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\chapter*{Prologue}
\section{Before the beginning}
This course on galaxies runs over 14 lectures. The aim is to give you
an overview of the types of galaxies we know of, and to try to
understand why galaxies look like they do, based on the physics we
think that applies. In contrast to the beauty of stellar physics,
you'll see that the galactic physics is often much more murky and
complex -- a bit like weather prediction: we (think we) know the
physics, but even so it still is difficult to make accurate
calculations. And often, we don't know the physics very well -- or even
at all. On the bright side, galaxies are beautiful. And we can in
principle probe physics on scales which are impossible to reach within
the laboratory and so learn about really fundamental physics -- just
recall the recent excitement about the possible non-zero value for the
cosmological constant. Anyway, I hope you'll enjoy the lectures.

As to the organisation of the course, I will basically follow the
structure of Simon Morris, i.e. there will be hand-outs after each
lecture, but more up-to-date versions will appear on the web in
DUO. Look there as well for (most of) the pictures that I show during
the lectures.

My name is Tom Theuns (tom.theuns@durham.ac.uk, tel 8275), and my
office is room 307 in the Ogden Centre. This is the first time I teach
this course, and in case I go too fast (or too slow), I count on you to
tell me! Suggestions, and lists of errors, will be most appreciated.

And for those that are only interested in passing their exams: there
will be two questions on galaxies. The first consist of 5 very short
questions where you're asked to describe {\em very briefly} some
concepts that we have discussed. For example, what is the interstellar
medium, and how do we observe it. The second type of question usually
starts from a derivation that we've done. The end of the question is
something you have not seen, but is very similar to what we have done.

\section{My source of knowledge}
Although this field is continually evolving, much of the basics is of
course text-book stuff. The web provides a very valuable source of
information, not just for pictures. Often, satellite mission web-pages
have extremely well designed pages. Some links you might want to look
at (and where I got many of the pictures) are:

\begin{itemize}
\item http://www.seds.org/
\item http://www.aao.gov.au/images.html/
\item http://hubblesite.org/newscenter/archive/
\item http://teacherlink.ed.usu.edu/tlnasa/pictures/picture.html
\item http://astro.estec.esa.nl/GAIA/
\item http://space.gsfc.nasa.gov/astro/cobe/cobe\_home.html
\item http://astro.estec.esa.nl/Hipparcos/
\item http://astro.estec.esa.nl/GAIA/
\end{itemize}

and the books I've used are

\begin{enumerate}
\item[1] Carroll \& Ostlie, {\it Modern Astrophysics}, Addison-Wesley,
1996
\item[2] Zeilik \& Gregory, {\it Astronomy \& Astrophysics}, Saunders College Publishing, 1998
\item[3] Binney \& Merrifield, {\it Galactic Astronomy}, Princeton,
1998
\item[4] Binney \& Tremaine, {\it Galactic Dynamics}, Princeton, 1987
\end{enumerate}

(I'll refer to these as CO, ZG, BM, and BT.) The first two are clearly
more general, [3] provides a nice overview on galaxies in much more
detail than I'll do, and [4] of course concentrates on the dynamics of
galaxies. [1] contains everything you {\em need} to know, in [2] you
can often find the basics formulated in a very clear way, and [3] and
[4] are where you could start looking if you want to know more.


\chapter{Introduction}
\section{Historical perspective}
Galaxies were discovered in the 20th century. This is a bit surprising,
you can see many of them on a clear night with the naked eye. They
differ from stars in that, even with the naked eye, you can see that
they are extended. For example, the nearby Andromeda galaxy has an
extent of several degrees (compare to the diameter of full moon, which
is 0.5 degrees). And the Large and Small Magellanic clouds, are even
bigger (but you can only seem them in the Southern hemisphere). And the
nearest galaxy (which was only discovered in the last years, extends
over a significant fraction of the whole sky!

To distinguish them from stars, they were called {\it
nebulae}. However, there are other extra-solar objects which are also
extended, like Planetary Nebulae, star clusters, and super nova
remnants, for example. Since, before the 1920s, we didn't know the
distance to those objects, it wasn't clear what was the difference
between all these types. And so, many of the catalogues of nebulae,
like Messier's catalogue (objects denoted by an \lq M\rq~, like M31,
which is Andromeda), or NGC (New General Catalogue, Dreyer 1888),
contain a baffling variety of different types of objects.

Immanual Kant, the German philosopher, already figured out that we see
the faint band of stars across the night sky -- usually called the
Milky Way (MW for short) -- because, on a larger scale, most stars
around us are distributed in a disk, and the sun is located in that
disk. And so, when you look out at night in the plane of the disk, you
see a faint band of light, which Galileo already had resolved into
stars!  And most of these other nebulae you see, were just other
galaxies. Though this was a correct hypothesis of Kant's, it took until
the 1920s before the matter was settled. The historical introduction in
BM is well worth reading.

As a final point: why do we study galaxies? Are they in some sense the
most important building blocks in the Universe? I like BMs reasoning:
stars are apparently huddled together in galaxies. And since (not
surprisingly if you think about it) we (humans) can {\em see} stars, we
are naturally inclined to study them. But, for example, although a
cluster of galaxies contains thousands of galaxies, they are by no
means the main constituent. Most of the {\em mass} is in dark matter --
which we cannot see at all. But most of the baryons is in hot gas,
which we can {\em observe} with X-ray satellites, but we cannot {\em
see}. And so the choice to study galaxies is a bit human centred: our
eyes can see them.

\section{Bringing order to the zoo of galaxies}
Before we start doing any calculations, it might be a good idea to look
around us and make an inventory of what types of galaxies are out
there. There are many good web pages to start, for example I like
http://www.taas.org/gnto/astropic.html, because it shows pictures made
with a small telescope -- so those are galaxies as you might have seen
them yourself, if your an amateur
astronomer. http://www.aao.gov.au/images.html/ is very nice as well, it
shows professional images. If you look at the more professional pages
(such as the HST archive for example), beware of an amusing bias:
professional astronomers tend to observe (and make picture of) unusual
galaxies.

\subsection{Vocabulary}
{\bf Colours.} These images are usually made through a broad-band
filter, typically\footnote{The names give an indication of the
corresponding wavelength region, \B\, for blue, \V\ for visual, \R\ for
red, \U\ for UV (in the direction of the UV-that is), \I\ for IR.}  \V,
\B or \R, and so should be just black and white. The blackness being a
measure of how much, or how little of the light in that wavelength band
fell on that part of the photograph/CCD. One often combines images in
different filters to produce a colour image, which tries to mimic the
colour your eye would see. Most of the galaxy images you'll see are of
this type. However, often one of the filters is a narrow-band filter,
e.g. in H$\alpha$ or O[III]. Combing such an narrow-band image with
images made with others filters gives a false colour image, i.e. the
colour of the object is {\em not} how your eye would see it, but the
colour coding has been chosen to bring out a particular feature. Those
really pretty colour pictures of Planetary Nebulae are false colour
images.

{\bf Surface brightness.} The brightness of the images depends, apart
from the size and quality of the optics of the camera, the integration
time, and the transparency of the atmosphere (all of which we should be
able to calibrate), on the apparent brightness of the galaxy. So this
is what we want to characterise. Since galaxies are extended, the
brightness varies over the image, and so what we want to measure is the
surface brightness (SB). The SB measures how much light we receive from
the galaxy, per unit solid angle (usually per arcsec$^2$) so it should
have units of flux (e.g. in W), per unit wavelength range, per unit
solid angle on the sky. So for example in units $L_{\V\odot}\arcsecsq$,
in units of the solar \V luminosity. Usually, SB is expressed in
magnitudes, by taking $-2.5\log_{10}$ of the previous number -- we say
(incorrectly, or at least confusingly) that we measure SB in magnitudes
per square arc seconds. (It's confusing, because this suggests we
divide the magnitude by \arcsecsq, but no: we first divide by the solid
angle, and then take the logarithm). As is usual with magnitudes, we
add a constant to it.

The number that characterises the {\rm intrinsic} brightness
distribution of the galaxy (as opposed to apparent) is how much light
the galaxy emits per unit {\rm surface area}, so in for example units
of $L_{\odot\V}\kpc^2$. Astronomers do not use separate names for
these two types of SB, so check the units!

{\bf Star counts} To quantify this a bit more, assume you observe a
distribution of stars, all of the same absolute luminosity $L$, and
with uniform number density $n$ (in stars per \pc$^3$, say). The flux
you receive from a single star, when at distance $r$, is $L/4\pi r^2$
(in W\,m$^{-2}$). The number of stars at distance between $r$ and
$r+dr$, and within a solid angle $d\Omega$ on the sky, is
$dN(r,d\Omega)=nr^2drd\Omega$. For the whole sky, $d\Omega=4\pi$, and
so the number is $dN(r)=4\pi nr^2dr$, which is the product of the volume
of the shell, $4\pi r^2dr$, with the density $n$ of stars. Combining
these two, the luminosity of all stars within $r$ and $r+dr$, and
within $d\Omega$, is
\begin{equation}
dL = LdN(r,d\Omega)={L n \over 4\pi}d\Omega dr\,.
\end{equation}

The surface brightness is $\propto dL/d\Omega=L ndr/4\pi$.  For an
external galaxy, you could integrate $\int ndr$ along the line of sight
through the galaxy, to get the surface density of stars, $\int
ndr=\Sigma$ of stars. The SB is then $L\Sigma$. The important thing to
notice is that it is {\em independent} of the distance $r$.  Going back
through the calculation, you can see what has happened: the luminosity
from a single star at distance $r$ {\em decreases} $\propto 1/r^2$, but
the number of stars within $d\Omega$ {\em increases} $\propto r^2$. And
since the SB is proportional to the product of these two, it is
independent of $r$\footnote{In a cosmological setting, this argument
does not hold, and the SB of very distant galaxies decreases with
distance, an effect called surface brightness dimming.}. (You might
object saying we've assumed all stars to have the same luminosity. So
just redo the calculation with the {\em average} luminosity then!)


When looking-out into a stellar distribution -- for example when
counting stars in the Milky Way -- we should be able to infer $n(r)$ by
counting how many stars we see as function of their luminosity. Here is
how to do it: the number of stars within a solid angle $d\Omega$, and
with distance between $r$ and $r+dr$, $dN(r,d\Omega)=nr^2drd\Omega$, as
we had before. Now again assume them to have all the same intrinsic
luminosity $L$. The trick is to change variables from distance $r$ to
apparent magnitude $m$, using
\begin{equation}
m=-2.5\log(L/r^2)=M+5\log(r)\,,
\end{equation}
up to a constant. Recall where this came from: since the observed
luminosity $\propto L/r^2$, the apparent magnitude
$m=-2.5\log(L/r^2)=-2.5\log(L)-5\log(r)=M-5\log(r)$, up to a
constant. (Also recall that, by convention, $m=M$ for a distance of
$r=10\pc$. And hence the proper equation is $m-M=5\log(r)-5$, when $r$
is expressed in \pc. $m-M$ is called the {\em distance modulus}.) Now
we're almost done, since rearranging
\begin{equation}
r\propto 10^{0.2(m-M)}\,,\nonumber
\ee
and by differentiating
\be
dr/dm\propto 10^{0.2(m-M)}\propto r\,.\nonumber
\ee

Substituting this in the expression for $dN$ gives
\be
dN(m,d\Omega)\propto n(m) 10^{0.6(m-M)} dm\,.
\ee

For example, if the number density is constant (and hence also $n(m)$
is constant), then, $dN(m+1,d\Omega)/dN(m,d\Omega)=10^{0.6}\approx
4$. In other words: by going one magnitude fainter, we get around 4
times as many stars. If we get fewer, then either the more distant stars
are intrinsically fainter, or light has been lost on the way, or the
density of stars drops (or some combination).

As a last exercise: we found that the luminosity of all stars in a
shell, within a given solid angle, was independent of the radius of the
shell. So, integrating over all radii, we get an infinite
luminosity. Which clearly conflicts with the fact that the night sky is
dark. This is called Olber's paradox. What is wrong with this
reasoning?

\subsection{Galaxy properties}
Looking at the different images of galaxies, it becomes immediately
clear that there are two basic types, elliptical and spiral galaxies.

Ellipticals (also called early type galaxies)
\begin{itemize}
\item have smooth light distributions, where the isophotes (lines
of constant SB) have (nearly) elliptical shapes. They are denoted as
E$n$, where $n=1-b/a$, and $a$ and $b$ are the semi-major and
semi-minor axis of the ellipse. So, an E0 is round, and an E7 is
flattened, $n=7$ is about how flattened E's get. A 3D ellipsoidal
shape, with semi-axes $a$, $b$ and $c$ (ordered as $a\ge b\ge c$), is
called oblate when $a=b>c$ (a pancake), prolate when $a>b=c$ (a cigar),
and tri-axial when $a>b>c$ (a squashed American football). We think
that E's are probably tri-axial.
\item have no current star formation
\item have old stars
\item have little gas and dust
\item occur preferentially in groups and clusters of many Es
\item have little evidence for rotation, so the stars must have large
random motions to keep the system from collapsing (also called a hot
stellar system)
\end{itemize}

Spirals (or late type galaxies)
\begin{itemize}
\item have most of the light coming from a thin disk. Seen face on, it
is nearly circular, seen edge on, it is nearly a line. In their centre,
they often have a nearly elliptical stellar system, similar in shape to
a small elliptical, called a bulge. In addition, some have a bar as
well.
\item have light profiles with a characteristic spiral pattern
\item have star formation going on, especially in their spiral arms
\item have both old and young stars
\item have gas and dust
\item seem to avoid dense groups, and are very rare in clusters
\item have large rotational velocities, enough to keep their stars on
circular orbits. 
\end{itemize}

We will see later that the SB as a function of radius, for these two
types, are quite different.

These are the archetypal properties of such systems. For example,
some ellipticals do have a lot of dust, as well as evidence for star
formation. Some even have a small disk. Given that some spirals have a
very large bulge, and some elliptical have a disk, the distinction
between the two types is not always as clear as you might
think. Clearly it would be interesting to know what processes are
responsible for forming an elliptical vs a spiral.

Both types are seen to have hundreds and sometimes thousands of {\em
globular clusters} (GCs for short). These are very dense, spherical,
gravitationally bound systems of typically $10^5$ to $10^6$ low mass
stars, and they are usually found in a spherical distribution around
their parent galaxy. The Milky Way has about 150 of them, but some of
the larger Es have several tens of thousands.

