Transcript Document

Lecture 3
Initial Mass Function and
Chemical Evolution
Essentials of Nuclear Structure
The Liquid Drop Model
An interval  0.3 around M1
would thus correspond to a
range in masses M1 / 2 to M1  2
See Shapiro and Teukolsky
for background reading
just a fit
Salpeter used
 (log M )  C M 
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
with   1.35
But Figer (2005) gets
 =-0.9 in a young
supercluster.
Warning. Salpeter
IMF not appropriate
below about 0.5
solar masses. Actual
IMF is flatter. Used
MS here.
d log M 
dM
M
Since  = -1.35
sensitive to
choice of ML
Use MS Table instead.
For Salpeter IMF
2 solar masses/Gyr/pc2
45 solar masses/pc2
current values - Berteli and Nasi (2001)
solar neighborhood
9
Upper mass limit: theoretical predictions
Ledoux (1941)
radial pulsation, e- opacity,
H
100 M
Schwarzchild & Härm (1959)
radial pulsation, e- opacity,
H and He, evolution
65-95 M
Stothers & Simon (1970)
radial pulsation, e- and atomic
Larson & Starrfield (1971)
pressure in HII region
50-60 M
Cox & Tabor (1976)
e- and atomic opacity
Los Alamos
80-100 M
Klapp et al. (1987)
e- and atomic opacity
Los Alamos
440 M
Stothers (1992)
e- and atomic opacity
Rogers-Iglesias
120-150 M
80-120 M
Upper mass limit: observation
R136
Feitzinger et al. (1980)
250-1000 M
Eta Car
various
120-150 M
R136a1
Massey & Hunter
(1998)
136-155 M
Pistol Star
Figer et al. (1998)
140-180 M
Damineli et al. (2000)
~70+? M
LBV 1806-20
Eikenberry et al. (2004)
150-1000 M
LBV 1806-20
Figer et al. (2004)
130 (binary?)
M
HDE 269810
Walborn et al. (2004)
150 M
WR20a (binary)
Bonanos et al. (2004)
Rauw et al. (2004)
82+83 M
Eta Car
each +- 5 Msun
What is the most massive star (nowadays)?
The Arches Supercluster
Massive enough and young
enough to contain stars of 500
solar masses if extrapolate Salpeter
IMF
Figer, Nature, 434, 192 (2005)
Kim, Figer, Kudritzki and Najarro
ApJ, 653L, 113 (2006)
Lick 3-m (1995)
Keck 10-m (1998)
HST (1999)
Initial mass function
Introductory Nuclear
Physics;
Liquid Drop
Model
Each nucleus is a bound collection of
N neutrons and Z protons. The mass
number is A = N + Z, the atomic number is Z
and the nucleus is written
AZ
E.g.
12
C, 13C, 14 C are isotopes of carbon all
with Z  6 and neutron numbers
N = 6, 7, 8
The neutrons and protons are bound together by
the strong or color force
In fact, the neutron and proton are themselves
collections of smaller fundamental quarks.
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TIFF ( Un co mpr es sed ) d eco mp res so r
ar e n eed ed to s ee this pi ctu re.
p
QuickT i me™ and a
T IFF (Uncompressed) decompressor
are needed to see this picture.
4He
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
In addition there is
a collection of bosons
whose exchange
mediates the four
fundamental forces.
g, W+-, Z0, gluon,
graviton
Only quarks and gluons
experience the “color”
force and quarks are
never found in isolation
In the standard model ….
Hadrons are collections of three quarks (baryons) or a
quark plus an anti-quark (mesons). This way they are able
to satisfy a condition of color neutrality. Since there are three
colors of quarks, the only way to have neutrality is to
have one of each color, or one plus an antiparticle of the same
(anti-)color.
The gluons also carry color (and anti-color) and there are
eight possible combinations, hence 8 gluons.
The color force only affects quarks and gluons.
The color force binds the quarks in the hadrons
A red quark emits a red-antigreen gluon which
is absorbed by a green quark making it red.
The weak interaction allows heavier quarks and
leptons to decay into lighter ones. E.g.,
n  p  e   e
udd uud
and if energy is provided
p  e  n   e
For background on all this, please read
http://particleadventure.org
d  u means a charge of -1/3 goes to
a charge of + 2/3. The W- is necessary to
conserve charge.
intermediate stages not observable
Mesons are quark antiquark pairs and this carry
no net spin. The lightest two mesons consist only
of combinations of u, d, u, and d are
Name

