Phys 321: Lecture 8 Stellar Remnants
Prof. Bin Chen, Tiernan Hall 101, [email protected] Evolution of a low-mass star
Planetary Nebula Main SequenceEvolution and Post-Main-Sequence track of an 1 Stellar solar Evolution-mass star
Post-AGB PN formation Superwind First He shell flash White dwarf
TP-AGB
Second dredge-up He core flash )
Pre-white dwarf L /
L E-AGB (
10 He core exhausted
Log RGB Ring Nebula (M57) He core burning First Red: Nitrogen, Green: Oxygen, Blue: Helium dredge-up H shell burning SGB Core contraction H core exhausted
ZAMS
To white dwarf phase 1 M
Log (T ) 10 e Sirius B FIGURE 4 A schematic diagram of the evolution of a low-mass star of 1 M from the zero-age main sequence to the formation of a white dwarf star. The dotted ph⊙ase of evolution represents rapid evolution following the helium core flash. The various phases of evolution are labeled as follows: Zero-Age-Main-Sequence (ZAMS), Sub-Giant Branch (SGB), Red Giant Branch (RGB), Early Asymptotic Giant Branch (E-AGB), Thermal Pulse Asymptotic Giant Branch (TP-AGB), Post- Asymptotic Giant Branch (Post-AGB), Planetary Nebula formation (PN formation), and Pre-white dwarf phase leading to white dwarf phase.
and becomes nearly isothermal. At points 4 in Fig. 1, the Schönberg–Chandrasekhar limit is reached and the core begins to contract rapidly, causing the evolution to proceed on the much faster Kelvin–Helmholtz timescale. The gravitational energy released by the rapidly contracting core again causes the envelope of the star to expand and the effec- tive temperature cools, resulting in redward evolution on the H–R diagram. This phase of evolution is known as the subgiant branch (SGB). As the core contracts, a nonzero temperature gradient is soon re-established because of the release of gravitational potential energy. At the same time, the temperature and density of the hydrogen-burning shell increase, and, although the shell begins to narrow significantly, the rate at which energy is generated by the shell increases rapidly. Once again the stellar envelope expands, absorbing some of the energy produced by the shell
Death of low-mass stars
• Crash course by Phil Plait The discovery of Sirius B
• In 1844, Friedrich Bessel found the brightest star in the night sky, Sirius, Sirius A has a companion, using precise parallax observations for ten years • In 1862, Alvan G. Clark discovered the “Pup” at its predicted position using his father’s new 18-inch refractor Sirius B • The Pup is very faint (~0.03 Lsun) so, probably cool and red? • In 1915, Walter Adams discovered that, however, the surface temperature is ~27,000 K! A hot, blue-white star!
Chandra X-ray image Appendix: Astronomical and Physical Constants
APPENDIX
APPENDIX Astronomical and Physical Constants
Physical Constants 11 2 2 Gravitational constant G 6.673(10) 10− Nm kg− APPENDIX = × 8 1 Speed of light (exact) c 2.99792458 10 ms− Astronomical and≡ Physical7 × 2 Constants Permeability of free space µ0 4π 10− NA− ≡ × 2 Permittivity of free space ϵ0 1/µ0c ≡ Astronomical Constants8.854187817 ... 10 12 Fm 1 = × − − Electric charge e 1.602176462(63) 1030 19 C Solar mass Estimating1M =radius1.9891 of ×Sirius10 −kg B Electron volt 1 eV ⊙ 1.602176462= (63×) 103 19 J 2 Solar irradiance S 1.365(2) 10 −Wm− Astronomical and= = Physical×× 34 Constants Planck’sSolar constant luminosity h 1L 6.626068763.839(52(5)) 1010−26 WJs =⊙ = ×× 15 Solar radius • Luminosity of 1RSirius4. 13566727B is 60.03.95508(16) (L26sun10) − 10eV8 m s ! = ⊙ = × 2×1/4 Solar effective temperature Te, h/2π L /(4πσR ) ≡⊙ ≡ ⊙ ⊙ 34 • Effective temperature1.054571596 is5777 27,000(822))K 10K− Js = = × 16 Astronomical Constants6.58211889(26) 10− eV s • Mass is ~1.05 solar= mass × 3 Planck’sSolar constant absolute bolometricspeed of light magnitduehc Mbol 1.239841864(.1674)30 10 eV nm Solar mass × 1M = 1=.9891 10 ×kg Solar apparent bolometric magnitude ⊙ mbol= 1240 eV nm26×.83 3 2 Solar irradiance • What is itsS radius?