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Introduction to Relativistic Gases
Takeshi Kodama
Instituto de Física - Universidade Federal do Rio de Janeiro
INTRODUCTION
Some of the observed quantities in relativistic heavy-ion collisions, such as ratios of pro-
duced particles, transverse spectra, etc. [1], are often well described in terms of simple
thermal models. However, this fact does not necessarily mean that the whole system is in
thermal equilibrium, and the interpretation of these analyses should be done with care.
In this lecture, I would like to introduce some fundamental concepts of statistical physics
to derive thermodynamical properties of a relativistic gas, with the aim of improving the
understanding of the forthcoming lectures on advanced models discussed in this school.
Several expressions for thermodynamical quantities frequently used in thermal models
are derived, and it is shown how these quantities can be calculated in practice. Therefore,
this lecture note is a kind of short digest of more complete books on statistical physics
[2, 3].
GRAND CANONICAL ENSEMBLE
First, consider a small portion of hadronic matter (or quark-gluon plasma) formed in
a relativistic heavy-ion collision. When the time scale of space-time evolution of the
system is relatively slow compared to the microscopic time scale (such as the mean
collision time), a lot of different physical configurations appear within a significant
global time scale of the system. Thus, physical properties of this segment are basically
determined by the statistical average over all the microstates which appear within the
macroscopic time-scale.
Let us suppose that a microstate of our system is specified by the occupation numbers
α � �n1�n2� � � � �ni� � � � � � � � � �� � (1)
where ni is the number of particles which are in the i� th “single particle state”. For an
ideal gas in a box of volume V , we can take plane wave states, and the index i can be
identified as the wavenumber vector,
i��k
Note that neither α nor i are simple numbers, but (infinite) sets of numbers.
For one microstate α , we can write the energy Eα and the total number of particles
Nα as
Eα � ∑
i
ni εi� (2)
Nα � ∑
i
ni� (3)
where the sum extends over all single-particle states of the system.
The macroscopic state of the system, which is determined by a small number of
parameters, such as temperature, pressure, total energy, etc., usually does not specify
the microstate α . Microscopically, a lot of interactions among particles will take place
within a macroscopic time-lapse τh. Almost an infinite set of microstates appear and dis-
appear in this time scale. This (infinite) set of microstates relevant for one macroscopic
state is called ensemble. The basic question in equilibrium statistical mechanics is to
determine the probability of a given microstate α to appear in the ensemble.
Let pα be the probability corresponding to a microstate α . To determine pα , we argue
as follows. First, let us build the ensemble by preparing � systems of the same gas,
all of them in the same macroscopic state. Here, we assume� � 1 (in fact, we take
� � ∞). Each system should be in some microstate α . Let�α be the total number of
systems which are in the microstate α (also�α � 1 ). Obviously,
� � ∑
α
�α � (4)
and the probability pα is then given by
pα �
�α
�
� (5)
We have
∑
α
pα � 1� (6)
Depending the way such an ensemble of systems is prepared, they are not always
identical. We can construct an ensemble, for example, in which every system is exactly
in the same microstate α � α0. In this case, every�α is equal to zero except�α0 �� .
Another extreme is to prepare the ensemble in such a way that every microstate occurs
exactly the same number of times. In this case, we have the same number�α for all α .
Thus, the distribution of�α depends on how the system is prepared. We may specify
the state of the ensemble by this distribution of�α ,�
�α
1
��α
2
� � � � ��αi� � � �
�
�
where the ensemble contains �α
1
systems in the microstate α1, �α2 systems in the
microstate α2, etc.
Now, let us introduce a basic hypothesis. We suppose that in the equilibrium state of
the system, every microstate has the same a-priori probability. That is, every microstate
can appear in the same way as others, just as the numbers of an unbiased die. So,
we are considering a kind of die that has an infinite number of surfaces, each surface
corresponding to a microstate α . As we know very well, even if the probability is equal
for any number of a die, when we throw 2 dice, then the probabilities for the sum of
numbers of the 2 dice are not uniform. The probability of having a total number 7 is
larger than the probability of having 2. This is because with two numbers from 1 to
6 there are more ways to compose 7 than 2. Certain configurations can occur more
frequently than others just because there are more ways to realize such configurations.
In our case, the number of different ways to have a specific partition
�
�α
1
��α
2
� � � �
�
of an integer� is given by
W �
� !
�α
1
!�α
2
! � � ��αi! � � �
� (7)
We then expect that the equilibrium should correspond to the configuration of the
ensemble for which W is maximum. This state of the ensemble is more probable than
others.
• In equilibrium, the configuration
�
�α
1
��α
2
� � � � ��αi� � � �
�
is obtained by maxi-
mizing W .