o

     + 
Made of
uu  dd
2
ud , du
Charge
Mass
 (sec)
0
135 MeV
8.4(-17)
1
139.6
2.6(-8)
 0  2g , occasionally e + + e-
There are many more mesons. Exchange of these lightest mesons give
rise to a force that is complicated, but attractive. But at
a shorter range, many other mesons come into play, notably
the rho meson (776 MeV), and the nuclear force becomes repulsive.
There are two ways of thinking of the
strong force - as a residual color interaction
or as the exchange of mesons. Classically
the latter was used.
E t 
Mc 2 / c 
The nuclear force at large
distances is not just small,
it is zero.
Repulsive at short distances.
Nuclear density nearly
constant.
･The nuclear force is only felt among hadrons.
･At typical nucleon separation (1.3 fm) it is a
very strong attractive force.
･At much smaller separations between nucleons
the force is very powerfully repulsive, which keeps
the nucleons at a certain average separation.
･Beyond about 1.3 fm separation, the force
exponentially dies off to zero. It is greater than the
Coulomb force until about 2.5 fm
･The NN force is nearly independent of whether the
nucleons are neutrons or protons. This property is
called charge independence or isospin independence.
･The NN force depends on whether the spins of the nucleons
are parallel or antiparallel.
･The NN force has a noncentral or tensor component.
Since the nucleons are fermions they obey
FD statistics
n = 0.17 fm-3

per nucleon
Nuclear density is
a constant.
Deformation is an
indication of nuclear
rotation
RA1/ 3
nuclear force is
spin dependent
Nuclear binding energy is the (positive)
energy required to disperse a bound nucleus,
AZ, into N neutrons and Z protons separated
by a large distance.
BE(n) = BE(p) = 0
It is the absolute value of the sum of the
Fermi energies (positive), electrical energy
(positive), and strong attractive potential
energy (negative). A related quantity is the
average binding energy per nucleon
BE/A
Coulomb Energy
• The nucleus is electrically charged with total charge Ze
• Assume that the charge distribution is spherical and
compute the reduction in binding energy due to the
Coulomb interaction
Ze
ECoulomb  
0
Q(r )
dQ
4 0 r
Q(r )  Ze(r / R)3 dQ  3Zer 2 / R3dr
to change the integral to dr ; R=outer radius of nucleus
R
ECoulomb
3(Ze)2 r 5
(Ze)2

dr  (3/ 5)
6
4 0 r R
4 0 R
0
… and remember R=R0A-1/3
includes self interaction of last
proton with itself. To correct this
replace Z2 with Z*(Z-1)
Z *( Z  1)
BCoulomb ( Z , A)  d
A1/ 3
Mirror Nuclei
• Compare binding energies of mirror nuclei (nuclei with np).
Eg 73Li and 74Be.
• If the assumption of isospin independence holds the mass
difference should be due to n/p mass difference and Coulomb
energy alone.
• From the previous page
3 e2
3 e2
Ecoulomb (Z , Z  1) 
[Z (Z 1)  (Z 1)(Z  2)] 
2(Z  1)
5 4 0 R
5 4 0 R
Z ~ A / 2 ; R  R0 A1/3 to find that EC (Z , Z 1)  A2/3
• Now lets measure mirror nuclei masses, assume that the model
holds and derive ECoulomb from the measurement.
• This should show an A2/3 dependence
“Charge
symmetry”
nn and pp
interaction same
(apart from
Coulomb)
ECoul  A2/3
More charge symmetry
Energy Levels of two mirror nuclei for a number of
excited states. Corrected for n/p mass difference and
Coulomb Energy
Ecorrected
42
Semi-Empirical Mass Formulae
• A phenomenological understanding of nuclear
binding energies as function of A, Z and N.
• Assumptions:
– Nuclear density is constant.
– We can model effect of short range attraction due to strong
interaction by a liquid drop model.
– Coulomb corrections can be computed using electro
magnetism (even at these small scales)
– Nucleons are fermions at T=0 in separate wells (Fermi gas
model  asymmetry term)
– QM holds at these small scales  pairing term
– Nuclear force does not depend on isospin
Liquid Drop Model
• Phenomenological model to understand binding energies.
• Consider a liquid drop
– Ignore gravity and assume no rotation
– Intermolecular force repulsive at short distances, attractive at intermediate
distances and negligible at large distances  constant density.
– n=number of molecules, T=surface tension, BE=binding energy
E=total energy of the drop, a,b=free constants
E=-an + 4R2T

