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Some notes on the ideal fermion gas

Anthony [email protected]

27.04.2010

Abstract. These are notes on the physics of the ideal fermion gas collected from various sources.The text provides a deeper understanding of the behaviour of degenerate fermion gases in order toapply this to white dwarfs and neutron stars.

The discussion in this note concerns the academic case of a pure (consisting of one type of particle)ideal fermion gas. What is neglected are the interactions between the particles (e.g., the Coulombforce in the case of electrons) and the fact that in practice the fermions will be embedded in a gasconsisting of a number of different kinds of particles (think of the carbon, oxygen, etc nuclei inwhite dwarfs and the neutrons and protons in neutron stars). This will introduce the importantforce of gravity of the heavier particles and alter the equation of state. Lastly, processes arisingfrom nuclear physics are completely ignored.

Revision History

Issue Rev. No. Date Author Comments2 0 27.04.2010 AB Slight reordering of text and addition of classical dilute gas exam-

ple. Text made available on ‘Sterren’ website.1 1 30.03.2009 AB Improvements in text following reading of Weiss et al. (2004), al-

ternative development of density and pressure in terms of Fermi-Dirac integrals.

1 0 17.03.2009 AB Update to include more material from statistical physics in the non-relativistic regime. Fermi-energy derived in terms of relativistic ki-netic energy. Equation of state for degenerate fermion gas derivedfor non-relativistic and ultra-relativistic case.

0 0 22.03.2008 AB Document created.

1

REFERENCES 2

Contents

1 The ideal gas 31.1 The ideal non-relativistic quantal gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 The classical dilute gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 The degenerate non-relativistic fermion gas . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Particle energies and momenta in special relativity 7

3 The relativistic description of the Fermion gas 8

4 The fermion gas phases 9

5 The equation of state for the fermion gas 11

6 Density and pressure in terms of Fermi-Dirac integrals 13

7 The energy density of the fermion gas 15

8 The full fermion phase diagram 16

References

Gradshteyn I.S., Ryzhik I.M., 2007, Table of Integrals, Series, and Products, seventh edition, A. Jeffrey & D.Zwillinger, eds., Academic Press

Mandl F., 1988, Statistical Physics, second edition, The Manchester Physics Series, John Wiley & Sons ltd,Chichester

Ostlie D.A., Carroll B.W., 2007, An Introduction to Modern Stellar Astrophysics, Second Edition, Addison-Wesley, San Francisco,

Phillips A.C., 1998, The Physics of Stars, second edition, The Manchester Physics Series, John Wiley & Sonsltd, Chichester

Press W.H., Teukolsky S.A., Vetterling W.T., Flannery B.P., 2007, Numerical Recipes: The Art of ScientificComputing, Third Edition, Cambridge University Press (Cambridge)

Weiss A., Hillebrandt W., Thomas H.-C., Ritter H., 2004, Cox & Giuli’s Principles of Stellar Structure, extendedsecond edition, Cambridge Scientific Publishers (Cambridge)

1.1 The ideal non-relativistic quantal gas 3

1 The ideal gas

In this section1 the properties of ideal gases are reviewed. The ideal gas is often also called a perfect gasand is defined in (Mandl, 1988, chapter 7) as follows: The perfect gas represents an idealization in which thepotential energy of interaction between gas particles is negligible compared to their kinetic energy of motion.This definition implies that one can write down a set of private energies εi for each particle, which simplifies theapplication of statistical physics to gases enormously. In practice one can indeed often neglect the interactionenergies of particles.

However, quantum effects cannot be neglected so readily and one needs to take the particle concentrationinto account. The ideal gas usually treated in elementary textbooks is therefore the classical ideal gas forwhich the condition is that the probability is very small that any single-particle state is occupied by more thanone particle. An ideal gas for which this last condition does not hold is an ideal quantal gas. These kindsof gases have to be treated with quantum statistics. In the book by Mandl (1988) this is done extensively forthe non-relativistic regime, while in Phillips (1998) the ideal quantal gas is treated in both the relativistic andnon-relativistic regimes. The next section summarizes the results for the non-relativistic regime from Mandl(1988).

1.1 The ideal non-relativistic quantal gas

Chapters 9 and 11 in Mandl (1988) discuss statistical physics of quantal gases, with chapter 11 focusing onfermions and bosons. For these particles the mean occupation number fi for the i-th single-particle state, withenergy εi, is given by:

fi =1

eβ(εi−µ) ± 1, (1)

where the plus sign is for fermions, giving the Fermi-Dirac (FD) distribution, and the minus sign is for bosons,for which the Bose-Einstein (BE) distribution holds. In this equation β = 1/kT and µ is the chemical potential.The latter is a thermodynamic quantity relevant for systems with a variable number of particles (e.g., chemicalreactions). From here on we concentrate on fermions for which equation (1) shows that fi is never larger than1. This reflects the Pauli exclusion principle which states that no more than one fermion can occupy a givenquantum state.

