Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Nuclei • Protons and neutrons in nuclei separately fill their energy levels: 1s, 1p, 1d, 2s, 2p, 2d, 3s…………… (we’ll see in 461 their ordering and split by total J) • often easier to analyze as 2 Fermi gases of (mostly) non-interacting particles • density ~1/fm3 slightly higher for neutrons large A p # p 4 3 R 3 Nuc n # n 4 3 R Nuc 3 • proton shifted higher due to Coulomb repulsion. Both p,n fill to top with p<->n coupled by Weak interactions so both at ~same level (Fermi energy for p impacted by n) n p e n p n e p e n 238 U 92 p,146n p P460 - Quan. Stats. III 1 Nuclei • gives Fermi momentum ~same for all except H pF 2mEF 200 MeV / c • if p,n were motionless, then the energy thresholds for some neutrino interactions are: e p n e E 1.8 MeV n p E 120 MeV • but Fermi momentum allows reactions to occur at lower neutrino energy. p n e Ethrsh 1.8MeV mp Ep pp 1.4MeV P460 - Quan. Stats. III 2 Nuclei: Fermi Suppression • But also have filled energy levels and need to give enough energy to p/n so that there is an unfilled state available. Simplest to say “above” Fermi Energy • similar effect in solids. Superconductivity mostly involves electrons at the “top” of the Fermi well e p n e n p e e • at low energy transfers (<40 MeV) only some p/n will be able to change states. Those at “top” of well. • Gives different cross section off free protons than off of bound protons. • In SN1987, most observed events were from antineutrinos (or off electrons) even though (I think) 1000 times more neutrinos. Detectors were water….. P460 - Quan. Stats. III 3 Fermi Gases in Stars • Equilibrium: balance between gravitational pressure and “gas” (either normal or degenerate) pressure • total gravitational Energy: GM (r ) Mass (r r ) E r R r (r ) 2 E G 4r dr 4r 2 (r )dr 0 0 r E 3 M 2G pressure cons tan t V 5 R • density varies in normal stars (in Sun: average is 1 g/cm3 but at r=0 is 100 g/cm3). More of a constant in white dwarves or neutron stars • will have either “normal” gas pressure of P=nkT (P=n<E>) or pressure due to degenerate particles. Normal depends on T, degenerate (mostly) doesn’t • n = particle density in this case P460 - Quan. Stats. III 4 Degenerate Fermi Gas Pressure • Start with p = n<E> • non-relativistic relativistic density states : D( p) p 2 both cases D( E ) E 1/ 2 N EF 0 E2 1/ 2 AE dE EF n 2 / 3 E EF 0 AE 2 dE n1/ 3 1/ 2 E E dE 1/ 2 E dE E 53 EF K1n 2 / 3 P K1n n 2 / 3 n 5 / 3 2 E E dE 2 E dE 34 EF K 2 n1/ 3 K 2 n n1/ 3 n 4 / 3 • P depends ONLY on density • Pressure decreases if, for a given density, particles become relativistic P460 - Quan. Stats. III 5 Older Sun-like Stars • Density of core increases as H-->He. He inert (no fusion yet). Core contracts • electrons become degenerate. 4 e per He nuclei. Electrons have longer wavelength than He Ee EHe thermal equilibriu m pe me mHe pHe e mHe mH He • electrons move to higher energy due to Pauli exclusion/degeneracy. No longer in thermal equilibrium with p, He nuclei • pressure becomes dominated by electrons. No longer depends on T Ptotal Pe PH PHe Pe PH PHe • allows T of p,He to increase rapidly without “normal” increase in pressure and change in star’s equilibrium. • Onset of 3He->C fusion and Red Giant phase (helium flash when T = 100,000,000 K) P460 - Quan. Stats. III 6 White Dwarves • Leftover cores of Red Giants made (usually) from C + O nuclei and degenerate electrons • cores of very massive stars are Fe nuclei plus degenerate electrons and have similar properties • gravitational pressure balanced by electrons’ pressure which increases as radius decreases ---> radius depends on Mass of star • Determine approximate Fermi Energy. Assume electron density = 0.5(p+n) density N M Sun 1 1 5 1035 e / m3 V m p 2 volume Earth EF (non rel ) EF (rel ) hc h2 N 3 2 / 3 8m V N 3 1/ 3 V 8 0.3 MeV 0.8 MeV • electrons are in this range and often not completely relativistic or non-relativistic---> need to use the correct E2 = p2 + m2 relationship P460 - Quan. Stats. III 7 White Dwarves + Collapse • If the electron energy is > about 1.4 MeV can have: e p n EThreshold 1.4MeV • any electrons > ET “disappear”. The electron energy distribution depends on T (average E) #e’s EF ET • the “lost” electrons cause the pressure from the degenerate electrons to decrease • the energy of the neutrinos is also lost as they escape ---> “cools” the star • as the mass increases, radius decreases, and number of electrons above threshold increases P460 - Quan. Stats. III 8 White Dwarves+Supernovas • another process - photodisentegration - also abosrbs energy “cooling” star. Similar energy loss as e+p combination 56Fe 134 He 4n He 2 p 2n 4 • At some point the not very stable equilibrium between gravity and (mostly) electron pressure doesn’t hold • White Dwarf collapses and some fraction (20-50% ??) of the protons convert to neutrons during the collapse • gives Supernovas LSN (light ) 109 LSun (light ) LSN (neutrinos ) 100(1000 ?) LSN (light ) P460 - Quan. Stats. III 9 Neutron Stars-approx. numbers • Supernovas can produce neutron stars - radius ~ 10 km - mass about that of Sun. always < 3 mass Sun - relative n:p:e ~ 99:1:1 • gravity supported by degenerate neutrons N 2M Sun V mn 1 44 3 3 6 10 / m . 6 / F 4 (10km) 3 3 separation 1.2 F • plug into non-relativistic formula for Fermi Energy -----> 140 MeV (as mass =940 MeV, non-rel OK) • look at wavelength h h 1240MeVF 2F p 2mE 2 940MeV 140MeV • can determine radius vs mass (like WD) • can collapse into black hole P460 - Quan. Stats. III 10 Neutron Stars • 3 separate Fermi gases: n:p:e p+n are in the same potential well due to strong nuclear force • assume independent and that p/n = 0.01 (depends on star’s mass) EF ( p) EF (n) VN 2/3 EF (e) EF (n) VN 14525MeV 6MeV 2 / 3 mn me 14525MeV 940 .5 10,000MeV ( wrong ) • need to use relativistic for electrons EF (e) N 3 1/ 3 V 8 hc 500MeV • but not independent as p <---> n n p e p n e p e n n e p • plus reactions with virtual particles • free neutrons decay. But in a neutron star they can only do so if there is an available unfilled electron state. So suppresses decay P460 - Quan. Stats. III 11 Neutron Stars • Will end up with an equilibrium between n-p-e which can best be seen by matching up the Fermi energy of the neutrons with the e-p system • neutrons with E > EF can then decay to p-e-nu (which raises electron density and its Fermi energy thus the balance) E F (e) E F ( p ) E F ( n ) 3n p 3ne 2 hc m p c 8 8 1/ 3 3nn 8 2/3 2/3 h2 2m p h2 mn c 2 2mn • need to include rest mass energies. Also density of electrons is equal to that of protons • can then solve for p/n ratio (we’ll skip algebra) • gives for typical neutron star: 2 1017 kg / m 3 nn 1 10 44 / m 3 ne n p nn / 200 P460 - Quan. Stats. III 12