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Milky Way Kinematics, or how we discovered the geometry and dynamics of our own galactic environment. The name “galaxy” comes from the Greek word for milk. The Galaxy appears to us like a river of milk, and hence its name, the h Milky Milk Way. W In the northern hemisphere, we can see the stars in the direction of the galactic center only in the summer time. [Fig. 12.2 from Shu] 1 Milky Way in Sagittarius The Visible Sky 2 Mosaic of Milky Way Map local stellar environment by means of star counts. First plot positions on sky of stars brighter than some level 1. Then plot positions of stars between brightness levels 1 and 2. And so on. If distribution of stars in space is uniform, each plot will be also. Number of stars per square degree in each successive plot will increase in a fashion that we could predict. Find stars concentrated in band of sky, the galactic equator. Find that numbers of fainter stars do not increase away from the galactic equator so rapidly (as uniform distribution would imply). This true also toward galactic equator, although shortfall is smaller. Led to “Kapteyn Universe” with us in center. 3 Starlight blocked by clouds of dust in galactic disk. In 1917, after Henrietta Levitt’s discovery of the period-luminosity relation for the Cepheid variable stars in the Small Magellanic Cloud, Harlow Shapley proposed that these stars were pulsating. Shapley applied Levitt Levitt’ss relation to RR-Lyrae RR Lyrae variable stars in globular clusters, to get their distances. Got systematic overestimates, but right geometrical distribution. (Globular clusters old, with few heavy elements.) Globular clusters have spherical distribution centered on a position about 30,000 light years from us (the modern, corrected distance) in the direction of the galactic equator. Thus we live in a disk, far from its center. Globular clusters orbit center in all directions, disk stars all go around in one direction and in a single plane. 4 Typical orbit of disk star. Vertical velocity is small compared to orbital velocity. Oscillates up and down about center plane of disk. Stars in the central bulge swarm around the center in all directions. 5 The Milky Way as observed in infrared light 6 Jan Oort observed bright red giant stars near sun. Got positions and line-of-sight velocities (Doppler shifts). Concluded these stars are in disk. Positions gave disk thickness and velocities gave mass in disk. This mass was twice what you get from adding up visible matter. Thus “mass-to-light ratio” of disk stars near us 5 times solar value. Mostly have faint dwarfs, so no surprise. Visible matter in galactic halo has only few % of disk mass. Dynamical arguments say halo 10 times disk mass. So mass-to-light ratio in halo is likely to be huge. Nearby disk stars go around galactic center in almost circular orbits. Typical orbit does not close on itself. We will come back to this. 7 See small number of “high velocity stars” near us moving rapidly relative to rest of material in disk. All appear to be going in about same direction. Lindblad said they are a population with little rotation about galaxy, andd hhence with i h large l radial di l velocities. l ii We are rotating past this small population of stars. 8 Hard to see rotation of galaxy, because we are revolving with it. But stars closer to center revolve faster; stars farther out go slower. Like planets going around sun (remember Kepler’s laws?) So our galactic disk is rotating differentially. Rotation curve: plot orbital velocity trend with radius. For solid body, like merry-go-round, get straight line. For solar system, with Kepler’s laws and nearly all mass at center, get rapidly falling curve. For galactic disk, encompass ever more mass as go out, so curve does not drop off. If look toward and away from galactic center, we see that stars generally orbit faster and slower, respectively. 9 10 We can measure this rate of shear in disk star motions (how much faster or slower stars orbit with distance toward or away from galactic center). Also need to measure orbital velocity at radius of sun. Li dbl d figured Lindblad fi d out how h to do d this hi last l thing. hi Measure dispersion of stellar velocities in two directions, along orbit and in direction of galactic center. 2 dispersions would be equal if galaxy rotated like a solid body. Measure amount that they are unequal, and then can calculate orbital period of stars near us. Answer is about 230 million years. Use sun’s orbital velocity to “weigh” the galaxy inside sun’s orbit. Result is 130 billion solar masses. 