Quick Review. To a good first approximation, universe consists of H, He, plus heavier elements or metals such as C, O, Fe.

Transcription

1 Quick Review To a good first approximation, universe consists of H, He, plus heavier elements or metals such as C, O, Fe. Bulk of He produced by nuclear reactions which occurred during the first few minutes of the universe - big bang nucleosynthesis. Brief history of the Universe. Need to explain that the universe is expanding and that the whole of space is filled with thermal radiation at a temperature of about 3K - CMB. Consistent with a big bang - a sudden decompression that occurred everywhere in space at a definite point in time. Nanoseconds after BB, universe was filled with a gas of fundamental particles: quarks, antiquarks, leptons and antileptons, neutrinos and antineutrinos plus gluons and photons. Temperature falling. When it falls below K, quarks, anitquarks plus gluons disappeared, forming less massive particles. Because the number of quarks slightly exceeded the number of antiquarks, a few quarks were left to form protons and newtrons present in today s universe. Heavier leptons and antileptons also annihalated as temperature fell. For 10 3 s < t < 1s, universe consistedn of neutrons, protons, electrons plus positrons, neutrinos, antineutrinos and photons. As T falls below about K, neutrinos stopped interacting effectively with matter - expansion of universe means this neutrino gas now has a temperature of about 2K Not detected as yet. After t = 100s, netutrons combined with protons to fomr light nuclei leading to a universe of about 75% H, 25% He. After t = 300, 000 years, T < 4000K, low enough for the formation of stable atoms. H and He nuclei combined with electrons to form H, He atoms. Photons no longer interacted strongly with matter - universe became transparent to EM radiation. This radiation has now cooled to about 3K because of the expansion of space - CMB. Slightly warmer than the undetected neutrino background. Universe continued to expand and cool unitl it reached its present lumpy state. He synthesis 1

2 What happened to the gas of neutrons and protons as the universe expanded and cooled from K to 10 9 K. Normally, a neutron beta decays with a mean life of about 15 minutes to proton, electron and anti-neutrino: n > p + e + ν e. However, in conditions of high temperature and density, neutrons and protons can be transformed into one another with ν e + n < > e + p, ν e + p < > e + + n. Becaue neutrons are more massive than protons, more energy needs to be taken from gas to make a neutron than a proton. Therefore neutrons outnumbered by protons. Ratio of neutrons to protons is given by Boltzmann factor: N n N p = exp[ mc 2 /kt], where m is the neutron-proton mass difference: 1.3MeV/c 2. Dynamic equilibrium between numbers of neutrons and protons due to these reactions. But as the universe s temperature and density decreased, numbers of neutrons and protons ceased to change rapidly. Neutron-proton ratio becomes frozen to a ration of about 0.2 when T < K. In fact, ration continued to decrease because of neutron beta decay. After a few minutes, the ratio does freeze out at about 1/7. Lower temperatures means that a gas of neutrons and protons can be transformed into a He 4, plus deuterons, He 3, Li 7 plus left over protons. BB makes a clear cut prediciton about the abundance of He 4 - precitions for other nuclei are less certain. - dependent on the uncertain density of the universe. He 4 abundance determined by the neutron-proton ratio just before nucleosynthesis. This ratio was 1/7. Consider 2 neutrons and 14 protons. These formed a single He 4 nucleus consisting of 2 protons and 2 neutrons leaving 12 protons. Thus 16 atomic mass units of neutrons and protons produced one atomic mass unit of 4. Thus the He 4 abundance is Gravitational Contraction 2

3 Gravity is the driving force in stellar evolution. Leads to the compression of matter and then to the formation of stars, galaxies etc. Consider a spherical system of mass M and radius R - the only forces acting are self gravity and internal pressure. Denote the desnity and pressure at distance r by ρ(r), P(r). Matter enclosed by a spherical shell of radius r is m(r) = r 0 ρ(r )4πr 2 dr. This acts as a gravitational mass situated at the center giving rise to an inward gravitational acceleration of g(r) = Gm(r) r 2. There is a force arising from the pressure gradient. This is [P(r) + dp dp r P(r)] A = dr dr r A. Then the inward acceleration of any element due to any element of mass at distance r from the center is dr dt = g(r) + 1 dp 2 ρ(r) dr. To oppose gravity, pressure must increase toward the center. Free Fall What if there is no pressure gradient to oppose gravitational collapse? Now each mass element at radius r moves toward the center with an acceleration g(r) = Gm(r)/r 2. A shell of matter enclosing a mass m 0 collapses under gravity with an inward acceleration Gm 0 /r 2 - the kinetic energy of the shell increases as its gravitational p.e. decreases. Need to find the inward velocity of the shell when its radius is r - assume each shell is initially at rest at radius r 0 and it encloses a mass which remains constant during collapse. Conservation of energy equation: [dr dt ] = Gm 0 Gm 0. r r 0 3

