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ASTR2050 Spring 2005
Lecture 10am 15 February 2005
Please turn in your homework now!
In this class we will discuss the structure of
Main Sequence Stars:
Spheres in static equilibrium
•
• Stellar energy sources: Nuclear Physics
• Stellar structure and stellar models
• Outstanding issues: Solar Neutrinos
Note: No class next Tuesday, but see lab for Friday
Also: See updated HW assignment (Ch.9) due in two weeks
1
The Main Sequence: “Normal” stars
40M!
1M!
2
0.2M!
Example: The Pleiades
See lab this Friday.
3
Example: The old cluster M67
4
Spheres in static equilibrium
R
Build a star in thin spherical shells:
dV =4!r dr
dM="(r)dV
2
=4!r "(r)dr
2
dr
r
Example: Mass of a sphere
with constant density
M=
Z
R
dM=4!"
0
Z
0
5
R
4 3
r dr = !R "
3
2
Gravitational potential energy
Bring two masses together from infinity
∞
m1
U=0
m2
r
U=−Gm1m2/r
“Negative work” is done in order to bring the two
masses together. In other words, energy is made
available thanks to the “gravitational collapse”.
6
Application: The gravitating sphere
Assemble the sphere in shells. How much energy
becomes available from gravitational collapse?
M(r)dM
m1 = M(r) and m2 = dM so dU = −G
r
Make assumption of constant density:
!
"
%
&2
$1
4 3 # 2
4!
4
dU = −G !r " 4!r "dr = −3G
" r dr
3
r
3
!
"2
Z R
2
3
4!
3 GM
5
U=
dU = − G
" R =−
5
3
5 R
0
7
Example: How long would the Sun live if its energy
was from gravitational collapse?
Answer: Estimate available energy from gravitation
and use the known luminosity to get a time
3
2
GM
/R
5
48
2 × 10 erg
14
t=
=
= 5 × 10 sec
33
L
4 × 10 erg/sec
7
=2 × 10 years
This sounds like a long time, but it isn’t. We know
that the solar system is over four billion years old.
i.e. The Sun gets its energy some other way!
8
Suppose this energy was turned into heat? How
hot would this make the Sun? (Make a rough guess.)
Consider thermal kinetic energies of particles in the Sun:
! "
33
3
3 M
3 1.99 × 10 gm
K= nkT =
kT =
kT
−24
2
2 m
2 1.67 × 10
gm
=1.8 × 1057kT = 2 × 1048 erg
(Assuming the Sun is made of hydrogen atoms.)
Solve for Temperature:
T = 0.81 × 10 K ≈ 8M K
7
This is much hotter than the surface of the Sun, but
the center of the Sun would be much higher pressure.
This is the key to how the Sun and other stars shine!
9
Nuclear Physics: Power source of stars
First the nomenclature:
AZ
Z = “atomic number” = number of protons
A = “atomic weight” = Z+N = protons + neutrons
Examples:
1H
= hydrogen (also “p”)
2H = deuterium (also “d”)
4He = helium (also “α”)
3He = helium-3
12C = carbon-12
10
Hydrogen fusion: The primary reaction
We will argue that nuclear fusion is the power source
behind stars. There are many possible fusion reactions.
However, the most important boil down to the following:
1
H + H + H + H → He + 2e + 2!e
1
1
1
4
−
Energy is released, which is carried off by electrons, which
heat up the star, and neutrinos which travel into space.
The different series of reactions which boil down to this
result are called the “pp chain” and the “CNO cycle”. See
Kutner Section 9.3 for details.
11
The Proton-Proton Chains
Proton-Proton Chain
CNO
Cycle
The
CNO
Cycle
One of the two ways to burn hydrogen.
p+p !
p+e-+p! 2H+"
2H+e!+"
e
(Carbon-12 nucleus
"#
acts!He,asbutauses
“catalyst”.)
C as a “catalyst”.
Still !p!
The other main way to burn hydrogen.
e
“pep”
PPI
2
Proton in
"#C$p
3
p+ H ! He+#
! "%N$"
Helium out
3He+3He
!
4He+p+p
3He+4He
!
"%N
"'
N$p ! "#C$!He
7Be+#
Proton in
7
Be+e$ ! 7Li+"e
7Li+p
! 4He+4He
7
Be+p ! 8B+#
8B
"'O
!&"'N$e$$#
PPIII
Proton in
"%C$p
! "!N$"
! 8Be+e!+"e
"!N$p
PPII
!&"%C$e$$#
8Be
! 4He+4He
! "'O$"
Proton in
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Energy from Nuclear Reactions
Binding energy per
nucleon (MeV)
Stars get energy from fusion. (Nuclear reactors use fission.)