≃ 1=.365−(2) 10 Wm23 − 1 Boltzmann’sSolar apparent constant ultraviolet magnitudek U = 1.380650325(24×.91) 10 JK− Solar luminosity Astronomical1L = Constants3=.839−(5) 10× 26 −W Solar apparent blue magnitude B 8.617342326(153.10) 10 5 eV K 1 Solar mass ⊙ 1M== = 1−.9891× ×1030 −kg8 − Solar radiusStefan–BoltzmannSolar apparent constant visual magnitude 1Rσ V 2π65.k955084/(1526c(226.h753)) 10 m ⊙ ⊙ = × 3 2 Solar irradiance S≡= = 1−.365(2)2×110/48 Wm−2 4 Solar effectiveSolar temperature bolometric correction Te, BC 5.670400L= /(4(πσ400.)08R ×10) − 26Wm− K− Solar luminosity ⊙ 1L=≡ =⊙ 3.839− (5×)⊙ 10 W Radiation constant a ⊙ 4σ5777=/c (2) K × 8 EarthSolar mass radius Solar Radius: 1M1R== 65.95508.9736(2610) 241610kg m 3 4 ⊙ 7.565767= (54) 102−×1/4Jm− K− Solar effective temperature T= ⊕ = L /(4πσ××R ) 6 AtomicEarth mass radius unit (equatorial) Earth Radius:1u 1Re, 1.660538736.378136(13) 101027mkg ⊙⊕ ≡ ⊙ ⊙ − Solar absolute bolometric magnitdue Mbol = = 4.577774 (2) K×× 2 = 931=.494013(37) MeV/c2 4 Solar apparentAstronomical bolometric unit magnitudeUsing Stefanmbol -Boltzman’s1= AU Law26.83:1.4959787066 31 1011 m Electron mass me = 9.10938188−= L(72=) 4π10R−×σkgT Solar apparentLightSolar ultraviolet (Julian) absolute year magnitudebolometric magnitdueU 1M= lybol 25.919.4607304724.74 × 41015 m 5.485799110== This means(12) compressing10×− u the mass of the ParsecSolar apparent bolometricWe have magnitude R ~ 5.51m x= pcbol =106 m − 20626426.83 .806× AU27 Solar apparentProton mass blue magnitude BmWDp 1.67262158==26.−10 (13) 10− kg Solar apparent ultraviolet magnitude U== − entire325.0856776 .Sun91 within× 10 a 16volumem < Earth! Solar apparent visual magnitude V 1.00727646688==26.−75 (13)×u Solar apparent blue magnitude B== − 326.2615638.10 ly (Julian27 ) Neutron mass mn 1.67492716= − (13) 10− kg Solar bolometricSolar correction apparent visual magnitudeBC V= = 0.0826.75 × = 1.00866491578− (55) u = − h m s SiderealSolar bolometric day correction BC= 230.5608 04.090530927 Hydrogen mass mH 1.673532499== − (13) 10− kg Earth mass Solar day 1M = 5.97368640010 s24×kg 1.00782503214= (35) u SiderealEarth mass year ⊕ 1M== 53.9736.15581450× 10624 kg107 s Earth radiusAvogadro’s (equatorial) number 1RN 6.022141996.378136(47)10 10m23 mol 1 A ⊕ ⊕ == × ×6 − Earth radius (equatorial) 1R== 6365.378136.256308× × 10 d1 m 1 Gas constant R ⊕ 8.314472== (15) J mol× − K− 7 Tropical year = 2 3.1556925192 1110 s AstronomicalBohr radius unit 1a0 AU, 4πϵ1=.04959787066! /mee 10× 11m Astronomical unit ∞ 1≡ AU= 1365.4959787066.2421897× d 1011 m 5.291772083== (19) 10×157 m Light (Julian)JulianLight year year (Julian) year 1 ly 1= ly 9.46073047293.460730472.1557600×101010− 15ms m a = (m=≡/µ)a ××× Parsec Parsec 10 pc,H 1 pc 206264e 2062643650,.806.25.806 d AU AU ≡ ≡ ∞ 11 Gregorian year = 5.294654075= 3.1556952(20) 16101016−7 sm = 3.408567763.08567763 2!3 10× 10 mm Rydberg constant R me=≡e /64π ϵ0 c ×× ∞ ≡= 3365.2615638.2425× lyd (Julian) 1.09737315685493=≡.2615638 ly ((Julian83) )107 m 1 Note: Uncertainties in the last digits are indicated== in parentheses. For instance, × − RH (µ/m8 e)R h m s theSidereal solar radius, day 1 R , has an uncertainty of 0.≡00026 10 m.23∞56 04.0905309 Sidereal day ⊙ ± 1×.0967758323= h56m04(13.0905309) 107 ms 1 Solar day = 86400 s × − Note: Uncertainties in the last digits are indicated in parentheses.= = For instance, the universal Solar day Sidereal year 864003.15581450 s 107 s 11 2 2 gravitational constant, G, has an uncertainty of 0.010= 10 =Nm kg− . × 7 Sidereal year ± × −3.15581450365.25630810 d s = = × Tropical year 365.2563083.155692519 d 107 s = × = 365.2421897 d 7 Tropical year 3=.155692519 10 s Julian year = 3.1557600× 107 s 365≡ .2421897 d× = 365.25 d ≡ 7 Julian year Gregorian year 3.15576003.15569521010s7 s ≡ ≡ × × 365.25365 d.2425 d ≡ ≡ 7 Gregorian yearNote: Uncertainties in the last digits are indicated in parentheses.3.1556952 For instance,10 s ≡ × the solar radius, 1 R , has an uncertainty of 0.00026 365108 m..2425 d ⊙ ± ≡ × Note: Uncertainties in the last digits are indicated in parentheses. For instance, From Appendix A o f An In tro du ctio n to Mo dern Astro p hysics, Second Edition, Bradley W. Carroll, Dale A. Ostlie. Copyright © 2 007 the solar radius, 1 R , has an uncertainty of 0.00026 108 m. by Pearson Education, Inc. Published by ⊙Pearson Addison-Wesley.± All× rights reserved.
