Download Sources of energy and the origin of the chemical elements

Sources of energy and the
origin of the chemical
elements
Origin of the elements
Could the Sun be powered purely by gravitational contraction?
Sphere of uniform density
(underestimate?) has gravitational
potential energy:
3 GM 2
Ug = −
5 R
Change in radius liberates potential energy, but 50% of this (from Virial
Theorem) is required to maintain pressure support (e.g. heat):
Available energy = 0.5 U g ~ 10 41 J
LSun ~ 4 ×10 26 W
Timescale to maintain solar luminosity:
t KH ~
0.5 U g
Lsun
~ 107 yr
Note that every factor of 2 collapse would release as much energy again
Gravity as source
ASIDE: Note the curious behaviour of a self-gravitating body under its
own pressure support: it has negative heat capacity:
The Virial Condition for equilibrium state of self-gravitating system is:
2K + W = 0, i.e. E = -K = 0.5W
K = kinetic energy = 3/2 NkT
W = potential energy
E = total energy
3
E = − NkT
2
dE
3
= − Nk
dT
2
Self-gravitating system heats up when it loses energy.
Same effect as an orbiting satellite speeding up under drag forces
Virial theorem
Could the Sun be powered purely by gravitational contraction?
Number of particles
M sun
n~
~ 1057
mp
Chemical energies are typically eV (10-19 J) per particle, i.e. 1038 J
available. Gives lifetime of order only 104 years!
Nuclear reactions are typically MeV (10-13 J) per particle, i.e. 1044 J
available. Gives lifetime of order 1010 years
Chemistry as source
Fusion energetics
mp = 1.6726 × 10-27 kg
mn = 1.6749 × 10-27 kg
a.m.u.= 1.6605 × 10-27 kg (= 931 MeV)
mHe = 4.0026 a.m.u.
Making 4He out of 2n + 2p
results in 0.7% of the mass
(i.e. 6.5MeV) being lost, due
to increased negative binding
energy of the nucleus.
Nuclear fusion reactions
(a)  provide the energy source in
the Sun, and
(b)  produce the diversity of
chemical elements out of
initial H+He produced in the
Big Bang.
Fusion as source
Nuclear energy levels
•  Pauli exclusion principle ensures np ~ nn
•  Proton levels are more widely spaced because of electrostatic
repulsion
•  Gross shape of binding energy curve is a result of the finite range
of the strong force (attraction saturates as n increase too far).
Fusion as source
Fusion energetics
fusion
releases
energy
Fusion as source
Formation requires
energy, fission
releases energy
Fusion reaction rates:
Difficulty of fusion arises
from coulomb repulsion
between positively charged
nuclei
Potential energy of two
charges separated by r
Point of closest approach
if initial velocity is v
Ue =
rclose
q1q2
4πε 0 r
2q1q2
=
4πε 0 mv 2
Naively setting r ~ 10-15 m and mv2 = 3kT would require T ~ 1010K >> Tsun
Coulomb repulsion
•  Quantum tunneling through potential barrier
•  Velocity distribution at a given T
Velocity in center of mass frame of two particles of reduced m = m1m2/(m1+m2)
& mv 2 #
!! dv
n(v) dv ∝ exp$$ −
% 2kT "
Tunneling probability through distance r given by de Broglie λ:
& 4π 2 q1q2 #
& 2π 2 rclose #
!!
!! ∝ exp$$ −
P ∝ exp$$ −
λ "
%
% 4πε 0 hv "
Probability of reaction given by product of these two, plus usual number of
encounters vσdt, where σ is cross section
Tunneling and Maxwell distribution
( mv 2 πq1q2 %
## dv dt
dN ∝ n1n2σ v exp && −
−
' 2kT ε 0 hv $
(
)
∝ exp − αE − βE −1 / 2 dEdt
Highly peaked function with maximum dN/dE
= 0 when the two terms are equal (plus a factor
of 2 from differentiating)
v max
$ πq1q2 kT '1/ 3
=&
)
ε
hm
% 0
(
Rate at this peak is given by
2/3
1/ 3 #
1/ 3
€ &
&
#
3
π
q
q
m
T
&
#
&
#
dN / dT ∝ exp$ − $$ 1 2 !! $ ! ! ∝ exp − $ 0 !
$ 2 % hε 0 " % kT " !
