Download PHYSICS OF THE SUN The Sun is a main

PHYSICS OF THE SUN
The Sun is a main-sequence star... still burning its initial supply of hydrogen.
In this course, we won’t talk much about earlier/later phases:
Mostly, we’ll concentrate on how the Sun currently maintains itself in a
quasi-steady state, with
Solar mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solar radius . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Solar luminosity . . . . . . . . . . . . . . . . . . . . . . .
Solar core temperature . . . . . . . . . . . . . . . . .
Solar effective temperature . . . . . . . . . . . . .
Age of Sun & solar system . . . . . . . . . . . . .
M⊙ = 1.989 × 1030 kg
R⊙ = 6.963 × 108 m
L⊙ = 3.83 × 1026 W
Tcore = 15.7 million K
Teff = 5770 K
t⊙ ≈ 4.6 Gyr
For context, main sequence stars have M∗ between about 0.1 and 100 M⊙ .
Brown dwarfs (not massive enough to burn hydrogen) have M∗ between about
0.01 to 0.1 M⊙ . Less than that: planets!
Also, two key derived quantities:
4
{surface flux} = σTeff
=
L∗
4πR∗2
,
{surface gravity} = g =
5.1
GM∗
.
R∗2
What we will do:
In this part of the course, we’ll try to answer two basic questions about the
basic physics of a stellar interior:
• Where does the energy come from?
– Gravitational contraction
– Nuclear burning (fusion)
– Radioactive decay (fission) ?
– Tidal heating (if it has a massive neighbor & eccentric orbit) ?
• How is energy transported out through the interior?
– radiation (slow diffusion of photons)
– heat conduction (slow diffusion of particles)
– convection (unstable bubbling)
After that, we’ll talk about the solar atmosphere, which is the part we
actually “see” (photons stop random walking; start freely escaping).
Sources of Energy
First, gravity. Sir Arthur Eddington, in early 1900s, postulated that stars
generated their L∗ by slowly contracting, and converting gravitational energy
into heat & light.
We now know that this isn’t how the Sun generates energy, but this basic
process should still be explored.
Thus, let’s look at the effects of gravity on the solar interior.
Remember our MHD momentum equation, for B = 0,
∂u
1
+ u · ∇u = g − ∇P
∂t
ρ
A static star is held together via hydrostatic equilibrium, in which all
u terms = 0.
We solved for ρ(r) for the simple case of both g and T being constant, but
that’s not true in general.
5.2
For spherical symmetry, a point a distance r from the origin only cares about
the mass interior to r, which we’ll call Mr (r). (Effects of mass outside r cancel
out.) We can treat the interior mass as if it’s a point-mass at the origin.
Thus,
g = −êr
GMr
r2
and
∂P
GMr ρ
= − 2
∂r
r
We also write a conservation equation for Mr by looking at thin, infinitesimal
shells. It’s kind of just “bookkeeping:”
Mass contained: dMr
Volume: 4πr2 dr
Alternate estimate of mass contained:
ρ (4πr2 dr)
∂Mr
= 4πr2 ρ
∂r
Z r
Z Mr
dMr′ =
Mr =
dr′ 4πr′2 ρ(r′)
Thus,
Enclosed mass:
0
0
This equation essentially takes the place of the mass conservation equation
(again, in the limit of u = 0 and ∂/∂t = 0).
These two equations are basic building blocks for much of what we’ll do
regarding models of stellar interiors.
5.3
However... first, we’d like to know something even more basic than how static
objects are maintained. Let’s posit some thought experiments.
..............................................................................
If the “pressure gradient support” got turned off somehow, how fast would
gravity alone be able to make changes to the system and reconfigure the mass?
We can derive the stellar “dynamical time” tD in several ways.
Here’s a simple way to estimate the free-fall time:
Imagine a test particle at distance R∗ from a point-mass
M∗ at the center.
Here, let’s assume that |a| = GM∗ /R∗2 stays constant over
the time we’re considering.
