INTRODUCTORY ASTRONOMY Lecture 6. Final stages of stellar evolution. Evolution after the red giant phase. If the star has reached the end of the Giant Branch, there is only Hydrogen-Shell-burning seat. However, the central part pulls together under its own gravity, causing the temperature to rise again. When the temperature has risen high enough (∼ 108 K) Helium burning begins. Two helium nuclei can form a beryllium-8 nu- cleus. But this is not stable isotope. 1 However, in such a stellar core, conditions are such that the decay time is relatively long (about 3 × 10−16 sec). As a result, an additional new helium nucleus is captured and carbon-12 is formed. In practice, part of the carbon will be `burnt on' to oxygen. 4He +4 He $ 8Be + γ 8Be +4 He ! 12C + γ 12C +4 He ! 16O + γ The stars then ae located on the Horizontal Branch and climb during the textcolormagentaHelium- burning back up along the textcolormagentaAsymptotic Branch. This can be seen very well in the H-R diagrams of globular clusters. 2 3 In globular clusters the Horizontal Branch ex- tends far to the blue. That's because these stars have very little heavy elements. `Heavy' means all elements heavier than helium. Stars near the Sun contain more heavy elements (in the sun about 2% of the total and mainly in carbon, nitrogen and oxygen). For such stars, the entire Horizontal Branch lies almost exactly on the Giant Branch. It is called the Clump. After helium-burning, the process may repeat irself. The core cotracts and the temperature rises even more. Around it, shell-burning of he- lium and hydrogen takes place. If the star is heavy enough, carbon-burning takes place. After this the core contracts so intense and the star explodes as a Supernova. 4 The most spectacular, recent supernova is SN 1987A in the Great Magellanic Cloud (a com- panion of our own galaxy), which could be seen with the naked eye in February 1987. 5 A supernova can reach an absolute magnitude of -20 during maximum. The outer parts are being expelled at velocitys of a few thousand km/sec. ejected. Below SN1987A some 7 and 10 years after the explosion included with the Hubble Space Tele- scope. 6 In the end, an expanding sphere or shell of gas remains, like the Crab Nebula, which is the rem- nant of a supernova explosion that took place in 1054. The distance is about 2 kpc. 7 Stars, which are not heavy enough, never be- come a supernova. They emit their outer layers and become a Plan- etary Nebula 8 A good example of this is the Helix nebula. What is left of such stars is a compact object, a White Dwarf, which will slowly extinguish. For heavier stars (after a supernova explosion) what remains is a Neutron Star or a Black Hole. 9 Degeneration pressure. When there is no more energy production, the contraction is no longer stopped. This will con- tinue until the matter becomes degenerate. We have the Heisenberg uncertainty relationship ∆~p∆V ≈ h3; where ~p is the imomentum and V a volume and h the Planck constant. So ∆~p is a volume in the impulse space and∆ V in real space. Suppose that in terms of momentum there is a sphere in phase space with radius Pmax. 10 3 That volume will then be (4=3)πpmax. The number of possible volumes in which parti- cles are indistinguishable in that case is 4 p3 3πp3 π max = max∆V 3 ∆~p 4h3 But according to thePauli Exclusionary Principle two particles can never be in the same state. We talk here of electrons or neutrons and they have a spin, which can be up or down. 11 So in every volume∆ V only two particles may be present. So if n is the number of particles per unit volume, then 2 × number options 8π n = = p3 ∆V 3h3 max So 3n1=3 pmax = h 8π Non-relativistic. The energy of each particle with mass m is p2 E = 2m So the average energy of the particles is p2 hEinr = h i 2m Z pmax p2 Z pmax −1 = 4πp2dp 4πp2dp 0 2m 0 3p2 = max 5 2m 12 Relativistic. The energy of each particle with mass m is 2 2 2 2 4 E = p c + m◦c 2 In the highly relativistic case, pc m◦c , so E = pc Then the average energy of the particles is hEirel = hpci Z pmax Z pmax −1 = pc4πp2dp 4πp2dp 0 0 3c = pmax 4 In a gas, the pressure is P = (2=3)nhEi, so 2 5=3 π h 3n A 5=3 5=3 Pnr = = n / ρ 15 m 8π m 1=3 hc 3 4=3 4=3 4=3 Prel = n = Areln / ρ 2 8π The temperature is missing in these equations! That is, there is no longer any connection be- tween T and the kinetic energy. 