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PHY216 PHY472 Data Provided: Data Provided: andconstants physical constants FormulaFormula sheet andsheet physical DEPARTMENT OF PHYSICS DEPARTMENT OF PHYSICS & Spring&Semester 2015-2016 ASTRONOMY ASTRONOMY Autumn Semester 2009-2010 GALAXIES 2 hours 2 hours ADVANCED QUANTUM MECHANICS Answer question ONE (Compulsory) and TWO questions, Answer ALL questions from Section A, and TWO questions fromother Section B. Please one each from clearly indicate theA question numbers section and section B. on which you would like to be examined on the front cover of your answer book. Cross through any work that you do not want to All be examined. questions are marked out of ten. The breakdown on the right-hand side of the paper is meant as a guide to the marks that can be obtained from each part. Questions from Section A are marked out of ve, whereas those from Section B are marked out of fteen. The breakdown on the right-hand side of the paper is meant as a guide to the marks that can be obtained from each part. PHY216 TURN OVER 1 PHY216 SECTION A – answer ALL questions 1. A Cepheid variable in a nearby galaxy has a mean apparent magnitude in the V band of hmV i = 24.45 magnitudes, a period of P = 8 days, and suffers extinction of AV = 0.35 magnitudes due to foreground dust in the Milky Way. If Cepheid variables follow a V-band period-luminosity relation of form hMV i = −2.76 log10 P(days) − 4.16, estimate the distance to the galaxy. What are the main uncertainties involved when using Cepheid variables as distance indicators? [5] 2. Briefly explain each of the following. - The blue colours of reflection nebulae. - The detection of stars moving at very high velocities (>1000 km s−1 ) close to the Galactic Centre. - The low metallicities and high velocities relative to the Sun of stars in the Galactic Halo. - The red colours of elliptical galaxies. - The fact that the colours of spiral galaxies become bluer along the Hubble sequence from type Sa to Sc. [5] 3. If the stellar bulge of a nearby spiral galaxy has a mass of 109 M and an effective radius of 2 kpc, estimate its velocity dispersion (σ ). [5] 4. HI 21cm observations show that a spiral galaxy has a flat rotation curve at radial distances between 5 and 30 kpc from its nucleus, with an inclination corrected rotation velocity of 250 km s−1 . Estimate the mass density in the dark matter halo at a distance of 15 kpc from the nucleus. [5] PHY216 CONTINUED 2 PHY216 SECTION B — answer TWO questions 5. Write an essay on the determination of the Hubble Constant (H0 ). Your answer should include reference to the following: - the cosmic distance ladder and the steps required to produce an accurate value of H0 ; - the various primary and secondary distance indicators required (give a brief description of two of each); - the problems caused by peculiar galaxy velocities relative to the Hubble Flow; - how estimates of the Hubble Constant have changed since the first determinations in the 1920s; - the reasons for the early inaccuracy in the Hubble Constant; [15] - the impact of the Hubble Space Telescope. 