BE=an-bn2/3
• Analogy with nucleus
surface area ~ n2/3
– Nucleus has constant density
– From nucleon-nucleon scattering experiments we know:
• Nuclear force has short range repulsion and is attractive at intermediate distances.
– Assume charge independence of nuclear force, neutrons and protons have same
strong interactions check with experiment (Mirror Nuclei!)
Volume and Surface Term
• If we can apply the liquid drop model to a nucleus
– constant density
– same binding energy for all constituents
• Volume term:
• Surface term:
BVolume ( A)  aA
BSurface ( A)  bA2/3
a ~ 15 MeV
b ~ 17 MeV
• Since we are building a phenomenological model in which
the coefficients a and b will be determined by a fit to
measured nuclear binding energies we must include any
further terms we may find with the same A dependence
together with the above
There are additional important correction terms to the
volume and surface area terms, notably the Coulomb
repulsion that makes the nucleus less bound, and the
symmetry energy, which is a purely quantum mechanical
correction due to the exclusion principle.
ECoul
Z2
Z2
 const
 const ' 1/3
R
A
Asymmetry Term
• Neutrons and protons are spin ½ fermions  obey Pauli
exclusion principle.
• If all other factors were equal nuclear ground state would have
equal numbers of n & p.
neutrons
protons
•
•
•
•
•
Illustration
n and p states with same spacing .
Crosses represent initially occupied states in ground
state.
If three protons were turned into neutrons
the extra energy required would be 3×3 .
In general if there are Z-N excess protons over
neutrons the extra energy is ((Z-N)/2)2 . relative to
Z=N.
correction
A
(1   )
2
A
N '  (1   )
2
Z '
(1  ) n 1 n 

n(n  1) 2
 ...
2!
N ' Z '
A
The proportionality constant is about 28 MeV
So far we have
2
N

Z
Z


BE  a A  b A2/3  c
 d 1/3
A
A
2
purely quantum mechanical
corrections to the liquid
drop model
Adding a nucleon increases
the nuclear binding energy
of the nucleus (no direct
analogue to atomic physics).
If this is nucleon is added
to a lower energy state,
more binding is obtained.
A low state might be one
where there is already an
unpaired nucleon.
Pairing Term
Neutron separation energy
[MeV] in Ba isotopes
56+N
• Nuclei with even number of n or
even number of p more tightly
bound then with odd numbers.
• Only 4 stable o-o nuclei but 153
stable e-e nuclei.
Neutron number
56Ba
Pairing Term
• Phenomenological fit
BPairing ( A)  
d
e-e
e-o
o-o
+
0
-
d
1/ 2
A
Note: If you want to plot binding
energies versus A it is often best to
use odd A only as for these the
pairing term does not appear
Putting it all together:
Pairing increases the
binding energy of nuclei
with even numbers of
neutrons and/or protons
Experiment
Liquid drop
Volume E/A = const
Surface E/A  A -1/3
Coulomb E/A  Z2 / A 4/3
Symmetry E/A  (N-Z) 2 / A 2
Evans 3.5
Semi Empirical Mass Formula
Binding Energy vs. A for beta-stable odd-A nuclei
Iron
Fit parameters in
MeV
a 15.56
Not smooth because Z
not smooth function of A
b 17.23
c 23.285
d 0.697
d +12 (o-o)
d 0 (o-e)
d -12 (e-e)
Utility
• Only makes sense for A greater than about 20
• Good fit for large A (<1% in most instances)
• Deviations are interesting - shell effects
• Explains the “valley of beta-stability” (TBD)
• Explains energetics of nuclear reactions
• Incomplete consideration of QM effects (energy levels
not all equally spaced)
http://128.95.95.61/~intuser/ld3.html
Given A, what is the
most tightly bound Z?
N = A-Z
N-Z = A-2Z
Only the Coulomb and
pairing terms contained Z
explicitly
xxxxxxxxxxxxxxa A2 / 3  4a
3
4
xxxx
symmetry dominates
A
Zs
20
9.6
40
18.6
60
27.3
N  Z stable  A  2Z stable


4a4
 A 1 
a3 A 2 /3  4a4 

 a3 A 2 /3 
 A

 a3 A 2 /3  4a4 

a3 5 /3
A  0.0064 A 5/3
4a4
a3  0.717 a4  28.1
Evans 3.4