From Appendix A o f An In tro du ctio n to Mo dern Astro p hysics, Second Edition, Bradley W. Carroll, Dale A. Ostlie. Copyright © 2 007 by Pearson Education, Inc. Published by Pearson Addison-Wesley. All rights reserved.
From Appendix A o f An In tro du ctio n to Mo dern Astro p hysics, Second Edition, Bradley W. Carroll, Dale A. Ostlie. Copyright © 2 007 by Pearson Education, Inc. Published by Pearson Addison-Wesley. All rights reserved. APPENDIX
AstronomicalAppendix: Astronomical and and Physical Physical Constants Constants
Properties of white dwarfs
AstronomicalDensity Constants Solar mass 1M 1.9891 1030 kg ⊙ = × 3 2 Solar irradiance Sirius B SR ~ 5.5 x 106 m1.365(2) 10 Wm− PhysicalWD Constants= × 26 Solar luminosity 1L 3.839(5) 1110 2W 2 Gravitational constant G ⊙ = 6.673(10) 10×− Nm kg8 − Solar radius 1R = 6.95508× (268) 101 m Speed of light (exact) c 2.99792458 10 ms− ⊙ ≡ = 7 × 2 2×1/4 Solar effectivePermeability temperature of free space Tµe,0 9 4π L-103 /(− 4NAπσ−R ) ρ ⊙~ 3 x≡ 10≡ kg m× ⊙2 ⊙ Permittivity of free space ϵ0 1/µ57770c (2) K ≡ 12 1 = 8.854187817 ... 10− Fm− = × 19 Electric charge Surfacee gravity: Phil1.602176462 said 100,000x(63) that10− ofC the Earth’s gravity = × 19 Solar absoluteElectron bolometric volt magnitdue M1 eVbol 1.6021764624.74(63) 10− J = = × 34 Solar apparentPlanck’s bolometric constant magnitude mh bol 6.6260687626.83(52) 10− Js = = − × 15 Solar apparent ultraviolet magnitude U 4.1356672725.91(16) 10− eV s ! = M × = h/2π− 6 -2 Solar apparent blue magnitude B g≡ G 26.10 ~ 4.6 x 3410 m s = 1.0545715962 (82) 10− Js = = − × Solar apparent visual magnitude V 6.58211889R26.75(26) 10 16 eV s = × − Planck’s constant speed of light hc = 1.23984186− (16) 103 eV nm Solar bolometric correction× BC = 0.08 × = 1240− eV nm ≃ Boltzmann’s constant k 1.3806503(24) 10 23 JK 1 = × 24− − Earth mass 1M 8.61734235.9736(153)1010 kg5 eV K 1 ⊕ = = × × − − Earth radiusStefan–Boltzmann (equatorial) constant 1Rσ 2π 56k4.378136/(15c2h3) 106 m ≡ ⊕ = 5.670400(40) ×10 8 Wm 2 K 4 = × − − − Radiation constant a 4σ/c 11 Astronomical unit 1 AU = 1.495978706616 103 m4 7.565767(54) 10− Jm− K− = = × ×27 15 Light (Julian)Atomic year mass unit 11u ly 1.660538739.460730472(13) 10− 10kg m = = × × 2 Parsec 1 pc 931.206264494013(37.806) MeV AU/c = 31 Electron mass me = 9.10938188(72) 10− 16kg = 3.0856776× 10 4 m = 5.485799110(12) ×10− u = × 27 Proton mass mp 1.672621583.2615638(13) ly10(Juliankg ) = = × − 1.00727646688(13) u = 27 Neutron mass mn 1.67492716h m(13) 10 kgs Sidereal day = 23 56 04.×0905309− = 1.00866491578(55) u Solar day = 86400 s 27 Hydrogen mass mH 1.673532499(13) 10 kg = = × − 7 Sidereal year 1.007825032143.15581450(35) u 10 s = = ×23 1 Avogadro’s number NA 6.02214199365.256308(47) 10 d mol− = × 1 1 Gas constant R = 8.314472(15) J mol− K− 7 Tropical year = 3.1556925192 2 10 s Bohr radius a0, = 4πϵ0! /mee × ∞ ≡ 5.291772083365.2421897(19) 10 d 11 m = = × −7 Julian year a0,H (me3/µ)a.15576000, 10 s ≡ ≡ ∞ × 11 5.294654075365.25( d20) 10− m = 4 3 2!3 × Rydberg constant R ≡ mee /64π ϵ0 c 7 Gregorian year ∞ ≡ 3.1556952 10 s 1.0973731568549(83) 107 m 1 = ≡ × × − RH (µ/m365e)R.2425 d ≡ ≡ ∞ 7 1 Note: Uncertainties in the last digits are indicated in parentheses.1.09677583 For instance,(13) 10 m− = × the solar radius,Note: 1 Uncertainties R , has an in uncertainty the last digits of are indicated0.00026 in parentheses.108 m. For instance, the universal 11 2 2 gravitational⊙ constant, G, has an uncertainty± of 0.010× 10 Nm kg− . ± × −
From Appendix A o f An In tro du ctio n to Mo dern Astro p hysics, Second Edition, Bradley W. Carroll, Dale A. Ostlie. Copyright © 2 007 by Pearson Education, Inc. Published by Pearson Addison-Wesley. All rights reserved. Central conditions of WDs
• Using the temperature gradient equation due to radiative energy transfer (p. 160 of textbook), the central temperature of WDs is ~ several times 107 K • Hot enough for H fusion, sometimes also He • Not enough for triggering carbon and oxygen fusion • WDs consist primarily of completely ionized helium/carbon/oxygen/neon/magnesium nuclei
Initial mass of the star: (< 0.5 Msun) => He white dwarf (0.5 - 5 Msun) => Carbon/Oxygen white dwarf (5 -7 Msun) => Oxygen/Neon/Magnesium white dwarf The Degenerate Remnants of Stars
