Download Milky Way Kinematics, or how we discovered the geometry and

Milky Way Kinematics, or how we discovered the geometry and
dynamics of our own galactic environment.
The name “galaxy” comes from the Greek word for milk.
The Galaxy appears to us like a river of milk, and hence its name,
the
h Milky
Milk Way.
W
In the northern hemisphere, we can see the stars in the direction of
the galactic center only in the summer time.
[Fig. 12.2 from Shu]
1
Milky Way in Sagittarius
The Visible Sky
2
Mosaic of Milky Way
Map local stellar environment by means of star counts.
First plot positions on sky of stars brighter than some level 1.
Then plot positions of stars between brightness levels 1 and 2.
And so on.
If distribution of stars in space is uniform, each plot will be also.
Number of stars per square degree in each successive plot will
increase in a fashion that we could predict.
Find stars concentrated in band of sky, the galactic equator.
Find that numbers of fainter stars do not increase away from the
galactic equator so rapidly (as uniform distribution would imply).
This true also toward galactic equator, although shortfall is smaller.
Led to “Kapteyn Universe” with us in center.
3
Starlight blocked by clouds of dust in galactic disk.
In 1917, after Henrietta Levitt’s discovery of the period-luminosity
relation for the Cepheid variable stars in the Small Magellanic
Cloud, Harlow Shapley proposed that these stars were pulsating.
Shapley applied Levitt
Levitt’ss relation to RR-Lyrae
RR Lyrae variable stars in
globular clusters, to get their distances.
Got systematic overestimates, but right geometrical distribution.
(Globular clusters old, with few heavy elements.)
Globular clusters have spherical distribution centered on a position
about 30,000 light years from us (the modern, corrected distance) in
the direction of the galactic equator.
Thus we live in a disk, far from its center.
Globular clusters orbit center in all directions, disk stars all go
around in one direction and in a single plane.
4
Typical orbit of disk star. Vertical velocity is small compared to
orbital velocity. Oscillates up and down about center plane of disk.
Stars in the central bulge swarm around the center in all directions.
5
The Milky Way as observed in infrared light
6
Jan Oort observed bright red giant stars near sun.
Got positions and line-of-sight velocities (Doppler shifts).
Concluded these stars are in disk.
Positions gave disk thickness and velocities gave mass in disk.
This mass was twice what you get from adding up visible matter.
Thus “mass-to-light ratio” of disk stars near us 5 times solar value.
Mostly have faint dwarfs, so no surprise.
Visible matter in galactic halo has only few % of disk mass.
Dynamical arguments say halo 10 times disk mass.
So mass-to-light ratio in halo is likely to be huge.
Nearby disk stars go around galactic center in almost circular orbits.
Typical orbit does not close on itself. We will come back to this.
7
See small number of “high velocity stars” near us moving rapidly
relative to rest of material in disk.
All appear to be going in about same direction.
Lindblad said they are a population with little rotation about galaxy,
andd hhence with
i h large
l
radial
di l velocities.
l ii
We are rotating past this small population of stars.
8
Hard to see rotation of galaxy, because we are revolving with it.
But stars closer to center revolve faster; stars farther out go slower.
Like planets going around sun (remember Kepler’s laws?)
So our galactic disk is rotating differentially.
Rotation curve: plot orbital velocity trend with radius.
For solid body, like merry-go-round, get straight line.
For solar system, with Kepler’s laws and nearly all mass at center,
get rapidly falling curve.
For galactic disk, encompass ever more mass as go out, so curve
does not drop off.
If look toward and away from galactic center, we see that stars
generally orbit faster and slower, respectively.
9
10
We can measure this rate of shear in disk star motions
(how much faster or slower stars orbit with distance toward or away
from galactic center).
Also need to measure orbital velocity at radius of sun.
Li dbl d figured
Lindblad
fi
d out how
h to do
d this
hi last
l thing.
hi
Measure dispersion of stellar velocities in two directions, along
orbit and in direction of galactic center.
2 dispersions would be equal if galaxy rotated like a solid body.
Measure amount that they are unequal, and then can calculate
orbital period of stars near us.
Answer is about 230 million years.
Use sun’s orbital velocity to “weigh” the galaxy inside sun’s orbit.
Result is 130 billion solar masses.