It is easy to see that if we don’t have any more physical constraints, the maximum of W
is given simply by
�α
1
��α
2
�� � � ���αi � � � � �
that is, every microstate appears with the same probability. However, if we are consid-
ering a system where the average value of the total energy and the number of particles
are fixed, for example, then we have to maximize W (or equivalently lnW ) taking into
account two constraints. We have
δ lnW � 0� (8)
together with
δ�E�� 0� (9)
δ�N�� 0� (10)
where the average total energy is
�E�� 1
�
∑
α
Eα�α � (11)
and the average total number of particles is
�N�� 1
�
∑
α
Nα�α � (12)
Furthermore, we have to fix the total number of systems in the ensemble,
� � ∑
α
�α � (13)
Using the Lagrangian multiplier method, the configuration
�
�α
1
��α
2
� � � � � � �
�
is de-
termined by the condition
δ
�
1
�
lnW
�
�β δ�E��λ δ�N�� 0� (14)
or equivalently
δ
�
lnW �β ∑
α
Eα�α �λ ∑
α
Nα�α �γ�
�
� 0 � (15)
where β � λ and γ are Lagrange multipliers, and the variation should be taken with
respect to the change of the configuration ��α
1
��α
2
� � � � � � �	� that is, with respect to
the numbers of the systems in each microstate α in the ensemble.
Now, for�α � 1, we can make an approximation,
lnW � ln� !�∑
α
ln�α !
C�∑
α
�α �ln�α �1� � (16)
where C �� �ln� �1� is a constant and we have used the Stirling formula, lnN! 
N �lnN�1�. We get
∑
α
δ�α �ln�α �βEα �λ Nα �γ� � 0 (17)
for all δ�α so that
ln�α �βEα �λ Nα �γ � 0� (18)
Equivalently, we have
�α ��0 e
�βEα�λ Nα � (19)
where
�0 � e
�γ� (20)
Since
∑
α
�α �� � (21)
we have
�0 ∑
α
e�βEα�λ Nα �� � (22)
We define the partition function by
Z �V�β �µ� �∑
α
e�βEα�λ Nα � (23)
Here, the quantity V is the volume of the system. The dependence on V appears,
because the number of single particle-states depends on the volume of the system.
In the thermodynamical limit, V � ∞, this number is proportional to V (see the later
discussion).
We can determine β and λ from the conditions
�0
�
∑
α
Nα e
�βEα�λ Nα � �N� � (24)
�0
�
∑
α
Eα e
�βEα�λ Nα � �E� � (25)
In this way, in equilibrium, the probability of finding a given microstate α in the
ensemble is given by
pα �
�α
�
�
1
Z
e�β�Eα�µNα �� (26)
where
µ �
λ
β
� (27)
Remember that α is the index for distinguishing a microstate as introduced in Eq.(1),
and not a single number. In the equations above, the sum over α is in fact a sum over all
single particle configurations, �
n1�n2� � � � �ni� � � �
�
�
Consequently, we have
∑
α
� ∑
n1
∑
n2
∑
n3
� � �∑
ni
� � � � (28)
Equation (14) can be rewritten as
δ�E�� 1
β
δ
�
1
�
lnW
�
�µ δ�N�� (29)
The above relation shouldhold for any change ��α	 � �� �α	, fixing the parameters
β and µ . Therefore, this equation should be compared with the well-known thermody-
namical law
δ�E�� T δ�S��µ δ�N�� (30)
for fixed volume (δV � 0). We thus identify the temperature T� the chemical potential
µ , and the entropy S, respectively. Thus,
kT �
1
β
� (31)
S � k
�
1
�
lnW
�
� (32)
where k is the Boltzmann constant (here, it is introduced to adjust the dimension of T ).
If we use the relation
pa �
�α
�
� (33)
we have
S
 k 1
�
�
� �ln� �1��∑
α
pα� �ln�pα� ��1�
�
��k∑
α
pα ln pα �Const � (34)
We may take the constant above as zero, so that if the system is in a particular microstate
α � α0� that is,
pα �
	
1�
0�
α � α0
otherwise � (35)
we have zero entropy.
We can reformulate the variational principle in terms of the probability distribution
pα . Eq.(29) reads
δ∑
α
�pα Eα � kT pα ln pα �µNα pα �γpα	� 0� (36)
for arbitrary variation δpα . The last term in this equation is added to incorporate the
restriction
∑
α
pα � 1� (37)
Eq.(36) is read as
δ�E��T δ�S��µ δ�N��γ δ�1�� 0� (38)
where
�E�� ∑
α
pα Eα �
etc., and the last term
δ�1� � δ
�
∑
α
pα
�
is the change of the overall normalization of the total probability. We may interpret
Eq.(38) as minimizing the free energy,
F � E�T S�µN�
under the fixed temperature and chemical potential and the normalization. Of course, the
original form of Eq.(38) is
δ�S��β �δ�E��µ δ�N��γ δ�1�� � 0� (39)
that is, to maximize the entropy fixing the total average energy and total average number
of particles, together with the constraint of conserving the total probability.