. − × . ×
2 WHITE DWARFS
white dwarf
Classes of White Dwarf Stars Central Stellar pressure PulsationStellar Pulsation of WDs From hydrostatic equilibrium, assuming a constant density DA (which white is unrealistic) dwarfs dP DBGM whiter ρ dwarfsG πr ρ ρ dP GMr ρ G πrπGρρ ρr. DC whitedr dwarfs= − r = − r = − π Gρ r. ! " dr DQ= white− r dwarfs= − ! r " = − DZ white dwarfs P = Integrate r from r to r = R (surface), where P (r = R) = 0, the pressureP becomes Central Conditions in White Dwarfs = r P (r) πGρ R r . = − M R ! " r The central pressure Pis (r) πGρ R r . = = − R R dr dr ! " Pc πGρ R . − , Π ≈ ≈ × , ≈ vs ≈ # # γπGρ R r About 1.5 million timesR larger than theR pressure− at the center of the Sun dr$ dr Π ! " , ≈ v ≈ # s # γπGρ R r − π $ Π . ! " ≈ % γ Gρ
π Π . period– mean density relation ≈ % γ Gρ . period– mean density relation± M R = ⊙ = ⊙ Π ≈ Radial Modes of Pulsation . ± radial modes standing waves M R = ⊙ = ⊙ Π ≈ node antinode fundamental mode Radial Modes of Pulsation first overtone radial modes standing waves first harmonic node antinode fundamental mode first overtone first harmonic
The physics of degenerate matter
• The Pauli Exclusion Principle allows at most one fermion to occupy each quantum state • Everyday gas at room temperature and pressure, 1 of every 107 quantum states is occupied by a gas particle • Degeneracy is insignificant • Pressure is dominated by thermal pressure • When all the electrons (fermions) are squeezed together due to high density and pressure, electron degeneracy pressure becomes important The Degenerate Remnants of Stars
N Nx Ny Nz = = = N Nx Ny Nz N N > N > =N > + + x y !z
Ne πN . = " #" # N The Degenerate Remnants of Stars / Ne N . N Nx Ny Nz The Degenerate Remnants= of Starsπ = = = " # N N N N x y z N N x N y Nz N N > N > =N > + + = = = x y !z ! / N N N N x y z εF π n , = +Ne + πN . N = m Nx > Ny > Nz > = ! " #" # $ % N m n Ne/L ≡ / εF N Ne πN . N e . = = π " #" # " # The Condition N for DegeneracyFermi Energy / εF ! N / e εF π n , N . = m • Fermi energy= π : the maximumT> energy of any " #εF $ % electron in a completely m degenerate gasn atNe/L T = 0 The Degenerate Remnants≡ of Stars εF ! T / εF π n The, Condition for Degeneracy = m εF $ % Z ρ ne , = = A mH T> m " Mass of electron#"n NNumbere/L density# "of electron# ≡ εF Z A If expressed in mass density ρ of fully T mH ionized gas / # of protons per nuclei The Condition for Degeneracy F ! / Z ρ FIGURE 5 ε ZT ρ ε π . = F ε with εF neT> εF , = m A m ≤ = = A mH e & " # H ' " #" # " # # of nucleons per nuclei Z T>A mH L L L λx , λy , λz , / = Nx = Ny = Nz Nx Ny Nz ! Z ρ / ε π . F hN hN hN = mx e yA mH x px ,p& y " ,p# y ' . = L = L = L p Z ρ ε , = m ne , = = A pm Hp p p " #" # " # = x + y + z h h N ε (N N N ) , Z A = mL x + y + z = mL mH N N N N N ≡ x + y + z (Nx ,Ny ,Nz) / Nx Ny Nz ms / ! / =± Z ρ Nx Ny Nz four εF π . = m A m e & " # H '
Condition for degeneracy
The Degenerate Remnants of Stars
If the average thermal energy of an electron 3/2kT, is smaller than kT the k kT < ε Fermi Energy F The Degenerate Remnants of Stars The Degenerate3 Remnants of Stars / kT < ε ! F Z ρ kT kT < π , 2 kT km e A m H kT < ε ! " # $ F k kT < εF An average electron is unable to make a transition to an unoccupied quantum state to “break” the degeneracy, and the electron gas will stay ! / T π Z / / degenerate < − ! Z ρ ρ / m !k m ZA ρ = / kT < π , kT < e ! πH " #$ , m A m m A m e ! " # H $ e ! " # H $ Z/A . = / − , ! / ! / T For π Z/A Z= 0.5 / T π D ≡Z < / − < ρ / − m k m A ρ / m k m A = e H = e ! H " #$ ! " #$ Z/A . Z/A . = = T / The condition for degeneracy becomes < / . where − , ρ / − D, D ≡ D ≡ T/ρ / T T < . < . Example 3.1. ρ / D ρ / D Tc . ρc . = × = × − T/ρ / T/ρ /