4 Time for free fall to the center of the sphere is t FF = 0 r 0 dt dr dr = [ 3π 32Gρ ] 1/2. In practice energy releases by the gravitational field of the collapsing system usually dissipated into random thermal motion of the constituents - creating a pressure which opposes further collapse. A molecular cloud can collapse rapidly as long as it is transparent to its own radiation. But the gravitational energy releases in an opaque cloud of ionized H will be trapped as internal thermal motion - which increases the pressure and slows down further collapse. Cloud then approaches hydrostatic equilibrium. Hydrostatic equilibrium An element of matter at a distance r from the center of a spherical system will be in hydrostatic equilibrium if dp dr = Gm(r)ρ(r) r 2. The whole system is in equilibrium if this is valid at all radii r. Then can derive a relation between the average internal pressure and the gravitational potential energy of the system. Using the above equation, we have R 0 4πr 3dP R dr = 0 Gm(r)ρ(r)4πr 2 dr. r The r.h.s is the gravitational potential energy of the system, m=m E GR = m=0 Gm(r) dm, r where dm is the mass between r and r + dr - which is? The l.h.s can be integrated by parts to give R [P(r)4πr 3 ] R 0 3 P(r)4πr 2 de. The first term is zero because the pressure on the outside surface at r = R is zero. The second term is equal to 3 < P > V, where V is the volume of the system and < P > is the volume averaged pressure. 0 4

5 The average pressure needed to support a system with gravitational energy E GR and volume V is This is the Virial Theorem. Pressure < P >= 1 E GR 3 V. Consider relation between the pressure and internal energy density due to the translational motion of the particles. Consider a gas of N particles in a cubical box of volume L 3 with its edges orientated along the x, y, z axes. Gas particle has velocity v = (v x, v y, v z ) and momentum p = (p x, p y, p z ). The rate at which it strikes one of the sides perpendicular to the z axis is v z /2L and in so doing it transfers a momentum 2p z with each strike. Thus the rat of momentum transfer to unit area of the side is p z v z /L 3. Now consider all N particles - the pressure due to these particles on a side perpendiucalr to the z axis is P = N L 3 < p zv z >, where the brackets denote an average over all particles. If the gas is isotropic, all directions of motion are equally likely and < p x v x >=< p y v y >=< p z v z >=< p.v > /3, where p.v = p x v x + p y v y + p z v z. Thus, the pressure on each side of the box is the saqme and equal to P = n 3 < p.v >, where n is the number of particles per unit volume. This expression derived using classical physics but is valid when quantum effects are important and in degenerate gases and in relativistic gases. General relation between the energy ɛ p and momentum, p of a particle of mass m is ɛ 2 p = p62c 2 + m 2 c 4, and the velocity of the particle is v = pc 2 /ɛ p. 5

6 Non-relativistic limit is p << mc, so that ɛ p = mc 2 + p 2 /2m and v = p/m. Ultra-relativistic limit is p >> mc so that ɛ p = pc and v = c. General expression for the pressure in an ideal gas takes the forms: For a gas of non-relativistic particles of mass m, p.v = mv 2 and the pressure is P = 2 3 n < 1 2 mv2 >. This is 42/3 of the translational kinetic energy density. For a gas of ultra-relativistic particles p.v = pc, the pressure becomes P = 1 3 n < pc >. This is 1/3 of the translational kinetic energy density. This change has an important effect on the hydrostatic equilbrium of gases under gravity. Consider a gas of volume V held together by gravity. if the gas is ideal and the gas particles are non-relativistic, then the average pressure is < P >= 2 E KE 3 V, where E KE is the kinetic energy due to the translational motion of all the particles in the entire gas. The gravitational and kinetic energues of an ideal gas of non-relativistic particles in hydrostatic equilibrium under their own gravity is 2E KE + E GR = 0. But also know that E TOT = E KE +E GR. Then we also have E TOT = E KE and E TOT = 1 2 E GR. Tightly bound gas clouds habe high kinetic energies ie. they are hot. If the system evolves and remains close to hydrostatic equilibrium, then changes in gravitational and kinetic energies are simply related to the change in total energy, e.g. a 1% decrease in the total energy is accompanied by a 2% decrease in the gravitational energy and 1% increase in the kinetic energy. Such changes characterize the behaviour of many astrophysical systems. 6

