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.

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

• 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?

• 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)

3/ 20 0 .N ff Af

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

•Actual star counts at low fluxes are less than predicted by this relationship and the numbers at larger fluxes.

3/ 20 0 .N ff Af

0log f

3/ 20f

0

logN

ff

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

•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

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

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

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

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

R

mGM

R

vm

FF gc

R

GMv 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 +

c

C

b

B

a

A

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

lR

Rsincos 0

Therefore,

lRRR

lR

R

llR

RvR

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 lRvR sin00

GC

d

R

R0

l

0

l

l

90-90 +

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

T =

si

n

0T =

0 co

sll vT = T – 0T = ·sin – 0·cosl

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 dlRvT cos00

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

dlRvT cos00 dlRvT cos00 lRvR sin00 lRvR sin00

Now we will make an approximation.

dlRvT cos00 dlRvT cos00 lRvR sin00 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

!2

1

!2

1

!2

1

RRdR

RdRR

dR

Rd

RRRdR

RdRR

dR

RdR

RRdR

RdR

dR

RdR

RRR

RRR

RRRR

RRRR

RRRR

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

lRRRdR

d

lRRRRdR

d

Rv

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

1

RdR

d

R

R

R

dR

d

dR

Rd

R

RRRR

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

00

0

RRdR

RdRR

RR

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

dcos(l)

R

d

l

R0

ldRR cos0 ldRR cos0

llddR

d

R

lRRRdR

dv

R

RR

sincos

sin

0

0

0

0

00

0

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

lddR

d

Rv

RR 2sin

00

0

We may abbreviate the relation to

ldAvR 2sin ldAvR 2sin

00

0

2

1

RdR

d

RA

00

0

2

1

RdR

d

RAwhere

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

dlRvT cos00 dlRvT cos00

dllddR

d

R

dlRRRRdR

d

R

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

dlddR

d

Rv

RT

12cos2

1

00

0

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

dd

0

00

Therefore,

BdlAd

ddR

d

RlAd

dR

ddR

d

Rld

dR

d

Rv

R

RRT

2cos

2

12cos

2

12cos

2

1

0

00

0

0

0

0

0

0

0

0

where BlAdvT 2cos BlAdvT 2cos

00

0

2

1

RdR

d

RB

00

0

2

1

RdR

d

RB

Summarizing,

where

BlAdvT 2cos BlAdvT 2cos

00

0

2

1

RdR

d

RB

00

0

2

1

RdR

d

RB

ldAvR 2sin ldAvR 2sin

00

0

2

1

RdR

d

RA

00

0

2

1

RdR

d

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

74.4

2cos BlAl

74.4

2cos BlAl

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.

ld

vAv Rcalculate

R

1-1- kpcskm 8.20.122cos Measure 2. lAd

vBv Tcalculate

T

BA 0 Calculate 3.

)(

get weB, andA of definition theFrom 4.

0

BAdR

d

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