Lecture Notes on Stellar Corpses | ASTR 1020, Study notes of Astronomy

Download Lecture Notes on Stellar Corpses | ASTR 1020 and more Study notes Astronomy in PDF only on Docsity! ASTR-1020: Astronomy II Course Lecture Notes Section VIII Dr. Donald G. Luttermoser East Tennessee State University Edition 4.0 Abstract These class notes are designed for use of the instructor and students of the course ASTR-1020: Astronomy II at East Tennessee State University. Donald G. Luttermoser, ETSU VIII–3 b) The core becomes stable due to degenerate electron pressure. c) When this happens, the core is now called a white dwarf (WD). d) White dwarfs range in mass from 0.6 M to 1.4 M . e) The radius of a WD scales as R ∝ 1/M1/3 =⇒ the more massive a white dwarf, the smaller it is. f) Most of the WDs are about the size of the Earth giving an enormous density for these objects (ρ = 109 kg/m3) =⇒ 1 teaspoon of WD material on the surface of the Earth would weight 5.5 tons – as much as an elephant! 5. White dwarfs are in HSE since electron degeneracy pressure balances gravity. However, if the mass of a WD exceeds 1.4 M , this degeneracy pressure will be too weak to counterbalance gravity! a) This mass limit is called the Chandrasehkar limit. b) Due to this limit, all white dwarfs have M < 1.4M . c) If M > 1.4M , the core will continue to collapse to the next type of stellar corpse — the neutron star. 6. The nearest white dwarfs to the solar system are Sirius B and Procyon B, each a companion of the brighter star we see in the night sky. VIII–4 ASTR-1020: Astronomy II B. Neutron Stars. 1. Stars more massive than 8 M on the main sequence will be able to successively burn heavier and heavier elements as their core collapses. 2. The final type of stellar burning to go on in a massive core is Si (silicon) burning. The ash from this burning is Fe (iron). a) Fe is the most stable of all the chemical elements =⇒ you can neither get energy out of an Fe nucleus from either nuclear fusion (i.e., bringing lighter elements together to form heavier ones) or nuclear fission (i.e., the break- ing apart of heavier elements into lighter ones). b) Once an Fe core forms, its collapse cannot be halted by further nuclear reactions. c) The Fe core become degenerate as it collapses but is too massive for the electron degeneracy pressure to hold up the weight of the star. d) Since the Pauli Exclusion Principle dictates that elec- trons cannot exist in the same state, the only place for them to go is into the Fe nuclei! 3. The electrons interact with protons in an inverse β-decay: e− + p −→ n + ν, forming a stellar core of pure neutrons ! 4. Once the neutrons are formed, they stop the collapse of the core via the strong nuclear force =⇒ neutron degeneracy pressure holds up the weight of the core =⇒ a neutron star (NS) is born! Donald G. Luttermoser, ETSU VIII–5 a) This halting of the collapse of the core is rather sudden and causes the core to bounce and rebound a bit. This bounce sets up a shock wave which propagates outward and blows apart the outer envelope of the star in a Type II supernova explosion (see §VII.D). b) A tremendous amount of energy is released during this explosion over a fraction of a second, increasing the lu- minosity of the star by a factor of 108! c) However this luminosity increase only corresponds to 1% of the total energy released, most of the energy comes out in the form of neutrinos! d) As we have seen in the last section, after the explosion, the outer envelope continues to expand, creating a su- pernova remnant which can last for a few million years (e.g., the Crab nebula). 