Rotation of the Galaxy

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°

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

b Galactic latitude(b) is shown here

Galactic Coordinate System: l b

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

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)

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

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

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.

Keplerian Model for [l = 0  and 180  ]: GC R R0R0 l = 0  l = 180  00 d 22 11 v R = GC R R0R0 l = 0  l = 180  00 d 22 1 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 

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  +  

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

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

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

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

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

Let us study  (R): GC R0R0 l = 0  l = 90  l = 180  l = 270 

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

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

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

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

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

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

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

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:

Therefore, where

Summarizing, where The units for A and B are or

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

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

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)

Keplarian Rotation curve

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