It’s Physics 101...
dvr
d2 r
=
= a,
dt2
dt
so v = at (starting from rest at v0 = 0, t0 = 0)
Integrating again,
2
r − r0 =
at
2
(r − r0 = R∗ ) ,
and thus
tD ≈
s
2R∗3
GM∗
We could do better by considering the actual collapse of a spherical
distribution of gas... i.e., as it collapses (r ↓), the gravitational force increases.
However, the end-result is only off by a bit... replace factor of 2 above by
(π 2 /8) ≈ 1.23. For the Sun, tD ≈ 30 minutes.
..............................................................................
The above is nice to know, but in most situations we know that spherical,
self-gravitating bodies are supported by gas pressure, too.
The time it would take gravity to “fight” against the real reservoir of
thermal energy is going to really be much longer than tD .
5.4
To estimate it, we first need to derive a fundamental relationship between the
two main energy components of a star:
The Virial Theorem
Again, there are several ways to go about this. We’ll do it by building an
analytic model of the solar interior that obeys the mass & momentum
equations that we derived above.
The big assumption in this model is that the gas maintains a constant
density [ρ(r) = ρ0 ] throughout the whole interior.
This is a notoriously bad approximation, but amazingly the physics insight it
gives us is huge....
dMr
= 4πr2 ρ0
dr
Start with mass conservation:
It’s (relatively) easy to solve because it’s separable:
Z R∗
Z M∗
dr r2
dMr = 4πρ0
0
0
M∗ = 4πρ0
R∗3
3
i.e.,
ρ0 =
M∗
= hρi
(4/3)πR∗3
so if we know any two of {ρ0, M∗ , R∗} this gives the third.
If we stop the integration part-way, we get
Mr (r) =
4
πρ0 r3
3
and this (black curve) often doesn’t look
very much like what one gets numerically
for more centrally condensed stars (red
curve).
Putting aside that bit of incongruity, we can solve the hydrostatic equilibrium
equation for P (r),
dP
4πGρ20
GMr
= − 2 ρ0 = −
r
dr
r
3
(again, a separable diff. eqn.)
5.5
Keeping track of limits,
Z P (r)
Z
4πGρ20 r ′ ′
′
dP = −
dr r
3
Pc
0
(though we don’t know Pc yet)
2
P (r) − Pc = − π Gρ20 r2
3
If we set P ≈ 0 at r = R∗ , we can solve for the central pressure:
2π 2 2
3 GM∗2
Pc =
Gρ0 R∗ =
3
8π R∗4
And also,
P (r) = Pc
r2
1− 2
R∗
Note that these results are independent of the equation of state.
It’s true for an ideal gas... also true for weird crystalline matter in the cores of
white dwarfs!
However, we do know that the plasma in the solar interior obeys an ideal gas
equation of state.
So if we know P (r) and ρ0 , we can use the equation of state to solve for T (r),
ρ0 kB T (r)
r2
kB T c
GM∗
Pc
P (r) =
, so... T (r) = Tc 1 − 2
=
=
µmH
R∗
ρ0
µmH
2R∗
Note: for the Sun’s ionized interior, µ ≈ 0.6, we get Tc = 7 million K.
A more realistic solar model would replace the 1/2 on the RHS by a factor
closer to ∼1, to get the actual value of ∼15 million K. But we’re close!
Note that we’ve solved for the core temperature of the Sun without knowing
anything about nuclear fusion.
..............................................................................
The above relationship looks very fundamental. Note that both sides have
units of velocity squared: “thermal speed” ≈ “escape speed” in interior.
5.6
Another way to express this is by looking at the total energy components of
the star. If there’s NO rotation, mass motions, magnetic fields, etc.,
Total energy:
Etot = EK + EG
(in actual Joules)
EK is the thermal energy of an entire system, integrated over its volume.
EG is the system’s integrated gravitational potential energy.
We’d like to know how EK and EG relate to one another.
For the thermal energy,
Z
Z
EK =
dV U =
dV
Also,
P
γ−1
for an ideal gas.
4 3
πr , dV = 4πr2 dr = d3r .
