Transcript Static Stellar Structure

Static Stellar Structure
Static Stellar Structure
Most of the Life of A Star is Spent in Equilibrium


Evolutionary Changes are generally slow and can
usually be handled in a quasistationary manner
We generally assume:



Hydrostatic Equilibrium
Thermodynamic Equilibrium
The Equation of Hydrodynamic Equilibrium
d r
Gm( r )   P
 2 

2
dt
r
r
2
Static Stellar Structure
2
Limits on Hydrostatic Equilibrium


If the system is not
“Moving” - accelerating
Gm( r )   P dP
in reality - then d2r/dt2 = 


2
r
 r dr
0 and then one recovers
the equation of
hydrostatic equilibrium:
If ∂P/∂r ~ 0 then
2
d
r
Gm
(
r
)
which is just the freefall

2
2
condition for which the
dt
r
time scale is tff
(GM/R3)-1/2
Static Stellar Structure
3
Dominant Pressure Gradient

When the pressure gradient dP/dr
dominates one gets (r/t)2 ~ P/ρ


This implies that the fluid elements must
move at the local sonic velocity: cs = ∂P/∂ρ.
When hydrostatic equilibrium applies
V << cs
 te >> tff where te is the evolutionary time scale

Static Stellar Structure
4
Hydrostatic Equilibrium

Consider a spherical star







Shell of radius r, thickness dr and density ρ(r)
Gravitional Force: ↓ (Gm(r)/r2) 4πr2ρ(r)dr
Pressure Force: ↑ 4r2dP where dP is the pressure
difference across dr
Equate the two: 4πr2dP = (Gm(r)/r2) 4πr2ρ(r)dr
r2dP = Gm(r) ρ(r)dr
dP/dr = -ρ(r)(Gm(r)/r2)
The - sign takes care of the fact that the pressure
decreases outward.
Static Stellar Structure
5
Mass Continuity


r
m(r) = mass within a shell = m( r )  0 4 r2  (r)dr
This is a first order differential equation which needs
boundary conditions


We choose Pc = the central pressure.
Let us derive another form of the hydrostatic
equation using the mass continuity equation.


Express the mass continuity equation as a differential:
dm/dr = 4πr2ρ(r).
Now divide the hydrostatic equation by the masscontinuity equation to get: dP/dm = Gm/4πr4(m)
Static Stellar Structure
6
The Hydrostatic Equation in Mass
Coordinates

dP/dm = Gm/4πr4(m)
The independent variable is m
 r is treated as a function of m
 The limits on m are:
 0 at r = 0
 M at r = R (this is the boundary condition on the
mass equation itself).


Why?
Radius can be difficult to define
 Mass is fixed.

Static Stellar Structure
7
The Central Pressure


Consider the quantity: P + Gm(r)2/8πr4
Take the derivative with respect to r:
d 
Gm( r )2  dP Gm( r ) dm Gm( r )2
P




4 
4
dr 
8 r  dr
4 r dr
2 r 5



But the first two terms are equal and opposite so the
derivative is -Gm2/2r5.
Since the derivative is negative it must decrease outwards.
At the center m2/r4 → 0 and P = Pc. At r = R P = 0 therefore
Pc > GM2/8πR4
Static Stellar Structure
8
The Virial Theorem
dP
mG

dm
4 r ( m ) 4
mG
3 dP
4 r

dm
r
d
dr
mG
3
2
(4 r P )  4 r 3P

dm
dm
r
M 3P
M mG
3
M
(4 r P ) |0  
dm   
dm
0
0

r
Remember: 4 r 2 rdr  dm
Static Stellar Structure
9
The Virial Theorem



The term (4 r3P) |0M is 0: r(0) = 0 and P(M) = 0
Remember that we are considering P, ρ, and r as
variables of m
For a non-relativistic gas: 3P/ = 2 * Thermal energy
per unit mass.


M
0
3P

dm  2U
for the entire star
GM
 
dm   the gravitional binding energy
r
Static Stellar Structure
10
The Virial Theorem




-2U = Ω
2U + Ω = 0
Virial Theorem
Note that E = U + Ω or that E+U = 0
This is only true if “quasistatic.” If
hydrodynamic then there is a modification
of the Virial Theorem that will work.
Static Stellar Structure
11
The Importance of the Virial
Theorem

Let us collapse a star due to pressure
imbalance:


If hydrostatic equilibrium is to be maintained
the thermal energy must change by:


This will release - ∆Ω
∆U = -1/2 ∆Ω
This leaves 1/2 ∆Ω to be “lost” from star

Normally it is radiated
Static Stellar Structure
12
What Happens?




