Some notes on the ideal fermion gas

Transcription

1 Some notes on the ideal fermion gas Anthony Brown Abstract. These are notes on the physics of the ideal fermion gas collected from various sources. The text provides a deeper understing of the behaviour of degenerate fermion gases in order to apply this to white dwarfs 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 Coulomb force in the case of electrons) the fact that in practice the fermions will be embedded in a gas consisting of a number of different kinds of particles (think of the carbon, oxygen, etc nuclei in white dwarfs the neutrons protons in neutron stars). This will introduce the important force of gravity of the heavier particles alter the equation of state. Lastly, processes arising from nuclear physics are completely ignored. Revision History Issue Rev. No. Date Author Comments AB Slight reordering of text addition of classical dilute gas example. Text made available on Sterren website AB Improvements in text following reading of Weiss et al. (24), alternative development of density pressure in terms of Fermi- Dirac integrals AB Update to include more material from statistical physics in the nonrelativistic regime. Fermi-energy derived in terms of relativistic kinetic energy. Equation of state for degenerate fermion gas derived for non-relativistic ultra-relativistic case AB Document created.

2 REFERENCES 2 Contents The ideal gas 3. The ideal non-relativistic quantal gas The classical dilute gas The degenerate non-relativistic fermion gas Particle energies 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 6 Density pressure in terms of Fermi-Dirac integrals 3 7 The energy density of the fermion gas 5 8 The full fermion phase diagram 6 References Gradshteyn I.S., Ryzhik I.M., 27, Table of Integrals, Series, Products, seventh edition, A. Jeffrey & D. Zwillinger, eds., Academic Press Ml F., 988, Statistical Physics, second edition, The Manchester Physics Series, John Wiley & Sons ltd, Chichester Ostlie D.A., Carroll B.W., 27, An Introduction to Modern Stellar Astrophysics, Second Edition, Addison- Wesley, San Francisco, Phillips A.C., 998, 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., 27, Numerical Recipes: The Art of Scientific Computing, Third Edition, Cambridge University Press (Cambridge) Weiss A., Hillebrt W., Thomas H.-C., Ritter H., 24, Cox & Giuli s Principles of Stellar Structure, extended second edition, Cambridge Scientific Publishers (Cambridge)

3 . The ideal non-relativistic quantal gas 3 The ideal gas In this section the properties of ideal gases are reviewed. The ideal gas is often also called a perfect gas is defined in (Ml, 988, chapter 7) as follows: The perfect gas represents an idealization in which the potential 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 ε i for each particle, which simplifies the application of statistical physics to gases enormously. In practice one can indeed often neglect the interaction energies of particles. However, quantum effects cannot be neglected so readily one needs to take the particle concentration into account. The ideal gas usually treated in elementary textbooks is therefore the classical ideal gas for which the condition is that the probability is very small that any single-particle state is occupied by more than one particle. An ideal gas for which this last condition does not hold is an ideal quantal gas. These kinds of gases have to be treated with quantum statistics. In the book by Ml (988) this is done extensively for the non-relativistic regime, while in Phillips (998) the ideal quantal gas is treated in both the relativistic non-relativistic regimes. The next section summarizes the results for the non-relativistic regime from Ml (988).. The ideal non-relativistic quantal gas Chapters 9 in Ml (988) discuss statistical physics of quantal gases, with chapter focusing on fermions bosons. For these particles the mean occupation number f i for the i-th single-particle state, with energy ε i, is given by: f i = e β(εi µ) ±, () where the plus sign is for fermions, giving the Fermi-Dirac (FD) distribution, the minus sign is for bosons, for which the Bose-Einstein (BE) distribution holds. In this equation β = /kt µ is the chemical potential. The latter is a thermodynamic quantity relevant for systems with a variable number of particles (e.g., chemical reactions). From here on we concentrate on fermions for which equation () shows that f i is never larger than. This reflects the Pauli exclusion principle which states 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 number has to be multiplied by the so-called density of states, which basically describes in how many ways a particular energy state can be attained for a particle. In terms of momentum p the density of states g(p) is (see Phillips, 998, chapter 2): V g(p)dp = g s h 3 p2 dp, (2) where g s is the degeneracy of each state (for example g s = 2 for particles with two possible polarizations or spins for the same momentum). The particles are confined to the volume V h is the Planck constant. The number of particles at a certain energy is then given by N(ε i ) f i g(p i ), where p i = p(ε i ). De distribution of the particles over the different energies can then be written in terms of the momentum p as: V p 2 N(p) dp = g s h 3 e β(ε µ) dp, (3) + where the function f i is now written in its continuum form f(ε). This expression is entirely general for ideal fermion gases as we have not yet specified the relation between momentum 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: The material in this section is extracted from Ml (988) g(ε) dε = g s 2πV h 3 (2m)3/2 ε /2 dε. (5)

