The Probability and Effects of an Asteroid Impact with Earth

1
10. The Interiors of Stars Goals 1. Develop
the basic equations describing equilibrium
conditions applying in stellar interiors. 2.
Estimate the physical conditions that must exist
at the centre of a typical star, the Sun. 3.
Outline potential sources of energy generation in
stars and investigate the basics of nuclear
reactions as a means of providing a
self-sustained energy source. 4. Develop the
equations of energy transport in stars.
2
Fact or Fiction? Consider an inhabited planet
completely enshrouded by clouds, upon which
astronomy is not a scientific discipline since it
is not possible to see into space from the
planets surface. Yet, one can construct abstract
mathematical models of massive spheres of hot gas
in equilibrium and deduce their properties from
what is known about the physics of matter. Would
it come as a surprise to scientists on such a
planet if the constant canopy of surrounding
clouds one day parted and stars came into view?
3
Models of Stars The parameters used for studying
and modeling stellar interiors include r radia
l distance from the centre of the
star M(r) mass interior to r T(r) temperatur
e at r P(r) pressure at r L(r) luminosity
at r e(r) energy generation at
r ?(r) opacity at r ?(r) density at r In
modern models mass M is used as the dependent
variable rather than radial distance r, but it is
more informative to initiate the study of stellar
interiors using the geometrical variable r.
4
At the natural boundaries of the star the
corresponding values are At the centre At the
surface r 0 r R M(r) 0
M(r) M T(r) Tc T(r) 0 (or
Teff) P(r) Pc P(r) 0 L(r) Lc
L(r) L ?(r) ?c ?(r) 0 Rotation and
magnetic fields are usually ignored in most
models (i.e. spherical symmetry is imposed), as
well as any temporal changes (i.e. radial
pulsation). Let us examine the equations of
stellar structure.
5
Hydrostatic Equilibrium For balance at any point
in the interior of a star, the weight of a block
of matter of unit cross-sectional area and
thickness dr must be balanced by the buoyancy
force of the gas pressure, i.e. Mass of block
density volume ? dr Weight of the block
mass local gravity ?g dr But local gravity, g
GM(r)/r2 Buoyant Force pressure difference
(top bottom) dP So
6
Example Obtain a crude estimate for the
pressure at the centre of the Sun. Assume 1 M?
1.9891 1033 gm, 1 R? 6.9598 1010 cm
and Solution (instructor) Convert the
equation of hydrostatic equilibrium into a
difference equation, evaluate it at r ½R,
assume that the mean density and M ½M applies
there as well. Then Set dP Ps Pc 0
Pc Pc and dr rs rc R 0 R ?
7
So, for the Sun Pc GM?2/ R?4 6.6726
108 (1.989 1033)2/ (6.9598 1010)4
5.63 1015 dynes/cm2 The textbook method of
solution gives a value of Pc 2.7 1015
dynes/cm2 A more rigorous solution involving
integration with the standard solar model gives a
more reliable estimate of Pc 2.5 1017
dynes/cm2 two orders of magnitude larger! Since
1 atmosphere 1.013 106 dynes/cm2, the
pressure at the centre of the Sun is equivalent
to 2.5 1011 atmospheres!
8
Conservation of Mass The mass of a star must
increase uniformly from the interior to the
surface, there cannot be any holes! For each
element of thickness dr the volume must increase
by the mass of the shell encompassed, i.e. by the
spherical area thickness dr. The resulting
equation is
9
Ideal Gas Law The ideal gas law is introduced in
PHYS1211 PV NkT, where P is the gas
pressure, V its volume, N the number of particles
in the volume, T is the temperature (on the
absolute scale), and k 1.3807 1016 ergs/K is
the Boltzmann constant. It is also expressed
using the gas constant R as PV nRT, where n
is the number of moles of gas. The relationship
can be derived from first principles through the
kinetic theory of gases, which specifies
that (i) gas consists of small particles
(molecules, atoms) that are smaller than the
distances separating them, (ii) particles are in
constant motion and make perfectly elastic
collisions with container walls, (iii) the
motions of particles are random, i.e. ? are
moving in any specific direction.