Star gets hotter
Energy is radiated into space
System becomes more tightly bound: E
decreases
Note that the contraction leads to H burning
(as long as the mass is greater than the critical
mass).
Static Stellar Structure
13
An Atmospheric Use of Pressure


We use a different form of the equation of
hydrostatic equilibrium in an atmosphere.
The atmosphere’s thickness is small compared
to the radius of the star (or the mass of the
atmosphere is small compared to the mass of
the star)

For the Sun the photosphere depth is measured in
the 100's of km whereas the solar radius is 700,00
km. The photosphere mass is about 1% of the Sun.
Static Stellar Structure
14
Atmospheric Pressure


Geometry: Plane Parallel
dP/dr = -Gm(r)ρ/r2 (Hydrostatic Equation)




R ≈ r and m(R) = M. We use h measured with respect to
some arbitrary 0 level.
dP/dh = - gρ where g = acceleration of gravity. For
the Sun log(g) = 4.4 (units are cgs)
Assume a constant T in the atmosphere. P = nkT and
we use n = ρ/μmH (μ is the mean molecular weight)
so
P = ρkT/μmH
Static Stellar Structure
15
Atmospheric Pressure Continued




Or
P=
dP =
dP =
dP/P =
ρkT/μmH
-g ρ dh
-g P(μmH/kT) dh
-g(μmH/kT) dh
Integrate : P  P0e

   0e

gmH  h
kT

gmH  h
kT
Where: P0 = P and ρ0 = ρ at h = 0.
Static Stellar Structure
16
Scale Heights

H (the scale height) is defined as kT / μmHg


It defines the scale length by which P decreases
by a factor of e.
In non-isothermal atmospheres scale
heights are still of importance:

H ≡ - (dP/dh)-1 P = -(dh/d(lnP))
Static Stellar Structure
17
Simple Models
To do this right we need data on energy generation and energy transfer but:


The linear model: ρ = ρc(1-r/R)
Polytropic Model: P = Kργ
K and γ independent of r
 γ not necessarily cp/cv

Static Stellar Structure
18
The Linear Model
r
   c (1  )
R
dP
Gm ( r )
r

 c (1  )
2
dr
r
R



Defining Equation
Put in the Hydrostatic Equation
Now we need to deal with m(r)
Static Stellar Structure
19
An Expression for m(r)
dm
r
 4 r 2  c (1  )
dr
R
4 r 3 c
2
 4 r  c 
R
4 r 3 c  r 4  c
m( r ) 

3
R
 R 3 c
Total Mass M  m( R ) 
3
 4r 3 3r 4 
So m( r )  M  3  4 
R 
R
Static Stellar Structure
Equation of Continuity
Integrate from 0 to r
20
Linear Model
• We want a
particular
(M,R)
• ρc = 3M/πR3
and take Pc =
P(ρc)
• Substitute
m(r) into the
hydrostatic
equation:
dP  G 2  4r 3 r 4  
r
 2 c 
  1  
dr
r
R 
R
 3
2
3


4
r
7
r
r
2
  G  c  
 2
 3 3R R 
So
2
3
4


2
r
7
r
r
2
P  Pc   G  c 


2 
3
9
R
4
R


where Pc is the central pressure.
Static Stellar Structure
21
Central Pressure
At the surface r  R and P  0
5 G  R
P( R )  Pc 
0
36
2 2
2
2
5 G  c R
5 GR 9 M
 Pc 

2 6
36
36  R
2
5GM
Pc 
4
4 R
2
c
Static Stellar Structure
2
22
The Pressure and Temperature
Structure

The pressure at any radius is then:
2
3
4


5
24
r
28
r
9
r
2 2
P( r ) 
G c R 1 

 4
2
3
36
5R
5R
5R 


Ignoring Radiation Pressure:
mH P
T
k
2
3


5 G mH
r
19
r
9
r
2

c R 1   2  3 
36 k
5R 
 R 5R
Static Stellar Structure
23
Polytropes


P = Kργ which can also be written as
P = Kρ(n+1)/n


N ≡ Polytropic Index
Consider Adiabatic Convective Equilibrium
Completely convective
 Comes to equilibrium
 No radiation pressure
 Then P = Kργ where γ = 5/3 (for an ideal
monoatomic gas) ==> n = 1.5

Static Stellar Structure
24
Gas and Radiation Pressure







Pg = (N0k/μ)ρT = βP
Pr = 1/3 a T4 = (1- β)P
P= P
Pg/β = Pr/(1-β)
1/β(N0k/μ)ρT =1/(1-β) 1/3 a T4
Solve for T: T3 = (3(1- β)/aβ) (N0k/μ)ρ
T = ((3(1- β)/aβ) (N0k/μ))1/3 ρ1/3
Static Stellar Structure
25
The Pressure Equation
P
N 0k  T


