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\title{\Huge{Stars and {\bf Galaxies}}}
\author{Tom Theuns}
\date{February 12, 2003}
\maketitle
% title page
\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, 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
galactic physics is often much more murky and complex -- a bit like
weather predictions: 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
%\htmladdnormallink{http://www.aao.gov.au/images.html/}{http://www.aao.gov.au/images.html/}
\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
\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.

A truly excellent web site where much of what I'll discuss is explained
really well with many pictures, and by someone who knows what he's
talking about, is by {\bf Chris Mihos}, who is at the department of
astronomy, Case Western University. You'll find it at:
\hbox{http://burro.astr.cwru.edu/Academics/Astr222/index.html}, and
you'll recognise some of the picture in my course as originating from
here.

\chapter{Introduction}
\section{Historical perspective}
Galaxies were discovered in the 20th century. This is a bit surprising,
since 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! The
reason that they don't stand-out very clearly is not because they are
{\em faint}, it is because they are so {\em large}. Since they are
large, their light is distributed over a wide area, and our eyes do not
pick-out the increase in surface brightness over the night 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 (SN for
short, and SNe for the plural) 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, 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 arcsec$^2$, 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 $l=L/4\pi r^2$
(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
$dN(r,d\Omega)=nr^2drd\Omega$. Combining\footnote{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.}  these two, the luminosity of all stars within $r$ and
$r+dr$, and within $d\Omega$, is
\begin{equation}
dl = l\,dN(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\footnote{On cosmological scales this argument does not
hold, and the SB of very distant galaxies decreases with distance, an
effect called surface brightness dimming.} of $r$. (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.\footnote{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\,d\Omega\,.
\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 are quite
different for these two types.

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.

\begin{figure*}
\center\resizebox{.9\textwidth}{!}{\includegraphics{TuningFork.pdf}}
\caption{The Hubble sequence.}
\label{fig:Hubble}
\end{figure*}

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}
(Fig.~\ref{fig:Hubble}) 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 as much as $10^{13}\msun$, 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 now that this is not the way galaxies
evolve, yet these 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?

\pagebreak
\section{Summary}
After having studied this lecture, you should be able to 

\begin{itemize}
\item Describe the main galaxy types, Es and Ss, and list five
characteristics of each
\item Describe the main properties of Globular Clusters.
% and of Open Clusters.
\item Define surface brightness, and show it to be independent of
distance.
\item Derive the relation between the surface density of stars in a
galaxy, and its surface brightness.
\item Derive the relation between the stellar density and star counts
\item Know where to find web-pages with images of galaxies.
\end{itemize}


\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, and the size of the MW. There were 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 this debate, hoping we can learn from the mistakes. Before
we do this, let's describe the main observables, and observing
techniques.

\section{The main observables}
\noindent{\bf Star clusters} Two types of star cluster were (and are)
known. We already described {\bf Globular Clusters} (GCs), spherical,
bound systems of $10^5--10^6$ stars. The MW has about 150 GCs, 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 {\bf Open Cluster}. These typically contain
of order $10^4$ stars. They are of great importance for studies of
stellar evolution, since we know all the stars in such a cluster have
(more or less) the same age, distance and initial composition, and
hence they provide us with a sample of stars in which the only (or at
least the main) difference is the stellar mass.\\

\noindent{\bf Star counts} Previously we described how star counts can
be used to probe 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, with axes ratio about 5:1.\\

\noindent{\bf 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 $\Delta m$ and the magnitude $m$ of these Cepheids. Since
all these stars are at (nearly) the same distance, this must mean it is
actually a relation between the {\em absolute} magnitude $M$ and
$P$. There are many types of variable stars, but Cepheids have many
advantages (a) the shape of their light curve (i.e. luminosity as
function of time) is a very characteristic sawtooth pattern, (b) they
are luminous - and so can be seen out to large distances and (c) the
$P(M)$ 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{\bf 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 have moved with
respect to the background. You've done (your first?) parallax
measurement. If the wall is sufficiently distant then there is a
relation between how much your finger appeared to have moved (in
degrees, say), the length of your arm, and the distance between your
eyes.

For astronomical measurements, 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\ \footnote{An \arcsec\ is 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 1\pc/1\AU=1\arcsec in radians\footnote{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. So it appears as if the Sun is near
the centre of the MW. Since we've made this type of error once before
(assuming the earth was at the centre of the Universe), this made
several people slightly uncomfortable. And as we'll discuss later, it
turned out to be 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? There was no clear way how to
reconcile these different data, and so the issue remained unresolved.

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. van
Maanen himself, but also {\em Edwin Hubble} later redid the proper
motion measurements with better data (using a better telescope), and
both 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.

The 1920s were of course a period of great economic depression. But
from the scientific point of view, they must have been tremendously
exciting. In only a couple of years, the general theory of relativity
was formulated, quantum mechanics revolutionised physics, and the
Universe was found to be far bigger than realised before, and expanding
as well.

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 suggests that some of the other 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 with axes ratio 5:5:1
\item[1900-1920] Kapteyn counts stars, decides wrongly that extinction
is unimportant, and deduces the MW to be $5\kpc\times5\kpc\times1\kpc$
big, with the Sun at 650\pc\ 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. The spectra he takes suggests
presence of stars, not of gas. A clear indication this are not
proto-planetary structures in the MW, but other galaxies.
\item[1915] Shapley finds the centre of the MW's globular cluster
system to be 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
commissioned 100-inch telescope. Given the large inferred distance
means that M31 must be outside the MW. He also discovers Cepheids, and
the distance to M31 is estimated at 300\kpc. So Andromeda us indeed
another galaxy.
\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 with detect 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. Consider a ray of light with intensity $I$. The amount of light
of this ray absorbed per unit distance, 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 to estimate absorption. The amount of
absorption (the value of $A$) will then depend on $\lambda$, and so
$A_\V\ne A_\B$, say: 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\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.}  $A_\V\approx
1{\rm mag}\,\kpc^{-1}$. If we now measure $E_{\B-\V}$ for another star
of {\em known} colour (from stellar evolutionary models say) we can
estimate $r$.

\pagebreak
\section{Summary}
After having studied this lecture, you should be able to

\begin{itemize}
\item Describe the Great Debate, give the points in favour of each
camp, and discuss the problems with their arguments.
\item Explain what a standard candle is.
\item Explain why Cepheids and RR Lyrae variables are good standard
candles.
\item Explain what parallax is, and know how the distance to nearby
stars can be measured using their parallax.
\item Explain how parallax and Cepheids can be used to walk-up the
distance ladder.
\item Explain how Hubble's observations revolutionised our view of the
MW and the realm of the Nebulae.
\item Derive the effect of scattering by dust on the apparent magnitude
and colour of distant stars. Know the approximate value of the
interstellar absorption in the MW.
\end{itemize}

\chapter{The modern view of the Milky Way}
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. And 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).


\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:MW}
\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.

\pagebreak
\section{Summary}
After having studied this lecture, you should be able to

\begin{itemize}
\item Describe how radio and infra-red observations are being used to
clarify our picture of the MW.
\item Explain how the Hipparcos satellite was a major step in setting
the scale of the MW, by performing accurate parallax and proper motions
measurements, and fixing a reference frame with respect to distant
objects.
\item Describe the main components of the MW, thin and thick disk,
bulge and bar, stellar halo, and their typical properties such as their
density and luminosity profiles.
\item Explain what is meant by the metallicity of a star, and known
what is meant by the notation [Fe/H]=-1.
\item Know how the Galactic Coordinate system $(l,b)$ is defined, know
the direction to the galactic centre $l=b=0$, and the galactic plane
$|b|<5^o$.
\end{itemize}


\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. (Wouldn't 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, is far
bigger than stellar masses, and so we need to understand why stars are
so much smaller than GMCs.


\section{Interactions between stars and the ISM}
We've seen that stars form from the dense ISM gas, and enrich it with
metals. However, the interaction between stars and the ISM is much more
intricate than just this. Supernovae (SNe) and winds from massive stars
are thought the be important contributors to stirring-up the ISM and
regulating the fraction of the ISM in the different phases.

Consider for example the case where the ISM of a galaxy contains a lot
of dense, cold gas. This may lead to strong star formation in these
clouds, and the resulting massive stars will heat-up the dense gas,
both with their energetic stellar winds, their radiation, and their SNe
explosions. Alternatively, if a galaxy has no strong star formation,
the hot gas will eventually cool and form denser clouds. So you see
that star formation may act as a (negative) feedback process,
regulating the amount of cold gas and the rate of star formation in the
disk.

This intricate process is not well understood, but let me wave my hands
a bit and make the following argument. The energy of a SN depends on
the type of star, not the type of galaxy. Hence you can imagine that
SNe explosions (for example) have more effect on the gas distribution
in a small galaxy, with little mass (mostly in dark matter) to
constrain hot gas in it, than in a very massive galaxy. If there are
sufficient numbers of recent SNe in a galaxy with not too deep a
potential well, then such a galaxy may have trouble holding-on to the
hot gas generated by the SNe, and a {\em galactic wind} may develop.

And so the net effect of feedback in a small galaxy might be that a
significant fraction of the gas is removed from its potential well,
leaving mostly dark matter and some residual gas, and the rest of the
stars. Indeed, we find observationally that such {\em dwarf galaxies}
contain very little gas or stars, in comparison to the amount of dark
matter in them. Feedback from SNe is thought to be the culprit.

In more massive galaxies, the potential well is too deep for a galactic
wind to remove a lot of gas, and so more massive galaxies tend to
hold-on to most of the heated gas. In a galaxy like the MW, the effect
of SNe is to generate a {\em galactic fountain}. Gas heated by the
combined effect of several SNe rises perpendicular to the disk, and
starts to cool as it expands. When it has cooled sufficiently, it rains
back onto the disk. Crucially: it eventually comes back to the disk.

As I explained, the SNe not only heat gas, they also produce metals. If
the hot gas can escape the halo of a galaxy, then presumably so can the
metals. And so small galaxies cannot hold-on to the metals they
produce, but more massive galaxies can. At least, that is what we
expect. But indeed, it turns-out that the more massive galaxies tend to
have higher metallicities than the lower mass ones.

Let me make clear that this certainly is not the whole story. Magnetic
fields and energetic cosmic ray particles\footnote{The earth is
bombarded by an extra-solar flux of high energy particles, mostly
electrons, protons, and $\alpha$ particles (i.e. He nuclei), but other
nuclei as well, some of which have truly astounding energies.  The
highest energy particle detected so far had about the same kinetic
energy as a tennis ball when served by Venus Williams -- {\em in a
single elementary particle!} To obtain such energies in one of the
particle accelerators on earth, requires the accelerator to have an
astronomic size -- several \pc\,. And so it seems safe to say that this
will {\em never} be achieved on earth. They are detected for example by
their Cerenkov radiation as they enter the atmosphere. At least some
fraction of cosmic rays is produced in the shocks of SNe shells plowing
into the ISM, but it is not clear where the highest energy particles
originate from.} contribute significantly to the energy density in the
plane of the MW. So they are likely to be important in understanding
the dynamics of the ISM, yet they certainly make an already complex and
non-linear system even more intractable. And hence are usually ignored,
may be to our peril.

\pagebreak
\section{Summary}
After having studied this lecture, you should be able to 

\begin{itemize}
\item Describe how we know the properties of interstellar dust from
scattering and absorption of star light.
\item Explain why we find different ionisation states of interstellar
gas, depending on density, temperature, and ionising background. 
\item Derive the Str\"omgren radius for an HII region, and explain the
concept.
\item Explain the origin of the hydrogen 21-cm line, and explain its
importance in understanding the structure of the MW.
\item Explain the concept of Jeans mass, and its relation to
fragmentation of clouds.
\item Briefly explain the interaction between the ISM and stars.
\end{itemize}

\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. 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 may expect also most of the mass to be interior to the orbit. If
the star is on a circular orbit, 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 guarantees 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, so 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 $P\propto R^{3/2}$.