%T "
%
"
'3$
T0 = % "
&2#
Tunneling and Maxwell distribution
3
2
' πq1q2 $ ' m $
%%
"" % " ~ 4 ×1010 K
& hε 0 # & k #
Very strong T
dependence
Steep dependence
of fusion reaction
rate coupled with
negative heat
capacity of gas
provides natural
thermostat
This is why stars do
not explode like Hbombs
T4 at ~ 107K
Later stages of fusion require higher temperatures
because of larger nuclear charges
H to He
He to C, O
C to O, Ne, Na, Mg
Ne to O, Mg
O to Mg – S
Si to around Fe
1×107 K
1×108 K
5×108 K
1×108 K
2×108 K
3×109 K
Note also that flattening of binding energy curve implies that
energy release per reaction goes down in later stages
Question: Are the interiors of stars hot enough for fusion?
(c.f. Tsurface ~ 6000 K for the Sun)
Quite general arguments indicate yes:
GM 2
U grav = − f
R
with f of order unity
Hydrostatic support at radius r
dP
Gm
=− 2r ρ
dr
r
R
R
Gmr
3 dP
3
4
π
r
dr
=
−
4
π
r
dr
∫
∫
2
dr
0
0 r
R
−3 ∫ P(r) 4π r 2 dr = −U grav
0
P
Thermal
pressure
V
=−
1 U grav
3 V
ρ
P ~
kTint
m
Tint ~
GMm
~ 3×10 6 K
3kR
for Sun
Question: How does it get that hot?
Tint ~
GMm
~ 3×10 6 K
3kR
Overall energetics of gravitational collapse:
3M
GM 2
kT ~
2m
R
2 GMm
T~
3 kR
So there is enough energy in liberation of gravitational PE (but
without a lot to spare). What actually happens?
Initial gravitational P.E. goes into dissociating H2 molecules (4.5eV) and
ionizing H atoms (13.6eV) and proto-star collapses in free-fall.
When interior is ionized plasma (when R ~ 100 Rsun and T ~ 30,000 K),
mean free path of photons becomes very short and interior heats up,
establishing temperature gradient and hydrostatic support governed by
above equation.
for Sun
Are stars inevitable?
Tint ~
GMm
~ 3×10 6 K
3kR
for Sun
Proto-star collapses in quasi-hydrostatic equilibrium with Tinterior rising as R
shrinks.
Negative heat capacity causes star to increase in temperature as energy is lost
through radiation to exterior.
Virial condition implies ½ of Ugrav goes into internal heat energy, rest is
radiated.
When Tinterior reaches point where fusion can balance heat loss at surface, a
static stable structure (= star) is formed, with constant R.
This ignition point will inevitably be reached at some R unless another source
of pressure (e.g. degenerate electron pressure) can support the object before T
gets high enough for fusion.
Tint ~
Minimum mass of stars
de Broglie wavelength of thermal electron
λ~
GMm
~ 3×10 6 K
3kR
h
1/2
( mekT )
Critical density above which QM
effects dominate
m
( m kT )
ρ~ 3 ~m e 3
λ
h
We had before, for general
hydrostatic support…..
kT ~
3/2
GMm
~ GmM 2/3ρ1/3
3R
so the temperature increases as ρ1/3.
What is maximum temperature
reached before degeneracy kicks in?
! G 2 m8/3me $ 4/3
kT ~ #
&M
2
h
"
%
For the Sun, temperature could rise to 107K.
Fusion will not occur for M < 0.08 Msun ( = “brown dwarfs”)
Conclusion: for all more massive objects, stars are “inevitable”
for Sun
Mass-luminosity relation
Details not critical
Tint ~
GMm
~ 3×10 6 K
3kR
for Sun
Transport of energy out of the (opaque) star by diffusion due to random walk of photons
Energy density and SHC of radiation
Heat transport due to diffusion of
photons in T gradient set by m.f.p..
Net luminosity passing out of radius
r (assuming no fusion at this radius)
Approximate for order of magnitude
To get luminosity of star
Uγ = aT 4
C = 4aT 3
jx = −K r
dT
dx
4
4
1
K r = caT 3 l = caT 3
3
3
κρ
4
1 dT
L(r) = 4π r 2 jr = 4π r 2 caT 3
3
κρ dr
dT Tint
~
dr
R
L = ( 4π )
2
4 caG 4 1 3
M
4
4
3 k κ
Lifetimes of stars
L = ( 4π )
2
4 caG 4 1 3
M
4
4
3 k κ
This is a strong function of the mass of a star. Actually, putting in dependence
of κ(T), it is more like M3.5
Implication: Since the amount of fuel for fusion will be (more or less)
proportional to mass, expect lifetime of stars to strongly depend on M-2.5.
e.g. 30 Msun star has lifetime < 107 years, c.f. 1010 years for Sun.
