Transcript PH607 The Physics of Stars

PH607
The Physics of Stars
Dr J. Miao
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We will cover the following materials:
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Equations of Stellar Structure
The physics of stellar interiors
Sun’s model
The Structure of Main-sequence Stars
Stellar evolution: Star birth
Evolution of stars
The death of stars
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Chapter one The Equations of Stellar Structure
(A Warming up)
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 conservation.
dM (r )
  (r )4r 2
dr
r
r + dr
(1.1)
3
2.
Equation of hydrostatic support
The balance between gravity and internal (thermal) pressure is known as
hydrostatic equilibrium
r+dr
PA
gM
(P+P)M
r
The gravitational mass
situated at the centre gives
rise to an inward
gravitational acceleration
equal to :
g (r ) 
Gm(r )
r
2
(1.2)
the inward force on the element due to the pressure gradient is
dP
dP
M   (r )rA,
[ P(r ) 
r  P(r )]A 
rA
dr
dr
the inward acceleration of any element of mass at
distance r from the centre due to gravity and pressure is

d 2r
dt
2
 g (r ) 
1 dP
 dr
(1.3)
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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
2
Gm0 Gm0
1  dr 




2  dt 
r
r0
Kinetic energy = change in
potential energy).
It follows that the time for free fall to the centre of the sphere is given by
t FF
0
1/ 2
(See Appendix)
 12
 2Gm0 2Gm0 
dt
  dr    

 dr
dr
r
r0 
r0
r0 
0
 3 
t FF  t d  

 32G 
1
R3  2

~

GM


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
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ii) What will happen if a star is in an equilibrium state?
an element of matter at a distance r from the centre
dP
will be in hydrostatic equilibrium if the pressure
gradient at r is
dr
Gm (r )  (r )

(1.4)
2
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 Viral Theorem
If m is chosen as the independent space variable rather than r, (1.4) 
dP
Gm

dm
4r 4
Ps
Ms
Pc
0
4r dP  (Gm / r )dm  3  VdP   
3
(1.5)
Gm
dm
r
Integrating the left-hand side of above by parts, the equation can be written
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Vs
3PV   3  PdV  
Ms
s
c
0
Using the symbol:
M
Gmdm
  
r
0
4r Ps  3
3
P

 (Gm / r )dm
0
Noting that dm=dV, so we
have
dm  
If the star were surrounded by a vacuum, its surface pressure would
be zero
3
P

dm    0
(1.6)
This is the general, global form of the Virial Theorem and used very often
later on, it relates the gravitational energy of a star to its thermal energy.
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2.3 Estimate the minimum pressure at the centre of a star:
Integrating eq.(1.5) from the centre to the surface of the star
M
Gm dm
P( M )  P(0)   
4
4

r
0
Pc  P(0),
P( M )  0.
On the right-hand side we may replace r by the stellar radius to
obtain a lower limit for the central pressure:
M
M
Gm dm
Gm dm
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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2.4 Estimate the minimum mean temperature of a star:
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
)kT
u
3 kT
3 P

2 mp
2 
2U    0
This is also the Virial theorem in another form
(1.8)
the system is stable
and bound at all points
because the total energy (binding energy) E =  + U =  / 2 ,
 and E are always negative.
In a contracting gas (protostar ):
The energy for radiation is provided by the balance of these terms;
- E = U = -  / 2.
When there is no energy from contraction, the radiation is provided by
thermonuclear reactions.
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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
Gm dm GM 2
  

R
2R
0
M
U 
 udm 
0
2U    0
3 kT
3 k
dm

TM
 2 mp
2 mp
T 
GMmp
6kR
(1.9)
 
T ~
M
R
(1.8)
3M
4R 3
2
3
T ~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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2.5 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
 1030 / 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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2.6 How accurate is the Hydrostatic Assumption?
From
Suppose:

d 2r
2
dt2
 g (r ) 
d r
 0
2
dt
1 dP
 dr
(1.3)
1 dP
a  g (r ) 
 g
 dr
i.e:
If the 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 is that it would take a star to collapse if the forces are out of
balance by a factor 
Fossil and geological records indicate that the properties of the Sun
have not changed significantly for at least 109 years(31016s)
  < 10-27
most stars are like the sun and so we may conclude that: the equation
of hydrostatic support must be true to a very high degree of accuracy !12
2. 7 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, however, and for these 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
M
It is a likely source of the stellar energy and has the form    
0
The total energy of the system :
E 
Gmdm
r
1

2
Assuming a constant density distribution  
3M
4R 3
the gravitational potential energy:
(4 ) 2 R 4
3GM 2
 ~ G
 r dr ~ 
0
3
5R
the total mechanical energy of the star is:
3GM 2
E~
10R
What can this tell us?
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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 1041 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 )
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3.8 10
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 !
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3.2 Nuclear reaction
the total energy equivalent of the mass of the Sun, . E
 M c 2  1.791047 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 1020 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 flux 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 )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
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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 r =constant
(1.12)
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
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By using P/r =constant and the second assumption
  (  / P)P

 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

P T





(1.13)
P

T

P
T
Substitute (1.12) and (1.13) into (1.11)
T   1 P

T
 P
(1.14 )
The critical temperature gradient for convection is given by
dT   1 T dP

dr
 P dr
(1.15)
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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 or
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,
the flux of energy flow
dT
f (r ) ~
dr
(1.17 )
Which of the two - conduction and radiation - is the more dominant in
stars to transport energy?
particles
Energy:
Numberdensity:
mean free path:
3
kT
2
nparti
parti ~ 10-10 m
photons
~
hc

nphoton
>
<<
photon~ 10 2m
Photons can get 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 larger transport of energy.
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Conduction is therefore negligible in nearly all main sequence stars and
radiation is the dominant energy transport mechanism 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
dT
 rad
dr
(1.18)
where the minus sign indicates that heat flows down the temperature
gradient. Assuming that all energy is transported by radiation
The radiative conductivity measures the readiness of heat to flow
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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 Tayler) 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  
 dr
3


rad


(1.21)
Lr  4r 2 Fr
3 rad Lr
dT

2 3
dr
16acr T
the equation of radiative transport
(1.22 )
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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 



  
 



3

 dr   rad  cond 
 3   dr
1

Then (1.22) can be
written as

1
 rad

(1.23)
1
 cond
dT
3k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
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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, and also plays a central role in events like supernovae
associated with the death of massive stars.
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Summary:
1. Based on two fundamental assumptions:
we derived the four equations of stellar structure
dM ( r )
 4r 2 
dr
dP
G M (r )
 

2
dr
r
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
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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 2

 
0
r
R
3
M
 
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
T ~ M
(1.7 )
2
3

1
3
(1.16 )
(1.9)
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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
Course work: ( hand in before 5.pm each Friday from week 17)
1. Assuming that a star of mass M is devoid of nuclear energy sources, find the
rate of contraction of its radius, if it maintains a constant luminosity L.
(hint: The rate of change of the energy, as given by
dE
 L
dt
Then use the Viral theorem to relate E  find a formula for dR/dt )
2. For a star of mass M and radius R, the density decreases from the centre
to the
2
surface as a function of radial distance r, according to    [1   r  ]
c
where c is the central density constant.
 
 R
(a) Find m( r ).
(b) Derive the relation between M and R and show that the average density of the star is 0.4c .
(c) Find the central pressure and check the validity of inequality
for the given density distribution
Pc 
GM 2
8R 4
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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