AU/Mpc km/au s 1 (2) = s 1 (3) n(t 0 ) = ɛ(t 0) mc 2 (7) m(h) = m p = kg (8)

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1 Cosmology Solutions Useful constants: 1AU = m 1pc = 206, 265AU G = kg 1 m 3 s 2 M sun = kg m proton = kg 1. Cosmic density (a) Calculate the energy density of particles today (giving your answer in SI units) Calculate the energy density of particles today (giving your answer in SI units) and hence calculate the number density of the particles today if the particles are: Since H 0 given to 2 sig fig answers should be given to same precision H 0 = 72kms 1 Mpc 1 (1) = AU/Mpc km/au s 1 (2) = s 1 (3) ɛ(t 0 ) = 3c2 H 2 0 8πG = 3(3 108 ms 1 ) 2 ( s 1 ) 2 (5) (8π( m 3 kg 1 s 2 ) = kgm 1 s 2 (2sf) (6) (b) Hence calculate the number density of the particles today if the particles are: i. Hydrogen atoms (giving your answer in atoms m 3 ). ii. Stars, the same mass as our Sun (giving your answer in stars kpc 3 ). iii. Galaxies, the same mass as the Milky Way (giving your answer in galaxies Mpc 3 ). i. Hydrogen atoms (giving your answer in atoms m 3 ). (4) Assume n(t 0 ) = ɛ(t 0) mc 2 (7) m(h) = m p = kg (8) n H (t 0 ) = kgm 1 s 2 ( kg)( ms 1 ) 2 (9) = 5.6atoms m 3 (2sf) (10) Cosmology - 1

2 ii. Stars, the same mass as our Sun (giving your answer in stars kpc 3 ). m Sun = kg (11) n stars (t 0 ) = ( kgm 1 s 2 ( kg)( ms 1 ) 2 (12) = m 3 (13) = 137stars kpc 3 (2sf) (14) iii. Galaxies, the same mass as the Milky Way (giving your answer in galaxies Mpc 3 ). m(mw ) = M sun (15) 137starskpc 3 n galaxies (t 0 ) = stars/mw (103 Mpc/kpc) 3 Mpc 3 (16) = 0.24galaxies Mpc 3 (2sf) (17) 2. Observing distance galaxies {Expanded from Harwit: Astrophysical Concepts} This question deals with what limits the observation of distant galaxies - is it their size, brightness or whether the universe has been around long enough for light from them to have reached us? The smallest extragalactic sources resolved with currently available telescopes are of the order of 0.05 of arc in diameter. While the faintest observable objects are of about 30th magnitude. (a) Find the distance to i. the farthest resolved galaxies Careful with units here - don t forgot to change arcsec to radians Either calculate distance in same units as diameter (in which case you need to change angle in arcsec to RADIANS). Distance, D, to galaxy in kpc is related to diameter,d, in kpc and resolution, θ in radians. θ(rad) = o 2πrad o (18) = rad (19) D(kpc) = d(kpc) θ(radians) (20) = rad (21) = kpc (22) = 181Gpc (23) Or use analogy to trigonometric parallax to use angle in arcseconds directly (BUT remember diameter is then measured in AU and distance is in pc!). Cosmology - 2

3 Distance, D, to galaxy in pc is related to diameter,d, in AU and resolution, θ, in arcsec. In Euclidean D(pc) = d(au) θ(arcsec) (24) = pc AU/pc (25) 0.05 = pc (26) = 182Gpc (27) ii. The faintest obesrvable galaxies Let magnitude be m gal, and F M31/gal be the observed flux from M31/ galaxy, m gal m M31 = 2.5 lg F M31 F gal (28) = 2.5 lg L M31 L gal D 2 gal D 2 M31 (since F = L/4πD 2 ) (29) Since we are assuming they have the same luminosity m gal = lg ( Dgal ) 778kpc (30) finally we want the galaxy to be the faintest, so m gal = 30 so that = 5 lg ( ) Dgal D gal = 103Gpc (31) 778kpc (b) Giving a sensible value of H 0, what would you estimate the size of the observable horizon to be today? H 0 = 72kms 1 Mpc 1 (32) d Hubble = c H 0 (33) = kms 1 72kms 1 Mpc 1 (34) = 4.2Gpc(2 sig f ig) (35) Note that this would imply that the galaxies we are interested in are receding from us at superluminal speeds (if Hubble s law holds to those distances) H 0 = AU/Mpc km/au s 1 (36) Cosmology - 3