4 The particle number N(ε) per energy interval now becomes:.2 The classical dilute gas 4 2πV ε /2 N(ε) dε = f(ε)g(ε) dε = g s h 3 (2m)3/2 e β(ε µ) dε. (6) + The total number of fermions is now obtained by integrating (6) which results in: N = g s 2πV h 3 (2m)3/2 ε /2 dε e β(ε µ) +. (7) In terms of the particle concentration, or number density, n the last two equations can be written as: 2π ε /2 n(ε) dε = f(ε)g(ε) dε = g s h 3 (2m)3/2 e β(ε µ) dε, (8) + 2π ε /2 dε n = g s h 3 (2m)3/2 e β(ε µ) +. (9) Equations (7) (9) provide an implicit definition of the chemical potential for a fixed number of fermions in a fixed volume (eq. 6), or for a fixed number density (eq. 9). For a fixed number volume the chemical potential is a function of temperature only µ = µ(t ) (cf. Ml, 988, section.5.)..2 The classical dilute gas If the condition e βµ () is satisfied the term + in f i in equation () can be ignored the average occupation of each state is much lower than. In this case the fermion gas obeys Maxwell-Boltzmann statistics is said to behave like a dilute classical gas. This can be verified by starting from equation (8) using the assumption above in which case the following expression for n(ε) holds: which in terms of momentum (using expression 4) becomes: n(ε) dε = e µ/kt g s 2π h 3 (2m)3/2 e ε/kt ε /2 dε, () or in terms of particle velocities: n(p) dp = e µ/kt g s h 3 e p2 /2mkT p 2 dp, (2) n(p) dp = e µ/kt g s ( m h ) 3 e mv 2 /2kT v 2 dv. (3) Equations () (3) represent of course the familiar Maxwell-Boltzmann distribution of particle energies momenta for the ideal classical gas. In this case the chemical potential can be determined from: n = The resulting chemical potential is: n(ε) dε where n Q is the so-called quantum concentration: ( ) n µ = kt ln, g s n Q e x x /2 dx = π 2. ( ) 2πmkT 3/2 n Q =. h 2

5 .3 The degenerate non-relativistic fermion gas N 2 gas at T = 3 K, P = 5 N m f(ε) n(ε) [ 45 J m 3 ] ε/kt ε/kt Figure : The mean occupation number f(ε) (left panel) the corresponding number density of particles n(ε) (right panel) as a function of energy for a nitrogen (N 2 ) gas at room temperature atmospheric pressure. The energy is expressed in units of kt. Substituting this value of the chemical potential in () one correctly obtains the expression: n(ε) dε = 2 π /2 (kt ) 3/2 e ε/kt ε /2 dε Figure shows the mean occupation number for particle states with energy ε the particle concentration as a function of energy for the example of a classical N 2 gas at room temperature atmospheric pressure. Note the very small, O( 8 ), values of the average occupation number, showing that this is indeed a dilute classical gas..3 The degenerate non-relativistic fermion gas We return now to the quantal gas define the Fermi energy ε F as the value of the chemical potential at the absolute zero temperature: ε F µ(). (4) It is shown in Ml (988) that ε F must be positive. As T β which means that exp[β(ε µ)] tends to exp( ) for ε < ε F ε > ε F, respectively. That is, f(ε) tends to or for ε < ε F ε > ε F respectively. The fermion gas in this state is said to be completely degenerate. The distribution of particles over the momenta then becomes: { g s p 2 dp p p n(p) dp = h 3 F, (5) p > p F while the distribution over energies becomes: { 2π g s (2m) 3/2 ε /2 dε n(ε) dε = h 3 ε ε F. ε > ε F (6)