13 White dwarves. In a white dwarf, the electrons are degenerate. So for m we should take the mass of the electron me and n = ρ/mp, where mp is the mass of the proton. So in the nonrelativistic case we get 5=3 A −5=3 P = Aeρ pos(192; 235) with Ae = mp me Now look at the structure of the star in a sim- plistic way. 14 The pressure halfway the radius R is the weight of the column above GM(1R) P (1R) = ρ × 1R × 2 2 2 1 2 2R 1 1 Here M(2R) is the mass within 2R and that is M=8 if M is the total mass and the density constant. Then GMρ P (1R) = 2 4R GMρ P (1R) = 2 4R So take the average pressure as M P¯ ∼ Gρ¯ R With M = (4=3)πρR¯ 3 we then get s 4π P = GM2=3 ρ¯4=3 3 Then 4π G ρ¯ = M2 3 Ae 15 2 3 3 = Ae R = M−1=3 4π G For a white dwarf with a mass of 0.5 M this gives¯ρ ≈ 5 × 106 g cm−3 and R ≈ 5 × 103 km. When the mass is larger, ρ also increases and so does pmax. So we are getting closer to the 2 relativistic case. In the end, pmaxc m◦c and 4=3 −4=3 4=3 P = Areln = Arelmp ρ So s −4=3 4π A m ρ4=3 = GM2=3 ρ4=3 rel p 3 From this, ρ can be solved and so M deter- mined. This is a maximum mass, because the electrons have all possible velocities between 0 and the speed of light and that is all. Then 3=2 s Arel 3 −2 Mcrit = m = 1:75M G 4π p A detailed calculation gives Mcrit = 1:44M . 16 Neutronstars. If the mass of the star was originally above about 5 M and the supernova blast is over, what re- mains is a compact object with a mass too large to become a white dwarf. The degeneration pressure from the electrons can not stop further contraction. The protons and electrons then merge (as it were) into each other and the remaining neu- trons now become relativistic. Neutronstars of of1 :5M have a radius R ≈ 5 km and a density ρ ≈ 1015 g cm−3. Because of the contraction, these stars are spin- ning very fast. If they emitting along their mag- netic axis (radio) radiation we see this as a Pul- sar. 17 The period of pulss frabges rom seconds to mil- liseconds. 18 We can estimate the density of a pulsar from the centrifugal force. At the equator of the pulsar s V 2 GM GM = or V = R R2 R 4π 3 The period is P = 2πR=V and with M = 3 R ρ we get 2πR3=2 2πR3=2 P = p = = 3:8 × 105ρ−1=2 GM q4π 3 3 GR ρ with P in seconds and ρ in kg cm−3. For a period P of 2 ms a density ρ follows of 4×1016 kg cm−3. The period of pulsars is decreasing. This is be- cause the pulsar radiates energy and energy in the pulsar is mainly in the form of rotation. For a rotating sphere with mass M, radius R and angular velocity ! the rotational energy 1 E = MR2!2 rot 5 19 The period is P = 2π=! and the decrease is then dE 4 d 1 8 MR2 dP rot = π2MR2 = − π2 dt 5 dt P 2 5 P 3 dt Suppose that the radiated energy per second of the pulsar L = −dErot=dt. Then dP 5 LP 3 P_ = = dt 8π2 MR2 For example, for the Crab-pulsar, the following applies: P = 0:03 sec and L ≈ 1031 Watt. With our previous values of M ≈ 1:5 M and R ≈ 5 km then follows P_ ≈ 2 × 10−13sec. This is also roughly observed. We can also estimate the lifetime of the pulsar as P t = P_ For the Crab-pulsar we find ≈ 1011 sec or ≈ 4 × 103 years. 20 Black holes Aso for neutron stars there is a maximum mass, although not quite well known. One suspects somewhere in the neighborhood of 5 M . Above that no known mechanism can stop the collapse and we get a Black Hole. For a body with mass M and radius R the es- q cape velocity at the surface is 2GM=R. If it becomes equal to the speed of light c, we get the Schwarzschild radius 2GM RSch = c2 For a mass of 1 M this is 2.7 km.