1. 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.)

2. The Milky Way Summary of major visible components and structure The Galactic Rotation Dark Matter and efforts to detect it

3. Brief History Late 1700s - Herschels counted stars in 683 regions of sky, assumed all are equally luminous. Concluded that Sun at center of a flattened system. What is the Shape of the Milky Way?

4. 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: (Note: This formula was derived on the board in class)

5. 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 f 0, is given by Actual star counts at low fluxes are less than predicted by this relationship and the numbers at larger fluxes. Star counts in directions of the Milky Way disk (blue), and in the perpendicular directions (red).

6. 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

7. 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

8. 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 Globular clusters

9. Distances from Variable Stars

10. Morphology of Galaxy

12. Disk Young thin disk Old thin disk Thick disk

13. 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  10 10 M sun M dust+gas = 0.5  10 6 M sun (scale height 0.16) Average stellar mass ~ 0.7 M sun L B ~ 1.8  10 10 L sun 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

14. Thick disk Diameter ~ 50 kpc Scale height = 1.4 kpc M * = 2-4  10 9 M sun L B ~ 2  10 8 L sun [Fe/H] ~ -1.6 - -0.4 (less metal rich than thin disk) Age ~14-17 Gyrs

15. Gas and Dust

16. Spheroidal Components Central bulge Stellar Halo Dark Matter Halo

17. Central Bulge Diameter ~ 2 kpc Scale height = 0.4 kpc M * = 1  10 10 M sun L B ~ 0.3  10 10 L sun [Fe/H] ~ -1.0 - +1.0 (less metal rich than thick and thin disk) Age ~10-17 Gyrs

18. Stellar Halo Diameter ~ 100 kpc Scale height = 3 kpc number density distribution ~  r -3.5 M * = 0.1  10 10 M sun L B ~ 0.1  10 10 L sun [Fe/H] ~ -4.5 - -0.5 (metal poor) Age ~14-17 Gyrs

19. 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?)

20. 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?)

21. Determining the rotation when we are inside the disk rotating ourselves 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. 23.5° 39.1°

22. The Galactic midplane is inclined 62.6° with the plane of the celestial equator. We will introduce the Galactic coordinate system. l l=0° l=180° l=90° l=270° Galactic longitute (l) is shown here

23. b Galactic latitude(b) is shown here

24. Galactic Coordinate System: l b

25. GC d R R0R0 l b Let us introduce the following coordinate system

26. Assumptions: 1.Motion is circular  constant velocity, constant radius 2.Motion is in plane only (b = 0  )  no expansion or infall GC d R R0R0 l l = 0  l = 90  l = 180  l = 270  00  00  R 0 Radius distance of  from GC RRadius distance of  from  dDistance of  to   0 Velocity of revolution of   Velocity of revolution of   0 Angular speed of   Angular speed of  R 0 Radius distance of  from GC RRadius distance of  from  dDistance of  to   0 Velocity of revolution of   Velocity of revolution of   0 Angular speed of   Angular speed of   (rad/s)

27. Keplerian Model for [l = 0 , 180  ]: GC R R0R0 l = 0  l = 180  00 d 22 11 v R = 0 

28. Keplerian Model for [l = 45 , 135  ]: GC d R0R0 45  l = 0  l = 90  l = 180  l = 270  00 22 00 22 45  R > R 0 R < R 0 GC d R0R0 45  l = 0  l = 180  00 00 45  R > R 0 R < R 0 Star moving toward sun Star moving away from sun  0R -  1R = v R < 0 11 11  1R  2R  0R  0R -  2R = v R > 0

29. Inner Leading Star Outer Star Leading Inner Star (moving away from Sun) Lagging Outer Star (moving towards Sun) Leading Star At Same Radius Inner Leading Star Lagging Outer Star (moving away From Sun) Leading Inner Star (moving towards Sun) Lagging Star At Same Radius Keplerian Model for [l for all angles]: R < R 0 R = R 0 R > R 0 R = R 0 R < R 0 At 90 and 270 , v R is zero for small d since we can assume the Sun and star are on the same circle and orbit with constant velocity.