6 .3 The degenerate non-relativistic fermion gas 6 f(ε) n(ε) ε/ε F ε/ε F Figure 2: The Fermi-Dirac mean occupation number f(ε) energy distribution n(ε) for a non-relativistic fermion gas at zero temperature: T = K. The energy is expressed in units of the Fermi energy ε F. The quantity p F is the Fermi momentum corresponding to the Fermi energy. The interpretation of this particle distribution is as follows. At T = K the gas is in its lowest energy state but because of the Pauli 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 = K. The energy distribution at T = K will be proportional to ε /2 with a sharp cutoff at ε = ε F. Figure 2 shows the occupation number energy distribution for the completely degenerate fermion gas. The Fermi energy can now be derived from equation (6): which leads to: n = g s 2π h 3 (2m)3/2 εf ( ) ε F = h2 3 2/3 n 2/3 = 2m g s 2m ε /2 dε = g s 2π h 3 (2m)3/2 2 3 ε3/2 F, (7) ( ) 6π 2 2/3 n 2/3. (8) From this result it is clear that ε F depends on the fermion mass m the particle concentration n. The Fermi temperature is defined as: ε F = kt F. (9) In terms of momentum one can write for the degenerate case: n = g s h 3 pf g s p 2 dp, (2) where p F is now the Fermi-momentum. Working out the integral above solving for p F gives: ( ) 3 /3 ( ) 3 /3 p F = h n /3 = g s g s 2π 2 n /3. (2) So far the (ideal) fermion gas has been treated in the non-relativistic regime. However, for the application to stars the possibility of relativistic energies has to be taken into account this is done in Phillips (998). The following material is extracted from that text further worked out. It will become clear that the energy distribution for a fermion gas will be slightly different when the full relativistic formulation for particle energies is taken into account.

7 2 Particle energies momenta in special relativity 7 2 Particle energies momenta in special relativity I first recall the full expressions for energy momentum for the particles in a gas according to special relativity. The energy ɛ p of a particle of mass m in quantum state with momentum p is then given by: 2 where c is the speed of light. The speed of the particle v p is given by: which is consistent with the familiar form: ɛ 2 p = p 2 c 2 + m 2 c 4, (22) p = v p = pc2 ɛ p, (23) mv p (vp /c) 2. (24) For later use the following expressions for the energy momentum are useful: [ ɛ ( p p ) ] 2 /2 mc 2 = +, (25) mc [ p ( mc = ɛp ) ] 2 /2 mc 2. (26) These expressions give the energy momentum in a form normalized to the rest-mass energy the dividing line between non-relativistic relativistic momenta. In addition the following relation between energy momentum is useful. dɛ p dp = pc2 = v p or dp = ɛ p ɛ p pc 2 dɛ p = dɛ p, (27) v p For most applications it is the kinetic energy of the gas that is important as only gas particles in motion can exert a pressure, transport energy, etc. The kinetic energy ɛ kin is simply given by: Similar useful relations to the ones for ɛ p p can be derived in this case: ɛ kin = ɛ p mc 2. (28) [ ɛ ( kin p ) ] 2 /2 mc 2 = +, (29) mc [ p ( mc = ɛkin ) ] 2 /2 mc 2 +, (3) dɛ kin dp = pc 2 ɛ kin + mc 2 or dp = ɛ kin + mc 2 pc 2 dɛ kin. (3) 2 The variable ɛ rather than ε is used here for energy in order emphasise that the relativistic formulation is used.

8 3 The relativistic description of the Fermion gas 8 3 The relativistic description of the Fermion gas With the energy of the fermions ɛ p defined in the previous section we can now write for the mean occupation number: f(ɛ p ) = e (ɛp µ)/kt + = e (ɛ kin (µ mc 2 ))/kt +, (32) so that the condition for a classical gas now becomes: e (mc2 µ)/kt (33) The density of states is still given by equation (2) so for the energy distribution we can write: n(p) dp = f(ɛ p )g(p) dp, (34) n = f(ɛ p )g(p) dp. (35) In addition to classical or quantal the gas can now be ultra-relativistic, in which case for a significant fraction of the particles p mc, or non-relativistic, if for most of the particles p mc. This leads to four possible extreme states of the fermion gas: classical non-relativistic, classical ultra-relativistic, degenerate non-relativistic, degenerate ultra-relativistic. At this point it should be realized that when labelling the energies of the states that the fermions can occupy there is no point in counting the rest mass energy mc 2. This is the same for all particles cannot be used to distinguish quantum states. Therefore the equation for f n can be written in terms of ɛ kin, the kinetic energy of the fermions (i.e., the energy over above the rest mass energy). The way to formally account for the rest mass energy is to define a modified chemical potential: µ = µ mc 2. (36) The basic equations are straightforward to write down as we are simply relabelling the energy levels from ɛ p to ɛ kin the chemical potential from µ to µ: f(ɛ kin ) = e (ɛ kin µ)/kt +, (37) n(p) dp = f(ɛ kin )g(p) dp. (38) In order to derive the distribution of particles over kinetic energy I now work out equation (38) I make use of: n(p) dp = n(ɛ kin ) dɛ kin or n(ɛ kin ) = n(p) dp dɛ kin. (39) Starting from: n(p) dp = f(ɛ kin )g s h 3 p2 dp. (4) the distribution in energy can be found by using the relation above eq. (3). The expression for the number density then becomes: n(ɛ kin ) dɛ kin = f(ɛ kin )g s h 3 p2 dp dɛ kin dɛ kin = f(ɛ kin )g s h 3 pɛ kin + mc 2 c 2 dɛ kin = f(ɛ kin )g s h 3 p mc m c mc2 (ɛ kin /mc 2 + ) dɛ kin. (4)