11
Determining the mass of an object (review from earlier lecture):
1. Use small orbiting object.
2. Kepler’s third law:
P2 = a3
3. Newton’s law of gravitation:
P2 = a3 [4π2 / G (m1 + m2)]
4. If one mass dominates, plot P2 against a3 and slope gives M:
P2 = (4 π2 / GM) a3
5. Solve for M in terms of the other quantities:
M = (4 π2 / G) (a3/P2)
6. Measure G in the laboratory, then put in a and P for a
planet to get M for the sun.
7. If measure a as an angle on the sky, then need to use the
distance to the object M to convert a into distance units.
12
13
Use stars in disk to measure mass of Galaxy inside their orbits.
Galaxy is very flat (with small bulge in center).
Assume circular orbits and axisymmetric distribution of mass.
(Axisymmetric means that the mass density is constant along
every circle centered on the Galaxy’s rotation axis. Thus, if
you were to spin the Galaxy a bit, an observer would see no
difference, since an axisymmetric Galaxy looks the same as
you go out along any radial line.)
Also assume orbits of all stars lie in same plane (flat Galaxy).
In infrared light that penetrates dust clouds, our galaxy looks really flat.
To get a rough measure of the mass from velocities of orbiting stars, we
need only assume circular orbits and an axisymmetric distribution of
mass. We will also assume all orbits lie within a single plane.
The Milky Way, photographed in diffuse infrared radiation by the COBE satellite.
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We assume that the mass of the galaxy is distributed evenly around each ring near a
given radius (we assume that the galaxy is axisymmetric). Then chunks A and B of the
typical ring shown
A
B
have equal masses. They pull on the star equally in the directions indicated by the
arrows. If we add up these two forces, the components pulling up and down on the
page cancel, leaving only the component directed toward the galactic center.
Each chunk, A, of a given ring has an equal-mass mirror chunk, B, so that the total
force exerted on the star from each whole ring of matter is directed toward the galactic
center. We get this total force by adding up the contributions from all the chunks, A.
A
B
Therefore the total force on the star from all the matter closer than the star to the galactic center is
directed toward the galactic center. With a computer, or using the calculus, we could add up all
these force contributions, from all the chunks in all the rings, to get the total.
15
Similarly, we can show that the total force on our star from each ring located outside its
radius is directed away from the galactic center. However, it takes only a much smaller
mass located at the center of the disk to cancel this outward pull.
A
B
To find the mass contained in each annulus of our galaxy, given a series of measured
stellar revolution velocities at increasing radii from the center, we must solve what is
called an “inverse problem.”
Given a set of choices for the masses of the various annuli (rings), Newton’s laws and some calculus
(or some computer-enabled algebra) allow us to calculate the accelerations, and hence the rotation
velocities of all the stars whose orbits define the annuli.
The rotation
velocity of this
star determines
the mass in the
shaded ring.
B t what
But
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d is
i to
t compute
t things
thi
the
th other
th way around.
d We
W know
k
the
th stellar
t ll
velocities but not the masses of the annuli. Again, we can use a computer to make thousands or
even millions of trials in a matter of minutes, giving us the answer.
16
4
2
0
-2
-4
-6
-5
-3
-1
1
3
5
Ring #
5
4
3
2
1
-1
-2
-3
-4
-5
-6
Radius
1.667
1.533
1.400
1.267
1.133
0.867
0.733
0.600
0.467
0.333
0.200
Accel.
0.248
0.330
0.472
0.766
1.693
-2.248
-1.127
-0.709
-0.471
-0.305
-0.173
Ratio
-0.178
-0.257
-0.402
-0.722
-1.783
1.783
3.096
1.834
1.411
1.204
1.092
1.033
If we suppose that the density in this galactic disk is constant, then we can get a sense of the relative
importance of the mass inside and outside of a particular radius. We take the radius of Ring 0 to be
unity and the density of the disk to be unity. We give the “Ratio” of the mass we would need to
place at the center of the disk to obtain this same acceleration to the actual mass in the annulus.
Are our assumptions true for other galaxies?
Nearby spiral galaxy M83 seen nearly face-on.
Prominent spiral arms. Not very axisymmetric.
But we see light distribution, not mass distribution.
Spiral arms traced out by bright, massive stars.
Th are incredibly
They
i
dibl bbright
i h for
f their
h i mass.
If look in infrared, get better idea of mass distribution.
