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EQUATION OF STATE FOR STARS
NOTES CH 3
LOCAL THERMODYNAMICAL EQUILIBRIUM=LTE
1.The interior of a star is a very optically thick medium:
where is the mean free path of photons between scatterings and/or absorptionlph
⌧ ' R
lph� 1 e.g. ⌧Sun ' RSun
< lph >⇡ 1011
In a region of several mean free path but << R matter and radiation are in thermal equilibrium and can be described by the same local temperature
~1 cm
Note: thermal equilibrium LTE6=
we can calculate microscopic thermodynamical properties as a function of the local temperature, density and composition
mixing locally is very effective: well defined local composition
In each position within the star, radiation has a black body spectrum with that local temperature
EOSTHE EQUATION OF STATE
• Def: A relation between the pressure exerted by a system of particles of known composition, temperature and density:
P = P (T, ⇢, X)
EOSIDEAL OR PERFECT GAS
• Def: an ensemble of free, non-interacting particle so that the internal energy is given by the sum of the particles’ kinetic energy
• Good description for ionised gas in a star with T~106 K. One can show that the ratio
of Coulomb interaction energy per particle and mean kinetic energy per particle is
• Note: Cristallization can happen in cooling white dwarfs (3.6.1)
✏Ckb < T >
⇡ 1% see 3.1 in Prialnik’s book and section 3.6.1 notes
• Note: gas is completely ionised in the hot stellar interiors but in cooler (<10 6 K) atmospheric layers there maybe partial ionization (3.5)
THERMODYNAMICAL QUANTITIES
definition of perfect gas
definition of n(p) =momenta distribution
P= flux of momentum
We need the particle energy ✏p = ✏�mc2 and velocity vp =pc2
✏
Relation between U & P
NR limit: p ⌧ mc
< pvp >=
⌧p2
m
�= 2 < ✏p > P =
2
3n < ✏p >=
2
3U
vp = c
ER limit p � mc
}< pvp >=< pc >=< ✏p >✏p = pc P =
1
3U
”
“n(p)dp
LET’S REFRESH OUR KNOWLEDGE OF STATISTICAL MECHANICS
TO COMPUTE THE EOS WE NEED
CLASSICAL IDEAL GASConsider:
•Gas of identical, non-interacting particles (perfect gas), with mass “m”, non relativistic (p << mc)•At LTE, at temperature T
the momentum distribution is a Maxwell-Boltzmann distribution:
vp =p
mwith
P = nkbT
NOTE: true also for a relativistic Maxwellian distribution
MIXTURE OF IDEAL NON DEGENERATE GASES
ne = ⌃iZini = ⌃iZiXi
Ai
⇢
mu⌘ 1
µe
⇢
mu
mean molecular weight per free electron
nion
= ⌃ini = ⌃iXi
Ai
⇢
mu
⌘ 1
µion
⇢
mu
mean atomic mass per ion
mu =1
12m12C
Pgas
= Pion
+ Pe
=
✓1
µion
+1
µe
◆R⇢T
R = kb/mu = the universal gas constant
=R
µ⇢T
MEAN MOLECULAR WEIGHT
Fully ionised hydrogen Z=A =1 gas while fully ionised He and metals Z/A ~1/2
1
µ=
1
µion
+1
µe= ⌃i
(Zi + 1)Xi
Ai
1
µ⇡ 1
2X + 32Y + 1
2Z⇡ 0.6
it is the inverse sum of the mean atomic mass per ion and the mean molecular weight per electron
X=0.7, Y=0.28, Z=0.02
QUANTUM MECHANICAL STATISTICAL DESCRIPTION OF GAS
•Perfect gas, with mass “m” and spin “s”•At, at LTE temperature T•Fermions
More generally, then in the previous case, we state that the distribution of its numerical density as a function of its linear
momentum is
total energy
μ chemical potential = is the work necessary to change the particle number by dN: dW/dN
Fermi-Dirac distribution
g = 2s+1 energy level degeneracy
dn(p)
dp=
4⇡g
h3
p2
e✏�µkbT � 1
✏ = (pc)2 +m2c4
DONEC QUIS NUNC
==> classical Maxwell-Boltzmann distribution
WDS ARE MADE OF “COLD” GAS
Degenerate gas of fermions at T—>0
Here the chemical potential is called Fermi energy
The corresponding Fermi momentum
Fermi temperature
“zero” temperature distribution can be assumed if a gas has T << TF (i.e. KT << μ)
dn
dp=
4⇡g
h3p2⇥{ 0, ✏ > µ
1, ✏ µ
pF = mc
s✓EF
mc2
◆2
� 1
kbT = EF �mc2
particle density determines Fermi Energy
simply counting particles in a sphere with
radius pF
THERMODYNAMICAL PROPERTIES: NUMBER DENSITY
n =
Zn(p)dp
THERMODYNAMICAL PROPERTIES: PRESSURE
let’s now specialise to WD typical conditions...
DEGENERATE ELECTRON PRESSURE IN WDS&
for typical density and number of free electron per nucleon (Ye), electron are mildly relativistic XF ~1.
DEGENERACY TEMPERATURE FOR ELECTRONS
Degeneracy if T is << TF
Typical central temperature ~107 K
electrons are fully degenerate in a WD
NUCLEONS ARE NON DEGENERATE IN WDS
TF / n2/3/m
TF,e ⇠ 2000⇥ TF,p/n
inversely proportional to mass
In a WD the condition Te << TF,e is attained well before Tn/p << TF,e/p .
In fact, in WD only electrons are degenerate
TF,p/n = 3⇥ 106K < TWD ⌧ TF,e ⇡ 6⇥ 109K
DEGENERATE ELECTRON PRESSURE IN WDtwo extreme regimes:
xF � 1
xF ⌧ 1
⇢ � 106g cm�3
⇢ ⌧ 106g cm�3
P ⇡ 3⇥ 1020 Ba (Ye/0.5)5/3
✓⇢
106g/cm3
◆5/3
P ⇡ 5⇥ 1020 Ba (Ye/0.5)4/3
✓⇢
106g/cm3
◆4/3
WD’s pressure does not depend on temperature => does have to be hot to be in hydrostatic equilibrium
P / ⇢�
RADIATION
photon are bosons with gs = 2 and µ = 0
✏p = pc = h⌫
and completely relativistic
Planck function
Bose-Einstein distribution
BLACK BODY RADIATION
A MIXTURE OF GAS AND RADIATION
P = Prad
+ Pion
+ Pe�
Pgas
= Pion
+ Pe� = �P
Prad = (1� �)P
RADIATION PRESSURE
Prad =1
3U =
1
3aT 4
U = aT 4 energy density
T/⇢1/3 ⇡ 107µ�1/3Prad = Pgas,classical
SUMMARY
from O.R. Pols
P rad=P ga
s
Pe,d,N
R=
Pe,d,U
R
P gas=P e
,d,N
R