7 Equilibrium of a gas of relativistic particles. Consider a gas of volume V held together by gravity. If the gas is ideal and the gas particles are relativistic, then the average pressure is < P >= 1 E KE 3 V. Now E KE + E GR = 0. That is hydrostatic equilibrium is only possible if the binding energy is zero - the system is on the cusp of being bound and unbound. The system is easily disrupted. Equilbrium and the adiabatic index Stability of hydrostatic equilibrium is often described in terms of the adiabatic index, γ of the gas. The adiabatic index γ describes the relation between the pressure and the volume of a gas during an adiabatic compression or expansion. For such a process, PV γ is constant, that is for small adiabatic changes in the volume and pressure, d(pv ) = PdV + V dp, d(pv ) = PdV + V dp dv dv, d(pv ) = (γ 1)PdV. In an adiabtic change, no heat transfer and the change in the internal energy of the system is determined solely by the work done. Let E IN denote the internal energy due to the translational kinetic energy and the excited internal degrees of freedom of the gas particles, then de IN = PdV, and, de IN = 1 d(pv ). γ 1 If the adiabatic index γ is constant, then, E IN = 1 γ 1) PV. 7

8 Consider a self-gravitating gas with adiabatic index γ in hydrostatic equilibrium. Then from the above, the average pressure, < P >, is using the virial theorem. < P >= (γ 1) E IN V = 1 E GR 3 V, Then a self-gravitating gas with adiabatic index γ is in hydrostatic equilibrium if 3(γ 1)E IN + E GR = 0. Previous equations can be obtained from this by seting γ = 5/3 for non-relativistic particles and γ = 4/3 for ultra-relativistic particles. Also in a gas with no excited internal degrees of freedom, E IN = E KE. The total energy of a gas with adiabatic index γ in hydrostatic equilibrium is E TOT = E IN + E GR = (3γ 4)E IN. The gas is bound if E TOT < 0, i.e. if γ > 4/3. If γ is close to 4/3, the binding energy is small and then a small change in the total energy is accompanied by a much larger change in the internal and gravitational energies e.g. if γ is 1% bigger than 4/3, a 1% decrease in the total energy is accompanied by a 25% increase in the internal kinetic energy and a 26% decrease in the gravitaional potential energy. The stability of such a system is precarious. The adiabatic index can also approach 4/3 when there are processes which provide new ways of absorbing heat such as the ionization of atoms (pulsating stars). Star Formation Stars formed from clouds of gas by gravitational contraction. But how exactly? This is currently a very important topic in Astrophysics. Now we discuss some general features of this process. Conditions for gravitational collapse A cloud of gas, in order for it to contract or condense into a cluster of stars, must be sufficiently compact - why? So that gravitational forces can overcome internal pressure. Cloud becomes bound if the magnitude of the gravitational potential energy is larger than the internal kinetic energy. Consider a cloud of radius R, mass M containing N particles with average mass m at a uniform temperature T. ASsume the cloud is just hydrogen for simplicity. 8

9 Then the gravitational p.e. is E GR = f GM2 R, where f is a numerical factor depening on the density distribution of the cloud. What is it for a spherical cloud of uniform density? Each particle contributes 3 kt to the thermal kinetic energy of the 2 cloud and so E KE = 3 2 NkT. The critical condition for the onset of condensation is E GR > E IN. Show that this implies a cloud of radius R can condense if its mass exceeds M J = 3kT 2Gm R. Show it also implies a cloud of mass M can condense if its average density exceeds ρ J = 3 3 3kT 4πM2( 2Gm ). These are the Jeans mass and density. For example, a cloud of molecular hydrogen at a temperature of 20K with a mass of kg could condense if its desnity reaches kgm 3. This is quite low. Star formation takes place in several stages. A massive extended gas cloud contracts and when its density exceeds the Jeans density, smaller parts of it will be able to contract independently. Perhaps the cloud will fragment to form a cluster of protostars. Contraction of a protostar The equation for the jeans density suggests that when a cloud at a temperature of 20K reaches a density of kgm 3, a fragment with a mass comparable to with the solar mass is capable of contracting independently. This fragment forms a protostar with a radius of about meters - about a million times larger than the Sun. It collapses freely, unopposed by internal pressure, if the gravitational energy released is not converted into random thermal motion - why? For this to occur, much of the gravtational energy released by contraction must go into the dissociation and ionization of hydrogen molecules and atoms respectively. 9