So stars around the sun will each move on their own circular orbit, and
so there should be a relation between the relative velocity of the
star, it's distance, and it's direction in the disk. This relation is
described by Oort's constants.

\subsection{Oort's constants}
\begin{figure*}[!ht]
\center\resizebox{.9\textwidth}{!}{\includegraphics{oort.pdf}}
\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 $r$, and galactic
longitude $l$, on circular orbit with velocity $\Theta$ and radius
$R$. In the text, $r$ is called $d$ and the circular velocities are
$V_c$ and $V_{c,0}$ for the star and the Sun, respectively.}
\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 in.

For the geometry, look at Figure~(\ref{fig:Oort})\footnote{We are
describing the motion of stars {\em in the plane of the disk}, hence
the galactic coordinate $b=0$}. 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(l) + R\cos(\beta)\approx d\cos(l) +
R\,{\hbox{\rm when $d\ll R_0$}}\,,
\end{eqnarray}

where $\beta$ is the angle Sun-MW Centre-Star (see Fig.22.24 in
OC). 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 $\Omega=V_c/R$ the angular velocity of the star. If $|R-R_0|$ is
small, we can perform 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
example with $d=1\kpc$, $V_t\approx 2.4\kms$.

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 $R_0\approx 8.5\kpc$ 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} = -13.0\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 suggests 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)\footnote{Note that this is not an inertial reference frame!},
and, for example the Sun moves with a velocity of $\sim 16 \kms$ with
respect to its LSR. This is called the {\em solar mation}. To determine
$A$ and $B$, we need to measure $V_r$ and $V_t$ as a function of $d$
and $l$ {\em statistically}, and take into account that we also need to
determine the solar motion.

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, they are stars in the MW halo.
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 HI~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

\be
V_{r,{\rm max}}=V_c-V_{c,0}\sin(l)\,,
\ee

for any $l$ with $|l|<\pi/2$, which occurs for $\alpha=0$, at which
point

\be
R=R_0\sin(l)\,,
\ee

is a minimum, and $d=R_0\cos(l)$.\footnote{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.}
Combining these, we find

\begin{equation}
{dV_c(R)\over dR}={dV_{r,{\rm max}}(l)\over dl}\,/R_0\,\cos(l)+V_{c,0}/R_0\,.
\end{equation}

Performing this analysis essentially confirmed Oort's measurements,
i.e. $|dV_c/dR|$, 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.

\section{Rotation curves and dark matter}
\label{sect:rotcurve}
If the rotation curve is flat, it means there must be more mass in the
outskirts of spirals than we've assumed so far. 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. And so, although the {\em light
density} drops significantly in the outer parts of the MW, the {\em
mass density} drops much slower with increasing $R$.

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 galaxy 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 mysterious type of
mass in the laboratory.

\section{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 in the disk, even at the position of the Sun.

Here is a simplified way how to do this. First let me prove that for a
uniform disk, the force on a star 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 centre 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.

\section{Spiral arms}
Before leaving the subject of spiral galaxies, we should address what
is in fact probably the first question you'd like to know about them:
{\em why} do they have spiral arms? So let's start by reviewing what we
know about spiral arms.

Firstly, spiral arms are really evident when we look at massive stars
and HII regions -- these really delineate the spiral structure. But
there is more: we also detect the arms in HI, and even better in
molecular gas. In fact, GMCs are probably the best tracers of the
pattern. So may be they have something to do with star formation?

However, they are also present in the \KB\, band where we are probing
{\em old} stars. So first conclusion: it's really the whole galaxy that
takes part in the spiral pattern. However, the {\em density contrast}
(i.e. the relative difference in density at given $r$ from the centre,
within vs outside of the arm) is not very large, probably less than a
factor of two. The contrast in {\em surface brightness} can be quite a
bit higher though, since the young stars in the arms are massive and
hence bright.\\

{\em The winding problem} Given that the disk is in differential
rotation, a spiral arm cannot be a material structure (meaning that it
is always the same stuff in the arm rotating around the disk) --
because of it were, we would have a {\em winding problem}. To realise
why this is so, consider the following thought experiment.

Suppose you paint all stars along a radius in the disk green, and
follow how those green stars move (see figure~(\ref{fig:winding}). If
the circular velocity is a constant, $V_c$, then the period of an orbit
is inversely proportional to its radius, $P\propto 1/R$. Now
concentrate on two green stars, one at distance $R$ (star 1), the other
at distance $2R$ (star 2), having periods $P$ and $2\,P$. After a time
$P$, star 1 will be at the original position, whereas star 2 will be at
the other side of the galaxy. After time $2P$, both stars will be back
at their original position, and so the green line of stars in between
them will have circled the galaxy, i.e. {\em the initially straight
line of green stars now encircles the galaxy}. Applying this to stars
originally in a spiral arm, it is clear now that very quickly, the
spiral arm will wind tighter and tighter. Considering how loosely some
spiral arms are, this would imply they are all recently formed, placing
us at a peculiar instant in time where spiral structure forms.

In cases where we were able to determine the sense of rotation of the
pattern, it turned out that the spiral arms are {\em trailing}, that
is, at least they are rotating in the same sense as the spiral pattern
of our green stars.

\begin{figure*}
\setlength{\unitlength}{1cm} \centering
\begin{picture}(9,4)
\put(-4.5, 0){\special{psfile=wind1.pdf hscale=80 vscale=80}}
\put( 3.5, 0){\special{psfile=wind3.pdf hscale=80 vscale=80}}
\end{picture}
\caption{A spiral pattern made out of stars will rapidly wind-up in a
disk undergoing differential rotation. }
\label{fig:winding}
\end{figure*}

So what's the solution? May be it's good to realise that there is quite
a variety in spiral patterns. {\rm Grand Design} spirals have two
really well defined relatively open arms. Some spirals have {\em
flocculent} arms, which are not very long (i.e. you can't trace them
for very long), but often there are many in the disk. So may be we're
looking at more than one possible origin.

The theory of {\em spiral density waves} suggest that a density wave
rotates with a constant pattern speed in the disk. You can think of
such a wave as a natural mode of oscillation of the disk, much as you
have sound waves in air, or harmonic waves in a violin string. If the
rotation period of a star is shorter than of the pattern, then the star
will occasionally overtake the wave. The gravitational force of the
matter in the wave will make the star linger a bit near the pattern,
before it leaves the arm again. An analogy often made is that of a slow
lorry on the motor way causing a pile-up of faster cars behind
it. Although most cars go fast, the speed with which {\em the pile-up}
moves is determined by the slow lorry.

Another process which we {\em know} to occur, is the triggering of
spiral arms by {\em galaxy encounters}. When two galaxies come close
together, they will exert tidal forces on each other.\footnote{The
tidal force exerted by the moon causes the tides in the earth's
oceans.} This effect can be convincingly demonstrated with numerical
simulations. And so, although these arms may quickly disappear when
left on their own (winding problem), the satellite galaxy may
continually re-excite the spiral arms. Another hint comes from the fact
that in barred galaxies, the spiral arms emanate from the ends of the
bar (have a look at NGC~1365 if you don't believe me). It's is thought
that Grand Design spirals are either triggered by a tidal encounter, or
by the rotating bar.

Goldreich and Lynden-Bell said about flocculent spirals: \lq a swirling
hoch-potch of spiral arms is a reasonably apt description\rq\ . So here
we're probably faced by some kind of gas-dynamical instability in the
disk.

So the conclusion about spiral arms, is that there may be several
processes that give rise to spiral arms, and depending on the
particular spiral galaxy, one of these may dominate. So tidal
encounters are probably responsible for the beautiful Grand Design
spiral arms. Whereas the more messy pattern of flocculent spirals is
probably due to a gas-dynamical instability in the disk.


\pagebreak
\section{Summary}
After having studied this lecture, you should be able to

\begin{itemize}
\item Derive the rotation curve for a Keplerian disk
\item Derive the equations for the radial and tangential velocity of stars
on circular orbits in a disk in differential rotation, and derive
expressions for Oort's constants.
\item Compute Oort's constants $A$ and $B$ for a Keplerian
disk. Explain how $A$ and $B$ are measured in the MW.
\item Explain how the 21-cm emission line can be used to estimate the
rotation curve of the MW.
\item Explain why both Oort's constants, and the rotation curve
measured from 21-cm emission, suggest the presence of dark matter in
the outer parts of the MW.
\item Explain how Oort derived a limit on the amount of dark matter in
the MW disk from the vertical motion of stars.
\item Describe why spiral arms cannot all be material structures by
explaining the winding problem.
\item Discuss solutions to the winding problem.
\item Explain the density wave theory of spiral arms.

\end{itemize}

\chapter{Elliptical galaxies. I }

\section{Luminosity profile}
In contrast to spirals galaxies, elliptical have smooth surface
brightness profiles ellipsoidal in shape.

The SB profile of some elliptical are shown in
Figure~(\ref{fig:2D}). Note how in the centre the isophotes are well
defined and elliptical, whereas 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.pdf}}
\caption{(Taken from astro-ph/0206097) Lines of constant surface
brightness (isophotes) in the \KB\ band for four elliptical
galaxies. In successive isophotes, the surface brightness 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, is 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 straight line, as in Figure~(\ref{fig:3379}). 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.pdf}}
\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.pdf}}
\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.pdf}}
\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.pdf}}
\caption{Image of NGC~1399, with an $r^{1/4}$ profile subtracted, from
Bridges et al., AJ 101, 469 (1991). The centre 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 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\,. Even though the
distance to Fornax is large, it means that the extent on the sky of
NGC~1399 is about the size of the moon! This gigantic size has prompted
the suggestion that may be we should associate such a halo with the
cluster of galaxies, rather than with any 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
galaxy 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 luminosity 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 might produce a
characteristic density profile\footnote{rather, a characteristic {\em
distribution} function, the density of stars in six-dimensional
position and velocity space.}. He termed this process {\bf violent
relaxation}. And so the suggestion is that, as (predominantly spiral)
galaxies fall into galaxy clusters, they will violently interact with
other galaxies, transforming them somehow 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 constraints 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.pdf}}
\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.pdf}}
\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}
Elliptical galaxies emit X-rays. I want to briefly discuss why, how
X-rays are observed, and what we can learn from them.

Fortunately for us, the earth's atmosphere blocks X-rays. Unfortunately
for astronomers, this means we need to be above the atmosphere, in a
balloon, rocket or satellite, to observe in the X-ray spectrum.

How does one built an X-ray telescope?  X-rays do not reflect off
mirrors the same way that visible light does.  Because of their high
energy, X-ray photons penetrate into the mirror in much the same way
that bullets slam into a wall. Likewise, just as bullets ricochet when
they hit a wall at a grazing angle, so too will X-rays ricochet off
mirrors. These properties mean that X-ray telescopes must be very
different from optical telescopes. Basically, they contain many almost
parallel plates, that gently nudge the incoming X-rays into the new
direction you want them to go in. See
http://chandra.harvard.edu/xray\_astro/ for a nice diagram and more
detailed info.

The sun (we're really close!) was the first source to be detected in
the X-rays, in 1949. The technology improved in great leaps, and it
came somewhat as a surprise to detect X-rays from Es that were not
associated with point sources.

Why am I talking about point sources? Recall from the stellar part of
the course the concept of {\em X-ray binary}. In these binary stars,
hot gas associated with mass transfer radiates in the X-ray band. Such
(binary) stars are also seen in Es. But subtracting these sources from
the X-ray image, we still detect {\em extended X-ray emission} coming from
the galaxy.

