Transcript Chapter 10 powerpoint presentation

Chapter 10
Stellar Interiors
Revised 10/8/2012
The laws of stellar structure that
lead to L=M3
1.
2.
3.
4.
5.
Hydrostatic Equilibrium
Ideal gas law
Energy transport
Mass conservation
Energy conservation
Hydrostatic Equilibrium
Mathematically……..
The pressure gradient across a slab of area A supports the weight,
mg, of the layer so that
mg = (Pl – Pu ) A
For a spherical shell, A = 4pr2, and the mass of a shell dm = 4pr2rdr
Thus,
dP/dr = -gr
Where g is the “local value” of g inside the star
g = GM(r)/r2
Where M(r) is the mass enclosed by the radius r
Continuing ….
So,
dP/dr = - GM(r) r(r) /r2
“equation
of Hydrostatic equilibrium”
which may be integrated to determine Pc, exactly, with some
simplifying assumptions;
P(0) = Pc, and P(R) = 0
and constant density r(r) ~ r ~ Mo/ (4pR3o /3)
Then,
Pc ~ G Mo r/2 Ro
Which upon substituting standard values for
the constants yields,
Pc ~ 1.4 x 1014 N/m2
Which is the wrong answer! More detailed calculations, that
allow for the fact that the density increases towards the
center of the Sun, yield
Pc ~ 2.7 x 1016 N/m2
The problem here is that the Sun’s density is not constant!
Ideal Gas Law
What provides the Pressure? - the thermal energy of the hot gas,
which we assume is a “perfect gas” governed by the ideal gas
law;
P = n(r) k T(r)
Where n(r) is the number density (atoms/m3), k is Boltzmann’s
constant and T(r) is the temperature gradient.
We can express n(r) in terms of the density r(r) since
n(r) = r(r) /mH where mH is the mass of a hydrogen atom
P(r) = r(r) k T(r) /mH
which may be differentiated to get the temperature gradient,
or, we can use the same trick as before to get an estimate
for the central temperature since re-arranging,
Tc = Pc mH / r k
Which upon substituting the values for the constants, (including
the wrong Pressure from the previous example!), yields
Tc ~ 1.1 x 107 K
(which is close to the correct answer !)
At these temperatures hydrogen is dissociated (ionized) into
protons and electrons, an electrically neutral mixture called
a plasma.
Mean Molecular Weight
This version of the ideal gas law, P(r) = r(r) k T(r) /mH, only
works for a star composed entirely of Hydrogen, which, of course,
they are not.
So, to accommodate different chemical compositions, astronomers
introduce the concept of mean molecular weight, denoted by the
symbol .
Then the gas law becomes,
P(r) = r(r) k T(r) /  mH,
With the understanding that  is the average mass of an atom, in
units of the hydrogen mass, so that the average mass of a stellar
atom is m, where m =  mH
Energy Generation & Transport
The temperature is only
high enough for fusion
reactions to occur in the
very center of the Sun, a
region enclosing ~ 10%
of the Sun’s total mass
called the “core”.
The primary mode of
energy transport is
radiation.
Radiative Transfer Equation
The Sun’s central temperature is much higher than the surface
temperature. Thus, heat flows from hot to cold, down a temperature
gradient.
The fact that the surface temperature of the Sun is much lower than
The core temperature means that the photon’s are loosing a lot of
energy on their journey out of the star. Thus, there must be some
source of opacity.
The primary source of opacity is ionization. Protons and electrons will
recombine into H atoms unless they are continually ionized by photons. A
secondary source of opacity is electron scattering, where photons are scattered
by electrons. (see Chapter 9)
dPrad
d
=
F/c
(Note there are now 2 pressures; gas and
radiation. Prad is the latter).
where  =  r dr, is the radial optical depth and F = L / 4p r2 is the outward
flux, expressed in terms of the luminosity, L.
Also, from the second moment of the R.T.E., we have Prad = 4 T4/ 3c which can
be differentiated w.r.t.r. Following some algebra one finds,
L = 16  T3 4p r2 dT
3 r
dr
Important note
In this and similar equations, it is important
to note that T, r, r are “interior” values, themselves a function
of r, so that T = T(r), r = r(r). L=L(r), etc.