1
3
 N k 4 3 1    4
  0 
 3
   a  




True for each point in the star
If β ≠ f(R), i.e. is a constant then
P = Kρ4/3
n = 3 polytrope or γ = 4/3
Static Stellar Structure
26
The Eddington Standard Model n = 3




For n = 3 ρ  T3
The general result is: ρ  Tn
Let us proceed to the Lane-Emden Equation
ρ ≡ λφn





λ is a scaling parameter identified with ρc - the central
density
φn is normalized to 1 at the center.
Eqn 0: P = Kρ(n+1)/n = Kλ(n+1)/nφn+1
Eqn 1: Hydrostatic Eqn: dP/dr = -Gρm(r)/r2
Eqn 2: Mass Continuity: dm/dr = 4πr2ρ
Static Stellar Structure
27
Working Onward
r 2 dP
Solve 1 for m( r ) : m( r )  
 G dr
Put m( r ) in 2 :
1
r2
 d  r 2 dP  
 
   4 G 
 dr   dr  
n 1
dP
d
 K  n ( n  1) n
dr
dr
n 1

1  d  r2
n d 
n
n
K

(
n

1)



4

G

 

r 2  dr   n
dr  
Static Stellar Structure
28
Simplify
1
r2
n 1
 d  r2

n d 
n
n
  n K  ( n  1)
   4 G
dr  
 dr  
1
n
K  ( n  1)
K
1 n
n
1
r2
( n  1)
 d  2 d  
n
 dr  r dr    4 G

 
1 1  d  2 d  
n
r
   
2 
4 G r  dr  dr  
1
2


K  ( n  1) 
Define a  


4 G


r

so ad   dr
a
1 d  2 d 
n
So :




 2 d   d  
1 n
n
Static Stellar Structure
29
The Lane-Emden Equation
1 d  2 d 
n

 


2
 d  d 

Boundary Conditions:





Center ξξ = 0; φ = 1 (the function); dφ/d ξ = 0
Solutions for n = 0, 1, 5 exist and are of the form:
φ(ξ) = C0 + C2 ξ2 + C4 ξ4 + ...
= 1 - (1/6) ξ2 + (n/120) ξ4 - ... for ξ =1 (n>0)
For n < 5 the solution decreases monotonically and φ → 0 at
some value ξ1 which represents the value of the boundary
level.
Static Stellar Structure
30
General Properties

For each n with a specified K there exists a family of
solutions which is specified by the central density.

For the standard model:
 N k  3 1   
K   0 
4 
   a  
4



1
3
We want Radius, m(r), or m(ξ)
Central Pressure, Density
Central Temperature
Static Stellar Structure
31
Radius and Mass
R  a1
1
2
1 n
 (n  1) K  2 n

 1

 4 G 
 
d 
M ( )  4 a 3  d   2

0
 d 
d
 4 a 3 2
d
Now substitute for a and evaluate at   1 (boundary )
a
M ( )   4 r 2  dr
0


 4 a
3
1 d
 2 d
 2 d 
n




 d 


0
 n 2d
M ( )  4 a  
3

0
d
d
3
2
3 n
 (n  1) K  2 n
M  4 


4

G


 2 d 
  d 

 1
 2 d 
  d  d


Static Stellar Structure
32
The Mean to Central Density
c  
3
2

M
4
 R3
3

3 n
 ( n  1) K  2 n
4 


 4 G 
3
2
 2 d 
  d 

 1
4  ( n  1) K 



3  4 G 
3(1 n )
2n
13
 d 
    
1  d   1
3

3  d 
 
c
1  d   
1
Static Stellar Structure
33
The Central Pressure



At the center ξ = 0 and φ = 1 so Pc = Kλn+1/n
This is because P = Kρn+1/n = Kλ(n+1)/nφn+1
Now take the radius equation:
1
2
1 n
 ( n  1) K  2 n
R
 1

 4 G 
1
2
 ( n  1)12   1nn 

K 

 4 G  

So
K
1 n
n
1
2
4 GR 2

n  1)12
Static Stellar Structure
34
Central Pressure
Pc  K 
1 n
n
 2  K
1 n
n
 c2
2


1
1

 2
 3  d / d   1 
4
16 2 6 2
3
2
M   R  so M   R 
3
9
4 GR

( n  1) 2
2
But


1
1


 3  d / d   1 
2


GM 2
1



4
4 R ( n  1)   d / d    
1

2
4 GR 2
Pc 
(n  1) 2
Static Stellar Structure
 9M 2 
 16 2 R 6 


35
Central Temperature
Pg 
N 0k

 cTc   c Pc
which means
 c Pc 
Tc 
N 0k  c
Static Stellar Structure
36