons mixed
with alpha particles without heavy nuclei is referred to as uniform matter. By comparing
9the free energy of non-uniform matter with that of uniform matter, we determine the mostfavorable 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 nonuniform 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 , andthen fix 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 minimization• 2 perform minimization for uniform matter at low density• 3 perform 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
get good numerical accuracy in doing linear approximation. We calculate this input tablein 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 andthe entropy per baryon S. The free energy per baryon can be obtained by F = E − T S.
We note that the data at nn + np > 0.16 fm−3 is not necessary for the minimization procedure, while the ideal-gas approximation is used to describe neutrons and protons whennn + np < 10−5 fm−3 .
• 2 At low density and finite temperature, the thermodynamically favorable state might
be a 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
10

the minimum free energy. Since nB = nn + np + 4nα and Yp = (np + 2nα )/nB are fixed in thiscase, there is only one independent parameter, the alpha-particle fraction Xα = 4nα /nB , inthis 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 find out the starting density at T , and then calculate the non-uniform matter results inthis range by using the minimization codes with respect to some independent parameters
as described in Sec. IV. Actually, we have to use a few codes to do this calculation becauseof the numerical difficulty in the minimization procedure. As the density ρB increases with
fixed T and Yp , the heavy nucleus fraction increases, while the free nucleon fractions andalpha-particle fraction decrease. When one of them decreases to a small value of about 10−5 ,it brings difficulty in the minimization code, and lose good accuracy in the results. In thiscase, we use another code 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 all codes used in this procedure in Table II. We determine whichcode 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 freeneutrons (Yp > 0.45) quickly after the non-uniform matter appears. The alpha particles
will disappear as the density increases, and then 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 alphaparticles. The alpha-particle fraction Xα decreases as the density ρB increases. We switch
off the alpha-particle 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.
• 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 II. 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, allquantities that are needed for the finial EOS table are included.
• 6 We calculate pressure and chemical potentials through the following thermodynamical

11

TABLE II: The minimization codes for the non-uniform matter phase
code compositions of non-uniform matternameparameter
numbernpah free neutrons, free protons,

alpha particles, heavy nuclei

8pahalpha particles, heavy nuclei

7alpha particles, heavy nuclei

7

nph free neutrons, free protons,

heavy nuclei

7

nh free neutrons,

heavy nuclei

6

heavy nuclei

6ahalpha particles, heavy nuclei

6hheavy nuclei

5

free protons,

nah free neutrons,phfree protons,

relations]n2B (∂F/∂nB ) T,Yp ,(42)µn = [ ∂(nB F )/∂nn ]T,np ,(43)µp = [ ∂(nB F )/∂np ]T,nn ,

(44)

p =

[

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 [8]. nn = (1 − Yp )nB and np = Yp nB are the neutron
and proton number densities, respectively. The numerical derivatives are calculated by afive-point differentiation method. In the high density range where alpha particles have beenswitched off, we take the exact results calculated in the RMF theory instead of the numericaldifferentiations. 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 II for the non-uniform
matter phase, some fractions might be equal to zero at some densities in the file EOS.TAB.
This is because these compositions have been switched off for getting good accuracy. However, we know 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 andwrite it in the finial EOS table,out
Xi = (nout)/(nB Vcell ),
i V

(45)

3
RA
is the volume
where Vcell is the cell volume as defined in Eq. (29), while V out = Vcell − 4π
3outside the heavy nucleus with RA being the maximum of Rp and Rn considered as theis the free neutron number density (i = n) or freeboundary of the heavy nucleus. nouti

12should be four times of the
proton number density (i = p). For alpha particles (i = α), noutialpha-particle number density outside the heavy nucleus, because there are two protons andtwo neutrons in an alpha particle. The number densities outside the heavy nucleus can beobtained through the relations,
(
nout
n

= 2
(

nout
p

= 2
(

nout
α

= 8

MT
2π
MT
2π
MT
2π

)3/2
exp
)3/2
exp
)3/2

(µ )
n

T
(µ )
p

(
exp

T

,

(46)

,

(47)

µα + Bα
T

)
,

(48)

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 ̸= 0 cases. (1) At T = 0, there are no alpha particles and free protons(or free neutrons) outside the heavy nucleus, so we need not calculate so many codes listed
in Table II. The starting density of the non-uniform matter phase is below 105 g/cm3 at
T < 0.4 MeV, so we can skip step 2 in this case. (2) The ideal-gas approximation is notavailable at T = 0, so we use a function Ckf2 based on the Fermi-gas model to express thenonrelativistic kinetic energy at low density. Note that the free energy at T = 0 is equal tothe internal energy, because the entropy 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 thiscalculation. 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 of alpha particles. In the high temperature range 14 MeV < T < 100 MeV, wetake into account the alpha-particle contribution from ρB = 1010 g/cm3 to the density whereXα 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 ideal-gas approximation and the RMF results is mainly due to therelativistic effect, and it increases as the temperature increases. We use the RMF theory to
perform the calculation in the whole density 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 thewhole density range, and get the final EOS table for Yp = 0.