There are many schemes for classifying galaxies, based on their
morphology. What do you expect from a good classification? Ideally, it
should be unambiguous, and have some physical basis behind it. Also,
since for most galaxies you don't know the distance, it should be
independent of distance. The {\em Hubble sequence} is still very
popular. It divides galaxies in ellipticals and spirals. Spirals are
divided in two strands, according to whether they are barred (SB), or
not (S).  The sequence has the shape of a tuning fork -- the Hubble
tuning fork. Along the handle of the tuning fork, the range goes from
E0 to E7. The sequence then forks, toward the top are unbarred spirals
-- denoted S, along the bottom are barred spirals, denoted SB. Both
strands are further divided along the sequence (away from the handle),
as Sa toward Sc, and analogously SBa to SBc, depending on the relative
strength of bulge to disk, and, correlated with this, how
tightly/loosely wound the arms are. Sa's and SBa's have a large bulge
and tightly wound arms, Sc's and SBc's have a much smaller bulge, and
more loosely wound arms. Finally, any galaxy that does not fit into the
sequence, is called irregular (Irr).

The physical scale of galaxies is huge: E's can have masses from as
little as $10^7\msun$ to $10^{13}\msun$, with linear sizes ranging from
a few tenths of a \kpc\ to hundreds of \kpc. In contrast, spirals tend
to be more homogeneous, with masses $10^9$--$10^{12}$\msun, and disk
diameters from 5 to 100\kpc\ or so.

Hubble thought (incorrectly) that this sequence was due to evolution,
where galaxies evolve from being E0s, toward Sa or Sba, and later
evolving in the more loosely wound Sc's or SBc's. Therefore these types
are also called early type and late type. So E's are also called early
type, and Ss are called late type. And an Sc is a late-type spiral, and
Sa an early type spiral. We know that this is not the way galaxies
evolve, yet the names are still widely used.

There are other reasons why this is not a particularly good scheme. For
example, galaxies may look quite different, depending on which filter
you use.

Another type you'll often come across are cD galaxies, which stands
for {\bf c}entral {\bf D}ominant: these are very large ellipticals, found
in the centres of clusters, with a faint but very large outer halo of
stars.

Another type are called S0 (or SB0 when barred) or lenticulars: they
are the divide in Hubble's sequence between E and S. They have disks
without gas or dust, and no recent star formation. 

Galaxies can fail to fit in the sequence, for example because they
recently had a tidal encounter with another galaxy. This typically
leads to spectacular tidal arms, like the case of e.g. the Antennae.

Now, most smaller galaxies are of the Irregular type: no disk, no nice
spherical distribution of stars, but more a loose aggregate of stars,
like for example the Magellanic Clouds. Since there are far more small
than large galaxies, most galaxies are in fact of type Irr. The aim of
these lectures is to understand which types of observation can be used
to describe these different types a bit better (the \lq biology bit of
the lectures\rq), and also try to understand the physical processes
behind them. May be you'll be surprised to learn that, for many of the
most basic questions you might think of, such as -- why are some
galaxies of type E, and others of type S?, or, why do Ss have spiral
arms? -- we are only just beginning to understand the underlying
physics, and we have by no means a well defined, generally accepted
textbook answer. May be something for you to work on, when you start
doing astronomy research yourself?



\chapter{The discovery of the Milky Way and of other galaxies}
Around the 1920s, a great debate was raging about the nature of the
nebulae, with two camps. According to one camp, the nebulae were
structures within the Milky Way, probably proto-planetary systems or
something similar, and the Sun was near the centre of MW. The other
camp claimed that many of the nebulae were galaxies outside of the MW,
and the Sun was at large distance from the centre of the MW. The reason
they couldn't decide was mainly because they didn't know how to
determine distances to the nebulae. The reason the first group got it
wrong was not because they were stupid or anything like that: their
mistake was quite interesting. And so we'll go through it, hoping we
can learn from their mistakes. Before we do this, let's describe the
main observables, and observing techniques.

\section{The main observables}
\noindent{\em Star clusters} Two types of star cluster were (and are)
known. We already described {\em Globular clusters}, which are
spherically distributed around the MW. Stars form in the MW disk out of
large molecular clouds. Sometimes, after the gas has been dispersed,
the newly formed stars are gravitationally bound to each other, in what
is called an {\em Open clusters}. They typically contain of order
$10^4$ stars. They are of great importance for studies of stellar
evolution, since we know all the stars in it have (more or less) the
same age, distance and initial composition.

\noindent{\em Star counts} Previously we described how star counts can
be used to probed the number density of stars, as a function of
position. In the early days, William Herschel and his sister, counted
the numbers of stars as function of their magnitude in many (700!!)
directions in the sky, by spending (most?) nights gazing through his
telescope! Later, photographic plates were used for the same
purpose. These confirmed that the stars around the Sun are distributed
in a disk.

\noindent{\em Standard candles} are objects with a known absolute
property, for example a known size, or known luminosity. So measuring
their angular size size, or apparent luminosity, you can determine
their distance. A most important example is that of {\bf Cepheid
variables}. Henrietta Leavitt studied variable stars in the Magellanic
Cloud in 1912. She found a relation between the period $P$ of the
variation, and the luminosity of Cepheid. Since all these stars are at
the same distance, this must mean it is a relation between the {\em absolute}
luminosity and $P$. There are many types of variable stars, but
Cepheids have many advantages (a) the shape of their light curve is
a very characteristic sawtooth pattern, (b) they are luminous - and so
can be seen to large distances and (c) the $P(L)$ relation has little
scatter. Unfortunately, they are also relatively rare. {\bf RR Lyrae}
are similar to Cepheids, but occur in a different type of star. They
are also used as standard candles. Of course, to get an absolute
distance, we need somehow to find the distance to some Cepheids or RR
Lyrae using another method, for example using the parallax. 

\noindent{\em Parallax} Stretch your arm, point your index finger
upward, and look with your left eye alone toward a distant wall. Now,
look with your right eye alone: your finger seems to move with respect
to the background. You've done (your first?) parallax measurement. If
the wall is sufficiently distant, there is a relation between how much
your finger appeared moved, the length of your arm, and the distance
between your eyes.

In astronomy, you can increase the distance between your eyes, to twice
the distance earth-Sun, by looking at the same object half a year
apart. Since one can relatively easily measure angles to a fraction of
an \arcsec\ (a 60th of an arc minute, or 1/3600th of a degree, now we
can do much better!), we have a new distance unit: the \pc. An object
at 1\pc distance, has a parallax of 2\arcsec. Let's compute how much
this really is. Consider an equi-lateral triangle, with two sides of
length 1\pc, third side 1\AU. By definition of \pc\, 1\pc/1\AU=1\arcsec
in radians, or 1/206265 radians (approximately). (We've used the fact
that this angle is very small). Hence \be 1\pc={180\times 3600\over
\pi}\AU\approx 206265\AU\,.  \ee

Unfortunately, this literally doesn't get use very far: the distance to
even the nearest stars is of this order. We'll come back to the
distance scale later.

\section{The main players, their measurements, and their mistakes}
The famous Dutch astronomer, {\em Jacobus Kapteyn}, and his
collaborators, started a major project of counting stars around the
1900s, to determine once and for all the structure of the MW. So what
he did is count stars on photographic plates as function of their
luminosity, in several directions.

His analysis basically confirmed Herschel's picture: the MW is a
flattened elliptical system, with stellar density $n$ decreasing away
from the centre. $n$ drops to half its central value at 150\pc\
perpendicular to the MW plane, and 800\pc\ in the galactic plane. The
Sun is at 650\pc\ from the centre. As we'll discuss later, this is
wrong!

In our derivation of counts of stars as function of distance $r$, we
assumed that the luminosity drops as $1/r^2$. One reason this could be
wrong, is because of absorption or scattering of the stellar light. For
Rayleigh scattering off atoms, there is a colour dependence. Of course,
not taking absorption into account would give you the wrong answer, and
Kapteyn was fully aware of this. If absorption were important, one
would expect fainter (more distant stars) to be on average
redder. Kapteyn found this indeed to be the case, yet the amount of
reddening was very small, and so absorption was no thought to be an
issue. The mistake was that the absorption is due to {\em dust}, not
Rayleigh scattering. Dust absorption is much less colour dependent,
and so the small reddening measured by Kapteyn actually implied a much
greater dimming of the distant stars.

The American astronomer {\em Harlow Shapley}, estimated the distances
to the brightest globular clusters using their RR Lyrae variables, and
found they were not centred around the Sun, but around a point 15\kpc\
away in the direction of Sagittarius. This was in conflict with the
Kapteyn universe -- it seemed difficult to reconcile the Sun to be near
the centre of the MW, yet the centre of the globular cluster system not
to coincide with the centre of MW?

Meanwhile, another famous Dutch astronomer, {\em van Maanen}, found
that some of the nebulae had proper motions, implying they were nearby,
within the MW. (He compared the positions of the nebulae on plates
taken several years apart). It was generally accepted these nebulae
were probably proto-planetary systems, or something like that. {\em
Edwin Hubble} later redid the proper motion measurements with better
data (using a better telescope), and failed to reproduce the proper
motions. van Maanen probably just made a mistake! In the 1930s, Hubble
made his even more famous discovery that the spectral lines of atoms in
other galaxies are shifted to the red, by an amount proportional to the
distance to the galaxy. The universe is expanding!

So it was a combination of mistakes (van Maanen), and not taking into
account the importance of dust (Kapteyn), that led to the confusion.
In the end, Hubble's better data clinched the matter. He resolved
Cepheids in Andromeda, one of the bigger nebulae, and found a distance
of 300\kpc\, and so clearly outside of the MW. And so the size of the
universe all of a sudden increased dramatically.

It was not until the 1930s that {\em Trumpler} found what had been
wrong with Kapteyn's results. He studied Open Clusters in the MW
plane. He assumed them to be all of about the same size, and so got an
independent relative distance for them. He used another estimate of
their distance, based on \lq spectroscopic parallaxes\rq\, by observing
the spectral type of some stars in the cluster, and using a calibration
between spectral type and absolute magnitude. He showed that the more
distant clusters (as judged from their angular size), appeared to have
unusually faint stars in them. Or equivalently, that the more distant
clusters (as judged from their photometric parallax) were
larger. Clearly, something was wrong here. He correctly interpreted
this, as due to the fact that the light from these more distant stars
was strongly attenuated. By comparing the reddening with the
attenuation, it was clear that a given amount of reddening corresponded
to much more dimming than you'd expect for Rayleigh scattering. And
hence Kapteyn's scales were wrong.

Before I summarise this in a time-line, just think of the fact that
Einstein formulated his theory of general relativity, on which the
current cosmological models are still based, before this puzzle was
resolved. In fact, our view of the Universe around Einstein's time, was
just plain wrong. I guess he rightly paid little attention to what the
astronomers of his days told him!

\subsection{Time-line}
\begin{enumerate}
\item[1610] Galileo resolves the MW into stars
\item[1750] Immanual Kant suggest some of Nebulae are other galaxies,
similar to the MW
\item[end of 1700s] Messier and Herschel catalogue hundreds of
Nebulae. Herschel counts stars, and deduces that the Sun lies near the
centre of an elliptical distribution, axes ratio 5/1
\item[1900-1920] Kapteyn counts stars, decides wrongly that extinction
is unimportant, and deduces the MW is $5\kpc\times1\kpc$ big, with the
Sun at 650\kpc\ from the centre.
\item[1912] Leavitt discovers the $P(L)$ relation for Cepheids.
\item[1914] Slipher measures large (1000s\kms) velocities for some
Nebulae, and finds evidence for rotation. Spectra suggest presence of
stars, not of gas.
\item[1915] Shapley find centre of globular clusters far away from
Kapteyn's MW centre.
\item[1920] Van Maanen claims (erroneously) that some spiral nebulae
have a large proper motion, suggesting they are within the MW.
\item[1920] Shapley and Curtis debate publicly over the size of the
MW, but the matter is not settled.
\item[1923] Hubble resolves M31 (Andromeda) into stars, using the newly
commissioned100-inch telescope. So M31 must be outside the MW. He also
discovers Cepheids, and the distance to M31 is estimated at 300\kpc.
\item[1926] Lindblad computes that Kapteyn's MW is so small, it cannot
gravitationally bind its Globular Clusters. But Shapley's much bigger MW could.
\item[1927] Jan Oort shows that several aspects of the local motion of
stars can nicely be explained if the Sun (and the other nearby stars),
is on a nearly circular motion around a position 12\kpc\ away in the
direction of Sagittarius. Nearly the same position as found by Shapley,
and implying a much larger MW than Kapteyn's.
\item[1927] larger MW picture, where many of the nebulae are
extra-galactic MWs, gains general acceptance.
\item[1929] Hubble discovers his expansion law. His derived value is a
factor of 10 too large!
\item[1930] Trumpler uses open clusters to show the importance of
extinction, and explains why Kapteyn's measurement were faulty
\item[1930-35] Hubble's new data confirm the modern picture of
galaxies, and demonstrates van Maanen's measurements must have been
wrong.
\end{enumerate}

And so in a very short time indeed, three sweeping changes in our view
of the visible universe took place:
\begin{itemize}
\item an 8-fold increase in the size of the MW
\item the acceptance that many of nebulae were external galaxies
\item the universe is expanding
\end{itemize}

\section{Absorption, scattering and reddening}
What is the effect of dust on the light from distant stars? Suppose the
wavelength of the light is much smaller than the size of the dust
grains (bricks, say) -- then you could compute the dimming just from
the geometric cross-section of the bricks along the line of sight. Now
consider a ray of light, with intensity $I$. The amount of light of this
ray, absorbed per unit length, will satisfy
\begin{equation}
{dI\over dr}=A\,I\,,
\end{equation}
which expresses the fact that each distance $dr$ will absorb a constant
{\em fraction} $dI/I=A\,dr$ of the light. This is what we want since,
if the light source were twice as bright, then the bricks will stop
twice as many photons. $A$ is some constant, which will depend on the
size of the bricks, and their number density. This equation is easy to
integrate, and the solution is $I=I_0\exp(Ar)$. Taking the $\log$ of
both sides, we find that the effect on the magnitude will be $\Delta m=A\,r$.
So the relation between apparent ($m$) and absolute magnitude ($M$) will
change from $m-M=5\log(r)-5$ to
\begin{equation}
m-M=5\log(r)-5+A\,r\,,
\label{eq:dm}
\end{equation}
due to the effect of absorption. So the unit of $A$ is magnitudes per
\kpc\,. (Careful readers will notice that I have slightly changed the
definition of $A$ along the way.)

