Computational Applications in Nuclear Astrophysics using JAVA

Size: px

Start display at page:

Download "Computational Applications in Nuclear Astrophysics using JAVA"

Anastasia Lynch
5 years ago
Views:

1 Computational Applications in Nuclear Astrophysics using JAVA Lecture: Friday 10:15-11:45 Room NB 6/99 Jim Ritman and Elisabetta Prencipe Computer Lab: Friday 12:15-13:45 NB 6/99 Michael Kunkel 1

2 Lecture dates 1) JR 2) EP 3) EP 4) JR 5) JR 6) EP 7) JR 8) EP 9) JR 10) EP 11) EP 12) JR 13) EP 14) JR 2

3 Contents Introduction 1 Basics of nuclear physics 2 Big Bang 1 BB-Nuclear synthesis 1 Stellar atmospheres 3 H-burning cycles 1 He-burning Supernova 1 s,r,rp, αp processes 1 Calc. of abundances for Z>Fe/Ni 2 3

4 Literature 4

5 Grades To successfully complete the course there is the 2 SWS lecture and the 2 SWS computer lab. The computer lab will have regular homework that needs to be handed in. During the semester each student will develop a library of code to calculate networks of reactions. At the end of the semester each student should make a brief (15-20 minute) presentation on their code library. 5

6 Introduction 6

7 Nuclear Astrophysics Nuclear astrophysics attempts to understand nuclear processes that take place in the universe that contribute to the energy production in stars and to the nuclear synthesis of elements. 7

8 Nuclear Synthesis Nuclear synthesis addresses the generation and distribution of chemical elements in our universe There are many different chemical elements From H (Z=1) to U (Z=92) Typical atomic binding energies are ev (T=10 4 K) Energy scale for nuclear transformations is however MeV (T=10 10 K) Such energies can only be found in the center of stars and shortly after the Big Bang. 8

9 View into the Milky Way 9

10 Map of Galactical 26 Al (COMPTEL) 10

mass is He O is the most important Metal (circa 50%) C, N

11 Element Abundances Very simplified we have: Nearly the same distribution everywhere 70% of the mass is H 28% of the mass is He O is the most important Metal (circa 50%) C, N and Fe are also important All stable elements have been found 11

Cl Fe Mars This figure shows a comparison of the

photosphere of the sun compared to what is measured

Stones on the earth have a similar relative

12 Cl Fe Mars This figure shows a comparison of the abundance of elements that one measures in the photosphere of the sun compared to what is measured in meteorites. NB the logarithmic axes. Stones on the earth have a similar relative abundance to the cosmic abundances. However hydrogen and helium are exceptions.

13 B 2 FH In 1957 Burbidge, Burbidge, Fowler and Hoyle came up with the following picture that is still valid today: H and He are mostly produced after the Big Bang, everything else is produced in stars He is also produced in stars Light elements up to Si are generated in normal stars up to the end of their life cycles Heavy elements are produced in special phases of stars with especially heavy masses Super heavy element are from supernovae 13

14 B 2 FH 14

15 Nuclear Synthesis Processes Contributions to the synthesis of elements Possible SN II (ν-process) contributions to 11B, 19F, 138Le, 180Ta, 15

16 Cosmic Element Machine 16

17 Hoyle s Cosmic Cycle 17

18 Energy and Lifetime of the Sun The energy radiated by the sun is very large (L = 3.826x10 33 erg/s). There is a direct relationship between the lifetime and the total amount of energy available Time = Energy / Luminosity If we consider the potential sources of energy, we can estimate the lifetime of the sun. - Chemical (EM interaction) - Gravitational (Gravity) - Nuclear power (Strong interaction) (what about the weak interaction??)

19 Chemical Reaction Time Scale Chemical processes are based on the interaction of shell electrons in atoms. Typically 1-10 ev per Atom is released. Assuming that the sun is made of pure hydrogen, and that 10 ev per atom is released, then the lifetime is: τ Chem = = = E 10eV L Sonne 10eV M L 33 Sonne s = Sonne g erg s 33 5 yr m H 24 g Since the earth is many 10 9 years old, there is no way that chemical reactions could be powering the sun.

Gravitational Heating The potential energy U of a point-like mass m at a distance r from a big mass M is: U = G M m r U will become more negative as r between M and m gets smaller.

