Today in Astronomy 142: the Milky Way
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Today in Astronomy 142: the Milky Way The shape of the Galaxy Stellar populations and motions Stars as a gas: • Scale height, velocities and the mass per area of the disk • Missing mass in the Solar neighborhood Wide-angle photo and overlay key of the Sagittarius region of the Milky Way. The very center of the Milky Way lies behind particularly heavy dust obscuration. (By Bill Keel, U. Alabama.) 19 March 2013 Astronomy 142, Spring 2013 1 The number of stars brighter than f0 Suppose stars are uniformly distributed in space, with number density n, and have typical luminosity L. How many are brighter (that is, have flux greater) than some value f0? Presuming that there is no extinction: to the flux f0 corresponds a distance r0 . 2 fLr0=44ππ 00 ⇒= r L f 0 −3/2 4π ∝ f N( f>= f) rn3 log N 0 003 3 4π nL2 − = ∝ 32 f0 log f 34π f0 0 19 March 2013 Astronomy 142, Spring 2013 2 What is the shape of the Milky Way? Herschel (1785), and later Kapteyn (1922), used this fact log N −3/2 to characterize the shape of ∝ f0 the Milky Way. Their idea ( ff> 0 ) was that if the Milky Way has edges, there would be points past which N would decrease log f −3/2 0 faster than f0 . And indeed actual star Star counts in directions counts at low fluxes are of the Milky Way disk less than predicted by this (blue), and in the relationship, and the perpendicular directions numbers at larger fluxes. (red). 19 March 2013 Astronomy 142, Spring 2013 3 Herschel’s Milky Way “Section of our sidereal system” (Herschel 1785). The long axis of the figure runs roughly from 20h22m, +35° in Cygnus (left) to 8h20m, -35° in Puppis (right); the short axis points h m ° toward 12 24 ,+58 (Ursa Major). Sun The Great Rift in the summer Milky Way, interpreted as a dearth of stars – referred to as “an opening in the heavens,” since he was writing in English this time. 19 March 2013 Astronomy 142, Spring 2013 4 Kapteyn’s “universe” Kapteyn (1922) knew the distances to many nearby stars, and could therefore calibrate the star counts in terms of the stellar density n, and a shape of the stellar assemblage in terms of hydrostatic structure. This enabled him to hand-wave a position for the Sun based upon the stellar density in the Solar neighborhood: r = 650 parsecs, z = 38 parsecs. Sun 19 March 2013 Astronomy 142, Spring 2013 5 Shapley and Galactocentric distance Harlow Shapley thought that globular clusters, massive as they are, would indicate Galactic structure better than stars. He was also armed with Henrietta Leavitt’s discovery of the Cepheid period-luminosity relation (6 April 2013), which Dots: cluster positions he applied to the RR Lyr stars. projected onto the plane This led to a Solar distance of the Milky Way. from the center of the cluster Circles: radii in integer multiples of 10 kpc. From distribution of 100 parsecs Shapley 1919. (Shapley 1918). 29 March 2011 Astronomy 142, Spring 2011 6 The Milky Way as an “island universe” All these methods got the shape, and the Sun’s distance within from the center, quite wrong. As we have seen, interstellar extinction is substantial in the plane of the Milky Way, and obscures most of the distant starlight. Both Herschel and Kapteyn were therefore seeing only to the edge of the extinction, not to the edge of the stars. By Kapteyn’s time many astronomers were on the right track by analogy: that the Milky Way is like the spiral nebulae, and we live well off the center, as Shapley found. • But not nearly as far: modern measurements of parallax of water-vapor masers in molecular clouds near the Galactic center give 8.4 kpc for our Galactocentric radius. (Reid et al. 2009). 19 March 2013 Astronomy 142, Spring 2013 7 Halo, bulge and disk The globular clusters seem to trace a spherical halo around the galaxy, in which we are immersed ourselves. One can see non-cluster stars out there too, if one looks hard enough. Once we know where to look for the Galactic center, we notice a couple of other features: the bulge, a thicker, brighter concentration of stars surrounding the center… Bruno Gilli/ESO 19 March 2013 Astronomy 142, Spring 2013 8 Halo, bulge and disk …and the disk, a belt of stars and extinction that passes through the center – this is the Milky Way proper. Since the belt seems not to have ends, we are also immersed within it. Distribution of stars thicker than that of dust everywhere along the belt: there is a thick disk and a thin disk. Much like many other galaxies. Bruno Gilli/ESO 19 March 2013 Astronomy 142, Spring 2013 9 The Milky Way’s central regions, in starlight (near- infrared), from the NASA COBE DIRBE experiment. NGC 891, also in starlight (near- infrared), from the 2MASS survey (U. Mass./NASA). 