some notes on the fermion gasbrown/college_sterren/...results for the non-relativistic regime from...

Some notes on the ideal fermion gas

Anthony [email protected]

08.04.2013

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

The discussion in this note concerns the academic case of a pure (consisting of one typeof particle) ideal fermion gas. What is neglected are the interactions between the particles(e.g., the Coulomb force in the case of electrons) and the fact that in practice the fermionswill be embedded in a gas consisting of a number of different kinds of particles (think of thecarbon, oxygen, etc nuclei in white dwarfs and the neutrons and protons in neutron stars).This will introduce the important force of gravity of the heavier particles and alter theequation of state. Lastly, processes arising from nuclear physics are completely ignored.

Revision History

Issue Rev. No. Date Author Comments

2 1 08.04.2013 AB Typo fixed and definition of n added.

2 0 27.04.2010 AB Slight reordering of text and addition of classical dilute gasexample. Text made available on ‘Sterren’ website.

1 1 30.03.2009 AB Improvements in text following reading of Weiss et al. (2004),alternative development of density and pressure in terms ofFermi-Dirac integrals.

1 0 17.03.2009 AB Update to include more material from statistical physics inthe non-relativistic regime. Fermi-energy derived in terms ofrelativistic kinetic energy. Equation of state for degeneratefermion gas derived for non-relativistic and ultra-relativisticcase.

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& Sons ltd, Chichester

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

Weiss A., Hillebrandt W., Thomas H.-C., Ritter H., 2004, Cox & Giuli’s Principles of Stellar Structure,extended second 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 perfectgas and is defined in (Mandl, 1988, chapter 7) as follows: The perfect gas represents an idealization inwhich the potential energy of interaction between gas particles is negligible compared to their kineticenergy of motion. This definition implies that one can write down a set of private energies εi for eachparticle, which simplifies the application of statistical physics to gases enormously. In practice onecan indeed often neglect the interaction energies of particles.

However, quantum effects cannot be neglected so readily and one needs to take the particle concen-tration into account. The ideal gas usually treated in elementary textbooks is therefore the classicalideal gas for which the condition is that the probability is very small that any single-particle state isoccupied by more than one particle. An ideal gas for which this last condition does not hold is an idealquantal gas. These kinds of gases have to be treated with quantum statistics. In the book by Mandl(1988) this is done extensively for the non-relativistic regime, while in Phillips (1998) the ideal quantalgas is treated in both the relativistic and non-relativistic regimes. The next section summarizes theresults 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 focusingon fermions and bosons. For these particles the mean occupation number fi for the i-th single-particlestate, with energy 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 isfor bosons, for which the Bose-Einstein (BE) distribution holds. In this equation β = 1/kT and µ isthe chemical potential. The latter is a thermodynamic quantity relevant for systems with a variablenumber of particles (e.g., chemical reactions). From here on we concentrate on fermions for whichequation (1) shows that fi is never larger than 1. This reflects the Pauli exclusion principle whichstates that no more than one fermion can occupy a given quantum state.

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

g(p)dp = gs4πV

h3p2 dp , (2)

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

N(p) dp = gs4πV

h3p2

eβ(ε−µ) + 1dp , (3)

where the function fi is now written in its continuum form f(ε). This expression is entirely generalfor ideal fermion 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)

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

1.2 The classical dilute gas 4

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

g(ε) dε = gs2πV

h3(2m)3/2ε1/2 dε . (5)

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

N(ε) dε = f(ε)g(ε) dε = gs2πV

h3(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 = gs2πV

h3(2m)3/2

∫ ∞0

ε1/2 dε

eβ(ε−µ) + 1. (7)

In terms of the particle concentration, or number density n (number of particles per unit volume), thelast two equations can be written as:

n(ε) dε = f(ε)g(ε) dε = gs2π

h3(2m)3/2

ε1/2

eβ(ε−µ) + 1dε , (8)

and

n = gs2π

h3(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 offermions in a fixed volume (eq. 6), or for a fixed number density (eq. 9). For a fixed number andvolume the chemical potential is a function of temperature only µ = µ(T ) (cf. Mandl, 1988, section11.5.1).

