Initial Mass Function and

1
Lecture 3 Initial Mass Function and Chemical
Evolution Essentials of Nuclear Structure The
Liquid Drop Model
2
See Shapiro and Teukolsky for background reading
3
just a fit
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
But Figer (2005) gets ? -0.9 in a young
supercluster.
4
Warning. Salpeter IMF not appropriate below about
0.5 solar masses. Actual IMF is flatter. Used
MS here.
5
Since G -1.35
sensitive to choice of ML Use MS Table instead.
For Salpeter IMF
6
(No Transcript)
7
solar neighborhood
2 solar masses/Gyr/pc2 45 solar
masses/pc2 current values - Berteli and Nasi
(2001)
8
9
9
(No Transcript)
10
(No Transcript)
11
(No Transcript)
12
(No Transcript)
13
Upper mass limit theoretical predictions
14
Upper mass limit observation
(binary)
each - 5 Msun
15
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)
16
Lick 3-m (1995)
17
Keck 10-m (1998)
18
HST (1999)
19
Initial mass function
20
Introductory Nuclear Physics Liquid Drop Model
21
(No Transcript)
22
In fact, the neutron and proton are
themselves collections of smaller fundamental
quarks.
4He
p
23
(No Transcript)
24
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
25
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.
26
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.
27
The weak interaction allows heavier quarks and
leptons to decay into lighter ones. E.g.,
For background on all this, please
read http//particleadventure.org
28
intermediate stages not observable
29
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.
30
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.
31
The nuclear force at large distances is not just
small, it is zero.
Repulsive at short distances.
Nuclear density nearly constant.
32
?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.
33
(No Transcript)
34
Since the nucleons are fermions they obey FD
statistics
n 0.17 fm-3
per nucleon
35
Nuclear density is a constant.
Deformation is an indication of nuclear rotation
36
nuclear force is spin dependent
37
(No Transcript)
38
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
39
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

to change the integral to dr Router radius of
nucleus
includes self interaction of last proton with
itself. To correct this replace Z2 with Z(Z-1)
and remember RR0A-1/3
40
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

to find that

• Now lets measure mirror nuclei masses, assume
  that the model holds and derive DECoulomb from
  the measurement.
• This should show an A2/3 dependence

41
Charge symmetry
nn and pp interaction same (apart from Coulomb)
42
More charge symmetry

• Energy Levels of two mirror nuclei for a
  number of excited states. Corrected for n/p mass
  difference and Coulomb Energy

DEcorrected
43
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 T0 in separate wells
  (Fermi gas model ? asymmetry term)
• QM holds at these small scales ? pairing term
• Nuclear force does not depend on isospin

44
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.
• nnumber of molecules, Tsurface tension,
  BEbinding energy
• Etotal energy of the drop, a,bfree constants
• E-an 4pR2T ? BEan-bn2/3

• Analogy with nucleus
• 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!)

45
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
• 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

a 15 MeV b 17 MeV
46
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.
47
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.

• 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 33 ?.
• In general if there are Z-N excess protons over
  neutrons the extra energy is ((Z-N)/2)2 ?.
  relative to ZN.

48
(No Transcript)
49
correction
50
The proportionality constant is about 28 MeV
51
So far we have
52
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.
53
Pairing Term

• 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.

54
Pairing Term

• Phenomenological fit

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
55
Putting it all together
Pairing increases the binding energy of
nuclei with even numbers of neutrons and/or
protons
56
Experiment
Liquid drop
Evans 3.5
57
Semi Empirical Mass Formula Binding Energy vs. A
for beta-stable odd-A nuclei
Iron
58
Utility

• Only makes sense for A greater than about 20
• Good fit for large A (lt1 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)

59
(No Transcript)
60
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
61
xxxxxxxxxxxxxx
xxxx
62
Evans 3.4
63
(No Transcript)
64
(No Transcript)