In order to obtain the distribution of particles as a function of energy the mean occupation number has to bemultiplied by the so-called density of states, which basically describes in how many ways a particular energystate can be attained for a particle. In terms of momentum p the density of states g(p) is (see Phillips, 1998,chapter 2):

g(p)dp = gs4πVh3

p2 dp , (2)

where gs is the degeneracy of each state (for example gs = 2 for particles with two possible polarizations orspins for the same momentum). The particles are confined to the volume V and h is the Planck constant. Thenumber of particles at a certain energy is then given by N(εi) ∝ fig(pi), where pi = p(εi). De distribution ofthe particles over the different energies can then be written in terms of the momentum p as:

N(p) dp = gs4πVh3

p2

eβ(ε−µ) + 1dp , (3)

where the function fi is now written in its continuum form f(ε). This expression is entirely general for idealfermion gases as we have not yet specified the relation between momentum and energy.

In the non-relativistic regime considered here one can write:

ε =p2

2m, (4)

where m is the mass of the fermion. The density of states in terms of energy then becomes:

g(ε) dε = gs2πVh3

(2m)3/2ε1/2 dε . (5)

1The material in this section is extracted from Mandl (1988)

1.2 The classical dilute gas 4

The particle number N(ε) per energy interval now becomes:

N(ε) dε = f(ε)g(ε) dε = gs2πVh3

(2m)3/2 ε1/2

eβ(ε−µ) + 1dε . (6)

The total number of fermions is now obtained by integrating (6) which results in:

N = gs2πVh3

(2m)3/2

∫ ∞0

ε1/2 dε

eβ(ε−µ) + 1. (7)

In terms of the particle concentration, or number density, n the last two equations can be written as:

n(ε) dε = f(ε)g(ε) dε = gs2πh3

(2m)3/2 ε1/2

eβ(ε−µ) + 1dε , (8)

and

n = gs2πh3

(2m)3/2

∫ ∞0

ε1/2 dε

eβ(ε−µ) + 1. (9)

Equations (7) and (9) provide an implicit definition of the chemical potential for a fixed number of fermionsin a fixed volume (eq. 6), or for a fixed number density (eq. 9). For a fixed number and volume the chemicalpotential is a function of temperature only µ = µ(T ) (cf. Mandl, 1988, section 11.5.1).

1.2 The classical dilute gas

If the conditione−βµ � 1 (10)

is satisfied the term +1 in fi in equation (1) can be ignored and the average occupation of each state is muchlower than 1. In this case the fermion gas obeys Maxwell-Boltzmann statistics and is said to behave like a diluteclassical gas. This can be verified by starting from equation (8) and using the assumption above in which casethe following expression for n(ε) holds:

n(ε) dε = eµ/kT gs2πh3

(2m)3/2e−ε/kT ε1/2 dε , (11)

which in terms of momentum (using expression 4) becomes:

n(p) dp = eµ/kT gs1h3e−p

2/2mkT 4πp2 dp , (12)

or in terms of particle velocities:

n(p) dp = eµ/kT gs

(mh

)3e−mv

2/2kT 4πv2 dv . (13)

Equations (11)–(13) represent of course the familiar Maxwell-Boltzmann distribution of particle energies andmomenta for the ideal classical gas. In this case the chemical potential can be determined from:

n =∫ ∞

0n(ε) dε and

∫ ∞0

e−xx1/2 dx =√π

2.

The resulting chemical potential is:

µ = kT ln(

n

gsnQ

),

where nQ is the so-called quantum concentration:

nQ =(

2πmkTh2

)3/2

.

1.3 The degenerate non-relativistic fermion gas 5

0 1 2 3 4 5ε/kT

0

1

2

3

4

5

6

7

8

9

108×f

(ε)

0 1 2 3 4 5ε/kT

0.0

0.5

1.0

1.5

2.0

2.5

3.0

n(ε

)[1

045J−

1m−3

]

N2 gas at T = 300 K, P = 105 N m−2

Figure 1: The mean occupation number f(ε) (left panel) and the corresponding number density of particles n(ε)(right panel) as a function of energy for a nitrogen (N2) gas at room temperature and atmospheric pressure. Theenergy is expressed in units of kT .

Substituting this value of the chemical potential in (11) one correctly obtains the expression:

n(ε) dε =2

π1/2

1(kT )3/2

e−ε/kT ε1/2 dε

Figure 1 shows the mean occupation number for particle states with energy ε and the particle concentration asa function of energy for the example of a classical N2 gas at room temperature and atmospheric pressure. Notethe very small, O(10−8), values of the average occupation number, showing that this is indeed a dilute classicalgas.

1.3 The degenerate non-relativistic fermion gas

We return now to the quantal gas and define the Fermi energy εF as the value of the chemical potential at theabsolute zero temperature:

εF ≡ µ(0) . (14)

It is shown in Mandl (1988) that εF must be positive. As T → 0 β → ∞ which means that exp[β(ε − µ)]tends to exp(∓∞) for ε < εF and ε > εF, respectively. That is, f(ε) tends to 1 or 0 for ε < εF and ε > εF

respectively. The fermion gas in this state is said to be completely degenerate.The distribution of particles over the momenta then becomes:

n(p) dp =

{gs

4πh3 p

2 dp p ≤ pF

0 p > pF

, (15)

while the distribution over energies becomes:

n(ε) dε =

{gs

2πh3 (2m)3/2ε1/2 dε ε ≤ εF

0 ε > εF

. (16)

1.3 The degenerate non-relativistic fermion gas 6

f(ε)

ε/εF

0

1

0 0.2 0.4 0.6 0.8 1.0

n(ε)

ε/εF

00 0.2 0.4 0.6 0.8 1.0

Figure 2: The Fermi-Dirac mean occupation number f(ε) and energy distribution n(ε) for a non-relativisticfermion gas at zero temperature: T = 0 K. The energy is expressed in units of the Fermi energy εF.