Is there a maximum mass for a star?
Yes. If internal pressure dominated by radiation pressure rather than
thermal pressure then it is likely to be unstable. Mmax ~ 60-100 Msun
Self-gravity vs. thermodynamics
•  Ordinary stars (like our Sun) are self-gravitating and therefore have to be
hot inside to sustain the thermal pressure that resists the inward pull of
gravity
•  The interstellar medium, however, is typically much colder and energy
flows continuously from the star to the exterior.
•  No thermodynamic equilibrium is possible under this circumstances,
losing energy the star will be forced to contract and to get hotter and
hotter
•  Thermonuclear production of energy keeps the star structure stable for a
limited amount of time. When it runs out of fuel a major structural reassessment has to happen.
What supports a star against self-gravity?
X mass fraction of H
Y mass fraction of He
Z mass fraction of metals
Available sources of pressure
•  Non-degenerate plasma: P = nkT = (ρkT/mH)(2 X+0.75 Y+0.5 Z)
•  Radiation pressure:
P=aT4/3
•  Non-relativistic degenerate plasma:
P∝ [0.5 (1+X) ρ]5/3
•  Ultra-relativistic degenerate plasma:
P∝ [0.5 (1+X) ρ]4/3
•  Atomic and molecular interactions
•  Quark and hadronic interactions
•  Rotation and magnetic fields The Hertzsprung-Russell diagram
Brown Dwarfs
A brown dwarf is a self-gravitating gaseous object composed
mainly of hydrogen and helium, whose mass is too small to ignite
stable thermonuclear hydrogen fusion in its interior (but they do
produce some helium isotopes by fusing protons with deuterium,
lithium and berillium).
Theoretical expectations: M < 0.08 M¤.
Incapable of generating substantial amounts of nuclear energy, the
gravitational contraction of a brown dwarf takes place until the
pressure of the degenerated electrons in its interior stops it.
Young BDs can release substantial amounts of gravitational
energy, while older BDs radiate remnant internal heat only.
Gliese 229a: cool, main sequence red dwarf
(0.46 M¤, 0.53 R¤, 0.016 L¤), approximately 3 Gyr old.
Gliese 229b: brown dwarf
(25-65 MJ, 0.9-1.1 RJ, T~1100 K) orbiting at about 39 AU from Gliese 229a and
showing abundant methane in its atmosphere.
Main-sequence stars
Objects with M > 0.08 M¤ ignite stable thermonuclear reactions in their
centres and become stars. In the initial phase of their lives they are
characterized by chemical homogeneity and core-hydrogen burning.
Stars with M < 2 M¤ fuse hydrogen into helium by the p-p chain, while
more massive stars do it by the CNO cycle.
In the HR diagram they populate an approximately one-dimensional locus
along which positions are mainly determined by the stellar mass.
Vogt-Russell Theorem: The mass and chemical composition of a star
uniquely determine its radius, luminosity, and internal structure, as well as
its subsequent evolution.
Gives approximate scaling relations:
L ∝ M3.5
E∝M
t ∝ M-2.5
R∝M
Te ∝ M0.5
The Hertzsprung-Russell diagram
What happens when Hydrogen fuel is exhausted in the core?
Inert He core surrounded by shell still burning H to He.
Changes structure of star: x1000+ increase in L, plus 100x increase
in size (with decrease in temperature) à Red Giant Star
(N.B. Sun will consume Earth)
Temperature rise in core
may ignite He to C fusion
à Horizontal Branch Star
Followed by He shell
burning and so on à
Asymptotic Giant Branch
Star
If ever the core gets dense enough to be supported by
degenerate electron pressure, it will cease contracting and
raising its temperature. Therefore no new fusion reactions
will occur. e.g. the Sun will never fuse C because it will
stabilize its core below the ignition temperature of C.
White Dwarfs
When the triple-α process in a red-giant is complete, the star core starts
contracting. For a main sequence mass of M < 4 M¤ there is not enough
energy to ignite the carbon fusion process, thus the collapse keeps going
until it is halted by the pressure arising from electron degeneracy. From
the stage of PN the star loses its extended envelope. The resulting hot
and small (relative to the Sun) star is called a White Dwarf.
