the milky way galaxy 02/03/2005. the milky way summary of major visible components and structure the...
DESCRIPTION
Morphology of GalaxyTRANSCRIPT
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