The quantity pF is the Fermi momentum corresponding to the Fermi energy. The interpretation of this particledistribution is as follows. At T = 0 K the gas is in its lowest energy state but because of the Pauli exclusionprinciple not all N fermions can crowd in to the single-particle state of lowest energy. The Fermi energy then isthe topmost energy level occupied at T = 0 K. The energy distribution at T = 0 K will be proportional to ε1/2

with a sharp cutoff at ε = εF. Figure 2 shows the occupation number and energy distribution for the completelydegenerate fermion gas.

The Fermi energy can now be derived from equation (16):

n = gs2πh3

(2m)3/2

∫ εF

0ε1/2 dε = gs

2πh3

(2m)3/2 23ε

3/2F , (17)

which leads to:

εF =h2

2m

(3

gs4π

)2/3

n2/3 =~

2m

(6π2

gs

)2/3

n2/3 . (18)

From this result it is clear that εF depends on the fermion mass m and the particle concentration n. The Fermitemperature is defined as:

εF = kTF . (19)

In terms of momentum one can write for the degenerate case:

n = gs4πh3

∫ pF

0p2 dp , (20)

where pF is now the Fermi-momentum. Working out the integral above and solving for pF gives:

pF = h

(3

gs4π

)1/3

n1/3 = ~(

3gs2π2

)1/3

n1/3 . (21)

So far the (ideal) fermion gas has been treated in the non-relativistic regime. However, for the applicationto stars the possibility of relativistic energies has to be taken into account and this is done in Phillips (1998).The following material is extracted from that text and further worked out. It will become clear that the energydistribution for a fermion gas will be slightly different when the full relativistic formulation for particle energiesis taken into account.

2 Particle energies and momenta in special relativity 7

2 Particle energies and momenta in special relativity

I first recall the full expressions for energy and momentum for the particles in a gas according to specialrelativity. The energy εp of a particle of mass m in quantum state with momentum p is then given by:2

ε2p = p2c2 +m2c4 , (22)

where c is the speed of light. The speed of the particle vp is given by:

vp =pc2

εp, (23)

which is consistent with the familiar form:

p =mvp√

1− (vp/c)2. (24)

For later use the following expressions for the energy and momentum are useful:

εpmc2

=[( p

mc

)2+ 1]1/2

, (25)

andp

mc=[( εpmc2

)2− 1]1/2

. (26)

These expressions give the energy and momentum in a form normalized to the rest-mass energy and the dividingline between non-relativistic and relativistic momenta. In addition the following relation between energy andmomentum is useful.

dεpdp

=pc2

εp= vp or dp =

εppc2

dεp =1vpdεp , (27)

For most applications it is the kinetic energy of the gas that is important as only gas particles in motion canexert a pressure, transport energy, etc. The kinetic energy εkin is simply given by:

εkin = εp −mc2 . (28)

Similar useful relations to the ones for εp and p can be derived in this case:

εkin

mc2=[( p

mc

)2+ 1]1/2

− 1 , (29)

andp

mc=[( εkin

mc2+ 1)2− 1]1/2

, (30)

anddεkin

dp=

pc2

εkin +mc2or dp =

εkin +mc2

pc2dεkin . (31)

2The variable ε rather than ε is used here for energy in order emphasise that the relativistic formulation is used.

3 The relativistic description of the Fermion gas 8

3 The relativistic description of the Fermion gas

With the energy of the fermions εp defined in the previous section we can now write for the mean occupationnumber:

f(εp) =1

e(εp−µ)/kT + 1=

1e(εkin−(µ−mc2))/kT + 1

, (32)

so that the condition for a classical gas now becomes:

e(mc2−µ)/kT � 1 (33)

The density of states is still given by equation (2) so for the energy distribution we can write:

n(p) dp = f(εp)g(p) dp , (34)

andn =

∫ ∞0

f(εp)g(p) dp . (35)

In addition to classical or quantal the gas can now be ultra-relativistic, in which case for a significant fractionof the particles p � mc, or non-relativistic, if for most of the particles p � mc. This leads to four possibleextreme states of the fermion gas: classical and non-relativistic, classical and ultra-relativistic, degenerate andnon-relativistic, degenerate and ultra-relativistic. At this point it should be realized that when labelling theenergies of the states that the fermions can occupy there is no point in counting the rest mass energy mc2. Thisis the same for all particles and cannot be used to distinguish quantum states. Therefore the equation for f andn can be written in terms of εkin, the kinetic energy of the fermions (i.e., the energy over and above the restmass energy).