11 Determining the mass of an object (review from earlier lecture): 1. Use small orbiting object. 2. Kepler’s third law: P2 = a3 3. Newton’s law of gravitation: P2 = a3 [4π2 / G (m1 + m2)] 4. If one mass dominates, plot P2 against a3 and slope gives M: P2 = (4 π2 / GM) a3 5. Solve for M in terms of the other quantities: M = (4 π2 / G) (a3/P2) 6. Measure G in the laboratory, then put in a and P for a planet to get M for the sun. 7. If measure a as an angle on the sky, then need to use the distance to the object M to convert a into distance units. 12 13 Use stars in disk to measure mass of Galaxy inside their orbits. Galaxy is very flat (with small bulge in center). Assume circular orbits and axisymmetric distribution of mass. (Axisymmetric means that the mass density is constant along every circle centered on the Galaxy’s rotation axis. Thus, if you were to spin the Galaxy a bit, an observer would see no difference, since an axisymmetric Galaxy looks the same as you go out along any radial line.) Also assume orbits of all stars lie in same plane (flat Galaxy). In infrared light that penetrates dust clouds, our galaxy looks really flat. To get a rough measure of the mass from velocities of orbiting stars, we need only assume circular orbits and an axisymmetric distribution of mass. We will also assume all orbits lie within a single plane. The Milky Way, photographed in diffuse infrared radiation by the COBE satellite. 14 We assume that the mass of the galaxy is distributed evenly around each ring near a given radius (we assume that the galaxy is axisymmetric). Then chunks A and B of the typical ring shown A B have equal masses. They pull on the star equally in the directions indicated by the arrows. If we add up these two forces, the components pulling up and down on the page cancel, leaving only the component directed toward the galactic center. Each chunk, A, of a given ring has an equal-mass mirror chunk, B, so that the total force exerted on the star from each whole ring of matter is directed toward the galactic center. We get this total force by adding up the contributions from all the chunks, A. A B Therefore the total force on the star from all the matter closer than the star to the galactic center is directed toward the galactic center. With a computer, or using the calculus, we could add up all these force contributions, from all the chunks in all the rings, to get the total. 15 Similarly, we can show that the total force on our star from each ring located outside its radius is directed away from the galactic center. However, it takes only a much smaller mass located at the center of the disk to cancel this outward pull. A B To find the mass contained in each annulus of our galaxy, given a series of measured stellar revolution velocities at increasing radii from the center, we must solve what is called an “inverse problem.” Given a set of choices for the masses of the various annuli (rings), Newton’s laws and some calculus (or some computer-enabled algebra) allow us to calculate the accelerations, and hence the rotation velocities of all the stars whose orbits define the annuli. The rotation velocity of this star determines the mass in the shaded ring. B t what But h t we really ll wantt to t do d is i to t compute t things thi the th other th way around. d We W know k the th stellar t ll velocities but not the masses of the annuli. Again, we can use a computer to make thousands or even millions of trials in a matter of minutes, giving us the answer. 16 4 2 0 -2 -4 -6 -5 -3 -1 1 3 5 Ring # 5 4 3 2 1 -1 -2 -3 -4 -5 -6 Radius 1.667 1.533 1.400 1.267 1.133 0.867 0.733 0.600 0.467 0.333 0.200 Accel. 0.248 0.330 0.472 0.766 1.693 -2.248 -1.127 -0.709 -0.471 -0.305 -0.173 Ratio -0.178 -0.257 -0.402 -0.722 -1.783 1.783 3.096 1.834 1.411 1.204 1.092 1.033 If we suppose that the density in this galactic disk is constant, then we can get a sense of the relative importance of the mass inside and outside of a particular radius. We take the radius of Ring 0 to be unity and the density of the disk to be unity. We give the “Ratio” of the mass we would need to place at the center of the disk to obtain this same acceleration to the actual mass in the annulus. Are our assumptions true for other galaxies? Nearby spiral galaxy M83 seen nearly face-on. Prominent spiral arms. Not very axisymmetric. But we see light distribution, not mass distribution. Spiral arms traced out by bright, massive stars. Th are incredibly They i dibl bbright i h for f their h i mass. If look in infrared, get better idea of mass distribution. Then see it is nearly axisymmetric. See that edge-on galaxies are extremely flat. Can also use companion galaxies to “weigh” whole galaxies. Could use two small companions of Andromeda Galaxy. Could use Magellanic Clouds to weigh our Galaxy. Weighing an irregular galaxy like the Large Magellanic Cloud would require that we generalize our methods, since it is neither axisymmetric nor flat. 17 Clearly, if our galaxy is like M83, shown here, our assumption of mass spread out evenly around concentric rings is in trouble. Happily, what we see in this picture is the distribution of light in M83, not the distribution of mass. We have built our model of our own galaxy partly by looking at other ones in order to get a better perspective. This one is M83, 14 million light years away from us. This galaxy looks pretty symmetrical, and pretty thin. NGC 891, a spiral galaxy in the constellation Andromeda that is seen edge on, probably looks pretty much the way our galaxy would look when viewed from outside and edge on. 18 This galaxy has two companions, that we could use to help determine its mass The Andromeda Galaxy, which is probably quite like our own. Its nuclear bulge is about 12,000 light years across. These two companions of our own galaxy should have motions that we could relate to our galaxy’s mass. The Magellanic Clouds as viewed from Australia 19 The Large Magellanic Cloud, a satellite of our own Milky Way We could use the velocities of stars in this irregular galaxy to help to determine its mass, but we would have to generalize our method. Mapping the Galaxy using neutral hydrogen gas: Henk van de Hulst realized that neutral hydrogen gas emits radio waves of 21 cm wavelength. This radiation goes clear across the Galaxy without absorption. Small radio telescope left in Holland by German army was then used to map the entire galactic disk. This gave full “rotation curve” and revealed spiral arms as well. At 21 cm, you can see forever, but result is confusion. Consider single line of sight on diagram. diagram Only aid in handling confusion is different Doppler shifts of radiation from different regions along line of sight. Gives a “line profile,” and is an art to interpret one. 20 To interpret line profile, use model of gas motions in galactic disk. Simplest is circular orbits everywhere. Then is single point along line of sight, the so-called tangent point, where gas has maximum velocity relative to us. This maximum velocity measures speed of galactic rotation at one radius, radius of tangent point. Whole series of such measurements gives the “rotation curve.” 21 Rest of the line profile: 2 locations on line of sight can produce each other velocity. Radiation from near location would extend farther above and below galactic plane than radiation from far location. Radiation from tangent point gives thickness of disk at that radius. From all tangent points, get idea of disk thickness trend in radius. This helps untangle near and far point problem. This done in 1968 by Frank Kerr. S Some guess work, k bbut result l ffairly i l clear. l We live in a spiral galaxy. No doubt about it. 22 This sometimes allows us to choose one of the locations unambiguously. Sometimes not. 23 The 21-centimeter Sky Maps of the Spiral Structure of our Galaxy 24 Fig. 20.11 – Milky Way Galaxy (from above) currently known or estimated features 25 Can do same thing with CO emission from much denser, colder gas. Get much flatter distribution. Also see a “molecular ring” in inner disk, where spiral arms strongest. 26 Ring of Molecules around the core of the Milky Way Molecular Clouds 27 On CO diagram, see rotation curve traced by maximum velocities. As near center, max velocity does not go up, but goes down. Not like rotation curve for solar system. So all mass in central region not concentrated at center. In CO diagram, see ridges of bright emission along diagonal lines. These are the spiral arms. It takes a complex model to understand the shapes of these ridges. This is a big, big deal. The spiral arms are the crests of a WAVE. Kind of like white caps on a lake. Explained by Lin and Shu in 1968. Believed by everyone 10 years later. Mosaic of Milky Way 28 The Visible Sky Infrared Sources 29 The 21-centimeter Sky Molecular Clouds 30 IRAS image – Milky Way Center 31 The Radio Sky The Gamma-ray Sky 32 Mapping the distribution of neutral hydrogen gas in the disk of our galaxy reveals that we live in a spiral galaxy. We cannot see it in all its spiral glory, because we are inside it. So let’s look at a few other spiral galaxies to get an idea of how our own galaxy l would ld appear to people l living li i in, i for f example, l the h Andromeda Galaxy. We believe that spiral structure of galaxies are waves in the disks of these galaxies. Individual