19 March 2013 Astronomy 142, Spring 2013 10 Schematic structure of the Milky Way Astronomy Today Figure: Chaisson Chaisson and Figure: McMillan, 19 March 2013 Astronomy 142, Spring 2013 11 Halo, bulge and disk (continued) As we will be seeing this week and next, the different components of the Galaxy owe their distributions to differences in motion. The disk is dominated by rotation. Objects which belong to the disk have randomly-directed and rotational components of motion, but the rotational components are much larger. The bulge and halo are dominated by random motion, with little to no trace of rotation. Though our immersion within two of the components makes things complicated, we can use the motions to weigh the various components, and the Galaxy itself. 19 March 2013 Astronomy 142, Spring 2013 12 Halo, bulge and disk (continued) The visible mass of the galactic halo is small compared to that of the disk and bulge. As we will see, there are strong reasons to think that the true mass of the halo is similar to the others, leading to hypotheses of dark matter. Dynamics and composition of stars are correlated: Population I: small dispersion of velocities (i.e. small random velocities), absorption lines of heavy metals, confined to a very thin plane. Relatively young. Population II: large dispersion of velocities, can lie further from the Galactic plane. Population I lies predominantly in the disk, less in the bulge, not in the halo. Population II can be found in all three components. 19 March 2013 Astronomy 142, Spring 2013 13 Motions of stars in the Galaxy, and their use in determining its mass distribution How much does the Galaxy weigh? Stars move about in response to the gravitational potential of the rest of the stars in the galaxy, so… their systematic motions (rotation) can be used with Newton’s Laws to measure masses within the Galaxy. • This works even better with interstellar gas than with stars. Stars are too massive to be influenced by the pressure of the interstellar medium, but they collide inelastically very rarely, so their random motions can be used with thermodynamics to measure masses within the Galaxy - the stars in this sense can be thought of as particles in a gas. 19 March 2013 Astronomy 142, Spring 2013 14 Mass per unit area of the Galaxy’s disk, in the Sun’s neighborhood The radial structure of the disk is determined in the usual manner from centrifugal support: balancing the force on a test particle at radius r from the mass M(r) contained in interior orbits, with centrifugal force. Just like a protoplanetary disk, or any other astro-disk. mV2 r V GM( r) m r2 19 March 2013 Astronomy 142, Spring 2013 15 Mass per unit area of the Galaxy’s disk, in the Sun’s neighborhood (continued) The vertical structure is determined by hydrostatic equilibrium, as usual, but there are two main differences from protoplanetary disks: much of a galactic disk is self gravitating: the weight is from the disk itself, not from a “star” in the middle. the pressure is that of the stars’ motions. Guess: kinetic energy 1 2 PA Pv≈≈ρ stars stars volume 2 stars random V madisk 19 March 2013 Astronomy 142, Spring 2013 16 Mass per unit area of the Galaxy’s disk, in the Sun’s neighborhood (continued) Weight. If the disk is thin and self gravitating, we can regard it locally as an infinite plane, and work out the weight of stars above and below the plane accordingly. m h µ dF Mass per unit r disk area, not to dr be confused with mean particle mass in Gm(2πµ rdr ) h a gas. dF = 22 hr+ hr22+ 19 March 2013 Astronomy 142, Spring 2013 17 Mass per unit area of the Galaxy’s disk, in the Sun’s neighborhood (continued) m h µ dF r dr ∞∞2rdr du F= πµ G mh = πµG mh ∫∫0 3/2 h2 3/2 (hr22+ ) u ∞ u−1/2 1 = πµGmh = 22πµGmh= πµ Gm −1/2 2 h2 h 19 March 2013 Astronomy 142, Spring 2013 18 Mass per unit area of the Galaxy’s disk, in the Sun’s neighborhood (continued) Pressure. Recall formula for pressure in terms of particle number density, speed and momentum (12 February): F 11dp nAδ z Number of particles that hit Pp≅= ≅ z A A dt Aδ t wall 2 Typical momentum per particle =nvzz p = ρ v z Time interval in which they hit Consider a certain class of stars to be gas particles, and consider the component of their motion perpendicular to the Galactic plane.