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Consider a box of N particles of mass m, where ?N
are moving in the x-direction. The average force
on the right-hand wall (shaded) is given by the
rate of change of momentum Favg ?(mv)/?t A
particle takes time ?t to complete a trip from
one wall of the box back to the same wall, i.e.
?t 2l/v, where l is the dimension of the
box. ?(mv) mv(before) mv(after) mv (mv)
2mv The resulting pressure (force/unit area)
on the shaded wall is given by the total force
exerted by all particles in the box on the end
wall, i.e.
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But l 3 V, the volume of the box, so Since
Nm ?V (density volume) But the average
kinetic energy of a particle ½mv2 3/2 kT,
so The average kinetic energy of a gas
molecule at room temperature is
12
Pressure Equation of State The ideal gas law is,
once again PV NkT, where P gas
pressure V volume of the gas N number of
particles in the gas T temperature (K) k
1.3807 1016 ergs/K is the Boltzmann
constant. The number density for the gas can be
written as n N/V, so P nkT, and since the
number density can also be written as n ?/µmH,
where ? is the actual density (gm/cm3), mH is
the mass of a hydrogen atom, and µ is the mean
molecular weight. So the gas pressure can be
written as
13

• Mean Molecular Weight
• As the name implies, the mean molecular weight is
  the average mass of a gas particle, but in units
  of the mass of a hydrogen atom, i.e.
• where is the average mass of a gas
  particle.
• Consider possible examples
• Hydrogen neutral,
• ionized,
• molecular,

14
Helium neutral, ionized, Heavy
element neutral, ionized, where Aj is the
atomic number. In stellar interiors the value of
µ for a fully-ionized gas is desired. How does
one take into consideration the contributions
from all elements?
15
Let X fractional abundance by mass of
hydrogen, Y fractional abundance by mass of
helium, and Z fractional abundance by mass of
all heavy elements, where X Y Z 1, by
definition. In stellar interiors the value of µ
for a fully-ionized gas is desired. How does one
take into consideration the contributions from
all elements? In a cubic centimeter of gas of
density ? there is X? of hydrogen, Y? of helium,
and Z? of heavy elements by mass. Each of the
elements contributes different numbers of
electrons to the mix, under the assumption of
complete ionization 1 for hydrogen, 2 for
helium, and 3, 4, 5 for the heavy elements. The
number of particles per cubic centimeter
is X?/mH 2 2X?/mH for hydrogen Y?/4mH 3
3Y?/4mH for helium, and Z?/2AjmH (Aj 1)
Z?/2mH for the heavy elements.
16
The total number of particles per cubic
centimeter is therefore given by So or fo
r fully ionized gas, and for neutral
gas, or, since X Y Z 1 When Z is
negligible
17
Example What is the mean molecular weight for
gas in the Sun, where X 0.75, Y 0.23, and Z
0.02? Solution (instructor) Consider the case
for gas that is fully ionized, which
gives i.e., slightly larger than ½,
the value for a pure hydrogen gas.
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Example Obtain an estimate for the temperature
at the centre of the Sun. Assume, for simplicity,
µi 0.5944, i.e. fully-ionized gas. Solution
(instructor) We can use the perfect gas law if
we know the density at the centre of the Sun.
Approximate the value using estimates for
conditions at r ½R (as before), where we assume
that ? . Assume also that P ½Pc at that
point, that k 4/3 1016 erg/K, mH 3/2
1024 gm, and µi 0.5944. Then
19
If the average temperature ½Tc , then Tc 2.4
107 K, fairly close to the actual value of 1.6
107 K. The central density is found
from whereas the actual value is 82
gm/cm3. Since the mass of a hydrogen atom is
1.6726 1024 gm, a density of 1.8 gm/cm3
corresponds to a number density of n ?/µmH
1.81 1024 /cm3. The volume occupied by a
particle is 4/3pr3 1/n, so the average particle
radius is r 5.1 109 cm 0.51 Å, compared
with the Bohr radius of 0.523 Å. With an actual
density for the Sun of 82 gm/cm3, the average
particle radius is r 1.4 109 cm 0.14 Å,
compared with the Bohr radius of 0.523 Å. Such
highly compressed matter cannot exist as bound
atoms, and is referred to as pressure-ionized gas.