3
For our constant-density star, made up of ideal gas, we can compute
Z
Z R∗
3 GM∗2
3
dr 4πr2 P (r) = ❀ ❀ =
EK =
dV U =
2
10 R∗
0
V =
What about EG ? For test particles of mass m in the field of a gravitating body
of mass M∗ , we know that
GM∗ m
EG = −
r
where r is the distance between the two bodies. So, for a star in its own
potential well, it’s analogous to write
Z
GMr
EG = − dMr
.
r
We can solve for this using our constant-density model, too:
Z
GMr
3 GM∗2
EG = − dMr
=❀❀= −
= −2EK !
r
5 R∗
The general version of the virial theorem is:
EG = 3(1 − γ)EK
and for an ideal gas with γ = 5/3,
EG = −2EK
and
Etot =
5.7
EG
EG −
2
=
EG
2
This actually holds not just for the “constant-density” model, but for more
general stellar models (as long as it’s made out of ideal gas).
Note that since EG is negative, then the total energy Etot is negative, too.
(The star is bound.)
..............................................................................
Remember Sir Arthur Eddington and his theory about collapse generating
energy.
If R∗ is decreasing, then EG decreases (i.e., becomes more negative) in time,
and thus Etot keeps decreasing, too.
Eddington assumed total energy is “lost” in the form of radiation; i.e., that
L∗ = −
dEtot
1 dEG
= −
> 0.
dt
2 dt
If all that’s changing is R∗, then we can take the derivative,
3 GM∗2 d
3 GM∗2 dR∗
EG dR∗
3 GM∗2
1
1 d
=
= −
=
.
−
L∗ = −
2 dt
5 R∗
10 dt R∗
10 R∗2 dt
2R∗ dt
Note that L∗ > 0 because both EG < 0 and dR∗ /dt < 0.
How long could a star survive before “collapsing to zero?” i.e., we want to solve
dR∗
∆R∗
R∗
≈
≡ −
dt
∆t
tK−H
for the so-called Kelvin-Helmholtz timescale tK−H . Using the above, we get
tK−H ≈
|EG |
2L∗
This is also roughly the same as the time it would take for the star to lose all of
its thermal energy by radiating it away at a given luminosity (tK−H ∼ EK /L∗).
Thus, tK−H is sometimes called the thermal diffusion timescale.
5.8
For the Sun’s present-day values of M∗ , R∗, L∗, we get tK−H ∼ 107 years.
(“only” 10 million years)
Of course, carbon dating, moon rocks, and lots of other evidence point to the
age of the solar system as ∼ 4 × 109 years, so there must be some other major
energy source besides gravity!
We know this other source is nuclear burning.
However, the Sun’s core temperature is NOT set by nuclear burning! We saw
earlier how it’s set by the balance between gravity & gas pressure. Fusion
allows this high Tc to be maintained for much longer than would otherwise be
possible!
tD
∼30 minutes
≪
tK–H
∼107 years
≪ tMS life
∼1010 years
Stellar interiors are thus one big balance...
• If the momentum equation is unbalanced (i.e., gravity only), the star
collapses almost immediately.
• If the momentum equation is balanced (∇P = ρg), but the energy
equation is unbalanced (i.e., “cooling” via L∗ without anything else on
the RHS), the star takes millions of years to collapse.
• However, if we balance the energy equation with “heating” from nuclear
burning, we get something much more stable!
5.9
Nuclear energy generation
Let’s delve into the different types of nuclear reactions that can occur when
nuclei are bombarding one another.
Basic ingredients of nuclear reactions:
Baryons/Nucleons: protons (p+), neutrons (n0 )
Leptons:
electrons (e−), electron neutrinos (νe)
Anti-particles:
p− , n0 , e+ , νe
Photons:
γ (“raw” energy, makes up for E = mc2 imbalances)
Nuclear reactions are similar to chemical reactions: “ingredients” come
together and get transformed into “products.”