If the size of the particles is of order of the wavelength $\lambda$ of
the light, then it is not a good approximation to use the geometric
cross-section of the dust grains. The amount of absorption (the value
of $A$) will then depend on $\lambda$, and so $A_\V\ne A_B$: absorption
in the \B\ band will differ from that in the \V\ band, and so
absorption will also lead to a colour change (reddening, since the
longer wavelengths get {\em less} absorbed than the shorter ones).

Applying Eq.(\ref{eq:dm}) to two bands, we find
\begin{eqnarray}
(m-M)_\B&=&5\log(r)-5+A_\B\,r\,\nonumber\\
(m-M)_\V&=&5\log(r)-5+A_\V\,r\,\nonumber\\
E_{\B-\V}&\equiv&(m_\B-m_\V)-(M_\B-M_\V)=(A_\B-A_V)\,r\,.
\label{eq:colour}
\end{eqnarray}
The quantity $E_{\B-\V}$ is called the colour excess, note that it is
the difference between the observed and intrinsic colour of the star,
\begin{equation}
E_{\B-\V}=(\B-\V)_{\rm observed}-(\B-\V)_{\rm intrinsic}\,.
\end{equation}

Trumpler's measurements, and also laboratory measurements, show that
for interstellar dust grains
\begin{equation}
E_{\B-\V}\approx {1\over 3}A_\V\,r\,.
\end{equation}

This is a crucial result. Reddening and hence $E_{\B-\V}$, is easy to
measure, and so if we do this for stars of known distance, we find
$A_\V\approx 1{\rm mag}\kpc^{-1}$\footnote{The amount of dust is not
the same everywhere: there are regions where the absorption is much
stronger, not surprisingly called dark clouds, and some directions
along which the absorption is much less, a well known direction is
called Baade's window}. If we now measure $E_{\B-\V}$ for another star,
we can estimate $r$! (Well, we need to know its intrinsic colour from
models.

\chapter{The modern view of the Milky Way}
Hipparcos/GAIA
COBE/DIRBE
Radio observations

Our observational view of the MW has improved due to steady hard work,
but occasionally in great leaps, when new technology was applied to old
problems. Let's quickly describe some of these.

\section{New technology}
\subsection{Radio-astronomy}
Radio astronomy used the new technology developed during the war (for
other purposes ...), and really
flourished. Roto-vibrational\footnote{Most transitions you know, which
occur in the visible and UV-part of the spectrum, are {\em electronic}
transitions: in emission, they occur because an electron changes from a
higher energy state to a lower one, e.g. from $n=2\rightarrow 1$ for
the hydrogen Lyman-$\alpha$ transition, and from $n=3\rightarrow 2$ for
the hydrogen H$\alpha$ transition. An in absorption, they go the other
way around. But molecules also have excited levels, due to their
rotation or vibration. Since these are also quantum-mechanical, these
energy levels are quantised. And transitions between them correspond to
rotational or vibrational transitions. Since the energy-levels are
lower, these correspond to longer wavelengths -- typically mm and cm:
in the radio.} transitions in molecules have low excitation levels,
which are ideally suited to probe the low temperatures in interstellar
clouds. Because the wavelengths of these transitions are long, one can
build very large dishes to observe them, and also it is more
straightforward to build interferometers.\footnote{One way to think of
this is to realise that for a good \lq mirror\rq\, it's departure from
the ideal shape should not be larger than some small fraction of the
wavelength of the radiation you plan to observe. So for optical
telescopes, it should be of the order of 5000\AA say, but for radio,
fractions of a mm, and up to a m. There is a down-side. The resolution
you get, when making diffraction limited observations, is of order
$\lambda/D$, where $\lambda$ is the wavelength, and $D$ the diameter of
the telescope. And so, radio-telescopes have to be much bigger to
obtain a similar resolution. In an interferometer, instead of building
one giant dish -- which would be very expensive, and difficult to move
-- one builds many smaller dishes, put far apart, and combines their
output. The result is that the effective $D$ is not the size of a dish,
but rather the biggest distance between them. What you loose, is that
the collecting area is only the sum of the sizes of all dishes, not $D^2$.}

Today, one regularly combines data from radio-observatories across the
earth to get Very Long Baseline Interferometry, which gives superb
angular resolution ($10^{-6}\arcsec$!). Another great advantage of
radio over optical observations, is that radio waves are not absorbed
by dust. (By now you know why: it's because the wavelengths are much
larger than the size of the dust grains.) And so, yo can use
radio-waves to probe into very dense regions, for example where star
formation occurs. One of the bigger problems facing radio-astronomy
today, is mobile phones ...

\subsection{Infrared observations}
IR astronomy took-off relatively recently, with the advent of rockets
and satellites to get the instruments above the atmosphere. The
construction of the camera for near IR observations is
challenging. They need to be cooled, since otherwise the main source of
signal is thermal radiation from the camera itself. I suspect that,
again, a major driver behind their construction was not really
astronomical, but over the past 10 years or so, it seems that much of
the technology has becomes available to the astronomers.

IR-observations share with radio the great advantage not to be too
bothered about dust. In fact, much of the visible and UV-light absorbed
by dust grains, is re-emitted in the IR and sub mm. And so we can use IR
to probe the dust and look deep inside star forming regions. We thank
the DIRBE instrument on the COBE (Cosmic Background Explorer) satellite
for one of the best views of the Milky Way.

\subsection{Star-counts, again}
Although counting stars seems a bit boring, the Hipparcos satellite
mission\footnote{http://astro.estec.esa.nl/Hipparcos/} was really a
tremendous revolution, due to the incredible accuracy of its
measurements. This satellite was launched in 1989 and specifically
designed to measure very accurately the positions of stars, several
times during the satellite's lifetime. For distant stars (some of which
were in fact very bright, extra-galactic sources called quasars), this
made it possible to construct an absolute reference system, that can be
used define the positions of extra-solar objects. But for more nearby
stars, Hipparcos could not only measure their parallax and hence their
distance, but also, by comparing the positions at two different times,
the {\em proper motion} of the stars with respect to this \lq
fixed\rq\ reference system.

The parallax is the ultimate step in creating a distance scale. We
first need to determine the parallax to standard candles, like
e.g. Cepheids, before we can use them to look further. 

The proper motion, i.e. the dance of nearby stars around us, tells us
about the distribution of mass. For example, the oscillation of stars
perpendicular to the disk of the MW tells us how much mass there is
locally in the disk.

GAIA\footnote{http://astro.estec.esa.nl/GAIA/} is a planned mission to
be launched before 2012. It will measure positions of stars up to 15-th
magnitude with a resolution of 10$^{-5}\arcsec$. Gaia's expected
scientific harvest is of almost inconceivable extent and implication
(according to the web page). In total, about $10^9$ MW stars will be
measured (as compared to $10^4$ measured by Hipparcos). Gaia will be so
accurate that it can even measure proper motions of some of the nearest
globular clusters and galaxies.

\section{The components of the Milky Way}
We've already briefly discussed what the major constituents of the MW
are: stars, gas and dust. Now we'll describe how they're
distributed. We think the MW is a pretty much ordinary spiral galaxy,
of which we know several million others, but we know this one in a bit
more detail. Even so, bare in mind that some of the details are
uncertain. We will discuss the {\em disk} (there is a thin and a thick
disk), {\em bulge} and {\em halo}.

\subsection{The disk}
Most of the MW's stars are in a thin disk, which represents about 70 per
cent of the total star light of the MW. The centre of the disk is in
the direction of Sagittarius, $\alpha=17^h\,42^m\,29.3^s$,
$\delta=-29^{o}\,59'$, at a distance of $R_0$. The latter is called the
solar galacto-centric distance, and there is quite some discussion to
how large it really is. $R_0=8.5\kpc$ is usually quoted, but some
studies find $R_0=8.0\kpc$. These studies try to determine the centre
of a set of objects, for example globular clusters, variable stars such
as RR Lyrae or Mira variables, or O and B stars. An $H_2O$ maser source
is located near the centre of the MW. The maser emission is thought to
come from a spherical shell, and one can measure both the radial and
transverse speed of the shell. If the shell is expanding spherically,
then one can determine its distance.

Apart from stars, the disk also contains gas and dust. On top of the
smooth disk are spiral arms, traced by young stars, molecular clouds,
and ionised gas. The disk stars are in (nearly) circular motion around
the centre. For the sun, the circular velocity is $\approx 220\kms$.

How old is the disk? We don't really know. But we can put limits on the
ages of the {\em stars} in the disk, but have to keep in mind that the
stars could possibly be older than the disk (i.e., they have formed
somewhere else). One neat age estimator are White Dwarfs. As you
recall, these don not undergo nuclear fusion anymore, but are presently
just cooling down. So, by measuring their current temperatures, and
cooling rates, we can compute how old they are. Some of the older White
Dwarfs are thought to be $\sim 10--12\times 10^9$ years old.

Let's describe the position of a star in the MW using cylindrical
coordinates $(R,\varphi,z)$, where $R$ is the distance to the centre,
$z$ the distance to the plane, and $\varphi$ an angle. The density of
stars in the disk goes approximately like $n(R,\varphi,z)\propto
\exp(-R/R_h)\exp(-z/z_h)$. Since the disk is (nearly) axis-symmetric,
$n(R,\varphi,z)$ is independent of $\varphi$. Away from the centre, $n$
decreases exponentially, and at distance $R_h\approx 3.5\kpc$, it has
fallen to $1/e$ of it's central value. Also perpendicular to the disk,
the fall-off is exponential, with $z_h\approx 0.3\kpc$. $R_h$ and $z_h$
are called the {\em scale length} and {\em scale height} of the disk,
respectively. These scale lengths really depend on the type of star
you're using. Young stars are born out of gas, which is much more
concentrated toward $z=0$. And so the scale height of young stars is
much smaller than of older stars. 

Note that there isn't really an edge to the disk. It can be traced to a
distance of around 30\kpc. With a height of 0.3\kpc, this is a ratio
100:1, which is thinner than a compact disc!

Given the exponential distribution in $n$, the luminosity profile (in
$L_\odot$ per unit volume) of the disk is often modelled as

\begin{equation}
L(R,z)=L_0 \exp(-R/R_h) {2\over \exp(z/z_h)+\exp(-z/z_h)}\,,
\label{eq:diskL}
\end{equation}
where $L_0$ is some normalisation constant. In the \B-band, the total
luminosity is $\sim 2\times 10^{10}\lsun$.

Because of the component I'll describe next, the disk is also sometimes
called the thin disk.

\subsection{The thick disk} About 4 per cent of the MW's stars belong
to a thicker disk, which is aligned with the (thin) disk, but has a
scale height of $z_h\approx 1\kpc$. It's existence has been discovered
only recently, in the 1980s. The difference with the thin disk is not
just the kinematics of it's stars: it turns out that thick disk stars
are usually older than the stars of the thin disk, and have a lower
metallicity [Fe/H]$\sim $-0.5, i.e. they have less Fe (and other metals
as well, but we can usually measure Fe accurately) for a given amount
of hydrogen (we'll see later what this notation means).


\subsection{The bulge} The bulge is a spheroidal stellar system with radius of
$\sim 1\kpc$, located at the centre of the MW. In fact, it's properties
are quite similar to that of an elliptical galaxy: it is as if there is
a small elliptical galaxy at the centre of each spiral. The surface
brightness profile (in $L_\odot$ per unit area) of this component goes like

\begin{equation}
I(r)=I_e \exp[-7.6(r/r_e)^{1/4}]\,.
\label{eq:r4}
\end{equation}

This is a famous fit to the SB-profile of elliptical galaxies,
introduced by {\em de Vaucouleurs} in 1948, and called the \lq de
Vaucouleurs\rq\ or \lq $r^{1/4}$\rq (r-to-the-one-quarter) profile.
($r=(R^2+z^2)^{1/2}$ is the 3D distance to the centre.)

$r_e$ is called the effective radius. Note that $I(r=r_e)=I_e$. The
factor 7.6 is chosen such that half of the light comes from stars
within $r_e$. For the MW, $r_e\approx 0.7\kpc$. 

Although the bulge is bright, we can't see much of it, due to the large
amount of obscuring dust (in particular, you can't see it with the
naked eye!). The total amount of extinction toward the centre is abut
28 magnitudes in the visual! In other spirals, we can see the bulge very well (and
recall that the classification of spirals is based on the bulge/disk
ratio), but for the MW we need to go the the IR to get a nice picture.
of it. The MW's bulge is rather elongated with axis ratio 5:3, with
strong evidence for a bar.

The stellar content of the bulge is a bit of a mixed bag, with both
very old, low metallicity stars, but also young, metal rich stars. The
mean metallicity is about {Fe/H]$\sim 0.3$, i.e. twice the solar value. 
The mass of the bulge is $\sim 10^{10}\msun$, with a \B-band luminosity
of $\sim 3\times 10^9$\lsun. 

\subsection{The stellar halo}
The final luminous component of the MW is the stellar halo, composed of
globular clusters (GCs), and \lq field stars\rq. These stars, just like
the stars in the globular clusters, are typically of very low
metallicity, [Fe/H]$\sim -0.8$. The current record holder of low
metallicity has [Fe/H]$\approx -5$.

Unlike the disk, the halo rotates very little, if at all. As a
consequence, halo stars all appear to move at high velocity (in fact,
rather it is the sun that is moving at $\sim 220\kms$), which is how
they were discovered initially.

We know about 150 GCs in the halo. There is some evidence that there
are two types, distinguished according to age, metallicity and
position. The younger, more metal rich ones have a relatively flattened
distribution, and may in fact be associated with the disk. The older,
more metal poor ones, are spherically distributed.

There is quite some discussion on the current ages of GCs, with
estimates for the oldest ones varying between 12 and 17Gyears
(1Gyear$\equiv 10^9$ years). For a long time, these estimates were
rather embarrassing for cosmologists, given that this was older than
the best estimates for the age of the {\em Universe}. This discrepancy
is now resolved, since the Universe is thought to be older now, and the
ages of the clusters have come down a bit.

If the absolute age of the GCs is difficult to estimate, the relative
ages are more secure. And so it is rather more secure to say that there
is a relatively large spread in the ages of the MW's GCs. Presumably,
this tells us something about how the MW formed.
 