4He
Atomic Mass A
13
A Crucial Test: The lifetime of the Sun
!
"
1
4
1
4m( H) − m( He) = 0.007 × 4m( H)
In other words, 0.7% of the proton mass is converted
to energy when hydrogen fuses to form helium.
Suppose that only 10% of the hydrogen in the Sun can
be used for fusion power. Then, the available energy is
E = 0.10 × 0.007 × M"c = 1.25 × 10 erg
2
51
and the lifetime of the sun would be
t = E/L = 3.13 × 10 sec ≈ 10 years
17
10
This makes sense! The earth is about four billion
years old, i.e. the Solar System is “middle aged”.
14
The Fusion Barrier and Temperature
Potential energy
Need to overcome electrical repulsion of nuclei
We rely on thermal
energy to force the
nuclei close enough
(every so often) so
that they undergo
nuclear reactions at
some rate.
Separation
Quantum mechanics is necessary for a full understanding!
15
The “Gamow Peak”
Combined effect of quantum mechanical “tunnelling”
and Maxwell-Boltzmann thermal distribution.
1
Maxwell distribution:
−E/kT ★
P!e
0.8
Tunnelling probability:
0.6
−ar/"
P!e
h
h
!= =
&
p mv
so P ! e
0.4
1 2
E = mv
2
−b/E 1/2
See Kutner Fig.9.5
also HW Prob.9.11
0.2
0
0
★
16
★ Product
1
2
E/kT
3
4
5
Stellar structure and stellar models
How do we go about “building” a (main sequence) star?
Our assumptions:
•
• Ideal gas law (is it consistent??)
• Spherical symmetry
• Hydrostatic equilibrium
• Some model of radiation transport
Some elemental composition (mainly 1H)
17
Hydrostatic Equilibrium
Consider pressure and the spherical shells:
Inward force = P(r+dr)dA
Outward force
= P(r)dA
Inward force slightly smaller
than outward force because
gravity also pulls inward!
r
Mass of the red slug is
dm = !(r)drdA
so, the gravitational force is
M(r)dm
FG = −G
r2
18
“Equilibrium” means that all forces balance:
Inward forces = Outward forces
P(r + dr)dA − FG=P(r)dA
M(r)dm
P(r + dr)dA − P(r)dA=−G
r2
M(r)!(r)dr
P(r + dr) − P(r)=−G
r2
!
"
dP
GM(r)
Now let dr→0:
=−
!(r)
dr
r2
“Differential Equation” for pressure as a function of radius.
19
This is not the first differential equation we’ve seen!
Recall: dM = 4!r "(r)dr
2
or
dM
= 4!r2"(r)
dr
We solved this (i.e. we found M(r), and then the total
mass M(R)) by making an assumption about the
density, namely that it was a constant.You can also find
the mass by making different assumptions. (See HW.)
We can solve for the pressure by
making similar assumptions.
20
Example: The central pressure of a star
Assume a constant density and integrate “in one big gulp”:
dr = R − 0 = R and dP = 0 − PC = −PC
!
"
!
"!
"
−PC
GM
GM
M
⇒
=−
!
=
−
R
R2
R2
4"R3/3
So, ignoring the numerical factors for a rough estimate
GM 2
PC ≈ 4
R
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The Missing Ingredient: The “Equation of State”
Need a relation between pressure, density, and temperature:
P = f (!, T )
For main sequence stars, the Ideal Gas Law works okay:
PV = nkT and ! = M/V = nm/V
so P = (!/m)kT
Example: The temperature at the center of a star
m
TC = PC
!k
22
Example: The Mass-Luminosity relationship
L(r) = 4!r f (r)
2
f (r) = !T (r)
4
where
Expect d f = !(r)"(r) f (r)dr
κ(r) = “opacity”
3
T
(r)
dT
df
dT
3
‫ ؞‬f (r) !
So
= !" f (r) = 4#T (r)
"(r) dr
dr
dr
r2T (r)3 dT
Therefore L(r) !
"(r) dr
L = L(R) !
R2TC3 TC
M/R3 R
Now, integrate in
“one big gulp” again!
! "4
R 4 R M
3
or
= TC !
L!M
M
M R
4
4
23
A more realistic
solar model:
Recall:
Compare Kutner Fig.9.11
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