9 4 The fermion gas phases 9 Now the value of p/mc can be substituted by using eq. (3) at the same time I express the energies in terms of mc 2. ( ɛkin ) ( ɛkin ) [ m ( ɛkin ) ] 2 /2 ( n mc 2 d mc 2 = f(ɛ kin )g s h 3 c mc 2 + mc 2 ɛkin + mc 2 ) mc 2 Note the extra factor mc 2 coming from d(ɛ kin /mc 2 ). The final result then is: n(ɛ kin ) = f(ɛ kin )g s ( mc 2 ɛkin ) d mc 2. ( mc ) 3 (ɛ h kin + ) ( (ɛ kin + )2 ) /2, (42) where ɛ kin =. Note that f(ɛ kin ) = f(ɛ kin ) because µ kt are also expressed in terms of mc2. The units of mc/h are m which means that n(ɛ kin ) is indeed a number density (for a given energy interval). The Fermi energy now becomes (in analogy to equation 7): n = g s ( mc ) 3 ɛf h ( ɛ kin + ) ( (ɛ kin + )2 ) ( /2 dɛ mc ) 3 ( kin = g s (ɛ h 3 F + ) 2 ) 3/2, (43) where ɛ F = ɛ F/mc 2. In the limit ɛ F /mc 2 the expression (7) is recovered. Thus the expression for the Fermi energy now becomes: ɛ F mc 2 = or with the particle density n normalized to g s (mc/h) 3 : ɛ F mc 2 = ( ( ) 3 n 2/3 ( ) ) h 2 /2 +, (44) g s mc ( ( ) ) 3n 2/3 /2 g s (mc/h) 3 +. (45) The Fermi-momentum can again be derived using equation (2) which leads to the same expression for p F. Normalizing to mc we obtain: ( ) p F mc = 3n /3 g s (mc/h) 3. (46) Note that the expression for the Fermi-energy could thus have been obtained more easily by starting from p F then using the now different relation between energy 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. 8) the relativistic expression (eq. 44) one can proceed to draw the T vs n diagram broadly indicate where the gas is degenerate/nondegenerate /or relativistic or non-relativistic. I first write the expressions for the Fermi energies (where NR sts for non-relativistic) as follows: ɛ F,NR mc 2 = ( 2 3n g s (mc/h) 3 ) 2/3 = ( ) n 2/3 (47) 2 n where the density normalization is: ɛ F mc 2 = ( ( ) ) n 2/3 /2 +, (48) n n = g s 3 ( mc ) 3. (49) h

10 4 The fermion gas phases Classical, Ultra-Relativistic P = nkt Classical, Non-Relativistic P = nkt T mc 2 /k 5 Degenerate, Ultra-Relativistic P = K UR n 4/3 Degenerate, Non-Relativistic P = K NR n 5/ n/n Figure 3: Phase diagram for fermions in the log T vs. log n plane, with T expressed in units of mc 2 /k n expressed in units of n. The equations of state for the four extreme phases are also shown. Classical, Ultra-Relativistic Classical, Non-Relativistic T [K] 5 Degenerate, Ultra-Relativistic Degenerate, Non-Relativistic n [m 3 ] Figure 4: Phase diagram for electrons in the log T vs. log n plane.