13

VI.

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) eos4.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(2) eos4.t00 (EOS at T = 0, the same range of Yp and ρB , size: 1.54 MB)(3) eos4.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://my.nankai.edu.cn/wlxy/sh en/list.htm,
http://user.numazu-ct.ac.jp/∼ sumi/eos/index.html,
and http://doi.org/10.5281/zenodo.3612487
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 are formed 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 variationof the heavy nucleus fraction between XA = 0 and XA ̸= 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. [8].
• (3) proton fraction: Yp
The proton fraction Yp of uniform matter is defined by
Yp =np + 2nα
np + 2nα=,nBnn + np + 4nα(49)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.

14

For non-uniform matter case, Yp is the average proton fraction defined by
Yp =

Np
,
NB

(50)

where Np is the proton number per cell, and NB is the baryon number per cell given
by
∫
Np =
[ np (r) + 2nα (r) ] d3 r,
(51)
∫cell
NB =
[ nn (r) + np (r) + 4nα (r) ] d3 r.
(52)
cell

Here, np (r) and nn (r) are the proton and neutron density distribution function given
by Eq. (30), and nα (r) is the alpha-particle density distribution function given by
Eq. (32).
• (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

(53)

• (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 asϵ− mu .
(54)
Eint =nB• (6) entropy per baryon: S [kB ]The entropy per baryon is related to the entropy density via
s
S=
.
nB
• (7) mass number of the heavy nucleus: A
The mass number of the heavy nucleus is defined by∫ RA
A=[ nn (r) + np (r) ] 4πr2 dr,

(55)

(56)

0

where nn (r) and np (r) are the density distribution functions given by Eq. (30), andRA 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∫ RAZ=
np (r) 4πr2 dr.
0

15

(57)

• (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 free nucleon 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

(58)

3
where Vcell is the cell volume as defined in Eq. (29), V out = Vcell − 4π
RAis the volume
3
outoutside the heavy nucleus, nn is the free neutron number density outside the heavy
nucleus, 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

(59)

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 ),where Nα is the alpha-particle number per cell obtained by∫Nα =nα (r) d3 r,

(60)

(61)

cell

and nα (r) is the alpha-particle density distribution function given by Eq. (32).
• (13) heavy nucleus fraction: XA
The heavy nucleus fraction is defined by
XA = A/(nB Vcell ),(62)

where A is the mass number of the heavy nucleus as defined in Eq. (56).
• (14) pressure: p [MeV/fm3 ]The pressure is calculated through the thermodynamical relation][p = n2B (∂F/∂nB ) T,Yp .
16

(63)

• (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 .

(64)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 .

(65)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.

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

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

p
.
nB

(68)

In general, these consistency relations can be satisfied within a few thousandths. Physicalconstants to convert units are taken from Ref. [8].

VII.

SUGGESTIONS FOR USING THE EOS TABLE

This relativistic EOS table is designed for use in supernova simulations, so we performthe calculation at each T , Yp , and ρB . Comparing with EOS2 based on the original TM1model with L = 110.8, the TM1e model with L = 40 MeV is adopted in the new EOS4. In
the TM1e model, the slope parameter L = 40 MeV and the symmetry energy Esym = 31.38
MeV at saturation density are obtained, which fall well within the constraints from various
observations.
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
17

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 II may bringsome fluctuations 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
again that this EOS table includes only baryon contributions, so you have to add the lepton
contributions when you use it.

[1] H. Shen, H. Toki, K. Oyamatsu, and K. Sumiyoshi, Prog. Theor. Phys. 100, 1013 (1998).
[2] H. Shen, H. Toki, K. Oyamatsu, and K. Sumiyoshi, Nucl. Phys. A637, 435 (1998).
[3] H. Shen, H. Toki, K. Oyamatsu, and K. Sumiyoshi, Astrophys. J. Suppl. 197, 20 (2011).
[4] S. S. Bao, J. N. Hu, Z. W. Zhang, and H. Shen, Phys. Rev. C 90, 045802 (2014).
[5] Y. Sugahara and H .Toki, Nucl. Phys. A579, 557 (1994).
[6] J. M. Lattimer and F. D. Swesty, Nucl. Phys. A535, 331 (1991).
[7] K. Oyamatsu, Nucl. Phys. A561, 431 (1993).
[8] C. Amsler et al. (Particle Data Group), Phys. Lett. B667, 1 (2008).

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