# Particle Yields in Heavy Ion Collisions and the Influence of Strong Magnetic Fields

Authors: M. G. de Paoli; D. P. Menezes
Journal: Advances in High Energy Physics
 (2014)

---

## Abstract

It is expected that the magnetic fields in heavy ion collisions are very high. In this work, we investigate the effects of a strong magnetic field on particle ratios within a thermal model of particle production. We model matter as a free gas of baryons and mesons under the influence of an external magnetic field varying from zero to30
m
π
2 through an  χ
2 fitting to some data sets of the STAR experiment. For this purpose, we use the Dirac, Rarita-Schwinger, Klein-Gordon, and Proca equations subject to magnetic fields in order to obtain the energy expressions and the degeneracy for spin 1/2, spin 3/2, spin 0, and spin 1 particles, respectively. Our results show that, if the magnetic field can be considered as slowly varying and leaves its signature on the particle yields, a field of the order of 6
m
π
2 produces an improved fitting to the experimental data as compared to the calculations without magnetic field.


---

## Body

## 1. Introduction

According to quantum chromodynamics, the quark-gluon plasma (QGP) phase refers to matter where quarks and gluons are believed to be deconfined and it probably takes place at temperatures of the order of 150 to 170 MeV. In large colliders around the world (RHIC/BNL, ALICE/CERN, GSI, etc.), physicists are trying to find a QGP signature looking at heavy ion collisions and, in the last years, it has become evident that a strong magnetic field dependence is present in all experimental processes. Moreover, it has also been shown that the QCD phase diagram is modified by the presence of a magnetic field. Its effects have been calculated both within relativistic models [1, 2] and lattice simulations [3].Possible experiments towards the search for the QGP are Au-Au collisions at RHIC/BNL and Pb-Pb collisions at SPS/CERN. The hadron abundances and particle ratios are normally used in order to determine the temperature and baryonic chemical potential of the possibly present hadronic matter-QGP phase transition and this calculation is done through thermal equilibrium models [4, 5] Particle ratios are convenient quantities to be analysed because after the chemical freezeout they remain practically unaltered.In previous papers, a statistical model under chemical equilibration was used to calculate particle yields [4, 5] and in these works the densities of particles were obtained from free Fermi and Boson gas approximations. To achieve a better description of the data, an excluded volume term was assigned to all particles with the aim of mimicking the repulsive interactions between hadrons at small distances. Also, after thermal production, resonances and heavier particles were allowed to decay through a systematic parameter that regulates the weak decay processes.More recently, relativistic nuclear models have been tested in the high temperature regime produced in these heavy ion collisions. In [6, 7], different versions of Walecka-type relativistic models [8] were used to calculate the Au-Au collision particle yields at RHIC/BNL and in [9] the quark-meson-coupling model [10–12] was used to calculate this reaction results and also Pb-Pb collision particle rations at SPS/CERN without the enforcement of the excluded volume approximation. In all cases, 18 baryons, pions, kaons, ρ’s, and K
*s were incorporated in the calculations and a fit based on the minimum value of the quadratic deviation was implemented in order to obtain the temperature and chemical potential for each model, according to a prescription given in [4]. For Au-Au collision (RHIC) these numbers lie in the range of 132
<
T
<
169 MeV and 30.5
<
μ
B
<
62.8 MeV and, for Pb-Pb collision (SPS), 99
<
T
<
156.1 MeV and 167.5
<
μ
B
<
411 MeV.On the other hand, the magnetic fields involved in heavy-ion collisions [13–15] can reach intensities even higher than the ones considered in magnetars [16–19]. As suggested in [13–15] and [20–22], it is interesting to investigate fields of the order of e
B = 5–30m
π
2 (corresponding to 1.7 × 1019–1020 Gauss) and temperatures varying from T = 120 to 200 MeV related to heavy ion collisions. In fact, the densities related to the chemical potentials obtained within the relativistic models framework, in all cases, are very low (of the order of 10−3 fm−3). At these densities, the nuclear interactions are indeed very small and this fact made us reconsider the possibility of free Fermi and Boson gases, but now under the influence of strong magnetic fields.In [23], the author studies the synchrotron radiation of gluons by fast quarks in strong magnetic fields produced in heavy ion collisions and shows that strong polarization of quarks and leptons with respect to the direction of the magnetic field is expected. The polarization of quarks seems to be washed out during the fragmentation but this is not the case of the leptons. The observation of lepton polarization asymmetry could be a proof of the existence of the magnetic field, which may last for 1-2 fm/c. The author concludes that the magnetic field created by fast heavy ions can be considered as approximately constant due to the high electric conductivity of the quark-gluon plasma. Recently, the same author revisited this subject in two more papers [24, 25] and emphasized the possibility that, after rapidly decreasing its magnitude during the first fm/c of the plasma expansion, the magnetic field may last as long as the quark-gluon plasma lives. These results are contradicted in another recent reference [26], where the authors claim that the lifetime of the strong magnetic field is not affected by the conductivity. Hence, whether nuclear matter plays or not a decisive role in the evolution of the magnetic field with time is a subject of intense debate.The purpose of the analysis we present in this paper is to check if the inclusion of strong magnetic fields can improve the fitting of experimental results. We start from the simplest possible calculation, based on free Fermi and Boson gases. Nor volume excluded terms, neither resonance decays are taken into account to avoid blurring the effects of the magnetic field. Moreover, we have assumed that the magnetic field is homogeneous, constant, and time-independent. According to the calculations done in [23–25], the magnetic field varies slowly or is almost constant. If this is the case, it could certainly leave its signature in the particle yields. In [27, 28], it was shown that the shape of the magnetic field presents a special nontrivial pattern and one can see that, after averaging over many events, one is left with just one of the components of the magnetic field. Nevertheless, the event-by-event fluctuation of the position of charged particles can induce another component of the magnetic field (perpendicular to the remaining one in the average calculation) and also an electric field, which is quite strong at low impact parameters. While the magnetic field remains quite high in peripheral collisions, the opposite happens with the electric field. To make our first analysis as simple as possible, we will restrict ourselves to data at centralities of the order of 80%; that is, high values of the impact parameter b
≃
11–13 fm, where we are more comfortable to disregard the electric field effects.In the present paper, we briefly revisit the formalism necessary for the calculation of particle densities subject to magnetic fields and the expressions used to implement aχ
2 fit to the experimental results. The results are then displayed and discussed.