\subsection{Thermal bremsstrahlung}
How are the X-rays produced in this hot gas? Since the gas temperature
is very high, most elements are highly ionised, and so the gas is really
a hot plasma. The negatively 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\footnote{German
for \lq braking radiation\rq.}. And as the electron slows down, the
plasma also looses energy, and so the gas cools.\footnote{Also the ions
collide of course, but since they are much more massive than the
electron, 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. This is all you need to know about this,
but if you want more details, here they come.

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 electron and ion
densities, and $T$ is the temperature. You could have guessed the first
bit, since we're talking about an encounter between {\em two}
particles. So the exercise is how to get the $T^{1/2}\propto v$
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 frequency $\nu$. The trick to obtain $E(\nu)$ is by Fourier
transforming $E(t)$,

\begin{eqnarray}
E(\nu)&\equiv&\int_{-\infty}^{\infty} E(t)\exp(2\pi\,i\nu\,t)
dt\nonumber\\
&=&{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 strength of different transitions. This
is how one determines the temperature $T$ of the hot gas.

\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 from hydrostatic equilibrium in stars!) By
modelling the spectrum as function of position, we can determine $T$ and
$\rho$ separately, and hence the rhs of Eq.(\ref{eq:HE}). From that,
one can reconstruct the gravitational potential, and hence infer the
mass distribution.

The temperature $T$ is observed to remain approximately constant, in
which case we can obtain an approximate expression for the density
profile. The pressure is

\begin{equation}
p={\boltz\,T\over \mu\,\mp}\,\rho\,.
\end{equation}

If we take $T$ constant then the rhs of Eq.~(\ref{eq:HE}) becomes
$-{\boltz\,T\over \mu\,\mp}\nabla\ln(\rho)$. The lhs of this equation
is just the gravitational force, $\nabla\Phi=GM(<r)/r^2$, with $M(<r)$
the mass interior to $R$. Combining
these two, we obtain after some rearranging

\begin{equation}
r^2\,{d\ln(\rho)\over dr}={G\mu\,\mp\over\boltz\,T}M(<r)\, \,.
\label{eq:HE3}
\end{equation}

Taking the derivative of both sides with respect to $r$, gives

\begin{equation}
{d\over dr}\,\left(r^2\,{d\ln(\rho)\over dr}\right) =
-{4\pi\,G\mu\,\mp\over \boltz\,T}\,r^2\rho\,,
\label{eq:HE2}
\end{equation}

since $dM(<r)/dr=4\pi\,r^2\,\rho$. Try a solution of the form
$\rho=C\,r^b$, where $C$ and $b$ are constants. Substituting this, the
lhs of Eq.~(\ref{eq:HE2}) is $b$, and the rhs is
$-(4\pi\,G\mu\mp/\boltz\,T)\,C r^{b+2}$. If this is to be a solution,
then $b=-2$, and

\begin{equation}
\rho(r) = {\boltz\,T\over 2\pi\,G\mu\mp}\,r^{-2}\,.
\end{equation}


This would be the density profile if the gas were to furnish the
gravitational potential required to keep it in hydrostatic
equilibrium. However, in addition to the gas, there is also the
gravitational potential due to stars. And so we should have used
$dM/dr=4\pi\,r^2(\rho_{\rm gas}+\rho_{stars})$. By estimating the mass
in stars, we can correct our density distribution to what it should be,
in the presence of stars. Since this extra gravity increases the rhs of
Eq.~(\ref{eq:HE3}), the density $\rho_{\rm gas}$ will need to increase
as well.

Given the corrected density profile, we can compute the expected X-ray
emissivity, since that depends on the density $\propto\rho_{\rm
gas}^2$. The emissivity is much higher than expected, and hence we
infer the presence of dark matter in ellipticals.

\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.

\pagebreak
\section{Summary}
After having studied this lecture, you should be able to 

\begin{itemize}
\item describe the surface brightness profiles of Es in terms of the de
Vaucouleurs profile.
\item explain why we think that this profile results from galaxy encounters.
\item recall that Es have typically hundreds of GCs.
\item explain why dust lanes in Es and shells of stars around Es are
thought to be evidence for galactic cannibalism.
\item contrast the stellar population and ISM of Es with those of Ss.
\item describe how X-rays are detected, and explain the process with
which X-rays are produced in the hot gas in Es. (But not be able to
recall the equation, or reproduce the calculation.)
\item explain how X-ray observations can be used to infer the
gravitational potential in Es.
\end{itemize}

\chapter{The Dark Halo}

Measurements of the rotation curve from HI 21-cm emission, and the
analysis of the motions of stars in the solar neighbourhood with Oort's
constants, and the Oort limit, all suggest the presence of a large
amount of invisible \lq dark matter\rq\ in the outskirts of the
MW. Given such a startling conclusion, it may be a good idea to look
for other evidence for dark matter in galaxy haloes.

\section{High velocity stars}
A number of high-velocity stars near the Sun have measured
velocities\footnote{The quoted velocity is wrt to the centre of mass
velocity of the MW. Do not confuse these with Oort's high velocity
stars, which are typically low mass, low metallicity stars in the {\em
Galactic Halo}. The velocities of Oort's stars are of order
200\kms. The present high velocity stars are typically $A$-type stars,
presumably born in the disk, that have acquired their high velocity
following a super nova explosion.}  up to $v_\star\approx
500\kms$. Their existence provides us with a probe of the galaxy's
mass, if we assume that these stars are still bound to the MW: it means
that {\em the escape speed\footnote{Stars travelling at the local
escape speed are just able to leave the potential well and escape to
infinite.} at the solar circle, $v_e<v_\star$}.

\subsection{Point mass model}
If we assume there is no dark halo, then it is easy to find a relation
between the circular velocity $V_c$ and the escape speed $v_e$. If we
assume all mass $M$ of the MW is interior to the solar radius $R_\odot$
\footnote{$R_\odot$ is the radius of the circular orbit around the MW,
at the position of the Sun -- the solar galacto-centric radius.}, then
$V_c^2=GM/R_\odot$, and the potential $\Phi=-GM/R_\odot=-V_c^2$. The
velocity of a star just able to escape from the gravitational well is
the {\em escape speed}, $v_e$. So since that star has zero total
energy,

\begin{equation}
0=E={1\over 2}v_e^2+\Phi={1\over 2} v_e^2-V_c^2\,.
\label{eq:ve}
\end{equation}

So in this case, $v_e=2^{1/2}\,V_c\approx 311\kms$ (Using
$V_c=220\kms$.) So for a point mass model of the MW, these high
velocity stars cannot be bound to the MW -- suggestingthe presence of
more mass in the outskirts of the MW than expected, given the
rapid decrease in the distribution of light.

\subsection{Parameters of dark halo}
Given the previous failure of our MW model, let's assume there to be a
dark halo. In addition, let us assume the MW to be spherical, with a
constant circular speed $V_c$ out to some radius $R_\star$. Since
$V_c^2=GM/R={\rm constant}$, we have that

\begin{eqnarray}
M(R) &=& {V_c^2\,R\over G}       \,\,\,\,\,\,\,\,\,\hbox{\rm when $R<R_\star$}\nonumber\\
     &=& {V_c^2\,R_\star\over G} \,\,\,\,\,\,\,\,\,\hbox{\rm when $R\ge R_\star$}\,.
\end{eqnarray}

Now the equation for the potential $\Phi$ is

\begin{equation}
{d\Phi\over dR} = {GM\over R^2} = {V_c^2\over R}\,,
\end{equation}

for $R\le R_\star$. Integration gives $\Phi={\rm
constant}-V_c^2\ln(R_\star/R)$. We can determine the value of the
constant at $R=R_\star$, since then $\Phi=-G\,M/R_\star=-V_c^2$. Hence

\begin{equation}
\Phi(R) = -V_c^2\,\left[1+\ln(R_\star/R)\right]\,.
\end{equation}

Using Eq.(\ref{eq:ve}) for the escape speed, we obtain

\begin{equation}
v_e^2 = 2\,V_c^2\left[1+\ln(R_\star/R)\right]\,.
\end{equation}

At the position of the Sun, we have $R\approx 8.5\kpc$, $V_c=220\kms$
and $v_e\ge 500\kms$. This implies that $R_\star\ge 41\kpc$, and hence

\begin{equation}
M(R=R\star)\ge 4.6\times 10^{11}\msun\,.
\label{eq:mwmass1}
\end{equation}

Since the total luminosity of the MW is of order $1.4\times
10^{10}\lsun$ (in the \V\ -band), it means\footnote{Remember: we've
only derived a lower-limit to the MW mass, hence a lower-limit to the
mass-to-light ratio.} that the mass-to-light ratio is at least
30. Given that the mass-to-light ratio of the MW {\em stars} is around
3 or so, it again suggest that a dominant fraction of the MW mass is
not in stars.

Another estimate of the MW mass comes from the motion of neighbouring
galaxies in the {\em Local Group}.

\section{The Local Group}

The MW is located in a rather average part of the Universe, away from
any dense concentrations of galaxies. The \lq Local Group\rq\ consist
of the MW, Andromeda (M31), and about 40 small, irregular galaxies,
gravitationally bound to each other. 

\subsection{Galaxy population}
The MW has a couple of small galaxies gravitationally bound to it, for
example the Large and Small Magellanic Clouds. The Magellanic stream is
a stream of gas which originates from the Large Magellanic Cloud, which
can be traced over a significant fraction of the whole sky. It may
result from the tidal forces exerted by the MW.

The Sagittarius Dwarf galaxy is the neighbouring galaxy closest to the
MW. It was discovered as recently as 1994, having eluded detection
since it happens to be at the other side of the MW bulge. And so it is
hiding behind a large amount of foreground stars. Sagittarius has
ventured so close to the MW that the tidal forces will probably tear it
apart in the near future. Its discovery was actually hampered by its
close proximity, since the large area on the sky ($\sim 8^o\times 2^o$)
made it difficult to pick out against the bright MW foreground. Even
away from the galactic plane (where dust obscuration limits our view),
new members of the LG continue to be identified. The study of these
small galaxies is exciting, since they allow us to constrain models for
galaxy formation and evolution.

The Andromeda galaxy M31 is about twice as bright as the MW, and is
the brightest member of the Local Group (LG). It is very similar to the
MW, and so also has its own set of small galaxies orbiting around
it. These galaxies, together with another 20 small galaxies, form the
LG. M31 and the MW are by far the most massive galaxies in the LG.

The LG is almost certainly gravitationally bound to other nearby
groups, and so does not really have a well defined edge. Galaxies in
the outskirts of the LG may in fact belong to other groups. 

The distribution of galaxies in the LG is rather untidy and lacks any
obvious symmetry. Most of the galaxies are found either close to the
MW, or close to M31. There is a third, smaller condensation of
galaxies, hovering around NGC~3109.

The motion M31 can be used to estimate the mass of the Local Group,
using the Local Group timing argument.

\subsection{Local Group timing argument}
The dynamics of M31 and the MW can be used to estimate the total mass
in the LG. From the Doppler shifts of spectral lines, one can measure
the radial velocity of M31 with respect to the MW\footnote{What one
measures is the radial velocity wrt to the {\em Sun}. Since the Sun is
on a (nearly) circular orbit around the MW, one needs to correct the
measured heliocentric velocity to obtain the radial velocity of
Andromeda wrt the MW.},

\begin{equation}
v=-118\kms\,.
\end{equation}

The negative sign means that Andromeda is moving {\em toward} the
MW. This may be surprising, given that most galaxies are moving apart
with the general Hubble flow. The fact that Andromeda is moving toward
the MW is presumably because their mutual gravitational attraction has
halted, and eventually reversed their initial velocities. Kahn and
Woltjer pointed out in 1952 that this leads to an estimate of the
masses involved.