Expressing the average values in terms of pα , we obtain from Eq.(38)
∑
α
δpα �Eα �µNα �γ� kT �ln pα �1�	� 0� (40)
and consequently,
pα �
1
Z
e��Eα�µNα ��kT � (41)
where
1
Z
� eγ�kT�1 (42)
is related to the normalization of the probability. Using Eq.(37), we obtain
Z � Z�V�T�µ� �∑
α
e��Eα�µNα ��kT � (43)
This function is known as the partition function for the grand canonical ensemble.
Once we know the partition function Z in terms of µ and T , we can calculate all the
thermodynamical quantities. We have
�E��� ∂ lnZ
∂β
µβ
� (44)
�N�� 1
β
∂ lnZ
∂µ
β
� (45)
�S���k �ln pα � (46)
�
1
T ∑α
�Eα �µNα � pα � k lnZ (47)
�
1
T
��E��µ�N��� k lnZ� (48)
Note that the partial derivative in Eq.(44) should be performed by fixing the quantity
λ � µβ . When the system is large enough and the general extensive thermodynamical
relation
�S�� 1
T
�E�� µ
T
�N�� 1
T
PV� (49)
is valid, then we identify
1
β
lnZ � PV� (50)
which is in fact the thermodynamical potential for a grand canonical ensemble.
Ideal Fermi Gas
Let us apply the above results to an ideal gas of Fermi-Dirac particles. For a Fermi
gas, due to the Pauli exclusion principle, the occupation number of particles ni for each
state i is limited to be 0 or 1. Thus, the partition function becomes
Z �V�β �µ� � ∑
α
e�β�Eα�µNα �
� ∑
n1
∑
n2
∑
n3
� � �∑
ni
� � �exp
�
�β ∑
i
ni
�
εi�µ

�
� ∏
i
∑
ni�0�1
e�βni�εi�µ�
� ∏
i
�
1� e�β�εi�µ�
�
� exp∑
i
ln
�
1� e�β�εi�µ�
�
� (51)
Reminding that the single particle state i for an ideal gas is taken as a plane wave state
of momentum�k, we may replace the sum over states i by an integral in�k�
∑
i
� gV
�2π��3
�
d3k �
where g is the statistical factor of the particle1. For simplicity, from now on, we switch
to the system of units where �� c � 1. For a spin 1�2 particle, this factor is 2. We get
lnZ �V�T�µ� �
gV
�2π�3
�
d3k ln
�
1� e�β�εk�µ�
�
� (52)
where εk is the energy of the state with momentum�k.
The total energy of the system is
�E��� ∂
∂β
lnZ �V�T�µ�
β µ
�
gV
�2π�3
�
d3k
εk e
�β�εk�µ�
1� e�β�εk�µ�
�
gV
�2π�3
�
d3k
εk
eβ�εk�µ� �1
(53)
and the total number of particles of the system becomes
�N�� 1
β
∂
∂µ
lnZ �V�T�µ�
β
�
gV
�2π�3
�
d3k
1
eβ�εk�µ��1
� (54)
1 Note that this is true in the thermodynamical limit, V � ∞. See Ref[5].
The above Eqs.(53) and (54) show that the average of occupation number for the energy
level εk in the Fermi gas is given by
f
�
εk
�
1
eβ�εk�µ� �1
� (55)
which is known as the Fermi distribution. The pressure is given by
P �
g
�2π�3
1
β
�
d3k ln
�
1� e�β�εk�µ�
�
� (56)
Finally, the entropy is calculated from
T �S�� �E��µ �N��PV�
Ideal Bose-Einstein Gas
For bosons, the sum over states differs from that of a Fermi gas. There is no restriction
for the occupation numbers ni, so we have to sum over all non-negative integers:
Z �V�β �µ� � ∑
α
e�β�Eα�µNα �
� ∑
n1
∑
n2
∑
n3
� � �∑
ni
� � �exp
�
�β ∑
i
ni
�
εi�µ

�
� ∏
i
∞
∑
ni�0
e�βni�εi�µ�
� ∏
i
1
1� e�β�εi�µ�
� exp
�
�∑
i
ln
�
1� e�β�εi�µ�
��
� (57)
where we have assumed
εi�µ � 0 (58)
to assure the convergence. Introducing again the integral over plane-wave states, we get
lnZ �V�T�µ� �� gV
�2π�3
�
d3k ln
�
1� e�β�εk�µ�
�
� (59)
In an analogous way as in the case of the Fermi gas, the expressions for the energy and
the particle number of a boson gas are found to be
�E�� gV
�2π�3
�
d3k
εk
eβ�εk�µ��1
� (60)
�N�� gV
�2π�3
�
d3k
1
eβ�εk�µ��1
� (61)
The pressure is given by
P �� g
�2π�3
1
β
�
d3k ln
�
1� e�β�εk�µ�
�
� (62)
and again
T �S�� �E��µ �N��PV� (63)
The probability of occupation of the energy level εk for a boson gas is
f
�
εk
�
1
eβ�εk�µ��1
� (64)
Note that we should have εk � µ for all k, so that
m � µ (65)
where m is the mass of the boson. In the limit µ �m, �E��V and �N��V diverge, and the
behavior of the equation of state changes qualitatively. This is known as Bose-Einstein
condensate (see the later discussion).