The Standard Solar Model
Numerically integrating the 5 equations of stellar structure reveal
that the pressure, temperature, and density all increase
rapidly towards the center of the Sun;
The Physical Basis for L a M3
Can be deduced from the first 3 equations governing stellar structure;
Pc ~ G Mo r/ Ro
Pc ~ r k Tc /mH
Lc ~
16 T3 4 p r2 dT
3r
dr
From which we can derive the following proportionalities,
Pc a Mo r/ Ro
Pc a r Tc
Lc a T4 r/r
The Physical Basis for L a M3 (continued)
Eliminating Pc , and T (since neither appear in the mass –luminosity
relationship) and writing r a M/r3, one can show that
L a M3
as observed !
Question: Using the proportionalities given on the previous page,
show that L a M3.
Energy Sources
What source of energy makes the Sun shine? This was a mystery
until quite recently when in 1938 Hans Bethe recognized that
the temperature at the core of the Sun was high enough to support
thermonuclear fusion reactions.
Prior to 1938 there were all sorts of ideas floating around including
The meteoric theory, whereby astronomers believed that the Sun
was powered by in-falling comets.
Comets do fall
into the Sun as
these pictures
show!
Plus, comets can yield a lot of energy….
The energy released by in-falling comets is that due to the gain in
kinetic energy, E, of the comets as they fall into the gravitational
potential well of the sun, where E = G M m /R and m is the mass
of the comet,
E
r
In-falling comets yield energy (continued)
If all that energy were converted into heat and light then the
rate of energy conversion, L, is related to the mass in-fall rate,
dm/dt, so that
L = G M dm
R dt
Which, assuming a 100% conversion of mass to energy,
the observed luminosity of the Sun, would require a mass in-fall
rate of 6.3 x 1022 kg/yr.
This may seem like a lot, but when expressed in terms of the
mass of the moon, it corresponds to a rate of only 1 moon per year!!
The Contraction Theory
The meteoric theory was never disproved, but rather supplanted
by the Contraction Theory. In this theory, the Sun has to contract in
order to loose heat. The problem with this theory is that the Sun is
not contracting, but back in the 19th Century, no one could tell.
The amount of energy released by a contracting self-gravitating
collection of particles may be calculated using the virial theorem,
which states that one half of the gravitational potential energy, U,
is radiated away, and the other half goes into heating up the star.
So, now, U/2 = 3 G M2 /10 R, is released, as the Sun shrinks
from infinity down to it’s current size.
But, the problem with the Contraction theory that if the Sun had been
radiating at it’s present rate, L, all the time, then the lifetime of the Sun,
t, would be
t ~ E/L = 3G M2 /10 L  R 
which, upon substituting the appropriate constants, yields a
lifetime of about 10 million years. But we know from carbon
dating of terrestrial and lunar rocks that the earth is at least 4.5
billion years old ! and one would expect the Sun to be
at least as old as the earth-moon system, if not older. So there’s
a major time discrepancy.
Nuclear Fusion
We now know that the real energy source for the Sun’s luminosity
is thermonuclear fusion reactions occurring in the Sun’s core.
Although the temperatures in the Sun’s core are high, they are not
high enough to overcome the coulomb repulsion force resulting from
two positively charged nuclei colliding under the laws of classical
physics. The solution to this problem is quantum mechanical tunneling.
The essence of this energy generating process is that 4 1H are
converted into one 4He, with the difference in mass-energy
being released mostly as heat according to Einstein’s famous
equation E = mc2.
Here are the details;
The total possible number of reactions, n = M  /4mH
The energy released in each reaction, e, is
e = (4 mH – mHe) c2
The total energy released, E, is
E=ne
E = M (4 mH – mHe) c2 /4mH
The quantity (4 mH – mHe) /4mH = 0.0071, thus
E = 0.0071 M  c2
But, only the core, 10% of the total mass, is involved
in fusion, so actually
E = 0.00071 M  c2
Solar Lifetime
Substituting the appropriate constants yields
E = 1.28 x 1044 J
Which, at the rate of the Sun’s present luminosity,
L = 4 x 1026 J/s, will last for a time, t
t = E/L = 3.2 x 1017 s
or
10 billion years !
There are several reactions, in order of decreasing probability;
The PPI chain
4 11H  42He + 2e+ + 2e + 2
The PPII chain
The PPIII chain
The CNO cycle
PPI Chain
Stellar Main Sequence Lifetimes
We can use the mass luminosity relationship to predict the
main sequence lifetimes for all other stars since
t
a E/L a M/L
and
L a M3
then,
t a 1/M2 , ie. the most massive stars have the shortest lifetimes