Lecture 3

Initial Mass Function andChemical Evolution

Essentials of Nuclear StructureThe Liquid Drop Model

See Shapiro and Teukolskyfor background reading

just a fitxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxBut Figer (2005) gets =-0.9 in a young supercluster.

Warning. SalpeterIMF not appropriatebelow about 0.5 solar masses. ActualIMF is flatter. Used MS here.

For Salpeter IMFSince G = -1.35sensitive to choice of MLUse MS Table instead.

solar neighborhood2 solar masses/Gyr/pc245 solar masses/pc2current values - Berteli and Nasi (2001)

9

Upper mass limit: theoretical predictions

Ledoux (1941)radial pulsation, e- opacity,H100 MSchwarzchild & Hrm (1959)radial pulsation, e- opacity,H and He, evolution65-95 MStothers & Simon (1970)radial pulsation, e- and atomic80-120 MLarson & Starrfield (1971)pressure in HII region50-60 MCox & Tabor (1976)e- and atomic opacityLos Alamos80-100 MKlapp et al. (1987)e- and atomic opacityLos Alamos440 MStothers (1992)e- and atomic opacityRogers-Iglesias120-150 M

Upper mass limit: observationeach +- 5 Msun(binary)

R136Feitzinger et al. (1980)250-1000 MEta Carvarious120-150 MR136a1Massey & Hunter (1998)136-155 MPistol StarFiger et al. (1998)140-180 MEta CarDamineli et al. (2000)~70+? MLBV 1806-20Eikenberry et al. (2004)150-1000 MLBV 1806-20Figer et al. (2004)130 (binary?) MHDE 269810 Walborn et al. (2004)150 MWR20aBonanos et al. (2004)Rauw et al. (2004)82+83 M

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)

What is the most massive star (nowadays)?

Lick 3-m (1995)

Keck 10-m (1998)

HST (1999)

Initial mass function

Introductory Nuclear Physics;Liquid Drop Model

In fact, the neutron and proton are themselvescollections of smaller fundamental quarks.4Hep

In addition there isa collection of bosonswhose exchange mediates the fourfundamental forces.

g, W+-, Z0, gluon, gravitonOnly quarks and gluonsexperience the color force and quarks are never found in isolation

In the standard model .

Hadrons are collections of three quarks (baryons) or aquark plus an anti-quark (mesons). This way they are ableto satisfy a condition of color neutrality. Since there are threecolors 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.

A red quark emits a red-antigreen gluon whichis absorbed by a green quark making it red.The color force binds the quarks in the hadrons

The weak interaction allows heavier quarks and leptons to decay into lighter ones. E.g., For background on all this, please readhttp://particleadventure.org

intermediate stages not observable

There are many more mesons. Exchange of these lightest mesons giverise to a force that is complicated, but attractive. But at a shorter range, many other mesons come into play, notablythe rho meson (776 MeV), and the nuclear force becomes repulsive.

There are two ways of thinking of the strong force - as a residual color interactionor as the exchange of mesons. Classicallythe latter was used.

The nuclear force at largedistances is not just small,it is zero.Repulsive at short distances.Nuclear density nearlyconstant.

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 statisticsn = 0.17 fm-3per nucleon

Nuclear density isa constant.Deformation is an indication of nuclear rotation

nuclear force isspin 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 ZeAssume that the charge distribution is spherical and compute the reduction in binding energy due to the Coulomb interactionto change the integral to dr ; R=outer radius of nucleusincludes self interaction of last proton with itself. To correct this replace Z2 with Z*(Z-1) and remember R=R0A-1/3

Mirror Nuclei

Compare binding energies of mirror nuclei (nuclei with np). 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 pageto find thatNow lets measure mirror nuclei masses, assume that the model holds and derive DECoulomb from the measurement. This should show an A2/3 dependence

Charge symmetrynn and pp interaction same (apart from Coulomb)

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

Semi-Empirical Mass FormulaeA 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 dropIgnore gravity and assume no rotationIntermolecular 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 energyE=total energy of the drop, a,b=free constantsE=-an + 4pR2T BE=an-bn2/3Analogy with nucleusNucleus has constant densityFrom 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 TermIf we can apply the liquid drop model to a nucleusconstant densitysame binding energy for all constituentsVolume 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 abovea ~ 15 MeVb ~ 17 MeV

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 mechanicalcorrection due to the exclusion principle.

Asymmetry TermNeutrons and protons are spin fermions obey Pauli exclusion principle.If all other factors were equal nuclear ground state would have equal numbers of n & p.

Illustrationn 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 Z=N.

correction

The proportionality constant is about 28 MeV

So far we have

purely quantum mechanical corrections to the liquiddrop modelAdding a nucleon increasesthe nuclear binding energyof the nucleus (no directanalogue to atomic physics).If this is nucleon is addedto a lower energy state, more binding is obtained.A low state might be onewhere there is already anunpaired nucleon.

Pairing TermNuclei 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.

Pairing Term

Phenomenological fitNote: 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

Pairing increases the binding energy of nucleiwith even numbers ofneutrons and/or protonsPutting it all together:

ExperimentLiquid dropEvans 3.5

Semi Empirical Mass Formula Binding Energy vs. A for beta-stable odd-A nucleiIron

Fit parameters in MeVa15.56b17.23c23.285d0.697d+12 (o-o)d0 (o-e)d-12 (e-e)

Utility Only makes sense for A greater than about 20 Good fit for large A (

N = A-ZN-Z = A-2ZGiven A, what is the most tightly bound Z?Only the Coulomb andpairing terms contained Zexplicitlyhttp://128.95.95.61/~intuser/ld3.html

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AZs209.64018.66027.3

Evans 3.4