In all these reactions, the conserved quantities (i.e., equal on left and right
sides) are:
•
•
•
•
total mass-energy (E = mc2 )
total charge
baryon number (i.e., # of nucleons)∗
lepton number (leptons = 1, anti-leptons = –1)
∗
(In strong interactions, the #s of protons & neutrons tend to be conserved
separately, too – which helps automatically conserve charge.)
There are 2 main types of reactions:
Strong interactions:
e.g., N1 + N2 −→ N3 + γ
These involve nucleons grouping up or splitting apart. Usually, N1 is an
arbitrary nucleus, and N2 is often just a p+ , n0 , or α particle. For two
positively charged nuclei, they overcome Coulomb repulsion only if they get
very close to one another.
The above is an exergonic (energy releasing) fusion reaction. If the γ is on the
left side, it’s endergonic (energy consuming). If N3 −→ N1 + N2 , it’s fission.
5.10
Weak interactions:
e.g., radioactive beta decay... n0 −→ p+ + e− + νe
These usually occur within an existing nucleus. There’s usually no γ produced.
We care about them because they help get a nucleus into a state in which
strong reactions can then happen.
All sorts of other combinations (of what goes on the left/right sides) are
possible, as long as the conserved quantities are conserved...
Thus, some other possible weak interactions are:
p+ −→ n0 + e+ + νe
p+ + e− −→ n0 + νe
n0 + νe −→ p+ + e−
37
Cl + νe −→ 37Ar + e−
(positron emission in proton-rich nuclei)
(inverse beta decay; electron capture in NS)
(used in chlorine neutrino detectors;
37
Ar has short half-life; detect its decay!)
A good way to think about weak reactions is to realize that a nucleus with Z
protons has an “ideal” number of neutrons needed to make it a stable isotope.
Too many neutrons, and it will undergo n → p beta decay.
Too few neutrons, and it will undergo p → n inverse beta decay.
Mini-chemistry refresher: (most stable isotopes of example elements)
Element Z (# protons)
Helium
2
Carbon
6
Oxygen
8
Sulfur
16
Chlorine
17
Iron
26
Silver
47
Gold
79
Uranium
92
N (# neutrons)
2
6
8
16
18
30
60
118
146
Each proton “wants” to repel away from its neighbor protons.
The strong force keeps nuclei together, but nuclei with more protons need
even more neutrons to help prevent the Coulomb repulsion due to all those
positive charges.
..............................................................................
5.11
We care about the energy released in all of the above types of reactions,
both by photons (γ) – which are rapidly thermalized – and by neutrinos.
Why is energy released? The output nuclei are in a lower mass-energy state
than the input nuclei.
The relative energies are measured in terms of B, the nuclear binding energy.
It’s energy that needs to be added to break a nucleus apart into its component
protons & neutrons:
X
B
mnucleus + 2 =
mp , mn
c
If the reaction is going ←−, that B energy is released!
Expressing masses in units of MeV/c2 , we can compare
mp = 938.26 MeV/c2 , mn = 939.55 MeV/c2 , me = 0.511 MeV/c2
to some example binding energies:
For 1 H,
For 2 D,
For 4 He,
For 56Fe,
B
B
B
B
=0
≈ 2.2 MeV
≈ 27 MeV
≈ some big #
(though the bound electron contrib’s 13.6 eV)
(∼1.1 MeV/nucleon)
(∼7 MeV/nucleon)
(∼8.8 MeV/nucleon)
Handout: Binding energies per nucleon (B/A) are shown as a function of
atomic mass A. Unstable isotopes have lower binding energies than the stable
ones, so they’re more apt to break apart.
Being “higher up” on this diagram means it’s more stable. Thus, more input
energy would be needed to break it apart.
Favorable reactions go from lower to higher binding energy.
ր is fusion ... տ is fission.
Note: One gets the most “bang for the buck” out of converting H to He. Less
from further stages of fusion? This is why a star’s main sequence (H-burning)
lifetime is so long, compared to successive stages of nuclear burning.