Tc / − > . Example 3.1. / = Example 3.1. D ρc Tc . ρc . Tc . ρc . = × = × = × = × − − Tc / Tc − > . / evolve, electron degeneracy / will become increasingly important /(Fig. 6). The Sun will− > . ρc = D = D develop a degenerate helium core while on the red giant branchρc of the H–R diagram, leading eventually to a core helium flash. Later, on the asymptotic giant branch, the progenitor of a carbon–oxygen white dwarf will form in the core to be revealed when evolve,the Sun’selectron surface degeneracy layers will are become ejected increasingly evolve,as a planetary electron important nebula. degeneracy (Fig. 6). will The become Sun will increasingly important (Fig. 6). The Sun will develop a degenerate helium core while ondevelop the red a giant degenerate branch heliumof the H–Rcore diagram,while on the red giant branch of the H–R diagram, leading eventually to a core helium flash.leading Later, oneventually the asymptotic to a core giant helium branch, flash. the Later, on the asymptotic giant branch, the progenitor of a carbon–oxygen whiteTc dwarf will form in the core to be revealed when progenitor of a/ carbon–oxygen, white dwarf will form in the core to be revealed when the Sun’s surface layers are ejected as / a planetary nebula.− ρc =the Sun’s surface≪ layersD are ejected as a planetary nebula. T c / T / − , c / ρc = ≪ D / − , ρc = ≪ D
The Degenerate Remnants of Stars
kT k kT < εF
! Z ρ / kT < π , me A mH ! The" Degenerate# $ Remnants of Stars kT ! / T π Z / T ! π Z / T < / / − ρ < . mek mH A = ρ / D ! " #$ Z/A . = / T/ρ / − , D ≡ Example 3.1. Tc . ρc . = ×T = × − < . ρ / D Tc / − > . / = D Degenerateρ c T/ρ / or not Example 3.1. Tc . ρc . evolve, electron degeneracy will become increasingly important (Fig. 6). =The Sun× will = × develop a degenerate helium core − while on the red giant branch of the H–R diagram, leading eventually to a core helium flash. Later, on the asymptotic giant branch, the Tc / progenitor of a carbon–oxygen white dwarf will form / in the core to be− revealed> . when ρ = The Degenerate RemnantsD of Stars the Sun’s surface layers are ejected as a planetary nebula.c For Sirius B: evolve,T celectron degeneracy / will become increasingly important (Fig. 6). The Sun will − , / developρc a =degenerate helium core≪ D while on the red giant branch of the H–R diagram, leading eventually to a core helium flash. Later, on the asymptotic giant branch, the progenitor of a carbon–oxygen white dwarf c will form in the core to be revealed when T the Sun’s surface layers are ejected as a planetary nebula. Degeneracy in Tthec Sun’s / − , center as it evolves / = ≪ D ρc c FIGURE 6 Ap. J. 311 Electron Degeneracy Pressure !x !px !, ≈ p !p p P nepv, ≈ ne ne The Degenerate Remnants of Stars The Degenerate Remnants of Stars c T The Degenerate Remnants of Stars c T N Nx Ny Nz = = = N Nx Ny Nz N N > N > =N > + + x y !z c Ne πN . = FIGURE 6 " #" # Ap. J. 311 N c / Electron Degeneracy Pressure Ne N . FIGURE 6 = π The Degenerate Remnants of Stars Ap. J. 311 " # / ne− Electron Degeneracy Pressure ! / / The Degenerate Remnants of Stars εF π n , !x ne− Electron = m degeneracy pressure ≈ $ % / ne− m n Ne/L! ! x !p ≡x , εF / ≈ !!x ne− ≈ / px !p x !n e ≈ ≈ !x ≈ The Condition for Degeneracy p !p The Degenerate Remnants of Stars !x ! px !, ! εF / ≈ px !px !n / ≈ ≈ !x ≈ e ne− For a degeneracy gas, electrons are T> Pressure in a gas associated packed as tightly as possible, so the p px py pz , / p !p = ! x= ne− with particle’s momentum separation between electrons≈ is just 1/3 ne . Heisenberg’s uncertainty principle P nepv, p p p , ≈ gives x = y = z ! n ! / e p 1D: px !px ne Z ρ p ≈px p≈y !xpz≈ px , ne , = + + = = = A mH " ne #" # " # P nepv,3D: p p p p p , Z A ≈ = x + y + z = x mH ne or p √ p . / p p xp , x = y = z p √ p . ! / n e x Z ρ = εF π . = m A m e & " # H ' / Z ρ pp √ p! p p p / ,. ≈ = x + Ay +Z mz =ρ x p √ !"! # H $ . ≈ A mH !" # $ p √ px . p = p v =v me = me Z ρ / √√ !!! p √ / / . ≈ neneA mH ≈ ≈meme !" # $ / √ √! ! ZZ ρρ / p . . v≈ ≈m me AA mmH = mee !"!" ## H $$ √ ! / ne ≈ me ! Z ρ / P √! ! Z ρ // . ≈ m Z A ρm P e !" # H $ .. ≈≈ mme AA mmH e !" !" ## H $$ P / P ! / ! π Z ρ / P P ne ,. = / m ≈ me % A& !m He π!" # $/ P ne , = % & m e P / ! π / P ne , = % & me The Degenerate Remnants of Stars The Degenerate Remnants of Stars / − c ne T / !x ne− ≈ The Degenerate Remnants of Stars N Nx Ny Nz ! = = = / px !px !n e N N N N ≈ ≈ !x ≈ = x + y + z N Nx > Ny > Nz > ! c Ne πN . FIGURE 6 = px py pz , " #" # Ap. J. 311 = = N N / N e . Electron Degeneracy Pressure TheThe Degenerate Degenerate Remnants Remnants= of of Stars πStars " # p p p p p , x y / / The z Degeneratex Remnants of Stars = + n−ne−+ = e / ne− ! / / !x ne− / ε!F x ne− π n , = ≈ ≈m p √ p x . / = !x ne− $ % ≈ m n Ne/L !