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PH607 The Physics of Stars
Dr J. Miao
2
• 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
We will cover the following materials:
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Chapter one The Equations of Stellar Structure (A Warming up)
How does a star exist?
Force of gravitation Internal pressure gradient
Two fundamental assumptions:
a) Stars are spherical and symmetric about their centers
b) Stars are in hydrostatic equilibrium
1. Equation of Mass conservation.
)1.1(4)()( 2rr
dr
rdM
Four equations of stellar structure
rr + dr
4
2. Equation of hydrostatic support
The balance between gravity and internal (thermal) pressure is known as hydrostatic equilibrium
)2.1()(
)(2r
rGmrg
The gravitational mass situated at the centre gives rise to an inward gravitational acceleration equal to :
the inward force on the element due to the pressure gradient is
Ardr
dPArPr
dr
dPrP )]()([ ,)( ArrM
the inward acceleration of any element of mass at distance r from the centre due to gravity and pressure is
)3.1(1
)(2
2
dr
dPrg
dt
rd
r
r+dr
PAgM (P+P)M
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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
0
00
2
2
1
r
Gm
r
Gm
dt
dr
Kinetic energy = change in
potential energy).
It follows that the time for free fall to the centre of the sphere is given by
drr
Gm
r
Gmdr
dr
dtt
rr
FF
21
00
0
0
000 22
(See Appendix)2
132/1
~32
3
GM
R
Gtt dFF
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
2.1 What can we know from the equation of hydrostatic support
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ii) What will happen if a star is in an equilibrium state?
)4.1()()(
2r
rrGm
dr
dP
)5.1(4 4r
Gm
dm
dP
an element of matter at a distance r from the centre will be in hydrostatic equilibrium if the pressure gradient at r is
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)
dmr
GmVdPdmrGmdPr
s
c
MPs
P 0
3 3)/(4
Integrating the left-hand side of above by parts, the equation can be written
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)6.1(03 dmP
ss MV
sc dmrGmPdVPV
00
)/(33
M
r
Gmdm
0
Using the symbol: Noting that dm=dV, so we have
dmP
Pr s 34 3
If the star were surrounded by a vacuum, its surface pressure would be zero
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:
)7.1(~4
2
R
MPc
Integrating eq.(1.5) from the centre to the surface of the star
M
r
GmdmPMP
044
)0()(
),0(PPc .0)( MP
On the right-hand side we may replace r by the stellar radius to obtain a lower limit for the central pressure:
42134
2
04
04
)()(104.4844 R
R
M
M
R
GMP
R
Gmdm
r
GmdmP c
MM
c
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:
)8.1(02 U
the equation of state of a classical gas is known as
nkTPgas kTm
Pp
)(
the internal energy per unit mass is P
m
kTu
p 2
3
2
3
)6.1(03 dmP
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 ,
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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R
MT ~
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
R
GM
R
Gmdm
0
2
2
TMm
kdm
m
kTudmU
pp
M
2
3
2
3
0
)8.1(02 U
)9.1(6kR
GMmT p 3
1
3
2
~ MT
34
3
R
M
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 4
3
1aTPrad
nk
aT
P
P
gas
rad
3
3
with T =4 106 K and 330 /10 m
mn
H
( =1.4 103 kgm-3, a =7.5510-16 Jm-3K-4), 410
gas
rad
P
PWe have:
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?
)3.1(1
)(2
2
dr
dPrg
dt
rd
From
Suppose: 02
2
dt
rd gdr
dPrga
1)(i.e:
If the element starts from rest with this acceleration its inward displacement s after a time t :
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2
1
2
1gtats 2
1
2
1
)/2()/1( gst 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
2
)()(
r
rGmrg 2
12
132
1
/~)/2()/1( fftGmrt
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(31016s)
< 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 !
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2. 7 How valid is the spherical symmetry assumption?
Departure from spherical symmetry may be caused by rotation of the star.