The density of halo stars (and of GCs as well) falls spherically as
\begin{equation}
n(r)=n_0 (r/r_0)^{-3.5}\,,
\end{equation}
and extremely distant field stars have been detected out to
$r=50\kpc$. The total mass is actually quite low, $M\sim 10^9\msun$ and
luminosity $\sim 10^9\lsun$, or only 10 per cent of that of the disk.

\subsection{The dark matter halo}
We'll discuss evidence for dark matter later, but it appears to be the
case that most, may be as much as 90 per cent, of the mass in galaxies
is actually invisible. We don't know what it is made of, but this
matter does not emit or absorb any light -- hence {\em dark} matter,
yet there seems to be a lot of it.

\section{Metallicity of stars}
Astronomers call all elements more massive than Helium\footnote{The Big
Bang produced hydrogen, helium, and trace amounts of more massive
elements} \lq metals\rq, and denote them by $Z$ ($X$ and $Y$ being the
hydrogen and helium abundance by mass, respectively). These are
produced in stars, but for some elements, like e.g. Carbon, we don't
really know which stars are the dominant source. A lot of metals are
produced during super novae (SNe) explosions. There are two basic types
of SNe. SNe of type II are explosions of very massive stars. Nuclear
fusion during the explosion is responsible for many elements of the
$\alpha$-type, like Oxygen and Silicon. SNe of type I, on the other
hand, are thought to occur during mass-transfer in binary stars --
these are the dominant producers of Iron, Fe. The chemical mix of the
stars is therefore a fossil relict of the relative fractions of type I
and type II SNe. Typically though, one measures just one metallicity,
for example the Fe abundance since that is often the easiest to
measure, and then assumes that the other elements scale with Fe. But we
know for a fact that some stellar systems have a rather different
relative abundance pattern of the elements.

For the sun, the total amount of metals by mass, is about
$Z_\odot=0.02$, or 2 per cent of the mass in the solar system is {\em
not} hydrogen or helium. This is inferred not just from observations of
the sun, but also from the composition of comets.

For other stars, one usually compares the metallicity in units of the
solar value, on a logarithmic scale,
\begin{equation}
[Fe/H]\equiv log_{10}\left[(Fe/H)/(Fe/H)_\odot\right]\,.
\end{equation}

So if a star has [Fe/H]=0, it has the same Iron abundance as the Sun,
and for [Fe/H]=-1, it has one tenth the solar value.

One might expect the metallicity of a star to be related to its
age. Indeed, very old stars formed before there had been many
generations of stars\footnote{The Sun is thought to be a third
generation star.}, and hence before many SNe exploded, and so would
have been formed from gas containing mostly hydrogen and helium. But
stars forming now, will contract from gas that has already been
polluted by SNe, and hence will be more metal rich. Within the MW, this
seems to be born-out, at least to some extent. Stars in old GCs, for
example, typically have [Fe/H]$\sim -1$, so quit a bit below the solar
value. And most of the old stars in the halo also have low
metallicity. These are called {\em population II} stars. In contrast,
stars in the disk usually have higher metallicity, and are called {\em
population I}. There is no strict divide between those, some disk stars
also have low $Z$ for example.

Recently, there has been quite a bit of interest in the very first
generation of stars that formed after the Big Bang. Those would have
$Z=0$! They are called {\em population III}. The halo star with then
lowest metallicity currently known, has [Fe/H]$\approx -5$, and might
well be one of the first stars to form in the MW.

The evolution of $Z$ within the MW, or within galaxies in general, is
called their chemical evolution (which is a misnomer, since the
elements are produced in nuclear reactions which do not involve
chemistry).


Table of numbers for the MW
\label{sect:simulation}
\begin{table}
 \hrule
 \vskip0.1truecm
 \hrule
 \centering \begin{minipage}{120mm} \caption{Disks}
 \vskip0.1truecm
 \hrule
 \vskip0.05truecm
 \begin{tabular}{llll} 
        & Neutral Gas    & Thin Disk           & Thick Disk \\[1pt]
$M/10^{10}\msun$ & 0.5	& 6	& 0.2 to 0.4\\
$L_B/10^{10}\lsun$ &	& 1.8 & 0.02\\
$M/L_\B$ (\msun/\lsun) & 	& 	3 &\\
Diameter (\kpc) &	50	&	50	& 50\\
Distribution &	$\exp(-z/0.16\kpc)$ &	$\exp(-z/0.325\kpc)$ & $\exp(-z/1.4\kpc)$ \\
$[Fe\/H]$ & $>0.1$ &	-0.5 -- 0.3 & -1.6 -- -0.4\\
Age (Gyr) &	0 -- 17 & $<12$ & 14 -- 17\\
\label{table:runs}
\end{tabular}
 \vskip0.05truecm
 \hrule
\end{minipage}
\end{table}


\label{sect:simulation}
\begin{table}
 \hrule
 \vskip0.1truecm
 \hrule
 \centering \begin{minipage}{120mm} \caption{Spheroids}
 \vskip0.1truecm
 \hrule
 \vskip0.05truecm
 \begin{tabular}{llll} 
        & Central Bulge    & Stellar Halo        & Dark Matter Halo \\[1pt]
$M/10^{10}\msun$ & 1	& 0.1	& 55\\
$L_B/10^{10}\lsun$ &0.3	& 0.1 & 0\\
$M/L_\B$ (\msun/\lsun) &3	& 	$\sim 1$ & -\\
Diameter (\kpc) &	2	&	100	& $>200$\\
Distribution &	bar &	$r^{-3.5}$ & $(a^2+r^2)^{-1}$ \\
$[Fe\/H]$ & -1 -- 1 &	-4.5 -- -0.5 & -1.6 -- -0.4\\
Age (Gyr) &	10 -- 17 & 14 -- 17 & 17\\
\label{table:runs}
\end{tabular}
 \vskip0.05truecm
 \hrule
\end{minipage}
\end{table}

\subsection{Galactic Coordinates}
When describing the position of a star within the MW, it is useful to
use Galactic Coordinates $(l,b)$, $0^o\le l\le 360^o$, and $-90^o\le
b\le 90^o$. From the vantage point of the Sun, the direction to the
galactic centre is taken to be $l=0$, $l=180^o$ is the ante-centre, and
$b=0$ corresponds to the galactic plane. $b=\pm 90^o$ to the galactic
poles. Knowing these direction in terms of right ascension $\alpha$ and
declination $\delta$, one uses spherical trigonometry to convert
between them. Since the plane of the MW is inclined with respect to the
celestial equator, the planes $\delta=0$ and $b=0$ do not coincide.


\chapter{The Interstellar Medium}
The interstellar medium (ISM) is the stuff between the stars, basically
gas, dust, magnetic fields and cosmic rays. Stars form in regions where
the ISM is very dense, such as in molecular clouds. During their
lifetimes they return some of this material back to the ISM in the form
of stellar winds and Planetary Nebulae, and at their deaths during a
super nova explosion. The detailed interplay between stars and the ISM
is certainly very complicated and presently not well understood. Here
we will discuss briefly how the gas and dust can be observed, describe
the Jeans criterion for the collapse of gas into stars, and describe
the physics of ionised regions of gas called HII regions.

\section{Interstellar dust}
Dust particles interact with light both through {\em scattering} and
{\em absorption}. In both cases, there is a reduction in the amount of
starlight you receive, described by Eqs.~(\ref{eq:colour}).

Before I describe the physics, let me tell you that you already know
most of this. The molecules in the earth's atmosphere scatter light
from the sun. Because of the quantum mechanical properties of the
molecules, they scatter blue light more efficiently than red light.
Now the sun light you see in the sky {\em away} from the sun, you can
see only because it is scattered toward you -- and so this makes the
sky blue. At sun set or sun rise, the sun light passes through a lot
more atmosphere than at midday. The increased path length causes the
strong reddening of the sun. This is partly due to scattering, partly
due to absorption on little dust grains. For example, volcanic
eruptions spew lots of dust grains into the atmosphere, which cause
amazingly red sun sets. So dust particles also cause reddening. The
same process of scattering and absorption also occur in the ISM.

{\em Scattering} changes the direction of the incoming photon, but not
it's energy. This type of reflection can also induce polarisation of
the light, if the dust grains are aligned in a given direction, for
example due to a magnetic field. You know that scattering can cause
polarisation: expensive sunshades are polarised, so as to block
incoming polarised light reflected from the ocean, or road surface for
example. So dust grains may scatter some of the light coming from a
distant star out of the line-of-sight, thereby reducing the amount you
detect, and hence the apparent luminosity of the star. Given the
typical size of interstellar dust grains, blue light is scattered more
than red light, and hence scattering also leads to reddening. Now if
you look at a dust cloud, you may actually detect some of the light
scattered off dust from a nearby bright star for example. And so such a
{\em reflection nebulae} is typically blue (so for the same reason that
the sky is blue, except it's scattering by dust (for the reflection
nebula) vs by molecules (for the earth's atmosphere)).

When a dust grain {\em absorbs} a photon, it can sometimes undergo a
change in structure, for example a molecular bond will be broken, or an
atom or molecule gets into an excited state. The energy of the photon
is converted into another form of energy, and is lost to us: the photon
is {\em absorbed}. So one effect of absorption is to decrease the
amount of light we detect from the star. Very often, the dust grain
will emit other photons, but at much lower wavelengths, typically in
the infrared. For example in star forming regions where the young stars
produce large amounts of dust, and so, even though these stars may be
very bright, we cannot see them at visual wavelengths because of the
large amount of absorption by dust. Conversely, these regions are very
bright in the infrared, because of the reprocessing. If the wavelength
of the light is much larger than the size of the dust grains, then
absorption is much reduced. And so you understand that the newly planned
space-telescope which will work in the infrared, will teach us a lot
about star forming regions.\footnote{Many molecules in the earth's
atmosphere, for example water, absorb infrared light in their
molecular bands, which is why it is difficult to perform infrared
observations from the ground.}

The scattering and absorption by dust grains may seem to be just a
nuisance for astronomers, given that they decrease the amount of star
light we observe. But because the absorbing properties depend on the
size and composition of the dust, we can learn about dust properties by
studying their effect on star light. And so this is how we infer the
typical size and composition of the grains. Although the amount of
absorption typically decreases with increasing wavelength, there is
much more absorption at some specific {\em resonances}, which teach us
that some grains have graphite in them (graphite, like diamond, is a
phase of carbon), and also silicates (we detect the resonance at the
energy of the Si-O chemical bond).

\section{Interstellar gas}
We detect hydrogen gas in the ISM in the form of neutral, ionised and
molecular form. In astronomy, these are denoted as HI (neutral), HII
(ionised), and H$_2$ (molecular), which is far from ideal. You would
think HII to be double ionised -- impossible of course -- it means
H$^+$. And when speaking, you cannot distinguish HII from H$_2$ -- both
are pronounced as H-two. So, CIV is {\em triply} ionised carbon. I'm
sorry about this!

How much is there of each type, and more importantly, {\em why} do we
sometimes find one form, sometimes the other? 

\subsection{Collisional processes}
Suppose you have a collision between two H$_2$ molecules. If the
collision is sufficiently violent, you can imagine that it will break
the molecular bond, and you would have converted H$_2\rightarrow
2$HI. Similarly, if you collide two hydrogen atoms (or a HI with an
electron) with enough speed, it may lead to ionisation,
HI+e$\rightarrow $HII+2e. Since the energy required to break the
molecular bond (11eV) is lower than to ionise hydrogen(13.6eV), the
second collision needs more energy than the first.  And so you expect
that {\em the higher the temperature of the gas -- and hence the more
energetic the collisions -- the more you will shift from molecular, to
atomic, to ionised gas}.

Incidentally, the reverse process is very important as well: when an HI
atom collides with another HI, or with an electron and gets ionised,
then the system has converted kinetic energy into ionisation energy.
When the ion recombines, it can be that the photon escapes from the
cloud, in which case there has been a net loss of energy for the
gas. Since the particles are now moving slower, the temperature of the
gas $T\propto \langle v^2\rangle$ is lower, or the gas has cooled
down. Not surprisingly, this process is called {\em radiative cooling},
and it is very important to understand the process of star formation.

So we understand now why colder gas will tend to be molecular, whereas
hotter gas is more likely to be atomic or even ionised. Now, if the
{\em pressure} between these phases is nearly the same, then the colder
gas (in which H$_2$ is favoured) will need to be denser than the hotter
gas (atomic or, at higher temperature, ionised). And so you expect to
find dense, cold clouds to be molecular, and hot, rarefied gas to be
ionised -- which is indeed what we observe, except that there is one
more important process, photo-ionisation.

\subsection{Photo-ionisation}
Interaction between a photon, from a nearby star say, and a molecule or
atom, may lead to dissociation of the molecular bond, and/or ionisation
of the atom. And so the process of molecule formation, which makes a
cloud cool, get denser, and eventually form stars, can be completely
reversed when the newly born star starts to ionise and heat the gas.

Recall from the first part of the course that massive stars, like O and
B-type stars, are hot and hence emit lots of hydrogen ionising
photons. These photons have a dramatic effect on the surrounding gas,
and convert the hydrogen into it's HII form. And so massive bright
stars are often surrounded by {\em HII regions}. These regions
delineate spiral arms, because star formation occurs predominantly in
spiral arms and massive stars don not live for very long.

HII regions are important, not only because they are beautiful to look
at\footnote{One of the more famous HII regions is the Orion nebula,
M42.}, but because their physics is reasonably well understood. And by
observing the relative strengths of the emission lines that occur in
their spectra, one can deduce the properties of the HII region, such as
it's temperature, density, and the relative abundance of the elements.