≡ ! / εF px !px !n / px !px !n e Z !ρ / ≈≈! ≈ ≈! !xx≈≈ e !!x !px , / p √ p.x !px !n ≈ A ≈m ≈ ≈ ! x ≈ e !" #TheH Condition$ for Degeneracy εF p x p y p z , px =py =pz , p p !p = = T> v p p Thep Degenerate , Remnants of Stars = m x y z e = = Electron degeneracy / pressure √ ! ne− / p p p p p , ne x y z x p m p =px +py + pz = px , / ≈ e = + + = !x ne− p p p p p , ≈ / = x + y + z = x Z ρ √ ! Z ρ ne , P nepv, . = √ = A mH " p #" px . # " # ≈ m≈e A mH p =√ px . !" # $ = ! Z A √ p ! p !n / n p px . x x e e mH = ≈ ≈ ! x ≈ / / Z ρ ! / ne p √ Z ρ . / ≈√ ! ! A m / ! p / HZ . ρ Z ρ ≈Z εF ρ !"A # πmH $ . P p. √ ! =!" me #. $A m H ≈ m A m ≈ A m H & p"x #py ' pz , e !" # H $ !" # $ = = p The Degenerate Remnants of Stars v p v = me Exact calculation gives: P p v = me √ ! / = me /p px py pz px , ! √ ! ne = + + = / π / / ≈ me / P π ! Z ρn , √ ! ne P e . ≈ / me = me n / = % me & A mH e ! ! " #$ % & ≈ me √ Z ρ √ ! p / √. p . ≈ m ZA ρm x ! e / H = . Z/A . !~ #$/& √ ≈ Z m ρ !" A #m $ = For nonrelativistice . caseH ≈ m A m !" # $ e H . − !" # $ / × Z ρ Electron degeneracy pressure is responsible for maintaining hydrostatic √ ! p / . equilibrium in a white dwarf. ! Z≈ ρ A mH P /!" . # $ ≈!Pm KρZ A/ ρm e / H n . ! P Z =ρ !" # $ . ≈ me A mH = P !" #. $ ≈ me A mH p !" # $ v P = me P / ! ! Pπ √ / / /n P ! ne e, = π ≈ mme e π P / ! % & n / , = / m e 4 THE CHANDRASEKHAR LIMIT P % ne& , √e ! Z ρ / = % & me . ≈ me A mH !" # $ there is a maximum mass for white dwarfs ! Z ρ / P . The Mass–Volume Relation ≈ m A m e !" # H $ R M P / / π ! π ! Z ρ / P n / , πGρ R . = m e = m A m % & e ! " e #$ % H & ρ M / πR = ( π) / ! Z / R . ≈ / A m GmeM #$ % H & R . ⊙ ≈ × M R = M V . = The Degenerate Remnants of Stars / π ! Z ρ / P . = m A m ! " e #$ % H & Z/A . = The Degenerate Remnants of Stars . − × Electron degeneracy pressure is responsible for maintaining hydrostatic The Degenerate Remnants of Stars equilibrium in a white dwarf. / / π ! Z ρ / P . P Kρ = m A m = n . ! " e #$ % H & = / π ! Z ρ / The DegenerateZ/A PRemnants . of Stars . = = m A m ! " e #$ % H & . − × Z/A . Electron / degeneracy pressure is responsible for maintaining hydrostatic = π ! Z ρ / equilibrium inP a white dwarf. . = m A m . ! " e #$ % H & P Kρ / − = × Electron n degeneracy . pressure is responsible for maintaining hydrostatic 4 THE CHANDRASEKHAR Z/A LIMIT . = equilibrium in= a white dwarf. / . − P Kρ × Electronn . degeneracy pressure is responsible for maintaining hydrostatic = equilibrium in a white= dwarf. / P Kρ n . there is a maximum mass for white = dwarfs = 4 THE CHANDRASEKHAR LIMIT The Mass–Volume Relation 4 THE CHANDRASEKHAR LIMIT R there is a maximumM mass for white dwarfs 4 THE CHANDRASEKHAR LIMIT The Mass–Volume Relation there is a maximum mass for white dwarfs there/ is a maximum mass for / white dwarfs π ! R Z ρ M πG ρ R . Mass -Volume= Relationme A mH TheThe Mass–Volume Mass–Volume Relation Relation ! " #$ % & ρ M / πR R R / M M = π ! Z ρ / By setting the central pressure π Gρ R to equal to electron . = m A m degenerate pressure: ! " e #$ % H & / / ! / / ( π) π ! / ZZ ρ / ρ MπG ρ/ R π R ! . R = π Z ρ . πGρ R= / me A mH . ≈ Gm ! eM" #$A % mH& = ! "#$me% A &mH ρ M / πR #$ % & / ( π) / ! Z = Use ρ M / πR RR . / . ⊙ = ≈ Gm M A mH / ≈ × e / ( π) ! Z #$ % & M R R . / = ≈ Gm M A mH / we have a mass -volume( relation:π) / e ! #$ R% Z .& ⊙ R ≈ × . / M R ≈ R Gm . eM A mH = ⊙ ≈ × #$ % & M V . M R = = R . ⊙ ≈M V× . = M R M V . = More massive white dwarfs= are smaller! M V . = The Degenerate Remnants of Stars smaller mass–volume relation ρ M ∝ The Chandrasekhar Limit − • Packing more mass onto the WD will decrease the volume of the WD • Density increases significantly ! / Z ρ v . − , ≈ m A m = × e !" # H $ • v goes toward the speed of light -> relativistic effect kicks in • Mass-volume relation in the non -relativistic limit breaks WDs collapse to smallerzero volume at M Nonrelativistic ~ 1.4 M sun Chandrasekhar Limit DynamicalRelativistic Instability No WDs > 1.4 Msun have been found so far ρ < − P Kρ / K γ / = nonrelativistic limit. As we discussed in Section 14.3, this means that the white= dwarf is dynamically stable. If it suffers a small perturbation, it will return to its