22 RM
RGM /2
GM
R32~ 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
It is a likely source of the stellar energy and has the form M
r
Gmdm
0
The total energy of the system : 2
1E
Assuming a constant density distribution 34
3
R
M
the gravitational potential energy:
R
GMdrrG
R
5
3~
3
)4(~
2
0
42
the total mechanical energy of the star is: R
GME
10
3~
2
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
JR
GMEEEE fifg
412
101.110
3)(
Further assuming Lsun is a constant throughout its lifetime, then it would emit energy at that rate for approximately
)10(~103~108.3
101.1 714
26
41
yrsL
Et g
th
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, . JcME 472 1079.1
If all this energy could be converted to radiation, the Sun could continue shining at its present rate for as long as
yrsL
Etn
1320 104.1sec106.4
is called nuclear timescale
The Sun just have consumed its mass:
MMMt
tM
n
sun 420
16
10~106.4
103
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:
drrrrrdLrLdrrL 24)()()()()(
where (r) is the density; (r) is the energy production rate per unit mass
)10.1()()(4)( 2 rrr
dr
rdL
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.
valuecriticaladr
dT
Consider a convective element of stellar material a distance r from the centre of the star
r+r
r
T+TP+P+
T, P, T, P,
T+T, P+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:
)11.1( 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;
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
PV r =constant (1.12)
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By using P/r =constant and the second assumption
)12.1(11
)/(P
P
P
PPP
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:
)13.1(T
T
P
P
T
T
P
P
Substitute (1.12) and (1.13) into (1.11) )14.1(1
P
P
T
T
The critical temperature gradient for convection is given by
)15.1(1
dr
dP
P
T
dr
dT
21
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: )16.1(1
dr
dP
P
T
dr
dT
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
b) the temperature gradient is very steep
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
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 )17.1(~)(dr
dTrf
Which of the two - conduction and radiation - is the more dominant in stars to transport energy?
particles photons
Energy: kT2
3
hc~
Numberdensity:nparti nphoton
>
mean free path:parti ~ 10-10 m 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.
23
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,
)18.1(dr
dTF radrad
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
24
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
)19.1(3
4 3
radrad
acT
where a is the radiation density constant and c is the speed of light
Combining the above equations we obtain:
)20.1(3
4 3
dr
dTacTF
radr
Recalling that flux and luminosity are related by rr FrL 24
)21.1(3
16 32
dr
dTTacrL
radr
the equation of radiative transport
)22.1(16
332Tacr
L
dr
dT rrad
25
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.
)23.1(3
1611
3
16 3232
dr
dTTacr
dr
dTTacrLLL
condradcondr
condrad 111
)24.1(16
332Tacr
Lk
dr
dT
Then (1.22) can be written as
Clearly, the flow of energy by radiation/conduction can only be determined if an expression for is available
26
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.
27
Summary:
1. Based on two fundamental assumptions:
we derived the four equations of stellar structure
32
2
2
2
16
3
4
)(
4)(
Tacr
Lk
dr
dT
rdr
dLr
rGM
dr
dP
rdr
rdM
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 equationsP: 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
28
2. From the most important hydrostatic equilibrium equation:
)4.1()()(
2r
rrGm
dr
dP
-- Drive the global form of Viral theorem. )6.1(03 dmP
R
GM
r
GmdmM2
0 With the gravitational potential energy of a stat
If the density of the system is a constant,5
3
Drive another form of Viral theorem: )8.1(02 U
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 : )7.1(~4
2
R
MPc
-- estimate the minimum mean temperature of a star: 3
1
3
2
~ MT
3. Criteria for convection:
td << tth << tn. (1.9)
)16.1(1
dr
dP
P
T
dr
dT
4. Three important time scale:
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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
Ldt
dE
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 surface as a function of radial distance r, according to ]1[
2
R
rc
where c is the central density constant.
(a) Find m( r ).(b) Derive the relation between M and R and show that the average density of the star is 0.4c .
(c) Find the central pressure and check the validity of inequality
for the given density distribution
4
2
8 R
GMPc
30
Appendix: drr
Gm
r
Gmdr
dr
dtt
rrFF
21
00
0
0
000 22
This may be simplified by introducing the parameter
0/ rrx to give
2
12
1
0
30
2
1
0
30
2
0
22
0
2
11
0
22
11
0
2
1
0
30
32
3
23
4
4
3
2
1
22
2sin2cossin
cos
sin2
1
)cossin2,sin(12
Gm
r
GGm
rt
dddxx
x
ddxxdxx
x
Gm
rt
FF
FF