There are two things I would like you to remember about this, one is to
understand the physics that determines the properties of the spectrum
that you get from such a nebula, and the second one to understand what
sets their size (the so-called Str\"omgren radius}

When an ion (for example HII, OIII, ...) recombines with an electron
(and form HI, respectively OII), the electron does not necessarily have
to fall directly to the lowest possible energy level, i.e. the ground
state. Taking the example of HI, the ground state would correspond to
the electronic $n=1$ state. Typically, the electron will cascade down
to $n=1$, and the relative probability for the intermediate steps can
be computed from quantum mechanics. However, {\em this does not
translate directly into the spectrum you observe}. For example, suppose
the electron makes a transition $n=2\rightarrow n=1$, and emits the
corresponding photon. As this photon starts to move toward us, it may
actually interact with another neutral hydrogen atom, and cause the
electron of that atom to be excited from the $n=1$ to $n=2$ level. In
which case the photon does not actually exit from the nebula! Suppose
on the other hand, the photon makes an $n=3\rightarrow n=2$
transition. This photon has much more chance of leaving the nebula,
since most of the neutral HI will be in the $n=1$ state, and not so
much in the $n=2$ state. And so that photon will escape from the
nebula. So curiously, it may be that lines with a low quantum
mechanical probability, dominate a nebular spectrum, since photons from
more likely transitions are unable to escape from the nebula. This is
especially true for Planetary Nebulae spectra.\footnote{The strongest
line from a Planetary Nebula is O[III], which is a {\em forbidden}
transition. This means that the quantum mechanical probability is zero,
and such lines are not observed in laboratory environments. The reason
that the line does occur is because a collision with another particle
makes the transition possible. But because the transition is forbidden,
once the photon is produced, it has no trouble escaping the nebula.}

For an HII region, the dominant wavelength photon that escapes results
from the $n=3\rightarrow n=2$ transition, denoted as
H$\alpha$\footnote{Recall that transitions to $n=1$ are called the
Lyman series and to $n=2$ the Balmer series. Furthermore, a transition
$n+k$ to $n$ are \lq counted\rq\ with the Greek alphabet, and so
$n=3\rightarrow n=2$ is called H$\alpha$, $n=4\rightarrow n=2$ is
H$\beta$, $n=2\rightarrow n=1$ is Lyman$\alpha$, $n=3\rightarrow n=1$
is Lyman$\beta$, etc.}. This red line is the reason HII regions appear
to fluoresce red. And by observing galaxies through a filter that only
lets H$\alpha$ light through, one can easily find HII regions.

Suppose a source of ionising photons such as a hot star starts emitting
ionising photons at a rate ${\dot N}$, in photons per second, and
assume the source is surrounded by a homogeneous cloud of atomic
hydrogen, with density $n$. The source will quickly ionise all hydrogen
close to it.  Let $R(t)$ be the radius of the {\em ionisation front},
within which most of the hydrogen is ionised, and outside of which the
gas is mostly neutral. As $R$ increases between $R$ and $R+\Delta R$, the
number $N$ of atoms that need to be ionised is 

\begin{equation}
(4\pi/3)n[(R+\Delta R)^3-R^3]\approx 4\pi nR^2 \Delta R\,.
\end{equation}

Since it takes the source a time $\Delta t=N/\dot N$ to produce this
many photons, we find that the speed of the front is

\begin{equation}
v={\Delta R\over\Delta t}={\dot N\over 4\pi nR^2}\,.
\end{equation}

Of course, this speed cannot be faster than the speed of light, and you
see that, as the HII regions grows in size, the speed with which it
grows decreases $\propto 1/R^2$. However, eventually some of the HII
ions inside the ionisation front will start to recombine. Since extra
photons are needed to re-ionise these, the speed of the front will slowly
down even more. Eventually, an equilibrium is reached, in which the
number of photo-ionisations within $R$ equals the number of
recombinations. The radius of the HII region has grown to its maximal
size: the Str\"omgren radius, $R_S$. To compute $R_S$, consider a small
volume of the HII region. Since a recombination is an interaction between an
electron and an HII ion, the rate at which HII ions recombine is
proportional to the electron and the ion density:

\begin{equation}
\hbox{\rm recombination rate=} \alpha n_e N_{\rm HII}\,.
\end{equation}

Since the recombination rate is in ions s$^{-1}$ volume$^{-1}$, we find
that the recombination rate $\alpha$ must have dimensions of volume
s$^{-1}$. If the gas is composed purely of hydrogen, and is very highly
ionised, then $n_e\approx N_{\rm HII}\approx n_{\rm H}$, where $n_{\rm
H}$ is the density of hydrogen, either HII or HI. The total number of
recombinations within radius $R$ is then $(4/\pi/3) \alpha n_{rm H}^2
R^3$. In equilibrium, this is the rate $\dot N$ at which the source
produces ionising photons. Equating these, we find that

\begin{equation}
R_S = ({3\dot N\over 4\pi\alpha})^{1/3} n_{\rm H}^{2/3}\,.
\end{equation}

For example, assume $n_{\rm H}=5\times 10^3{\rm cm}^{-3}$ for the
density of the cloud, and $\dot N=10^{49}{\rm s}^{-1}$ for the
ionisation rate of the star. Then $R_S\approx 0.21\pc$, since
$\alpha\approx 3.1\times 10^{-13}$ cm$^{3}$ s$^{-1}$ at a temperature
of $T=8000K$ typical of HII regions.
 
\subsection{21-cm radiation}
We have just seen that you can observe ionised gas in HII regions
because of its recombination radiation. So how can we observe HI?

A hydrogen atom is composed of a proton and an electron. Both these
particles have a spin degree of freedom. Spin is a purely quantum
mechanical concept, but there is some analogy with a bar magnet. So in
a hydrogen atom, the spins of proton and electron can either be aligned
or anti-aligned. In the bar-magnet analogy: either the north-poles of
both magnets point in the same direction (aligned) or in the opposite
direction (anti-aligned).\footnote{Two bar magnets can of course have
any angle between them. But for quantum mechanical spins, this is
impossible, and they are either fully aligned, or fully anti-aligned,
with nothing in between! So the analogy only goes so far.} And just as
in the bar-magnet case, there is a small energy difference between the
two states, with the aligned state having higher energy. Transitions
between states of different spins are also called hyper-fine
interactions.\footnote{Interactions between the spin and the {\em
orbital} angular momentum of the electron also result in slightly
different energy levels and transitions.}

When an HI atom with aligned spins is left on its own, it has some
small probability for a spontaneous spin reversal to the lower energy
state. When it does so, it emits a photon with wavelength 21-cm. The
mean time it takes an HI to perform such a spin reversal is several
million years. This makes it almost impossible to observe this
transition in the laboratory, because collisions between the particles
will de-excite the aligned spin well before there has been a
spontaneous transition. But in the ISM, densities are far lower than
can be generated in the lab, and hence the time between collisions such
greater, and the 21-cm line is of great observational value. First of
all, there is enough HI to observe the line. And once the line is
produced, it's low probability becomes an advantage, since the 21-cm
radiation can pass through a lot of HI gas, without it being absorbed
again.

An external magnetic field shifts the energy levels (called the Zeeman
effect). And so 21-cm radiation can be used to estimate the strength of
the magnetic field in the ISM.

It was the Dutch astronomer Jan Oort who suggested one of his students,
van der Hulst, to compute the wavelength expected for this
transition. (Would you like to be given such a project?). When it was
clear that it could be observed, he asked the government for money to
build a radio-telescope to go an observe it.

\subsection{Other radio-wavelengths}
We've seen how we expect gas to become mostly molecular in high density
regions. And so we cannot observe these with 21-cm observations, since
also the hydrogen will become molecular. Fortunately, many molecules
also emit radiation at radio-wavelengths. Typically these are caused by
rotational transitions.\footnote{Energy can be stored in rotation of
the molecule. In a quantum mechanical description, the amount you can
store is quantised. Rotational transitions are transitions between
different rotation speeds of the molecule.}

Which transition is strongest depends a lot on the density of the
cloud. The most commonly studied example is the CO molecule, with
wavelength 2.6cm. At higher densities, other molecules such as CH, OH
and CS become observable.

The result of many years of investigation is that the properties within
gas clouds can vary widely. In most of the outer parts of the disk,
densities are low, and most gas is in the form of HI, and typical
density of order 1\cm$^{-3}$. Closer in, we find a variety of clouds
within the HI gas, which differ in total mass, and density. {\em Giant
Molecular Clouds}, or GMCs for short, are enormous complexes of gas and
dust, with total masses up to $10^6\msun$, temperatures $T\sim 20$K,
and densities 100-300\cm$^{-3}$. These clouds often have
sub-condensations which are much denser. Thousands of GMCs are known in
the MW, mostly in the spiral arms.

It is thought that most, if not all, star formation in the MW occurs in
GMCs. To understand what determines whether a cloud can remain stable,
or will become unstable and undergo star formation, brings us to the
next concept: the Jeans mass.

\subsection{The Jeans mass}
I trust you have already encountered the concept of Jeans mass when
studying stars. Let me just phrase the same thing slightly differently,
starting with the difference between stable and unstable equilibrium.

When a system is in equilibrium, forces balance. In a {\em stable}
equilibrium, small changes remain small, whereas in an {\em unstable}
equilibrium, small changes are amplified and grow.

For example, consider small perturbations in an infinite,
self-gravitating fluid, with density $\rho$ and temperature $T$. If the
scale $\lambda$ of the perturbation is small, then you know what
happens: the perturbation will be a sound wave. This is why you can
hear what I am saying. But what happens for perturbations on larger
scales? I'll explain why eventually gravity will become dominant.

Consider the material inside one perturbation length $\lambda$. The
thermal energy $K$ enclosed within it, is of order $K\sim M\boltz\,T$, where
$M=\rho\lambda^3$ is the mass within the perturbation (I'm neglecting
constants of order unity, like the $4\pi/3$ and such -- we'll do better
later). The gravitational energy $U$ within the perturbation, is of
order $U\sim G\,M^2/\lambda\sim G\,M\rho\lambda^2$. For sufficiently small
$\lambda$, the gravity energy is small with respect to the thermal
energy, and can be neglected: these are the usual sound waves. But for
larger $\lambda$, the ratio $U/K\sim \lambda^2$ gets bigger, and so
eventually, gravity will dominate! The Jeans mass is the critical mass
above which gravity dominates. For perturbations below the Jeans mass,
pressure forces dominate, and so the perturbation will re-expand when
being compresses. But for perturbation more massive than the Jeans
mass, gravity dominates, and so a perturbation will collapse even
further when compressed, leading to run-away collapse.

Now the proper way to derive the criterion where gravity dominates
requires some more mathematics that I don't want to go into. What you
should do, is compute the {\em dispersion relation} for sounds waves,
i.e. the relation between the wave-length $\lambda$ and the sound speed
$c_s$. For small wavelengths, gravity can be neglected, and $c_s$ is
independent of $\lambda$. However, as I explained, for larger and
larger $\lambda$, eventually gravity becomes important and $c_s$ starts
to depend on $\lambda$, and for the critical wave-length $\lambda_J$,
the sound speed becomes zero. Now, gravity dominates, and as the wave
stalls its amplitude will grow, and the perturbation will collapse.  I
hope you will have the chance to investigate this nice problem in more
detail somewhere else.

Here we will do a different derivation that follows CO.\footnote{In
different books, you'll find slightly different numerical pre-factors
in the definition of the Jeans length and Jeans mass.} Putting in all
the constants, we find that $K$ and $U$ are given by

\begin{eqnarray}
K &=& M{\boltz T\over (\gamma-1)\mu\mH}\,\nonumber\\
U &=& -{3\over 5} {GM^2\over\lambda}\,.
\label{eq:UK}
\end{eqnarray}

Here, $\boltz$ is Boltzmann's constant, $\gamma$ the ratio of specific
heats of the gas (e.g. $\gamma=5/3$ for a mono-atomic gas), and $\mu$
the mean molecular weight ($\mu=1$ for a pure neutral hydrogen gas,
$\mu=0.5$ is the hydrogen is fully ionised). $K$ is the product of the
mass of the cloud, $M$, with the thermal energy per unit mass,
$u=(\boltz T)/((\gamma-1)\mu\mH)$. The potential energy $U\sim
GM^2/\lambda$. The factor in front, $3/5$ is appropriate for a spherical,
homogeneous density perturbation. Now, in virial equilibrium, $2K=|U|$,
So whenever $|U|$ is larger, gravity dominates, and so the critical
length $\lambda_J$ is whenever

\begin{equation}
2K=|U|\,.
\label{eq:ve}
\end{equation}
Now we can use $M=(4\pi/3)\rho\lambda^3$ in
Eqs.(\ref{eq:UK}+\ref{eq:ve}) to obtain the critical Jeans length
$\lambda_J$, and the Jeans mass $M_J$ as

\begin{eqnarray}
\lambda_J &=& \left({5\boltz T\over 2\pi(\gamma-1)\mu\mH\rho
G}\right)^{1/2}\\
M_J &\equiv&{4\pi\over 4}\rho\lambda_J^3=\left({10\boltz T\over
3(\gamma-1)\mu\mH G}\right)^{3/2}\, \left({3\over
4\pi\rho}\right)^{1/2}\,.
\end{eqnarray}

So note that in a homogeneous fluid, {\em the Jeans mass is the mass
contained in a volume with radius the Jeans length. The latter is such
that perturbations larger than the Jeans length will collapse under
their own gravity.}\\

\noindent{\em Fragmentation} A nice application of this concept is to
investigate what happens during collapse of a cloud. Let's say that the
Jeans mass is given by $M_J=A T^{3/2}/\rho^{1/2}$, where I've assembled
all constants into a new constant $A$. Now assume you have a cloud with
mass $M=M_J$ that starts to collapse, and hence $\rho$ increases.  In
general $T$ will rise as well, and so $M_J$ will change. Now suppose
$M_J$ {\em decreases} to $M_J'$. Following our earlier discussion, this
would mean that, if there were smaller perturbations within the cloud
already (substructure), then those with masses larger than $M_J'$ will
start collapsing on their own -- the cloud may fragment. Given our
expression for the Jeans mass, $M_J'< M_J$ will happen whenever
$T^{3/2}$ increases slower than $\rho^{1/2}$. Let's parametrise the
dependence of $T$ on $\rho$ as $T\propto\rho^{\gamma-1}$. Then $M_J'<
M_J$ requires $\gamma<4/3$. For adiabatic, mono-atomic gas,
$\gamma=5/3$ and you don't expect fragmentation. But if radiative
cooling can keep the gas isothermal, $\gamma=1$ and you expect
fragmentation.

So you've learnt how important cooling is for the formation of
stars. After all, the mass of the GMCs in which stars form, if far
bigger than stellar masses, and so we need to understand why stars are
so much smaller than GMCs.


\chapter{Dynamics of galactic disks}
Gravity keeps stars on their orbits around the centre of the MW. Given
this, we should be able to predict the motions you expect. Especially
in the outer parts of the MW disk, observations do not follow this
expectation at all. This is also true in other spiral disks.

So I'll describe the expectations first, then move on to the
observations, which will result in a rather startling conclusion.

