boson and fermion “gases”

P461 - Quan. Stats. II 1

Boson and Fermion “Gases” • If free/quasifree gases mass > 0 non-relativistic

P(E) = D(E) x n(E) --- do Bosons first

• let N(E) = total number of particles. A fixed number (E&R use script N for this)

• D(E)=density (~same as in Plank except no 2 for spin states) (E&R call N)

• If know density N/V can integrate to get normalization. Expand the denominator….

dEee

EDdEEDEnN

kTE

0 0 / 1

)()()(

1

)2(4/

2/12/13

0 3

kTEee

dEEm

h

VN

....)2

11(

)2(2/33

2/3 ee

h

VmkTN


Boson Gas • Solve for e by going to the classical region (very

good approximation as m and T both large)

• this is “small”. For helium liquid (guess) T=1 K, kT=.0001 eV, N/V=.1 g/cm3

• work out average energy

• average energy of Boson gas at given T smaller than classical gas (from BE distribution ftn). See liquid He discussion

2/3

3

)2( mkT

h

V

Ne

1)001.42(

)1240(/10

2/3

3322

eVGeV

eVnmcme

......)1(

)()(/)()(

2/3

3

2/5 )2(21

23

00

mkTh

VNkTE

dEEDEndEEDEEnE


Fermi Gas • Repeat for a Fermi gas. Add factor of 2 for S=1/2.

Define Fermi Energy EF = -kT change “-” to “+” in distribution function

• again work out average energy

• average energy of Fermion gas at given T larger than classical gas (from FD distribution ftn). Pauli exclusion forces to higher energy and often much larger

......)1(

)()(/)()(

2/3

3

2/5 )2(21

23

00

mkTh

VNkTE

dEEDEndEEDEEnE

)1(

)2(8)()( /)(3

2/12/13

kTEE Feh

EmVEDEn


Fermi Gas • Distinguishable <---> Indistinguishable

Classical <----> degenerate

• depend on density. If the wavelength similar to the separation than degenerate Fermi gas (“proved” in 460)

• larger temperatures have smaller wavelength --> need tighter packing for degeneracy to occur

• electron examples - conductors and semiconductors - pressure at Earth’s core (at least some of it) -aids in initiating transition from Main Sequence stars to Red Giants (allows T to increase as electron pressure independent of T) - white dwarves and Iron core of massive stars

• Neutron and proton examples - nuclei with Fermi momentum = 250 MeV/c - neutron stars

3/1 nseparationph

particle


Degenerate vs non-degenerate


Conduction electrons• Most electrons in a metal are attached to individual

atoms.

• But 1-2 are “free” to move through the lattice. Can treat them as a “gas” (in a 3D box)

• more like a finite well but energy levels (and density of states) similar (not bound states but “vibrational” states of electrons in box)

• depth of well V = W (energy needed for electron to be removed from metal’s surface - photoelectric effect) + Fermi Energy

• at T = 0 all states up to EF are filled

W

EF

V

Filled levels


Conduction electrons • Can then calculate the Fermi energy for T=0 (and it

doesn’t usually change much for higher T)

• Ex. Silver 1 free electron/atom

3/22

0

2/1

3

2/13

0

0

3

8

2

)2(82)(

)()(

V

N

m

hE

dEEh

mVdEED

dEEDEnN

F

EE FF

n

E

T=0

eVmMeV

eVnmEor

J

eV

mkg

jouleE

cmelectronsfree

atom

e

cm

g

moleg

moleatoms

V

N

F

F

5.5109.53

511.8

)1240(

106.1

1109.53

101.98

sec)106.6(

/109.5

15.10

/108

/1002.6

3/2

3

282

19

3/2

3

28

31

234

322

3

23


Conduction electrons

• Can determine the average energy at T = 0

• for silver ---> 3.3 eV

• can compare to classical statistics

• Pauli exclusion forces electrons to much higher energy levels at “low” temperatures. (why e’s not involved in specific heat which is a lattice vibration/phonons)

FF

FE

E

E

E

EE

E

dEE

dEEE

dEEnED

dEEnEDEE

F

F

F

F

5

3

)()(

)()(

322/3522/5

0

2/1

0

2/1

0

0

KKeV

eV

k

eVT o000,40

/106.8

3.33.35


Conduction electrons


Conduction electrons• Similarly, from T-dependent

• the terms after the 1 are the degeneracy terms….large if degenerate. For silver atoms at T=300 K

• not until the degeneracy term is small will the electron act classically. Happens at high T

• The Fermi energy varies slowly with T and at T=300 K is almost the same as at T=0

• You obtain the Fermi energy by normalization. Quark-gluon plasma (covered later) is an example of a high T Fermi gas

....