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Example (from Mechanics) Calculate the
gravitational self-energy (energy of assembly
piece-wise from infinity) of a uniform sphere of
mass M and radius R. Solution Think of
assembling the sphere a shell at a time (r 0 to
r R). For a shell of radius r the incremental
potential energy is dV fdm, where dm is the
mass of the shell and f is the gravitational
potential for the mass already assembled, which
is a sphere. Since the mass being assembled forms
a spherical shell, we have So the
potential self-energy of the mass is
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where So the potential self-energy of
the sphere is
22
Stellar Energy Sources As noted the
gravitational potential energy required to
contract a star to its present size is given
by But, of the potential energy lost by a
star, according to the Virial Theorem (ASTR
2100), one half is transformed into an increase
in the kinetic energy of the gas (heat) and the
remainder is radiated into space. The radiation
lost by a star upon contraction to the main
sequence is therefore given by For the Sun,
at its present mass (1.9891 ? 1033 gm) and radius
(6.9598 ? 1010 cm), the amount of energy radiated
through contraction is
23
The present luminosity of the Sun is L? 3.851 ?
1033 ergs/s, so if it had been shining at the
same luminosity for the entire duration of its
contraction (clearly erroneous) then the time
scale for contraction is given by the
following where tKH is the Kelvin-Helmholtz
time scale (more properly the Helmholtz-Kelvin
time scale, although it originated with the
Scottish physicist John James Waterston,
1811-1883, whose papers on the subject were
rejected by the Royal Society of London!).
Here or 107 years. The actual value should
be smaller because the Suns luminosity was
greater during the contraction phase, but the
main point is that the estimate is much shorter
than the estimated age of the solar system of
4.6 ? 109 years.
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Nuclear Energy Sources Consider the rest masses
of the fundamental nuclear particles Proton 1.6
72623 1024 gm Neutron 1.674929 1024
gm Electron 9.109390 1028 gm Atomic mass
unit, 1 u 1.660540 1024 gm 931.49432
MeV, for E mc2. The original nucleon symbolism
was where A mass number number of
nucleons Z number of protons (usually
omitted) X chemical symbol of the element as
specified by Z. i.e., is redundant,
since indicates the same thing.
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Typical masses 1H 1.007825 u 938.78326
MeV 2H 2.014102 u 1876.12457 MeV 4He
4.002603 u 3728.40196 MeV 5Li 5.0125 u
4669.115279 MeV 8Be 8.005305 u
7456.89614 MeV The major reaction in astronomy
converts 4 hydrogen nuclei (protons) into a
helium nucleus (4He). But 4 1H 1.007825 u
4 4.031280 u and 1 4He 4.002603 u 4.002603
u Difference 0.028677 u 0.0071 of 4
1H The energy released mc2 0.028677 u
1.660540 1024 gm c2 26.71 MeV. The lifetime
of a star via nuclear reactions depends upon how
much of its hydrogen content is converted to
energy via nuclear reactions.
26
For the Sun we can estimate At L? 3.851
1033 ergs/s, Also, if tnuclear Enuclear/L
XMc2/M4 then tnuclear Xc2/M3 1010
yrs/(M/M?)3 . 1 M?, tnuclear 1010 years 2
M?, tnuclear 109 years (A-star) 4.6 M?,
tnuclear 108 years 10 M?, tnuclear 107 years
(B-star) 21.5 M?, tnuclear 106 years
(O-star) 0.5 M?, tnuclear 1011 years gt 1/H0
(estimated age of the universe) The lifetime of
the Sun and stars via nuclear reactions is
consistent with the nuclear ages of meteorites,
as well as the oldest rocks on the Earth and the
Moon.
