Transcript PH507 The Hydrostatic Equilibrium Equations for Stars Dr J. Miao

PH507
The Hydrostatic Equilibrium
Equations for Stars
Dr J. Miao
1
We will cover the following topics:
1.
2.
3.
4.
Equation of Mass distribution
Equation of hydrostatic support
- Virial Theorem
-Centre pressure and mean temperature of a star
-Different time scales for stars
-How do we know we are right?
Equation of Energy generation
Equation of Energy transportation
2
The Equations for stars
How does a star exist?
Internal pressure gradient
Force of gravitation
a)
Stars are spherical and symmetric
about their centers
b)
Stars are in hydrostatic equilibrium
Two fundamental assumptions:
Four equations of stellar structure
1. Equation of mass distribution
dM (r )
  (r )4r 2
dr
r
r + dr
(1.1)
3
2.
Equation of hydrostatic support
r+dr
gM
PA
(P+P)A
r
The gravitational mass m(r)
situated at the centre gives
rise to an inward gravitational
acceleration equal to :
g (r ) 
Gm(r )
r
2
(1.2)
d 2r
[ PA  gM  ( P  P)A]  M
dt 2
Newton’s second law:
d 2r
[ gM  PA]  M 2
dt
M   (r )rA,
2
d
r

 g (r )rA  PA   (r )rA 2
dt
2
d 2r
P
d 2r

P
d
r
  g (r )r  P   (r )r 2   g (r ) 
  (r ) 2   g 
 2
r
dt
dt
 (r )r dt

d 2r
dt
2
 g (r ) 
1 dP
 dr
(1.3)
4
2.1 What can we know from the equation of hydrostatic support
i) What will happen if there is no pressure gradient to oppose the gravity?
Each spherical shell of matter converges on the centre  free fall of the star
The mass of the thin shell is M with an initial radius of r0
r0
The mass included in the sphere of radius r0 is m0,
r
m0
When the thin shell collapses to the distance r: KE = GE
M
2
2
Gm 0 Gm 0
 dr 


M
[

] 
 dt 
r
r0
It follows that the free fall time to the
centre of the sphere is given by
1/ 2
(See Appendix) t FF
 3 
 td  

 32G 
0
t FF  
1
3 2
R

~

 GM 
r0
2
Gm 0 Gm 0
1  dr 


2  dt 
r
r0
 2Gm0 2Gm0 
dt
dr    


dr
r
r
0

r0 
0
 12
dr
For the sun, tff ~ 2000s
In fact, collapse under gravity is never completely unopposed. During the
process, released gravitational energy is usually dissipated into random
thermal motion of the constituents, thereby creating a pressure which
opposes further collapse The internal pressure will rise and slow down the
rate of collapse. The cloud will then approach hydrostatic equilibrium
5
ii) What will happen if a star is in hydrostatic equilibrium state?
an element of matter at a distance r from the
centre will be in hydrostatic equilibrium if the
pressure gradient at r is (d2r/dt2 =0 in eq.1.3)
dP
Gm(r )  (r )

(1.4)
2
dr
r
The whole system is in equilibrium if this equation is valid at all radii.
* Eq. (1.4) implies that the pressure gradient must be negative, or in other
words, pressure decreases from the inner central region to the outer region
* The three quantities m, r,  are not independent
2.2 From Hydrostatic equilibrium equation to Virial Theorem
If m is chosen as the independent space variable rather than r,
Divide [4r2] into two sides of (1.4)
dP
Gm

dm
4r 4
dP
Gm (r )  (r )


4r 2  (r )dr
4r 2  (r )r 2
(1.4)
(1.5)
Ps

Pc
4r dP  (Gm / r )dm  3 VdP  
3
Ms

0
Gm
dm
r
6
i.e. 
Ps

Pc
3 VdP  
Ms

0
Gm
dm
r
d ( PV )  PdV  VdP  VdP  dPV  PdV 
Integrating the left-hand side of
above by parts, the equation
can be written:
Ps

Pc
Ps

Pc
3 VdP  3 [d ( PV )  PdV ]  
3PV   3 PdV  
Vs

0
s
c
Using the symbol:
Ms

0
Gm (r )dm

r
M

0
Note: Vc=0, and dm = dV
If the star were surrounded by a
vacuum, its surface pressure
would be zero
Gm
dm
r
(Gm (r ) / r )dm
Which is gravitational
potential energy of the star
4r Ps  3
3
3
Ms