5.12
Top: Binding energies of atomic nuclei (B/A) in MeV, shown as a function of mass number A. Data
for all known isotopes (black points) come from Audi & Wapstra (1993, Nucl. Phys., A565, 1). The red
curve connects the stable isotopes. Bottom: Solar elemental abundances, from Asplund et al. (2009,
Ann. Rev. Astron. Astrophys., 47, 481).
5.13
Anyway, if we manage to fuse 4 × 1 H → 4 He, there’s ∼27 MeV “left over” (i.e.,
released as γ).
Is it plausible to power a star in this way?
In other words, how much 1 H would need to fuse into 4He in order to support
the Sun over it’s lifetime so far?
L⊙ =
∆E
,
∆t
solve for ∆E = L⊙∆t
For L⊙ ∼ 4 × 1026 W, and
∆t ∼ 4 × 109 years × (3 × 107 seconds/year) ≈ 1.2 × 1017 s
Thus, ∆E ≈ 5 × 1043 J ≈ 3 × 1056 MeV.
If it’s ∼30 MeV per reaction, then ∼1055 reactions are needed.
4 × (mH ≈ 2 × 10−27 kg) ≈ 10−26 kg go into each reaction
Thus, we need 105510−26 ≈ 1029 kg of hydrogen to power the Sun over its
lifetime so far.
Since M⊙ ≈ 2 × 1030 kg, this is ∼ 1/20 of a solar mass.
This is a plausible order of magnitude, since we know it’s just the core that
undergoes nuclear fusion.
Problem: How can these reactions
ever really occur? Strong interactions
involve two positively charged nuclei...
But here’s where the strong nuclear force comes into play. It turns attractive
for tiny distances:
5.14
V12(r) is the potential energy of nucleon 1 as seen by nucleon 2. (V12 = V21)
Positive potential: repulsive force.
Negative potential: attractive force (like a “gravity well”).
In the outer part, for two positive charges,
V12(r) ≈
k e q1 q2
> 0 .
r
But, if the two + charges manage to get close enough together, the
potential goes negative and they stick! (This inner part doesn’t care about
charge: pp, pn, nn are similar.)
To penetrate the barrier, a particle’s kinetic energy, E = mv 2 /2, must exceed
max(V12). (I used nonrelativistic E here because even though the electrons
may be relativistic in stellar cores, the ions & nuclei rarely go so fast.)
Let’s compare E and V12 with one another...
Because N1 and N2 can have unequal masses (m1 , m2) and charges
(q1 = Z1e, q2 = Z2 e), it makes the most sense to use a center-of-mass
coordinate system.
This is handy; you’ve probably seen from classical mechanics how this makes
the calculations easier.
1
1
1
m12 |v12|2
Total energy:
E = m1 |v1 |2 + m2 |v2 |2 =
2
2
2
5.15
where
m12 =
m1 m2
(reduced mass) ,
m1 + m2
v12 = |v1 − v2 | (relative velocity) .
The Coulomb potential (for r > rn ) is given by V12 (r) = ke Z1 Z2 e2 /r.
In order for random thermal motions to overcome the barrier, we want
3
ke Z1 Z2e2
2ke Z1 Z2e2
1
2
= kB T =
m12 v12
which means T must exceed
.
2
2
rn
3kBrn
Big question: what’s the value of rn ? A typical nucleus has a size of ∼1 fm
≈ 10−15 m. (Five orders of magnitude smaller than an atom; 1 Å ≈ 10−10 m.)
For protons (Z1 = Z2 = 1) and the assumption of rn ≈ 1 fm, we get
T ≈ 1010 K .
This is kind of high for stellar interiors.
It’s telling us that we’re missing something...
Remember, however, that weird quantum stuff (wave-particle duality) happens
when we get down to the size of an object’s de Broglie wavelength λdb .
It doesn’t make much sense to talk about smaller sizes.
Recall that, in quantum mechanics, all particles exhibit wave-like
characteristics. de Broglie showed that a kind of uncertainty principle is
obeyed by particles,
∆x ∆p >
∼h
If we know a particle’s momentum p, it will exhibit “matter waves” with
wavelength
h
λdb =
p
and particles will produce diffraction patterns when shot at targets, flow
through narrow slits, etc.