4 = s 1 (37) v = H 0 D (38) = ( s 1 )( pc AU/pc ms 1 ) ms 1 (39) = ms 1 (40) = 27.8c (41) In Special Relativity it is not possible to accelerate matter to the speed of light or beyond. However in General Relativity, in which space is expanding, it is possible for matter to move away from us at superluminal speeds. To get some idea of this, consider two ants moving on the surface of a balloon initially 2cm apart from one another, with the ants having a maximum speed of 1cms 1 (as an analogy to the speed of light). If the balloon is not being blown up then the ants could approach each other and meet 1 second later. However if the ballon is being blown up, then the rate of increase in the distance separating the ants is in no way related to their maximum walking speed, but rather this rate of expansion of space (the balloon s surface) which is not related in any way to their maximal speed. (c) So what is the main limitation on distant galaxy observation? Based on the (simplistic) analysis above, the main limitation on observing these distant galaxies, would be that the galaxies lie outside the observable horizon (roughly taken to be the Hubble length c/h 0 ). We ve not taken into account the fact that the Hubble parameter H(t) varies over time 3. Cosmological and kinematical redshift Consider light from a galaxy at cosmological redshift, z cosmo, and with a peculiar recession velocity (the difference between the total velocity and the Hubble expansion at the position of the galaxy), v pec, purely along the line of sight. (a) Show that the combined redshift due to these two effects (kinematical and cosmological) is (1 + z tot ) = (1 + z cosmo )(1 + z kin ) (42) where the kinematical redshift, z kin, is related to the peculiar velocity through the Doppler formula. [Hint: compare with situation for an observer close to the source]. In a static universe the observed redshift of light from a galaxy receding with velocity v pec will be observed with wavelength λ received,static = (1 + z kin )λ emitted (43) z kin v pec c Cosmology - 4 (v pec c) (44)

5 This redshift is independent of the distance of the observer from the source, only the relative velocity. In a non-expanding universe this would be the only effect observed. The redshift caused by the expansion of the universe is then an additional, separate effect distorting the photon already stretched by the peculiar velocity. This effect is dependent on how much the universe has expanded as the light travels to the observer from the source. λ received,expanding = (1 + z cosmo )λ received,static (45) = (1 + z cosmo )(1 + z kin )λ emitted (46) (1 + z total ) = λ received,expanding λ emitted (47) = (1 + z cosmo )(1 + z kin ) (48) You can visualize this separation of the 2 redshifts by imagining an observer right next to the source, which is receeding from them at speed v pec then the expansion of the universe will not alter his observed wavelength of light. This redshifted light as it passed to a second observer on earth today will be cosmologically redshifted with respect to the first observer in the form above. (b) Assuming H 0 = 72Mpc 1 kms 1 and a typical galactic peculiar velocity of 500 kms 1 : i. At what distance does the cosmological redshift exceed that caused by the peculiar motion? ii. Comment on whether peculiar velocities are going to be an important consideration for measurements of the universe s expansion using galaxies i) within our Local Group of galaxies, ii) in galaxies within the Coma cluster. [Feel free to Google to get any necessary information!] Assuming Hubble s law v cosmo = H 0 d for inferred cosmological recession ve locity, v cosmo at a distance d then for v cosmo > v pec d > v pec H 0 500km/s 75Mpc 1 km/s 6.94Mpc (49) So this would be outside our Local Group of galaxies (diameter 3Mpc) but still within the Local Supercluster and closer than the Coma cluster of galaxies ( 100M pc away). Peculiar velocities therefore would be an issue when using Local Group galaxies but not a dominant effect on cosmological scales e.g. to Coma. Even so they would remain a source of statistical error. 4. Curvature {Slightly modified from Ryden Ch. 3 Qu. 3.2 and 3.3} Consider yourself a 2-dimensional being living on the surface of the Earth, modeled as a perfect sphere radius R = 6371km. In this 2D world, light travels on the surface of the Earth. Cosmology - 5

6 (a) The metric is ds 2 = dr 2 + R 2 sin 2 ( r R ) dθ 2 (50) (b) Sketch the variation of the angular width dθ you measure for an object of length ds = 10km( R) (lying perpendicular to your line of sight), as its distance away from you, r, varies. Clearly give a value on the sketch for any minimum or maximum value of dθ in arcseconds. Since object is lying perpendicular to your line of sight (i.e. wholly in the θ direction) dr = 0 ( ) r ds = R sin dθ (51) R dθ(rad) = 10km km sin(r(km)/6371km) As r increases from 0 to πr/2 the angular diameter decreases, this is because the distance separating the two lines of sight to the ends of the object first increase as you move towards the equator, and then decrease as you move towards the opposite pole. As you get to the pole all lines of sight converge to a single point i.e. all lines of sight will be able to see the opposite pole, and the object is of infinite angular extent (the concept of angular diameter is ill defined). (52) d! d! min =5.40 (3sf) d! min r/"r Cosmology - 6

