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Ph 12c Lecture 7.1
May 12th, 2015
Contents
1 White Dwarf Stars as Fermi Gas 1
1 White Dwarf Stars as Fermi GasWiens law; maxT = constant.
Absolute luminosity Labs = 4piR
2J , J = T 4. Hertzsprung-Russelldiagram. Demonstrates how the
lnL for stars varies inversely as max. White dwarfs are on the
lowerleft corner of that plot.
MWD Msun 2 1033 gRWD 2 109 cmRsun 7 1010 cmWD = 6 104 g/cm3
Note: core 107 g/cm3. Models of WD give core temperature of 107
K giviing a 800eV. Letsconstruct a white dwarf assuming uniform
density.
Star is stabilized by gravitational contraction versus pressure.
Therefore (ergo/3 dots):
PdV = dE
P (4piR2)dR = d
(3
5
M2G
R
)=
3GM2
RdR
P =3
5
GM2
4piR4
Where does this come from? Consider pressure of ideal gas at =
800 eV.
P =N
V=MWD/mFe
4/3piR2= 2.6 1017 dyne
cm2
but
P =3
5
GM2
4piR2= 1.3 1021 dyne
cm2
Something else must supply pressure quantum gas.Consider gas of
ionized e 1 e for each boundproton.
Ne =M
2mp= 6 1056
Then
ne = NeV
= 2 1028/cm3
1
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This gives EF = 30000 eV, and EF >> , so this is a
degenerate Fermi gas. In such a density coreEF 1 MeV, we have a
fairly relativistic Fermi gas and these electrons prevent stellar
collapse. Whatis the pressure caused by the electrons? We need U
because P = U
V ,N. When I squeeze a Fermi gas,
the entropy doesnt change much because its held in the number of
states, which is fixed since everystate has 1 e. We need a
relativistic form of U .
Recall
U =2
8
nmax0
ENT4pin2T dnT
but E2e = p2ec
2 + m2ec4. But quantum states still satisfy pe = ~k0, ke =
nTpi/L, nT = Lpe/pi~ and
dnT = (L/pi~)dpe.
U = pi
(L
pi~
)3 pF0
(p2ec2 +m2ec
4)1/2p2edpe
with pF =pi~Lnmax = (3pi
2n)1/3~ since nmax = (3N/pi)1/3 and L3 = V .
Let x = pe/mec, dx = dpe/mec,
U =
(V
pi2~3
)mec
2(mec)3
xmax0
(1 + x2)1/2x2 dx
For highly relativistic xmax >> 1 we have (1 + x2)1/2x2 =
x3(1 + 1
x2)1/2 x3 + x/2 + . . .. Therefore,
U (m4ec
5V
4pi2~3
)(p4Fm4ec
4+
p2Fm2ec
2
)Write
pFmec
=
V 1/3
with
=(3pi2N)1/3~
mec
And we conclude
U V(
4
V 4/3+
2
V 2/3
)P =
(U
V
)=
4
3V 4/3
2
3V 2/3
Noting that V = 4/3piR3 we have
P 1R4
2
R2
Now apply equilibrium condition, which has
1/R4 2/R2 = 3GM
2
20piR4
R2 =12
(1 3GM
2
20pi1
)Thus R = 0 if
3M2G
20pi1= 1
2
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Protons dont work because 1. theyre not relativistic, so youd
need to recalculate the pressure withoutrelativity. 2. electrons
are dominating the pressure.
EFeF = (peF )
2/2mFe
EF =~2
2m(3pi2n)2/3
EFeF EeF
memfe
1
261/3 15 eV
vs. the 800 eV for electrons. Fe is classical in comparison.
1 =m4ec
5
4pi2~3
((3pi2Ne)
1/3~mec
)41
3(4pi/3)4/3=
~c12pi2
(9pi/4)4/3N4/3e
maxMWD = (2pi/3G)1/2 (~c/12pi2)1/2 (9pi/4)4/6 (1/2mp)2/3M2/3
maxMWD =
(5~c9piG
)3/2(9pi
8mp
)2= 3.4 1033 g = 1.7Msun
A more complete derivation uses correct density and pressure
dependence on radius:
dP (r)
dr= G(r)M(r)
r2
M(r) =
r0
4pir2(r) dr
Solving above gives maxMWD = 1.4Msun.
If MWD > maxMWD, then its note a white dwarf, electron
pressure fails to stop contraction. Star heatsup with Fe 28n+ 26p+
26e, pressure causes p+ e n+ , then neutron Fermi gas
(degenerate)stops collapse, and it becomes a neutron star.
3
