Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

no text concepts found

Transcript

Today in Astronomy 328: the Milky Way Image: wide-angle photo and overlay key of the Sagittarius region of the Milky Way, showing vividly the effect of obscuration by dust clouds. The very center of the Milky Way lies behind particularly heavy dust obscuration. (By Bill Keel, U. Alabama.) The Milky Way • Summary of major visible components and structure • The Galactic Rotation • Dark Matter and efforts to detect it Brief History What is the Shape of the Milky Way? • Late 1700s - Herschels counted stars in 683 regions of sky, assumed all are equally luminous. Concluded that Sun at center of a flattened system. • 1920 - Kapteyn used a greater number of star counts and came to roughly the same conclusion Star Counts: If stars are distributed uniformly in space, then in any patch of sky, the total number of stars with flux less than a limiting flux, f is: N f f 0 Af 03 /2 . (Note: This formula was derived on the board in class) What is the shape of the Milky Way? •First answer, due to Kapteyn, came from star counts. If stars were distributed uniformly in the universe, then the number counted in any patch of the sky, with flux larger than f0, is given by N f f 0 Af 03 /2 . •Actual star counts at low fluxes are less than predicted by this relationship and the numbers at larger fluxes. log N f f0 f 03 /2 log f0 Star counts in directions of the Milky Way disk (blue), and in the perpendicular directions (red). Conclusion: stellar density not uniform but decreases with distance from Sun; faster in direction perpendicular to Milky Way and slower in the direction of the Milky Way Milky Way is a highly flattened disk • 1919 - Shapley studied globular clusters; used distance derived from pulsating stars to determine that Sun is not at center of Milky Way. These were found at great distances above and below the plane of the Galaxy, where extinction effects are much less than that found along the Milky Way Figure: Chaisson and McMillan, Astronomy Today Globular clusters •Definitely bound by gravity •Contain large numbers of stars in a very small volume: 20,000-1,000,000 stars in a volume 20 pc in diameter •very round and symmetrical in shape - very old -among the first stellar complexes formed in the galaxy Distances from Variable Stars Morphology of Galaxy Disk • Young thin disk • Old thin disk • Thick disk Thin disk • Diameter ~ 50 kpc • Young thin disk scale height = 50 pc • Old thin disk scale height = 325 pc • Contains youngest stars, dust, and gas • Contains Sun, which is 30 pc above midplane • M* = 6 1010 Msun • Mdust+gas = 0.5 106 Msun (scale height 0.16) • Average stellar mass ~ 0.7 Msun • LB ~ 1.8 1010 Lsun • Population I stars in the Galactic plane • Contains ~ 95% of the disk stars • [Fe/H] ~ -0.5 - +0.3 • Age ~ < 12 Gyr • Spiral structure seen in neutral H, HII regions, young O and B stars Thick disk • Diameter ~ 50 kpc • Scale height = 1.4 kpc •M* = 2-4 109 Msun •LB ~ 2 108 Lsun • [Fe/H] ~ -1.6 - -0.4 (less metal rich than thin disk) • Age ~14-17 Gyrs Gas and Dust Spheroidal Components • Central bulge • Stellar Halo • Dark Matter Halo Central Bulge • Diameter ~ 2 kpc • Scale height = 0.4 kpc •M* = 1 1010 Msun •LB ~ 0.3 1010 Lsun • [Fe/H] ~ -1.0 - +1.0 (less metal rich than thick and thin disk) • Age ~10-17 Gyrs Stellar Halo • Diameter ~ 100 kpc • Scale height = 3 kpc • number density distribution ~ r-3.5 •M* = 0.1 1010 Msun •LB ~ 0.1 1010 Lsun • [Fe/H] ~ -4.5 - -0.5 (metal poor) • Age ~14-17 Gyrs Dark Matter Halo? Rotation of Galaxy implies that there is a lot of mass in our Galaxy that we don’t see (ie, if we count up the mass from the stars that emit visible