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Properties of nuclear particles Particle Baryon
Lepton Spin Charge proton ? ?
½? 1 neutron ? ? ½? 0 electron ? ?
½? 1 positron ? ? ½? 1 neutrino ? ?
½? 0 muon ? ? ½? 1 Other baryons, the
hyperons xi (?) ? ? ½? 0, sigma (S) ? ?
½? 0, 1 lamda (?) ? ? ½? 0 Middle family,
the mesons ?-mesons ? ? 0 or 1? 0,
1 p-mesons ? ? 0 or 1? 0, 1 photons ?
? 0 0 Nuclear reactions conserve (i) number of
nucleons, (ii) number of leptons, (iii)
electronic/nuclear charge, (iv) particle spin.
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Example What is the missing particle in the
following reaction? 37Cl ?e ? 37Ar
? The collision of an electron neutrino with a
chlorine-37 nucleus produces an argon-37 nucleus
plus a to-be-identified particle. Solution 37Cl
is element 17 with 17 protons, 20 neutrons. 37Ar
is element 18 with 18 protons, 19
neutrons. Nucleons in 17 20 37. Nucleons
out 18 19 37. ? Leptons in 1. Leptons
out 0, so missing particle is a lepton.
? Charge in 17 (17 protons). Charge out
18 (18 protons), so missing particle has a
charge of 1. ? Spin in 19? (17 protons, 20
neutrons, 1 neutrino). Spin out 18½? (18
protons, 19 neutrons), so missing particle has
spin ½?. ? The only lepton with spin ½? and
charge of 1 is the electron. i.e., 37Cl
?e ? 37Ar e
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The proton-proton reaction The first step in
the proton-proton cycle is stymied by the fact
that the colliding particles are small and both
positively charged. Although the nuclear strong
force takes over at small separations of the
particles, at larger distances they are blocked
from interacting by mutual Coulomb repulsion.
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How is the Coulomb barrier overcome? Consider the
potential energy of a Coulomb barrier From
statistical mechanics we know that kinetic energy
and thermal energy are related through the
reduced mass of a particle so which is
much higher than the Suns central
temperature. But, by the Heisenberg uncertainty
principle, the uncertainty in a particles
momentum, ?px, is related to the uncertainty in
its position, ?x, via ?px?x ½? h/4p
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The quantum mechanical result is Two protons
can approach each other only to within ?, the de
Broglie wavelength, the wave functional size of
a particle. Energy, E mc2 pc h? hc/?, so
p hc/?c h/?, and ? h/p. So and which
leads to for the proton-proton reaction.
Note that it is necessary to increase Tquantum
by 1, 2, or 3 orders of magnitude for reactions
involving heavier nuclei.
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Reaction Rates The rates of individual nuclear
reactions depend upon a variety of factors 1.
Atomic nuclei must have sufficient energy to
penetrate the Coulomb barrier of the target
nucleus, only nuclei with specific energies
specified by the high velocity tail of the
Maxwell-Boltzmann distribution can react with the
target nuclei. 2. The cross-section for the
reaction, s(E), must be substantial. The
restriction from (1) implies that energetic
nuclei capable of reacting with the target nuclei
are described by the MB distribution,
i.e., where KE ½µmv2 is the kinetic energy
of the particle. The restriction from (2)
produces
33
Denote dNE as the number of particles of velocity
vE (2E/µm)½ that can strike a target nucleus in
time dt, i.e. So For nx
targets/unit volume, the total reaction rate per
unit volume per unit time is
34
But we still need to evaluate s(E). The
radius of an atomic nucleus can be estimated as r
? ( h/p), the de Broglie wavelength. The
size of the Coulomb barrier VC is also important,
so
35
Presumably the function describes the
main type of variability, although an additional
term that is a slowly varying function of E
cannot be excluded. The reaction rate therefore
becomes Where the MB relation has been
used as a substitute for nE. The resulting
functional dependence of the reaction rate
resembles a Gaussian, and is referred to as the
Gamov peak.
36
In actual cases there may be
resonant peaks superposed because of resonances
with excited energy levels in the nucleus
(analogous to excited energy levels for atoms).