)2(2

11

2

32/3

3

3/5 mkTV

NhkTE

82012

3 kTE

)0()300( FF EE


Fermi Gases in Stars • Equilibrium: balance between gravitational

pressure and “gas” (either normal or degenerate) pressure

• total gravitational Energy:

• density varies in normal stars (in Sun: average is 1 g/cm3 but at r=0 is 100 g/cm3). More of a constant in white dwarves or neutron stars

• will have either “normal” gas pressure of P=nkT (P=n<E>) or pressure due to degenerate particles. Normal depends on T, degenerate (mostly) doesn’t

• n = particle density in this case

tconsR

GM

drrrrdrr

rGE

r

rrMassrMGE

rR

tan5

3

)(44)(

)()(

2

0

22

0

V

Epressure


Degenerate Fermi Gas Pressure• Start with p = n<E>

• non-relativistic relativistic

• P depends ONLY on density

3/43/12

3/53/21

3/124

33/215

3

2

2

2/1

2/1

3/13/2

0

2

0

2/1

22/1

2

)(

)(:

nnnKnnnKP

nKEnKEE

dEE

dEEE

dEE

dEEEE

nnE

dEAEdEAEN

EEED

casesbothppDstatesdensity

FF

F

EE FF


Degenerate Fermi Gas Pressure non-relativistic relativistic

• P depends ONLY on density

• Pressure decreases if, for a given density, particles become relativistic

• if shrink star’s radius by 2 density increases by 8 gravitational E increases by 2

• if non-relativistic. <E> increases by (N/V)2/3 = 4

• if relativistic <E> increases by (N/V)1/3 = 2 non-relativistic stable but relativistic is not. can

collapse

3/43/12

3/53/21

3/124

33/215

3

3/13/2

nnnKnnnKP

nKEnKEE

nnE

FF

F


Older Sun-like Stars • Density of core increases as H-->He. He inert (no

fusion yet). Core contracts

• electrons become degenerate. 4 e per He nuclei. Electrons have longer wavelength than He

• electrons move to higher energy due to Pauli exclusion/degeneracy. No longer in thermal equilibrium with p, He nuclei

• pressure becomes dominated by electrons. No longer depends on T

• allows T of p,He to increase rapidly without “normal” increase in pressure and change in star’s equilibrium.

• Onset of 3He->C fusion and Red Giant phase (helium flash when T = 100,000,000 K)

Hemm

eHemm

e

Hee

H

He

He

e pp

mequilibriuthermalEE

HeHeHeHetotal PPPPPPP


White Dwarves• Leftover cores of Red Giants made (usually) from

C + O nuclei and degenerate electrons

• cores of very massive stars are Fe nuclei plus degenerate electrons and have similar properties

• gravitational pressure balanced by electrons’ pressure which increases as radius decreases radius depends on Mass of star

• Determine approximate Fermi Energy. Assume electron density = 0.5(p+n) density

• electrons are in this range and often not completely relativistic or non-relativistic need to use the correct E2 = p2 + m2 relationship

MeVhcrelE

MeVrelnonE

meEarthvolumem

M

V

N

VN

F

VN

mh

F

p

Sun

8.0)(

3.0)(

/1051

2

1

3/1

83

3/238

335

2


White Dwarves + Collapse• If the electron energy is > about 1.4 MeV can have:

• any electrons > ET “disappear”. The electron energy distribution depends on T (average E)

• the “lost” electrons cause the pressure from the degenerate electrons to decrease

• the energy of the neutrinos is also lost as they escape “cools” the star

• as the mass increases, radius decreases, and number of electrons above threshold increases

MeVEnpe Threshold 4.1

#e’s

EF ET


White Dwarves+Supernovas• another process - photodisentegration - also

absorbs energy “cooling” star. Similar energy loss as e+p combination

• At some point the not very stable equilibrium between gravity and (mostly) electron pressure doesn’t hold

• White Dwarf collapses and some fraction (20-50% ??) of the protons convert to neutrons during the collapse

• gives Supernovas

npHe

nHeFe

22

4134

456

)(?)1000(100)(

)(10)( 9

lightLneutrinosL

lightLlightL

SNSN

SunSN


White Dwarves+Supernovas


Neutron Stars-approx. numbers• Supernovas can produce neutron stars

- radius ~ 10 km - mass about that of Sun. always < 3 mass Sun - relative n:p:e ~ 99:1:1

• gravity supported by degenerate neutrons

• plug into non-relativistic formula for Fermi Energy 140 MeV (as mass =940 MeV, non-rel OK)

• look at wavelength

• can determine radius vs mass (like WD)

• can collapse into black hole

Fseparation

Fmkmm

M

V

N

n

Sun

2.1

/6./106)10(

12 33443

34

FMeVMeV

MeVF

mE

h

p

h2

1409402

1240

2


Neutron Stars• 3 separate Fermi gases: n:p:e p+n are in the same

potential well due to strong nuclear force

• assume independent and that p/n = 0.01 (depends on star’s mass)

• so need to use relativistic for electrons

• but not independent as p <---> n

• plus reactions with virtual particles

• free neutrons decay. But in a neutron star they can only do so if there is an available unfilled electron state. So suppresses decay

)(000,10)()(

6)()(

5.940

251453/2

251453/2

wrongMeVnEeE

MeVnEpE

MeVmm

VN

FF

MeVVN

FF

e

n

MeVhceE VN

F 500)( 3/1

83

pennep

enpepn


Neutron Stars• Will end up with an equilibrium between n-p-e

which can best be seen by matching up the Fermi energy of the neutrons with the e-p system

• neutrons with E > EF can then decay to p-e-nu (which raises electron density and its Fermi energy thus the balance)

• need to include rest mass energies. Also density of electrons is equal to that of protons

• can then solve for p/n ratio (we’ll skip algebra)

• gives for typical neutron star:

223/2

23/2

23/1

28

3

28

3

8

3

)()()(

cmm

hn

m

hncmhc

n

nEpEeE

nn

n

p

pp

e

FFF

200//101

/102344

317

npen nnnmn

mkg

boson and fermion “gases”

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t larger

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