5. There is another type of lower-energy stellar explosion associate with neutron stars. These are the X-ray bursters which are similar to classical novae, except matter is dumped on a neutron star instead of a white dwarf (as is the case for a nova). 6. All stars rotate while on the main sequence. As the core of a star collapses, it spins faster and faster due to the conservation of angular momentum. a) By the time a stellar core reaches NS size (e.g., RNS = 30 km, the size of a city), it is spinning hundreds of times a second! b) Pulsars are rapidly spinning neutron stars whose mag- netic axis is not aligned with its polar axis. VIII–8 ASTR-1020: Astronomy II observer in motion, and v is the velocity of the object. b) Length Contraction: If you, the stationary observer, was to measure a ruler of length L◦ traveling at v = 0.99c, you would measure its length L following the Fitzgerald-Lorentz contraction equation: L = L◦ √ 1 − (v/c)2. (VIII-2) or L = 1 ft × √ 0.0199 = 0.14 ft = 1.7 in. This is not due to an optical illusion, the ruler has actually contracted by this amount. c) Mass Increase without Bound: Mass also increases for the object that is traveling close to the speed of light! Einstein showed that M = M◦√ 1 − (v/c)2 . (VIII-3) Note that as v → c, that M → ∞! One cannot acceler- ate mass to the speed of light since it would require an infinite amount of energy to push an infinite amount of mass. 4. These effects have been observed in the laboratory with sub- atomic particles! Example VIII–1. The Starship Enterprise has a mass of 2.0×106 kg and a length of 500 m. What is the length and mass of the Enterprise as it accelerates from 1/2-sublight (v = 0.5c), 90%- sublight (v = 0.9c), 99%-sublight (v = 0.99c), 99.9%-sublight (v = 0.999c) as viewed by stationary observer on Starbase 12 as the Enterprise departs? For every day that passes on the Enterprise, how much time passes on the Starbase? Donald G. Luttermoser, ETSU VIII–9 For the passage of time: ∆t(0.5c) = 1 day √ 1 − (0.5c/c)2 = 1 day √ 1 − (0.5)2 = 1 day√ 1 − 0.25 = 1 day√ 0.75 = 1 day 0.866 = 1.15 days ∆t(0.9c) = 1 day √ 1 − (0.9c/c)2 = 1 day√ 1 − 0.81 = 1 day√ 0.19 = 1 day 0.436 = 2.29 days ∆t(0.99c) = 1 day √ 1 − (0.99c/c)2 = 1 day√ 1 − 0.98 = 1 day√ 0.02 = 1 day 0.141 = 7.07 days ∆t(0.999c) = 1 day √ 1 − (0.999c/c)2 = 1 day√ 1 − 0.998 = 1 day√ 0.002 = 1 day 0.0447 = 22.4 days For every day that passes on the Enterprise, the people on the Starbase age at 1.15, 2.29, 7.07, and 22.4 days as the Enterprise travels at velocities 0.5c, 0.9c, 0.99c, and 0.999c, respectively. For the length contraction: L(0.5c) = (500 m) √ 1 − (0.5c/c)2 = (500 m) √ 1 − (0.5)2 = (500 m) √ 1 − 0.25 = √ 0.75(500 m) = 0.866 × 500 m = 433 m L(0.9c) = (500 m) √ 1 − (0.9)2 = (500 m) √ 1 − 0.81 = √ 0.19(500 m) = 0.436 × 500 m = 218 m L(0.99c) = (500 m) √ 1 − (0.99)2 = (500 m) √ 1 − 0.98 = √ 0.02(500 m) = 0.141 × 500 m = 70.7 m VIII–10 ASTR-1020: Astronomy II L(0.999c) = (500 m) √ 1 − (0.999)2 = (500 m) √ 1 − 0.998 = √ 0.002(500 m) = 0.0447 × 500 m = 22.4 m The Enterprise’s length would be 433 m, 218 m, 70.7 m, and 22.4 m as measured from the Starbase as it traveled at 0.5c, 0.9c, 0.99c, and 0.999c, respectively. Finally, the mass of the Enterprise would increase to: M(0.5c) = 2.0 × 106 kg √ 1 − (0.5c/c)2 = 2.0 × 106 kg √ 1 − (0.5)2 = 2.0 × 106 kg√ 1 − 0.25 = 2.0 × 106 kg√ 0.75 = 2.0 × 106 kg 0.866 = 1.15 × 2.0 × 106 kg = 2.3 × 106 kg M(0.9c) = 2.0 × 106 kg √ 1 − (0.9)2 = 2.0 × 106 kg√ 1 − 0.81 = 2.0 × 106 kg√ 0.19 = 2.0 × 106 kg 0.436 = 4.6 × 106 kg M(0.99c) = 2.0 × 106 kg √ 1 − (0.99)2 = 2.0 × 106 kg√ 1 − 0.98 = 2.0 × 