equilibrium structure instead of collapsing. However, in the extreme relativistic limit, the electron speed v = c must be used instead of Eq. ( 10) to find the electron degeneracy pressure. The result is / π Z ρ / P !c = A m % & !" # H $ γ / dynamical instability = − γ / Fate of WDs The Degenerate Remnants of Stars • No energy source in WDs • The luminosity of WDs is from their residual internal energy L L • In time, WDs cool down, and luminosity decreases • Reaches its “eternal death”: a cool, dark, Earth-sized crystallized carbon/oxygen FIGURE 9 ⊙ “Diamonds in the sky” Ap. J. Lett. 315 Crystallization L /L − ⊙ ≈ kT The Fate of Massive Stars FIGURE 10 Supernovae and neutron stars iron core • At the end of the evolution of high mass stars iron peak (M > 8 Msun), stars will burn their fuel down to The Fate of Massive Stars iron (recall Lecture 7) ⊙ • The core’s mass exceeds M > 1.4 Msun (Chandrasekhar Limit), electron degeneracy is The Fate of Massive Stars no longer enough to keep gravity at bay photodisintegration • Tc ρc − Matter is crushed to form neutrons ∼ × ∼ ⊙ γ n Photodisintegration + → + γ p+ n. + → + FIGURE 10 Neutron formation p+ e− n νe. + → + • Supernova occurs, left with a bare neutron star . . ⊙ × iron core × . . iron⊙ peak ⊙ ⊙ ⊙ ⊙ photodisintegration − γ n + → + γ p+ n. + → + Example 3.2. R ⊕ . . ⊙ ⊙ ⊙ ⊙ GM U . = − R E = Rf R = ≪ ⊕ E R M f . . ≃! G ≃ × ≃ ⊙ Neutron Stars • A 1.4 solar-mass neutron star consists of 1057 neutrons packed together by gravity and supported by the neutron degeneracy pressure • It is literally a giant nucleus with a mass number A ~ 1057! • Uranium is a pretty heavy element, right? • A = 235 for U 235 and A = 238 for U 238 • Radius of neutron stars Homework problem For a 1.4 solar-mass neutron star, this yields 4.4 km. The actual radius is ~10 km Neutron stars • For a 1.4 solar-mass neutron star with 10-km radius: • Average density: 6.65 x 1017 kg m-3 • Comparing to typical density of an atomic nucleus, 2.3 x 1017 kg m-3 • Even denser than the atomic nucleus! • All neutrons are “touching” each other, bounded by gravity For white dwarfs How much does it weigh on neutrons stars (in tons)? Comparing it to the weight of Mt. Everest ~ 1 billion tons • Surface gravity is g ~ 1.86 x 1012 m s-2, 190 billion times stronger than Earth’s surface gravity Chandrasekhar Limit for neutron stars • Like white dwarfs, neutron stars also obey a mass-volume relation • However, it fails for more massive neutrons stars as there is a point beyond which neutron degeneracy pressure can no longer support the star • Mns < ~2.2 Msun for static neutron star • Mns < ~2.9 Msun for rapidly-rotating neutron star Most massive neutron star found • More massive neutron stars would to date is ~2 solar mass collapse and become black holes Discovery of Pulsars Jocelyn Bell Burnell The first pulsar discovered by Jocelyn Bell When first discovered, Bell dubbed it “Little Green Man 1” Pulsar characteristics • Short periods: ~0.25–2 s • Pulse periods are extremely well defined. Some has the accuracy that challenges the best atomic clocks • Pulse period increase very gradually in time (slow down), typically dP/dt ~ 10-15. The characteristic “spin-down” time P/(dP/dt) ~ a few x 107 years • Possible pulsar models: • Binary stars Was the world’s first all- • Pulsating stars electronic digital watch (1970) • Rotating stars Pulsars as rotating neutron stars Maximum angular velocity a fast spinning star can achieve If ω > ωmax, the star would fly apart! Centripetal Gravitational acceleration acceleration Shortest period is • Pmin ~ 7 s for Sirius B – too long to account for the observed pulsar periods! -> white dwarfs would be torn apart simply from this rapid rotation! -4 • Pmin ~ 5 x 10 s for 1.4 solar-mass neutron stars -> neutron stars are safe! Question: Why a rotating neutron star can account for the extremely precise periods? Pulsar and Crab Nebula • Crab Nebula is known as a supernova remnant (recall the AD 1054 Chinese record on “Guest Star”) • A neutron star (or none) is expected as the remnant of the supernova • A pulsar was discovered in 1968 in the Crab Nebula (PSR 0531-21) with P ~ 0.0333 s • No other stars can survive at this rotation period other than