10 Energy needed to dissociate a hydrogen molecule is ɛ D = 4.5eV. Energy needed to ionize a hydrogen atom is ɛ I = 13.6eV. Eenergy needed to dissociate and ionize all the hydrogen in a protostar of mass M is M 2m H ɛ D + M m H ɛ I, where m H is the mass of the hydrogen atom. If this energy is supplied by the protostar shrinking from a radius R 1 to radius R 2, then GM 2 R 2 GM2 R 1 M 2m H ɛ D + M m H ɛ I. For example, the energy needed to dissociate and ionize the hydrogen in a protostar of 1 solar mass is about J. Such a protostar collapses freely from its radius of R m to R m in about years and a density ρ kgm 3. When most of the hydrogen is ionized, the protostar starts to become increasingly opaque to its own radiation - explain this. The gravitational energy released by contraction now is converted to random thermal energy of the electrons and ions. The internal pressure therefore rises and the collapse of the protostar is slowed down as hydrostatic equilibrium is approached. Whats the average internal temperature at this stage? We have E KE M m H 3kT, the thermal kinetic energy of the hydrogen ions and electrons in the protostar at an internal temperature T. The gravitational energy at the end of the rapid collapse, if R 1 >> R 2, is E GR GM2 R 2 ( M 2m H ɛ D + M m H ɛ I ). The Virial theorem says 2E KE + E GR = 0. Thus a protostar approaches hydrostatic equilbrium at a temperature given by kt 1 12 [12ɛ D + 2ɛ I 2.6eV. This is an average internal temperature of about 30000K, independent of the mass of the protostar. Subsequent slow contraction of the protostar governed by the opacity of the ionized interior - what is opacity? 10

11 Time scale fo contraction is of the order of 10 7 to 10 8 years. The Virial theorem (because the protostar contracts close to hydrostatic equilibrium) states that half the gravitational energy released is lost from the surface, the other half is stored as kinetic energy. The temperature and pressure at the center of the protostar increase until thermonuclear fusion of hydrogen starts. Energy released by fusion lessens the need for the release of gravitational energy and the protostar ceases to contract. Its a star when nuclear reaction rate is sufficient to supply the radiant energy lost from the surface and maintain hydrpstatic equilbrium. Conditions for stardom Not all self-gravitating bodies achieve stardom. Consider this now. Electron wavelength, lambda = h/p. K.E. of an electron in a classical gas at temperature T is kt, momentum (m e kt) 1/2, typical de Broglie wavelength is λ h (m e kt) 1/2. Classical mechanics valid if wave functions of the electrons do not overlap. That is the average separation between the electrons large compared to λ. This is so if the density of the ionized gas satisfies ρ << m λ 3 (m ekt) 3/2 h 3, where m is the average mass of the particles in the ionized gas e.g. for ionized hydrogen gas, m = 0.5amu. The internal temperature for the protostar initially rises as the internal density increases because kt GMm 3R GmM2/3 ρ 1/3. The temperature is proportional to ρ 1/3, as long as the electrons are governed by classical mechanics ie.e the density is low enough. When the density reaches the value 11 ρ m (m ekt) 3/2 h 3,

12 The Sun quantum mechanics becomes important and the electron gas starts to be degenerate. Then the temperature no longer increases markedly if it is compressed. Can estimate the temperature at which the electron gas in a contracting protostar becomes degenerate by or kt GmM 2/3 m 1/3(m 3kT) 1/2 kt [ G2 m 8/3 m e h 2 M 4/3. This is an expression for the maximum internal temperature reached by a contracting protostar. Its dominated by the mass, M of the protostar. For a solar mass, this is about 10 6 K - the average internal temperature - so the central temperature is even higher. This is sufficient to trigger nuclear fusion reactions in the core. Minimum mass needed to initiate nuclear fusion is about 0.08 solar masses. Protostars with masses less than this evolve into objects where gravity is countered by the pressure of degenerate electrons - brown dwarfs. Also need to consider radiation pressure which can effect hydrostatic equilibrium. Stars with masses greater than solar masses are rare. Sun is the nearest star - can model it in detail - Standard Solar Model. Look up Bahcall (1989) and the latest versions. Table 1.2, p. 19 gives some basic parameters. Whats the free fall time for the Sun? What can you conclude from this calculation? The luminosity of the Sun is given by the Stefan Boltzmann law L = 4πR 2 T 4 e. Solar luminosity is cupplied by a chain of thermonulcear reactions called the proton-proton chanin: p + p > d + e + + ν e, p + d > 3 He + γ, 3 H e + 3 He > 4 He + p + p, where d denotes a deuteron or 2 H. 12 h