Since M31 and the MW are by far the most luminous members of the LG we
can neglect in the first instance the others, and treat the two
galaxies as an isolated system of two point masses. Since M31 is about
twice as bright as the MW, and given that they are so similar, it is
presumably also about twice as massive. If we further assume the orbit
to be radial, then Newton's law gives for the equation of motion

\begin{equation}
{d^2r\over dt^2} = -{GM_{\rm total}\over r^2}\,,
\end{equation}

where $M_{\rm total}$ is the sum of the two masses. Initially, at $t=0$, we can
take $r=0$ (since the galaxies were close together at the Big Bang).

The solution can be written in the well known parametric form

\begin{eqnarray}
r &=& {R_{\rm max}\over 2} (1-\cos\theta)\nonumber\\
t &=& \left({R^3_{\rm max}\over 8\,G\,M_{\rm total}}\right)^{1/2}\,
(\theta-\sin\theta)\,.
\end{eqnarray}

The distance $r$ increases from $0$ (for $\theta=0$) to some maximum
value $R_{\rm max}$ (for $\theta=\pi/2$), and then decreases again. The
relative velocity

\begin{equation}
v = {dr\over dt} = {dr\over d\theta}/{dt\over d\theta} =
\left({2\,G\,M_{\rm total}\over R_{\rm max}}\right)^{1/2} \left({\sin\theta\over
1-\cos\theta}\right)\,.
\end{equation}

The last three equations can be combined to eliminate $R_{\rm max}$, $G$
and $M_{\rm total}$, to give

\begin{equation}
{v\,t\over r} = {\sin\theta\,(\theta-\sin\theta)\over
(1-\cos\theta)^2}\,.
\label{eq:lgt}
\end{equation}

$v$ can be measured from Doppler shifts, and $r\approx 710\kpc$ from
Cepheid variables. For $t$ we can take the age of the
Universe. Current estimates of $t$ are quite accurate\footnote{From
properties of the micro-wave background radiation.}, but even using
ages of the oldest MW stars, $t\sim 15\Gyr$ gives an interesting
result. Using these numbers, we can solve the previous equation
(numerically) to find $\theta=4.32$ radians, {\em assuming M31 is on
its first approach to the MW\footnote{Equation~(\ref{eq:lgt}) has no
unique solution for $\theta$, since it describes motion in a periodic
orbit. On its first approach, $\theta$ should be the {\em smallest}
solution to the equation.}.}

Substituting yields $M_{\rm total}\approx 3.66\times 10^{12}\msun$, and
hence for the MW mass, $M\approx M_{\rm total}/3$
\begin{equation}
M\approx 1.2\times 10^{12}\msun\,,
\end{equation}
comfortably higher than our lower limit Eq.(\ref{eq:mwmass1}).

Since the luminosity of the MW (in the \V\ band) is $1.4\times
10^{10}\lsun$, the corresponding mass-to-light ratio for the MW is
around $\Gamma=100$. Furthermore, the estimate of $M$ is increased if
the orbit is not radial, or M31 and the MW have already had one (or
more) pericenter passages since the Big Bang.

If all stars in the MW and M31 were solar mass stars, we would expect
$\Gamma=1$. Now even in the solar neighbourhood, most stars are less
massive than the sun, and so the mass-to-light ratio for the {\em
stars} is about 3 or so\footnote{Recall that the luminosity of stars
scales with their mass quite steeply, $L\propto M^\alpha$, with
$\alpha\approx 5$, and hence $M/L\propto M^{1-\alpha}$. See your notes
on stars, p.~34. }. So the very large mass inferred from the LG
dynamics strongly corroborates the evidence from rotation curves and
Oort's constants, that most of the mass in the MW (and presumably also
in M31) is dark.

From these numbers, we can also estimate the extent $R_\star$ of such a putative
dark halo. If the the circular velocity $V_c=220\kms$ out to $R_\star$, then
from $V_c^2=GM/R_\star$ we find

\begin{equation}
R_\star={GM\over V_c^2} \approx {G\,10^{12}\msun\over (220\kms)^2} \approx
100\kpc\,.
\end{equation}

If, as is more likely, the rotation speed eventually drops below
220\kms, then $R$ is even bigger. Hence the extent of the dark matter
halo around the MW and M31 is truly enormous. Recall that the size of
the stellar disk is of order 20\kpc\ or so, and the distance to M31
$\sim 700\kpc$. So the dark matter haloes of the MW and M31 may almost overlap.

\pagebreak
\section{Summary}
After having studied this lecture, you should be able to

\begin{itemize}
\item Show that in a point mass model of the MW, the high velocity
stars are not bound.
\item Estimate the parameters of a dark halo, assuming the high
velocity stars are bound to the MW.
\item Describe the properties of the Local Group in terms of the
galactic content.
\item Estimate the mass and extent of the dark halo of the MW from the
Local Group timing argument.
\end{itemize}

\chapter{Elliptical galaxies. II}
This chapter is a bit harder, with more mathematics than I would like
to use. I {\bf do not} expect you to be able to reproduce all these
derivations, only the ones stressed in the Summary. A little note on
notation: if in one of the following equations 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 relate the stellar velocity dispersion,
$\sigma_{ij}^2=\langle (v_i-\langle v_i\rangle)(v_j-\langle
v_j\rangle)\rangle$ with 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 such an approach makes
sense, I first need to introduce the concept of relaxation time.

\section{Relaxation in stellar systems}
\label{sect:relax}
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
$T_E\rightarrow \infty$.\\

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\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 relaxation time increases with $N$ as
$T_{\rm relax}\propto N/\ln(N)$. Put differently, the effect of
encounters decreases with increasing particle numbers as $N/\ln(N)$.

For a Globular Cluster, $N=10^5$ say, $r=10\pc$, and $v=10\kms$, hence
$T_{\rm cross}\approx 1\Myr$. Putting in the numbers\footnote{A handy
approximate relation is $1\km/s\approx 1\pc/\Myr$.}, we find $T_{\rm
relax}\approx 10^9$ years, much smaller than the ages of the
stars in the GC. Hence we expect stellar encounters to be important in
shaping the structure of GCs.

However, for a big elliptical galaxy, $N=10^{11}$, 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} larger 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 conservation 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
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 length 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, $f(\bfx,\dot\bfx,t)$, 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
rather 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 \partial t}\,d\bfv = {\partial\over\partial t}\int
f(\bfr,\bfv,t)\,d\bfv\equiv {\partial n(\bfr,t)\over\partial t}\,.
\ee

The integral of the distribution function over velocities, $ n\equiv
\int\,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( n\bar v_i)\over\partial x_i}\,,
\ee

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

\be
 n\,\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 n\over\partial t}+{\partial ( n\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 proceeds exactly as for the density, and the result is
after a bit of algebra:

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


This is a more general version for the equation of hydrostatic
equilibrium. To see this, note that the matrix $\sigma_{ij}$ is
symmetric: $\sigma_{ij}=\sigma_{ji}$. For such symmetric tensors, we
can always rotate the coordinate system such that the tensor becomes
diagonal, i.e. $\sigma^2_{ij}=0$ for $i\ne j$. Now, suppose that the
velocities are {\em isotropic}, so that
$\sigma_{xx}=\sigma_{yy}=\sigma_{zz}\equiv\sigma^2$. Then we find that

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

which is {\em exactly} the same as the equation for hydrostatic
equilibrium, if $ n\sigma^2$ is identified with the pressure $p$.
Because of this analogy, the tensor $n\,\sigma^2_{ij}$ is called the
{\bf pressure tensor}.\\

In conclusion: the behaviour of {\rm collisionless} stellar systems is
similar to that of self-gravitating gas spheres. The role of
temperature is taken over by that of the stellar velocity dispersion.
It is the high velocity dispersion of the stars in an elliptical (and
also in the bulge of spirals), which balances the gravitational
pull. Since the required velocities are so high, such systems are
called \lq hot stellar systems\rq. Recall that in disk galaxies, it is
the ordered motion of the stars -- i.e. the rotation of the disk --
which supports the system against gravity. Such systems are dynamically
cold, i.e. the stellar velocity dispersion is small compared to
the streaming motions. For disks, the \lq velocity dispersion\rq\
refers to the small velocities that stars have with respect to their
\lq local standard of rest\rq\ (10s of \kms), versus the rotational
velocity of the disk, 200\kms.

In a gas the pressure is isotropic and hence stars are spherical.  In
contrast in stellar systems, the pressure tensor $n\,\sigma^2_{ij}$ is
in general {\em not} isotropic, and so the pressure gradient can be
larger in one direction ($x$, say) than in another direction ($y$). So
even if the potential is spherical, the stellar distribution may be
more extended in $x$ than in $y$, and the system will be elliptical in
shape.

\pagebreak
\section{Summary}
After having studied this lecture, you should be able to 

\begin{itemize}
\item to derive an estimate for the stellar dynamical relaxation time
in a spherical system of $N$ stars.
\item explain why we can describe the dynamics of elliptical galaxies
in terms of fluid equations.
\item explain the difference in dynamics between an elliptical (hot
stellar system) and a spiral (cold, rotating stellar system)
\item explain what is a continuity equation, and derive the
collisionless Boltzmann equation, Eq.(\ref{eq:ce2}), stating the
underlying assumptions.
\item explain that the ellipticity of ellipticals does not results from
rotation, but from the anisotropy of the pressure tensor.
\end{itemize}


\chapter{Groups and clusters of galaxies}

Galaxies are not distributed homogeneously in the Universe. Some
regions in the Universe have a galaxy density (i.e. the number of
galaxies per \Mpc$^{-3}$) significantly lower than the mean, and some a
significantly higher density. The main reason for this is that gravity
amplifies small inhomogeneities: if a region is slightly denser than
its surroundings (for example a group of galaxies), it will tend to
attract more matter than the surrounding regions, and hence become even
more massive (turn into a galaxy cluster, say). This is the essence of
hierarchical structure formation: small structures merge together to
make more massive one.

The galaxies huddle together in groups of a few galaxies -- such as the
MW and M31 in the Local Group -- and some are in clusters with hundreds
of galaxies. The groups and clusters themselves are clustered in super
clusters (i.e. clusters of clusters). But there are no clusters of
super clusters: on sufficiently large scales the Universe is not a
fractal but becomes homogeneous.

\section{Introduction}
Recall that the MW, M31, and some 40 small irregular galaxies within
2\Mpc\ or so from the MW, are part of a gravitationally bound system,
called the Local Group. Several other such bound groups occur at larger
distances from the MW, and searches over larger volumes have revealed
the existence of 1000s of such groups of galaxies.\\

Within about 20\Mpc\ from the MW are several clusters of galaxies,
containing 10s of large galaxies, and hundreds of smaller ones
each. The {\em Virgo} cluster of galaxies contains the big elliptical
M87. We already discussed the big cD galaxy NGC1399 at the centre of
the {\rm Fornax} galaxy cluster. The typical total mass of these
clusters is estimated at around $10^{14}\msun$. At 90\Mpc\ we find the
Coma cluster of galaxies, with a mass of may be as large as
$10^{15}\msun$. Clusters such as Coma contain the most massive known
galaxies.

In the 1950s, George Abell and collaborators eye balled photographic
plates to look for rich clusters of galaxies, by finding regions with a
large over density of galaxies. He found some 4500, and ranked them
from poor to rich, according to the number of galaxies they had. Paul
Hickson later performed a similar exercise looking for \lq Hickson\rq\
compact galaxy groups.