20 Gravitational Heating The potential energy U of a point-like mass m at a distance r from a big mass M is: U = G M m r U will become more negative as r between M and m gets smaller. Collisions between the atoms in M will convert part of the released energy into heat, which will be radiated into space. As a result: as the star contracts it will radiate photons from a hot source How long can that go on? Be careful! Only half of the released gravitational energy can be radiated, the remainder is used to heat up the star This is a result of the Virial theorem of classical mechanics

21 The Virial Theorem Consider a set of particles interacting via some force F. The system has the moment of inertia I, the kinetic energy E and the potential energy U. I = 2 m i r i 1 E = 2 m r! i i U = F i r i 2 di dt = 2 m r! i i r i 1 d 2 I m!! r i i r i + m i!r 2 i dt = 2 d 2 I dt = Fr + 2 i i d 2 I dt 2 = U + 2E 2 m i!r i If the system is periodic, then di/dt t=0 = di/dt t=τ and thus: 0 2 d I 2 dt = E U = = U E If F is the gravitational force, then U is negative for a bound system Equilibrium condition: I = constant Result: 50% of the gained energy is converted into heat (kinetic energy), 50% must be radiated away

22 Example: orbit with eccentricity=0.5 2 I r 2 d I 2 dt t 2 d I dt dt d I 2 dt = 0

23 Implications of the Virial Theorem for Stellar Formation Assume that a star is formed as a result of the gravitational collapse of a large nebula In equilibrium, the star must obey the Virial Theorem The potential energy of the star must have changed from an initial value of nearly zero (large distance between particles) to its negative static value Energy conservation requires that gravitational energy must have been radiated into space during the collapse half of the energy is converted into internal energy (heat) the second half is radiated away into space in the process of star formation

interaction of every possible pair of particles!

24 Gravitational Potential Energy For a system of 2 particles with masses M and m, the gravitational potential energy is given by Gravitational energy of a star: Ø This requires the consideration of the interaction of every possible pair of particles! Ø The force on a point mass dm i located outside a spherically symmetric mass M r is df = G M r i 2 dm r i M m U= G r Potential energy of the shell: 2 Mr 4πr ρ du = G dr r

25 Integration over all mass shells: R = π r ρ 0 U 4 G M rdr In principle, we need to know the ρ=ρ(r), but an approximate value can be obtained by assuming the density to be constant and equal to its average value: ρ ρ = R 4 3 M πr 3 M r = πr ρ R U= 4πG πr ρ rdr= πgρ r dr= πgρ R

26 or, by using the mass M of the star: M = πr ρ GM U= πgρ R = 15 5 R 2 We still have to apply the Virial Theorem to obtain the total mechanical energy of the star: E = 2 3GM 10 R Example: How much internal energy was created when the sun was formed (from an infinitely large cloud): E = GMsun R sun 48 erg

27 Kelvin-Helmholtz Timescale Assume that gravity is the only source of energy for the sun.. Also assume that the luminosity is roughly constant for the lifetime of the star How long can the sun radiate at this luminosity? t KH = GM Sun R Sun L Sun erg erg/s 107 yr t KH is called the Kelvin-Helmholtz Timescale. This must be compared with how long the there is life on earth (10 9 years). Gravity can not be the source of the solar energy! There must be another much stronger source for the energy!

28 Nuclear Timescale Differences of the binding energy of nucleons (protons and neutrons) to different nuclei can be used to release energy by fusion or fission. (E=mc 2 ) If protons (M=938 MeV) are fused (combined) to light nuclei with B =7-8 MeV/A, then about 0.7 % of the mass will be radiated. Main sequence stars typically burn about 10 % of their total mass during their lifetime. E nuclear = M Sun c erg This enables a nuclear timescale of: t nuclear = E nuclear L Sun yr That is plenty for the10 9 that there is already life on earth: Nuclear energy powers the stars!