The way to formally account for the rest mass energy is to define a modified chemical potential:

µ̄ = µ−mc2 . (36)

The basic equations are straightforward to write down as we are simply relabelling the energy levels fromεp to εkinand the chemical potential from µ to µ̄:

f(εkin) =1

e(εkin−µ̄)/kT + 1, (37)

andn(p) dp = f(εkin)g(p) dp . (38)

In order to derive the distribution of particles over kinetic energy I now work out equation (38) and I makeuse of:

n(p) dp = n(εkin) dεkin or n(εkin) = n(p)dp

dεkin. (39)

Starting from:

n(p) dp = f(εkin)gs4πh3p2 dp . (40)

the distribution in energy can be found by using the relation above and eq. (31). The expression for the numberdensity then becomes:

n(εkin) dεkin = f(εkin)gs4πh3p2 dp

dεkindεkin

= f(εkin)gs4πh3pεkin +mc2

c2dεkin

= f(εkin)gs4πh3

p

mc

m

cmc2(εkin/mc

2 + 1) dεkin .

(41)


Now the value of p/mc can be substituted by using eq. (30) and at the same time I express the energies in termsof mc2.

n( εkin

mc2

)d( εkin

mc2

)= f(εkin)gs

4πh3

m

c

[( εkin

mc2+ 1)2− 1]1/2

mc2

(εkin +mc2

mc2

)mc2 d

( εkin

mc2

).

Note the extra factor mc2 coming from d(εkin/mc2). The final result then is:

n(ε′kin) = f(ε′kin)gs4π(mch

)3 (ε′kin + 1

) ((ε′kin + 1)2 − 1

)1/2, (42)

where ε′kin = εkin/mc2. Note that f(εkin) = f(ε′kin) because µ̄ and kT are also expressed in terms of mc2. The

units of mc/h are m−1 which means that n(ε′kin) is indeed a number density (for a given energy interval).The Fermi energy now becomes (in analogy to equation 17):

n = gs4π(mch

)3∫ εF

0

(ε′kin + 1

) ((ε′kin + 1)2 − 1

)1/2dε′kin = gs4π

(mch

)3 13((ε′F + 1)2 − 1

)3/2, (43)

where ε′F = εF/mc2. In the limit εF/mc2 � 1 the expression (17) is recovered. Thus the expression for the

Fermi energy now becomes:

εFmc2

=

((3

4πn

gs

)2/3( h

mc

)2

+ 1

)1/2

− 1 , (44)

or with the particle density n normalized to gs4π(mc/h)3:

εFmc2

=

((3n

gs4π(mc/h)3

)2/3

+ 1

)1/2

− 1 . (45)

The Fermi-momentum can again be derived using equation (20) which leads to the same expression for pF.Normalizing to mc we obtain:

pF

mc=(

3ngs4π(mc/h)3

)1/3

. (46)

Note that the expression for the Fermi-energy could thus have been obtained more easily by starting from pF

and then using the now different relation between energy and momentum as given by equation (25). Keep inmind the subtraction of the rest-mass energy.

4 The fermion gas phases

With the expressions for the Fermi energy in the non-relativistic case (eq. 18) and the relativistic expression(eq. 44) one can proceed to draw the T vs n diagram and broadly indicate where the gas is degenerate/non-degenerate and/or relativistic or non-relativistic. I first write the expressions for the Fermi energies (where NRstands for non-relativistic) as follows:

εF,NR

mc2=

12

(3n

gs4π(mc/h)3

)2/3

=12

(n

n0

)2/3

(47)

andεFmc2

=

((n

n0

)2/3

+ 1

)1/2

− 1 , (48)

where the density normalization is:

n0 = gs4π3

(mch

)3. (49)


Tmc2/k

n/n0

10−15 10−10 10−5 100 105 1010

10−10

10−5

100

Classical, Ultra-RelativisticP = nkT

Classical, Non-RelativisticP = nkT

Degenerate,Non-RelativisticP = KNRn

5/3

Degenerate,Ultra-RelativisticP = KURn

4/3

Figure 3: Phase diagram for fermions in the log T vs. log n plane, with T expressed in units of mc2/k and nexpressed in units of n0. The equations of state for the four extreme phases are also shown.

T [K]

n [m−3]

1020 1025 1030 1035 1040 1045

100

105

1010

Classical, Ultra-Relativistic

Classical, Non-Relativistic

Degenerate,Non-Relativistic

Degenerate,Ultra-Relativistic

Figure 4: Phase diagram for electrons in the log T vs. log n plane.


For the Fermi-momentum the expression becomes:

pF

mc=(n

n0

)1/3

. (50)

The T vs. n plane can now be divided in four regions containing degenerate (D) or non-degenerate (ND)gas and non-relativistic (NR) or ultra-relativistic (UR) gas. These regions overlap leading to the combinationsD-NR, D-UR, ND-NR, and ND-UR. The dividing line between the NR and UR cases can be found be realizingthat the fermion gas becomes ultra-relativistic when (3/2)kT � mc2 or εF � mc2. The dividing lines arethen given by:

32kT ≈ mc2 ∨ εF

mc2≈ 1 , (51)

which can be written as:

T =23mc2

k∨ n = 33/2n0 . (52)

The latter expression follows from equation (48), setting εF/mc2 to 1.The dividing line between the degenerate and non-degenerate fermion gas can be found by setting the

temperature equal to the Fermi temperature, i.e.:

32kT ≈ εF , (53)

which leads to:

T =23mc2

k

(( n

n0

)2/3

+ 1

)1/2

− 1

. (54)

The latter expression can be simplified in the non-relativistic (n� n0) and ultra-relativistic regimes (n� n0)to:

TNR =13mc2

k

(n

n0

)2/3

and TUR =23mc2

k

(n

n0

)1/3

. (55)

Figure 3 shows the phase diagram for fermions in the T vs. n plane, with T expressed in units of mc2/kand n expressed in units of n0. The three dividing lines derived above are indicated and the various regions arelabelled. Note that setting n/n0 = 33/2 in expression (54) leads to T = (2/3)mc2/k, meaning that the variousdividing lines meet in one point.

For fermions of a given rest mass and with known value of gs figure 3 can be translated to actual physicalunits. This is shown in figure 4 for electrons. For these fermions gs = 2 and m = 9.10938215 × 10−31 kg,which leads to log n0 = 35.77 (kg m−3) and log(mc2/k) = 9.77 (K).

5 The equation of state for the fermion gas

From the distribution of particles over energy one can also derive expressions for the pressure as function of theparticle density and temperature, i.e. the equation of state. In chapter 2 of Phillips (1998) it is shown that forthe classical regime one always obtains:

P = nkT . (56)

Thus the well-known classical ideal gas equation of state holds for the non-relativistic and ultra-relativisticregimes.

In order to derive the equation of state for the degenerate fermion gas I first write down the general expres-sion for the pressure of the gas:

P =13

∫ ∞0

n(p)pvp dp

= gs4π3

1h3

∫ ∞0

pvpf(εkin)p2 dp .

(57)


This result is derived in Weiss et al. (2004, chapter 10) and can also be found in Ostlie & Carroll (2007, section10.2). This equation can be worked out further using equations (23), (30), and (31):

P = gs4π3

1h3

∫ ∞0

f(εkin)vpp3 dp

= gs4π3

1h3

∫ ∞0

f(εkin)pc2

εp(mc)3

(( εkin

mc2+ 1)2− 1)3/2 εkin +mc2

pc2dεkin

= n0

∫ ∞0

f(εkin)(( εkin

mc2+ 1)2− 1)3/2

mc2 d( εkin

mc2

).

(58)

In terms of the rest-mass energy normalized kinetic energy ε′kin = εkin/mc2 the expression for the pressure

becomes:P

n0mc2=∫ ∞

0f(ε′kin)

((ε′kin + 1)2 − 1

)3/2dε′kin . (59)

The pressure is expressed in terms of the rest-mass energy density at particle concentration n0. For completelydegenerate fermion gases this expression reduces to:

P

n0mc2=∫ ε′F

0

((ε′kin + 1)2 − 1

)3/2dε′kin . (60)

This expression could in principle be worked out to obtain the equation of state for the completely degeneratefermion gas. However it is easier to proceed via the momentum distribution, writing:

P = gs4π3

1h3

∫ pF

0vpp

3 dp . (61)

By switching to x = p/mc as integration variable and using equation (23) for vp the integral becomes:

P

n0mc2=∫ pF/mc

0

x4 dx√1 + x2

. (62)

The solution can be found using formula 2.273-3 from Gradshteyn & Ryzhik (2007) and realizing that ln(x +√1 + x2) = arcsinhx. The expression becomes:

P

n0mc2=

18

(z(1 + z2)1/2(2z2 − 3) + 3 arcsinh z

), (63)

where z = pF/mc = (n/n0)1/3 (equation 50).This is the general equation of state for the completely degenerate Fermion gas. Note that it does not

depend on temperature, but only on density (as expected, since T = 0 K). From this general expression theequations of state for the non-relativistic and ultra-relativistic regimes can be derived. Starting with the latter,for which z � 1, the leading term in eq. (63) is the only one that remains. This can be seen by considering thatarcsinh z = ln(z +

√z2 + 1). This means that the arcsinh term tends to ln(2z) which means that the last term

in eq. (63) will be negligible compared to the leading term which tends to 2z4 as z → ∞. The expression forthe pressure in the ultra-relativistic case then becomes:

PUR

n0mc2=

14

( pF

mc

)4/3=

14

(n

n0

)4/3

. (64)

For the non-relativistic limit (z � 1) a Taylor expansion of the above equation for the pressure can be used3

but it is easier to return to the integral (60) and note that for ε′F = εF/mc2 � 1 (i.e. x� 1) the integrand tends

3The Taylor expansion should be done up to and including the 5th power of z.


to (1 + 2x− 1)3/2 = 23/2x3/2. Thus the expression for the pressure now becomes:

PNR

n0mc2=∫ ε′F

023/2x3/2 dx

=[23/2 2

5x5/2

]ε′F0

.

(65)

Using expression (47) for the non-relativistic Fermi energy the pressure in the non-relativistic case is:

PNR

n0mc2=

15

(n

n0

)5/3

. (66)

Writing out the normalized expressions for the pressure one obtains the following equations of state for thenon-relativistic and ultra-relativistic degenerate fermion gas:

PNR =h2

5m

(3

4πgs

)2/3

n5/3 = KNRn5/3 , (67)

PUR =hc

4

(3

4πgs

)1/3

n4/3 = KURn4/3 . (68)

For an electron gas log(n0mc2) = 22.68 (N m−2).