Clusters of galaxies are the most massive virialised structures in the
Universe. Their crossing time\footnote{If $v$ is the typical velocity
of a galaxy in a cluster of galaxies, and $R$ is the radius of the
cluster, then the crossing time $T_{\rm cr}=R/v$, i.e. it is the
typical time a galaxy takes to cross the cluster ones.} is sufficiently
small that each galaxy has had the opportunity to cross the system
several times, and hence the system has had time to dynamically relax
to a near equilibrium situation, i.e. the system is close to being in
virial equilibrium. Although more massive bound systems exist, these
have crossing times which are so large that, given the age of the
Universe, they have not had time to become dynamically relaxed.

\subsection{Evidence for dark matter from galaxy motions}
Evidence for the existence of dark matter in galaxy clusters dates from
the 1930s. The Swiss astronomer Fritz Zwicky used the velocity
dispersion of galaxies in clusters as determined from Doppler shifts to
estimate their dynamical mass. Suppose all galaxies have the same mass
$m$, then the kinetic energy of the system of $N$ galaxies is

\begin{equation}
K={1\over 2}\sum m\,v^2={1\over 2}\,M\,\sigma^2\,,
\end{equation}

where $M=N\,m$ is the total mass of the cluster galaxies, and the
velocity dispersion $\sigma^2$ is defined as

\begin{equation}
\sigma^2 = {\sum m\,v^2\over M}\,.
\end{equation}

The potential energy in the system is of order

\begin{equation}
U=-{G\,M^2\over R}\,,
\end{equation}

where $R$ is a measure of the size of the system. When the system is in
{\em virial equilibrium}, $2K=|U|$ hence

\begin{equation}
M={R\,\sigma^2\over G}\,.
\end{equation}

Note that the {\em measured} rms velocity is just the line-of-sight
one, say $\sigma_z^2=\langle(v_z-\langle v_z\rangle)^2\rangle$, but if
the velocities are isotropic, then $\sigma^2=3\sigma_z^2$. So from
that, and from an estimate of $R$, one can estimate the dynamical mass
$M$.

Zwicky estimated the masses of the galaxies, and compared the result
with the dynamical mass. Not only was the dynamical mass much larger,
the velocities of the cluster galaxies was so high that, given the
amount of mass inferred to be in the galaxies, the cluster was not even
a gravitationally bound system. Left on its own, it would cease to be a
high density region of galaxies on a short time scale because the
galaxies would fly apart very rapidly. So either clusters were just
chance superpositions of galaxies, or there was much more mass in them
than was inferred from just the galaxies. Zwicky concluded that most of
the mass in clusters must be invisible.

\section{Evidence for dark matter from X-rays observations}
Just as elliptical galaxies, clusters emit X-rays due to thermal
bremsstrahlung produced in the highly ionised gas bound by the
gravitational potential well of the cluster. This can be used to
estimate the mass required to bind this gas, as we did when deriving
Equation~(\ref{eq:HE3}) from assuming hydrostatic equilibrium in an
isothermal gas. This type of analysis again strongly suggests the
presence of dark matter. From the pictures of the X-ray gas, it is also
immediately clear that the X-rays come from all over the cluster, they
are not obviously associated with individual galaxies.\\

From the X-ray intensity distribution, one can also estimate the amount
of hot gas present in the cluster. From modelling the colours of the
galaxies, one can estimate the mass in stars. Comparison of these two
shows that {\em hot gas is the dominant baryonic mass by a large
factor}. So most of the baryonic mass in a cluster of galaxies is
actually in hot gas and not in galaxies at all.\\

The temperature of the gas is so high that it balances the gravitational
force of the dark matter in the cluster. It gets to this high
temperature by {\em shock heating}. This is how it works: as the
cluster is accreting matter from its surroundings, newly accreted gas
falls into the cluster at high velocity. The velocity is high, because
the gas has been accelerated by the gravitational force from the
tremendous amount of mass in the cluster. The high velocity gas slams
into the stationary hot gas and converts most of its kinetic energy
into thermal energy: this is {\em an accretion shock}.\\

Suppose a parcel of gas starts at infinity with zero velocity, and
falls into a cluster with mass $M$ and radius $R$. By the time it
reaches the outskirts of the cluster at radius $R$ it will have an
in fall velocity

\begin{equation}
{1\over 2}v_{\rm infall}^2 = {G\,M\over R}\,,
\end{equation}

where I've used energy conservation. When it hits the cluster gas, it
will convert this kinetic energy into thermal energy, hence it will
be heated to a temperature $T$

\begin{equation}
{3\over 2}{\boltz T\over \mu}={1\over 2}\mp\,v_{\rm infall}^2\,.
\end{equation}

Here, $\mu$ is the mean molecular weight of the gas. Combining the last
two equations, we find that the temperature of the cluster will be

\begin{equation}
{3\over 2}\,{\boltz T\over \mu\mp}={G\,M\over R}\,.
\end{equation}

This temperature is called the {\em virial temperature}.\\

Where does all this gas come from? The X-rays we observe from clusters
imply that the hot gas is loosing energy and hence is cooling. So from
the observed X-ray luminosity, we can estimate the cooling time of the
gas $t_{\rm cool}$, 

\begin{equation}
{dT\over dt} \equiv {T\over t_{\rm cool}}\,.
\end{equation}

Given the present cooling rate, $dT/dt$, the gas will have lost it's
thermal energy after a cooling time, $t_{\rm cool}$. The measured
values of $t_{\rm cool}$ turn out to be longer than the age of the
Universe for most clusters\footnote{Except in the inner parts of some
{\rm cooling flow} clusters}. This means that {\em once the gas gets
hot it will remain hot}, and cannot form into galaxies anymore. So
clusters of galaxies are in fact regions where galaxy formation is
strongly suppressed because the gas is too hot to cool and form
stars.\\

But why did it not form galaxies {\em before} it became hot? This is
actually a rather hotly debated issue at the moment, but here's a hint:


\section{Metallicity of the Intra-cluster medium.}
The X-ray spectrum of cluster gas consists of the usual featureless
thermal bremsstrahlung continuum, with additional emission lines mostly
from inner-shell transitions of the Fe ion, see
Figure~\ref{fig:xrayspec}. By modelling the spectrum, one can estimate
the amount of Fe present in the gas. Comparison with the total amount
of gas gives a mean metallicity $M_{\rm Fe}/M_{\rm gas}\approx {1/3}$
of the solar value\footnote{Recall that the sun contains a mass
fraction of 0.02 of metals.}.\\

\begin{figure*}
\center\resizebox{.8\textwidth}{!}{\includegraphics{xrayspec.pdf}}
\caption{X-ray spectrum of the central region of the Centaurus cluster,
taken by the ASCA X-ray satellite. It shows some emission lines and an
absorption edge (which are labelled) due to different components of the
hot intra cluster medium.}
\label{fig:xrayspec}
\end{figure*}

This is highly surprising: it means that the gas in clusters has a
higher metallicity than for example globular cluster {\em stars}, or
even the stars and gas in most dwarf galaxies. The only way we know how
to produce Fe is in stars -- in fact we think that most of the Fe is
produced during the thermo-nuclear deflagration of a Super
Nova. Somehow Fe produced in stars, themselves presumably inside a
galaxy, was able to escape from that galaxy and end-up in the
intra cluster medium. If the metals were able to escape, then may be
also some of the explosion energy was able to escape the galaxy, and
heat the surrounding gas, preventing gas cooling and galaxy
formation. So the presence of hot metal enriched gas in clusters,
suggests that star formation in proto-clusters provided a negative
feedback mechanism, which kept most of the gas at temperatures too high
for efficient galaxy formation.

\section{The dark matter density of the Universe}
\label{sect:dmU}
Clusters provide a very nice way to estimate the total amount of dark
matter in the Universe. The argument goes like this.

Suppose the Universe starts-out smooth, with a (nearly) constant ratio
$\omega$ of dark matter to gas:

\begin{equation}
\omega = \rho_{\hbox{\rm dark matter}}/\rho_{\hbox{\rm gas}}\,.
\end{equation}

As a cluster starts to form, the density of dark matter and of gas will
increase. But the potential well of the cluster is so deep, that it is
probably a good approximation to assume that none of the dark matter,
nor the gas, can ever escape the gravitational pull of the
cluster. This means that the ratio of the total dark matter to total
gas mass {\em in the cluster}

\begin{equation}
{M_{\hbox {\rm cluster-in-dark-matter}}\over M_{\hbox{\rm cluster-in-gas}}} \approx \omega\,.
\end{equation}

We could do slightly better by also including the mass in stars -- but
that is a small correction anyway.

So we can find $\omega$ by determining $M_{\hbox{\rm
cluster-in-dark-matter}}$ from dynamics, and $M_{\hbox{\rm
cluster-in-gas}}$ from the X-ray emissivity. The result is
$\omega\approx 6$.\\

Next we need an estimate of the mean baryonic density, $\rho_{\hbox{\rm
gas}}$. An elegant way is by determining the {\em Deuterium} abundance
of gas\footnote{Recall that the nucleus of a Deuterium atom differs
from that of ordinary Hydrogen in that it has a proton and a neutron.}.
Deuterium is produced in the Big Bang and destroyed in stellar
burning. So if we find Deuterium, we know it is left over from the Big
Bang. Analysis of the spectra of quasars (we will discuss quasars soon)
allows us to measure the Deuterium abundance quite accurately. How does
this help us? The missing link is that {\em the gas density
$\rho_{\hbox{\rm gas}}$ determines how much Deuterium is produced in
the Big Bang}. Figure~\ref{fig:bbn} shows how the abundance (with
respect to hydrogen) of elements produced in Big Bang nucleo-synthesis,
as function of the baryon density. The result is that the mean baryon
density corresponds to of order $\rho_{\rm gas}/\mp\approx 2.2\times
10^{-7}$ hydrogen atoms per cm$^{-3}$ (i.e. $3.75\times 10^{-31}$ g
cm$^{-3}$. So the density of paper/density of the screen you are
reading this on, is about $10^{30}$ times higher than the mean!

\begin{figure*}
\center\resizebox{.9\textwidth}{!}{\includegraphics{altbbn.pdf}}
\caption{The production of various elements during Big Bang
nucleo-synthesis as a function of the baryon density, taken from
http://astron.berkeley.edu/\~{}mwhite/darkmatter/bbn.html. }
\label{fig:bbn}
\end{figure*}


And so we're done: the Deuterium abundance determines the gas density
through the known nuclear reactions that occurred during Big Bang
nucleo-synthesis. And the dark matter density is
$\omega\,\rho_{\hbox{\rm gas}}$ with $\omega\sim 6$ determined from
clusters.\\

\section{Summary}
After having studied this lecture, you should be able to 

\begin{itemize}
\item Define what is a cluster and a group of galaxies by listing some
of their properties.
\item Explain how galaxy motions suggest the presence of dark matter in clusters.
\item Explain how X-ray observations suggest the presence of dark
matter in clusters.
\item Explain the origin of the high temperature of the cluster gas.
\item Explain how we know that most of the baryons in a cluster are in
hot gas, and not in stars.
\item Explain why the high observed metallicity of the cluster gas
suggests that some kind of stellar feedback mechanism prevented most of
the gas turn into stars before it fell into the cluster.
\item Explain how clusters have been used to estimate the mean dark
matter density of the Universe.
\end{itemize}


\chapter{Galaxy statistics}

Having studied some properties of spiral and elliptical galaxies in
detail, we now take a step back to examine the mix and properties of
galaxies as a whole.