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 stateis much lower than 1. In this case the fermion gas obeys Maxwell-Boltzmann statistics and is said tobehave like a dilute classical gas. This can be verified by starting from equation (8) and using theassumption above in which case the following expression for n(ε) holds:

n(ε) dε = eµ/kT gs2π

h3(2m)3/2e−ε/kT ε1/2 dε , (11)

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

n(p) dp = eµ/kT gs1

h3e−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 en-ergies and momenta for the ideal classical gas. In this case the chemical potential can be determinedfrom:

n =

∫ ∞0

n(ε) dε and

∫ ∞0

e−xx1/2 dx =

√π

2.

The resulting chemical potential is:

µ = kT ln

(n

gsnQ

),

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

045

J−1

m−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 ofparticles n(ε) (right panel) as a function of energy for a nitrogen (N2) gas at room temperature andatmospheric pressure. The energy is expressed in units of kT .

where nQ is the so-called quantum concentration:

nQ =

(2πmkT

h2

)3/2.

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

n(ε) dε =2

π1/21

(kT )3/2e−ε/kT ε1/2 dε

Figure 1 shows the mean occupation number for particle states with energy ε and the particle con-centration as a function of energy for the example of a classical N2 gas at room temperature andatmospheric pressure. Note the very small, O(10−8), values of the average occupation number, show-ing that this is indeed a dilute classical gas.

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 potentialat the absolute 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 andε > 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

4πh3p2 dp p ≤ pF

0 p > pF, (15)

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-relativistic fermion gas at zero temperature: T = 0 K. The energy is expressed in units of the Fermienergy εF.

while the distribution over energies becomes:

n(ε) dε =

{gs

2πh3

(2m)3/2ε1/2 dε ε ≤ εF0 ε > εF

. (16)

The quantity pF is the Fermi momentum corresponding to the Fermi energy. The interpretation of thisparticle distribution is as follows. At T = 0 K the gas is in its lowest energy state but because of thePauli exclusion principle not all N fermions can crowd in to the single-particle state of lowest energy.The Fermi energy then is the topmost energy level occupied at T = 0 K. The energy distribution atT = 0 K will be proportional to ε1/2 with a sharp cutoff at ε = εF. Figure 2 shows the occupationnumber and energy distribution for the completely degenerate fermion gas.

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

n = gs2π

h3(2m)3/2

∫ εF0

ε1/2 dε = gs2π

h3(2m)3/2

2

3ε

3/2F , (17)

which leads to:

εF =h2

2m

(3

gs4π

)2/3n2/3 =

~2m

(6π2

gs

)2/3n2/3 . (18)

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

εF = kTF . (19)

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

n = gs4π

h3

∫ pF0

p2 dp , (20)

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

pF = h

(3

gs4π

)1/3n1/3 = ~

(3

gs2π2

)1/3n1/3 . (21)

So far the (ideal) fermion gas has been treated in the non-relativistic regime. However, for theapplication to stars the possibility of relativistic energies has to be taken into account and this is donein Phillips (1998). The following material is extracted from that text and further worked out. It willbecome clear that the energy distribution for a fermion gas will be slightly different when the fullrelativistic formulation for particle energies is 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 tospecial relativity. The energy �p of a particle of mass m in quantum state with momentum p is thengiven by:2

�2p = p2c2 +m2c4 , (22)

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

vp =pc2

�p, (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:

�pmc2

=

[( pmc

)2+ 1

]1/2, (25)

andp

mc=

[( �pmc2

)2− 1]1/2

. (26)

These expressions give the energy and momentum in a form normalized to the rest-mass energy andthe dividing line between non-relativistic and relativistic momenta. In addition the following relationbetween energy and momentum is useful.

d�pdp

=pc2

�p= vp or dp =

�ppc2

d�p =1

vpd�p , (27)

For most applications it is the kinetic energy of the gas that is important as only gas particles inmotion can exert a pressure, transport energy, etc. The kinetic energy �kin is simply given by:

�kin = �p −mc2 . (28)

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

�kinmc2

=

[( pmc

)2+ 1

]1/2− 1 , (29)

andp

mc=

[( �kinmc2

+ 1)2− 1]1/2

, (30)

andd�kindp

=pc2

�kin +mc2or dp =

�kin +mc2

pc2d�kin . (31)