0
P

P

dm  
dm    0
(1.6)
This is the general, global form of the Virial Theorem and will be used very
often later on. It relates the gravitational energy of a star to its thermal
energy.
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2.3 What can Virial theorem tell us for classical idea gas system ?
the equation of state of a classical gas is known as
Pgas  nkT
P (
the internal energy per unit mass is
P
3 dm    0 (1.6)



mp
u
)kT
3 kT
3 P

2 mp
2 
2udm    0  2U    0 (1.8)
This is also the Virial theorem in another form
the system is stable
and bound at all points
Because the total energy (binding energy) E =  + U =  -  / 2
=
/2,
 and E are always negative, if a star is in stable.
Virial theorem tells that in a contracting gas system (protostar ),
the energy for radiation is provided by half of the decreased gravitational
potential energy: E =  / 2.
When there is no energy from contraction, the radiation of a star is
provided by thermonuclear reactions.
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2.4 Estimate the minimum pressure at the centre of a star:
Integrating eq.(1.5) from the centre to the surface of the star
Gmdm P  P(0),
c
Ps ( M )  P (0)  
4
4r
M

0
Ps ( M )  0.
On the right-hand side we may replace r by the stellar radius to
obtain a lower limit for the central pressure: i.e, 1/r > 1/R
M
M
Gmdm
Gmdm
GM 2
13 M
2 R 4
Pc  

 Pc 
 4.4 10 (
) ( )
4
4
4
4r
4R
8R
M
R
0
0
M2
Pc ~
R4
(1.7)
The pressure at the centre of the Sun exceeds 450 million
atmospheres
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2.5 Estimate the minimum mean temperature of a star:
We can use this result to estimate the average internal temperature of a star
In the gravitational potential energy expression, r is less than R everywhere
M
Gmdm
GM 2
  

R
2R
0
M
U 
 udm 
0
2U    0
3 k
GM 2
2
TM 
0
2 mp
2R
3 kT
3 k
dm

TM
 2 mp
2 mp
 
T 
GMm p
6kR
(1.9)
(1.8)
Tmin
M
~
R
3M
4R 3
2
3
Tmin ~ M 
1
3
 between two stars of the same mass, the denser one is also hotter.
For the Sun, Eq. (1.9) gives us T > 4 106 K if the gas is assumed to be
atomic hydrogen
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2.6 Estimate the importance of the radiation pressure:
the corresponding expression for radiation pressure is
Prad
aT 3

Pgas
3nk
( =1.4
103
kgm-3,
with T =4 106 K and
a
=7.5510-16
Jm-3K-4),
Prad 
1
aT 4
3

n
 10 30 / m 3
mH
We have:
Prad
 10  4
Pgas
Therefore it certainly appears that radiation pressure is unimportant at an
average point in the Sun!
This is not true of all stars, however. We shall see later that radiation
pressure is of importance in some stars, and some stars are much denser
than the Sun and hence correction to the idea gas are very important.
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2.7 How accurate is the Hydrostatic Assumption?
From
Suppose:

d 2r
2
dt2
 g (r ) 
d r
 0
2
dt
1 dP
 dr
(1.3)
a  g (r ) 
i.e:
1 dP
 g
 dr
If a mass element starts from rest with this acceleration, its inward
displacement s after a time t :
1
1
1
1
2
2
s
at 2 
 gt 2
t

(
1
/

)

(
2
s
/
g
)
2
2
if we allow the element to fall all the way to the centre of the star, we
can replace s in the above equation by r and then substitute
Gm( r )
g (r ) 
r2
1
2
1
2
t  (1 /  )  (2r / Gm) ~ t ff / 
3
1
2
The time t is that it would take a star to collapse if the forces are out of
balance by a factor 
But fossil and geological records indicate that the properties of the
Sun have not changed significantly for at least 109 years(31016s)
If t > 3*1016s   < (tff / t)2
< 10-27
most stars are like the sun and so we may conclude that: the equation of
hydrostatic equilibrium must be true to a very high degree of accuracy !12
2. 8 How valid is the spherical symmetry assumption?
Departure from spherical symmetry may be caused by rotation of the star.
M 2 R 2
2
GM / R

 2 R3
GM
~ 2  10-5
 of the Sun is about 2.5 10 -6
Departures from spherical symmetry due to rotation can be neglected.
This statement is true for the vast majority of stars. There are some stars
which rotate much more rapidly than the Sun. For these stars, the
rotation-distorted shape of the star must be accounted for in the
equations of stellar structure. r (r, , )
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3. Energy generation in stars
3.1 Gravitational potential energy
It is a likely source of the stellar energy and has the form
M

0
The total energy of the system :
E 
1

2
3M
4R 3
Assuming a constant density distribution  
the gravitational potential energy:
(4 ) 2
 ~ G
3
m( r ) 
4
r 3
3
3GM
 r dr ~ 
0
5R
R
Gm (r )dm
r
dm  4r 2  dr
2
4
the total mechanical energy of the star is:
3GM 2
E~
10R
What can this tell us?
14
Assuming that the Sun were originally much larger than it is today
how much energy would have been liberated in its gravitational collapse?
If its original radius was Ri, where Ri >> R, then the energy radiated
away during collapse would be
3 GM 2
E g  ( E f  Ei )   E f 
 1.1  10 41 J
10 R
Further assuming Lsun is a constant throughout its lifetime, then it
would emit energy at that rate for approximately
t th 
E g
L
1.11041
14
7