In recent year, there have been a number of galaxy redshift surveys,
for example the two degree-field survey (2dF), and the Sloan Digital
Sky survey (SDSS). In a galaxy redshift survey, one first identifies
galaxies from images of the sky, and then takes a spectrum for each of
them. The shift of the spectral lines from the laboratory frame to what
is observed -- the redshift -- allows one to determine the position
along the line-of-sight. This is because the Universe is exapanding,
and hence there is a relation between how much the line is shifted, and
how distant the galaxy is. Doing this for all galaxies in the survey,
we obtain a three dimensional map of the positions of the
galaxies. Such surveys provide data on several hundred-thousands of
galaxies, and so enable one to investigate in great detail statistical
properties of the galaxy population. So what kind of general trends do
they uncover?\\

The {\em distribution} of galaxies probably traces to some extent the
distribution of dark matter in the Universe. For example, we've seen
that there is evidence for a lot of dark matter in the Universe as a
whole (from the cluster argument), but also in individual galaxies,
groups and clusters. And so, since we find that galaxies are
distributed very inhomogeneous in groups and clusters, presumably
also the dark matter is distributed in an inhomogeneous fashion. How?
And what process determines this in the first place?\\

We have some good idea of how the dark matter is distributed on a
cosmological scale, from some general considerations, but also
backed-up from observations of the micro-wave background (i.e. the sea
of photons left-over from when the Universe was still hot). Given this
input, we can use numerical simulations to predict the growth of
structure (at least in the dark matter). The nice part of the story is
that this theory is highly predictive, and does predict structures in
the galaxy distribution very similar to what we observe.\\

\section{The density-morphology relation}
In the previous chapter we saw how some galaxies are in low density
environments, some are in higher density groups, and others in dense
clusters of many hundred galaxies. Figure~\ref{fig:densmorp} shows that
the {\em fraction} of elliptical and S0 galaxies increases rapidly with
increasing galaxy density. In other words, in a region of high galaxy
density, such as a dense group or cluster of galaxies, most galaxies
tend to be S0 or E, and very few are of type S. Vice versa, in regions
of low galaxy density, most galaxies tend to be of the type S.  This is
called {\bf the density-morphology relation}, since it {\bf relates the
density of galaxies in a given region with their typical morphology}
(i.e. S, E, S0). Clearly this is telling us something on the nature of
galaxy formation and evolution.\\

Recall from the movie I showed of the motion of galaxies in a cluster,
that encounters between galaxies are very frequent in a region of high
galaxy density. Such tidal encounters could be responsible for
converting S-type galaxies into S0 and E, but other processes might
operate as well. Recall that we already discussed that this cannot be
the whole story, since the stellar populations of Ss and Es tend to
differ as well.

\begin{figure*}
\center\resizebox{.9\textwidth}{!}{\includegraphics{density-morphology.pdf}}
\caption{The fraction of Elliptical (E), S0, and Spiral + Irregular
(S+I) galaxies as function of the logarithm of the projected density,
in galaxies \Mpc$^{-2}$. The data shown are for all cluster galaxies in
the sample and for the field. Also shown is an estimated scale of true
space density in galaxies \Mpc$^{-3}$. The upper histogram show the
number distribution of the galaxies over the bins of projected
density. Es and S0s tend to populated regions of high galaxy density,
and Ss regoins of low galaxy density. From Dressler, ApJ 236, p. 351
(1980).}
\label{fig:densmorp}
\end{figure*}

\section{Galaxy scaling relations}

\subsection{The Tully-Fisher relation in spirals}
The circular velocity as function of radius -- the rotation curve -- in
spirals tends to be flat, i.e. the rotation velocity is independent of
radius sufficiently far out in the disk. We argued in section
\ref{sect:rotcurve} that this was evidence for the presence of dark
matter in spiral disks, since it implies lots of mass in the outer
parts, whereas we see very little light there. Figure\ref{fig:tf}
compares the measured rotation velocity $V_c$ (written $W$ in the
figure) with the total absolute magnitude $M=-2.5\log(L)$+constant, for
a sample of spiral galaxies. {\bf Brighter spirals} (more negative $M$)
{\bf rotate faster} (larger $W$). Since the figure shows there to be a
linear relation between $M$ and $V_c$, it implies a power-law relation
between the intrinsic luminosity $L$ and the circular velocity $V_c$,
called the Tully-Fisher relation:

\begin{equation}
L\propto V_c^\alpha\,,\,\,\,\,\,\,\,\alpha\approx 4\,.
\label{eq:tf}
\end{equation}

The slope $\alpha$ of this relation depends somewhat on which band is
being used (i.e., does $L$ refer to a \V\ -band, or an \I\ -band
luminosity).\\


\begin{figure*}
\center\resizebox{.9\textwidth}{!}{\includegraphics{tfr.pdf}}
\caption{R-band absolute magnitude $M$ of spiral galaxies as function
of the maximum rotational velocity $W=V_c$ (in \kms). {\em Brighter}
spiral galaxies (more negative $M$; recall that magnitude $M$ is
related to luminosity $L$ as $M=-2.5\log(L)$+constant) rotate {\em
faster}.}
\label{fig:tf}
\end{figure*}

What is the origin of this relation? Remember that the circular
velocity $V_c$ is determined by the enclosed mass,

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

Let us introduce a global mass-to-light ratio, \mol, as

\begin{equation}
\mol \equiv {M\over L}\,,
\end{equation}

then

\begin{equation}
V_c^2\propto \mol\, {L\over R}\,.
\end{equation}.

The mass-to-light ratio \mol\ will depend on the type of stars in the
galaxy (which determines $L$), and the amount of dark matter (which
mainly determines $M$). The surface brightness $I$ is the luminosity
per unit area, $I=L/(\pi\,R^2)$, or solving for $R$

\begin{equation}
R\sim \left({L\over I}\right)^{1/2}\,.
\end{equation}

Combining the last two equations gives

\begin{equation}
L\propto {V_c^4\over\mol\,I}\,.
\end{equation}

So the Tully-Fisher relation is reproduced, with $\alpha=4$, when the
surface brightness $I$ and mass-to-light ratio \mol\ are {\em
independent} of the luminosity of the galaxy. Remark that this is not a
derivation of the Tully-Fisher relation: we are just trying to
understand what is required for spirals to follow this relation. The
fact that \mol\ and $I$ are independent of $L$ implies that somehow
stars and dark matter are closely linked, or in other words: the star
formation history is largely determined by the mass of the dark halo of
the galaxy.

\subsection{The Faber-Jackson relation in ellipticals}
The Faber-Jackson law, Figure~\ref{fig:fj}, relates the velocity
dispersion $\sigma$ of the stars in an E-type galaxy with the total
luminosity $L$,

\begin{equation}
L\propto \sigma^4\,.
\label{eq:FJ}
\end{equation}

Compare with the Tully-Fisher relation for spirals, eq.~(\ref{eq:tf}):
the velocity dispersion $\sigma$ takes the role of the circular
velocity $V_c$.\\

Comparison of Figure \ref{fig:fj} with Figure \ref{fig:tf} shows that
the {\em scatter} in the Faber-Jackson relation is quite a bit bigger
than the scatter in the Tully-Fisher relation: at a given value of
$\sigma$, there is a range of $\pm 2$ magnitudes in $M_{\rm B}$ (i.e. a
factor 2.5 in $L$) for a given value of $\sigma$, versus a few tenths
of a magnitude for a given value of $V_c$ in the Tully-Fisher
relation. So ellipticals follow a similar relation as spirals, with
more luminous and hence more massive, ellipticals having a large
velocity dispersion, but the relation is not very tight.


\begin{figure*}
\center\resizebox{.9\textwidth}{!}{\includegraphics{faberjackson.pdf}}
\caption{B-band absolute magnitude $M$ for a sample of elliptical
galaxies plotted versus the velocity dispersion $\sigma$ (in
\kms). {\em Brighter} ellipticals galaxies (more negative $M_{\rm B}$;
recall that magnitude $M$ is related to luminosity $L$ as
$M=-2.5\log(L)$+constant) have a {\em larger} velocity dispersion.}
\label{fig:fj}
\end{figure*}

From all the parameters we can measure for an elliptical, the total
luminosity $L$, the velocity dispersion $\sigma$, the effective radius
$R_e$ and the surface brightness $I_e$ (which both enter in the de
Vaucouleurs profile, Eq.~(\ref{eq:r4})), only three are independent. So
we can try to obtain a tighter relation by introducing a second
parameter in Eq.~(\ref{eq:FJ}), for example $R_e$. If the galaxy is in
virial equilibrium, then its kinetic energy $M\,\sigma^2$ will be
proportional to its potential energy $M^2/R_e$, and so we expect

\begin{equation}
\sigma^2\propto {M\over R_e}\,.
\end{equation}

Introducing again the mass-to-light ratio, \mol\, this can be written
as

\begin{equation}
\sigma^2 \propto \mol {L\over R_e} \propto {L^{1+\alpha}\over R_e}\,,
\end{equation}

where I've assume \mol\ to depend on $L$ as
\begin{equation}
\mol\propto L^\alpha\,.
\label{eq:mol}
\end{equation}
The surface brightness $I_e\propto L/R_e^2$, hence

\begin{equation}
R_e\propto \sigma^{2/(1+2\alpha)}\, I^{-(1+\alpha)/(1+2\alpha)}\,.
\label{eq:fp1}
\end{equation}

\begin{figure*}
\center\resizebox{.9\textwidth}{!}{\includegraphics{fp.pdf}}
\caption{Scale radius $R_e$ versus $\sigma^{1.24}\,I_e^{-0.82}$ for the
same sample of Es plotted in Figure~\ref{fig:fj}. The galaxies follow
this relation with much less scatter then the Faber-Jackson relation in
Fig.\ref{fig:fj}.}
\label{fig:fp}
\end{figure*}

Figure~\ref{fig:fp} shows that

\begin{equation}
R_e\propto \sigma^{1.24}\,I_e^{-0.82}\,,
\label{eq:fp2}
\end{equation}

for the same sample of ellipticals plotted in Figure~\ref{fig:fj}. The
galaxies follow this relation with very little scatter. Comparing the
last two equations shows that if we assume $\alpha\approx 0.25$, then
$R_e\propto \sigma^{1.33}\,I_e^{-0.82}$ -- very nearly the same to what
the figure suggests.  Put differently, if the mass-to-light ratio of
ellipticals depends on luminosity as $M/L\propto L^{0.25}$, then we can
understand why ellipticals follow the relation plotted in
Figure~\ref{fig:fp} so tightly.

The relation Eq.~(\ref{eq:fp2}) is called the {\bf fundamental plane}
of elliptical galaxies. In four dimensional $L$, $\sigma$, $R_e$ and
$I_e$ space, galaxies do not occupy the whole space, but are restricted
to a 3-dimensional surface defined by relation (\ref{eq:fp2}).

\section{Tully-Fisher and Fundamental plane relations as standard
candles}

The Tully-Fisher relation Eq.~(\ref{eq:tf}) relates the absolute
luminosity $L$ of a galaxy to its circular velocity. Now suppose we
were able to measure the proportionality constant, by determining the
absolute luminosity for galaxies with known distance. Then 
by just measuring the circular velocity of a spiral galaxy, we could
infer its absolute luminosity, and hence from its apparent magnitude
obtain the distance, since

\begin{equation}
m-M=5\log(r)\,.
\end{equation}

Therefore, the Tully-Fisher relation can be used as a standard candle,
since it allows us to determine the absolute luminosity from the
distance independent quantity $V_c$.\\

The fundamental plane relation plotted in Figure~\ref{fig:fp} can also
be used as a standard candle: all we need to do is measure the velocity
dispersion of the stars $\sigma$ from a spectrum, and determine the
surface brightness (recall that the surface brightness $I$ is distance
independent). By putting the E galaxy onto Figure~\ref{fig:fp}, we then
find $R_e$. And so from the apparent size of the galaxy, we can
estimate its distance.\\

Both these methods are widely used, since measuring velocities is
relatively easy from a good quality spectrum, and can be done even for
faint and/or distant galaxies. And since the scatter in the relations
is small, we can obtain a relatively accurate distance estimate.