## 2. Formalism

We model matter as a free gas of baryons and mesons under the influence of a constant magnetic field. We consider only normal and strange matter, that is, the baryons and mesons constituted byu, d and s quarks: the baryon octet (spin 1/2 baryons), the baryon decouplet (spin 3/2 baryons), the pseudoscalar meson nonet (spin 0 mesons), and the vector meson nonet (spin 1 mesons), which leaves us with a total of 54 particles (18 baryons, 18 antibaryons, and 18 mesons).We utilize natural units (ℏ
=
c
=
1) and define ϵ
0
=
μ
0
=
1. From the relation α
=
e
2
/
4
π
ϵ
0
ℏ
c, we obtain that the electron charge is e
=
4
π
α, where α
=
1
/
137 is the fine structure constant. The natural units with the electron charge in that form are known as Heaviside-Lorentz units [29].In this work, the magnetic field is introduced trough minimal coupling, so the derivatives become(1)
∂
μ
→
D
μ
=
∂
μ
+
i
q
A
μ
.We write the charge asq
=
ϵ
q
|
q
|, where ϵ
q
=
+
(
-
) corresponds to a particle with positive (negative) charge, and assume the gauge
(2)
A
μ
=
δ
μ
2
x
1
B
→
A
0
=
0
,
A
→
=
(
0
,
x
1
B
,
0
)
,
so
(3)
∇
→
·
A
→
=
0
,
∇
→
×
A
→
=
B
e
^
3
,
and the derivatives
(4)
D
μ
=
∂
μ
-
i
ϵ
q
|
q
|
B
x
1
δ
μ
2
.We search for solutions of the fieldsψ in the form
(5)
ψ
α
(
ϵ
)
=
{
C
α
(
ϵ
)
e
-
i
ϵ
E
t
+
i
ϵ
p
→
·
x
→
(
q
=
0
)
f
α
(
ϵ
)
(
x
1
)
e
-
i
ϵ
E
t
+
i
ϵ
p
2
x
2
+
i
ϵ
p
3
x
3
(
q
≠
0
)
,
where ψ
α are the components of the field ψ and ϵ
=
+
(
-
) corresponds to the states of positive (negative) energy.For the spin 1/2 baryons (Dirac field)ψ has 4 components, for the spin 3/2 baryons (Rarita-Schwinger field) ψ
μ has 16 components, for the spin 0 mesons (Klein-Gordon field) ψ has just one component, and for the spin 1 mesons (Proca field) ψ
μ has 4 components.Due to the use of statistical methods to deal with the system under consideration, we do not need the complete expression forψ, but just the form of the energy E for each one of the fields and the degeneracy of the energy levels γ.A detailed calculation of the solution of the Klein-Gordon and Dirac equations in the presence of a constant external magnetic field can be found in [30]. The Proca equation solution in the presence of a constant external magnetic field can be found in [31], where the authors did not calculate explicitly the energy expression, but it can be straightforwardly obtained. Finally, the calculations for the Rarita-Schwinger equation in the presence of an external magnetic field have been done recently in [32].


### 2.1. Spin 1/2 Baryons

The baryons with spin 1/2 are described by the Dirac Lagrangian density [34]
(6)
ℒ
D
=
ψ
¯
(
i
γ
μ
D
μ
-
m
)
ψ
,
which (after we apply the Euler-Lagrange equation) lead us to the equation of motion
(7)
(
i
γ
μ
D
μ
-
m
)
ψ
=
0
,
where γ
μ are the Dirac matrices.The solution of the equation of motion gives the following [30]:
(8)
E
=
{
p
→
2
+
m
2
(
q
=
0
)
p
3
2
+
m
2
+
2
ν
|
q
|
B
(
q
≠
0
)
,
where ν runs over the possible Landau Levels and the degeneracy for the energy states is given by
(9)
γ
=
{
2
(
q
=
0
)
2
-
δ
ν
0
(
q
≠
0
)
.


### 2.2. Spin 3/2 Baryons

The baryons with spin 3/2 are described by the Rarita-Schwinger Lagrangian density [35, 36]
(10)
ℒ
RS
=
-
1
2
ψ
¯
μ
(
ϵ
μ
ν
ρ
σ
γ
5
γ
ν
D
ρ
+
i
m
σ
μ
σ
)
ψ
σ
,
where γ
5
=
i
γ
0
γ
1
γ
2
γ
3 and σ
μ
ν
=
(
i
/
2
)
[
γ
μ
,
γ
ν
].The equation of motion reads(11)
(
i
γ
μ
D
μ
-
m
)
ψ
ν
=
0
with
γ
μ
ψ
μ
=
0
,
D
μ
ψ
μ
=
0
.The solution of the Rarita-Schwinger equation is not trivial and poses noncausality problems. To obtain the degeneracy of the energy states, we follow the prescription used in [34], which is given in detail for the Rarita-Schwinger equation in [32]. Observing the equation of motion, one can see that each component of ψ
μ obeys a Dirac type equation, so the energy must have the form
(12)
E
=
{
p
→
2
+
m
2
(
q
=
0
)
p
3
2
+
m
2
+
2
ν
|
q
|
B
(
q
≠
0
)
.Besides that,ψ
μ has 4 components, but two equations are constrained, which means that only 2 components of ψ
μ are really independent. So ψ
μ have 2 polarizations, but (because of the Dirac equation solution) each polarization is double degenerate. In the presence of a magnetic field, there is another constraint for the ν
=
0 and ν
=
1 energy levels, which leads to the following degeneracy for the energy states:
(13)
γ
=
{
4
(
q
=
0
)
4
-
2
δ
ν
0
-
δ
ν
1
(
q
≠
0
)
.


### 2.3. Spin 0 Mesons

The mesons with spin 0 are described by the Klein-Gordon Lagrangian density [37]
(14)
ℒ
KG
=
D
μ
ψ
*
D
μ
ψ
-
m
2
ψ
*
ψ
,
whose equation of motion is given by
(15)
(
D
μ
D
μ
+
m
2
)
ψ
=
0
,
with the energy satisfying the relation [30]:
(16)
E
=
{
p
→
2
+
m
2
(
q
=
0
)
p
3
2
+
m
2
+
(
2
ν
+
1
)
|
q
|
B
(
q
≠
0
)
.


### 2.4. Spin 1 Mesons

The mesons with spin 1 are described by the Proca Lagrangian density [31]
(17)
ℒ
P
=
1
2
(
D
μ
ψ
ν
*
-
D
ν
ψ
μ
*
)
(
D
μ
ψ
ν
-
D
ν
ψ
μ
)
-
m
2
ψ
ν
*
ψ
ν
.The equation of motion is(18)
(
D
μ
D
μ
+
m
2
)
ψ
ν
=
0
with
D
μ
ψ
μ
=
0
.Each component ofψ
μ obey a Klein-Gordon type equation, so that the energy states are [31]
(19)
E
=
{
p
→
2
+
m
2
(
q
=
0
)
p
3
2
+
m
2
+
(
2
ν
+
1
)
|
q
|
B
(
q
≠
0
)
;
ψ
μ has 4 components, but one of the equations is a compressed constraint equation, which means that only 3 components of ψ
μ are independent. So each energy state has 3 polarizations in the case with zero charge (or without magnetic field). If the charge is different from zero (and we have the presence of an external magnetic field), there is an additional constraint for the ν
=
0 energy level, which leads to the following degeneracy for the energy states:
(20)
γ
=
{
3
(
q
=
0
)
3
-
δ
ν
0
(
q
≠
0
)
.