37
Examples of potential resonant
cross-sections in the reaction rate functional
dependence.
38
Electrons can partially shield the positive
charges of nuclei, resulting in lower effective
Coulomb barriers to reactions, i.e. where
Ves(r) lt 0, is the contribution from electron
screening. When electron screening is ignored,
the integration results in a function that can be
approximated as where r0 is a constant, Xi
is the mass fraction of impacting particles for
the reaction, Xx is the mass fraction of target
particles for the reaction, and a' and ß are
exponents established by the integration using a
power-law expansion for the reaction rate
equations, which have messy integrals.
39
The energy released by nuclear reactions per gram
of stellar material is given by where a
a' 1. The units are ergs/gm/s. Energy
generated through nuclear reactions is
responsible for a stars luminosity, through the
equation of continuity where e enuclear
egravity, where the latter term is not always
negligible.
40
Nuclear Reaction Chains All must conserve
momentum, energy, spin, charge, etc. The
Proton-Proton Reaction. PPI (69) t½ 7.9
109 yr t½ 4.4 108 yr t½ 2.4 105
yr PPII (30.85) t½ 9.7 105 yr t½
3.9 101 yr t½ 1.8 105 yr
41
PPIII (0.15) t½ 9.7 105 yr t½ 6.6
101 yr instantaneous t½ 3.0 108
yr Net Reaction 26.73 MeV of
energy Energies of Neutrino Products From
1H Maximum Energy 0.42 MeV From 7Be Maximum
Energy 0.86 MeV From 8B Maximum Energy 14.0
MeV
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The nuclear energy generation rate for the
process can be written as where T6 units
of temperature in 106 K, fPP fPP(X, Y, ?, T)
1 is the electron screening factor, fPP fPP(X,
Y, T) 1 is a correction factor to account for
the various branches of the PP chain, and cPP 1
is a correction factor for higher order
terms. In simple form the relationship is
written as for temperatures near 1.5 ? 107
K. In other words, the energy generation rate for
the proton-proton chain varies as the local
density and the temperature to the fourth power,
e ?T4.
44
The CNO Bi-Cycle. t½ 1.3 107 yr t½
2.8 105 yr t½ 2.7 106 yr t½ 3.2
108 yr t½ 5.6 106 yr t½ 1.1 105
yr 99.96 of the time, or 0.04 of the
time t½ 3.0 106 yr
45
In the CNO cycle, discovered by Hans Bethe in
1938, isotopes of carbon (C), nitrogen (N), and
oxygen (O) act as catalysts for the reaction. The
element fluorine (F) is also involved. Although
the elements are not destroyed in the reactions,
they proceed at such different rates that the
isotopes of nitrogen (N) increase in abundance
while those of carbon (C) and oxygen (O)
decrease.
46
The proton-proton chain dominates for cool stars
like the Sun, the CNO bi-cycle for stars hotter
than the Sun, which have higher core
temperatures.
47
The Triple-Alpha Process. t½ 1.3 107
yr t½ lt 8.2 1024 yr Summary Reaction ?-
dependence X-dependence T-dependence PP
Chain ?1 X2 T64 CNO Bi-Cycle ?1 XXCNO T619.9
Triple-Alpha ?2 Y3 T841.0 Note the higher
temperature dependence of the CNO cycle. It is
dependent upon the CNO abundances, but dominates
over the PP chain for stars somewhat more massive
than the Sun where the central temperatures are
higher.
48
The triple-alpha reaction involves an unstable
isotope in the production of 8Be. The reaction
proceeds because, under the high density
conditions at the centers of evolved stars, a
third alpha particle (4He nucleus) can collide
with 8Be before it has time to decay. The
resulting production of energy has an extremely
strong temperature dependence.