11 For the Fermi-momentum the expression becomes: 5 The equation of state for the fermion gas ( ) p F n /3 mc =. (5) n The T vs. n plane can now be divided in four regions containing degenerate (D) or non-degenerate (ND) gas non-relativistic (NR) or ultra-relativistic (UR) gas. These regions overlap leading to the combinations D-NR, D-UR, ND-NR, ND-UR. The dividing line between the NR UR cases can be found be realizing that the fermion gas becomes ultra-relativistic when (3/2)kT mc 2 or ɛ F mc 2. The dividing lines are then given by: which can be written as: 3 2 kt mc2 T = 2 mc 2 3 k ɛ F mc 2, (5) n = 3 3/2 n. (52) The latter expression follows from equation (48), setting ɛ F /mc 2 to. The dividing line between the degenerate non-degenerate fermion gas can be found by setting the temperature equal to the Fermi temperature, i.e.: which leads to: T = 2 mc 2 3 k 3 2 kt ɛ F, (53) ( ( ) ) n 2/3 /2 +. (54) n The latter expression can be simplified in the non-relativistic (n n ) ultra-relativistic regimes (n n ) to: T NR = mc 2 ( ) n 2/3 T UR = 2 mc 2 ( ) n /3. (55) 3 k n 3 k n Figure 3 shows the phase diagram for fermions in the T vs. n plane, with T expressed in units of mc 2 /k n expressed in units of n. The three dividing lines derived above are indicated the various regions are labelled. Note that setting n/n = 3 3/2 in expression (54) leads to T = (2/3)mc 2 /k, meaning that the various dividing lines meet in one point. For fermions of a given rest mass with known value of g s figure 3 can be translated to actual physical units. This is shown in figure 4 for electrons. For these fermions g s = 2 m = kg, which leads to log n = (kg m 3 ) log(mc 2 /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 the particle density temperature, i.e. the equation of state. In chapter 2 of Phillips (998) 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 ultra-relativistic regimes. In order to derive the equation of state for the degenerate fermion gas I first write down the general expression for the pressure of the gas: P = 3 = g s 3 h 3 n(p)pv p dp pv p f(ɛ kin )p 2 dp. (57)

12 5 The equation of state for the fermion gas 2 This result is derived in Weiss et al. (24, chapter ) can also be found in Ostlie & Carroll (27, section.2). This equation can be worked out further using equations (23), (3), (3): P = g s 3 = g s 3 h 3 h 3 = n f(ɛ kin ) f(ɛ kin )v p p 3 dp ( ( (mc) 3 ɛkin ) ) 2 3/2 ɛ p mc 2 + ɛ kin + mc 2 pc 2 ( ( ɛkin ) ) 2 3/2 ( mc 2 + mc 2 ɛkin ) d mc 2. f(ɛ kin ) pc2 dɛ kin (58) In terms of the rest-mass energy normalized kinetic energy ɛ kin = the expression for the pressure becomes: P n mc 2 = f(ɛ kin ) ( (ɛ kin + )2 ) 3/2 dɛ kin. (59) The pressure is expressed in terms of the rest-mass energy density at particle concentration n. For completely degenerate fermion gases this expression reduces to: P ɛ n mc 2 = F ( (ɛ kin + ) 2 ) 3/2 dɛ kin. (6) This expression could in principle be worked out to obtain the equation of state for the completely degenerate fermion gas. However it is easier to proceed via the momentum distribution, writing: P = g s 3 h 3 pf v p p 3 dp. (6) By switching to x = p/mc as integration variable using equation (23) for v p the integral becomes: P pf /mc n mc 2 = x 4 dx. (62) + x 2 The solution can be found using formula from Gradshteyn & Ryzhik (27) realizing that ln(x + + x 2 ) = arcsinh x. The expression becomes: P n mc 2 = ( ) z( + z 2 ) /2 (2z 2 3) + 3 arcsinh z, (63) 8 where z = p F /mc = (n/n ) /3 (equation 5). 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 = K). From this general expression the equations of state for the non-relativistic ultra-relativistic regimes can be derived. Starting with the latter, for which z, the leading term in eq. (63) is the only one that remains. This can be seen by considering that arcsinh z = ln(z + z 2 + ). 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 2z 4 as z. The expression for the pressure in the ultra-relativistic case then becomes: P UR n mc 2 = ( pf ) ( ) 4/3 n 4/3 =. (64) 4 mc 4 n For the non-relativistic limit (z ) a Taylor expansion of the above equation for the pressure can be used 3 but it is easier to return to the integral (6) note that for ɛ F = ɛ F/mc 2 (i.e. x ) the integr tends 3 The Taylor expansion should be done up to including the 5 th power of z.