### 2.5. Thermodynamics

We next outline some of the basic steps to obtain the baryonic and mesonic densities. The partition function in the Grand Canonical formalism is given by(21)
Z
=
Tr
⁡
[
e
-
β
(
∑
h
H
^
h
-
∑
h
μ
h
N
^
h
)
]
=
Z
f
Z
b
,
where the sum in h is over all the hadrons in consideration, both fermions (f) and bosons (b), μ
h is the chemical potential of the hadron h, H
^
h and N
^
h are the hamiltonian and number operators, respectively, and β
=
1
/
T.To find the trace, we do the standard procedure, dividing the particles into fermions and bosons. The result is(22)
Z
f
=
e
β
∑
f
,
r
E
r
f
∏
f
,
r
‍
(
1
+
e
-
β
(
E
r
f
-
μ
f
)
)
∏
f
,
r
‍
(
1
+
e
-
β
(
E
r
f
+
μ
f
)
)
,
Z
b
=
e
β
∑
b
,
r
E
r
b
∏
b
,
r
‍
1
1
-
e
-
β
(
E
r
b
-
μ
b
)
∏
b
,
r
‍
1
1
-
e
-
β
(
E
r
b
+
μ
b
)
,
where E
r
h
,
h
=
f
,
b is the total energy of the particle h and r labels its quantum state; that is, r
=
s
,
p
x
,
p
y
,
p
z (spin and momentum in the x-, y-, and z-axis) for particles without charge (q
=
0) and r
=
n
,
s
,
p
y
,
p
z (energy level, spin, and momentum in the y-, and z-axis) for particles with charge (q
≠
0).The Grand Canonical potential isΩ
h
=
-
(
1
/
β
)
ln
⁡
Z
h for h
=
f
,
b so that
(23)
Ω
f
=
-
∑
f
,
r
‍
E
r
f
-
1
β
∑
f
,
r
‍
ln
⁡
(
1
+
e
-
β
(
E
r
f
-
μ
f
)
)
-
1
β
∑
f
,
r
‍
ln
⁡
(
1
+
e
-
β
(
E
r
f
+
μ
f
)
)
,
Ω
b
=
-
∑
b
,
r
‍
E
r
b
+
1
β
∑
b
,
r
‍
ln
⁡
(
1
-
e
-
β
(
E
r
b
-
μ
b
)
)
+
1
β
∑
b
,
r
‍
ln
⁡
(
1
-
e
-
β
(
E
r
b
+
μ
b
)
)
.Using theparticle in the box prescription:
(24)
∑
r
‍
=
∑
s
,
p
x
,
p
y
,
p
z
‍
⇒
V
γ
h
1
(
2
π
)
3
∫
-
∞
∞
‍
d
3
p
(
q
=
0
)
,
∑
r
‍
=
∑
n
,
s
,
p
y
,
p
z
‍
⇒
V
γ
h
|
q
|
B
(
2
π
)
2
∫
-
∞
∞
‍
d
p
z
,
(
q
≠
0
)
,
where V is the total volume occupied by the system, γ
h and |
q
| are, respectively, the degeneracy and the charge modulus of particle h, and B is the value of the external magnetic field.With the correct expressions for the energy(
E
h
) and degeneracy (
γ
h
) of the hadron h, doing ω
h
=
Ω
h
/
V and utilizing the relations,
(25)
ρ
h
=
-
∂
ω
h
∂
μ
h
,
P
h
=
-
ω
h
,
s
h
=
∂
P
h
∂
T
,
ϵ
h
=
ω
h
+
T
s
+
μ
h
ρ
h
,
we can obtain all the thermodynamical quantities of interest. The particle densities for the baryons are
(26)
ρ
b
=
{
γ
b
1
2
π
2
∫
0
∞
‍
f
(
E
b
-
μ
b
)
p
2
d
p
(
q
=
0
)
∑
ν
=
0
∞
‍
γ
b
|
q
b
|
B
2
π
2
∫
0
∞
‍
f
(
E
b
-
μ
b
)
d
p
(
q
≠
0
)
,
for the antibaryons are
(27)
ρ
a
b
=
{
γ
b
1
2
π
2
∫
0
∞
‍
f
(
E
b
+
μ
b
)
p
2
d
p
(
q
=
0
)
∑
ν
=
0
∞
‍
γ
b
|
q
b
|
B
2
π
2
∫
0
∞
‍
f
(
E
b
+
μ
b
)
d
p
(
q
≠
0
)
,
where γ
b gives the degeneracy of each particle. For the mesons, the particle densities read
(28)
ρ
m
=
{
γ
m
1
2
π
2
∫
0
∞
‍
b
(
E
m
-
μ
m
)
p
2
d
p
(
q
=
0
)
∑
ν
=
0
∞
‍
γ
m
|
q
m
|
B
2
π
2
∫
0
∞
‍
b
(
E
m
-
μ
m
)
d
p
(
q
≠
0
)
,
with f
(
x
)
=
(
e
x
/
T
+
1
)
-
1 and b
(
x
)
=
(
e
x
/
T
-
1
)
-
1, and γ
m stands for the degeneracy of each particle. Note that antimesons are not taken into account because these antiparticles are written as other particles. So, in our notation, the antiboson sector of the formalism drops out.The total baryonic particle density is(29)
ρ
B
=
∑
b
‍
(
ρ
b
-
ρ
a
b
)
,
and the total mesonic density is
(30)
ρ
M
=
∑
m
‍
ρ
m
.The energy density is given by the sum of the energy densities of each particle, so(31)
ϵ
=
∑
b
‍
(
ϵ
b
+
ϵ
a
b
)
+
∑
m
‍
ϵ
m
,
with
(32)
ϵ
b
=
{
γ
b
1
2
π
2
∫
0
∞
‍
E
b
f
(
E
b
-
μ
b
)
p
2
d
p
(
q
=
0
)
∑
ν
=
0
∞
‍
γ
b
|
q
b
|
B
2
π
2
∫
0
∞
‍
E
b
f
(
E
b
-
μ
b
)
d
p
(
q
≠
0
)
,
ϵ
a
b
=
{
γ
b
1
2
π
2
∫
0
∞
‍
E
b
f
(
E
b
+
μ
b
)
p
2
d
p
(
q
=
0
)
∑
ν
=
0
∞
‍
γ
b
|
q
b
|
B
2
π
2
∫
0
∞
‍
E
b
f
(
E
b
+
μ
b
)
d
p
(
q
≠
0
)
,
ϵ
m
=
{
γ
m
1
2
π
2
∫
0
∞
‍
E
m
b
(
E
m
-
μ
m
)
p
2
d
p
(
q
=
0
)
∑
ν
=
0
∞
‍
γ
m
|
q
m
|
B
2
π
2
∫
0
∞
‍
E
m
b
(
E
m
-
μ
m
)
d
p
(
q
≠
0
)
.
The pressure is given by
(33)
P
=
∑
b
‍
(
P
b
+
P
a
b
)
+
∑
m
‍
P
m
,
with
(34)
P
b
=
{
γ
b
1
6
π
2
∫
0
∞
‍
1
E
b
f
(
E
b
-
μ
b
)
p
4
d
p
(
q
=
0
)
∑
ν
=
0
∞
‍
γ
b
|
q
b
|
B
2
π
2
∫
0
∞
‍
1
E
b
f
(
E
b
-
μ
b
)
p
2
d
p
(
q
≠
0
)
,
P
a
b
=
{
γ
b
1
6
π
2
∫
0
∞
‍
1
E
b
f
(
E
b
+
μ
b
)
p
4
d
p
(
q
=
0
)
∑
ν
=
0
∞
‍
γ
b
|
q
b
|
B
2
π
2
∫
0
∞
‍
1
E
b
f
(
E
b
+
μ
b
)
p
2
d
p
(
q
≠
0
)
,
P
m
=
{
γ
m
1
6
π
2
∫
0
∞
‍
1
E
m
b
(
E
m
-
μ
m
)
p
4
d
p
(
q
=
0
)
∑
ν
=
0
∞
‍
γ
m
|
q
m
|
B
2
π
2
∫
0
∞
‍
1
E
m
b
(
E
m
-
μ
m
)
p
2
d
p
(
q
≠
0
)
;
the entropy density can be found through
(35)
s
=
ϵ
+
P
-
∑
b
‍
μ
b
(
ρ
b
-
ρ
a
b
)
-
∑
m
‍
μ
m
ρ
m
.Note that, in the above equations,p and E are the momentum and energy of the particles in the hadrons gas, which are integrated over all the possible values of the momentum. They are, obviously, not the measured experimental energies and transverse momenta.