49
The Helium Flash. When stellar core material is
electron degenerate, the local pressure does not
depend upon T. Thus, when He-burning is
initiated, the high T-dependence of the 3a
process means the energy is generated, raises T
locally, thereby increasing the reaction rate,
but does not produce a pressure or density
decrease to moderate the reaction. The result is
known as a helium flash. It only occurs in red
giants for stars roughly as massive as the Sun or
less, and in more advanced stages of other stars,
typically in the white dwarf stage. It is best
pictured using the pressure equation When
H-burning or He-burning occurs, the result is a
gradual increase in the mean molecular weight µ.
If T and ? remain unchanged, P decreases and
unbalances the previously-existing hydrostatic
equilibrium. The core of the star collapses so
that both P and ? increase. That enhances the
energy generation rate, making the star more
luminous. If Teff does not change, the star
becomes larger, since L 4pR2sTeff4.
50
Other Reactions. More advanced reactions involve
fusion of 12C to 16O, 23Mg (endothermic) or 20Ne,
23Na, 24Mg (exothermic), as well as fusion of 16O
to 24Mg (endothermic) or 28Si, 31P, 31S, 32S,
(exothermic). Various reactions are possible.
Consider the binding energy per nucleon
51
At low atomic weights the most stable nuclei are
1H, 2H, 3He, 6Li, 4He, 12C, 16O, 24Mg, 40Ca, and
56Fe. At high atomic weights the most stable
nuclei are 86Kr, 107Ag, 127I, 174Yb, 208Pb, and
238U. Such unusually stable nuclei are called
magic nuclei. The maximum binding energy per
nucleus occurs at the iron peak, and all other
fusion reactions producing heavier nuclei are
endothermic.
The problem of explaining the existence of heavy
elements in the universe can be restricted to
explaining the existence of nuclear reactions
that are endothermic in stars. The solution is
advanced stages of evolution in massive stars
(proton and alpha capture) and supernova
explosions (neutron capture).
Iron Peak
52
Energy Transport and Thermodynamics So far we
have developed the following equations of stellar
structure Equation of Continuity Hydrostatic
Equilibrium Energy Generation But what
about the temperature gradient, dT/dr?
53
Radiative Transport Energy can be transported
through a star by radiation, convection, or
conduction. Conduction is unimportant in most
gaseous stellar interiors, but the other two
processes are important. In stellar atmospheres
radiative energy transport is described
by But Prad ?aT4, so Thus But Frad
Lr/4pr2, giving, for radiative energy transport
54
Convective Transport Convection is a
three-dimensional process that must be
approximated by one-dimensional equations for
simple stellar interior models. It is a process
that is difficult to model correctly, but begins
with certain assumptions. The pressure scale
height HP is defined as or, if HP
constant, HP is the distance over which the gas
pressure P decreases by a factor of
1/e. Example Estimate a typical value for the
pressure scale height in the Sun. Solution At
the midpoint of the Sun we estimated ??
4M?/3pR?3. But dP/dr GM?/r2 GM2/2R5
Pc?/R?. So the pressure scale height can be
estimated as HP P/dP/dr ½Pc?/(Pc?/R?)
½R? More typical values in the Sun are of order
HP 0.1R?.
55
Thermodynamics According to the first law of
thermodynamics, the change in internal energy of
a mass element, per unit mass, is the difference
between the amount of heat added and the work
done by the element on its surroundings dU
dQ dW. The internal energy of the mass
element, U, is where nR k/µmH is the
universal gas constant.
56
The change in heat of a gaseous mass element is
usually expressed in terms of the specific heat C
of the gas, where C is the amount of heat
required to raise the temperature of a unit mass
by 1K, i.e. The amount of work done per
unit mass by a gas on its surroundings is dW. The
usual expression for work done is dW
PdV, so dU dQ PdV. When the volume does
not change, i.e., dV 0, we have
57
But If the pressure is held constant instead
so that the volume changes as heat is added,
then But PV nRT, so and Define a
s the ratio of specific heats.
58
For a monatomic gas When a gas undergoes
ionization, some of the heat dU that would
normally increase the average kinetic energy of
the gas is used instead for ionization. The
temperature of the gas therefore increases less
rapidly so dT is lower than otherwise. In such
instances so in such instances. The
effect is particularly pronounced in stellar
ionization zones.
