Transcript Stellar Interior

Stellar Interior
Solar Facts
• Radius:
– R = 7  105 km = 109 RE
• Mass :
– M = 2  1030 kg
– M = 333,000 ME
• Density:
– r = 1.4 g/cm3
– (water is 1.0 g/cm3, Earth is
5.6 g/cm3)
• Composition:
– Mostly H and He
• Temperature:
– Surface is 5,770 K
– Core is 15,600,000 K
• Power:
– 4  1026 W
Solar Layers
• Core
– 0 to 0.25 R
– Nuclear fusion region
• Radiative Zone
– 0.25 to 0.70 R
– Photon transport region
• Convective Zone
– 0.70 to 1 R
– Fluid flow region
Equilibrium
• A static model of a star can be
made by balancing gravity
against pressure.
– Need mass density and
pressure
Fg  mg   r (hA) g
Ft
Ft   P0 A
Fb  PA
Fb
Fg
PA  P0 A  r (hA) g  0
P  P0  rgh
Particles and States


k  (k x , k y , k z )  (nx , n y , nz )
L
3
L
g (k )d 3 k    dk x dk y dk z
 
 L  4 2
g (k )dk   
k dk
  8
• The particles in a star form a
nearly ideal fluid.
– Classical ideal gas
– Quantum fluid
3
4V 2
g ( p )dp  3 p dp
h
note:
1
p 
V
3
• The particles quantum states
can be found by considering the
particle in a box.
– Dimension L
– Wave vector (kx, ky, kz)
Internal Energy
• The internal energy depends on
the quantum states.
– Density of states g(p)dp
– Energy of each state ep
– Number in each state f(ep)
e p2  p 2c 2  m2c 4

E   e p f (e p ) g ( p)dp
0

N   f (e p ) g ( p)dp
0
• The distribution depends on the
type of particle
– Fermion or boson
– Reduces to Maxwell



) e
f FD (e p )  e
f BE (e p
f (e p )  e
e p    kT
e p    kT


 e p   kT

 1
1
1
1
Pressure
dE  TdS  PdV  dN
 de p
E
P
 
f (e p ) g ( p)dp
0 dV
V
de p dp  pc 2  p 
 





dV
dp dV  e p  3V 
de p
pe p

dV
3V
de p
P
N
pv p
3V
• The energy is related to the
thermodynamic properties.
– Temperature T
– Pressure P
– Chemical potential 
• The pressure comes from the
energy.
– Related to kinetic energy
density
Relativity Effects
• The calculation for the ideal gas
applied to both non-relativistic
and relativistic particles.
• For non-relativistic particles
• For ultra-relativistic particles
e p2  p 2c 2  m2c 4
e p2  p 2c 2
vp  p m
vp  c
2N p2
P
3V 2m
N p2
P
3V 2m
Ideal Gas
• A classical gas assumes that the
average occupation of any
quantum state is small.
– States are g(p)dp
– State occupancy gs
– Maxwellian f(ep)
• The number N can be similarly
integrated.
– Compare to pressure
– Equation of state
– True for relativistic, also
P
N
1
pv p 
3V
3V

P   pv p e


0

 e p   kT
0
kT  kT  e p
P
e  e
0
V
N e
 kT



0
P
e
e p kT
pv p f (e p ) g ( p)dp
4V 2
g s 3 p dp
h
4V 2
kT
g s 3 p dp
h
gs
4V 2
p dp
3
h
N
kT  nkT
V
Particle Density
• The equation of state is the
same for both non-relativistic
and relativistic particles.
– Derived quantities differ
• For non-relativistic particles
V
32

  mc 2  kT
N e
g s 3 2mkT 
h
 2mkT 
nQ  

2
h


32
 g s nQ 

  mc  kT ln 
 n 
2
• For ultra-relativistic particles
N  e  kT g s
 kT 
nQ  8  
 hc 
3
8V  kT 

3 
h  c 
 g s nQ 

  kT ln 
 n 
3
Electron Gas
pF
N   gs
0
4V 2
p dp
3
h
 3nh 3 
8V 3

p F  
N  3 pF
8



3h
pF
4V
E   e p g s 3 p 2 dp
0
h
13
 2 3 pF 2 

E  N  mc 
10m 

2
2n p 2
np
P
 F
3 2m
10m
23
h2  3  5 3
P
  n
5m  8 
• Electrons are fermions.
– Non-relativistic
– Fill lowest energy states
• The Fermi momentum is used
for the highest filled state.
• This leads to an equation of
state.
Relativistic Electron Gas
pF
N   gs
0
4V 2
p dp
3
h
8V 3
N  3 pF
3h
E
pF
0
EN
pcg s
13
 3nh 3 

p F  
8



4V 2
p dp
3
h
3 pF c
4
n p2
np c
P
 F
3 2m
4
13
hc  3 
P    n4 3
4  8 
• Relativistic electrons are also
fermions.
– Fill lowest energy states
– Neglect rest mass
• The equation of state is not the
same as for non-relativistic
electrons.
Electron Regimes
• Region A: classical, nonrelativistic
– Ideal gases, P = nkT
• Region B: classical, ultrarelativistic
– P = nkT
• Region C: degenerate, nonrelativistic
– Metals, P = KNRn5/3
• Region D: degenerate, ultrarelativistic
– P = KURn4/3
T(K)
1015
B
1010
A
105
C
D
n(m3)
1025 1030 1035 1040 1045
Hydrogen Ionization
ep = p2/2m
• Particle equilibrium is
dominated by ionized hydrogen.
• Equilibrium is a balance of
chemical potentials.
n=3
n=2
 g H n nQp 

 H n   mH n c  kT ln 
 nH 
n


2
n=1
 g p nQp 

  p   m p c  kT ln 

n
p


g
n


 e  me c 2  kT ln  e Qe 
 ne 
2
 H n    e    p 
Saha Equation
mH n c 2  mp c 2  mec 2  e n
g ( H n )  g n ge g p  4n2
n( H n ) g n e n

e
ne n p
nQe
kT
• The masses in H are related.
– Small amount en for
degeneracy
• Protons and electrons each have
half spin, gs = 2.
– H has multiple states.
• The concntration relation is the
Saha equation.