### 2.6.  Chemical Potential

The hadron chemical potential is(36)
μ
h
=
B
h
μ
B
+
I
3
h
μ
I
3
+
S
h
μ
S
,
where B
h, I
3
h, and S
h are, respectively, the baryonic number, the third isospin component, and the strangeness of the particle h, taken from the Particle Data Group [38].We impose the local conservation of the baryonic number, isospin, and strangeness. These impositions lead to the following equations:(37)
∑
h
‍
B
h
ρ
h
=
N
B
V
,
∑
h
‍
I
3
h
ρ
h
=
I
3
V
,
∑
h
‍
S
h
ρ
h
=
S
V
,
where N
B is the total baryonic number, I
3 is the total isospin, S is the total strangeness of the system, and V are the volume occupied by the system. The baryonic chemical potential μ
B is a free parameter of the system (the other is the temperature T). The isospin chemical potential μ
I
3 and the strangeness chemical potential μ
S are determined through their respective conservation laws. The charge conservation is automatically achieved.The baryonic number of an Au atom isN
B
=
(
N
+
Z
)
=
79
+
118
=
197, the isospin is I
3
=
(
Z
-
N
)
/
2
=
19.5, and for the deuteron (d) we have that N
B
=
1
+
1
=
2 and I
3
=
0. Hence, assuming that the total strangeness of the system is zero, we write the following table for the conserved quantities:(i)
Au + Au collision,N
B
=
394, I
3
=
-
39, S
=
0.
(ii)
d + Au collision, N
B
=
199, I
3
=
-
19.5, S
=
0.Hence, our code deals with5 unknowns (μ
B, μ
I
3, μ
S, T, V) and 3 constrained equations. We run over the values of μ
B and T (the chosen free parameters) in order to find the smallest χ
2.At this point, it is important to emphasize some of the drawbacks of our simple calculation. As shown in [39], the magnetic field should depend on the charges of the colliding nuclei and the number of participants should vary for different centralities. These constraints were not taken into account directly in our calculations. All the information we use as input come from the experimental particle yields and the magnetic field is modified until the best fitting is encountered. The number of different participants is reflected only in the resulting radii.


## 2.1. Spin 1/2 Baryons

The baryons with spin 1/2 are described by the Dirac Lagrangian density [34]
(6)
ℒ
D
=
ψ
¯
(
i
γ
μ
D
μ
-
m
)
ψ
,
which (after we apply the Euler-Lagrange equation) lead us to the equation of motion
(7)
(
i
γ
μ
D
μ
-
m
)
ψ
=
0
,
where γ
μ are the Dirac matrices.The solution of the equation of motion gives the following [30]:
(8)
E
=
{
p
→
2
+
m
2
(
q
=
0
)
p
3
2
+
m
2
+
2
ν
|
q
|
B
(
q
≠
0
)
,
where ν runs over the possible Landau Levels and the degeneracy for the energy states is given by
(9)
γ
=
{
2
(
q
=
0
)
2
-
δ
ν
0
(
q
≠
0
)
.


## 2.2. Spin 3/2 Baryons

The baryons with spin 3/2 are described by the Rarita-Schwinger Lagrangian density [35, 36]
(10)
ℒ
RS
=
-
1
2
ψ
¯
μ
(
ϵ
μ
ν
ρ
σ
γ
5
γ
ν
D
ρ
+
i
m
σ
μ
σ
)
ψ
σ
,
where γ
5
=
i
γ
0
γ
1
γ
2
γ
3 and σ
μ
ν
=
(
i
/
2
)
[
γ
μ
,
γ
ν
].The equation of motion reads(11)
(
i
γ
μ
D
μ
-
m
)
ψ
ν
=
0
with
γ
μ
ψ
μ
=
0
,
D
μ
ψ
μ
=
0
.The solution of the Rarita-Schwinger equation is not trivial and poses noncausality problems. To obtain the degeneracy of the energy states, we follow the prescription used in [34], which is given in detail for the Rarita-Schwinger equation in [32]. Observing the equation of motion, one can see that each component of ψ
μ obeys a Dirac type equation, so the energy must have the form
(12)
E
=
{
p
→
2
+
m
2
(
q
=
0
)
p
3
2
+
m
2
+
2
ν
|
q
|
B
(
q
≠
0
)
.Besides that,ψ
μ has 4 components, but two equations are constrained, which means that only 2 components of ψ
μ are really independent. So ψ
μ have 2 polarizations, but (because of the Dirac equation solution) each polarization is double degenerate. In the presence of a magnetic field, there is another constraint for the ν
=
0 and ν
=
1 energy levels, which leads to the following degeneracy for the energy states:
(13)
γ
=
{
4
(
q
=
0
)
4
-
2
δ
ν
0
-
δ
ν
1
(
q
≠
0
)
.


## 2.3. Spin 0 Mesons

The mesons with spin 0 are described by the Klein-Gordon Lagrangian density [37]
(14)
ℒ
KG
=
D
μ
ψ
*
D
μ
ψ
-
m
2
ψ
*
ψ
,
whose equation of motion is given by
(15)
(
D
μ
D
μ
+
m
2
)
ψ
=
0
,
with the energy satisfying the relation [30]:
(16)
E
=
{
p
→
2
+
m
2
(
q
=
0
)
p
3
2
+
m
2
+
(
2
ν
+
1
)
|
q
|
B
(
q
≠
0
)
.


## 2.4. Spin 1 Mesons

The mesons with spin 1 are described by the Proca Lagrangian density [31]
(17)
ℒ
P
=
1
2
(
D
μ
ψ
ν
*
-
D
ν
ψ
μ
*
)
(
D
μ
ψ
ν
-
D
ν
ψ
μ
)
-
m
2
ψ
ν
*
ψ
ν
.The equation of motion is(18)
(
D
μ
D
μ
+
m
2
)
ψ
ν
=
0
with
D
μ
ψ
μ
=
0
.Each component ofψ
μ obey a Klein-Gordon type equation, so that the energy states are [31]
(19)
E
=
{
p
→
2
+
m
2
(
q
=
0
)
p
3
2
+
m
2
+
(
2
ν
+
1
)
|
q
|
B
(
q
≠
0
)
;
ψ
μ has 4 components, but one of the equations is a compressed constraint equation, which means that only 3 components of ψ
μ are independent. So each energy state has 3 polarizations in the case with zero charge (or without magnetic field). If the charge is different from zero (and we have the presence of an external magnetic field), there is an additional constraint for the ν
=
0 energy level, which leads to the following degeneracy for the energy states:
(20)
γ
=
{
3
(
q
=
0
)
3
-
δ
ν
0
(
q
≠
0
)
.