2The variable � 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 �p defined in the previous section we can now write for the meanoccupation number:

f(�p) =1

e(�p−µ)/kT + 1=

1

e(�kin−(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(�p)g(p) dp , (34)

and

n =

∫ ∞0

f(�p)g(p) dp . (35)

In addition to classical or quantal the gas can now be ultra-relativistic, in which case for a significantfraction of the particles p � mc, or non-relativistic, if for most of the particles p � mc. This leadsto four possible extreme states of the fermion gas: classical and non-relativistic, classical and ultra-relativistic, degenerate and non-relativistic, degenerate and ultra-relativistic. At this point it shouldbe realized that when labelling the energies of the states that the fermions can occupy there is nopoint in counting the rest mass energy mc2. This is the same for all particles and cannot be used todistinguish quantum states. Therefore the equation for f and n can be written in terms of �kin, thekinetic energy of the fermions (i.e., the energy over and above the rest mass 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 energylevels from �p to �kinand the chemical potential from µ to µ̄:

f(�kin) =1

e(�kin−µ̄)/kT + 1, (37)

andn(p) dp = f(�kin)g(p) dp . (38)

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

n(p) dp = n(�kin) d�kin or n(�kin) = n(p)dp

d�kin. (39)

Starting from:

n(p) dp = f(�kin)gs4π

h3p2 dp . (40)

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

n(�kin) d�kin = f(�kin)gs4π

h3p2

dp

d�kind�kin

= f(�kin)gs4π

h3p�kin +mc

2

c2d�kin

= f(�kin)gs4π

h3p

mc

m

cmc2(�kin/mc

2 + 1) d�kin .

(41)


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

n( �kinmc2

)d( �kinmc2

)= f(�kin)gs

4π

h3m

c

[( �kinmc2

+ 1)2− 1]1/2

mc2(�kin +mc

2

mc2

)mc2 d

( �kinmc2

).

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

n(�′kin) = f(�′kin)gs4π

(mch

)3 (�′kin + 1

) ((�′kin + 1)

2 − 1)1/2

, (42)

where �′kin = �kin/mc2. Note that f(�kin) = f(�

′kin) because µ̄ and kT are also expressed in terms of

mc2. The units of mc/h are m−1 which means that n(�′kin) is indeed a number density (for a givenenergy interval).

The Fermi energy now becomes (in analogy to equation 17):

n = gs4π(mch

)3 ∫ �F0

(�′kin + 1

) ((�′kin + 1)

2 − 1)1/2

d�′kin = gs4π(mch

)3 13

((�′F + 1)

2 − 1)3/2

, (43)

where �′F = �F/mc2. In the limit �F/mc

2 � 1 the expression (17) is recovered. Thus the expressionfor the Fermi energy now becomes:

�Fmc2

=

((3

4π

n

gs

)2/3( hmc

)2+ 1

)1/2− 1 , (44)

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

�Fmc2

=

((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 expressionfor pF. Normalizing to mc we obtain:

pFmc

=

(3n

gs4π(mc/h)3

)1/3. (46)

Note that the expression for the Fermi-energy could thus have been obtained more easily by startingfrom pF and then using the now different relation between energy and momentum as given by equation(25). Keep in mind 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 relativisticexpression (eq. 44) one can proceed to draw the T vs n diagram and broadly indicate where the gas isdegenerate/non-degenerate and/or relativistic or non-relativistic. I first write the expressions for theFermi energies (where NR stands for non-relativistic) as follows:

�F,NRmc2

=1

2

(3n

gs4π(mc/h)3

)2/3=

1

2

(n

n0

)2/3(47)

and

�Fmc2

=

((n

n0

)2/3+ 1

)1/2− 1 , (48)


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/kand n expressed in units of n0. The equations of state for the four extreme phases are also shown.

where the density normalization is:

n0 = gs4π

3

(mch

)3. (49)

For the Fermi-momentum the expression becomes:

pFmc

=

(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 thecombinations D-NR, D-UR, ND-NR, and ND-UR. The dividing line between the NR and UR casescan be found be realizing that the fermion gas becomes ultra-relativistic when (3/2)kT � mc2 or�F � mc2. The dividing lines are then given by:

3

2kT ≈ mc2 ∨ �F

mc2≈ 1 , (51)

which can be written as:

T =2

3

mc2

k∨ n = 33/2n0 . (52)

The latter expression follows from equation (48), setting �F/mc2 to 1.