7 (c) If you draw a circle of radius r on the Earth s surface, the circle s circumference is ( ) r C = 2πR sin. (53) R If you could measure distances with an error of ±1m, how large a radius would a circle you draw on the Earth s surface have to be to convince yourself that the Earth is spherical rather than flat? Assume that R is exact to the nearest meter, and give your answer to the nearest meter. ( ) r C flat C sphere = 2πr 2πR sin 1m (54) R x sin x π(6371), (x = r R ) (55) Either solve by iteration or use small angle approximation for sin x, 5. Gravitational lensing sin(x) x x3 (for x 1) (56) 3! ( ) /3 x (57) 2π(6371) x (58) r km (to nearest m) (59) The angle by which a photon is deflected by a compact mass M, at an impact parameter b is given by α = 4GM c 2 b A light ray just grazes the surface of a neutron star (M = kg, R = km). Through what angle α is the light ray bent by the gravitational lensing? Give you answer in arcminutes. (60) α = 4GM (rad) (61) c 2 R = rad (62) = 2.55 (63) 6. Measuring dark matter from the dynamics of a cluster of galaxies It is 1933 and your professor, Fritz Zwicky, has just given you some new observations of a cluster of galaxies for which he wants you to do a quick analysis to estimate the cluster s mass based on their dynamics. He has measured the radial velocities (i.e. along the line of sight) of twelve galaxies: Cosmology - 7

8 1110 km/s 890 km/s 1140 km/s 860 km/s 750 km/s 1250 km/s 1320 km/s 670 km/s 810 km/s 1190 km/s 590 km/s 1410 km/s He reports that the galaxies all have luminosities of about L sun, and says it is reasonable for you to assume that the stars in the galaxies all have about the same mass and luminosity as the sun (we know this is not true, but on average it is a reasonable approximation). (a) Briefly describe how one might measure the radial velocity of a galaxy. (b) Calculate an estimate of the 1D velocity dispersion of the galaxies. (c) Based on these observations, and treating the galaxies as if they were uniformly distributed in a spherical cluster of constant density with radius R = 300kpc (a back of an envelope assumption), show that the virial theorem is not satisfied for this system. (d) If the virial theorem were not satisfied, (qualitatively & briefly) what would happen to the cluster of galaxies in the future? If you can, give a rough time scale for your answer. (e) Zwicky suggests that there might be some form of unseen dark matter that is responsible for making the virial theorem hold. How much more dark matter is required? You can assume that the dark matter is evenly distributed and has the same kinetic energy per unit mass as the galaxies and that M total = fm luminous, where f is a constant. Give your answer in terms of f. 7. Nuclear energy Suppose the Sun was made up of 100% carbon-12 and that energy was produced through a reaction 12 C + 12 C 24 Mg. (64) The atomic weight of 12 C is 12 and 24Mg is (a) How much energy, in ev, is released per carbon nucleus in this reaction? Change in binding energy, E, is given by E = [mass( 24 Mg) 2 mass( 12 C]c 2 (65) = ( )931.46M ev (66) = 14.0M ev (67) Negative because energy is released. Cosmology - 8

9 (b) Assuming the Sun could burn 10% of the Carbon, how long would its lifetime be? Change in Binding energy per mole of Magnesium formed E = ev J/eV nuclei/mol (68) = Jmol 1 (69) Let lifetime be T, and total energy emitted from burning 10% of the carbon be, E tot, T E tot L sun (70) L sun = W (71) To get E tot, note that 2 carbon nuclei are required to produce 1 magnesium nuclei, so E tot = 0.1 M sun 2 12g) Jmol 1 (72) = kg kg Jmol 1 (73) = J (74) T = J (75) J/s = s (76) = 930 million years (77) Cosmology - 9

Astronomy 102: Stars and Galaxies Review Exam 3

October 31, 2004 Name: Astronomy 102: Stars and Galaxies Review Exam 3 Instructions: Write your answers in the space provided; indicate clearly if you continue on the back of a page. No books, notes, or