Then see it is nearly axisymmetric.
See that edge-on galaxies are extremely flat.
Can also use companion galaxies to “weigh” whole galaxies.
Could use two small companions of Andromeda Galaxy.
Could use Magellanic Clouds to weigh our Galaxy.
Weighing an irregular galaxy like the Large Magellanic Cloud
would require that we generalize our methods, since it is
neither axisymmetric nor flat.
17
Clearly, if our galaxy is like
M83, shown here, our
assumption of mass spread out
evenly around concentric rings
is in trouble.
Happily, what we see in this
picture is the distribution of
light in M83, not the
distribution of mass.
We have built our model of our own galaxy partly by looking at other
ones in order to get a better perspective. This one is M83, 14 million
light years away from us.
This galaxy looks pretty
symmetrical, and pretty
thin.
NGC 891, a spiral galaxy in the constellation Andromeda that is seen edge on, probably looks
pretty much the way our galaxy would look when viewed from outside and edge on.
18
This galaxy has two
companions, that we could
use to help determine its mass
The Andromeda Galaxy, which is probably quite like our own.
Its nuclear bulge is about 12,000 light years across.
These two companions of our
own galaxy should have motions
that we could relate to our
galaxy’s mass.
The
Magellanic
Clouds
as viewed
from
Australia
19
The Large
Magellanic
Cloud,
a satellite of our
own Milky Way
We could use
the velocities
of stars in this
irregular
galaxy to help
to determine
its mass, but
we would have
to generalize
our method.
Mapping the Galaxy using neutral hydrogen gas:
Henk van de Hulst realized that neutral hydrogen gas emits radio
waves of 21 cm wavelength.
This radiation goes clear across the Galaxy without absorption.
Small radio telescope left in Holland by German army was then
used to map the entire galactic disk.
This gave full “rotation curve” and revealed spiral arms as well.
At 21 cm, you can see forever, but result is confusion.
Consider single line of sight on diagram.
diagram
Only aid in handling confusion is different Doppler shifts of
radiation from different regions along line of sight.
Gives a “line profile,” and is an art to interpret one.
20
To interpret line profile, use model of gas motions in galactic disk.
Simplest is circular orbits everywhere.
Then is single point along line of sight, the so-called tangent point,
where gas has maximum velocity relative to us.
This maximum velocity measures speed of galactic rotation at one
radius, radius of tangent point.
Whole series of such measurements gives the “rotation curve.”
21
Rest of the line profile:
2 locations on line of sight can produce each other velocity.
Radiation from near location would extend farther above and below
galactic plane than radiation from far location.
Radiation from tangent point gives thickness of disk at that radius.
From all tangent points, get idea of disk thickness trend in radius.
This helps untangle near and far point problem.
This done in 1968 by Frank Kerr.
S
Some
guess work,
k bbut result
l ffairly
i l clear.
l
We live in a spiral galaxy.
No doubt about it.
22
This sometimes allows us to choose one of the locations
unambiguously. Sometimes not.
23
The 21-centimeter Sky
Maps of the Spiral Structure of our Galaxy
24
Fig. 20.11 –
Milky Way Galaxy
(from above)
currently known or
estimated features
25
Can do same thing with CO emission from much denser, colder gas.
Get much flatter distribution.
Also see a “molecular ring” in inner disk, where spiral arms
strongest.
26
Ring of Molecules around the core of the Milky Way
Molecular Clouds
27
On CO diagram, see rotation curve traced by maximum velocities.
As near center, max velocity does not go up, but goes down.
Not like rotation curve for solar system.
So all mass in central region not concentrated at center.
In CO diagram, see ridges of bright emission along diagonal lines.
These are the spiral arms.
It takes a complex model to understand the shapes of these ridges.
This is a big, big deal.
The spiral arms are the crests of a WAVE.
Kind of like white caps on a lake.
Explained by Lin and Shu in 1968.
Believed by everyone 10 years later.
Mosaic of Milky Way
28
The Visible Sky
Infrared Sources
29
The 21-centimeter Sky
Molecular Clouds
30
IRAS image – Milky Way Center
31
The Radio Sky
The Gamma-ray Sky
32
Mapping the distribution of neutral hydrogen gas in the disk of our
galaxy reveals that we live in a spiral galaxy.
We cannot see it in all its spiral glory, because we are inside it.
