Introductory Astronomy

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 nucleus. 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 helium 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 become 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 particles 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 between 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 remains 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 neutrons 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 because 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.

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