## 2.5. Thermodynamics

We next outline some of the basic steps to obtain the baryonic and mesonic densities. The partition function in the Grand Canonical formalism is given by(21)
Z
=
Tr
⁡
[
e
-
β
(
∑
h
H
^
h
-
∑
h
μ
h
N
^
h
)
]
=
Z
f
Z
b
,
where the sum in h is over all the hadrons in consideration, both fermions (f) and bosons (b), μ
h is the chemical potential of the hadron h, H
^
h and N
^
h are the hamiltonian and number operators, respectively, and β
=
1
/
T.To find the trace, we do the standard procedure, dividing the particles into fermions and bosons. The result is(22)
Z
f
=
e
β
∑
f
,
r
E
r
f
∏
f
,
r
‍
(
1
+
e
-
β
(
E
r
f
-
μ
f
)
)
∏
f
,
r
‍
(
1
+
e
-
β
(
E
r
f
+
μ
f
)
)
,
Z
b
=
e
β
∑
b
,
r
E
r
b
∏
b
,
r
‍
1
1
-
e
-
β
(
E
r
b
-
μ
b
)
∏
b
,
r
‍
1
1
-
e
-
β
(
E
r
b
+
μ
b
)
,
where E
r
h
,
h
=
f
,
b is the total energy of the particle h and r labels its quantum state; that is, r
=
s
,
p
x
,
p
y
,
p
z (spin and momentum in the x-, y-, and z-axis) for particles without charge (q
=
0) and r
=
n
,
s
,
p
y
,
p
z (energy level, spin, and momentum in the y-, and z-axis) for particles with charge (q
≠
0).The Grand Canonical potential isΩ
h
=
-
(
1
/
β
)
ln
⁡
Z
h for h
=
f
,
b so that
(23)
Ω
f
=
-
∑
f
,
r
‍
E
r
f
-
1
β
∑
f
,
r
‍
ln
⁡
(
1
+
e
-
β
(
E
r
f
-
μ
f
)
)
-
1
β
∑
f
,
r
‍
ln
⁡
(
1
+
e
-
β
(
E
r
f
+
μ
f
)
)
,
Ω
b
=
-
∑
b
,
r
‍
E
r
b
+
1
β
∑
b
,
r
‍
ln
⁡
(
1
-
e
-
β
(
E
r
b
-
μ
b
)
)
+
1
β
∑
b
,
r
‍
ln
⁡
(
1
-
e
-
β
(
E
r
b
+
μ
b
)
)
.Using theparticle in the box prescription:
(24)
∑
r
‍
=
∑
s
,
p
x
,
p
y
,
p
z
‍
⇒
V
γ
h
1
(
2
π
)
3
∫
-
∞
∞
‍
d
3
p
(
q
=
0
)
,
∑
r
‍
=
∑
n
,
s
,
p
y
,
p
z
‍
⇒
V
γ
h
|
q
|
B
(
2
π
)
2
∫
-
∞
∞
‍
d
p
z
,
(
q
≠
0
)
,
where V is the total volume occupied by the system, γ
h and |
q
| are, respectively, the degeneracy and the charge modulus of particle h, and B is the value of the external magnetic field.With the correct expressions for the energy(
E
h
) and degeneracy (
γ
h
) of the hadron h, doing ω
h
=
Ω
h
/
V and utilizing the relations,
(25)
ρ
h
=
-
∂
ω
h
∂
μ
h
,
P
h
=
-
ω
h
,
s
h
=
∂
P
h
∂
T
,
ϵ
h
=
ω
h
+
T
s
+
μ
h
ρ
h
,
we can obtain all the thermodynamical quantities of interest. The particle densities for the baryons are
(26)
ρ
b
=
{
γ
b
1
2
π
2
∫
0
∞
‍
f
(
E
b
-
μ
b
)
p
2
d
p
(
q
=
0
)
∑
ν
=
0
∞
‍
γ
b
|
q
b
|
B
2
π
2
∫
0
∞
‍
f
(
E
b
-
μ
b
)
d
p
(
q
≠
0
)
,
for the antibaryons are
(27)
ρ
a
b
=
{
γ
b
1
2
π
2
∫
0
∞
‍
f
(
E
b
+
μ
b
)
p
2
d
p
(
q
=
0
)
∑
ν
=
0
∞
‍
γ
b
|
q
b
|
B
2
π
2
∫
0
∞
‍
f
(
E
b
+
μ
b
)
d
p
(
q
≠
0
)
,
where γ
b gives the degeneracy of each particle. For the mesons, the particle densities read
(28)
ρ
m
=
{
γ
m
1
2
π
2
∫
0
∞
‍
b
(
E
m
-
μ
m
)
p
2
d
p
(
q
=
0
)
∑
ν
=
0
∞
‍
γ
m
|
q
m
|
B
2
π
2
∫
0
∞
‍
b
(
E
m
-
μ
m
)
d
p
(
q
≠
0
)
,
with f
(
x
)
=
(
e
x
/
T
+
1
)
-
1 and b
(
x
)
=
(
e
x
/
T
-
1
)
-
1, and γ
m stands for the degeneracy of each particle. Note that antimesons are not taken into account because these antiparticles are written as other particles. So, in our notation, the antiboson sector of the formalism drops out.The total baryonic particle density is(29)
ρ
B
=
∑
b
‍
(
ρ
b
-
ρ
a
b
)
,
and the total mesonic density is
(30)
ρ
M
=
∑
m
‍
ρ
m
.The energy density is given by the sum of the energy densities of each particle, so(31)
ϵ
=
∑
b
‍
(
ϵ
b
+
ϵ
a
b
)
+
∑
m
‍
ϵ
m
,
with
(32)
ϵ
b
=
{
γ
b
1
2
π
2
∫
0
∞
‍
E
b
f
(
E
b
-
μ
b
)
p
2
d
p
(
q
=
0
)
∑
ν
=
0
∞
‍
γ
b
|
q
b
|
B
2
π
2
∫
0
∞
‍
E
b
f
(
E
b
-
μ
b
)
d
p
(
q
≠
0
)
,
ϵ
a
b
=
{
γ
b
1
2
π
2
∫
0
∞
‍
E
b
f
(
E
b
+
μ
b
)
p
2
d
p
(
q
=
0
)
∑
ν
=
0
∞
‍
γ
b
|
q
b
|
B
2
π
2
∫
0
∞
‍
E
b
f
(
E
b
+
μ
b
)
d
p
(
q
≠
0
)
,
ϵ
m
=
{
γ
m
1
2
π
2
∫
0
∞
‍
E
m
b
(
E
m
-
μ
m
)
p
2
d
p
(
q
=
0
)
∑
ν
=
0
∞
‍
γ
m
|
q
m
|
B
2
π
2
∫
0
∞
‍
E
m
b
(
E
m
-
μ
m
)
d
p
(
q
≠
0
)
.
The pressure is given by
(33)
P
=
∑
b
‍
(
P
b
+
P
a
b
)
+
∑
m
‍
P
m
,
with
(34)
P
b
=
{
γ
b
1
6
π
2
∫
0
∞
‍
1
E
b
f
(
E
b
-
μ
b
)
p
4
d
p
(
q
=
0
)
∑
ν
=
0
∞
‍
γ
b
|
q
b
|
B
2
π
2
∫
0
∞
‍
1
E
b
f
(
E
b
-
μ
b
)
p
2
d
p
(
q
≠
0
)
,
P
a
b
=
{
γ
b
1
6
π
2
∫
0
∞
‍
1
E
b
f
(
E
b
+
μ
b
)
p
4
d
p
(
q
=
0
)
∑
ν
=
0
∞
‍
γ
b
|
q
b
|
B
2
π
2
∫
0
∞
‍
1
E
b
f
(
E
b
+
μ
b
)
p
2
d
p
(
q
≠
0
)
,
P
m
=
{
γ
m
1
6
π
2
∫
0
∞
‍
1
E
m
b
(
E
m
-
μ
m
)
p
4
d
p
(
q
=
0
)
∑
ν
=
0
∞
‍
γ
m
|
q
m
|
B
2
π
2
∫
0
∞
‍
1
E
m
b
(
E
m
-
μ
m
)
p
2
d
p
(
q
≠
0
)
;
the entropy density can be found through
(35)
s
=
ϵ
+
P
-
∑
b
‍
μ
b
(
ρ
b
-
ρ
a
b
)
-
∑
m
‍
μ
m
ρ
m
.Note that, in the above equations,p and E are the momentum and energy of the particles in the hadrons gas, which are integrated over all the possible values of the momentum. They are, obviously, not the measured experimental energies and transverse momenta.