light, it’s much less than that implied by observing the motion of stars as a function of radius from the center of the Galaxy. How do we know that the stars in the disk rotate around the center of the Galaxy? How do we know the rotational velocity of the Sun? How do we know the rotation curve? (rotational velocity as a function of radius from the Galactic center?) Dark Matter Halo? Rotation of Galaxy implies that there is a lot of mass in our Galaxy that we don’t see (ie, if we count up the mass from the stars that emit visible light, it’s much less than that implied by observing the motion of stars as a function of radius from the center of the Galaxy. How do we know that the stars in the disk rotate around the center of the Galaxy? How do we know the rotational velocity of the Sun? How do we know the rotation curve? (rotational velocity as a function of radius from the Galactic center?) Determining the rotation when we are inside the disk rotating ourselves 23.5° 39.1° To determine the rotation curve of the Galaxy, we will introduce a more convenient coordinate system, called the Galactic coordinate system. Note that the plane of the solar system is not the same as the plane of the Milky Way disk, and the Earth itself is tipped with respect to the plane of the solar system. The Galactic midplane is inclined at an angle of 62.6 degrees from the celestial equator, as shown above. The Galactic midplane is inclined 62.6° with the plane of the celestial equator. We will introduce the Galactic coordinate system. l=0° Galactic longitute (l) is shown here l l=90° l=270° l=180° Galactic latitude(b) is shown here b Galactic Coordinate System: b l Assumptions: 1. Motion is circular constant velocity, constant radius 2. Motion is in plane only (b = 0) no expansion or infall l = 180 l = 270 w0 Q0 l l = 90 d w Q R0 R v Rw 2 R T w (rad/s) GC l = 0 R0 Radius distance of from GC R Radius distance of from d Distance of to Q0 Velocity of revolution of Q Velocity of revolution of w0 Angular speed of w Angular speed of Keplerian Model for [l = 0, 180]: mv 2 GMm R R2 l = 180 vR = 0 Q1 vR = 0 Q0 R0 d vR = 0 R GC l = 0 Fc Fg Q2 v GM enc R R v Keplerian Model for [l = 45, 135]: w1 Q1R Q -Q = v < 0 0R 1R R Q1 l = 180 Star moving toward sun l = 180 R > R0 R > R0 Q0R l = 270 w0 45 Q0 l = 90 w0 45 45 45 d Q0 Q0R d w2 Q2 R0 R0 Q2R Q0R-Q2R = vR > 0 R < R0 R < R0 GC GC l = 0 l = 0 Star moving away from sun Keplerian Model for [l for all angles]: Relative Radial Velocity, v R vR Angle, l (o ) 0 Inner Leading Star 45 90 Leading Leading Inner Star Star At Same Radius (moving away from Sun) R < R0 135 180 Lagging Outer Star (moving towards Sun) Outer Star Lagging Lagging Leading Outer Star Inner Star At Same Radius Star (moving away (moving From Sun) towards Sun) Inner Leading Star R > R0 R = R0 R < R0 R = R0 225 270 315 360 Star Star Sun Sun Sun Sun Star Sun Sun Star Star Sun Sun Star Star Galactic Center Sun Star Star Galactic Center Galactic Center Galactic Center Galactic Center Galactic Center Galactic Center Galactic Center At 90 and 270, vR is zero for small d since we can assume the Sun and star are on the same circle and orbit with constant velocity. Galactic Center What is the angle g? Q0 We have two equations: l d l d g d GC d + l + g = 180 (2) If we subtract (1) from (2), i.e. (2) – (1): g - a = 90 Q QR R (1) 90-a a d l R0 d + l + a = 90 QT g = 90 + a Now let us derive the speed of s relative to the , vR (radial component). Relative speed, vR = QR – Q0R = Q·cosa – Q0·sinl Q0 l l d We now can employ the Law of Sines l d 90-a 90 + a R0 c d l a Q A B a A B C sin a sin b sin c b C R R0 R sin 90 a sin l R0 R cosa sin l cosa GC R0 sin l R Therefore, R vR Q 0 sin l Q 0 sin l R R Q 0 