6. (a) Oort’s constants (A and B) are defined as follows: " " # # 1 1 dvc dvc A= , B = − ω0 + , ω0 − 2 dr R0 2 dr R0 where ω0 is the angular velocity of the local standard of rest (LSR) around the Galactic centre in km s−1 kpc−1 , vc is the circular velocity in km s−1 , and r is the radial distance in kpc. Currently the best estimates of Oort’s constants and the distance to the centre of the Milky Way are A = 14.8 ± 0.8 km s−1 kpc−1 , B = −12.4 ± 0.6 km s−1 kpc−1 , and R0 = 8.0 ± 0.5 kpc respectively. Estimate, with associated uncertainties, (i) the circular rotation speed of the LSR around the centre of the Galaxy in km s−1 , and (ii) the velocity gradient of the disk of the Galaxy in the vicinity of the Sun in km s−1 kpc−1 . [6] (b) Describe in detail the tangent point method for determining the inner rotation curve of the Milky Way. [4] (c) Why is it much harder to measure the outer rotation rotation curve of the Milky Way than its inner rotation curve? [2] (d) The rotation curves of the Milky Way and other spiral galaxies are observed to remain flat out to large radii — well beyond the main concentrations of starlight. Explain why this might imply that they contain large amounts of dark matter. [3] PHY216 TURN OVER 3 PHY216 7. (a) Describe the general features that the various types of active galactic nuclei (AGN) have in common. [4] Based on optical spectroscopic observations, how would you distinguish an AGN from the nucleus of a normal, non-active galaxy? [2] (b) Compare and contrast, in terms of observational features, the following pairs of objects: - Type 1 and Type 2 Seyfert galaxies; - quasars and Type 1 Seyfert galaxies; - FRI and FRII radio galaxies. (c) If a quasar in the distant universe has a bolometric luminosity of 1039 W, estimate its mass accretion rate. State any assumptions made. [2] [2] [2] [3] 8. Write detailed accounts of any three of the following. (a) The general differences between spiral and elliptical galaxies. [5] (b) Determining the masses of elliptical galaxies. [5] (c) The observational differences between boxy and disky elliptical galaxies. [5] (d) The light profiles of elliptical galaxies. [5] (e) Theories for the formation of elliptical galaxies. [5] END OF QUESTION PAPER PHY216 4 PHYSICAL CONSTANTS & MATHEMATICAL FORMULAE Physical Constants electron charge electron mass proton mass neutron mass Planck’s constant Dirac’s constant (~ = h/2π) Boltzmann’s constant speed of light in free space permittivity of free space permeability of free space Avogadro’s constant gas constant ideal gas volume (STP) gravitational constant Rydberg constant Rydberg energy of hydrogen Bohr radius Bohr magneton fine structure constant Wien displacement law constant Stefan’s constant radiation density constant mass of the Sun radius of the Sun luminosity of the Sun mass of the Earth radius of the Earth e = 1.60×10−19 C me = 9.11×10−31 kg = 0.511 MeV c−2 mp = 1.673×10−27 kg = 938.3 MeV c−2 mn = 1.675×10−27 kg = 939.6 MeV c−2 h = 6.63×10−34 J s ~ = 1.05×10−34 J s kB = 1.38×10−23 J K−1 = 8.62×10−5 eV