29 Basics of Nuclear Physics Stability of nuclei Constituents, Forces Mass formula Nuclear reactions Cross section, Q-Value Nuclear reactions Coulomb barrier Nuclear reactions in stars Gamov peak Astrophysical factor 29

30 Nomenclature Nuclide: Refers to a particular atom or nucleus with a specific number N of neutrons and number Z of protons. A is the mass number (= Z +N). Nuclides are either stable or radioactive. Approximately 3000 nuclides are known, but only about 10% of these are stable. Atomic Number, Z: The number of protons in the nucleus. Neutron Number, N: The number of neutrons in the nucleus Isotope: One of the nuclides that have the same Z but different N Isotone: One of several different nuclides having the same N Isobar: Nuclides with the same atomic mass number A (= Z+N) but with different values of N and Z e.g. 14 B, 14 C, 14 N. Isomer: Atoms with the same Z and A in different long-lived states of excitation the higher states being metastable with respect to the ground state. For example, an isomer of 99 Tc is 99 mtc where the m denotes the long-lived excited state. 30

31 Nuclide Chart Isotopes Isobars Isotones 31

Nucleons and Forces Nucleons (s=1/2), are simplified 3-quark

There needs to be an attractive force for nuclei to be stable:

32 Nucleons and Forces Nucleons (s=1/2), are simplified 3-quark states (p=uud, n=udd) Protons: m p = MeV/c 2 Neutrons: m n = MeV/c 2 Protons repel each other due to the Coulomb force. There needs to be an attractive force for nuclei to be stable: Nuclear force (remnant of the strong interaction) Short ranged force between the color neutral nucleons (similar to the Van-der-Waals force)

33 Nuclear Forces V 1fm R Repulsive core (e.g. ω exchange) Short ranged attraction (π exchange)

34 Stability of Nuclei Initially one would think that very heavy nuclei would only consist of neutrons, because there is no Coulomb repulsion. However, nature is different! Lets look systematically at all nuclei as a function of : A: Number of nucleons Z: Number of protons N: Number of neutrons Ø A = Z + N

35 Binding Energy Standard: 1 u = 1 / 12 m( 12 C) = 1.66 x g = MeV/c 2 = 1 atomic mass unit In these units we have M p = u = MeV/c 2 (R p, M p current research interest) M n = u = MeV/c 2 (lifetime current research interest) M e = x10-4 u = MeV/c 2 Accurate mass measurements show that the mass of an atom(-ic nucleus) is less than the sum of the constituents. Mass of an atom m(z,a) = Z M p + Z M e + (A-Z) M n B(Z,A)/c 2 = Z m H + (A-Z) M n B(Z,A)/c 2 Nuclear mass M(Z,A) = Z M p + (A-Z) M n B(Z,A)/c 2

36 Mass Defect From these equations we get a relation between the atomic and the nuclear masses M(Z,A) = m(z,a) - (Z M e B e (Z)/c 2 ) The binding energy of the shell electrons is approximately given by: B e (Z) = Z 7/3 ev This can be neglected compared to the term Z M e, so that B(Z,A) = B (Z,A) A further useful quantity is the mass defect (excess) Δ(Z,A) = M(Z,A) - A

Example: Hydrogen Burning in the Sun Hydrogen converts into 4 He via a number of reactions The mass of 4 H-atoms is 4.

37 Example: Hydrogen Burning in the Sun Hydrogen converts into 4 He via a number of reactions The mass of 4 H-atoms is u The mass of 4 He-atoms is u If we ignore positrons and neutrinos then the mass defect is u (0.7 % of u) Thus, the binding energy of 4 He 26.7 MeV

38 Q-Value This denotes the amount of energy released by (or needed to initiate) a reaction: Example: 6 3 Li + n à 4 2 He + 3 1H + 4,8 MeV Q-Value Q-value is positive for exothermic reactions Given by the difference in the kinetic energy of the particles in the initial to final states Also calculated from the mass difference 38

39 Binding Energies Fission 39

40 Binding Energies The systematics of the binding energy show: Binding energy per nucleon B/A is roughly constant by 8 MeV. That implies a saturation of the nuclear forces (equivalently a short range) B/A drops for light and heavy nuclei, thus energy can be won by fusion and fission At certain numbers there are anomalies. That hints at shell effects, similar to the atomic shell. There are no stable A=5 nuclei -> Origin of A>5 elements?!

41 End of first lecture 41

The Sun Closest star to Earth - only star that we can see details on surface - easily studied Assumption: The Sun is a typical star

The Sun Closest star to Earth - only star that we can see details on surface - easily studied Assumption: The Sun is a typical star Why is the Sun hot and bright? Surface Temperature of the Sun: T =