RELATIVISTIC IDEAL GASES
In the final stage of relativistic heavy ion collisions, a lot of hadrons are produced. Let
us assume that such a state can approximately be described as an ideal gas of hadrons.
Such an approximation will be valid if the thermal energy is sufficiently large compared
to the interaction energies between hadrons. This means that for low mass particles, we
have to treat their kinematics relativistically. We have to evaluate the integral in Eq.(59)
with2
εk �
�
k2 �m2� (66)
where m is the mass of the particle. Expressions for number density n, energy density ε
and pressure P are given by
n �
g
2π2
� ∞
0
dk k2
1
eβ�
�
k2�m2�µ��1
� (67)
ε �
g
2π2
� ∞
0
dk k2
�
k2 �m2
eβ�
�
k2�m2�µ��1
� (68)
P �� g
2π2
1
β
� ∞
0
dk k2 ln
�
1� e�β�
�
k2�m2�µ�
�
� (69)
where the double sign � correspond to the case of Fermions and Bosons, respectively.
2 We use the natural unit, �� c� 1.
Non-degenerate Case
First, let us evaluate the pressure. When
e�β�m�µ� � 1 � (70)
we can expand the integrand as
ln
�
1� e�β�εk�µ�
�
��
∞
∑
n�1
�
1�n�1
n
e�βn�εk�µ� � (71)
so that
P �
g
2π2
1
β
∞
∑
n�1
�
1�n�1
n
eβnµ Φ�βn�m� � (72)
where
Φ�βn�m��
� ∞
0
dk k2 e�βnεk
�
� ∞
0
dk k2 e�βn
�
k2�m2
� m3
� ∞
0
dxx2 e�z
�
x2�1 (73)
with z � βnm. Using the integral representation of modified Bessel functions,
Kν �z� �
�
π
Γ
�
ν � 12

 � z
2
�ν � ∞
0
e�z
�
x2�1 x
2ν
�
x2 �1
dx�
which holds for z � 0, Re ν � 1�2, we identify
Φ�βn�m� ��m3 2Γ
�3
2
�
π
�
d
dz
�
1
z
K1�z�
��
z�βnm
� m3
�
1
z
K2�z�
�
z�βnm
� (74)
Finally, we get
P �
g
2π2
m3
β
∞
∑
n�1
�
1�n�1
n
enβ µ
�
1
z
K2�z�
�
z�βnm
� (75)
Once P is expressed as a function of β and µ , we obtain the number density
n �
∂P
∂µ
β
�
gm3
2π2
∞
∑
n�1
�
1�n�1 enβ µ
�
1
z
K2�z�
�
z�βnm
(76)
and the energy density
ε �� ∂ �βP�
∂β
β µ
��gm
4
2π2
∞
∑
n�1
�
1�n�1 enβ µ
�
d
dz
�
1
z
K2�z�
��
z�βnm
�
gm4
2π2
∞
∑
n�1
�
1�n�1 enβ µ
�
1
z
�
3
z
K2�z��K1 �z�
��
z�βnm
� (77)
while the entropy density can be calculated as
T s � ε �P�µ n� (78)
The above expressions are only valid for
e�β�m�µ� � 1� (79)
The series converges very slowly for e�β�m�µ�� 1, and for e�β�m�µ� � 1 the sum does
not converge. For bosons, this last situation does not happen, but for fermions it can
occur for rather high density and low temperature.
For practical applications,it is important to know the limit of validity of the series
expansion. In the figure 1 below, we show the number density of a typical baryon (m �
mn � 938 MeV) as a function of temperature T , for µ � m. The series representation of
integrals shown above are valid only for the domain below this curve.
0 40 80 120 160 200
T (M eV)
0 .0
0 .4
0 .8
1 .2
1 .6
n 
(f
m
-3
))
m =µ
Figure 1: Nuclear density as function of temperature T for m � µ .
Boltzmann Limit
When
e�β�m�µ�� 1�
or equivalently m�µ is sufficiently larger than T , the above series expansion converges
very rapidly and in practice, only the first term gives a good approximation. In this limit,
there is no difference between bosons and fermions. Explicitly, we have
P� PBoltz �
g
2π2
m2T 2e
µ
T K2
�m
T
�
� (80)
n� nBoltz �
g
2π2
m2Te
µ
T K2
�m
T
�
� (81)
ε � εBoltz �
g
2π2
m3Te
µ
T
�
K1
�m
T
�
�
3T
m
K2
�m
T
��
� (82)
We can immediately see
PBoltz � nBoltzT�
which is the well-known classical ideal gas equation of state. Furthermore, for m � T ,
we can express the mean energy per particle
�ε
n
�
Boltz
� m
�
K1 �z�
K2 �z�
�
3
Z
�
z�mT
� m
�
1�
3
2z
�
15
8z2
� � � �
�
z�mT
� m�
3
2
T �
15
8m
T 2 � � � �
where we have used the asymptotic expansion of Bessel functions. The first term is the
rest mass and the second term is the classical formula for the mean kinetic energy of
ideal gas. Higher terms are relativistic corrections.