59
For adiabatic processes there is no net heat flow
into or out of a mass element, so dQ 0 and dU
dW PdV. From PV nRT we have PdV VdP
nRdT and dU CVdT, so From the
definition of specific heats we have so PdV
VdP (?1)PdV
60
That gives (1?1)PdV VdP or ?PdV VdP
or The result is the adiabatic gas law
PV? K, where K is a constant. But the perfect
gas law also gives PV nRT, so
61
The sound speed in a gas is related to the
incompressibility and inertia of the gas,
specifically For adiabatic sound
waves For convective energy transport
one assumes that the gas bubbles are adiabatic.
Also the specific volume V 1/? refers to the
volume per unit mass. So if P KV? , we must
also have P K?? .
62
The formula for the pressure gradient in stellar
interiors therefore becomes With the
perfect gas law so If there is no gradient
in mean molecular weight in a star (not
necessarily a valid assumption!), then from
the adiabatic expression for P and ?. The
resulting equation gives an expression for the
adiabatic temperature gradient.
63
Namely or But so
64
The resulting temperature gradient
is If temperature gradient is
superadiabatic. The temperature gradient is a
negative quantity since the temperature decreases
with increasing radius inside a star. If heat
will be transported by convection. Otherwise
the heat is transported outwards by radiation.
The deep interiors of stars are relatively simple
to understand. They have either convective or
radiative cores depending upon the temperature
gradient. They may also be semiconvective in the
region immediately outside a convective core.
Stellar atmospheres are more complicated since
heat may be transported by both methods.
65
The test for whether or not convective or
radiative transport describes the temperature
gradient inside a star is therefore to test
whether or not a displaced bubble of gas rises or
falls. It will rise, i.e. convection applies if
?i(bubble) lt ?i(surroundings). The condition
is or or See textbook, also for
mixing length model.
66
Stellar Models The complete set of differential
equations describing the interiors of stars is
therefore Equation of Continuity Hydrostatic
Equilibrium Energy Generation Temperature
Gradient
67
Stellar Models The results for a series of
stellar models at the start of hydrogen burning,
i.e. with their initial composition unchanged, is
depicted in the following figures. In each case
solid circles denote models with a solar
metallicity (X 0.73, Y 0.25, Z 0.02),
whereas open circles denote models of extremely
low metallicity (X 0.749, Y 0.25, Z 0.001).
The actual metallicity of the Sun is presently
under debate as a result of newer models
describing turbulence in the solar atmosphere.
Models are constructed numerically using the
differential equations as difference equations.
They are also rearranged so that the dependent
variable is mass rather than radius. Models are
then denoted by the number of mass cells they
contain. Spherical symmetry is also assumed,
although that assumption is relaxed in more
recent models.
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One of the basic tenets of stellar evolutionary
models is the Vogt-Russell Theorem, which states
that the mass and chemical composition of a star,
and in particular how the chemical composition
varies within the star, uniquely determine its
radius, luminosity, and internal structure, as
well as its subsequent evolution. A consequence
of the theorem is that it is possible to uniquely
describe all of the parameters for a star simply
from its location in the Hertzsprung-Russell
Diagram. There is no proof for the theorem, and
in fact, it does fail in some special
instances. A prime example of where ambiguities
arise occurs when one compares models for two
stars, one of which is spherically symmetric and
the other of which is flattened as a result of
rapid rotation. Both stars can occupy the same
location in the Hertzsprung-Russell diagram, at
different evolutionary ages and even for
different masses.
75
Insights into Stellar Evolution Consider the
sequence of events that happens as stars
evolve During hydrogen burning, the primary 41H
? 4He reaction converts two protons and two
electrons into two neutrons. The reaction chain
may suggest that two positrons are produced, but
they quickly self-annihilate through collisions
with electrons, so the net result is as
stated. The production of a-particles from
protons and electrons has two effects (i) the
mean molecular weight µ of the gas increases
slightly, and (ii) the gas opacity, which is
dominated by electron scattering in stellar
cores, decreases because of the depleted
abundance of electrons. Consequence (i) implies
that the gas pressure must decrease, since the
mass density of the gas is unaffected. Consequenc
e (ii) implies that radiation escapes the core of
the star more easily.
