Astrophysics PDF Notes

ASTROPHYSICS Mr. P.R.S. Tissera Department of Physical Sciences and Technology Faculty of Applied Sciences Sabaragamuwa University of Sri Lanka OVERVIEW What powers the stars? What supports the stars (from collapsing)? What determines the internal structure of the st...

ASTROPHYSICS Mr. P.R.S. Tissera Department of Physical Sciences and Technology Faculty of Applied Sciences Sabaragamuwa University of Sri Lanka OVERVIEW What powers the stars? What supports the stars (from collapsing)? What determines the internal structure of the stars? 2 Observable Intrinsic properties properties of of a star such as a star such as Mass, chemical Brightness, composition etc. color etc. 3 WHAT ARE STARS MADE OF? ▪ Stars are basically a giant sphere of hot gas made of relatively simple stuff ▪ Our Sun: 73% H, 26% Helium, and 1% of higher atomic number atoms (called “metals”) ▪ Quantities in terms of mass fraction X = 𝑚𝐻 𝑛𝐻 /ρ = density of hydrogen / total density Y = 𝑚𝐻𝑒 𝑛𝐻𝑒 /ρ = density of helium / total density Z = 𝑚𝑧 𝑛𝑧 /ρ = density of metals / total density 4 WHAT POWERS THE STARS? 5 WHAT POWERS THE STARS? Stellar energy sources Gravitational energy Chemical energy Nuclear energy 6 GRAVITATIONAL POTENTIAL ENERGY ▪ Stars shine simply due to the release of gravitational potential energy ▪ A collapsing star could convert its gravitational potential energy to heat, then glow as the heat radiated into space ▪ Gravitational potential energy is given by GMm U=− r 7 ▪ Imagine that we start with a spherical cloud of gas (its density must depend only on radius) ▪ The force on an element of mass dm within the cloud is 𝑀 𝑑𝑚 𝑑𝐹 = 𝐺 𝑟 2 𝑟 Where Mr is the total mass interior to the shell the mass dm is on ▪ The potential energy of this element of mass is 𝑀𝑟 𝑑𝑚 𝑑𝑈 = −𝐺 𝑟 The mass of an entire spherical shell is just 𝑑𝑚 = 4𝜋𝑟 2 𝜌𝑑𝑟, where 4𝜋𝑟 2 is the area of the shell, 𝑑𝑟 is its thickness, and 𝜌 is its mass density. 8 ▪ The potential energy of this entire shell is then, 𝑀𝑟 4𝜋𝑟 2 𝜌 𝑑𝑈 = −𝐺 𝑑𝑟 𝑟 ▪ Thus, the total potential energy of the cloud is just the integral over radius, out to the edge of the cloud at radius R ▪ The value of this integral depends on the manner in which the density varies with radius, but a “reasonable” approximation is to let density be constant, and equal to the average density of the cloud ▪ This will somewhat over-estimate the true value for a centrally condensed cloud 9 ▪ The average density is just total mass / total volume 𝜌 4 3 = 𝑀ൗ 𝜋𝑅 3 4 ▪ So we approximate 𝑀𝑟 = 𝜋𝑟 3 𝜌 , to get 3 𝑅 4 16 2 𝑈 = −4𝜋𝐺 𝜋 𝜌 න 𝑟 𝑑𝑟 = − 𝜋 𝐺 𝜌 2 𝑅5 2 4 3 0 15 ▪ Substituting the value of 𝜌 back into this result, we have 3 GM 2 U≈− 5 R where the approximation sign is to remind us that we are only estimating the density 10 VIRIAL THEOROM ▪ When averaged over time (which is what the brackets mean), a system that changes from one equilibrium to another gains an amount of kinetic energy that is half of the released potential energy ▪ The Virial Theorem, that for a system in equilibrium: 2K + U =0 1 1 ▪ This also means that E = K + U = − U +U= U 2 2 ▪ So another expression for the Virial Theorem: 1 𝐸 = 𝑈 2 11 ▪ By the Virial theorem, then, the maximum energy released by the gravitational collapse of a star is: 3 GM 2 E≈− 10 R ▪ In the case of the Sun, this is 3 2 1 1 ∆E = − G𝑀𝑅 − 10 𝑅𝑖 𝑅𝑓 Where, 𝑅𝑖 is the original radius and 𝑅𝑓 is the current radius 12 ▪ Assuming that the original