\section{Differential rotation}
\subsection{Keplerian rotation}
Suppose you want to describe the orbit of a star in the outer parts of
the MW disk. Since most of the light is then enclosed within the orbit,
you might expect also most of the mass to be interior to the orbit. If
the star is on a circular orbit, than Newton's law tells us that there
is a relation between the circular velocity $V_c$, the radius $R$ of
the orbit, and the mass $M$ of the MW,

\begin{equation}
{V_c^2\over R} = {GM\over R^2}\,.
\end{equation}

Newton's law guarantee us that the force $GM/R^2$ is independent of the
actual density distribution of the MW, as long as it is cylindrically
symmetric.

The run of circular velocity with radius, hence the function $V_c(R)$, is
called the {\em rotation curve} of the MW. And so we expect
$V_c(R)\propto R^{-1/2}$, or the disk is in differential rotation, with
$V_c$ decreasing with increasing $R$. Since the period $P$ of the orbit
$P=2\pi R/V_c$, we expect that $P\propto R^{3/2}$.

\subsection{Oort's constants}
\begin{figure*}[!ht]
\center\resizebox{.9\textwidth}{!}{\includegraphics{oort.ps}}
\caption{(Taken from http://www.csupomona.edu/\~{}jis/1999/kong.pdf) The
observer is moving on a circular orbit with velocity $\Theta_0$ and
radius $R_0$, and observes a star at distance $d$, and galactic
longitude $l$, on circular orbit with velocity $\Theta$ and radius $R$.}
\label{fig:Oort}
\end{figure*}

Suppose you're on a circular orbit with radius $R_0$ around the MW, and
you measure the velocities of stars around you. For simplicity, assume
all stars are also on circular orbits. Stars on orbits with radius
$r<R_0$ will go faster than you do, and so will over take you, and the
opposite for stars with $r>R_0$. Jan Oort computed how the velocities
you measure for these stars depend on their distance, and the direction
you see them.

For the geometry, look at Figure~(\ref{fig:Oort}). To keep the notation
the same, let's call the circular velocity at the position of the
observer $V_{c,0}$, and at the position of the star $V_c$. The radial
and tangential velocity of the star with respect to the observer (at
the position of the Sun) are

\begin{eqnarray}
V_r & = & V_c\cos(\alpha) - V_{c,0}\sin(l)\nonumber\\
V_t & = & V_c\sin(\alpha) - V_{c,0}\cos(l)\,.
\label{eq:oort1}
\end{eqnarray}

Now, in the rectangular triangle on the figure, you can convince
yourself that

\begin{eqnarray}
d+R\sin(\alpha)&=&R_0\cos(l)\nonumber\\
R\cos(\alpha) &=& R_0\sin(l)\nonumber\\
R_0&=& d\cos(\alpha) + R\cos(\beta)\approx d\cos(\alpha) +
R\,{\hbox{\rm when $d\ll R_0$}}\,.
\end{eqnarray}

Combing these equations, we get

\begin{eqnarray}
V_r&=&(\Omega-\Omega_0) R_0\sin(l)\nonumber\\
V_t&=&(\Omega-\Omega_0) R_0\cos(l)-\Omega\, d\,,
\label{eq:oort}
\end{eqnarray}

where $\Omega_0=V_{c,0}/R_0$ is the angular velocity of the observer,
and $\Omega$ the angular velocity of the star. If $|R-R_0|$ is small, we
can obtain a Taylor expansion to find
$\Omega\approx\Omega_0+(d\Omega/dR)(R-R_0)$ and substitute this in the
previous expression. Finally, we can introduce Oort's constants

\begin{eqnarray}
A &\equiv & -{1\over 2} \left[ {dV_c\over dR}|_{R_0} -
{V_{c,0}\over R_0} \right]\approx 14.4\pm 1.2\km\,\s^{-1}\,\kpc^{-1}\nonumber\\
B &\equiv & -{1\over 2} \left[ {dV_c\over dR}|_{R_0} +
{V_{c,0}\over R_0} \right]\approx -12.0\pm
2.8\km\,\s^{-1}\,\kpc^{-1}\,,
\label{eq:oortct}
\end{eqnarray}

We'll discuss in a moment how the measured values are found. With these
definitions we obtain, after some algebra,

\begin{eqnarray}
V_r &=& A\,d\,\sin(2l)\nonumber\\
V_t &=& A\,d\,\cos(2l)+Bd\,.
\end{eqnarray}

Let's see what this means. For a star toward the galactic centre or
anti-centre ($l=0$ and $l=180^o$ respectively), we find the radial
velocity to be zero, and the tangential one to be $(A+B)\,d$. For
$d=1\kpc$, we find $V_t\approx 2.4\kms$.

To measure $A$ and $B$, we need to measure the radial and tangential
velocities of stars of known distance. 

Finally, let's see what we would expect to find for our Keplerian disk,
for which $V_c\equiv V_{c,0} (R_0/R)^{1/2}$, and hence
$dV_c/dR=-(1/2)V_c/R$ At the position of the sun, we know
$V_{c,0}\approx 220\kms\,$ and $R_0=8.5\kpc$ and so

\begin{eqnarray}
A_{\rm Kepler} &=&  {3\over 4} {V_{c,0}\over R_0}= 19.4\kms\,\kpc^{-1}\nonumber\\
B_{\rm Kepler} &=& -{1\over 4} {V_{c,0}\over
R_0}=-6.5\kms\,\kpc^{-1}\,.
\end{eqnarray}

This does not compare well with the measured values in
Eqs.(\ref{eq:oortct}). Assuming we got $V_c/R$ correct, let's compare the
measured and computed values for $dV_c/dR$,

\begin{eqnarray}
{dV_c\over dR}|_{\rm measured}  &=& -(A+B) = -2.4\pm 3.0 \kms\,\kpc^{-1}\nonumber\\
{dV_c\over dR}|_{\rm Keplerian} &=& -{1\over 2}{V_c\over R} = -13.\kms\,\kpc^{-1}\,.
\end{eqnarray}

So what appears to be wrong is that the circular velocity, as computed
for Keplerian fall-off, decreases much more rapidly with increasing
$R$, than the values measured from Oort's constants. We've assumed
$R_0=8.5\kpc$, but for the two values to agree, we would require $R_0$ to
be larger by a factor 6! Clearly, the discrepancy is much larger than
the uncertainties in our values of $V_c$ or $R_0$. We shall see later
that there is other evidence that suggest a much more dramatic failure
of our underlying assumptions.

We've so far assumed stars to be on exactly circular orbits. This is of
course only an approximation, and real stars will have velocities which
differ slightly from $V_c(R)$ and have a {\em peculiar} velocity in
all three Cartesian directions. The reference system that is on a
circular orbit with velocity $V_c(R)$ is called the {\em local
standard of rest} (LSR), and, for example the Sun moves with a velocity
of several 10s of \kms with respect to its LSR. So to determine $A$ and $B$, we
need to measure $V_r$ and $V_t$ as a function of $d$ and $l$ {\em
statistically}.

When Oort did his measurements of the constants $A$ and $B$, he
discovered that some stars had very large deviations from the expected
values of $V_r$ and $V_t$. He called them {\em high velocity stars}. He
also correctly identified their origin: he thought they were associated
with the halo of the MW. Unlike the disk, the halo does not rotate, and
so the high velocities of these stars are due to the rotation velocity
of the Sun.

\subsection{Rotation curves measured from 21-cm emission}
The HI 21-cm emission is able to penetrate virtually the entire galaxy,
and so can be used to measure the rotation curve $V_c(R)$. The biggest
problem with this, is how to determine $d$, the distance to the cloud
that produces the absorption.

A solution is to recognise from Eqs.(\ref{eq:oort1}) that $V_r$ has a
maximum $V_r=V_c-V_{c,0}\sin(l)$ for any $l$ with $|l|<\pi/2$,
which occurs for $\alpha=0$, at which point $R=R_0\sin(l)$ is a
minimum, and $d=R_0\cos(l)$. (This only works when $|l|<\pi/2$,
i.e. toward the centre, since in the outer parts there is no unique
orbit that you can associate with a given velocity.) And so measuring
this maximum velocity as a function of $l$, one can determine
$dV_c/dR=(dV_c/dl)/(dR/dl)$.

Performing this analysis essentially confirmed Oort's measurements,
meaning that $|dV_c/dR|$, the rate of change of the circular velocity
with $R$, is far smaller than you'd expect for a Keplerian rotation
curve. In fact, from measurements of Cepheids with $R>R_0$ it became
apparent that $dV_c/dR\approx 0$, i.e. the rotation curve is nearly
flat, or $V_c(R)\approx {\rm const}$, as opposed to $V_c\propto
(M/R)^{1/2}$ for Keplerian fall-off -- it is as if there is
considerable mass outside the solar circle.

\subsection{Rotation curves and dark matter}
If the rotation curve is flat, it means there must be more mass in the
outskirts of spirals than we see. Let's compute how much more.

Suppose the density $\rho$ of the MW goes like
$\rho=\rho_0(R_0/R)^{\alpha}$. For such a density distribution, the
mass within a given radius $R$, $M(<R)=\int_0^R 4\pi \rho(R)
R^2dR\propto R^{3-\alpha}$. Computing $V_c(R)$, you easily find that a
flat rotation curve requires $\alpha=2$, hence $\rho(R)\propto 1/R^2$
is required for a flat rotation curve.

More recently, rotation curves have been measured using the 21-cm line
in many other spiral galaxies, with always the same result: although
the rotation curve rises from the centre outward, at sufficiently
large radius $R_f$ it invariably becomes flat, even though most of the
light is enclosed within $R_f$. So why don't we see this mass? Two
solutions seem possible. Firstly, for some strange reason, the
mass-to-light ratio of the stars outside $R_f$ increases very much, so
that the little amount of light we see, actually represents much more
mass than we think. Or secondly, there is some type of invisible mass
-- dark matter -- that dominates the mass-density in the outer parts of
spiral galaxies.\footnote{There is a third way out: may be gravity does
not behave as expected on these large scales, or for these small
accelerations. This is not so easy to dismiss as you might think: we
have no measurements on other scales, for example in the solar system,
or in the laboratory, that can test the small acceleration regime that
applies on galactic scales. The theory of {\bf Mo}dified {\bf
N}ewtonian {\bf D}ynamics (MOND) is able to provide very good fits to
measured rotation curves with a small modification of gravity that
cannot be probed in other regimes.} Both seem rather contrived, yet
evidence for dark matter seems to reoccur again and again in other
places as well. 

Already in the 1930s, the Swiss astronomer Fritz Zwicky said that the
galaxies in a cluster of galaxies move too fast for the amount of matter
he could account for in the galaxies themselves. In fact, they moved so
fast that gravity could not contain them to remain inside the
cluster. So Zwicky postulated that there must be some type of invisible
matter inside clusters. We'll discuss evidence for dark matter in
elliptical galaxies, in clusters, and finally in the Universe as a
whole. The case is sufficiently strong that experiments have been
set-up all over the world, to try to detect this missing mass in the
laboratory.

\subsection{The Oort limit}
Jan Oort performed another interesting measurement, namely, count stars
as function of their height above the disk. From this he was able to
estimate how much mass was in the plane of the MW. By counting stars in
the plane of the disk as well, he computed a mass-to-light ratio, which
was also a bit on the high side, suggesting the presence of dark
matter.

Here is a simplified way how to do it. First let me prove that, if all
the mass were in the disk of constant surface density, then the force
on a star out of the disk, is {\em independent} of the distance to the
disk.

As an intermediate step, let's compute the gravitational acceleration
on a star, at distance $d$ from the center of a ring of matter with
radius $\omega$ and thickness $d\omega$, and of surface density
$\Sigma$. From symmetry, the acceleration is toward the middle of the ring,
and it has magnitude

\begin{equation}
da(d,\omega) = -2\pi\,G\,\Sigma {\cos(\alpha)\omega\,d\omega\over
\omega^2+d^2}\,
\end{equation}

where $\alpha$ is the angle at the position of the star, between the
vertical and the ring, so $\tan(\alpha)=\omega/d$. Now the acceleration
due to the whole disk is found by summing over all such rings, hence

\begin{equation}
a(d) =- 2\pi\,G\,\Sigma\int_0^\infty {\cos(\alpha)\omega\,d\omega\over
\omega^2+d^2} = -
2\pi\,G\,\Sigma\,\int_0^{\pi/2}\,\sin(\alpha)d\alpha=-2\pi\,G\,\Sigma\,,
\end{equation}

independent of $d$! So the equation of motion for the star is

\begin{equation}
{d^2z\over dt^2} = -2\pi\,G\,\Sigma\equiv -f\,.
\end{equation}

This is just the same as that for a ball thrown vertically on the
earth's surface, so the solution is $z=z_0+\dot z_0
t-(1/2)f\,t^2$. From this, one can easily obtain the fraction of stars
you expect that have $z$ between $[z_1,z_1+\delta]$ and between
$[z_2,z_2+\delta]$, which depends only on $f$. So, if you measure this from
your data, you can determine $\Sigma$. The value Oort found is called
the \lq Oort limit\rq\, since the MW disk needs to have at least this
surface density.

\chapter{Elliptical galaxies. I }

\subsection{Luminosity profile}
In contrast to spirals galaxies, elliptical have smooth surface
brightness profiles, which are elipsoidal in shape.