5.16
One minor complication: the most probable momentum p depends on T ,
so λdb depends on T also.
In an ideal gas,
3
p2
kB T =
,
2
2m12
so ❀
λdb = √
h
3m12kBT
If λdb is bigger than the actual size of the nucleus, then the precise position of
the nucleus is “fuzzy.”
Thus, let’s try assuming rn = λdb . We get
2keZ1 Z2 e2
T =
3kB
√
3m12kB T
h
which requires us to solve for T 1/2 and square it...
T =
4 ke2 Z12Z22 e4 m12
≈ 107 K
2
3
kB h
which is much more realistic. Here, λdb ≈ rn ≈ 10−12 m, between 1 fm & 1 Å.
Thus, the interactions between nuclei are inherently quantum mechanical
(i.e., probabilistic and “fuzzy”).
5.17
Okay, we’ve shown that nuclear fusion can supply enough energy, and that
barrier penetration is plausible.
Computing the probability of “barrier penetration” is a nasty quantum
problem... George Gamow worked it out in the 1940s.
We need to know how often the reactions actually occur in a plasma of given ρ,
T , P ,... i.e., we want the reaction rate between input species 1 & 2,


 Total # of fusion reactions that 
(in units of #/m3/s)
r12 =
occur per unit volume, per unit


time
The reaction rate tells us how number densities change vs. time.
r12 ends up being a function of ρ, T , and composition (the mixture of elements
in the star).
For example, if the system contained nothing but the (N1 + N2 → N3) fusion
reaction, and we write the rate as r12,3, then
∂n1
= −r12,3 ,
∂t
In general,
∂n2
= −r12,3 ,
∂t
∂n3
= +r12,3
∂t
X
X
∂ni
rmn,i
= −
rij,k +
∂t
m,n
j
and for a hypothetical single reaction in which 4 protons converge
simultaneously to turn into a helium nucleus, we have
∂np
= −4r ,
∂t
5.18
∂n4He
= +r
∂t
And, conservation of total baryon number is approximately the same as total
mass conservation, so one can assume ∂ρ/∂t = 0 and use the mass fractions
X=
mp np
,
ρ
Y =
4mp n4He
ρ
to show that
∂X
∂Y
= −
∂t
∂t
i.e., as X goes down, Y goes up.
All that quantum mechanics (combined with assumption of a M–B distribution
of nuclei in the hot plasma) gives
r12 ≈ ρ2 × { complicated function of T }
and the ρ2 is really n1n2 ... i.e., it depends on how many particles of both input
species are present.
The T dependence is really a v dependence (or momentum p dependence), and
all the de Broglie, barrier penetration stuff is in there.
For stellar interiors, we will care about how much power is released, per kg of
material in the core, per second:
ǫ =
∂Lr
r12 Bnet
≡
∂Mr
ρ
(make sure units work out)
where Bnet is the net binding energy released by one full reaction.
Astronomers like to fit power-laws to things, and it’s often seen that
ǫ ≈ ǫ0 ρ T ν
works pretty well.
For the present-day Sun, ν ≈ 4.
5.19
Nuclear Burning Cycles
Note that ǫ depends very strongly on T .
Also, since Vmax ∼ Z1Z2 m12 , we know that heavier nuclei have higher Coulomb
barriers, so their reactions become “enabled” at higher temperatures.
Thus, a star with a given core T will usually be undergoing only a small
fraction of all possible reactions.
Let’s look at the most relevant sets of reactions for the Sun.
..............................................................................
Hydrogen Burning
What we want is:
4p+ −→ 2p+ + 2n0 −→ 4 He + energy
Having all 4 protons come together at once is hugely unlikely. Assembling
heavier nuclei happens more efficiently in stages. For the present-day Sun, it
happens in 3 steps.