13 Total energy released is 26MeV per 4 He nucleus formed. Energy must be released at a rate of W to pwer the solar luminosity. Power release is large but power density is huge i.e. its inefficient. Fusion of protons is accompanied by the emission of at least neutrinos per second. These interact only weakly with matter - solar neutrino problem. Solar thermostat. Stellar Nucleosynthesis Stellar evolution is essentially the release of gravitational potential energy through contraction with pauses where nuclear fuels are ignited to supply the energy flow from the surface of the star - nucleosynthesis. Usually the ashes of one set of reactions become the fule for the next. Due to the binding energy per nucleon verses atomic number graph, this implies that nuclear fusion stops at Iron - cannot burn Cr, Mn, Co, Ni, and Fe. Not all stars can achieve the temperatures needed to ignite every stage of thermonuclear fusion and progress to Iron. The mass of a contracting star determines the maximum temperature achievable and hence which thermonuclear fusion stages can be reached. Proton-proton chain in stars like the Sun. CNO cycle is important in more massive stars. When H burning ceases in the center of a star, He core contracts under gravity and grows hotter. Increased temperature promotes H burning in a shell around the core. Also leads to an increase in pressure and a large expansion of the outer layers of the star. More He is deposited in the core which becomes denser and hotter. If the star is massive enough, He burning in core starts. Release of energy from He burning in core causes the core to expand and cool which also leads to a partial contraction of the outer layers of the star. This is the red giant phase. Also get shell He burning. Stars with a mass greater than 8M can ignite C at a temperature of about K to form elements such as Ne, Na, Mg. 13

14 Stars with a mass greater than about 11M can achieve the high temperatures necessary for ignition of every stage oof thermonuclear fusion. Stellar Life Cycles BB led to a universe composed of H, He with traces of light elements. This was enriched with heavier elements by a cycle of stellar formation and evolution in which matter has been transferred back and forth between stars and intersetllar matter. Try to use this cycle to explain the abundances of the chemical elements in the universe today. Rates of Stellar evolution Luminosity of a star determines the rate at which it consumes its nuclear fuel and the luminosity of a star is a rapidly increasing function of its mass, L M α, where α = 3 for massive stars and about α 3.5 for stars less massive than the sun. Massive stars have shorter lives despite their greater fuel reserves. H burning lifetime M 2 for high mass stars and M 2.5 for low mass stars. Give examples Lifetime of the Universe is between billion years, ample time for many generations of massive stars but insufficient time for the evolution of stars with a mass much smaller than the sun. The end-points of stellar evolution Stellar evolution depends critically on the mass that remains in the central core when nuclear fusion can no longer maintain pressure needed to prevent gravitational contraction. When nuclear fusion cannot provide this support, star must rely on a gas of degenerate electrons - but this is precarious. There is an upper limit to the mass that can be supported by a degenerate electron gas - Chandresekhar mass: 1.4M. Mass in central core the mass when the star joins the main sequence. The Sun will develop a central core whose mass is smaller than this limit and it can support itself against gravitational contraction by the pressure created from a degenerate electron gas - a white dwarf: radius 10 7 m, ρ 10 9 kgm 3. 14