In Figs. 2a, b, and c, we show how the series expansions for the energy density,
pressure and entropy density converge to the exact values when the number of terms
N is increased. Here, we take for a boson gas with mass � 150 MeV, corresponding to
typically π-mesons. The temperature is taken to be 200 MeV. In these figures, the curves
indicated as N � 1 (Boltzmann) correspond to the Boltzmann approximation. For boson
gas, at µ � m, the energy density and particle density diverge, corresponding to the
Bose-Einstein condensate. However, as a function of the number density, the pressure
and entropy density tend to constant, and the energy per particle decreases, tending to
the rest mass energy. This is because, the increase of particle number of the system after
certain amount is just consumed up to fill lowest energy states and does not contribute
to the total energy and entropy. For more details of Bose-Einstein condensation, see the
standard text books [2].
0.05 0.1 0 .15 0.2 0 .25 0.3 0 .35
P artic le D ens ity (1 /fm 3)
200
300
400
500
600
700
E
ne
rg
y 
pe
r 
pa
rt
ic
le
 (
M
eV
)
T=200 M eV
M =150 M eV
N =1 (Bo ltzm ann)
N =2
N =5
E xact
Figure 2a: Energy per particle of a boson gas.
0.05 0.1 0 .15 0.2 0 .25 0.3 0 .35
P artic le D ens ity (1 /fm 3)
10
20
30
40
50
P
re
ss
ur
e 
(M
eV
/f
m
3 )
T=200 M eV
M =150 M eV
N =1 (Bo ltzm ann)
N =2
N =5
E xact
Figure 2b: Pressure of a boson gas
0.05 0.1 0 .15 0.2 0 .25 0.3 0 .35
P artic le D ens ity (1 /fm 3)
0.4
0 .5
0 .6
0 .7
0 .8
0 .9
E
nt
ro
py
 d
en
si
ty
(1
/f
m
3 )
T=200 M eV
M =150 M eV
N =1 (B o ltzm ann)
N =2 N =5
E xact
Figure 2c: Entropy density of a boson gas
For boson gas, at µ �m, the energy density and particle density diverge, correspond-
ing to the Bose-Einstein condensate. However, as a function of the number density, the
pressure and entropy density tend to constant, and the energy per particle decreases,
tending to the rest mass energy. This is because, the increase of particle number of the
system after certain amount is just consumed up to fill lowest energy states and does not
contribute to the total energy and entropy. For more details of Bose-Einstein condensa-
tion, see the standard text books[2].
In Figs. 3a, b and c, we plotted the behaviors of the energy per particle, pressure and
entropy density of a fermion gas as functions of particle density. Here, we show the
example of a typical baryon gas, with mass m � 900 MeV at the temperature 200 MeV.
0 0.4 0 .8 1 .2 1 .6
P artic le D ens ity (1 /fm 3)
800
1000
1200
1400
E
ne
rg
y 
p
er
 P
a
rt
ic
le
 (
M
eV
)
T = 200 M eV
M = 900 M eV
N on-re la tiv is tic B o ltzm ann
N =1 (B o ltzm ann)
N =2
N =3
N =4
N =5
E xact
Figure 3a: Energy per particle of a fermion gas
0 0.4 0 .8 1 .2 1 .6
P artic le D ens ity (1 /fm 3)
0
100
200
300
P
re
ss
u
re
 (
M
eV
/f
m
3
)
T = 200 M eV
M = 900 M eV
N =1 (B o ltzm ann)
N =2
N =3
N =4 N =5
E xact
Figure3b: Pressure of a fermion gas.
0 0.4 0 .8 1 .2 1 .6
P artic le D ens ity (1 /fm 3)
0
1
2
3
4
E
nt
ro
py
 D
e
ns
ity
(1
/f
m
3
)
T = 200 M eV
M = 900 M eV
N =1 (Bo ltzm ann)
N =2
N =3
N =4
N =5
E xact
Figure 3c: Entropy of a fermion gas.