76
Since the gas pressure is reduced, the core of
the star responds to the pressure imbalance by
contracting. But according to the Virial Theorem,
a contracting sphere of gas converts half of the
decreased potential energy of the system into
kinetic energy, i.e. heat, while the remainder
escapes as radiation. The consequence of an
increased temperature for the gas at the stellar
core is an increase in the nuclear reaction rate,
so, along with the radiation increase induced by
contraction of the core, the luminosity of the
stellar core increases sharply, raising the
photon flux from the interior. The decreased
electron scattering opacity of the gas near the
stellar core means that radiation escapes more
easily into the stellar envelope, where the main
sources of opacity (from atoms and ions) exist.
The increased photon flux on the gas transfers
more outwards-directed momentum to the gas
particles, resulting in an outwards expansion of
the envelope gases. So, as the core of the star
contracts, the envelope expands!
77
Stellar evolution thus results in an increased
luminosity as the star evolves. The resulting
change in the effective temperature of the star
is more complicated, since increased luminosity
can be accommodated by an increase in the stellar
radius or an increase in the effective
temperature. Since the radius of the star must
increase because of envelope expansion, it is not
clear what will happen to the stars effective
temperature. Stellar evolutionary models
indicate that, for massive stars, the effective
temperature decreases as the star evolves. For
low mass stars like the Sun, however, the
effective temperature actually increases during
the initial stages of hydrogen burning. The
difference presumably originates from the
differences in how energy is transmitted outwards
in the two types of stars. Massive stars have
convective cores and radiative envelopes, whereas
low-mass stars like the Sun have radiative cores
and convective envelopes. Convection mixes gas so
that any changes in chemical composition are
transmitted throughout the convective region,
which is clearly more important for high mass
stars M gt 1¼ M?.
78
Question. Air is mostly (80) composed of
nitrogen molecules, each of which consists of 28
nucleons (protons and neutrons), where the mass
of one nucleon is about one atomic mass unit
(1.6605402 1024 g). The radius of a typical
nitrogen molecule is about 1 Å, and the density
of air at sea level is roughly 1.2 103 g cm3.
Calculate the mean free path for collisions
between air molecules under such conditions. The
air is at room temperature, i.e. about 300K,
which allows you to estimate the
root-mean-squared speed, vrms, for air molecules.
From that information, calculate the average time
between collisions of atoms. Solution For
nitrogen, s p(2r)2 p(2 108)2 cm2 p 4
1016 cm2 The number density is n ?/m(N2) 1.2
103 g cm3/(28 1.6605402 1024 g) 2.581
1019 cm3 So the mean free path is l 1/ns?
1/(2.581 1019 p 4 1016 ) 3 105
cm vRMS (3kT/m)½ (3 1.38 1016 300/4.65
1023)½ 5.17 104 cm/s The average time
between collisions is t l/v (1.25 104
cm)/(5.17 104 cm/s) 2 109 s
79
Question. According to a standard model for the
Sun, the central density is 162 g cm3 and the
Rosseland mean opacity ? is 1.16 cm2 g1. a.
Calculate the mean free path for a photon at the
centre of the Sun. b. If the mean free path
remains constant for the photons journey to the
surface of the Sun, how long, on average, would
it take for photons to escape from the
Sun? Solution The mean free path for a photon
is l 1/??? 1/(1.16 162) 5 103 cm The
distance traveled by a photon between collisions
is d2 Nl2 In order to escape from the Sun a
photon has to travel a distance of 6.9598 1010
cm Number of collisions is N d2/l2 (6.9598
1010)2/(5 103)2 1.9376 1026, each taking t
l/c 5 103 cm/3 1010 cm/s 1.7 1013 s
Time to escape the Sun is Nt 1.9376 1026
1.7 1013 s) 3.23 1013 s 106 years