In the diagrams shown in figures 3 and 4 the contours of constant pressure form vertical lines in the degen-erate area, while in the classical regime the contours are lines at a 45 degree angle according to:

log(

T

mc2/k

)= log

(P

n0mc2

)− log

(n

n0

). (69)

The contour lines should smoothly change to vertical in the transition regime between classical and degenerategases.

6 Density and pressure in terms of Fermi-Dirac integrals

In this note the full expressions for the density and pressure in terms of the relativistically formulated fermionenergies was derived. For the density we have from eq. (42):

n

n0=∫ ∞

0

3 (ε′kin + 1)((ε′kin + 1)2 − 1

)1/2eβ

′(ε′kin−µ̄′) + 1dε′kin , (70)

where ε′kin = εkin/mc2, β′ = mc2/kT , µ̄′ = µ̄/mc2, and:

n0 = gs4π3

(mch

)3. (71)

For the pressure the following expression was found:

P

n0mc2=∫ ∞

0

((ε′kin + 1)2 − 1

)3/2eβ

′(ε′kin−µ̄′) + 1dε′kin . (72)

In the non-relativistic regime with ε′kin � 1 and ε = mc2ε′kin these expression reduce to:

n

n0=

3√

2(mc2)3/2

∫ ∞0

ε1/2

eβ(ε−µ̄) + 1dε (73)

andP

n0mc2=

2√

2(mc2)5/2

∫ ∞0

ε3/2

eβ(ε−µ̄) + 1dε (74)


In the expressions above the energy and chemical potential are normalized tomc2. In the literature however,the integrals are written in terms of x = εkin/kT and the parameters θ = kT/mc2 and η = µ̄/kT . The integralsabove can be rewritten in these terms by using:

x =εkin

kT=mc2

kT

εkin

mc2=ε′kin

θ. (75)

The expression for the density then becomes:

n

n0=∫ ∞

0

3(xθ + 1)((xθ + 1)2 − 1

)1/2θ dx

ex−η + 1. (76)

Making use of (xθ+ 1)2 = 2θx+ θ2x2 the numerator of the integrand can be factored differently, resulting in:

n

n0= 3√

2θ3/2

∫ ∞0

x1/2(1 + 1

2θx)1/2 (1 + θx) dx

ex−η + 1. (77)

Similarly the expression for the pressure becomes:

P

n0mc2= 2√

2θ5/2

∫ ∞0

x3/2(1 + 1

2θx)3/2

dx

ex−η + 1. (78)

These expressions are consistent with equations (24.91) and (24.92) in Weiss et al. (2004). For the non-relativistic regime we have xθ � 1 and then the expressions for density and pressure reduce to:

n

n0= 3√

2θ3/2

∫ ∞0

x1/2 dx

ex−η + 1, (79)

andP

n0mc2= 2√

2θ5/2

∫ ∞0

x3/2 dx

ex−η + 1. (80)

The integrals in terms of x above can all be expressed in terms of the so-called generalized Fermi-Diracintegral:

Fν(η, θ) =∫ ∞

0

xν(1 + 12θx)1/2

ex−η + 1dx , (81)

which is discussed in, e.g., Press et al. (2007, section 6.10). In the non-relativistic case θ = 0. Much work inthe literature has gone into finding accurate approximations to these integrals. Press et al. (2007) present oneparticular algorithm for evaluating the approximation.

For the density and pressure we can write:

n

n0= 3√

2θ3/2(F1/2(η, θ) + θF3/2(η, θ)

), (82)

andP

n0mc2= 2√

2θ5/2

(F3/2(η, θ) +

12θF5/2(η, θ)

). (83)

In the non-relativistic case we can write:

n

n0= 3√

2θ3/2F1/2(η) , (84)

andP

n0mc2= 2√

2θ5/2F3/2(η) , (85)

where Fν(η) = Fν(η, 0).

7 The energy density of the fermion gas 15

7 The energy density of the fermion gas

So far the energy density of the fermion gas has not been considered but it is and important quantity especiallywhen we want to understand why there is an upper limit to white dwarf and neutron star masses. The energydensity u as a function of momentum or kinetic energy is simply:

u(p) dp = n(p)εkin dp or u(εkin) = n(εkin)εkin dεkin . (86)

Using the normalizations introduced in the preceding sections we can write for the total energy density:

u

n0mc2=∫ ∞

0

3ε′kin (ε′kin + 1)((ε′kin + 1)2 − 1

)1/2eβ

′(ε′kin−µ̄′) + 1dε′kin , (87)

where equation (70) was used. In terms of Fermi-Dirac integrals this becomes:

u

n0mc2= 3√

2θ5/2

∫ ∞0

x3/2(1 + 1

2θx)1/2 (1 + θx) dx

ex−η + 1

= 3√

2θ5/2(F3/2(η, θ) + θF5/2(η, θ)

).