The dividing line between the degenerate and non-degenerate fermion gas can be found by settingthe temperature equal to the Fermi temperature, i.e.:

3

2kT ≈ �F , (53)


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.

which leads to:

T =2

3

mc2

k

(( nn0

)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 =1

3

mc2

k

(n

n0

)2/3and TUR =

2

3

mc2

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 ofmc2/k and n expressed in units of n0. The three dividing lines derived above are indicated and thevarious regions are labelled. Note that setting n/n0 = 3

3/2 in expression (54) leads to T = (2/3)mc2/k,meaning that the various dividing lines meet in one point.

For fermions of a given rest mass and with known value of gs figure 3 can be translated to actualphysical units. 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 asfunction of the particle density and temperature, i.e. the equation of state. In chapter 2 of Phillips(1998) it is shown that for the 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-relativistic regimes.


In order to derive the equation of state for the degenerate fermion gas I first write down the generalexpression for the pressure of the gas:

P =1

3

∫ ∞0

n(p)pvp dp

= gs4π

3

1

h3

∫ ∞0

pvpf(�kin)p2 dp .

(57)

This result is derived in Weiss et al. (2004, chapter 10) and can also be found in Ostlie & Carroll(2007, section 10.2). This equation can be worked out further using equations (23), (30), and (31):

P = gs4π

3

1

h3

∫ ∞0

f(�kin)vpp3 dp

= gs4π

3

1

h3

∫ ∞0

f(�kin)pc2

�p(mc)3

(( �kinmc2

+ 1)2− 1)3/2 �kin +mc2

pc2d�kin

= n0

∫ ∞0

f(�kin)

(( �kinmc2

+ 1)2− 1)3/2

mc2 d( �kinmc2

).

(58)

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

pressure becomes:P

n0mc2=

∫ ∞0

f(�′kin)((�′kin + 1)

2 − 1)3/2

d�′kin . (59)

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

P

n0mc2=

∫ �′F0

((�′kin + 1)

2 − 1)3/2

d�′kin . (60)

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

P = gs4π

3

1

h3

∫ pF0

vpp3 dp . (61)

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

P

n0mc2=

∫ pF/mc0

x4 dx√1 + x2

. (62)

The solution can be found using formula 2.273-3 from Gradshteyn & Ryzhik (2007) and realizing thatln(x+

√1 + x2) = arcsinhx. The expression becomes:

P

n0mc2=

1

8

(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 doesnot depend on temperature, but only on density (as expected, since T = 0 K). From this generalexpression the equations 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 that arcsinh 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 leadingterm which tends to 2z4 as z →∞. The expression for the pressure in the ultra-relativistic case thenbecomes:

PURn0mc2

=1

4

( pFmc

)4/3=

1

4

(n

n0

)4/3. (64)


For the non-relativistic limit (z � 1) a Taylor expansion of the above equation for the pressurecan be used3 but it is easier to return to the integral (60) and note that for �′F = �F/mc

2 � 1 (i.e.x � 1) the integrand tends to (1 + 2x − 1)3/2 = 23/2x3/2. Thus the expression for the pressure nowbecomes:

PNRn0mc2

=

∫ �′F0

23/2x3/2 dx

=

[23/2

2

5x5/2

]�′F0

.

(65)

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

PNRn0mc2

=1

5

(n

n0

)5/3. (66)

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

PNR =h2

5m

(3

4πgs

)2/3n5/3 = KNRn

5/3 , (67)

PUR =hc

4

(3

4πgs

)1/3n4/3 = KURn

4/3 . (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 inthe degenerate area, while in the classical regime the contours are lines at a 45 degree angle accordingto:

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 anddegenerate gases.

