* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Lecture 18
Survey
Document related concepts
History of Solar System formation and evolution hypotheses wikipedia , lookup
Dyson sphere wikipedia , lookup
Nebular hypothesis wikipedia , lookup
Advanced Composition Explorer wikipedia , lookup
Corvus (constellation) wikipedia , lookup
H II region wikipedia , lookup
Planetary habitability wikipedia , lookup
Formation and evolution of the Solar System wikipedia , lookup
Timeline of astronomy wikipedia , lookup
Future of an expanding universe wikipedia , lookup
Aquarius (constellation) wikipedia , lookup
Standard solar model wikipedia , lookup
Stellar evolution wikipedia , lookup
Transcript
ASTR 200 : Lecture 18 Solar Stars : Pre- and Postmain-sequence evolution 1 Midterm exam • • • • • One week from today. In lecture, October 28 Will contain short answer questions and problems Formula sheet will be provided No calculators Will cover lectures and associated textbook readings listed on the course web site up to and including TODAY. Friday Oct 21. • A previous year's midterm has been posted, but it is a different format and course topic order. No solution set will be provided. 2 Collapse phase of stars • We have seen that cold interstellar clouds of order 0.1 pc in scale collapse to form stars • The free-fall time was ~105 years, and the clouds must span 0.01-100 solar masses, with smaller masses being much more common • The contracting cloud forms a disk, with a central condensation called a protostar. 3 Collapse phase : energy source • As the clouds collapse to smaller sizes, they heat up due to Kelvin Helmhotz heating 2 If it was all hydrogen, (M/m )kT ~ GM /R • p – mp is proton mass, M and R are the cloud's mass & radius – T ∝M /R , although half potential energy--> rotation • So as the cloud shrinks by a factor of 10, T rises by a factor ~10 • 10K when R~10,000 au, to T~10,000K at 1 au 4 - (BUT, a lot of the heat is radiated away DURING the contraction, so T of central protostar is more ~3000 K) Protostars : `reasonably hot' and big • With T~3000 K, where is the peak of the blackbody? – Wein's law: peak~ 1000 nm ~ 1 micron....infrared 5 • As the solar mass protostar's T reaches 3000 K, it is also shrinking, with a radius ~0.1 au = 20 solar radii • Convince yourself that this implies a luminosity of about 30 solar luminosities (!) • Right: protostars glowing strongly in IR in clouds On the HR diagram, stars move 6 • Low mass stars contract initially and then reach main sequence after 0.1-1 Myr • 1 solar mass star reaches ignites core fusion and settles to internal equilibrium where surface T~6000 K After ~10 Gyr of stability, the star consumes the H in its core into He • For a solar mass star, the post main-sequence evolution results in the star leaving the main sequence during the phase where it burns He in core to carbon • During this phase, becomes giant briefly, with radius 100 solar radii and slightly cooler surface T 7 • With masses < about 3 solar masses, the stars mass is insufficient to cause a core T high enough to burn carbon This is not going to be.....good.... In ~5 Gyr the Sun will leave the main sequence and reach a radius of about 1 au The Earth will then be at roughly the solar surface of T~3000 K, and vaporize 8 This is not going to be.....good.... In ~5 Gyr the Sun will leave the main sequence and reach a radius of about 1 au The Earth will then be at roughly the solar surface of T~3000 K, and vaporize 9 The C core contracts, but won't ignite • Instabilities late in the red giant's life push the less dense outer layers (beyond the core) of the star away • In the meantime, the core is contracting. This contraction releases yet more gravitational energy, so much in fact that the outward pressure pushes all the layers outside of the core out into space, making a planetary nebula – (has nothing to do with planets!) 10 So if there is no fusion, there is no energy generation in the core... • Gas pressure gets less and less as the very hot core radiates away its heat...why doesn't the star collapse to a point (a 'singularity', or black hole?) – Because another pressure, besides gas pressure or radiation pressure, takes over • As the star contracts, it becomes so dense that electrons begin to exert a new form of 'pressure' 11 12 13 Degenerate electron pressure • These electrons exert pressure at the rate they transfer momentum to a unit surface area. – For one electron, this is (p/3) v/V, where v is the electron speed v=(p/me) if non-relativistic electrons – The 'number of states available' is V/h3, multiplied by 2 because two electrons can occupy each momentum state p 2 V p pv 8π 2 4 P= 3 ∫0 4 π p dp= 3 ∫0 p dp 3V h 3 h me f f • Integrating, one finds 2 /3 8π 1 3 5 P= pf = ( π ) 3 20 15 h m e 14 2 h 5/3 ne me Equation of state: Degenerate matter • The electrons thus exert a `quantum mechanical' pressure. If this pressure exceeds the gas pressure (from the ideal gas law) we say that the electrons are `degenerate' • The number density ne of electrons is proportional to the number of protons (n e ∝ρ ) so the (non-relativistic) equation of 5/3 state for degenerate matter has : P∝ρ – Different proportionality than ideal gas law: P∝ρT 15 • The degenerate equation of state has NO temperature dependence! Degenerate electrons create pressure even if the temperate cools arbitrarily low... • With this equation of state, and hydrostatic equilibrium, we have two equations and two unkowns (P and) so we can solve for the structure of the star Radius of a white dwarf: pressure • Hydrostatic equilibrium told us that an estimate of the central 2 pressure of the object will be: 3 GM P( 0)≈ 4 4π R • Now the degenerate electron pressure was: 2 /3 1 3 Pe = ( π ) 20 2 h 5/3 ne me • We need ne . How many electrons are there per unit volume? A white dwarf if composed mostly of carbon (and sometimes oxygen); both give 1 electron per 2 mH when ionized, so M 3M n e≃ = 2m H V 8 π m H R3 16 Radius of a white dwarf: Mass • Equating the central pressure to be electron degeneracy, and solving for R gives 2 −1 /3 M RWD ≃0.24 ( ) 2 G memH mH ℏ • This is a factor of 4 too low. A more accurate analysis gives 2 −1/ 3 M RWD ≃0.95 ( ) 2 G me mH mH ℏ 17 • Amazingly, once the total mass M is set, then given the mass of the electron and proton, the white dwarf's size is determined by Planck's constant • Note that R depends inversely on the cube root of the mass – Increase M by 8x, and the radius DROPS by a factor 2! Mass-radius relation • Evaluating all the constants yields −1/ 3 M RWD ≃0.009 ( ) M sun R sun • For reference, 0.009 Rsun ~ Rearth ! 36 3 • The mean electron number density ne ~ 10 electrons/m • Of course, the mass is in the protons and neutrons, not electrons – 1 solar mass in 1 earth radius is about 2 tonnes/cm3 18 The Sirius A and B system • 1844: German astronomer F. Bessel realized that Sirius A (brightest star in sky!) must have a companion, based on wobble in its proper motion • 1862: American astronomy A. Clarke directly detects the faint Sirius B • Surface temperatures are not too different, but LA ~ 10,000 LB Modern HST picture A – Thus RA ~ 100 RB 19 • The small luminosity limits how quickly the WD can radiate its leftover heat B White Dwarf cooling • Once they shed their planetary nebula, the central star is initially very hot, with T>40,000 K • However, it still obeys the usual luminosity relation and radiates into space • As T drops, the total luminosity does too, and so the WD gets cooler and less luminous • Takes billions of years 20 Central WD White Dwarves → Black dwarves • There is no end to this process, the white dwarves will eventually cool to nearly zero absolute temperature • Such a star is coined a `black dwarf' • However, none exist yet, because even the very oldest stars have not had time to cool their surface T<1000 K 21 White dwarf cooling Age (Gyr) after planetary nebula