13 6 Density pressure in terms of Fermi-Dirac integrals 3 to ( + 2x ) 3/2 = 2 3/2 x 3/2. Thus the expression for the pressure now becomes: P ɛ NR n mc 2 = F = 2 3/2 x 3/2 dx [ 2 3/2 2 5 x5/2 ] ɛ F Using expression (47) for the non-relativistic Fermi energy the pressure in the non-relativistic case is:. (65) P NR n mc 2 = ( ) n 5/3. (66) 5 n Writing out the normalized expressions for the pressure one obtains the following equations of state for the non-relativistic ultra-relativistic degenerate fermion gas: ( ) P NR = h2 3 2/3 n 5/3 = K NRn 5/3, 5m g s (67) P UR = hc ( ) 3 /3 n 4/3 = K URn 4/3. 4 g s (68) For an electron gas log(n mc 2 ) = (N m 2 ). In the diagrams shown in figures 3 4 the contours of constant pressure form vertical lines in the degenerate area, while in the classical regime the contours are lines at a 45 degree angle according to: log ( T mc 2 /k ) = log ( P n mc 2 ) log ( ) n. (69) The contour lines should smoothly change to vertical in the transition regime between classical degenerate gases. 6 Density pressure in terms of Fermi-Dirac integrals In this note the full expressions for the density pressure in terms of the relativistically formulated fermion energies was derived. For the density we have from eq. (42): where ɛ kin =, β = mc 2 /kt, µ = µ/mc 2, : n 3 (ɛ kin = + ) ( (ɛ kin + )2 ) /2 n e β (ɛ kin µ ) dɛ kin +, (7) n = g s 3 For the pressure the following expression was found: n ( mc ) 3. (7) h P ( (ɛ n mc 2 = kin + ) 2 ) 3/2 e β (ɛ kin µ ) dɛ kin +. (72) In the non-relativistic regime with ɛ kin ε = mc2 ɛ kin these expression reduce to: n = 3 2 n (mc 2 ) 3/2 P n mc 2 = 2 2 (mc 2 ) 5/2 ε /2 e β(ε µ) dε (73) + ε 3/2 e β(ε µ) dε (74) +

14 6 Density pressure in terms of Fermi-Dirac integrals 4 In the expressions above the energy chemical potential are normalized to mc 2. In the literature however, the integrals are written in terms of x = ɛ kin /kt the parameters θ = kt/mc 2 η = µ/kt. The integrals above can be rewritten in these terms by using: The expression for the density then becomes: x = ɛ kin kt = mc2 ɛ kin kt mc 2 = ɛ kin θ. (75) n 3(xθ + ) ( (xθ + ) 2 ) /2 θ dx = n e x η. (76) + Making use of (xθ + ) 2 = 2θx + θ 2 x 2 the numerator of the integr can be factored differently, resulting in: n = 3 2θ 3/2 n Similarly the expression for the pressure becomes: P n mc 2 = 2 2θ 5/2 x /2 ( + 2 θx) /2 ( + θx) dx e x η + x 3/2 ( + 2 θx) 3/2 dx e x η +. (77). (78) For the non- These expressions are consistent with equations (24.9) (24.92) in Weiss et al. (24). relativistic regime we have xθ then the expressions for density pressure reduce to: n = 3 2θ 3/2 x /2 dx n e x η +, (79) P n mc 2 = 2 2θ 5/2 x 3/2 dx e x η +. (8) The integrals in terms of x above can all be expressed in terms of the so-called generalized Fermi-Dirac integral: x ν ( + 2 F ν (η, θ) = θx)/2 e x η dx, (8) + which is discussed in, e.g., Press et al. (27, section 6.). In the non-relativistic case θ =. Much work in the literature has gone into finding accurate approximations to these integrals. Press et al. (27) present one particular algorithm for evaluating the approximation. For the density pressure we can write: In the non-relativistic case we can write: n n = 3 2θ 3/2 ( F /2 (η, θ) + θf 3/2 (η, θ) ), (82) ( P n mc 2 = 2 2θ 5/2 F 3/2 (η, θ) + ) 2 θf 5/2(η, θ). (83) n n = 3 2θ 3/2 F /2 (η), (84) where F ν (η) = F ν (η, ). P n mc 2 = 2 2θ 5/2 F 3/2 (η), (85)