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 formulatedfermion energies was derived. For the density we have from eq. (42):

n

n0=

∫ ∞0

3 (�′kin + 1)((�′kin + 1)

2 − 1)1/2

eβ′(�′kin−µ̄′) + 1

d�′kin , (70)

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

n0 = gs4π

3

(mch

)3. (71)

For the pressure the following expression was found:

P

n0mc2=

∫ ∞0

((�′kin + 1)

2 − 1)3/2

eβ′(�′kin−µ̄′) + 1

d�′kin . (72)

In the non-relativistic regime with �′kin � 1 and ε = mc2�′kin these expression reduce to:

n

n0=

3√

2

(mc2)3/2

∫ ∞0

ε1/2

eβ(ε−µ̄) + 1dε (73)

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


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 to mc2. In the literaturehowever, the integrals are written in terms of x = �kin/kT and the parameters θ = kT/mc

2 andη = µ̄/kT . The integrals above can be rewritten in these terms by using:

x =�kinkT

=mc2

kT

�kinmc2

=�′kinθ. (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 + 12θx

)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 + 12θx

)3/2dx

ex−η + 1. (78)

These expressions are consistent with equations (24.91) and (24.92) in Weiss et al. (2004). For thenon-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-Dirac integral:

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. Muchwork in the literature has gone into finding accurate approximations to these integrals. Press et al.(2007) present one particular 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) +

1

2θF5/2(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 quantityespecially when we want to understand why there is an upper limit to white dwarf and neutron starmasses. The energy density u as a function of momentum or kinetic energy is simply:

u(p) dp = n(p)�kin dp or u(�kin) = n(�kin)�kin d�kin . (86)

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

u

n0mc2=

∫ ∞0

3�′kin (�′kin + 1)

((�′kin + 1)

2 − 1)1/2

eβ′(�′kin−µ̄′) + 1

d�′kin , (87)

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

u

n0mc2= 3√

2θ5/2∫ ∞

0

x3/2(1 + 12θx

)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 =2

3u . (90)

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

u = gs4π

h3

∫ pF0

�kinp2 dp . (91)

Switching again to x = pF/mc we obtain:

u

n0mc2= 3

∫ pF/mc0

((x2 + 1)1/2 − 1

)x2 dx , (92)

for which the solution is:

u

n0mc2=

1

8

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

). (93)

In the non-relativistic limit (x � 1) this expression tends to 310x5 (which can be derived starting

from the integral for u or by expanding the arcsinh term to order x5) and in the ultra-relativistic case(x� 1) it tends to 34x

4. Thus the expressions for the energy density become:

uNRn0mc2

=3

10

(n

n0

)5/3, (94)

uURn0mc2

=3

4

(n

n0

)4/3. (95)

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

P =1

3u . (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 indicatein which regions the gas is ultra/non-relativistic and classical/degenerate. Green indicates a classical,non-relativistic gas, cyan a classical ultra-relativistic gas, yellow a degenerate non-relativistic gas, andwhite a degenerate ultra-relativistic gas.

8 The full fermion phase diagram

The Fermi-Dirac integrals from the previous section can be evaluated using, for example, the algorithmfrom Press et al. (2007, section 6.10). This means that one can map the density and pressure as afunction of chemical potential and temperature. However, the phase diagram in the T vs n planeis what is usually shown (and is easier to understand). The problem is that the chemical potentialcannot easily be obtained for a given choice of density and temperature. This requires solving thefollowing 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 solvednumerically and this can be done with the bisection method described in section 9.1 of Press et al.(2007). The root of the equation can first be bracketed with the zbrac function and then found withthe rtbis function. Because of the enormous range of values of η (many decades) some care has tobe taken in the precision demanded of the bisection method. In practice this was done by finding theroot in two steps. First a rough determination was done by demanding an accuracy of 1% relative tothe interval returned by zbrac. This rough determination was then used to scale the accuracy of η to10−7 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 thepressure contours indeed behave as predicted above, sloping at an angle of 45◦ in the classical regimeand becoming vertical in the degenerate region of the phase diagram. The contours of constant ηshow different slopes in the non-relativistic and ultra-relativistic regimes. The transition from non-degenerate to a degenerate gas occurs around the η = 0 contour. Note that large negative values of ηin the classical regime.