## 2.6.  Chemical Potential

The hadron chemical potential is(36)
μ
h
=
B
h
μ
B
+
I
3
h
μ
I
3
+
S
h
μ
S
,
where B
h, I
3
h, and S
h are, respectively, the baryonic number, the third isospin component, and the strangeness of the particle h, taken from the Particle Data Group [38].We impose the local conservation of the baryonic number, isospin, and strangeness. These impositions lead to the following equations:(37)
∑
h
‍
B
h
ρ
h
=
N
B
V
,
∑
h
‍
I
3
h
ρ
h
=
I
3
V
,
∑
h
‍
S
h
ρ
h
=
S
V
,
where N
B is the total baryonic number, I
3 is the total isospin, S is the total strangeness of the system, and V are the volume occupied by the system. The baryonic chemical potential μ
B is a free parameter of the system (the other is the temperature T). The isospin chemical potential μ
I
3 and the strangeness chemical potential μ
S are determined through their respective conservation laws. The charge conservation is automatically achieved.The baryonic number of an Au atom isN
B
=
(
N
+
Z
)
=
79
+
118
=
197, the isospin is I
3
=
(
Z
-
N
)
/
2
=
19.5, and for the deuteron (d) we have that N
B
=
1
+
1
=
2 and I
3
=
0. Hence, assuming that the total strangeness of the system is zero, we write the following table for the conserved quantities:(i)
Au + Au collision,N
B
=
394, I
3
=
-
39, S
=
0.
(ii)
d + Au collision, N
B
=
199, I
3
=
-
19.5, S
=
0.Hence, our code deals with5 unknowns (μ
B, μ
I
3, μ
S, T, V) and 3 constrained equations. We run over the values of μ
B and T (the chosen free parameters) in order to find the smallest χ
2.At this point, it is important to emphasize some of the drawbacks of our simple calculation. As shown in [39], the magnetic field should depend on the charges of the colliding nuclei and the number of participants should vary for different centralities. These constraints were not taken into account directly in our calculations. All the information we use as input come from the experimental particle yields and the magnetic field is modified until the best fitting is encountered. The number of different participants is reflected only in the resulting radii.