Problem: One might think the most basic building block would be the strong
interaction,
p+ + p+ −→ 2 He
but it is highly unstable. The “di-proton” 2He breaks up into 2 protons pretty
much instantaneously. All nuclei heavier than hydrogen need neutrons to be
stable. However, there aren’t any free neutrons flying around...
In 1939, Hans Bethe realized that the only way to get the process started was
with a weak interaction,
(1)
p+ + p+ −→ D+ + e+ + νe
(i.e., one proton decays into a neutron, when near another proton, and then
they can fuse). Note shorthand: p+ is 1H+ , and D+ is 2H+ (pn).
This weak interaction is very slow & infrequent: its rate is only about 10−20
times that of typical strong reactions.
It is the “bottleneck” that determines how fast the whole process goes!
5.20
Once we have some D+ , and the core T is at least ∼1 MK, there’s deuterium
burning:
p+ + D+ −→ 3He++ + γ
(2)
{strong reaction, so it’s fast}
Note how p + pn gives ppn.
Some brown dwarfs (M∗ = 0.01–0.1 M⊙ ) can be kept warm for a few million
years from this, but eventually they run out of their initial supply of D.
In the Sun, the third step happens once there’s a chance for there to be two
3
He nuclei in close proximity,
(3)
3
He + 3 He −→ 4He + p+ + p+ + γ
{also fast}
Verify nucleon conservation for strong interactions: each 3 He is ppn, and the
4
He nucleus is ppnn.
Note also that it would be unlikely for two 3 He nuclei to combine together to a
single nucleus with atomic mass 6. There are no stable nuclei with A = 6.
The 3 main boxed reactions together form the PP-I chain. Note there are
two other dominant ways of doing this, but they both require an existing 4He
to come in as a catalyst...
5.21
Involvement of heavier species in the PP-II and PP-III chains means that they
start to dominate at higher temperatures. Using equations from Clayton’s
Chapter 5...
...but note that above 20 MK, other (non-PP) reactions start to occur.
For stars just a bit more massive than the Sun, these other reactions start to
dominate over PP.
When Tcore starts to exceed ∼20 MK, and there’s some C, N, O present, they
can act as catalysts for other avenues of hydrogen burning:
the CNO cycle (or “bi-cycle”).
If we start with a
• The
12
12
C nucleus (6p+6n)...
C gradually captures 4 protons.
• 2 of them convert into neutrons (via weak interactions).
• Just before it settles into being a 16O (8p+8n), an α particle (2p+2n)
breaks off, and we’re left with another 12 C again.
• Net result: 4 protons convert into one 4He!
5.22
Alternatively,
• One
14
N (7p+7n) gradually captures 4 protons.
• 2 of them convert into neutrons (via weak interactions).
• Just before it settles into being a 18 F (9p+9n, which is unstable; the stable
version has 10n), an α particle breaks off, and we’re left with another 14N
again.
For even hotter cores (T >
100 MK, i.e., not the present-day Sun!), there’s
∼
4
helium burning: three He’s fusing into one 12C... the “triple alpha process.”
Even hotter (T >
∼ 500 MK), fusion proceeds by “alpha capture” reactions...
12
C→
16
→
O→
36
20
Ar →
Ne →
40
24
Mg →
Ca →
44
28
Ti →
Si →
48
32
Cr →
S→
52
Fe →
56
Ni
These sequences are the main reason for the so-called Oddo-Harkins rule:
elements with even Z, which also are often those with A divisible by 4, are the
most abundant. “Once you go even, you never go back.”
56
Ni undergoes 2 rapid weak (p → n) beta decays:
56
Ni −→
56
Co −→
56
Fe .
and we’re at the stable peak of the binding energy curve.
Fusing the elements beyond iron only happens in the most extreme
environments (e.g., supernova explosions).
That’s it for energy generation by nuclear fusion in stars.
At the beginning of the lecture on energy generation, I mentioned two other
sources (besides gravity & fusion):
• Radioactive decay (fission)
• Tidal heating
These are more important in planetary interiors than in stars; we won’t
discuss them further.
5.23