The degree of degeneracy D and the degree to which the gas is relativistic R can be judged fromthe values of kT/mc2 and µ̄/mc2 = θη as discussed in Weiss et al. (2004). However a nicer definitionof these quantities is:

D = 1n

∫ ε′F0

n(�′kin) d�′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(�′kin)d�

′kin . (99)

The colour scale in figure 5 shows the values of D and R. The degree of degeneracy of the gas isindicated with the progressive saturation of the red colour channel while the extent to which the gasis relativistic is indicated by progressive saturation of the blue channel. The green channel was set toits highest value. Where the colours mix a combination of states is indicated. Hence green means thatthe gas is classical and non-relativistic, cyan means it is classical and ultra-relativistic, yellow meansthe 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(�′kin) and n(�′kin) for each of the four

regions in figure 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(�′kin)approaches a step function. At the limit T = 0 all the states with �′kin < µ̄/mc

2 = �F/mc2 are

occupied while none with energies above this limit are. In this case the fermions only occupythe lowest possible kinetic energy states above �′kin = 0, one fermion per single-particle stateaccording to the Pauli exclusion principle. Note that even at T = 0 the gas will exert a pressurebecause p 6= 0 for all fermions!

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

2 − 1)1/2

for0 < �′kin < µ̄/mc

2 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 diagramswith the dashed lines.

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

′kin) ∝

√2(�′kin)

1/2. Removing the normalization bymc2 to get n(�kin) d�kin then leads to the correct expression given by equation (8).

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

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


0 0.02 0.04 0.06

1×10−5

2×10−5

�kin/mc2

f(�kin)

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

�kin/mc2

n(�kin)

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(�kin) (top panel) and the correspondingdistribution of the number density of fermions as a function of kinetic energy n(�kin). All energies areexpressed in units of mc2, including µ̄ and kT . These diagrams represent a classical non-relativisticfermion gas according to figure 5.

0 20 40 60

5×10−6

1×10−5

1.5×10−5

�kin/mc2

f(�kin)

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

0 20 40 600

1×10−3

2×10−3

�kin/mc2

n(�kin)

n0


Figure 7: The average of occupation of the energy states f(�kin) (top panel) and the correspondingdistribution of the number density of fermions as a function of kinetic energy n(�kin). All energies areexpressed in units of mc2, including µ̄ and kT . These diagrams represent a classical ultra-relativisticfermion gas according to figure 5.


0 0.0005 0.0010

0.2

0.4

0.6

0.8

1

�kin/mc2

f(�kin)

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

0 0.0005 0.0010

0.05

0.1

�kin/mc2

n(�kin)

n0


Figure 8: The average of occupation of the energy states f(�kin) (top panel) and the correspondingdistribution of the number density of fermions as a function of kinetic energy n(�kin). All energies areexpressed in units of mc2, including µ̄ and kT . These diagrams represent a degenerate non-relativisticfermion gas according to figure 5.

0 2 4 60

0.2

0.4

0.6

0.8

1

�kin/mc2

f(�kin)

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

0 2 4 60

50

100

�kin/mc2

n(�kin)

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(�kin) (top panel) and the correspondingdistribution of the number density of fermions as a function of kinetic energy n(�kin). All energies areexpressed in units of mc2, including µ̄ and kT . These diagrams represent a degenerate ultra-relativisticfermion gas according to figure 5.


0 0.0005 0.001 0.00150

0.2

0.4

0.6

0.8

1

�kin/mc2

f(�kin)

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

�kin/mc2

n(�kin)

n0


Figure 10: The average of occupation of the energy states f(�kin) (top panel) and the correspondingdistribution of the number density of fermions as a function of kinetic energy n(�kin). All energiesare expressed in units of mc2, including µ̄ and kT . These diagrams represent a partially degeneratenon-relativistic fermion gas according to figure 5.

0 5 10 150

0.2

0.4

0.6

0.8

1

�kin/mc2

f(�kin)

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

0 5 10 150

50

100

150

200

�kin/mc2

n(�kin)

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(�kin) (top panel) and the correspondingdistribution of the number density of fermions as a function of kinetic energy n(�kin). All energiesare expressed in units of mc2, including µ̄ and kT . These diagrams represent a partially degenerateultra-relativistic fermion gas according to figure 5.

The ideal gasThe ideal non-relativistic quantal gasThe classical dilute gasThe degenerate non-relativistic fermion gas

Particle energies and momenta in special relativityThe relativistic description of the Fermion gasThe fermion gas phasesThe equation of state for the fermion gasDensity and pressure in terms of Fermi-Dirac integralsThe energy density of the fermion gasThe full fermion phase diagram

some notes on the fermion gasbrown/college_sterren/...results for the non-relativistic regime from...

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