\section{Galaxy luminosity function}
From a big galaxy redshift survey, we can make a histogram of the
number of galaxies with absolute luminosity between $L$ and $L+dL$,
where $L$ is either the total (or bolometric) luminosity, or the
luminosity in some band, e.g. the B-band, per unit
volume. Figure~\ref{fig:sf} shows that the number of faint galaxies
increases like a power law, whereas there is an abrupt cut-off toward
the brighter galaxies. This behaviour is well captured by the {\bf
Schechter} function

\begin{equation}
{dN\over dL/L_\star} = N_\star\,\left({L\over
L_\star}\right)^\alpha\,\exp(-L/L_\star)\,.
\end{equation}

This function has three parameters, $L_\star$, $\alpha\sim -1$ and
$N_\star$. For luminous galaxies, it has an exponential cut-off for
galaxies brighter than $L_\star$. For less luminous galaxies, the
number increases toward fainter galaxies, as $L^\alpha\sim
1/L$. Finally, $N_\star$ is a normalisation constant, which determines
the mean number density of galaxies.

Values are $N_\star\approx 0.007$\Mpc$^{-3}$, the steepness of the
faint-end slope $-1.3\le \alpha \le -0.8$, and the exponential cut-off
sets in at $L_\star\approx 2\times 10^{10}\lsun$ in the B-band. From
Table \ref{table:MW}, we find that the total B-band luminosity of the MW
is about $1.8\times 10^{10}\lsun$, so the MW is just slightly fainter
than an $L_\star$ galaxy.

If the faint-end slope is steeper than $\alpha=-1$, then the total
number of galaxies per unit volume, $n=\int_0^\infty (dN/dL)\, dL$
diverges, because there are so many faint galaxies. In reality, this
does not happen of course, since there must be some finite number of
small galaxies. The mean luminosity density can be found from integrating

\begin{equation}
l=\int_0^\infty L\, {dN\over dL}\, dL\,.
\end{equation}

This is the mean luminosity per unit volume. Combining this with the
mean density of the universe as obtained in section \ref{sect:dmU}, we
can estimate the mass-to-light ratio for the Universe as a whole!

\begin{figure*}
\center\resizebox{.9\textwidth}{!}{\includegraphics{schechter.pdf}}
\caption{Histogram depicting the number of galaxies per \Mpc$^3$\ with
given luminosity $L$ in a given colour band. At magnitudes fainter than
-18, the number increases as a power in $L$, but for magnitudes
brighter than -22, the number of galaxies cuts-of exponential. The
drawn line is the best-fit Schechter function.}
\label{fig:sf}
\end{figure*}

Finally, Figure\ref{fig:lf} shows how the Schechter luminosity function
is composed of contributions from the various galaxy types, E, S0 and
the different types of spirals, and irregular galaxies. Top panel is
for a sample of galaxies away from big clusters, bottom panel refers to
the Virgo cluster. Here we recognise the density-morphology once more:
most of the brighter galaxies in the top panel are Spirals, whereas in
the cluster sample, E an S0s play a much bigger role. 

\begin{figure*}
\center\resizebox{.8\textwidth}{!}{\includegraphics{lfs.pdf}}
\caption{This plot shows how the galaxy luminosity function for
galaxies in a cluster (bottom panel is for the Virgo cluster) or
galaxies not in a cluster (\lq field galaxies\rq, top panel), is
made-up from the mix of galaxy types.}
\label{fig:lf}
\end{figure*}


\pagebreak
\section{Summary}
After having studied this lecture, you should be able to 

\begin{itemize}
\item describe the density-morphology relation for galaxies
\item explain what is the Tully-Fisher relation, and derive what it
implies for the mass-to-light ratio and surface brightness of spirals
\item explain the Faber-Jackson and Fundamental Plane relations for
Ellipticals, and derive what the fundamental plane relation implies for
the mass-to-light ratio of ellipticals
\item explain what is the luminosity function of galaxies and sketch
it.
\end{itemize}

\chapter{Active Galactic Nuclei}
Some galaxies have in their centres immensely powerful sources of
radiation, detected from the longest (radio) to the shortest
wavelengths (X-rays). We think these are powered by accretion of matter
onto super massive black holes (SMBH), which have masses ranging from
$10^6$ to $10^9\msun$. In fact, we think that {\em most} or even all
reasonably massive galaxies have a SMBH lurking in their centres, but only a
small fraction of these are \lq active\rq. The evidence that the MW has
its own SMBH is particularly impressive.\\

Where do these SMBHs come from? Are they really black holes
(i.e. objects with an event horizon) or are they just very massive,
extremely dense objects (MDO)? Why are some associated with strong
radio-sources (radio-loud AGN), but others are not? How do radio-loud
AGN power the immense radio-lobes we observe? What is the effect of the
AGN on the galaxy itself? Most of these questions remain without a
clear answer, even today.\\

I will start by discussing some of the observations of AGN. Surely,
this must be one of the most exciting topics in Astronomy, or even
physics. Then I'll take you through the evidence that these really are
black holes, but leave it up to you to decide whether you believe it or
not. My own view is that the evidence for super massive and very dense
objects is very convincing -- but I have not seen evidence for an event
horizon (yet).

\section{Observational properties}
Carl Seyfert reported in 1943 that a small fraction of galaxies has a
very bright nucleus, which is a source of broad emission lines produced
by atoms in a wide range of ionisation stages. These nuclei are
stellar in appearance.

After World War II, radio surveys discovered a very bright radio
source, called Cygnus A. Given the accurate radio-position of the
source, the optical counterpart was identified with a peculiar-looking
cD-galaxy, whose centre is encircled by a ring of dust.

The spectral lines in the optical spectrum of Cygnus A are redshifted
to $\Delta\lambda/\lambda_{\rm laboratory}=0.057$. In terms of a
Doppler shift, this corresponds to a recession velocity of 16000\kms.
From this, one can estimate the distance to Cygnus A\footnote{The
redshifting of the spectral lines is due to the expansion of the
Universe. By measuring the expansion rate, one can convert redshift to
distance, assuming a cosmological model.}, $d\sim 240\Mpc$, and hence
compute the implied luminosity of the radio source, which turned out to
be roughly ten times the {\em total} energy output (i.e. over all
wavelengths) of the MW.\\

In 1960, Thomas Matthews and Alan Sandage were trying to find the
optical counterpart of another bright radio-source, called
3C~48\footnote{The C stands for Cambridge -- this is the third
Cambridge radio-survey.}. They found a 16th magnitude star like object,
with a very curious spectrum, displaying broad emission lines that they
could not identify with any known element or molecule. Sandage put it
very scientifically: \lq The thing was exceedingly weird\rq. In 1963,
another such weird object, 3C~273 turned-up. Given they looked like
stars, in that they were point like (unlike galaxies which are
extended), they were called quasi-stellar objects, or QSOs.\\

In the same year, the Dutch astronomer Maarten Schmidt recognised that
the pattern of lines in 3C~273 was similar to the Balmer
series\footnote{The Balmer series is the series of electronic
transitions in an atom which starts with the H$\alpha$ line
($n=3\rightarrow n=2$), then H$\beta$ ($n=4\rightarrow n=2$) etc.} of
the Hydrogen atom -- but only if they were redshifted to the
(improbably large ?) value of $\Delta\lambda/\lambda=0.16$, implying a
recession velocity of $0.16c$ and a distance of 630\Mpc. The distance
to 3C~48 was even greater: a recession velocity of $0.367c$ and a
distance\footnote{The \lq velocity\rq\ of such distant objects is not a
real velocity in the sense that something is moving. The velocity
results from space expanding, and it is often better to not convert
this to velocity at all -- just use the cosmological model to find the
relation between the redshift $z=\Delta\lambda/\lambda$ and distance.}
of 1800\Mpc.\\

For a distance of $d=630\Mpc$ to 3C~273, the distance modulus
$m-M=5\log(d/10)\approx 39$. The apparent magnitude of 3C~273 is
$m=12.8$ (in the V-band), hence the absolute magnitude is
$M=-26$. Since the Sun's absolute magnitude $M_\odot=+5.$-ish, it
implies that 3C~273 is $10^{0.4\times 31}\sim 10^{12}$ times brighter
than the Sun, or 100 times as bright as the entire MW. Give or take a
factor of a few. What kind of object can be so small yet so
tremendously powerful?\\

Since then, many more QSOs have been discovered, we know several
hundred thousands of them by now. The current\footnote{March 2003. The
text book we're using, OC, states we know 5000 of them. And that we
know \lq several\rq\ with redshift greater than 4. This just shows you
how rapidly this field is evolving.}  record holder has
$\Delta\lambda/\lambda=6.3$. In terms of recession velocity, this is 6
times the speed of light (!), showing that interpreting such a redshift
in terms of Doppler shift is not a good idea\footnote{You may argue we
should use special relativity. In fact, we need general relativity to
make proper sense of this.}. The luminosity of the brighter QSOs can
range up to $10^5$ times the luminosity of the MW.


\section{The central engine, and unified schemes}
QSOs certainly are very bright. But are they {\em unusually} bright?
Yes. The argument is neat, and based on the {\em variability} of
QSOs. The luminosity of QSOs varies on a range of time-scales -- hours,
days, and months, depending on the wavelength at which you observe
them. But variations in the X-ray properties tend to have very short
time-scales, of order of minutes. Now this implies that the X-rays come
from a small region -- of the order of light minutes. To come to this
conclusion, all we need is to say that the information about the
variation has to be able to travel across the object that is producing
the emission, and the fastest the information can travel is the speed
of light, $c$. Hence the engine is small, of the order of light
minutes. Recall that the distance to the Sun is 8 light
minutes\footnote{To do this properly, we need to take the relativistic
effects into account. But if you do everything properly, you get a
similar answer to what we found.}. And so we need to find a small
engine, of size of order a solar system, that can produce as much
energy as all the stars in 100 galaxies {\em combined}.\\

The most efficient way we know of for producing energy is by mass
accretion onto a compact object\footnote{Actually, annihilation of
matter--anti-matter is even better!}. Recall from the stellar part of
this course, that stellar X-ray binaries produce their energies this
way. The estimated efficiency (in terms of the rest mass energy $mc^2$
of the fuel converted into radiation) is 10 per cent, which you can
compare to the second most efficient way, nuclear fusion, at 0.7 per
cent.\\

Recall also from those lectures that there is a maximum luminosity --
the Eddington luminosity -- that a spherical object in equilibrium can
produce. If the luminosity is higher, than the radiation pressure
becomes larger than the gravitational force, and the object can no
longer remain bound. Comparing the Eddington luminosity with the QSO's
luminosity, gives us a lower limit to the masses involved, of order
$10^8\msun$. Given such a small size, and such a large mass, Donald
Lynden-Bell and Martin Rees suggested that accretion onto a massive
black hole must be the energy source powering QSOs.\\

There is a problem though. A black hole is characterised by just two
numbers: it's mass, and it's spin (i.e. it's rotational energy). But
there is a wide variety of QSOs out there. For example, some have jets,
and strong radio sources, others don't. The time variability is by no
means regular. How can that be, if there is only two parameters that
characterise the engine?\\

The gas is thought to accrete onto the SMBH by passing through an
accretion disk, where it has to loose enough angular momentum to be
able to accrete. This introduces a third parameter: the orientation of
the disk with respect to the observer. The most popular unification
schemes\footnote{i.e. theories that try to explain QSO activity in terms
of a single model -- accretion onto a SMBH.} suggest that orientation
effects determine many of the observed properties of the QSO. But there
is by no means a simple answer to understand the wide variety of
properties. 