(88)

In the non-relativistic limit (xθ � 1) we have:

u

n0mc2= 3√

2θ5/2

∫ ∞0

x3/2 dx

ex−η + 1

= 3√

2θ5/2F3/2(η) .

(89)

This last expression is the familiar result that for non-relativistic ideal gases:

P =23u . (90)

Analogous to the pressure we can write in the completely degenerate case:

u = gs4πh3

∫ pF

0εkinp

2 dp . (91)

Switching again to x = pF/mc we obtain:

u

n0mc2= 3

∫ pF/mc

0

((x2 + 1)1/2 − 1

)x2 dx , (92)

for which the solution is:

u

n0mc2=

18

(−8x3 + 3(x2 + 1)1/2(2x3 + x)− 3 arcsinhx

). (93)

In the non-relativistic limit (x � 1) this expression tends to 310x

5 (which can be derived starting from theintegral for u or by expanding the arcsinh term to order x5) and in the ultra-relativistic case (x� 1) it tends to34x

4. Thus the expressions for the energy density become:

uNR

n0mc2=

310

(n

n0

)5/3

, (94)

uUR

n0mc2=

34

(n

n0

)4/3

. (95)

From the latter expression we obtain the expected result for an ultra-relativistic ideal gas:

P =13u . (96)


−15 −10 −5 0 5 10log(n/n0)

−12

−10

−8

−6

−4

−2

0

2

log(kT/m

c2)

-25.0

-20.0

-15.0

-10.0

-5.0

0.0

5.0

10.0-30.0 -5.0

0.0

10.0

1.0e+

05

1.0e+10

Phase diagram for fermions, chemical potential expressed as η = (µ−mc2)/kT

Figure 5: Phase diagram for the fermion gas. The solid contours show lines of constant log(P/n0mc2) and

constant η is shown by the dashed contours. The colour coding and the dividing lines indicate in which regionsthe gas is ultra/non-relativistic and classical/degenerate. Green indicates a classical, non-relativistic gas, cyan aclassical ultra-relativistic gas, yellow a degenerate non-relativistic gas, and white a degenerate ultra-relativisticgas.

8 The full fermion phase diagram

The Fermi-Dirac integrals from the previous section can be evaluated using, for example, the algorithm fromPress et al. (2007, section 6.10). This means that one can map the density and pressure as a function of chemicalpotential and temperature. However, the phase diagram in the T vs n plane is what is usually shown (and iseasier to understand). The problem is that the chemical potential cannot easily be obtained for a given choiceof density and temperature. This requires solving the following equation for η:

3√

2θ3/2(F1/2(η, θ) + θF3/2(η, θ)

)− y/n0 = 0 , (97)

where y is the chosen density for a given temperature parameter θ. This equation has to be solved numericallyand this can be done with the bisection method described in section 9.1 of Press et al. (2007). The root of theequation can first be bracketed with the zbrac function and then found with the rtbis function. Because ofthe enormous range of values of η (many decades) some care has to be taken in the precision demanded of thebisection method. In practice this was done by finding the root in two steps. First a rough determination wasdone by demanding an accuracy of 1% relative to the interval returned by zbrac. This rough determinationwas then used to scale the accuracy of η to 10−7 of the first estimate.

The resulting phase diagram in the log θ vs. log(n/n0) plane is show in figure 5. The solid contoursshow lines of constant log(P/n0mc

2) and constant η is shown by the dashed contours. Note how the pressurecontours indeed behave as predicted above, sloping at an angle of 45◦ in the classical regime and becoming


vertical in the degenerate region of the phase diagram. The contours of constant η show different slopes in thenon-relativistic and ultra-relativistic regimes. The transition from non-degenerate to a degenerate gas occursaround the η = 0 contour. Note that large negative values of η in the classical regime.

The degree of degeneracy D and the degree to which the gas is relativisticR can be judged from the valuesof kT/mc2 and µ̄/mc2 = θη as discussed in Weiss et al. (2004). However a nicer definition of these quantitiesis:

D =1n

∫ ε′F

0n(ε′kin) dε′kin . (98)

For a completely degenerate gas (T → 0) the the value of D is 1. The fraction of particles with p > mcmeasures how non-relativistic the gas is:

R =1n

∫ ∞√

2−1n(ε′kin)dε′kin . (99)

The colour scale in figure 5 shows the values of D and R. The degree of degeneracy of the gas is indi-cated with the progressive saturation of the red colour channel while the extent to which the gas is relativisticis indicated by progressive saturation of the blue channel. The green channel was set to its highest value.Where the colours mix a combination of states is indicated. Hence green means that the gas is classical andnon-relativistic, cyan means it is classical and ultra-relativistic, yellow means the gas is degenerate and non-relativistic, and white means the gas is degenerate and ultra-relativistic.

Figures 6–9 (at the back of this note) show examples of f(ε′kin) and n(ε′kin) for each of the four regions infigure 5. The number densities in these figures are expressed in units of n0.

Some remarks on the average occupation of states and the energy distribution:

• As discussed for the non-relativistic case, as the gas becomes degenerate the function f(ε′kin) approachesa step function. At the limit T = 0 all the states with ε′kin < µ̄/mc2 = εF/mc

2 are occupied while nonewith energies above this limit are. In this case the fermions only occupy the lowest possible kinetic energystates above ε′kin = 0, one fermion per single-particle state according to the Pauli exclusion principle.Note that even at T = 0 the gas will exert a pressure because p 6= 0 for all fermions!