15 7 The energy density of the fermion gas 7 The energy density of the fermion gas 5 So far the energy density of the fermion gas has not been considered but it is important quantity especially when we want to underst why there is an upper limit to white dwarf neutron star masses. The energy density u as a function of momentum or kinetic energy is simply: u(p) dp = n(p)ɛ kin dp or u(ɛ kin ) = n(ɛ kin )ɛ kin dɛ kin. (86) Using the normalizations introduced in the preceding sections we can write for the total energy density: u n mc 2 = 3ɛ kin (ɛ kin + ) ( (ɛ kin + )2 ) /2 e β (ɛ kin µ ) dɛ kin +, (87) where equation (7) was used. In terms of Fermi-Dirac integrals this becomes: u n mc 2 = 3 2θ 5/2 x 3/2 ( + 2 θx) /2 ( + θx) dx e x η + = 3 2θ 5/2 ( F 3/2 (η, θ) + θf 5/2 (η, θ) ). (88) In the non-relativistic limit (xθ ) we have: u n mc 2 = 3 2θ 5/2 = 3 2θ 5/2 F 3/2 (η). x 3/2 dx e x η + (89) This last expression is the familiar result that for non-relativistic ideal gases: Analogous to the pressure we can write in the completely degenerate case: Switching again to x = p F /mc we obtain: for which the solution is: u n mc 2 = 8 u n mc 2 = 3 u = g s h 3 pf /mc P = 2 3 u. (9) pf ɛ kin p 2 dp. (9) ( ) (x 2 + ) /2 x 2 dx, (92) ( ) 8x 3 + 3(x 2 + ) /2 (2x 3 + x) 3 arcsinh x. (93) In the non-relativistic limit (x ) this expression tends to 3 x5 (which can be derived starting from the integral for u or by exping the arcsinh term to order x 5 ) in the ultra-relativistic case (x ) it tends to 3 4 x4. Thus the expressions for the energy density become: u NR n mc 2 = 3 ( ) n 5/3, (94) u UR n mc 2 = 3 4 n ( ) n 4/3. (95) From the latter expression we obtain the expected result for an ultra-relativistic ideal gas: n P = 3 u. (96)

16 The full fermion phase diagram 6 Phase diagram for fermions, chemical potential expressed as η = (µ mc 2 )/kt log(kt/mc 2 ) e e log(n/n ) Figure 5: Phase diagram for the fermion gas. The solid contours show lines of constant log(p/n mc 2 ) constant η is shown by the dashed contours. The colour coding the dividing lines indicate in which regions the gas is ultra/non-relativistic classical/degenerate. Green indicates a classical, non-relativistic gas, cyan a classical ultra-relativistic gas, yellow a degenerate non-relativistic gas, white 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 algorithm from Press et al. (27, section 6.). This means that one can map the density pressure as a function of chemical potential temperature. However, the phase diagram in the T vs n plane is what is usually shown ( is easier to underst). The problem is that the chemical potential cannot easily be obtained for a given choice of density temperature. This requires solving the following equation for η: 3 2θ 3/2 ( F /2 (η, θ) + θf 3/2 (η, θ) ) y/n =, (97) where y is the chosen density for a given temperature parameter θ. This equation has to be solved numerically this can be done with the bisection method described in section 9. of Press et al. (27). The root of the equation can first be bracketed with the zbrac function then found with the rtbis function. Because of the enormous range of values of η (many decades) some care has to be taken in the precision demed of the bisection method. In practice this was done by finding the root in two steps. First a rough determination was done by deming an accuracy of % relative to the interval returned by zbrac. This rough determination was then used to scale the accuracy of η to 7 of the first estimate. The resulting phase diagram in the log θ vs. log(n/n ) plane is show in figure 5. The solid contours show lines of constant log(p/n mc 2 ) constant η is shown by the dashed contours. Note how the pressure contours indeed behave as predicted above, sloping at an angle of 45 in the classical regime becoming

17 8 The full fermion phase diagram 7 vertical in the degenerate region of the phase diagram. The contours of constant η show different slopes in the non-relativistic ultra-relativistic regimes. The transition from non-degenerate to a degenerate gas occurs around the η = contour. Note that large negative values of η in the classical regime. The degree of degeneracy D the degree to which the gas is relativistic R can be judged from the values of kt/mc 2 µ/mc 2 = θη as discussed in Weiss et al. (24). However a nicer definition of these quantities is: D = n ε F n(ɛ kin ) dɛ kin. (98) For a completely degenerate gas (T ) the the value of D is. The fraction of particles with p > mc measures how non-relativistic the gas is: R = n 2 n(ɛ kin )dɛ kin. (99) The colour scale in figure 5 shows the values of D R. The degree of degeneracy of the gas is indicated with the progressive saturation of the red colour channel while the extent to which the gas is relativistic is 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 non-relativistic, cyan means it is classical ultra-relativistic, yellow means the gas is degenerate nonrelativistic, white means the gas is degenerate ultra-relativistic. Figures 6 9 (at the back of this note) show examples of f(ɛ kin ) n(ɛ kin ) for each of the four regions in figure 5. The number densities in these figures are expressed in units of n. Some remarks on the average occupation of states the energy distribution: As discussed for the non-relativistic case, as the gas becomes degenerate the function f(ɛ kin ) approaches a step function. At the limit T = all the states with ɛ kin < µ/mc2 = ɛ F /mc 2 are occupied while none with energies above this limit are. In this case the fermions only occupy the lowest possible kinetic energy states above ɛ kin =, one fermion per single-particle state according to the Pauli exclusion principle. Note that even at T = the gas will exert a pressure because p for all fermions! At this limit (T = ) the energy distribution becomes: n(ɛ kin ) (ɛ kin + ) ( (ɛ kin + )2 ) /2 for < ɛ kin < µ/mc2 zero elsewhere. In this limit of the completely degenerate fermion gas the number distribution in terms of p goes as n(p) p 2. The limiting case is shown in the diagrams with the dashed lines. In the limit that T the gas remains non-relativistic one can use that ɛ kin then the expression for n(ɛ kin ) converges to n(ɛ kin ) 2(ɛ kin )/2. Removing the normalization by mc 2 to get n(ɛ kin ) dɛ kin then leads to the correct expression given by equation (8). When T the gas is ultra-relativistic the expression for n(ɛ kin ) is proportional to (ɛ kin + )2 (ɛ kin = pc for an ultra-relativistic gas). Figures show the border cases for the non-relativistic ultra-relativistic fermion gas.