Q 0 sin l R Q Q 0 R0 sin l R R0 From v = Rw, we may substitute the angular speeds for the star and Sun, w vR w w0 R0 sin l Q Q ; w0 0 R R0 Now let us derive the speed of s relative to the , vT (tangential component). vT = QT – Q0T = Q·sina – Q0·cosl Q0 l l d l d 90-a 90 + a R0 d R GC l a Q R0 cosl d R sin a sin a R0 cosl d R Therefore, Q R0 cosl d Q0 cosl R Q Q R0 cosl d 0 R0 cosl R R0 vT l d 90 + a 90 - a R0 R a 90 - l GC w R0 cosl d w0 R0 cosl vT w w0 R0 cosl wd Summarizing, we have two equations for the relative radial and tangential velocities: vR w w0 R0 sin l vT w w0 R0 cosl wd Now we will make an approximation. vR w w0 R0 sin l vT w w0 R0 cosl wd We can work equally with w(R) or v(R) for the following approximation. Here we will work with w(R). wR w0 wR wR0 Let us write R=R0+DR. Then, the Taylor Expansion yields w R w0 w R w R0 w R0 DR w R0 1 d 2w R dw R 2 w R0 DR DR w R0 2 dR 2 ! dR R R0 R R0 1 d 2w R dw R 2 w R0 R R0 R R0 w R0 2 2! dR R R dR R R0 0 dw R 1 d 2w R 2 R R0 R R0 2 2! dR R R dR R R0 0 Here we make the approximation to retain only the first term in the expansion: dw R R R0 dR R R0 w R w R0 If we continue the analysis for speed, we would use the substitution: Q=Rw. Therefore, w=Q/R. The derivative term on the right-hand side of the equation must be evaluated after substitution by using the Product Rule. d QR dw R dR R R0 dR R R R0 1 dQ Q0 2 R0 dR R0 R0 Therefore, the radial relative speed between the Sun and neighboring stars in the galaxy is written as 1 dQ Q0 vR R R0 R0 sin l 2 R0 dR R0 R0 dQ Q0 R R0 sin l dR R R0 0 When d<<R0, then we can also make the small-angle approximation: R0=R+dcos(l). d dcos(l) R0 R d cosl l R0 R R R0 d cosl dQ Q0 vR R R0 sin l dR R R0 0 Q0 dQ d cosl sin l R dR R0 0 Using the sine of the double angle, viz. sin a cosa 12 sin 2a Q0 dQ vR d sin 2l R dR R0 0 We may abbreviate the relation to vR A d sin 2l 1 Q0 dQ where A 2 R0 dR R0 If we then focus our attention to the transverse relative speed, vT, we begin with vT w w0 R0 cosl wd Picking up on the lessons learned from the previous analysis, we write simply vT w w0 R0 cosl wd 1 dQ Q0 2 R R0 R0 cosl wd R dR R0 R0 0 Q 0 dQ d cosl cosl wd R dR R0 0 Using the cosine of the double angle, viz. cos2a 2 cos 2 a 1 1 Q0 dQ vT d cos2l 1 wd 2 R0 dR R0 Because RR0, ww0, which implies the last term is written as: wd w0 d Q0 d R0 Therefore, Q 1 Q0 dQ 1 Q0 dQ vT d cos2l d 0 d 2 R0 dR R0 2 R0 dR R0 R0 1 Q0 dQ Ad cos2l d 2 R0 dR R0 Ad cos2l Bd vT d A cos2l B where 1 Q0 dQ B 2 R0 dR R0 Summarizing, vR A d sin 2l vT d A cos2l B 1 Q0 dQ A 2 R0 dR R0 1 Q0 dQ B 2 R0 dR R0 where The units for A and B are km s or pc km s kpc We can define a new quantity that is unit-dependent. l A cos2l B 4.74 So that the transverse relative speed becomes vT 4.74 l d when [d] = parsec, [vT] = km/s. The angular speed of the Sun around the Galactic Center is found algebraically Q0 A B R0 Likewise, the gradient of the rotation curve at the Sun’s distance from the Galactic Center is w0 dQR A B dR R0 The quantities used can all be measured or calculated if the following order is obeyed. vR 14.4 1.2 km s -1 kpc-1 d sin 2l v 2. Measure vT calculate B T A cos2l 12.0 2.8 km s -1 kpc-1 d 1. Measure vR calculate A 3. Calculate w0 A B 4. From the definition of A and B, we get dQ ( A B ) dR R0 So, summarizing, for stars in the local neighborhood (d<<R0), Oort came up with the following approximations: Vr=Adsin2l Vt= =d(Acos2l+B) Where the Oort Constants A, B are: 1 Θ0 dΘ - 2 R0 dR 0 1 Θ0 dΘ B - 2 R0 dR 0 A w0=A-B dQ/dR |R = -(A+B) 0 Keplarian Rotation curve Dark Matter Halo • • • • M = 55 1010 Msun L=0 Diameter = 200 kpc Composition = unknown! 90% of the mass of our Galaxy is in an unknown form

Related documents