The Milky Way Galaxy 02/03/2005. The Milky Way Summary of major visible components and structure The Galactic Rotation Dark Matter and efforts to detect.

The Milky Way Galaxy

02/03/2005

The Milky Way

• Summary of major visible components and structure

• The Galactic Rotation• Dark Matter and efforts to detect it

Morphology of Galaxy

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

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:

lb

Assumptions:1. Motion is circular constant velocity, constant radius2. Motion is in plane only (b = 0) no expansion or infall

GC

d

R

R0

l

l = 0

l = 90

l = 180

l = 2700

0

R0 Radius distance of from GCR Radius distance of from d Distance of to 0 Velocity of revolution of Velocity of revolution of 0 Angular speed of Angular speed of

RT

Rv

2

(rad/s)

Keplerian Model for [l = 0, 180]:

GC

R

R0

l = 0

l = 180

0

d2

1vR = 0

vR = 0

vR = 0

2

2

RmGM

Rvm

FF gc

RGMv enc vR

Keplerian Model for [l = 45, 135]:

GC

d

R0

45

l = 0

l = 90

l = 180

l = 2700

2

0

2

45

R > R0

R < R0

GC

d

R0

45

l = 0

l = 180

00 45

R > R0

R < R0

Star movingtoward sun

Star moving awayfrom sun

0R-1R = vR < 01

11R

2R

0R

0R

0R-2R = vR > 0

Relative Radial Velocity, v R

0 45 90 135 180 225 270 315 360

Angle, l (o)

v R

InnerLeading

Star

OuterStar

LeadingInnerStar

(moving awayfrom Sun)

LaggingOuterStar

(movingtowards Sun)

LeadingStar

At Same Radius

InnerLeading

Star

LaggingOuterStar

(moving awayFrom Sun)

LeadingInnerStar

(movingtowards Sun)

LaggingStar

At Same Radius

Keplerian Model for [l for all angles]:

Star

Sun

Galactic Center

Star

Sun

Galactic Center

StarSun

Galactic Center

Star

Sun

Galactic Center

Star

Sun

Galactic Center

Star

Sun

Galactic Center

StarSun

Galactic Center

Star

Sun

Galactic Center

Star

Sun

Galactic Center

R < R0 R = R0 R > R0 R = R0 R < R0

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.

GC

d

R

R0

l

0

l

l

RT

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 +

cC

bB

aA

sinsinsin

GC

d

R

R0

l

0

l

l

90-90 +

Now let us derive the speed of s relative to the , vR (radial component).

R = cos

0R =

0 sinl

l Relative speed, vR = R – 0R

= ·cos – 0·sinl

We now can employ the Law of Sines

a b

cA

B

C

lRR

sin90sin0

lRR

sincos0

lRR sincos 0

Therefore,

lRRR

lRR

llRRvR

sin

sin

sinsin

00

0

00

00

From v = R, we may substitute the angular speeds for the star and Sun,0

00 ;

RR

lRvR sin00

GC

d

R

R0

l

0

l

l

90-90 +

Now let us derive the speed of s relative to the , vR (tangential component).

T =

sin

0T = 0

cosl

l vT = T – 0T = ·sin – 0·cosl

We will use trigonometry similar to that used when looking at the energy conservation of a pendulum.

GC

d

R

R0

l

90 +

90 -

90 -l

Rcos

Rsin

R0 sin(90-l)=R

0 cosl

sincos0 RdlR R

dlR

cossin 0

Therefore,

lRdlR

lRR

dlRR

ldlRR

vT

coscos

coscos

coscos

000

00

00

00

dlRvT cos00

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

dlRvT cos00 lRvR sin00

Now we will make an approximation.

dlRvT cos00 lRvR sin00

We can work equally with (R) or v(R) for the following approximation. Here we will work with (R).

00 RRR

Let us write R=R0+R. Then, the Taylor Expansion yields

202

2

0

02

02

2

00

02

2

2

0

00

00

00

00

00

!21

!21

!21

RRdR

RdRRdR

Rd

RRRdR

RdRRdR

RdR

RRdR

RdRdR

RdR

RRRRRR

RRRR

RRRR

RRRR

Here we make the approximation to retain only the first term in the expansion:

lRRRdR

d

lRRRRdR

dR

v

R

RR

sin

sin1

00

0

0020

0

0

0

0

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.

20

0

0 0

00

1RdR

dR

RR

dRd

dRRd

R

RRRR

Therefore, the radial relative speed between the Sun and neighboring stars in the galaxy is written as

000

RRdR

RdRRRR

When d<<R0, then we can also make the small-angle approximation: R0=R+dcos(l).

dcos(l)

R

d

l

R

ldRR cos0 ldRR cos0

llddRd

R

lRRRdR

dv

R

RR

sincos

sin

0

0

0

0

00

0

Using the sine of the double angle, viz. 2sincossin 21

lddRd

Rv

RR 2sin

00

0

We may abbreviate the relation to

ldAvR 2sin

00

0

21

RdRd

RAwhere

If we then focus our attention to the transverse relative speed, vT, we begin with

dlRvT cos00

dllddRd

R

dlRRRRdR

dR

dlRv

R

R

T

coscos

cos1

cos

0

0

0

0

0020

0

0

00

Picking up on the lessons learned from the previous analysis, we write simply

Using the cosine of the double angle, viz. 1cos22cos 2

dlddRd

Rv

RT

12cos21

00

0

Because RR0, 0, which implies the last term is written as: dR

dd

0

00

Therefore,

BdlAd

ddRd

RlAd

dR

ddRd

Rld

dRd

Rv

R

RRT

2cos

212cos

212cos

21

0

00

0

0

0

0

0

0

0

0

where BlAdvT 2cos

00

0

21

RdRd

RB

Summarizing,

where

BlAdvT 2cos

00

0

21

RdRd

RB

ldAvR 2sin

00

0

21

RdRd

RA

The units for A and B are

pcs

km

kpcs

kmor

We can define a new quantity that is unit-dependent.

So that the transverse relative speed becomes

dv lT 74.4

74.42cos BlA

l

The angular speed of the Sun around the Galactic Center is found algebraically

when [d] = parsec, [vT] = km/s.

BAR

0

00

Likewise, the gradient of the rotation curve at the Sun’s distance from the Galactic Center is

BAdR

Rd

R

0

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

1-1- kpcskm 2.14.14

2sin Measure 1.

ldvAv Rcalculate

R

1-1- kpcskm 8.20.122cos Measure 2. lAdvBv Tcalculate

T

BA 0 Calculate 3.

)(

get weB, andA of definition theFrom 4.

0

BAdRd

R

So, summarizing, for stars in the local neighborhood (d<<R0), Oort came up with the following approximations:

00

0

00

0

dRdΘ

RΘ

21- B

dRdΘ -

RΘ

21 A

Vr=Adsin2l

Vt= =d(Acos2l+B)

Where the Oort Constants A, B are:

0=A-B

d/dR |R0 = -(A+B)

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

This could be a topic for your final project