The SB profile of some elliptical are shown in
Figure~(\ref{fig:2D}). Note how in the centre the isophotos are well
defined, and are elliptical, in the outer parts they become less smooth
(noisier) because the SB decreases. Also note the presence of
fore-ground MW stars.

\begin{figure*}
\center\resizebox{.9\textwidth}{!}{\includegraphics{ellipticals_2D.ps}}
\caption{(Taken from astro-ph/0206097) Lines of constant surface
brightness (isophotes) in the \KB\ band for elliptical galaxies. In
succesive isophotes, the surface brigthness increases by 0.25
magnitudes.}
\label{fig:2D}
\end{figure*}

A plot of the average surface brightness as function of radius for
these same ellipticals is shown in Figure~(\ref{fig:1D}). The drawn
line which fits the data well, in an $r^{1/4}$ fit, Eq.~(\ref{eq:r4}).
Often, these profiles are plotted as SB vs $r^{1/4}$ in which case the
profile is a straigth line, as in Figure~(\ref{fig:1D}). For most
ellipticals, this profile shape fits their SB-distribution well,
although some people have suggested to use a more general form,
$I(r)\propto \exp[-(r/r_e)^{1/n}]$, where $n=4$ corresponds to the de
Vaucouleurs $r^{1/4}$ fit.

\begin{figure*}
\center\resizebox{.8\textwidth}{!}{\includegraphics{ellipticals_1D.ps}}
\vskip8truecm
\caption{(Taken from astro-ph/0206097) Surface brightness as function
of distance to the centre. The drawn line is an $r^{1/4}$ fit.}
\label{fig:1D}
\end{figure*}

\begin{figure*}
\center\resizebox{.8\textwidth}{!}{\includegraphics{3379.ps}}
\caption{$r^{1/4}$ profile for galaxy NGC 3379, taken from de
Vaucoleurs \& Capaccioli, ApJS 40, 1979.}
\label{fig:3379}
\end{figure*}

\begin{figure*}
\center\resizebox{.8\textwidth}{!}{\includegraphics{1399.ps}}
\caption{SB profile for NGC~1399 from Schombert, ApJS, 1989. The outer
parts of the profile do not follow the $r^{1/4}$ fit (bottom panel),
but are nearly a straight line in a SB-$\log(r)$ (i.e. a $\log-\log$)
plot (top panel), meaning the profile is close to a power-law.}
\label{fig:1399}
\end{figure*}

\begin{figure*}
\center\resizebox{.8\textwidth}{!}{\includegraphics{1399_GCS.ps}}
\caption{Image of NGC~1399, with an $r^{1/4}$ profile subtracted, from
Bridges et al., AJ 101, 469 (1991). The center of NGC~1399 is at the
left, the object toward the top right is another galaxy in the galaxy
cluster. The other extended objects in the image are als other
galaxies. Clearly seen are hundreds of high SB unresolved objects,
which are GCs in the halo of NGC~1399.}
\label{fig:1399GCs}
\end{figure*}

Some of the giant elliptical galaxies that are invariably found in the
centres of big clusters of galaxies have an extended halo that is not
fit well by the de Vaucouleurs profile, see for example the profile of
NGC~1399, which is the cD galaxy in the nearby Fornax cluster of
galaxies (Fig.~(\ref{fig:1399}). This very extended halo of stars is
truly enormous: it can be traced to around a \Mpc\,, prompting the
suggestion that may be we should associate such a halo with the cluster
of galaxies, rather than with an individual galaxy. This suggestion has
some support from the observations of some clusters where we observe a
faint halo, but {\em without} a galaxy in the centre!

Es also have many globular clusters. Figure~(\ref{fig:1399GCs}) is an
image of NGC~1399, where an $r^{1/4}$ fit to the SB-profile of the
agalxy has been subtracted. Clearly visible are 100s of high SB
objects, indistinguishable on this plate from foreground stars, which
are in fact GCs in the halo of the galaxy. The GC luminosity function
(i.e., the number of clusters as function of their magnitude) has been
used as a standard candle: by measuring the luminosty function of GCs
of a galaxy, and comparing it to the luminosity function of a galaxy
with known distance, one can obtain a distance estimate.

Why do ellipticals have an $r^{1/4}$ SB-profile? As I explained
earlier, there seems to be a strong connection between the density of
galaxies (in the sense of how many galaxies you find per \Mpc$^{3}$ in
a given region, say) and their types: high density regions invariably
contain a much larger fraction of Es than low density regions, where Ss
dominate. So, may be interactions between galaxies have something to do
with it?

Recent numerical simulations of collisions between two spiral galaxies
do produce objects with SB-profiles not too different from $r^{1/4}$
appropriate for an E. And although this is a good hint that we're on
the right track, in fact, also in the simulations we don't really know
{\em why} this is. In a famous paper, Donald Lynden-Bell suggested that
during a galaxy merger, the gravitational potential $\Phi$ which the
stars feel, changes very rapidly and this would produce a
characteristic desity profile (rather, a characteristic {\em
distribution} function, the density of stars in six-dimensional
position and velocity space). He termed this process violent
relaxation. And so the suggestion is that, as (predominantly spiral)
galaxies fall into galaxy clusters, they will violenetly interact with
other galaxies, which transforms them into Es.  And although this is
certainly an important piece of the puzzle, it can't be the whole
story, because the stellar populations of Es and Ss differ.

\section{Stellar populations and ISM}
\noindent {\em Stars} in Es tend to be older and more metal rich than those in Ss, and
as a consequence Es are redder than Ss. You can easily understand why
this is so. If the atmosphere of a star contains more metals, then the
dust and the gas will scatter the blue light more, and this will redden
the star. That's one. Secondly, if you have young stars in a stellar
population, then the more massive stars may still be
alive\footnote{Recall that massive stars have short lifetimes.}, and
these are hot and hence blue. So when a {\rm given} stellar population
gets older, it will become redder as well. And in an E galaxy, both
effects occur.

The fact that a stellar population can be bluer because it is young, or
because it is metal poor, is called the {\em age-metallicity degeneracy}:
just observing the colour of a population, you cannot tell a young but
metal rich population apart from an old but metal poor one. To make
sure, you have to have independent contraints on $Z$, for example from
stellar spectra, or, the astronomer's favourite, from HII regions.

\noindent{\em Dust and gas} Although I said that a characteristic
difference between Es and Ss is that Es don't contain dust or gas, and
have no star formation, this is of course not complete true, they
usually have some, but much less than Ss. Many Es contain strong dust
lanes, the Sombrero galaxy is a beautiful example(Fig.~\ref{fig:sombrero}).  Often, there is
almost no connection between the orientation of the dust lane, and any
other of the galaxy's parameters. It is thought that such dust lanes
are evidence that the E recently swallowed a small galaxy. In fact, one
sometimes observes very faint rings of stars around an E, and also this
could be the result of \lq galactic cannibalism\rq{}. A good example is
shown in Figure~(\ref{fig:shellgalaxy}). 

\begin{figure*}
\center\resizebox{.8\textwidth}{!}{\includegraphics{sombrero.ps}}
\caption{Image of the \lq Sombrero\rq\ galaxy, with its striking dust lane.}
\label{fig:sombrero}
\end{figure*}

\begin{figure*}
\center\resizebox{.8\textwidth}{!}{\includegraphics{shellgalaxy.ps}}
\caption{Image of NGC~3923 taken by David Malin. Several faint shells
of stars appear in the outer parts of the galaxy.  (see
http://www.ast.cam.ac.uk/AAO/images/general/ngc3923.html)}
\label{fig:shellgalaxy}
\end{figure*}


\section{X-rays}
\subsection{Thermal bremsstrahlung}
How are the X-rays produced in this hot gas? Since the gas temperature
is very high, most elements are higly ionised, and so the gas is really
a hot plasma. The negativel charged electrons in this plasma experience
a force when they pass close to an ion, and an accelerating charge
emits electro-magnetic dipole radiation. This is the radiation that we
observe. Since the electron looses energy in the encounter, it will
slow down. Hence the name thermal bremsstrahlung (German for \lq
braking radiation\rq). And as the electron slow down, the plasma also
looses energy, and so this hot gas cools.\footnote{Also the ions
collide of course, but since they are much more massive than the
elctron, their acceleration is far smaller, and hence their radiation
is negligible.} The electron is not bound to any particular ion
throughout the interaction. This type of radiation is therefore also
called free-free radiation.

Let's try to obtain an expression for the radiated power. The aim of
this derivation is to show that the power emitted is $\propto
n_e\,n_i\,T^{-1/2}$, where $n_e$ and $n_i$ are the elctron and ion
densities, and $T$ is the temperature. You could have guessed the first
bit, since we're talking about an encounter between these two
particles. So the exercise is how to get the $T^{-1}\propto 1/v$
dependence.

If the deflection angle is small, we can simple neglect it, and say
that the electron moves at constant speed $v$ along a straight line
with impact parameter $b$. The acceleration is then

\begin{equation}
\bfa(t) = {Ze^2\over m_e d^3}\bfd = {Ze^2\over
m_e(b^2+v^2t^2)}{\bfd/d}\,,
\end{equation}

where $Z$ is the charge of the ion in units of the electron charge $e$,
and $\bfd$ is the distance between electron and ion. $t=0$ is the time
of closest approach, and so the encounter lasts from $t\rightarrow
-\infty$ to $t\rightarrow +\infty$.

The dipole electric field at distance $r$ from the electron is

\begin{equation}
|{\bf E}(t)| = {e\,a\over c^2\,r}\sin(\theta)={Z\,e^3\sin(\theta)\over
 m_e\,c^2\,r\,(b^2+v^2t^2)}\,.
\end{equation}

This is the electric field as function of time $t$, but we want it as
a function of frequence $\nu$. The trick to obtain $E(\nu)$ is by taking
Fourier transforming $E(t)$,

\begin{eqnarray}
E(\nu)&=&\int_{-\infty}^{\infty} E(t)\exp(2\pi\,i\nu\,t)
dt={Ze^3\sin(\theta)\over m_ec^2\,r}\int_{-\infty}^{\infty}
{\exp(2\pi\,i\nu\,t)\over b^2+v^2t^2} dt \nonumber\\
&=& {Ze^3\sin(\theta)\over m_ec^2\,r}{1\over b\,v}
\int_{-\infty}^{\infty} {\exp(2\pi\,ix(\nu\,b/v))\over (1+x^2)}\,dx\nonumber\\
&=& {Ze^3\sin(\theta)\over m_ec^3b^2v^2}\,{\pi\over bv}\,\exp(-2\pi\nu\,b/v)\,.
\end{eqnarray}

Along the way, I've introduced the variable $x=vt/b$. The integral
$\int\exp(2\pi x\alpha)/(1+x^2)\,dx=$ some function of $\alpha$ is a
little tricky to perform, and I don't expect you to be able to do
it. What is important is to note that this change of variables gives
rise to a factor $1/bv$ .

Now we're done, because it means that the power radiated by this single
electron per unit frequency is

\begin{equation}
{dW\over d\nu}\propto \int E^2(\nu)\,d\Omega\propto {1\over
(b\,v)^2}\exp(-4\pi\nu\,b/v)\,.
\end{equation}

So the spectrum is independent of $\nu$ for low frequencies $\nu\ll
v/b$, and cuts off exponentially for large $\nu$. Now the number of
encounters with impact parameter between $b$ and $b+db$, in time $dt$
is $n_e 2\pi b\,db vdt$, and so this gives the power as

\begin{equation}
{dW\over d\nu\,dt\,dV}\propto n_e\,n_i\,v^{-1}\,\exp(-4\pi\nu\,b/v)\propto
n_e\,n_i\,T^{-1/2}\,\exp(-4\pi\nu\,b/v)\,.
\end{equation}

So the origin of the $1/v$ is that the energy radiated per encounter
$\propto 1/v^2$ whereas as the rate of encounters is $\propto v$.  

To obtain the final result, we still need to integrate over the impact
parameter $b$, and average over all impact velocities $v$ (for example
by assuming a Maxwell-Boltzmann distribution for $v$). The result is
that the total power emitted is
\begin{equation}
{\hbox{\rm power }} \propto n_e\,n_i\,T^{1/2}\,.
\end{equation}

Notice that the total power is much more dependent on the density of
the gas ($\propto $ density {\em squared}), than on its temperature
($\propto T^{1/2}$)

\subsection{Spectrum}
On top of the thermal bremsstrahlung emission spectrum, observed X-ray
spectra also exhibit spectral lines. Given the high temperature of the
gas, it won't come as a surprise that we observe lines of highly
ionised species -- Fe-lines are especially important in the wave-bands
observed by modern X-ray telescopes. And just as we did for HII
regions, we can try to constrain the physical conditions in the hot
gas, by modelling the relative strenght of different transitions. This
is how one determines the temperature $T$.

\subsection{Hydrostatic equilibrium}
An important application of X-ray observations is to probe the
gravitational potential of the galaxy. Indeed, if the system is in
equilibrium, then the gravitational pull will be balanced by the
gradient of the pressure in the hot gas,

\begin{equation}
\nabla\Phi=-\nabla p/\rho\,.
\label{eq:HE}
\end{equation}

(This should look familiar to what you saw about hydrostatic
equilibrium in stars!) By modeling the spectrum as function of
position, we can determine $T$ and $\rho$ seperately, and hence the rhs
of Eq.(\ref{eq:HE}). From that, one can reconstruct the gravitational
potential, and hence infer the mass distriution.

\subsection{Dark matter}
The mass distribution obtained from X-ray measurements can be compared
to the {\em light} distribution in Es, and so one can estimate the
variation of the mass-to-light ratio, $M/L$, through the galaxy. This,
together with the measurements of the orbits of stars, to be discussed
next, gives us strong evidence that also Es are dark-matter dominated,
at least in their outer parts.


\chapter{Elliptical galaxies. II}
This section is a bit harder, with more mathematics than I would like
to use. I {\bf do not} expect you to be able to reproduce these
derivations. A little note on notation: if in one of the following
equation an index occurs twice in a given term, like the index $i$ in
$x_i\,y_i$, then you have to sum over $i$. So $x_i\,y_i\equiv
\sum_i\,x_iy_i=\bfx\cdot\bfy$ is a short hand for the scalar product of
$\bfx$ and $\bfy$. 

We've seen that the stars in spiral disks are all on nearly circular
orbits. Although some of the smaller Es sometimes rotate as well, the
angular momentum of stars is not sufficient to balance gravity. So we
need to look for another mechanism to prevent collapse. Basically, it's
the {\em velocity dispersion} of the stars that balances gravity, much
like it is a pressure gradient in the gas that balances gravity in a
star. For this reason, Es (but also spiral bulges, and globular
clusters for example) are called {\em hot stellar systems}.

The Jean's equations describe the relation between the stellar velocity
dispersion, $\sigma_{ij}^2=\langle (v_i-\langle v_i\rangle)(v_j-\langle
v_j\rangle)\rangle$, and the gravitational potential of the system, and
are the stellar analogue of the equation of hydrostatic equilibrium
$\nabla\Phi=-\nabla p/\rho$ in a star. They do not describe the
properties of the orbits of a single star, but rather assume one can
use averaged properties. To investigate whether this makes sense, I
first need to introduce the concept of relaxation time.

\section{Relaxation in stellar systems}
As a star moves through a stellar system, it will feel the
gravitational force due to all other stars. Is the motion of this star
mainly determined by the average gravitational force of all other stars
combined, or is it mostly sensitive to the force of a few of the stars
nearest to it? 