neutron stars • Pulsar in the Crab Nebula is a neutron star Powering the Crab Nebula • The Crab Nebula is bright in radio, infrared, optical, to X-rays • Initial energy released in the supernova is not enough to account for its current emission • The Crab Nebula needs a replenishment of magnetic field and relativistic electrons • The total power needed for accounting for the current nebula 5 emission is ~10 Lsun Powering the Crab Nebula • Crab pulsar rotation period is P = 0.0333 s, its spin- down rate is dP/dt ~ 4 x 10-13 s/s, or 10-5 s/yr • It is losing its rotational kinetic energy! • This lost energy goes into powering the Crab Nebula Rotational kinetic energy Energy loss rate For Crab, assuming a uniform sphere with 1.4 solar mass and 10 km 31 5 Inserting , dK/dt ~ 5 x 10 W ~ 10 Lsun! Pulsar model Black Holes • Crash course by Phil Plait on black holes If a collapsing stellar core exceeds 3 solar mass, nothing, even the neutron degeneracy pressure, cannot stop the core from collapsing, and it becomes a black hole! Escape Velocity Total mechanical energy of a particle 1 Mm Initial state: E = mv2 − G i 2 i r If the particle can escape to infinitely large distance Final state: v > 0 r = ∞ Gravitational potential energy is zero 1 2 Mm 1 2 E f = mv f − G = mv f > 0 2 r∞ 2 1 Mm From energy conservation E = mv2 − G = E > 0 i 2 i r f 2GM So in order for the particle to escape v > v = i esc r Escape Velocity on different objects What if, the escape velocity = c? That means only light can barely escape! 2GM v = = c Wrong derivation, but correct result esc r r = R = 2GM / c2 = 3km(M / M ) Schwarzschild Radius s ⊙ If an one solar-mass star packs all its material under the Schwarzschild Radius (3 km), even light cannot escape from its surface! This is smaller than the radius of a neutron star. For this reason, a star that has collapsed down below R s is called a black hole It is enclosed by the event horizon, the spherical surface at r = R s , within which nothing can escape, and is completely out of reach Schwarzschild radius The curvature of spacetime Curved space by massive objects The shortest path connecting two points is not necessarily a straight line Space is stretched under the influence of a massive object The Principle of Equivalence All local, freely falling, nonrotating laboratories are fully equivalent for the performance of all physical experiments In other words, one cannot tell the difference between a free-falling reference frame from one “at rest” Space is curved under gravity • Light likes to take quickest path possible • Light path is curved according to the observer on the ground • Space(time) is curved! An elevator at rest A free-falling elevator The spacetime near Earth is only slightly curved, but not the case near black holes! Proofs that the spacetime is curved Apparent shift of star’s position during solar eclipse (experiment led by Arthur Eddington in 1919) This is a simulation Gravitational Lensing Time is also “messed up”! Gravitational redshift and time dilation • The observer on the ground thinks the meter should measure a blueshifted light • Yet the equivalence principle says the meter should measure exactly the light with the same frequency • So from the point of view of the ground- based observer, the light should be redshifted on its journey to the meter • This is the gravitational redshift: any photon escaping from the gravitational potential well would be redshifted and lose its energy Very small for small gh, but huge near the holes An elevator at rest A free-falling elevator Gravitational redshift and time dilation If a light beam escapes from an initial position of r0 outside a massive object to infinity, the light gets redshifted continuously The light observed by a distant observer would have a frequency Note time is just an inversion of frequency, so clock also slows down At the Schwarzschild radius 2 Δt0 RS = 2GM / c = 0 Time is frozen! Δt∞ ν ∞ = 0 Light is infinitely redshifted ν 0 Nothing escapes from within the event horizon Falling into a black hole Black holes: Observational evidence Supermassive black holes Stellar-mass black holes Cygnus A The Galactic Center Mass and size estimates Dynamic mass in binary systems