## 3. Results and Discussions

We have implemented anχ
2 fit in order to obtain the temperature and chemical potential. The particle properties (spin, mass, baryonic number, isospin, and strangeness) were taken from the Particle Data Group [38].In Tables1, 2, 3, and 4, we show our results corresponding to the temperature and chemical potential that give the minimum value for the quadratic deviation χ
2:
(38)
χ
2
=
∑
i
‍
(
ℛ
i
exp
⁡
-
ℛ
i
theo
)
2
σ
i
2
,
where ℛ
i
exp
⁡ and ℛ
i
theo are the ith particle ratio given experimentally and theoretically and σ
i represents the errors in the experimental data points.Table 1
Au + Au (70–80%)s
NN = 200 GeV. Results obtained for different values of the magnetic field. The lower χ
2 is given in boldface.
B (×1019 G)
0
0.1
0.5
1
2
5
10
STAR/RHIC [33]
e
B (m
π
2)
0
0.3
1.5
3
6
15
30
π
-
/
π
+ (Δ
%)
1.000 (0.30%)
1.000
1.000
1.000
1.000 (0.30%)
1.000
1.000
1.003 ± 0.044
K
-
/
K
+
0.993 (1.23%)
0.993
0.986
0.980
0.971 (0.99%)
0.966
0.939
0.981 ± 0.049
p
-
/
p
0.842 (0.13%)
0.843
0.834
0.839
0.837 (0.69%)
0.836
0.776
0.843  ±  0.048
K
-
/
π
-
0.169 (32.7%)
0.169
0.175
0.177
0.178 (40.1%)
0.182
0.218
0.127  ±  0.010
p
-
/
π
-
0.019 (77.8%)
0.019
0.026
0.033
0.037 (55.7%)
0.036
0.046
0.084  ±  0.007
K
+
/
π
+
0.170 (30.5%)
0.170
0.177
0.181
0.183 (40.9%)
0.189
0.232
0.130  ±  0.011
p
/
π
+
0.022 (77.9%)
0.023
0.031
0.039
0.044 (55.6%)
0.043
0.059
0.100  ±  0.013
T (MeV)
124
125
138
152
170
194
199
157.9  ±  3.9
μ
B (MeV)
11
11
13
14
16
19
20
14.1 ± 4.2
μ
S (MeV)
0.758
0.787
1.43
2.15
3.28
4.82
5.08
μ
I
3 (MeV)
−0.653
−0.660
−0.949
−1.22
−1.76
−3.24
−5.00
χ
2
/ndf
42.5
52.5
46.0
39.8
35.6
39.7
47.9
R (fm)
66.2
64.6
45.5
34.0
24.5
16.8
14.8
ρ
B (×10−2 fm−3)
0.0325
0.0349
0.100
0.239
0.643
2.00
2.91
ρ
Δ (×10−2 fm−3)
0.00813
0.00878
0.0276
0.0717
0.211
0.732
1.14
ρ
M (fm−3)
0.0918
0.0960
0.167
0.293
0.582
1.51
2.73
ρ
π (fm−3)
0.0636
0.0664
0.109
0.183
0.355
0.942
1.83
ϵ (MeV/fm3)
55.4
58.3
111
212
456
1203
1820
P (MeV/fm3)
11.2
11.8
22.6
43.7
97.4
283
490
s (MeV/fm3)
0.537
0.561
0.967
1.68
3.26
7.66
11.6Table 2
Au + Au (58–85%) collision ats
NN = 130 GeV. Results obtained for different values of the magnetic field. The lower χ
2 is given in boldface.
B (×1019 G)
0
0.1
0.5
1
2
5
10
STAR/RHIC [33]
e
B (m
π
2)
0
0.3
1.5
3
6
15
30
π
-
/
π
+ (Δ
%)
1.000 (0.40%)
1.000
1.000
1.000
1.000 (0.40%)
1.000
1.000
0.996 ± 0.066
K
-
/
K
+
0.985 (2.32%)
0.984
0.977
0.967
0.956 (0.71%)
0.947
0.952
0.963 ± 0.050
p
-
/
p
0.793 (0.35%)
0.783
0.788
0.787
0.792 (0.25%)
0.794
0.792
0.790 ± 0.043
K
-
/
π
-
0.196 (40.3%)
0.196
0.195
0.195
0.194 (38.53%)
0.199
0.201
0.140 ± 0.018
p
-
/
π
-
0.027 (67.6%)
0.027
0.033
0.040
0.045 (44.9%)
0.045
0.036
0.082 ± 0.010
K
+
/
π
+
0.199 (38.2%)
0.200
0.200
0.202
0.203 (40.9%)
0.210
0.211
0.144 ± 0.016
p
/
π
+
0.034 (67.4%)
0.035
0.042
0.050
0.057 (44.6%)
0.057
0.046
0.103 ± 0.012
T (MeV)
133
134
145
159
178
206
217
159-
7
+
11
μ
B (MeV)
16
17
18
20
22
26
29
19.9 ± 4.9
μ
S (MeV)
1.52
1.67
2.39
3.55
5.07
7.38
8.42
μ
I
3 (MeV)
−1.10
−1.18
−1.41
−1.79
−2.41
−4.26
−6.92
χ
2
/ndf
17.2
21.2
17.9
14.8
12.7
14.2
18.1
R (fm)
47.1
45.2
35.5
26.9
19.8
13.4
11.1
ρ
B (×10−2 fm−3)
0.0899
0.102
0.211
0.486
1.22
3.95
6.92
ρ
Δ (×10−2 fm−3)
0.0243
0.0273
0.0610
0.151
0.415
1.50
2.82
ρ
M (fm−3)
0.127
0.132
0.207
0.352
0.693
1.84
3.49
ρ
π (fm−3)
0.0813
0.0845
0.127
0.208
0.398
1.07
2.16
ϵ (MeV/fm3)
86.4
90.6
152
282
608
1704
2863
P (MeV/fm3)
17.0
17.8
30.2
56.6
126
385
729
s (MeV/fm3)
0.777
0.809
1.26
2.13
4.12
10.1
16.5Table 3
Au + Au (70–80%) collision ats
NN = 62.4 GeV. Results obtained for different values of the magnetic field. The lower χ
2 is given in boldface.
B (×1019 G)
0
0.1
0.5
1
2
5
10
STAR/RHIC [33]
e
B (m
π
2)
0
0.3
1.5
3
6
15
30
π
-
/
π
+ (Δ
%)
1.000 (1.19%)
1.000
1.000
1.000
1.000 (1.19%)
1.000
1.000
1.012 ± 0.031
K
-
/
K
+
0.992 (5.97%)
0.991
0.982
0.971
0.958 (2.34%)
0.959
0.972
0.936 ± 0.036
p
-
/
p
0.621 (0.27%)
0.624
0.623
0.625
0.625 (0.26%)
0.623
0.618
0.623 ± 0.047
K
-
/
π
-
0.128 (17.0%)
0.128
0.130
0.132
0.132 (20.9%)
0.131
0.129
0.109 ± 0.009
p
-
/
π
-
0.008 (87.1%)
0.008
0.011
0.014
0.016 (74.7%)
0.013
0.010
0.063 ± 0.007
K
+
/
π
+
0.129 (8.99%)
0.129
0.133
0.136
0.138 (16.6%)
0.137
0.132
0.118 ± 0.010
p
/
π
+
0.013 (87.1%)
0.013
0.018
0.023
0.026 (74.7%)
0.021
0.016
0.101 ± 0.013
T (MeV)
111
112
123
135
149
162
193
154-
6
+
8
μ
B (MeV)
27
27
30
33
37
42
45
37.7 ± 6.5
μ
S (MeV)
1.04
1.09
1.97
3.24
5.06
6.93
7.85
μ
I
3 (MeV)
−1.17
−1.20
−1.75
−2.49
−3.83
−7.28
−11.3
χ
2
/ndf
23.0
28.5
26.0
23.7
22.4
24.9
27.5
R (fm)
70.4
68.4
48.6
35.4
25.7
19.3
17.3
ρ
B (×10−2 fm−3)
0.0269
0.0295
0.0821
0.212
0.552
1.31
1.81
ρ
Δ (×10−2 fm−3)
0.00584
0.00643
0.0199
0.0566
0.162
0.430
0.648
ρ
M (fm−3)
0.0557
0.0587
0.103
0.183
0.361
0.871
1.59
ρ
π (fm−3)
0.0429
0.0451
0.0758
0.131
0.257
0.646
1.24
ϵ (MeV/fm3)
28.1
29.8
54.4
102
206
444
687
P (MeV/fm3)
5.97
6.33
11.9
22.8
48.9
122
214
s (MeV/fm3)
0.307
0.322
0.539
0.922
1.71
3.49
5.52Table 4
d + Au (40–100%) collision at s
NN = 200 GeV. Results obtained for different values of the magnetic field. The lower χ
2 is given in boldface.
B (×1019 G)
0
0.1
0.5
1
2
5
10
STAR/RHIC [33]
e
B (m
π
2)
0
0.3
1.5
3
6
15
30
π
-
/
π
+ (Δ
%)
1.000 (0.79%)
1.000
1.000
1.000
1.000 (0.79%)
1.000
1.000
1.008 ± 0.042
K
-
/
K
+
0.996 (1.98%)
0.996
0.992
0.988
0.981 (0.43%)
0.981
0.987
0.977 ± 0.037
p
-
/
p
0.843 (0.18%)
0.844
0.844
0.847
0.843 (0.22%)
0.842
0.841
0.841 ± 0.067
K
-
/
π
-
0.137 (14.3%)
0.137
0.140
0.143
0.146 (21.8%)
0.146
0.142
0.120 ± 0.011
p
-
/
π
-
0.011 (86.3%)
0.012
0.015
0.020
0.023 (71.8%)
0.019
0.014
0.082 ± 0.010
K
+
/
π
+
0.138 (11.0%)
0.138
0.141
0.145
0.149 (20.1%)
0.149
0.144
0.124 ± 0.012
p
/
π
+
0.013 (86.4%)
0.014
0.018
0.024
0.027 (71.0%)
0.023
0.017
0.098 ± 0.014
T (MeV)
114
115
126
139
155
170
171
159-
7
+
10
μ
B (MeV)
10
10
11
12
14
16
17
16.5 ± 6.5
μ
S (MeV)
0.443
0.464
0.807
1.32
2.18
3.00
3.26
μ
I
3 (MeV)
−0.467
−0.475
−0.671
−0.939
−1.49
−2.80
−4.33
χ
2
/ndf
18.2
22.5
20.5
18.6
17.5
19.4
21.7
R (fm)
71.5
69.5
50.2
36.4
25.6
18.9
17.1
ρ
B (×10−2 fm−3)
0.0130
0.0141
0.0375
0.0982
0.283
0.700
0.944
ρ
Δ (×10−2 fm−3)
0.00292
0.00320
0.00936
0.0270
0.0859
0.237
0.344
ρ
M (fm−3)
0.0627
0.0660
0.113
0.204
0.415
1.00
1.80
ρ
π (fm−3)
0.0472
0.0495
0.0818
0.142
0.283
0.715
1.36
ϵ (MeV/fm3)
33.0
34.9
62.9
121
259
573
857
P (MeV/fm3)
6.93
7.33
13.5
26.6
59.6
151
258
s (MeV/fm3)
0.350
0.367
0.606
1.06
2.06
4.25
6.52In Tables1, 2, 3, and 4, B is the magnetic field, T is the temperature, μ
B is the baryonic chemical potential, χ
2 is quadratic deviation, μ
I
3 is the isospin chemical potential, μ
S is the strangeness chemical potential, R is the radius of the “fire-ball”, ρ
B
=
∑
b
‍
(
ρ
b
-
ρ
a
b
) is the usual baryonic density, ρ
Δ
=
ρ
Δ
+
+
-
ρ
Δ
¯
+
+
+
ρ
Δ
+
-
ρ
Δ
¯
+
+
ρ
Δ
0
-
ρ
Δ
¯
0
+
ρ
Δ
-
-
ρ
Δ
¯
- is delta baryon density, ρ
M
=
∑
m
‍
ρ
m is the meson density, ρ
π
=
ρ
π
0
+
ρ
π
+
+
ρ
π
- is the pion density, ϵ is the energy density, P is the pressure, s is the entropy density, and ndf is the number of degrees of freedom. For B
=
0, ndf
=
5 (7 experimental values minus 2 free parameters, T and μ); for B