To see that this would make a huge difference, consider the case where
a single star (a test particle) moves through a smooth density
distribution with spherically symmetric density distribution $\rho(r)$,
where $r=|\bfr|$. As you know, in such a potential both the angular
momentum ${\bf L}$ and energy $E$ of the star will be conserved along
its orbit. Now suppose such a density distribution to be represented by
a finite number of stars. As our test star moves along its orbit, it
will be deflected by encounters with these other stars. In such a
situation, neither ${\bf L}$ nor $E$ will be conserved, and so the
properties of its orbit will change in time. Clearly, the longer the
star orbits in this density distribution, the more $E$ will differ from
its initial value. {\em The relaxation time $T_E$ is a measure of how
long the star remembers its initial energy}. We will see that the more
stars are there in the galaxy, the longer is the relaxation time. You
might have expected this, since for an infinite number of stars, the
density is smooth, ${\bf L}$ and $E$ are constant, and hence
$T_r\rightarrow \infty$.

To estimate $T_E$, we will compute the change in energy $\Delta E$ of a
star due to a single encounter with another star, as function of the
impact parameter $b$ and impact velocity $v$ (the velocity at infinity,
before the encounter started). You'll notice that this derivation is
very similar to what we did to obtain the expression for thermal
bremsstrahlung. This derivation follows BT (p.~188)

Let's computed the change in velocity $\delta\bfv$ of our star. To make
things simple, we'll approximate the orbit as a straight line,
traversed with constant velocity $v$. Now the force perpendicular to
the orbit is

\begin{equation}
\bfF_\perp = {Gm^2\over b^2+(vt)^2}\,\cos(\theta)\approx {Gm^2\over
b^2}\,\left[1+({vt\over b})^2\right]^{-3/2}\,,
\end{equation}

where I've assumed equal mass stars. The approximation
$\cos(\theta)=b/r\approx b/vt$ is valid for $b$ small. Now Newton's law
tells us that

\be
m{\bf {\dot v}}_\perp = \bfF_\perp\,.
\ee

Combining these two, we find the net change $|\delta\bfv_\perp|$ over a
full encounter:

\be
|\delta\bfv_\perp| \approx {Gm^2\over bv} \int_{-\infty}^{\infty}
(1+s^2)^{-3/2}\,ds\,={2Gm\over b\,v}\,.
\label{eq:dv}
\ee

Thus the change is roughly equal to the force at closest approach,
$Gm/b^2$, times the duration of this force, $\Delta t=b/v$.

Suppose our test star moves with velocity $v$ through a stellar system
with $N$ stars, and radius $R$. When it traverses the system once, the
number $\delta n$ of encounters with other stars, with impact parameter
between $b$ and $b+db$ is

\be
\delta n = {N\over \pi R^2}\, 2\pi b\,db = {2N\over R^2}\, b\,db\,.
\ee

Because of symmetry, the mean change $\delta\bfv_\perp$ is of course
zero. So to characterise the effect of encounters, let's compute

\be
\delta v^2_\perp = \left({2G\,m\over bv}\right)^2\, \delta n\,.
\label{eq:rel1}
\ee

To find the net change, we need to integrate over the impact parameter
$b$. A recurring problem in calculations of this kind, is that the
integral $\int_0^\infty db/b$ diverges, both for small and for large
impact parameters. Now for the largest $b$ we can simply use $R$, the
radius of the system. For the smallest $b$, we can take $b_{\rm
min}=Gm/v^2$, for which the {\em change} in velocity is comparable to the
{\em initial} velocity (cfr.Eq.~\ref{eq:dv}). Recall that our
derivation assumed small changes anyway.

With this caveat, we can integrate Eq.~(\ref{eq:rel1}) over $b$ to find
the net change in $v_\perp^2$ when the star moves through the system
once, as

\be
\Delta v_\perp^2 \equiv \int_{b_{\rm min}}^R \delta v^2_\perp \approx
8N\left({Gm\over Rv}\right)^2 \ln\Lambda\,,
\label{eq:rel2}
\ee

where

\be
\ln\Lambda \equiv \ln\left({R\over b_{\rm min}}\right)\,
\label{eq:relax3}
\ee

is the (logarithm of the) ratio of the largest over the smallest impact
parameter. Now if the system is in virial equilibrium, then the kinetic
energy $K=(1/2)Nmv^2$ of its stars, should be of order its potential
energy, $U=GM^2/R$, or neglecting factors of order unity, and using
$M=Nm$

\be
v^2\approx {GNm\over R}\,.
\label{eq:rel4}
\ee

This is the typical velocity of a star. Substituting this in
Eq.~\ref{eq:relax3}, using $b_{\rm min}\approx Gm/v^2$ gives

\be
\ln\Lambda\approx \ln N\,.
\ee

Finally eliminating $R$ between Eqs.~(\ref{eq:rel2}+\ref{eq:rel4})
gives

\be
{\Delta v_\perp^2\over v^2} = {8\ln(N)\over N}\,.
\ee

If the star makes many crossing of the galaxy, $v_\perp^2$ will change
by of order $\Delta v_\perp^2$ at each crossing, so that the number of
crossings $n_{\rm relax}$ that are required for its velocity to change
by of order itself is given by

\be
n_{\rm relax} = {N\over 8\ln(N)}\,.
\ee

Thus, the {\bf relaxation time} may be defined as $T_{\rm relax}=n_{\rm
relax}\times T_{\rm cross}$, where $T_{\rm cross}\equiv R/v$ is the {\bf
crossing time}, i.e. the time it takes the average star to cross the
system. So we find

\be
T_{\rm relax} = {N\over 8\ln(N)}\,T_{\rm cross}\,.
\ee

For systems of the same mass and size, and hence a {\em given}
crossing time, we find that the relxation time increases wth $N$ as
$T_{\rm relax}\propto N/\ln(N)$. Put differently, the effect of
encounters decreases with increasing particle number as $\propto N/\ln(N)$.

Take a star in a big E with $N=10^{11}$, say. The crossing time is of
order $20\kpc/200\kms\approx 10^2 M\yr$, and $T_{\rm relax}\sim
10^{13}M\yr$ which is {\em much} smaller than the age of the Universe
($\sim 10^{10}\yr$). Therefore we can completely neglect encounters
between stars, when describing the equation of stellar dynamics in a
galaxy. This is what we'll do next.

\section{Jeans equations}
\subsection{A continuity equation}
A continuity equation expresses conservations of a quantity. This is a
general concept, let's do a simple example first: conservation of
number of cars on a motor way.

Suppose you're standing on a hill above a tunnel on a motor way, and
you're counting the number of cars entering the tunnel on one side,
and at the same time, the number leaving the tunnel on the other
side.

If $n$ is the number density of cars in the tunnel (in cars per km
say), and $\Delta l$ is the lenght of the tunnel, then the number of
cars in the tunnel is obviously $N=n\,\Delta l$. The change in $N$
between two times is

\be
\Delta N=[n(l,t+\Delta t)-n(l,t)]\,\Delta l\approx {\partial
n(l,t)\over\partial t}\,\Delta l\,\Delta t\,.
\ee

Assuming there is no other exit, this is also the change between the
number of cars entering and leaving the tunnel,

\be
\Delta N=[n(l,t)\dot l(l,t)-n(l+\Delta l,t)\dot l(l+\Delta l,t)]\,\Delta
t\approx -{\partial n(l,t)\,\dot l(l,t)\over \partial l}\,\Delta l\,\Delta
t\,.
\ee

The first term $n(l,t)\dot l(l,t)\,\Delta t$ is the number of cars entering
the tunnel in time $\Delta t$, the second term is how many are leaving
in that time. $\dot l(l,t)$ obviously denotes the speed of the cars.

Combining these two equation, we get the following continuity equation:

\be
{\partial n(l,t)\over \partial t} + {\partial n(l,t)\,\dot l(l,t)\over \partial l}
= 0\,.
\label{eq:ce}
\ee

A little bit of thought shows that, in a N-dimensional case (spaceships
through a worm-hole?), the more general equation would be

\be
{\partial n(\bfl,t)\over \partial t} + {\partial n(\bfl,t)\,\dot\bfl(\bfl,t)\over \partial \bfl}
= 0\,,
\label{eq:ce}
\ee

so there is a partial derivative $\partial/\partial l_i$ for every
Cartesian coordinate $l_i$, and $n$ is multiplied by $\dot l_i$ in the
second term, which is the velocity of the spaceships in the
$i$-direction.

\subsection{Boltzmann's equation}
We're interested in finding an equation that describes the distribution
of stars in a galaxy. So we want to know is how many stars there are in
a small volume $d\bfx$ around a position $\bfx$, that have velocity in
a small interval $d\dot\bfx$ around a given velocity $\dot\bfx$. This density
of stars, $f(\bfr,\dot\bfx,t)$, is called the {\em distribution
function}. If we don't care about the velocity of stars, then we can
integrate $f$ over all velocities, to obtain the density of stars,

\be
n(\bfx,t)=\int f(\bfx,\dot\bfx,t)\,d\dot\bfx\,.
\ee

Now we only have to realise that stars acts as our cars on the motor
way, and hence $f$ should satisfy the continuity equation
(\ref{eq:ce}). The way to do it is to substitute for the $l$
coordinates in that equation the positions and velocities of the stars,

\begin{eqnarray}
\bfl     &\equiv & (\bfx,\bfv)\nonumber\\
\dot\bfl &\equiv & (\bfv,\dot\bfv) = (\bfv,-\nabla\Phi)\,,
\end{eqnarray}

where I've written $-\nabla\Phi$ for the gravitational acceleration
$\dot\bfv$ of the star.

So this $\bfl$ vector has 6 coordinates, $l_\alpha=\bfx$ for
$\alpha=1\rightarrow 3$, and $l_\alpha=\bfv$ for $\alpha=4\rightarrow
6$. Now these coordinates are very special, since

\begin{eqnarray}
\sum_{\alpha=1}^6 {\partial\dot l_\alpha\over \partial l_\alpha} 
&=&\sum_{i=1}^3 \left({\partial v_i\over\partial x_i}+{\partial
\dot v_i\over\partial v_i}\right)\nonumber\\
&=&\sum_{i=1}^3 -{\partial\over\partial v_i}\left({\partial\Phi\over
\partial x_i}\right)\nonumber\\
&=&=0\,.
\end{eqnarray}

Here, $\partial v_i/\partial x_i=0$ since the positions and velocities
of the stars are independent variables, and the second term is zero
because the gravitational acceleration $\partial\Phi/\partial x_i$ does
not depend on velocity.

We can use this equation to simplify the continuity equation, and
to obtain

\be
{\partial f\over\partial t}+\sum_{\alpha=1}^6\dot l_\alpha{\partial
f\over\partial l_\alpha}=0\,.
\ee

Rewriting in terms of positions and velocities, we obtain the {\bf
collisionless Boltzmann equation},

\be
{\partial f\over\partial t} + \sum_{i=1}^3\left(v_i{\partial
f\over\partial x_i}-{\partial\Phi\over\partial x_i}{\partial
f\over\partial v_i}\right)=0\,.
\label{eq:ce1}
\ee

or in vector notation
\be
{\partial f\over\partial t}+\bfv\cdot{\partial f\over\partial \bfr}-\nabla\Phi\cdot {\partial
f\over \partial\bfv}=0\,.
\label{eq:ce2}
\ee

\subsection{Moments of the Boltzmann equation}
Although the collisionless Boltzmann equation (CBE for short) provides
us with a complete description of the stellar dynamics problem, it is a
but unwieldy for practical applications. The Jeans equations, which
give a clearer physical picture of what is going on, are {\em moments}
of the Boltzmann equation. Here is how to obtain them.

Suppose we integrate Eq.~(\ref{eq:ce1}) over all velocities. The first
term will become

\be
\int {\partial f(\bfr,\bfv,t)\over t}\,d\bfv = {\partial\over\partial t}\int
f(\bfr,\bfv,t)\,d\bfv\equiv {\partial\nu(\bfr,t)\over\partial t}\,.
\ee

The integral of the distribution function over velocities, $\nu\equiv
f\,d\bfv$ is just the density of stars. And we're allowed to
interchange the integral over velocities, with the derivative wrt time,
since the range of velocities over which we integrate does not depend
on $t$.

Integrating the second term, gives

\be
\int v_i{\partial f\over\partial x_i}\,d\bfv = {\partial\over\partial
x_i}\,
\int v_i\,f\,d\bfv\equiv {\partial(\nu\bar v_i)\over\partial x_i}\,,
\ee

where I've introduced the mean velocity $\bar v_i$ as

\be
\nu\,\bar v_i=\int v_i\,f\,d\bfv\,.
\ee

Now the last term is zero, since

\be
\int_{-\infty}^\infty {\partial f\over\partial v_i}
dv_i=f|_{v_i=-\infty}^{v_i=+\infty}\,,
\ee

since there are no stars with infinite velocity.

And so the first moment of the CBE is a continuity equation for the
density,

\be
{\partial\nu\over t}+{\partial (\nu\bar v_i)\over\partial x_i} = 0\,.
\ee

Now the second moment is what we're after. The trick is to {\em first} multiply the
CBE with $v_j$, and {\em then} integrate over all velocities. The
calculation procedes exactly as for the density, and the result is
after a bit of algebra:

\be
\nu\,{\partial\bar v_j\over\partial t} + \nu\,\bar v_i\,{\partial \bar
v_j\over\partial x_i} = -\nu\,{\partial\Phi\over\partial x_j} - {\partial
\sigma^2_{ij}\over \partial x_i}\,.
\ee

where

\be
\sigma^2_{ij}\equiv  \overline{v_i v_j}-\bar v_i\bar v_j\,.
\ee

This was a lot of mathematics to come to a simple equation: suppose the
system is in a steady state, $\partial/\partial t$=0, and there are no
streaming motions, so that $\bar v_i=0$. Then the previous equation
simplifies to

\be
{\partial\Phi\over\partial x_j} = -{1\over\nu}{\partial
\nu\sigma^2_{ij}\over \partial x_i}\,.
\ee


This is a more general version for the equation of hydrostatic
equilibrium. Indeed, suppose the system is isotropic in its velocities,
so that $\sigma^2_{ij}=0$ for $i\ne j$, and equals $\sigma^2$ for
$i=j$. Then we find that

\be
{\partial\Phi\over\partial x_j} = -{1\over\nu}{\partial\nu\sigma^2\over \partial x_j}\,.
\ee

which is {\em exactly} the same as the equation for hydrostatic
equilibrium, if $\nu\sigma^2$ is identified with the pressure $p$. 

\end{document}