As we see from these figures, the Boltzmann approximation is not so bad at these
temperature and density values but the convergence of the series expansion becomes
catastrophic for particle density greater than 0�8 f m�3. For very high density, see the
section
Mixture of Particles and Chemical Equilibrium
It is easy to extend the formulation of the previous section to a system which contains
more than one kind of particles. Let us denote baryon number, strangeness, and electric
charge of the type t particle as bt �st�and et , respectively. The total baryon number Qb,
total strangeness Qs and total electric charge Qe of the system are
Qb � ∑
t
btNt � (83)
Qs � ∑
t
stNt � (84)
Qe � ∑
t
etNt � (85)
where Nt is the total number of particles of the type t. Suppose that these are the only
conserved quantum numbers of the system. Then we may ask “what is the probability
distribution of microstates at equilibrium, given the numbers of conserved quantum
numbers?” To find the answer, we extend Eq.(39) as
δ�S��β
�
δ�E��µb δ�Qb��µSδ�Qs��µeδ�Qe��∑
t
γ tδ�1�t
�
� 0� (86)
where
�S�� ∑
t
�St�� (87)
�E�� ∑
t
�Et�� (88)
are the entropy and total energy of the system. Here, �St� and �Et� represent the entropy
and energy of the particle t� Variation should be taken with respect to the probability
distribution
�
p�t�α
�
for each particle of type t, so that the last term in Eq.(86) represents
the constraints of the normalization of each
�
p�t�α
�
� Substituting Eqs.(83,84,85,87,88)
into Eq.(86), we have
∑
t
δ ��St��β �δ�Et��µt δ�Nt��γ tδ�1�t �	� 0�
where
µt � btµb � stµs � etµe�
Since
�
p�t�α
�
are independent for each particle type t, we get
δ ��St��β �δ�Et��µt δ�Nt��γ tδ�1�t�	� 0� (89)
for each t. This last equation shows that all the results for the unique particle-type
case can readily be generalized to a mixture of many different kinds of particles, just
substituting the chemical potential of the type t particle by
µ � µt � btµb � stµs � etµe�
thus introducing chemical potentials for each conserved quantity. The resulting formulas
describe the chemical and thermal equilibrium among particles.
SOME USEFUL APPROXIMATIONS
As we see from the above figure, the domain of applicability of the series expansion
for Fermi integrals is rather small. Particularly for light mass fermions, the situation
becomes worse. For example, if we extend our ideal gas description to quarks, then the
series expansion does not apply for most regions of interest. In this section, we will see
some useful analytical approximations of the above Fermi integrals[4].
When a fermion gas is in thermal equilibrium, its antiparticle also appears due to pair
production or other possible reaction channels. The chemical potential of the antiparticle
is just the opposite of the chemical potential of the particle. Since all thermodynamical
quantities in the Grand Canonical Ensemble are deduced from the thermodynamical
potential, let us consider here only the pressure. The pressure of a system of particles
and antiparticles in equilibrium is then
P�µ�T � �
gT
�2π�3
�
d3k
�
ln
�
e��E�µ��T �1
�
� ln
�
e��E�µ��T �1
��
�
g
6π2
� ∞
m
dE
�
E2�m2
3�2 � 1
e�E�µ��T �1
�
1
e�E�µ��T �1
�
� (90)
Let us consider the evaluation of the integral
F �a�b��
� ∞
a
dx
�
x2�a2
3�2 � 1
e�x�b��1
�
1
e�x�b��1
�
� (91)
The pressure P is proportional to a function of this form,
P �
gT 4
6π2
F �a�b� � (92)
where a and b are related to the mass and the chemical potential by
a �
m
T
� (93)
b �
µ
T
� (94)
with T measured in units of energy �k � 1�.
1) Degenerate case
At extremely high densities, the pressure of a Fermi gas is determined essentially
by the density and less dependent on the temperature. Such a situation occurs often
in astrophysical processes, for example, in the core of advanced stage of heavy stars,
white dwarfs, and neutron stars. Several peculiar processes like supernova explosion
are intimately associated to the degeneracy of Fermi gas. This is due to the change
of behavior of the pressure with respect to the temperature [6]. In our expression the
degeneracy corresponds a large chemical potential compared to the temperature
b �� 1� (95)
and for this case, we can safely approximate�
1
e�x�b��1
�
1
e�x�b��1
�
 1
e�x�b��1
� (96)
for all x between zero and infinity. Then, writing
x � bu � (97)
we get
F �a�b�
 b4
� ∞
a�b
du
�
u2� �a�b�2
�3�2 1
eb�u�1��1
� (98)
In general, for any function f �u�, we can write
� ∞
a�b
du f �u�
1
eb�u�1��1
�
� ∞
a�b
du f �u�
�
θ �1�u�� 1
eb�u�1��1
�θ �1�u�
�
�
� 1
a�b
du f �u��
� ∞
a�b
du f �u�
�
1
eb�u�1��1
�θ �1�u�
�
� I1 � I2 � (99)
Here the second term is
I2 �
� ∞
a�b
du f �u�
�
1
eb�u�1��1
�θ �1�u�
�
�
� 1
a�b
du f �u�
�
1
eb�u�1��1
�1
�
�
� ∞
1
du f �u�
1
eb�u�1��1
��
� 1
a�b
du f �u�
eb�u�1�
eb�u�1��1
�
� ∞
0
du f �u�1�
1
ebu �1
��
� 1
a�b
du f �u�
1
1� e�b�u�1�
�
� ∞
0
du f �u�1�
1
ebu �1
��
� 1�a�b
0
du f �1�u� 1
1� ebu
�
� ∞
0
du f �u�1�
1
ebu �1
(100)
For a� b, we may safely approximate (with an error the order of 
 e��b�a�)
I2 
�
� ∞
0
du f �1�u� 1
1� ebu
�
� ∞
0
du f �u�1�
1
ebu�1
�
� ∞
0
du
1
1� ebu
� f �u�1�� f �1�u�� (101)
Expanding the function f in a power series of u, we get
f �u�1� � f �1��
1
1!