15 This will slowly cool without an appreciable contraction because its mechanical support is insensitive to temperature. This is the fate for less massive stars. More massive stars can ignite more heavier elemenst for nuclear fusion and eventually develop an onion-like structure with a central core of Fe. Eventaully the mass of the core exceeds the Chandrekshar limit and the core collapses - gravitational energy released. Pulse of neutrinos carries away much of this energy. Some small fraction of this energy may be used to eject some of the stellar mass - supernova. Elements heavier than Fe are formed during these stages. Mass of the collapsed core then determined what happens next. Most likely result is a neutron star - compact star consisting of primarily degenerate neutrons. Maximum mass for a neutron star is 3M. If collapsed core mass exceeds this - black hole. One uncertainty in this is mass loss e.g. solar wind carries away about M away every year. INtermediate mass stars, when they exhaust their nuclear fuel, they shed their outer layers in an expanding cloud called a planetary nebula. More massive stars - supernova. This ejected material forms the raw material for a future generation of stars. Pop I and Pop II stars. Pop III? Abundance of the elements Cycle of star formation leads to an enrichment of primordial H, He with heavoer elements. Observed abundance of chemical elemens in the solar system is a result of nucleosynthesis during the big bang and during stellar evolution of an earlier generation of stars. Fig 1.5, p. 34 shows abundances plotted against atomic number Z. Dominance of H, He - a relic of the big bang. A lack of abundance between He and C: difficult to build H, He in the absence of stable mass 5 and mass 8 atomic nuclei. 15

16 Peaks corresponding to the major products of stellar nucleosynthesis: C, O, Ne, Si and elements near Fe. High abindance of N due to H burning in CN cycle. H, He, C, O, Ne, Si burning is responsible for the abundances of elements with atomic number 12 < Z < 30. Elements with atomic number Z > 30 owe their existence to neutron capture during the terminal stages of stellar evolution. The Hertzsprung - Russell (HR) diagram Most important diagram in astronomy. Luminosity Oberved brightness expressed as a magnitude - a logarithmic measure of brightness. If the energy flux received from two stars is f 1, f 2, the magnitudes differ by m 1 m 2 = 2.5 log(f 1 /f 2 ). Energy flux is amount of energy received per second per unit area on Earth in a particular wavelength range, e.g. m V is the visual band. Bright stars have lower magnitudes than dimmer stars. Magnitudes can be negative. Need a reference point. Define the visual magnitude of Vega to be zero. The absolute magnitude is the brightness of a star at 10 parsecs. Absolute and apparent magnitudes denoted by capital and lower case M, m, respectively. The bolometric magnitude is the brightness over all wavelengths. The parsec is the standard astronomical unit of distance: 1 parsec is the distance the star is at if its parallax angle if 1 second of arc light years. Determination of distances accurately is a fundamental problem in astronomy. Luminosity is the total amount of energy emitted by a star per second over all wavelength ranges. Then M BOL = 2.5 log(l/l ) Surface Temperature The effective surface temperature of a star T E is the temperature of a black body of the same size that would give the same 16

17 luminosity. For a star of luminosity and radius L, R: L = 4πR 2 σt 4 E, where σ is Stefan s constant. For the Sun, T E 5800K. Can deduce the surface temperature from a spectral analysis of the radiation from the star. The color of the star is the difference of its brightness at two wavelengths.e.g. B V. In principle, for a perfect black body, knowing the color leads to a precise determination of effective temperature. In practice, stars are not perfect black bodies and interstellar absorption effects determination of color. Is color dependent on distance? Study of stellar spectrum: absorption/emission lines: OABFGKM. temperature sequence. Main observational properties of a star L and T E are not uncorrelated: HR diagram, Fig 1.6, p. 36. Observational counterpart of this is the color-magnituda (CM) diagram. Explain Fig. 1.6, p. 36. An HR diagram for stars in a particular region of space provides a snapshot of stars at different stages of their evolution. Main sequence - hydrogen burning in core: 80-90% of observed stars are on the main sequence. Mass-luminosity relation, L M α, where α ranges between 3 and 3.5 on the main sequence. A star does NOT evolve along on the main sequence. It evolves onto the main sequence when a protostar contracts and ignites H. Star evolves of the main sequence, moving into the red giant region, when H in the core is depleted. When core H is exhausted, H bunring in a shell stars but the outer layers expand to form a star of high luminosity and low surface temperature - red giant region: e.g. Betelgeuse. Stars in the advanced stages of nulcear burning occupy the top right hand regions of the HR diagram. Time scales are very fast here (why?) so this region is not densely populated. Intermediate stars end their life by shredding their outer layers to form a planetary nebula which merges with the interstellar 17

18 medium to leave a remnant with low luminosity and high surface temperature in the white dwarf region of the HR diagram.e.g Sirius B. In time white dwarfs will cool and fade away. Use of HR diagrams in age and distance estimation and connection between observations and theory. 18