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. For Salpeter IMF Since  = -1.35 sensitive to choice of M L Use MS Table instead.

7. solar neighborhood 2 solar masses/Gyr/pc 2 45 solar masses/pc 2 current values - Berteli and Nasi (2001)

8. 9

13. 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 atomic80-120 M  Larson & Starrfield (1971)pressure in HII region50-60 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 

14. Upper mass limit: observation R136Feitzinger et al. (1980)250-1000 M  Eta Carvarious120-150 M  R136a1 Massey & Hunter (1998) 136-155 M  Pistol StarFiger et al. (1998)140-180 M  Eta CarDamineli et al. (2000)~70+? M  LBV 1806-20Eikenberry et al. (2004)150-1000 M  LBV 1806-20Figer et al. (2004)130 (binary?) M  HDE 269810Walborn et al. (2004)150 M  WR20a Bonanos et al. (2004) Rauw et al. (2004) 82+83 M  each +- 5 Msun (binary)

15. 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)?

16. Lick 3-m (1995)

17. Keck 10-m (1998)

18. HST (1999)

19. Initial mass function

20. Introductory Nuclear Physics; Liquid Drop Model

22. In fact, the neutron and proton are themselves collections of smaller fundamental quarks. 4 He p

24. In addition there is a collection of bosons whose exchange mediates the four fundamental forces. , W +-, Z 0, 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. A red quark emits a red-antigreen gluon which is absorbed by a green quark making it red. The color force binds the quarks in the hadrons

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.

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

38. Nuclear binding energy is the (positive) energy required to disperse a bound nucleus, A Z, 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 ; R=outer radius of nucleus includes self interaction of last proton with itself. To correct this replace Z 2 with Z*(Z-1) … and remember R=R 0 A -1/3

40. Mirror Nuclei Compare binding energies of mirror nuclei (nuclei with n  p). Eg 7 3 Li and 7 4 Be. 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  E Coulomb from the measurement. This should show an A 2/3 dependence

41. nn and pp interaction same (apart from Coulomb) “Charge symmetry”

42. 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  E corrected

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 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

44. surface area ~ n 2/3 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, ,  =free constants E=-  n + 4  R 2 T  BE=  n-  n 2/3 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. neutronsprotons 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.

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. Neutron number Neutron separation energy [MeV] in Ba isotopes 56+N 56 Ba

54. Pairing Term Phenomenological fit  e-e+ e-o0 o-o- 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. Pairing increases the binding energy of nuclei with even numbers of neutrons and/or protons Putting it all together:

56. Experiment Liquid drop Evans 3.5

57. Semi Empirical Mass Formula Binding Energy vs. A for beta-stable odd-A nuclei Iron Not smooth because Z not smooth function of A Fit parameters in MeV a15.56 b17.23 c23.285 d0.697  +12 (o-o)  0 (o-e)  -12 (e-e)

58. 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)

60. N = A-Z N-Z = A-2Z Given A, what is the most tightly bound Z? Only the Coulomb and pairing terms contained Z explicitly http://128.95.95.61/~intuser/ld3.html

61. xxxxxxxxxxxxxx xxxx AZsZs 209.6 4018.6 6027.3

62. Evans 3.4