f �1� �1�u�
1
2!
f �2� �1�u2 � � � � � (102)
f �1�u� � f �1�� 1
1!
f �1� �1�u�
1
2!
f �2� �1�u2��� � � (103)
and
I2 
∞
∑
k�0
2
�2k�1�!
f �2k�1��1�
� ∞
0
du
u2k�1
1� ebu
(104)
In addition, we have
� ∞
0
du
u2k�1
1� ebu
�
� ∞
0
du u2k�1
∞
∑
n�1
��1�n�1 e�bnu
�
1
b2k�2
∞
∑
n�1
��1�n�1
n2k�2
� ∞
0
dx x2k�1e�x
�
�2k�1�!
b2k�2
∞
∑
n�1
��1�n�1
n2k�2
�
�2k�1�!
b2k�2
�
1� 1
22k�1
�
ζ �2k�2� � (105)
so that
I2 
∞
∑
k�0
2
b2k�2
�
1� 1
22k�1
�
ζ �2k�2� f �2k�1��1�
�
1
b2
ζ �2� f �1� �1��
7
4b4
ζ �4� f �3� �1�� � � � (106)
Finally, for b� 1� a�
� ∞
a�b
du f �u�
1
eb�u�1��1
� 1
a�b
du f �u��
π2
6b2
f �1� �1��
7π4
360b4
f �3� �1�� � � � (107)
This is known as Sommerfeld expansion. For
f �
�
u2� z2
3�2 � (108)
where z� a�b, we have
f �1� �1� � 3
�
1� z2 (109)
and
f �3� �1� � 3
2�3z2
�1� z2�3�2
� (110)
so that
F �a�b�
 1
8
�
b
�
2b2�5a2
�b2�a2
1�2 �3a4 ln b�
�
b2�a2
a
�
�
π2b
�
b2�a2
2
�
7π4
120
b�2b2�3a2�
�b2�a2�3�2
� � � � (111)
This is valid for b� a�1. The approximation contains errors of the order of
� e��b�a��
2) Ultrarelativistic Limit
For a� 0, we can evaluate the integral as shown below. We have
F �a�b� �
� ∞
0
dx
�
x3� 3
2
a2x
��
1
e�x�b��1
�
1
e�x�b��1
�
�O�a3� (112)
In general,
� ∞
0
dx f �x�
�
1
e�x�b��1
�
1
e�x�b��1
�
�
� ∞
�b
dx f �x�b�
1
ex �1
�
� ∞
b
dx f �x�b� 1
ex �1
�
� ∞
0
dx � f �x�b�� f �x�b�� 1
ex �1
�
� 0
�b
dx f �x�b�
1
ex �1
�
� b
0
dx f �x�b� 1
ex �1
(113)
However,
� 0
�b
dx f �x�b�
1
ex �1
�
� b
0
dx f ��x�b� 1
e�x �1
�
� b
0
dx f ��x�b� e
x
ex �1
�
� b
0
dx f ��x�b�
�
1� 1
ex �1
�
� (114)
so that � ∞
0
dx f �x�
�
1
e�x�b��1
�
1
e�x�b��1
�
�
� ∞
0
dx � f �x�b�� f �x�b�� 1
ex �1
�
� b
0
dx f ��x�b�
�
� b
0
dx � f �x�b�� f ��x�b�� 1
ex �1
� (115)
In our case, f �x� � x3 � �3a2�2
x is an odd function of x, so that the last term just
vanishes. Following same steps as above,
� ∞
0
dx
xk
ex �1
� k!
�
1� 1
2k
�
ζ �k�1� �
and
F �a�b�
 1
4
b4� 3
4
a2b2 �2
� ∞
0
dx
�
x3 �3
�
b2� 1
2
a2
�
x
�
1
ex �1
�
1
4
b4� 3
4
a2b2 �2 �3
�
b2� 1
2
a2
�
1
2
ζ �2��2 �3! � 7
8
ζ �4�
�
1
4
b4� 3
4
a2b2 �
π2
2
�
b2� 1
2
a2
�
�
7π4
60
� (116)
Note that the first terms in the above expression coincide with the Taylor expansion
in a of the degenerate limit, Eq.(111). Therefore, we can express both cases, the ultra-
relativistic and extreme degeneracy limits as
F�a�b�
 b4F0�z��
π2
2
b2
�
1� z2 � 7π
4
120
�2�3z2�
�1� z2�3�2
� (117)
where z � a�b � m�µ and
F0�z� �
1
8
��
2�5z2
�1� z2
1�2 �3z4 ln
�
1�
�
1� z2
z
��
� (118)
This approximation is valid in the whole domain (0 � b � ∞� as far as z is sufficiently
smaller than unity. Finally, we get the expression for the pressure of an ultra-relativistic
fermion gas including the antiparticles as
P�µ�T �
 g
6π2 ��c�3
�
µ4
8
��
2�5z2
�1� z2
1�2 �3z4 ln
�
1�
�
1� z2
z
��
�
π2
2
µ2T 2
�
1� z2 � 7π
4T 4
120
�
2�3z2
�1� z2�3�2
�
� (119)
and the corresponding number density is
n
 g
6π2 ��c�3
�
µ3
�
1� z2
3�2 � π2
2
µT 2
2� z2�
1� z2 �
7π4T 4
40µ
z4
�1� z2�5�2
�
�
The approximation is valid for
z �
m
µ
� 1� (120)
and they are exact for both of T � 0 and m � 0 cases. In Fig. 4, we show the equation
of state p � p�n�T � for a Fermi gas with m � 900 MeV, for two different temperatures.