• At this limit (T = 0) the energy distribution becomes: n(ε′kin) ∝ (ε′kin + 1)((ε′kin + 1)2 − 1

)1/2 for0 < ε′kin < µ̄/mc2 and zero elsewhere. In this limit of the completely degenerate fermion gas thenumber distribution in terms of p goes as n(p) ∝ p2. The limiting case is shown in the diagrams with thedashed lines.

• In the limit that T → 0 and the gas remains non-relativistic one can use that ε′kin � 1 and then theexpression for n(ε′kin) converges to n(ε′kin) ∝

√2(ε′kin)1/2. Removing the normalization by mc2 to get

n(εkin) dεkin then leads to the correct expression given by equation (8).

• When T → 0 and the gas is ultra-relativistic the expression for n(ε′kin) is proportional to (ε′kin + 1)2

(εkin = pc for an ultra-relativistic gas).

• Figures 10 and 11 show the border cases for the non-relativistic and ultra-relativistic fermion gas.


0 0.02 0.04 0.06

1×10−5

2×10−5

εkin/mc2

f(εkin)

Fermion occupation of states, µ/mc2 = − 1.06×10−1, log(kT/mc2) = −2.00

0 0.02 0.04 0.060

1×10−6

2×10−6

3×10−6

4×10−6

εkin/mc2

n(εkin)n0

Fermion kinetic energy distribution: log(n/n0) = −7.00, log(εF/mc2) = −4.97

Figure 6: The average of occupation of the energy states f(εkin) (top panel) and the corresponding distributionof the number density of fermions as a function of kinetic energy n(εkin). All energies are expressed in unitsof mc2, including µ̄ and kT . These diagrams represent a classical non-relativistic fermion gas according tofigure 5.

0 20 40 60

5×10−6

1×10−5

1.5×10−5

εkin/mc2

f(εkin)

Fermion occupation of states, µ/mc2 = − 1.11×102, log(kT/mc2) = 1.00

0 20 40 600

1×10−3

2×10−3

εkin/mc2

n(εkin)n0


Figure 7: The average of occupation of the energy states f(εkin) (top panel) and the corresponding distributionof the number density of fermions as a function of kinetic energy n(εkin). All energies are expressed in unitsof mc2, including µ̄ and kT . These diagrams represent a classical ultra-relativistic fermion gas according tofigure 5.


0 0.0005 0.0010

0.2

0.4

0.6

0.8

1

εkin/mc2

f(εkin)

Fermion occupation of states, µ/mc2 = 1.08×10−3, log(kT/mc2) = −5.00

0 0.0005 0.0010

0.05

0.1

εkin/mc2

n(εkin)n0


Figure 8: The average of occupation of the energy states f(εkin) (top panel) and the corresponding distributionof the number density of fermions as a function of kinetic energy n(εkin). All energies are expressed in unitsof mc2, including µ̄ and kT . These diagrams represent a degenerate non-relativistic fermion gas according tofigure 5.

0 2 4 60

0.2

0.4

0.6

0.8

1

εkin/mc2

f(εkin)

Fermion occupation of states, µ/mc2 = 5.89×100, log(kT/mc2) = −1.50

0 2 4 60

50

100

εkin/mc2

n(εkin)n0

Fermion kinetic energy distribution: log(n/n0) = 2.50, log(εF/mc2) = 0.77

Figure 9: The average of occupation of the energy states f(εkin) (top panel) and the corresponding distributionof the number density of fermions as a function of kinetic energy n(εkin). All energies are expressed in unitsof mc2, including µ̄ and kT . These diagrams represent a degenerate ultra-relativistic fermion gas according tofigure 5.


0 0.0005 0.001 0.00150

0.2

0.4

0.6

0.8

1

εkin/mc2

f(εkin)

Fermion occupation of states, µ/mc2 = 1.07×10−3, log(kT/mc2) = −4.00

0 0.0005 0.001 0.00150

0.05

0.1

εkin/mc2

n(εkin)n0


Figure 10: The average of occupation of the energy states f(εkin) (top panel) and the corresponding distributionof the number density of fermions as a function of kinetic energy n(εkin). All energies are expressed in unitsof mc2, including µ̄ and kT . These diagrams represent a partially degenerate non-relativistic fermion gasaccording to figure 5.

0 5 10 150

0.2

0.4

0.6

0.8

1

εkin/mc2

f(εkin)

Fermion occupation of states, µ/mc2 = 7.25×100, log(kT/mc2) = 0.00

0 5 10 150

50

100

150

200

εkin/mc2

n(εkin)n0

Fermion kinetic energy distribution: log(n/n0) = 2.80, log(εF/mc2) = 0.88

Figure 11: The average of occupation of the energy states f(εkin) (top panel) and the corresponding distributionof the number density of fermions as a function of kinetic energy n(εkin). All energies are expressed in unitsof mc2, including µ̄ and kT . These diagrams represent a partially degenerate ultra-relativistic fermion gasaccording to figure 5.

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