So let’s look at a few other spiral galaxies to get an idea of how our
own galaxy
l
would
ld appear to people
l living
li i in,
i for
f example,
l the
h
Andromeda Galaxy.
We believe that spiral structure of galaxies are waves in the disks of
these galaxies. Individual stars or gas clouds enter the spiral arms,
travel for a ways along them, and then leave again. The spiral arms
are therefore not always made up of the same material. The very
bright stars that delineate the spiral arms are very massive. These
stars last only a short time, and therefore they must just recently
have formed in the spiral arms. They light them up and make them
prominent rather like foam makes the crest of a water wave on a
lake stand out to our view.
M81, on the next slide, looks like a good example of a symmetric,
two-armed spiral galaxy that has no close companions (although
it does indeed have companions). M81 was therefore chosen as
the subject of an in-depth study using Westerbork radio telescope
observations of neutral hydrogen for the purpose of detailed
comparisons with predictions of spiral wave theory.
theory The 21 cm
line of neutral hydrogen allows one to measure Doppler shifts,
and hence to find the departures from simple circular motions of
the gas that are associated with the spiral arm features. These
studies have shown the spiral arms to be a wave pehnomenon, as
was expected from theory.
The following slides show different exposures,
exposures in visible light,
light
which highlight different morphological features of this galaxy.
The slide showing the distribution of neutral hydrogen gas in M81
reveals the gravitational interaction of M81 with a companion
galaxy which is located at the end of one of its spiral arms.
33
M 81
M 81
34
Neutral
hydrogen in
M81 and
NGC 3077
with contours
of radial
velocity.
From the
thesis of J. M.
van der Hulst,
1977.
Now let’s look at some other classic spirals to see if they exhibit the
same classic structure.
35
Here is a beautiful,
symmetrical, face-on
spiral with prominent
dust lanes along the
inner edges of its
spiral
p
arms and with
giant regions of
ionized gas and newly
formed massive stars
just outside the dust
lanes. This is the
classic morphology
explained by spiral
wave theory.
Seyfert Galaxy; NGC 1566
Here’s another classic example:
Spiral Galaxy, type Sc, in ANTLIA (NGC 2997)
36
This classic spiral,
M83, is shown in
color, so that the
new and old stars
are easily
distinguished.
distinguished
M-83
Recent observations using radio telescopes and infrared light have
revealed that there appears to be a compact, non-luminous object of
about 3 million solar masses at the center of our galaxy.
We see that this object is there, because we observe the motion of stars
that are orbiting about it, but we can see no light from the object
itself.
It is thought that this object must be a black hole, that is, an object that
is so dense that its gravity prevents light from escaping from it.
Because its gravity is so strong, Einstein’s theory of general
relativity states that this gravity will distort space near the object,
and therefore light rays will be deflected.
On the following slides, the curvature of the space near the object
caused by its very strong gravity is illustrated by the curvature of a
deformable rubber-like surface, and the deflection of light paths is
illustrated by the deflection of objects rolling on this rubber surface.
37
Fig. 17.12 (a) A 2-D representation of “flat” space-time.
The circumference of each circle is 2π times its radius.
(b) A 2-D representation of the “curved” space-time around a black
hole. The black hole’s mass distorts space-time, making the
radial distance between two circles larger than it would be in a
“fl t” space-time.
“flat”
ti
38
At the left is an image, taken in visual light, of the region of the
Galactic center, and at the right is a radio image..
Radio close-up of the central region. We see a “hypernova”
remnant toward the left, and a spiral of gas centered on the center.
39
Still closer view of the central region. We see a “hypernova”
remnant toward the left, and a spiral of gas centered on the center.
Still closer view of the central region. We see a “hypernova”
remnant toward the left, and a spiral of gas centered on the center.
40
Stars in the very central region orbit the center, and Kepler’s laws
tell us how massive the region they orbit must be: about 3 million
solar masses.
Stars in the very central region orbit the center, and Kepler’s laws
tell us how massive the region they orbit must be: about 3 million
solar masses.
41
With high frequency radio observations, we can measure the size
of the central object, or at least a size that it cannot exceed.
With high frequency radio observations, we can measure the size
of the central object, or at least a size that it cannot exceed. It
turns out to be no larger than the diameter of the earth’s orbit
about the sun.
42