30. Keplerian Model for [l = 0  and 180  ]: GC R R0R0 l = 0  l = 180  00 d 22 11 v R = 0 1 0 2 GC R R0R0 l = 0  l = 180  00 d 22 11 1 0 2 R > R 0 R < R 0 t = 0 t > 0 l = 0  l = 180  aa bb Racing forward Falling backward CCW Rotation of INNER & OUTER local  s relative to 

31. GC d R R0R0 l 00   l l   RR TT  90-   What is the angle  ? We have two equations:  + l +  = 90  (1)  + l +  = 180  (2) If we subtract (1) from (2), i.e. (2) – (1):  -  = 90   = 90  +  

32. GC d R R0R0 l 00 l l    90-  90  +  Now let us derive the speed of  s relative to the , v R (radial component).   R =  cos   0R =  0 sinl l Relative speed, v R =  R –  0R =  ·cos  –  0 ·sinl We now can employ the Law of Sines a b c A B C

33. Therefore, From v = R , we may substitute the angular speeds for the star and Sun,

34. GC d R R0R0 l 00 l l    90-  90  +  Now let us derive the speed of  s relative to the , v T (tangential component).   T =  sin   0T =  0 cosl l v T =  T –  0T =  ·sin  –  0 ·cosl

35. GC d R R0R0 l 90  +  90  -   90  -  l Rcos  Rsin  R 0 sin(90-l)=R 0 cosl  Therefore,

36. Summarizing, we have two equations for the relative radial and tangential velocities:

37. Let us study  (R): GC R0R0 l = 0  l = 90  l = 180  l = 270  1 2 34

38. Quadrant  : R > R 0, 90  < l < 180  GC R0R0 l = 0  l = 90  l = 180  l = 270  Conclusion:Star is moving towards the Sun v R < 0 Always! R d l fixed R > R 0 d  0 R 0 sinl vRvR 1

39. Quadrant  : R > R 0, 180  < l < 270  GC R0R0 l = 0  l = 90  l = 180  l = 270  Conclusion:Star is moving away from the Sun v R > 0 Always! R d l fixed R > R 0 vRvR d  0 R 0 sinl 2

40. Quadrant  : R R 0, 0  < l < 90  GC R0R0 l = 0  l = 90  l = 180  l = 270  R d l fixed R < R 0 3 I II Case I: R<R 0 small Case II: R>R 0 large Star moving away! Star moving towards! vRvR d When R 0 and  >  0. When R> R 0, then v R <0 and  <  0. R > R 0

41. Quadrant  : R R 0, 270  < l < 360  GC l = 0  l = 90  l = 180  l = 270  l fixed 4 Quadrant  is the negative of Quadrant  ! vRvR d

42. Now we will make an approximation. We can work equally with  (R) or v(R) for the following approximation. Here we will work with  (R). Let us write R=R 0 +  R. Then, the Taylor Expansion yields

43. Here we make the approximation to retain only the first term in the expansion: If we continue the analysis for speed, we would use the substitution:  =R . Therefore,  =  /R. The derivative term on the right-hand side of the equation must be evaluated after substitution by using the Product Rule. Therefore, the radial relative speed between the Sun and neighboring stars in the galaxy is written as

44. When d<<R 0, then we can also make the small-angle approximation: R 0 =R+dcos(l). dcos(l) R d l R0R0  Using the sine of the double angle, viz. We may abbreviate the relation to where

45. If we then focus our attention to the transverse relative speed, v T, we begin with Picking up on the lessons learned from the previous analysis, we write simply Using the cosine of the double angle, viz. Because R  R 0,  0, which implies the last term is written as:

46. Therefore, where

47. Summarizing, where The units for A and B are or

48. We can define a new quantity that is unit-dependent. So that the transverse relative speed becomes The angular speed of the Sun around the Galactic Center is found algebraically when [d] = parsec, [v T ] = km/s. Likewise, the gradient of the rotation curve at the Sun’s distance from the Galactic Center is

49. The quantities used can all be measured or calculated if the following order is obeyed.

50. So, summarizing, for stars in the local neighborhood (d<<R 0 ), Oort came up with the following approximations: V r =Adsin2l V t = =d(Acos2l+B) Where the Oort Constants A, B are:  0 =A-B d  /dR | R 0 = -(A+B)

51. Keplarian Rotation curve

53. Dark Matter Halo M = 55  10 10 M sun L=0 Diameter = 200 kpc Composition = unknown! 90% of the mass of our Galaxy is in an unknown form