~
3

10
s
(~
10
yr )
3.8 10 26
Is the Kelvin-Helmholz
time scale
We have already noted that fossil and geological records indicate that
the properties of the Sun have not changed significantly for at least 109
years (3 × 1016 s)
But the Sun has actually lost energy: L * 3 * 1016 = 1.2 × 1043 J
Gravitational potential energy alone cannot account for the Sun’s
luminosity throughout its lifetime !
15
3.2 Nuclear reaction
the total energy equivalent of the mass of the Sun, . E
 M  c 2  1.79  10 47 J
If all this energy could be converted to radiation, the Sun could
continue shining at its present rate for as long as
tn 
E
 4.6  10 20 sec  1.4  1013 yrs
L
is called nuclear timescale
The Sun just have consumed its mass:
t sun
3  1016
4
M 
M 
M
~
10
M

20
tn
4.6  10
Hence, for most stars at most stages in their evolution, the following
inequalities are true
td << tth << tn.
(1.9)
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3.3
How do we include the energy source?
Define luminosity L (r) as the energy flow across any sphere of radius r.
The change in L across the shell dr is provided by the energy generated
in the shell:
L(r  dr )  L(r )  dL(r )
r
r+dr
 (r )dm   (r ) (r )4r 2 dr
where  (r) is the density;  (r) is the energy production rate per unit mass
dL( r )
 4r 2  ( r ) ( r )
dr
(1.10)
This is usually called the energy-generation equation
The energy generation rate depends on the physical conditions
of the material at the given radius.
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4. How is energy generated transported from center to outside?
4. 1 Convection.
Energy transport by conduction (and radiation ) occurs whenever a
temperature gradient is maintained in any body
But convection is the mass motion of elements
of gas, only occurs when.
dT
 a critical value
dr
Consider a convective element of stellar material
a distance r from the centre of the star
r+r
T+T
P+P
+
T+T, P+P, +
T, P, 
r
T, P, 
define P,  as the change in pressure
and density of the element
P,  , as the change in pressure and
density of the surroundings
18
If the blob is less dense than its surroundings at r+r then it will keep
on rising and the gas is said to be convectively unstable.
The condition for this instability is therefore:
  
(1.11)
Whether or not this condition is satisfied depends on two things:
a) the rate at which the element expands (and hence decreases in
density) due to the decreasing pressure exerted on it
b) the rate at which the density of the surroundings decreases with height.
We can make two assumptions about the motion of the element
1.
The element rises adiabatically, i.e. it moves fast enough to
ensure that there is no exchange of heat with its surroundings;
PV  =constant
2. The element rises with a speed < the speed of sound..
This means that, during the motion, sound waves have plenty of time
to smooth out the pressure differences between the element and its
surroundings and hence P = P at all times
19
By using P/  =constant
P c
r
 P  
 
 1
c
By the second assumption
P = P :
P
P



P
P
  1c  c 

1 P
1 P



 P
 P
(1.12)
For an ideal gas in which radiation pressure is negligible, we have:
P = kT / m,
 log P = log+ log T + constant.
This can be differentiated to give
P  T


relationship of the changes between P,
P
 T
T and  of the surrounding:
Substitute (1.12) and (1.13) into (1.11)
1 P P T


 P
P
T

T P 1 P


T
P  P
The critical temperature gradient
for convection is given by
   


 P T


(1.13)

P T
 



T   1 P

T
 P
dT   1 T dP

dr
 P dr
(1.11)
(1.14)
(1.15)
20
Note that the temperature and the pressure gradients are both negative in
this equation, we can use modulus sigh to express their magnitudes:
Eq.(1.15) can also be written as:
dT    1  T dP
 

dr    P dr
(1.16)
Convection will occur if temperature gradient exceeds a certain
multiple of the pressure gradient.
The criterion for convection derived above can be satisfied in two ways :
a) The ratio of specific heats, , is close to unity
In the cool outer layers of a star, the gas is only partially ionized,
much of the heat used to raise the temperature of the gas goes into
ionization and hence cv and cp are nearly same ~1.
A star can have an outer convective layer
b) the temperature gradient is very steep
a large amount of energy is released in a small volume at the centre
of a star, it may require a large temperature gradient to carry the
energy away  A star can have convective core.
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4.2 Conduction and radiation
Conduction and radiation are similar processes because they both
involve the transfer of energy by direct interaction,
f (r ) ~
the flux of energy flow
dT
dr
(1.17)
Which of the two - conduction and radiation - is the more dominant in
stars to transport energy?
particles
Energy:
Number density:
mean free path:
3
kT
2
nparti
parti ~ 10 -10 m
photons
~
hc
>
nphoton
<<