radius, 𝑅𝑖 >>> 𝑅𝑓 , then the energy released by its gravitational collapse would be 3 2 1 ∆E = G𝑀𝑅 10 𝑅𝑓 ▪ For a solar-type star where M = 𝑀𝑠𝑜𝑙𝑎𝑟 and 𝑅𝑓 = 𝑅𝑠𝑜𝑙𝑎𝑟 3 1 ▪ we find ΔE = 2 G𝑀𝑠𝑜𝑙𝑎𝑟 = 1.1 × 1041 J 10 𝑅𝑠𝑜𝑙𝑎𝑟 13 KELVIN-HELMHOLTZ TIME SCALE ▪ If the Sun were originally much larger than it is today, how much total energy would have been liberated in its gravitational collapse? How long does it take? Originally radius, 𝑅𝑖 >>> 𝑅𝑓 3 2 1 ΔE𝑔 = − 𝐸𝑓 − 𝐸𝑖 ≅ −𝐸𝑓 ≅ G𝑀⨀ ≅ 1.1 × 1041 J 10 𝑅⨀ ▪ Assuming the luminosity is constant throughout the lifetime: ΔE𝑔 t𝐾𝐻 = ≃ 107 years This is known as the Kelvin-Helmholtz timescale 𝐿⨀ 14 ENERGY FROM CHEMICAL REACTIONS ▪ A misconception: the Sun is a “burning fireball” ▪ Burning is a chemical process which releases energy by rearrangement of chemical bonds, which involves bound electrons ▪ Chemical reactions are based on the interactions of orbital electrons in atoms ▪ Typical electron transitions in hydrogen and helium rarely exceed ten electron volts 15 If every atom in the Sun were available to release, say, 10 eV of energy, how much total energy would that be? M⊙ ▪ Total number of hydrogen atoms: N = = 1.3 × 1057 atoms MH ▪ Each atom can release 10 eV ▪ Total energy available is 1.3 × 1058 eV How long does it last? E𝐶ℎ𝑒𝑚𝑖𝑐𝑎𝑙 T𝑐ℎ𝑒𝑚𝑖𝑐𝑎𝑙 = ≃ 1.6 × 105 years 𝐿⨀ ▪ Chemical reactions cannot be responsible for the energy output of stars 16 NUCLEAR ENERGY AS SOURCE OF ENERGY ▪ While chemical processes/transitions of electrons in atoms are of the order of 10 eV, nuclear processes involve energies millions of times larger (at the order of MeV) ▪ Energy contained in the nucleus is given by Einstein’s famous equation, E = mc 2 Where m is the rest mass ▪ Rest mass of proton, neutron and electron 𝑚𝑝 = 1.67262158 × 10−27 𝑘𝑔 = 1.00727646688 𝑢 𝑚𝑛 = 1.67492716 × 10−27 𝑘𝑔 = 1.00866491578 𝑢 𝑚𝑒 = 9.10938188 × 10−31 𝑘𝑔 = 0.0005485799110 𝑢 17 ▪ When protons and neutrons are combined into nucleus, the total mass is slightly less. The difference accounts for the nuclear binding energy ▪ The binding energy is the energy required to split the nucleus into its constituent parts ▪ When lighter atoms combine into larger atoms, sometimes their combined mass gets smaller, and releases energy, known as atomic fusion Binding Energy = 28.22 MeV 18 Fusion Reaction : Initial mass = 6.693 × 10−27 kg Final mass = 6.645 × 10−27 kg Mass defect = Δm = 0.048 × 10−27 kg E = Δm 𝑐 2 = 0.43 × 10−11 J 19 How long (in years) will the Sun last if it shines at its current rate and converts 10% of its mass from hydrogen to helium in nuclear fusion? ▪ 0.43 × 10−11 J joules may sound like an insignificant amount of energy ▪ The mass defect of the helium nucleus is m = 0.048 × 10−27 kg, or 0.7% of the mass of the constituent hydrogen atoms ▪ The total nuclear energy available is Enuclear = M⊙ 𝐶 2 x 0.007x 0.1 = 1.3 × 1044 𝐽 How long does it last? Nuclear timescale E𝑛𝑢𝑐𝑙𝑒𝑎𝑟 Enough to account for the energy output T𝑛𝑢𝑐𝑙𝑒𝑎𝑟 = ≃ 1010 years of stars 20 𝐿⨀ NUCLEAR REACTIONS IN STARS Proton-Proton chain (p-p chains) CNO cycle Triple-alpha reaction More reactions at higher temperatures Carbon burning; Oxygen burning; Silicon burning Neutron capture 21 PROTON – PROTON CHAIN (P-P CHAIN) ▪ Two hydrogen nuclei combine to make one deuterium nucleus (an isotope of hydrogen). A positron and neutrino are emitted ▪ A hydrogen nucleus combines with a deuterium nucleus to produce helium-3. A gamma ray is emitted ▪ Two helium-3 nuclei combine to make a helium-4 nucleus. Two hydrogen nuclei are emitted 22 23 PP I, PP II AND PP III CHAIN ▪ In the Sun, each helium-3 nucleus produced in these reactions exists for only about 400 years before it is converted into helium-4 ▪ Once the helium-3 has been produced, there are four possible paths to generate 4He ▪ In PP I, helium-4 is produced by fusing two helium-3 nuclei 24 ▪ The PP II and PP III branches fuse 3He with pre-existing 4He to form beryllium-7, which undergoes further reactions to produce two helium-4 nuclei PP II : PP III : 25 possible variations in the steps of the PP chain 26 ▪ The following conditions in the stellar interior will alter the branching ratios: ✓ Variations in the temperature ✓ Density and abundance of elements in the stellar interior 27 possible variations in the PP chain due to temperature variations in the stellar interior ▪ Although the branching ratios can vary, the energy released by any PP chain reaction will be the same ▪ The formation of each helium-4 nucleus via any branch of the PP chain liberates 26 MeV of energy 28 CNO CYCLE ▪ The ‘CNO cycle’ refers to the Carbon-Nitrogen-Oxygen cycle, a process of stellar nucleosynthesis in which stars fuse hydrogen into helium via a six-stage sequence of reactions 29 SIX STAGES SEQUENCE OF CNO CYCLE ▪ A carbon-12 nucleus captures a proton and emits a gamma ray, producing nitrogen-13 ▪ Nitrogen-13 is unstable and emits a beta particle, decaying to carbon-13 ▪ Carbon-13 captures a proton and becomes nitrogen-14 via emission of a gamma-ray ▪ Nitrogen-14 captures another proton and becomes oxygen-15 by emitting a gamma- ray ▪ Oxygen-15 becomes nitrogen-15 via beta decay ▪ Nitrogen-15 captures a proton and produces a helium nucleus (alpha particle) and carbon-12, which is where the cycle started 30 31 Since both the PP chain and the CNO cycle produce energy by converting hydrogen to helium, they compete with each other. How do we know which reactions take place within a given star? 1. The CNO cycle can only occur in stars where carbon, nitrogen or oxygen is present. However, as only a small quantity of these elements is required, this condition is often fulfilled 2. The reactions have very different temperature dependencies 32 33 TRIPLE ALPHA REACTION ▪ The combination or fusion of three alpha particles (helium nuclei 4He) to form a carbon nucleus (12C) is known as the triple alpha process ▪ If the central temperature rises to 108 K, six times hotter than the Sun's core, alpha particles can fuse fast enough to get past the beryllium-8 barrier and produce significant amounts of stable carbon-12 34 ▪ As a side effect of the process, some carbon nuclei fuse with additional helium to produce a stable isotope of oxygen 35 CARBON AND OXYGEN BURNING ▪ As the temperature of the stellar core increases, more advanced nuclear fusion reactions are possible ▪ If sufficient carbon and oxygen is generated by the triple alpha process, for example, these nuclei may fuse to create still heavier elements ▪ The temperature threshold for carbon burning is approximately 5 x 𝟏𝟎𝟖 K, and for oxygen burning, temperatures greater than 𝟏𝟎𝟗 K are required ▪ In general, these temperatures only occur in the interiors of massive stars during later stages of their evolution 36 ▪ The principal carbon fusion reactions are as follows: ▪ The principal oxygen fusion reactions are as follows: 37 END OF THE CHAPTER