stars or gas clouds enter the spiral arms, travel for a ways along them, and then leave again. The spiral arms are therefore not always made up of the same material. The very bright stars that delineate the spiral arms are very massive. These stars last only a short time, and therefore they must just recently have formed in the spiral arms. They light them up and make them prominent rather like foam makes the crest of a water wave on a lake stand out to our view. M81, on the next slide, looks like a good example of a symmetric, two-armed spiral galaxy that has no close companions (although it does indeed have companions). M81 was therefore chosen as the subject of an in-depth study using Westerbork radio telescope observations of neutral hydrogen for the purpose of detailed comparisons with predictions of spiral wave theory. theory The 21 cm line of neutral hydrogen allows one to measure Doppler shifts, and hence to find the departures from simple circular motions of the gas that are associated with the spiral arm features. These studies have shown the spiral arms to be a wave pehnomenon, as was expected from theory. The following slides show different exposures, exposures in visible light, light which highlight different morphological features of this galaxy. The slide showing the distribution of neutral hydrogen gas in M81 reveals the gravitational interaction of M81 with a companion galaxy which is located at the end of one of its spiral arms. 33 M 81 M 81 34 Neutral hydrogen in M81 and NGC 3077 with contours of radial velocity. From the thesis of J. M. van der Hulst, 1977. Now let’s look at some other classic spirals to see if they exhibit the same classic structure. 35 Here is a beautiful, symmetrical, face-on spiral with prominent dust lanes along the inner edges of its spiral p arms and with giant regions of ionized gas and newly formed massive stars just outside the dust lanes. This is the classic morphology explained by spiral wave theory. Seyfert Galaxy; NGC 1566 Here’s another classic example: Spiral Galaxy, type Sc, in ANTLIA (NGC 2997) 36 This classic spiral, M83, is shown in color, so that the new and old stars are easily distinguished. distinguished M-83 Recent observations using radio telescopes and infrared light have revealed that there appears to be a compact, non-luminous object of about 3 million solar masses at the center of our galaxy. We see that this object is there, because we observe the motion of stars that are orbiting about it, but we can see no light from the object itself. It is thought that this object must be a black hole, that is, an object that is so dense that its gravity prevents light from escaping from it. Because its gravity is so strong, Einstein’s theory of general relativity states that this gravity will distort space near the object, and therefore light rays will be deflected. On the following slides, the curvature of the space near the object caused by its very strong gravity is illustrated by the curvature of a deformable rubber-like surface, and the deflection of light paths is illustrated by the deflection of objects rolling on this rubber surface. 37 Fig. 17.12 (a) A 2-D representation of “flat” space-time. The circumference of each circle is 2π times its radius. (b) A 2-D representation of the “curved” space-time around a black hole. The black hole’s mass distorts space-time, making the radial distance between two circles larger than it would be in a “fl t” space-time. “flat” ti 38 At the left is an image, taken in visual light, of the region of the Galactic center, and at the right is a radio image.. Radio close-up of the central region. We see a “hypernova” remnant toward the left, and a spiral of gas centered on the center. 39 Still closer view of the central region. We see a “hypernova” remnant toward the left, and a spiral of gas centered on the center. Still closer view of the central region. We see a “hypernova” remnant toward the left, and a spiral of gas centered on the center. 40 Stars in the very central region orbit the center, and Kepler’s laws tell us how massive the region they orbit must be: about 3 million solar masses. Stars in the very central region orbit the center, and Kepler’s laws tell us how massive the region they orbit must be: about 3 million solar masses. 41 With high frequency radio observations, we can measure the size of the central object, or at least a size that it cannot exceed. With high frequency radio observations, we can measure the size of the central object, or at least a size that it cannot exceed. It turns out to be no larger than the diameter of the earth’s orbit about the sun. 42