18 8 The full fermion phase diagram 8 Fermion occupation of states, µ/mc 2 =.6, log(kt/mc 2 ) = f(ɛ kin) Fermion kinetic energy distribution: log(n/n ) = 7., log(ε F/mc 2 ) = n(ɛ kin ) n Figure 6: The average of occupation of the energy states f(ɛ kin ) (top panel) the corresponding distribution of the number density of fermions as a function of kinetic energy n(ɛ kin ). All energies are expressed in units of mc 2, including µ kt. These diagrams represent a classical non-relativistic fermion gas according to figure Fermion occupation of states, µ/mc 2 =. 2, log(kt/mc 2 ) =. f(ɛ kin) Fermion kinetic energy distribution: log(n/n ) =., log(ε F/mc 2 ) =.99 n(ɛ kin ) n Figure 7: The average of occupation of the energy states f(ɛ kin ) (top panel) the corresponding distribution of the number density of fermions as a function of kinetic energy n(ɛ kin ). All energies are expressed in units of mc 2, including µ kt. These diagrams represent a classical ultra-relativistic fermion gas according to figure 5.

19 8 The full fermion phase diagram 9 Fermion occupation of states, µ/mc 2 =.8 3, log(kt/mc 2 ) = 5..8 f(ɛ kin) Fermion kinetic energy distribution: log(n/n ) = 4., log(ε F/mc 2 ) = n(ɛ kin ) n.5.5. Figure 8: The average of occupation of the energy states f(ɛ kin ) (top panel) the corresponding distribution of the number density of fermions as a function of kinetic energy n(ɛ kin ). All energies are expressed in units of mc 2, including µ kt. These diagrams represent a degenerate non-relativistic fermion gas according to figure 5. Fermion occupation of states, µ/mc 2 = 5.89, log(kt/mc 2 ) =.5.8 f(ɛ kin) Fermion kinetic energy distribution: log(n/n ) = 2.5, log(ε F/mc 2 ) =.77 n(ɛ kin ) n Figure 9: The average of occupation of the energy states f(ɛ kin ) (top panel) the corresponding distribution of the number density of fermions as a function of kinetic energy n(ɛ kin ). All energies are expressed in units of mc 2, including µ kt. These diagrams represent a degenerate ultra-relativistic fermion gas according to figure 5.

20 8 The full fermion phase diagram 2 Fermion occupation of states, µ/mc 2 =.7 3, log(kt/mc 2 ) = 4..8 f(ɛ kin) Fermion kinetic energy distribution: log(n/n ) = 4., log(ε F/mc 2 ) = n(ɛ kin ) n Figure : The average of occupation of the energy states f(ɛ kin ) (top panel) the corresponding distribution of the number density of fermions as a function of kinetic energy n(ɛ kin ). All energies are expressed in units of mc 2, including µ kt. These diagrams represent a partially degenerate non-relativistic fermion gas according to figure 5. Fermion occupation of states, µ/mc 2 = 7.25, log(kt/mc 2 ) =..8 f(ɛ kin) Fermion kinetic energy distribution: log(n/n ) = 2.8, log(ε F/mc 2 ) =.88 2 n(ɛ kin ) n Figure : The average of occupation of the energy states f(ɛ kin ) (top panel) the corresponding distribution of the number density of fermions as a function of kinetic energy n(ɛ kin ). All energies are expressed in units of mc 2, including µ kt. These diagrams represent a partially degenerate ultra-relativistic fermion gas according to figure 5.