photon~ 10 2m
Photons can walk more easily from a point where the temperature is high
to one where it is significantly lower before colliding and transferring
energy, resulting in a higher transport of energy.
22
Conduction is therefore negligible in nearly all main sequence stars and
radiation is the dominant energy transport mechanism over conduction in
most stars.
4.3 Equation of Radiative transport
If we assume for the moment that the conditions for the occurrence
of convection is not satisfied
we can write down the fourth equation of stellar structure,
The energy carried by radiation in the flux Frad, can be expressed in
terms of the dT/dr and a coefficient of radiative conductivity, rad,
Frad  rad
dT
dr
(1.18)
where the minus sign indicates that heat flows down the temperature
gradient.
The radiative conductivity measures the readiness of heat to flow
23
Astronomers generally prefer to work with an inverse of the conductivity,
known as the opacity, which measures the resistance of the material to
the flow of heat. Detailed arguments (see Appendix 2 of Taylor’s book
‘stellar evolutions’) show that the opacity
 rad
4acT 3

3 rad
(1.19)
where a is the radiation density constant and c is the speed of light
Combining the above equations we obtain:
4acT 3 dT
Fr  
3 rad  dr
(1.20)
Recalling that flux and luminosity are related by
 16acr 2T 3  dT

Lr  

 3 rad   dr
(1.21)
Lr  4r 2 Fr
3 rad Lr
dT

2 3
dr
16acr T
the equation of radiative transport
(1.22)
24
It is the temperature gradient that would arise in a star if all the
energy were transported by radiation
It should be noted that the above equation also holds if a significant
fraction of energy transport is due to conduction, but in this case,
Lr Lr + Lcond.
L  Lr  Lcond
 16acr 2T 3  dT  1
 16acr 2T 3  dT
1 
 

  
 


 dr
3

dr


3


rad
cond





1

Then (1.22) can be
written as

1
 rad

(1.23)
1
 cond
dT
3 L

dr
16acr 2T 3
(1.24)
Clearly, the flow of energy by radiation/conduction can only be
determined if an expression for  is available.
25
4. 4 Radiation of Neutrinos
In massive stars late in their lives, the amount of energy that must
be transported is sometimes larger than either radiation of photons
or convection can account for .
In these cases, significant amounts of energy may be transported from
the center to space by the radiation of neutrinos.
This is the dominant method of cooling of stars in advanced burning
stages, which also plays a central role in events like supernovae
associated with the death of massive stars.
26
Summary:
1. Based on two fundamental assumptions:
we derived the four equations of stellar structure
dM ( r )
 4r 2 
dr
dP
GM ( r )
 

dr
r2
dL
 4r 2 
dr
dT
3kL
 
dr
16acr 2T
3
There are four primary variables M(r ), P(r), L(r ), T(r ) in these equations, all as a
function of radius
We also have three auxiliary equations
P: equation of state, P=P(,T, Xi)
: opacity (,T,Xi)
: nuclear fusion rate, (,T,Xi).
These are three key pieces of physics and we will discuss them in detail
27
2. From the most important hydrostatic equilibrium equation:
dP
Gm( r )  ( r )

dr
r2
-- Drive the global form of Viral theorem.3
(1.4)
P

dm    0
With the gravitational potential energy of a stat
If the density of the system is a constant,   
Drive another form of Viral theorem:
(1.6)
Gmdm
GM

 
0
r
R
3
M
2

5
2U    0
(1.8)
Which tells us: a star in hydrostatic equilibrium is stable and bound at all points
- E = U = -  / 2. .–only half of the released potential energy can be used
as radiation during the collapse process inside a star!
-- estimate the minimum center pressure in a star : Pc ~
-- estimate the minimum mean temperature of a star:
3. Criteria for convection:
4. Three important time scale:
   1  T dP
dT

  
 P dr
dr


td << tth << tn.
M 2
R4
(1.7)
T ~ M
2
3

1
3
(1.16)
(1.9)
28
5. Show that the radiation pressure is not important in Sun-like stars
6. Radiation is more efficient way to transport energy from place to place than
conduction
29
Appendix:
 12
0  2Gm
2Gm0 
dt
0
dr    

dr

r dr
r
r0 
 r
0
t FF  
0
0
This may be simplified by introducing the parameter
t FF
1
 2 1



2
Gm

0

r03
1
2
x  r / r0
to give
1
2
x
dx  ( x  sin 2  , dx  2 sin  cos  d )

0 1  x 



2 sin 
2
x

2
dx

2
sin

cos

d


2
sin

d






2
0 1  x 
0 cos 
0
1
t FF
1
2



2
Gm

0

r03
1
r03  2
 1
3 4




2
2
G
4


3
m

0




1
2
 3


2
 32G 
30