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```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 03 /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 03 /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 03 /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


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]:
vR
Angle, l (o )
0
Inner
Star
45
90
Inner
Star
Star
(moving away
from Sun)
R < R0
135
180
Lagging
Outer
Star
(moving
towards Sun)
Outer
Star
Lagging
Lagging
Outer
Star
Inner
Star
Star
(moving away
(moving
From Sun)
towards Sun)
Inner
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

cosa  sin l 
cosa  
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 cosl   d  R sin a 

sin a  
R0 cosl   d
R
Therefore,
Q
R0 cosl   d   Q0 cosl 
R
Q
Q
 R0 cosl   d   0 R0 cosl 
R
R0
vT 
l
d
90 + a
90 - a
R0
R
a
90 - l
GC
 w R0 cosl   d   w0 R0 cosl 
vT  w  w0 R0 cosl   wd
Summarizing, we have two equations for the relative radial and tangential velocities:
vR  w  w0 R0 sin l 
vT  w  w0 R0 cosl   wd
Now we will make an approximation.
vR  w  w0 R0 sin l 
vT  w  w0 R0 cosl   wd
We can work equally with w(R) or v(R) for the following approximation. Here we will
work with w(R).
wR  w0   wR  wR0 
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  QR  
 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 cosl 
l
R0
R

R  R0  d cosl 
 dQ 
Q0 
vR  
 
  R  R0   sin l 
dR
R
 R0

0 

 Q0  dQ  


   d cosl   sin l 
R
dR
 R0 
 0 
Using the sine of the double angle, viz.
sin a  cosa   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 cosl   wd
Picking up on the lessons learned from the previous analysis, we write simply
vT  w  w0 R0 cosl   wd
 1  dQ 
Q0 
 
  2   R  R0   R0 cosl   wd
R
dR
 R0 R0 
 0 
 Q 0  dQ  


   d cosl  cosl   wd
R
dR
 R0 
 0 
Using the cosine of the double angle, viz. cos2a   2 cos 2 a   1
1  Q0  dQ  
vT  

   d cos2l   1  wd
2  R0  dR  R0 
Because RR0, ww0, which implies the last term is written as: wd  w0 d 
Q0
d
R0
Therefore,
Q
1  Q0  dQ  
1  Q0  dQ  
vT  


   d cos2l   
 d  0 d
2  R0  dR  R0 
2  R0  dR  R0 
R0
1  Q0  dQ  
 Ad cos2l   

 d
2  R0  dR  R0 
 Ad cos2l   Bd
vT  d  A cos2l   B
where
1  Q0  dQ  
B    
 
2  R0  dR  R0 
Summarizing,
vR  A  d sin 2l 
vT  d  A cos2l   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 cos2l   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 
 dQR  

   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 cos2l   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:
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
```
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