These curves are obtained by Eqs.(119,120), eliminating the chemical potential µ . For
comparison, we also show the corresponding curves using the exact integrals. For higher
densities, the pressure tends to independent of the temperature T and in this region, the
approximation becomes asymptotically exact.
0 1 10 100 1000
n
1E +001
1E +002
1E +003
1E +004
1E +005
1E +006
p
A na ly tic A pprox.
E xact
T=500M eV
T=50M eV M =900M eV
Figure 4: Comparison of the analytic
approximation with the exact values for
the degenerate Fermi gas.
0 .1 1 10 100 1000
n
0
0.2
0 .4
0 .6
0 .8
1
m
/ µ
T=500M eV
T=50M eV
Figure 5: Behavior of chemical potential
as a function of particle density.
As we see, for the lower temperature, the approximation extends much more to the
domain of low particle densities compared to high temperature case. This is because,
when T becomes smaller, the value of z � m�µ stay smaller than unity for lower
densities as we can see from Fig. 5.
• Exercise: Derive the expression for the pressure in ultra-relativistic regime of a
boson gas.
Numerical Method
Although the analytical expressions above are useful to discuss the general properties
of a relativistic gas, none of them can cover the whole region of parameters µ and T .
For practical calculations, it is desirable to possess a simple and efficient method to
obtain precise values of thermodynamical quantities. In this sense, it is more effective
to evaluate directly the integrals Eqs.(67,68,69) using Gauss’ quadrature method. To do
this, we can rewrite the integrals as
n �
gT 3
2π2 ��c�3
� ∞
0
dx
x1�2 �x�a��x�2a�1�2
ea�b� e�x e
�x� (121)
ε �
gT 4
2π2 ��c�3
� ∞
0
dx
x1�2 �x�a�2 �x�2a�1�2
ea�b� e�x e
�x� (122)
P �
gT 4
6π2 ��c�3
� ∞
0
dx
x3�2 �x�2a�3�2
ea�b� e�x e
�x� (123)
where as before, a � m�T and b � µ�T � We note that all these integrals have the form
� ∞
0
dx xα e�x f �x� �
which can well be evaluated by the Gauss-Laguerre quadrature methods as
� ∞
0
dx xα e�x f �x�
N
∑
i�1
ωi f
�
xi
�
where xi and ωi are calculated in terms of orthogonal polynomials associated to confluent
hypergeometric functions. A subtle detail here is that, for the n-integration Eq.(121), for
example, we can take,
α � 1
2
�
f �x�� �x�a��x�2a�
1�2
ea�b� e�x �
but if a is much smaller than the first x value of Gauss-Laguerre quadrature, that is,
a� x1�
then the approximation becomes better choosing
α � 2�
f �x�� �1�a�x��1�2a�x�
1�2
ea�b� e�x �
Similarly, for the ε-integral,
α � 1
2
�
f �x�� �x�a�
2 �x�2a�1�2
ea�b� e�x
but if
a� x1�
then
α � 3�
f �x�� �1�a�x�
2 �1�2a�x�1�2
ea�b� e�x �
and similarlyfor the pressure. The Gauss-Laguerre quadrature abscissa
�
xi
�
and weights�
ωi
�
can readily be calculated by, for example, the program, “gaulag” in the Numerical
Recipes [7]. For practical uses, the number of abscissa points N can be taken to the order
of 30 to 40, within relative errors of the order of 10�6 for the most of necessary a and b
values, as far as b� xN . For extremely degenerate region of a Fermi gas, a larger value
of N is required.
The author acknowledges Drs. D. Gomez Dumm and A. Mihara for careful revision
of the text and useful suggestions. He also thanks J.Rafelski and H-Th. Elze for crit-
ical reading of the manuscript and suggestions. This work is supported by PRONEX
#41.96.0886.00, CNPq and FAPERJ.
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Plasma (Cambridge U. Press, 2002) for the most recent developments.
2. L.D. Landau and E.M. Lifshitz, Statistical Physics (Pergamon Press, 1959).
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H-Th. Elze and W. Greiner, Phys. Lett. B179 (1986) 385.
6. N.K.Glendenning, Compact Stars, Springer-Verlag, 2000.
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