Stars with M< 0.4 M¤ will not become red giants because they do not
become hot enough for He burning, but their lifetimes will be longer than
the present age of the Universe.
Typical C,O white dwarf specs:
•  mass of about 1 M¤ (on average)
•  size of about 1 Earth diameter
•  ~ 4% of solar luminosity
•  surface temperature of ~ 25000 K
What about Life?
Good news: Because they are so small
(and have such a low L) they will shine
for a very long time, even though they
are not producing more energy from
fusion
Bad news: All the products of fusion
are locked up in the interiors
Mass Loss and “Planetary” Nebulae
Shells of gas thrown off by Sun-like stars nearing the ends of their
lives. The action of radiation pressure on small dust grains formed in
the cool stellar atmospheres probably causes the mass loss.
The UV radiation emitted by the remaining hot (but quickly inert)
stellar core (white dwarf) causes the ejected gas to fluoresce as the
planetary nebula.
Stars with M < 8 M¤ loose most of their mass during this phase. But
this is largely unprocessed material with a composition similar to that
of the original gas cloud out of which the star formed.
But more massive stars will go all the way to an Fe core.
abundance
Final structure of a 25 Msun star
enclosed mass
Onion-like structure with successive
shells burning different fusion reactions
But time is running out:
Energy per reaction is dropping, required luminosity is increasing.
For a 25 Msun star:
–
–
–
–
–
Hydrogen burns in the core for 7 million years
Helium burns for 500,000 years
Carbon burns for 600 years
Oxygen burns for 6 months
Silicon burns for 1 day
What happens next?
But first, back to that Gamow Peak………
Key point: Gamow
peak means that
most individual
reactions have the
same energy… i.e.
the Δmc2 of the
reaction plus the
K.E. associated with
the Gamow Peak.
One consequence:
The existence of
well-placed energy
levels in product
nucleus leads to
resonances, i.e.
greatly increased σ
at particular
energies, leading to
wide variation of
reaction rates for
different reactions.
Reaction pathways for H → 4He fusion
Extremely slow due to
weak interaction
τ ~ 1010 yr in Sun
Slow, but less so than
above due to
spontaneous decay of
unstable nuclei
13N
(τ1/2 ~ 600s) and
15O (τ
1/2 ~ 120s)
Also, requires higher T
because of higher +ve
charge on nuclei
p-p chain
1H + 1H → 2H + e- + ν
2Η + 1Η → 3He + γ
3He + 3He → 4He + 1H + 1H
CNO cycle
12C + 1H → 13N + γ
13N → 13C + e- + ν
13C + 1H → 14N + γ
14N + 1H → 15O + γ
15O → 15N + e- + ν
15N + 1H → 12C + 4He
CNO cycle
Helium burning and the interesting case of Carbon:
Absence of more tightly bound
nuclei between 4He and 12C
Requires effectively a three
particle reaction
4He
+ 4He + 4He → 12C + 7.7 MeV
Which will be extremely rare
statistically
Carbon
•  Production of 8Be from two 4He requires 0.1 MeV of energy.
•  8Be is unstable with half life of 10-17 s
While it survives, 8Be can join with 4He to make an excited
state of 12C.
8Be
+4He has energy 7.33 MeV above ground state of 12C.
Hoyle (1952) correctly predicted that 12C must have a
resonant state just above 7.33 MeV for this reaction to occur
often enough to produce significant amounts of 12C.
Subsequently found at 7.65 MeV – a triumph of theoretical
astrophysics.
Most 12C* falls apart back to 8Be + 4He, but some de-excites
to ground-state through emission of γ-ray photons.
Carbon production
Gamow peak
C Resonance level
Three states of 4He, 8Be and
12C* are so similar in energy
that these three species will
be in equilibrium, described
by Saha equation.
Production of 12C is
effectively “leakage” out
of equilibrium from 12C*
excited state
Equilibrium
There is a potentially easy
route to burn 12C:
12C
+ 4He → 16O + 7.16 MeV
Absence of resonance
(“near-miss”) for 16O in
Gamow band hinders this
reaction, allowing 12C to
survive in reasonable
abundance (~ 1/3 O)
Destruction
of 16O also
hindered
Energy levels
Consequence: O and C are the most abundant
elements after H and He – good for organics,
water etc .
The relatively high abundances of these two
arise from coincidences in the nuclear energy
levels of these and other nuclear species
Abundances