106 kg√ 0.02 = 2.0 × 106 kg 0.141 = 1.4 × 107 kg M(0.999c) = 2.0 × 106 kg √ 1 − (0.999)2 = 2.0 × 106 kg√ 1 − 0.998 = 2.0 × 106 kg√ 0.002 = 2.0 × 106 kg 0.0447 = 4.5 × 107 kg The Enterprise’s mass would increase from 2.0 × 106 kg to 2.3 × 106 kg (15% more massive), 4.6 × 106 kg (130% more massive), 1.4 × 107 kg (600% more massive), and 4.5 × 107 kg (2150% more massive) as it traveled at a velocity of 0.5c, 0.9c, Donald G. Luttermoser, ETSU VIII–13 8. Starlight bending around the Sun during a solar eclipse in the 1920’s proved the validity of general relativity. D. Black Holes. 1. As a stellar core collapses, it gets denser and denser, and the escape velocity from the surface of the star goes up following: vesc = √√√√2GM R , (VIII-4) where G is the gravitational constant, M the mass of the core, and R the radius of the core (or star). a) At the Earth’s surface: R = R⊕, M = M⊕, so vesc = 11 km/s. b) At the Sun’s photosphere: R = R , M = M , so vesc = 620 km/s. c) At a WD surface: R = R⊕, M = M , so vesc = 6500 km/s = 0.02 c. d) At a NS surface: R = 30 km, M = 2M , so vesc = 230, 000 km/s = 0.77 c. 2. A black hole (BH) is defined when an object reaches a size such that its escape velocity equals the speed of light. Example VIII–2. How big would the Sun have to be in order to become a black hole? vesc = c, M = M v 2 esc = 2GM R R = 2GM c2 = 3 × 103 m = 3 km VIII–14 ASTR-1020: Astronomy II Our Space-Time Event Horizon SingularityEinstein-Rosen Bridge Distant part of Space-Time Hyperspace Figure VIII–2: The space-time continuum in the vicinity of a black hole. a) A black hole is defined as a region in space-time where vesc ≥ c. The outer boundary of this region where vesc = c is called the event horizon. b) The event horizon has a radius of RS = 2GMBH/c 2, (VIII-5) which is called the Schwarzschild Radius. c) The vast majority of the region within a black hole (i.e., within the event horizon) is empty space. The mass that was once a stellar core has no size (i.e., zero volume). This infinitely dense object at the center of a black hole is called a singularity. 3. Stellar cores will become black holes only if Mc > 3M . No known force in nature will prevent its collapse. Donald G. Luttermoser, ETSU VIII–15 4. BH’s bend space-time to such an extent that the BH actually rips a hole in the fabric of the Universe and reconnects with a distance part of the Universe or connects to a parallel Universe in a different space-time dimension (see Figure VIII-2). a) A connecting tunnel forms in space-time called an Einstein- Rosen bridge (also called a wormhole). b) Our 3-dimensional Universe bends into a 4th dimension, called hyperspace. c) Time proceeds slower and slower as you approach the event horizon of a BH. d) Tidal forces increase without bound as you approach the singularity =⇒ anything with size, even nuclear parti- cles, get ripped apart from these tidal forces as matter falls into the wormhole. 5. The only way to detect a BH is to observe the effect one has on a companion star in a binary star system (see the figure on Page VII-12 of these notes). a) As material is stripped away from a normal star by the BH, it spirals down to the BH and heats up to very high temperatures (due to friction). b) Before crossing the event horizon, the gas heats to such a high temperature that it emits X-rays. 6. Black hole candidates must satisfy the following observa- tional requirements before they are considered black holes. a) An X-ray source in a binary star system whose X-rays vary in brightness over the time period of seconds (hence emitting region must be very small in size).