K−1 c = 299 792 458 m s−1 ≈ 3.00×108 m s−1 ε0 = 8.85×10−12 F m−1 µ0 = 4π×10−7 H m−1 NA = 6.02×1023 mol−1 R = 8.314 J mol−1 K−1 V0 = 22.4 l mol−1 G = 6.67×10−11 N m2 kg−2 R∞ = 1.10×107 m−1 RH = 13.6 eV a0 = 0.529×10−10 m µB = 9.27×10−24 J T−1 α ≈ 1/137 b = 2.898×10−3 m K σ = 5.67×10−8 W m−2 K−4 a = 7.55×10−16 J m−3 K−4 M = 1.99×1030 kg R = 6.96×108 m L = 3.85×1026 W M⊕ = 6.0×1024 kg R⊕ = 6.4×106 m Conversion Factors 1 u (atomic mass unit) = 1.66×10−27 kg = 931.5 MeV c−2 1 astronomical unit = 1.50×1011 m 1 eV = 1.60×10−19 J 1 atmosphere = 1.01×105 Pa 1 Å (angstrom) = 10−10 m 1 g (gravity) = 9.81 m s−2 1 parsec = 3.08×1016 m 1 year = 3.16×107 s Polar Coordinates x = r cos θ y = r sin θ ∂ 1 ∂2 1 ∂ 2 r + 2 2 ∇ = r ∂r ∂r r ∂θ dA = r dr dθ Spherical Coordinates x = r sin θ cos φ y = r sin θ sin φ z = r cos θ dV = r2 sin θ dr dθ dφ 1 ∂ 1 ∂2 1 ∂ ∂ 2 2 ∂ ∇ = 2 r + 2 sin θ + 2 2 r ∂r ∂r r sin θ ∂θ ∂θ r sin θ ∂φ2 Calculus f (x) f 0 (x) f (x) f 0 (x) xn ex nxn−1 ex tan x sin−1 ln x = loge x 1 x sin x cos x cos x − sin x cosh x sinh x sinh x cosh x a cos−1 xa tan−1 xa sinh−1 xa cosh−1 xa tanh−1 xa cosec x −cosec x cot x uv sec x sec x tan x u/v sec2 x x √ 1 a2 −x2 − √a21−x2 a a2 +x2 √ 1 x2 +a2 √ 1 x2 −a2 a a2 −x2 0 0 u v + uv u0 v−uv 0 v2 Definite Integrals Z ∞ xn e−ax dx = 0 Z +∞ n! an+1 r (n ≥ 0 and a > 0) π a −∞ r Z +∞ 1 π 2 −ax2 xe dx = 2 a3 −∞ Z b b Z b du(x) dv(x) Integration by Parts: u(x) dx = u(x)v(x) − v(x) dx dx dx a a a −ax2 e dx = Series Expansions (x − a) 0 (x − a)2 00 (x − a)3 000 f (a) + f (a) + f (a) + · · · 1! 2! 3! n X n n−k k n n! n Binomial expansion: (x + y) = x y and = (n − k)!k! k k k=0 Taylor series: f (x) = f (a) + (1 + x)n = 1 + nx + ex = 1 + x + n(n − 1) 2 x + ··· 2! x 2 x3 + +· · · , 2! 3! sin x = x − ln(1 + x) = loge (1 + x) = x − Geometric series: n X rk = k=0 Stirling’s formula: (|x| < 1) x3 x5 + −· · · 3! 5! x2 x3 + − ··· 2 3 and cos x = 1 − x2 x4 + −· · · 2! 4! (|x| < 1) 1 − rn+1 1−r loge N ! = N loge N − N or ln N ! = N ln N − N Trigonometry sin(a ± b) = sin a cos b ± cos a sin b cos(a ± b) = cos a cos b ∓ sin a sin b tan a ± tan b 1 ∓ tan a tan b sin 2a = 2 sin a cos a tan(a ± b) = cos 2a = cos2 a − sin2 a = 2 cos2 a − 1 = 1 − 2 sin2 a sin a + sin b = 2 sin 21 (a + b) cos 12 (a − b) sin a − sin b = 2 cos 12 (a + b) sin 12 (a − b) cos a + cos b = 2 cos 12 (a + b) cos 12 (a − b) cos a − cos b = −2 sin 12 (a + b) sin 21 (a − b) eiθ = cos θ + i sin θ 1 iθ 1 iθ cos θ = e + e−iθ and sin θ = e − e−iθ 2 2i 1 θ 1 θ cosh θ = e + e−θ and sinh θ = e − e−θ 2 2 sin a sin b sin c Spherical geometry: = = and cos a = cos b cos c+sin b sin c cos A sin A sin B sin C Vector Calculus A · B = Ax Bx + Ay By + Az Bz = Aj Bj A×B = (Ay Bz − Az By ) î + (Az Bx − Ax Bz ) ĵ + (Ax By − Ay Bx ) k̂ = ijk Aj Bk A×(B×C) = (A · C)B − (A · B)C A · (B×C) = B · (C×A) = C · (A×B) grad φ = ∇φ = ∂ j φ = ∂φ ∂φ ∂φ î + ĵ + k̂ ∂x ∂y ∂z ∂Ax ∂Ay ∂Az + + ∂x ∂y ∂z ∂Ax ∂Az ∂Ay ∂Ax ∂Az ∂Ay − î + − ĵ + − k̂ curl A = ∇×A = ijk ∂ j Ak = ∂y ∂z ∂z ∂x ∂x ∂y div A = ∇ · A = ∂ j Aj = ∇ · ∇φ = ∇2 φ = ∂ 2φ ∂ 2φ ∂ 2φ + + 2 ∂x2 ∂y 2 ∂z ∇×(∇φ) = 0 and ∇ · (∇×A) = 0 ∇×(∇×A) = ∇(∇ · A) − ∇2 A