≠
0, ndf
=
4 (7 experimental values minus 3 free parameters, T, μ, and B). π
-
/
π
+, K
-
/
K
+, p
¯
/
p, K
-
/
π
-, K
+
/
π
+, and p
/
π
+ are the theoretical (the first 7 columns) and experimental (the last column) particle ratios [33]. The temperatures and baryonic chemical potentials from the statistical model with the excluded volume approximation obtained in [33] are also given in the last columns of all tables. It is important to point out that, had we considered excluded volume effects in our calculations, the χ
2 fitting would also give results close to 1. To make clear the improvement in the data fitting by the addition of the magnetic field, we calculate the relative percent deviation (
Δ
%
) with respect to the experimental values for B
=
0 and the best B
≠
0 (the bold columns in the tables) through
(39)
Δ
%
=
|
ℛ
theo
-
ℛ
exp
⁡
ℛ
exp
⁡
|
·
100
%
,
and show these values in parenthesis in all the tables. If one compares the relative deviation percentage of the results given in bold, obtained from the lower value of χ
2, with the ones obtained for zero magnetic field (the first columns in all tables), it is clear that, with very few exceptions, the inclusion of the magnetic field produces particle ratios much closer to the experimental results. It is worth mentioning that the larger the electric charge of the particle, the stronger the influence of the magnetic field on its density (see (26)–(28)). However, as we are mainly interested in particle ratios, this fact cannot be observed from the results displayed in Tables 1–4. Moreover, other properties of individual particles, as their spectrum for instance, result from a delicate balance between the filling of the Landau levels related to the particle mass [18, 19] and the interaction that governs the particle production. In this work, the strong interaction is neglected since we are dealing with free gases.In Figures1(a), 1(b), 2(a), 2(b), 3(a), 3(b), 4(a), and 4(b), we plot the experimental and theoretical ratios for B
=
0 and for the B value that gives the best fitting. In Figures 1(c), 2(c), 3(c), and 4(c), we show the χ
2 behavior again for B
=
0 and for the B value that gives the best fitting. In Figures 1(d), 2(d), 3(d), and 4(d), we show the χ
2 behavior in function of the magnetic field. One can notice that the best fitting is generally obtained for magnetic fields around 6
m
π
2, a little higher than what is expected for RHIC collisions (5
m
π
2). It is surprising to see that the best fitting results are always obtained for the same value of the magnetic field.Au + Au (70–80%) collision ats
N
N = 200 GeV. (a) Particle/antiparticle ratios. (b) Mixed ratios. (c) and (d) χ
2 behavior.
(a)
(b)
(c)
(d)Au + Au (58–85%) collision ats
N
N = 130 GeV. (a) Particle/antiparticle ratios. (b) Mixed ratios. (c) and (d) χ
2 behavior.
(a)
(b)
(c)
(d)Au + Au (70–80%) collision ats
N
N = 62.4 GeV. (a) Particle/antiparticle ratios. (b) Mixed ratios. (c) and (d) χ
2 behavior.
(a)
(b)
(c)
(d)d + Au (40–100%) collision at s
N
N = 200 GeV. (a) Particle/antiparticle ratios. (b) Mixed ratios. (c) and (d) χ
2 behavior.
(a)
(b)
(c)
(d)Our results show that, even for the free Fermi and Boson gas models, a strong magnetic field plays an important role. The inclusion of the magnetic field improves the data fit-up to a field of the order ofB
=
1
0
19 G. For stronger magnetic fields, it becomes worse again. This behavior is easily observed in Tables 1–4 and in Figures 1(d)–4(d). It is worth pointing out how the “fireball” radius R and the total density ρ vary with the magnetic field in a systematic way: R and ρ practically do not change between B
=
0 and B
=
1
0
18 G, but when the field increases even further, the density increases and the radius decreases. This behavior is common to all collision cases studied. This huge jump in the density explains why the ratios get worse for a magnetic field of the order of B
=
1
0
20 G, for which the densities are much higher than what is expected in a heavy ion collision.Our model gives a good description for the particle/antiparticle ratios, but fails to describe the relation between pions and other particles. This occurs because our model produces too many pions as shown explicitly in the particle densities. In all collision types, our model presents a baryon density (ρ
B) with more than 30% of Δ baryons and a meson density (ρ
M) with more than 60% of π (ρ
π). The relative percent deviations in the particle yields show clearly that some results improve considerably when the magnetic field is considered, while others remain unaltered or even get slightly worse. However, our figures also show that the behavior of the χ
2 changes drastically with the addition of the magnetic field and that the temperature and chemical potentials calculated with the statistical model lie within the 3
-
σ confidence ellipse obtained for the best χ
2 in some cases, but they are always outside the confidence ellipses obtained with zero magnetic field.We would like to comment that, when we first started these calculations, we were not aware of [27, 28] and we used data obtained for low centralities, that is, low impact parameters. In that case, the minimum χ
2 was generally smaller than the ones shown in this work and we believe this was so because of the larger error bars accompanying data at low centralities.It is fair to point out that our model overestimates the effect of the magnetic field in heavy ion collisions. In [40], the authors show that the magnetic field drops up to four orders of magnitude from the initial field of the order of 5
m
π
2 until the freezeout is reached and the particle rations are established. Also, in [24], a calculation of the time dependence of the magnetic field including effects of QGP electrical conductivity and longitudinal expansion is performed. The author concludes that the magnetic field, which has a value close to 6
m
π
2 in the initial time, decreases by a factor of 100 between the initial and final (5 fm/c) times. With our rather simplistic model, we find a field value very close to the expected real one at the beginning of the collisions. As discussed above, by the time the particle ratios are obtained, the fields are supposed to decay considerably and if this is really the case, our calculations are more academic than realistic.Further improvements on the presented calculations are under investigation, namely, the inclusion of electric fields at low impact parameters and the variation of both electric and magnetic fields with the number of participants in the collisions. The temporal evolution of the system is certainly the most important aspect to be considered. Other possible improvements are the inclusion of the anomalous magnetic moments and the description of pion-pion interactions. Volume excluded corrections and resonance decays can also be considered. We next intend to repeat these calculations for the ALICE/LHC data for the future Pb + Pb runs with all these improvements, so that our results become more realistic.



---

*Source: 479401-2014-01-20.xml*