\section{Evidence for a SMBH}
I\footnote{I will follow a nice review paper by Luis Ho, see\\
http://nedwww.ipac.caltech.edu/level5/March01/Ho/Ho.html for the full
text.} am going to take you through several lines of evidence that QSOs are
indeed powered by a SMBH. You may think that the fact that we need an
engine the size of the solar system that produces as much energy as 100
galaxies together, would seem enough of a proof on it's own.\\

So let me discuss just one alternative: suppose there is a dense
cluster of stars in the centre of a galaxy hosting a QSO, and what we
are seeing is the combined effect of several tens of SNe explosions. A
single SN at it's peak is about as bright as a whole galaxy, so we need
about 100 SNe and we're done. This model has withstood for several
decades, but it does have some pitfalls. For example, we don't know how
such SNe would generate these tremendous radio-lobes that some QSOs
have. But then, we don't know how a SMBH does that either. As Luis Ho
puts it: \lq our confidence that SMBHs must power AGN largely rests on
the implausibility of alternative explanations\rq. Most of the
arguments in the next sections just suggest the presence of a very
massive, dense object (MDO) in the centre of galaxies (not just in
galaxies harbouring a QSO -- in fact, some of the best evidence is in
galaxies which do not have an active QSO), but not really require the
MDO to be a {\em black hole}, i.e. an object with an event horizon.

\subsection{Photometry}
The density of stars near a MDO will rise very sharply in the region
where the MDO dominates the mass. This basically comes from Jeans'
equations expressing hydrostatic equilibrium, in the form

\begin{equation}
{G\,M_{\rm DO}\over r^2}=-{\nabla n\sigma^2\over\rho}\,.
\end{equation}

{\em If} the velocity dispersion does not rise very rapidly,
$\sigma^2\approx$ constant, then the density must rise very rapidly, 
$d\ln(\rho)/dr\propto -1/r^2$ hence $\rho\propto \exp(1/r)$. So a rapid
increase in density of light toward the centre could signify the
presence of a SMBH.

However, the region where the MDO dominates may be very small, of order
$r_{\rm MDO}\approx {GM_{\rm DO}/ \sigma^2}$. As viewed from Earth for
a QSO at distance $D$, the angular extent will be $\approx 1\arcsec
(M_{\rm DO}/2\times 10^8\msun)\,(\sigma/200\kms)^{-2}\,(D/5\Mpc)$. Such
a small extent is hard to see from the ground (due to atmospheric
seeing), and Hubble Space Telescope (HST) will be needed.

Now some galaxies do indeed show a dramatic increase in brightness -- a
cusp -- toward the centre, even at HST resolution. Unfortunately, it is
also known that the velocity dispersion $\sigma^2$ is not constant. So
we cannot claim this proves the presence of a MDO -- it's a good
indication though.

\subsection{Stellar kinematics}
The previous method was not perfect, since we did not take into account
that $\sigma$ may vary. Now close enough to the MDO where it dominates
the mass, we should be able to detect the Keplerian rise in velocity
dispersion $\sigma\propto r^{-1/2}$. So people tried to determine
$\sigma(r)$ close to the centre.

However again there is a loop hole: $\sigma(r)$ can grow as well in
case the velocity distribution becomes anisotropic, e.g. in spherical
coordinates $\sigma^2_r\ne\sigma^2_\theta\ne\sigma^2_\phi$. Again using the
Jeans' equations, the radial variation in mass can be related to the
stellar velocity dispersion as\footnote{I just state the result and
don't expect you to derive it.}

\begin{eqnarray}
M(<r) &=& {V^2\,r\over G} + {\sigma^2_r\,r\over G}\nonumber\\
&\times&\left[-{d\ln\nu\sigma^2_r\over
dr}+({\sigma^2_\theta+\sigma^2_\phi\over\sigma^2_r}-2)\right]\,.
\end{eqnarray}

$V$ is the circular velocity. Clearly, if we were allowed to {\em
choose} the three components of $\sigma^2$, we could balance a large
increase in density with the velocity tensor becoming more
anisotropic, even in the absence of a MDO.

Observed galaxies do indeed show a large increase in $\sigma$ close to
the centre, but we cannot rule out that this is due to anisotropic
velocities, and not to the presence of a MDO. So once more, the
evidence is a bit circumstantial.

\subsection{Stellar kinematics in the MW centre}
The MW centre has a weak AGN in its centre, identified with the radio
source Sagittarius A. 

Over the past couple of years, Genzel and collaborators measured the
kinematics of stars close to the MW centre from near-IR
spectra\footnote{Recall the need to go to the IR: there is a lot of
absorption by dust toward the MW centre but IR-light undergoes far less
absorption.} They also determined very accurate positions of the
stars. By doing this over several consecutive years, they were able to
measure not just the 3D proper motions of the stars, but also measure
{\em accelerations}. In fact they were able to follow one star over
nearly a whole orbit, when it swung passed the centre at velocities up
to 1000\kms.

Using Kepler's laws, they were able to put very stringent limits on the
mass of the central concentration, $M_{\rm DO}\approx 2.6\times
10^6\msun$, in a region smaller than 0.006\pc, implying an enormous
mass density higher than $2\times 10^{12}\msun/\pc^3$. This leaves
almost no room to escape the conclusion that the MDO is indeed a
SMBH.

The presence of the large mass is also supported by the presence of
other very high velocity stars. From that, and from the fact that the
putative centre, i.e. the radio source Sagittarius A itself does not
seem to move, Genzel derived a lower limit to the density of $3\times
10^{20}\msun/\pc^3$. 

Incidentally, they also showed that the velocity dispersion remains
nearly isotropic, and so unless the MW is peculiar, this then
strengthens the case for SMBHs in other galaxies, from the arguments in
the previous section.

\subsection{Methods based on gas kinematics}
Unlike the situation for stars where anisotropy clouds the
interpretation, gas kinematics is much easier to interpret if the gas
is in Keplerian motion\footnote{Whereas stars can freely move through
one another, gas motion must be ordered not to produce shocks.}.  There
are two caveats, however. There may be non-gravitational forces,
magnetic fields for example. Second, there is no {\em a priori} reason
for the gas to be in Keplerian motion. So we need to lucky then.\\

{\bf Optical emission lines} Several examples of nuclear disks around
centres of galaxies were found by HST. For example for M87 (the big
elliptical in the Virgo cluster), HST measurements found a Keplerian
rotation curve in the disk with a velocity of 1000\kms\ at a distance
of 19\pc\ from the centre\footnote{Don't get confused with evidence for
dark matter from {\em flat} rotation curves. Here we are looking close
to the centre, whereas for rotation curves we were looking at large
distances, $\sim 20\kpc$ from the centre.}. This implies a MDO with
mass $\approx 2.4\times 10^9\msun$. Similar evidence for a MDO now
exists for other galaxies as well.\\

{\bf Maser emission} Some AGN have luminous maser sources close to the
centre. Radio observations of these maser lines show that the masers
are in nearly Keplerian rotation, and from the speed one can place
tight constraints on the mass density of the MDO, $> 5\times
10^{12}\msun$.\\

{\bf Reverberation mapping} Another elegant way to determine the
extent of an AGN is \lq reverberation mapping\rq. Many of the emission
lines seen from an AGN are due to the light from the central source
being reprocessed by the surrounding material. As the central source
varies in luminosity, the emission lines follow, but with a time-lag
that corresponds to the light-travel time between the central source
and the line-emitting gas. So by measuring the time-lag, we can
estimate the size of the light-emitting region.

\subsection{Profile of X-ray lines}
The Fe K$\alpha$ line at 6.4keV is a common feature in the X-ray
spectra of AGN. It is thought to arise from fluorescence of the X-ray
continuum off cold material in the surroundings, possible associated
with the accretion disk around the SMBH.

This line is produced very close to the SMBH, and it exhibits Doppler
motions approaching relativistic speeds, $\sim 10^5\kms\approx
0.3c$. Now crucially, the line-profile displays asymmetries consistent
with a {\em gravitational redshift}. The photons lose energy as they
climb out of the potential well near the black hole. In doing so, their
wavelengths become longer -- a \lq gravitational redshift\rq.

For the AGN galaxy MCG-6-30-15, the best fitting accretion disk has an
inner radius of 6 Scharzschild radii, i.e. {\em very} close to the
event horizon. A similar analysis has now been performed for other AGN
as well. {\bf The shape of the line probably provides the best evidence
to date for the existence of a SMBH.} However, other mechanisms for
generating the line profile are possible, but implausible. With better
data, detailed modelling of the profile even has the potential to
determine the spin of the SMBH.

\section{Discussion}
So are you convinced of the existence of SMBH? I think the evidence
from stellar motions around the MW centre is truly impressive. The Fe
K$\alpha$ line comes very close to convincing me as well. But I think
we really need to see relativistic motions or phenomena originating
close to the Schwarzschild radius to clinch the matter.

So let me summarise. We have evidence that probably most, if not all,
galaxies have a SMBH in their centres. It also transpires that the mass
of the SMBH seems to correlate with the mass of the galaxy: the more
massive the galaxy, the more massive the SMBH. What does that tell us
about the formation of the galaxy or of the SMBH? We don't know (yet!).

Also, if most galaxies contain a SMBH, then why are there so many times
fewer QSOs than galaxies? Indeed, QSOs are rare, so only a very small
fraction of SMBH is active at any one time. The best guess is that
there is no gas accreting onto those SMBHs: these inactive QSOs are
starved for food. Still it may be a bit surprising that so many black
holes are so very black -- some gas is lost by the stars close to the
centre, which somehow must be accreted onto the black hole {\em
without} producing much radiation.

And finally: if a star ventures too close to a SMBH, then it may be
torn apart as a result of the tidal forces from the SMBH. This could
lead to a very bright flare, which would last for of order months. The
realisation that SMBHs may be quite common provides fresh motivation to
search for such flares. Upcoming missions designed to detect SNe will
find these flares -- if they exist.

\pagebreak
\section{Summary}
After having studied this lecture, you should be able to 

\begin{itemize}
\item describe observational properties of AGN.
\item describe briefly the current paradigm for energy generation in
AGN (mass accretion through a disk onto a SMBH)
\item describe three arguments that suggest the presence of SMBHs in
galaxies.
\end{itemize}


\chapter{Gravitational lensing}
According to the theory of General Relativity, a massive object
distorts the space around it. For example, the distortion in space due
to the the Sun makes the planets go around it. When light travels
through such a distorted space, it also gets deflected.\\

Therefore, in general, the gravitational field generated by a mass
distribution will distort the images you see of background sources,
pretty much like the hot air above a road distorts the view behind
it. This is in fact a good way to think about this phenomena: it is as
if space has an index of refraction which varies in space, because the
gravitational field also varies in space. (Whereas it is of course the
varying temperature that causes the refractive index to vary above the
hot road.) This effect is called \lq gravitational lensing\rq, although
\lq gravitational miraging\rq\ might be a better description since the
distortion is more \lq blurring\rq\ (like the hot air) than focusing
(like a proper lens).\\

A very good introduction to the subject is the recent review by Joachim
Wambsganss\footnote{see
http://www.livingreviews.org/Articles/Volume1/1998-12wamb}. Scientists
are not very good at reconstructing the history of a subject, but
Joachim claims that a German physicist in 1804 was the first to suggest
that gravitational lensing might be observable, for example in terms of
the deflection of star light passing close to the Sun.\\

There are several possible observable effects of gravitational lensing:
(1) deflection: the image position is displaced from the source
position, (2) amplification: the image may be become brighter or
fainter, (3) distortion: if the source is extended (for example lensing
of a galaxy by a galaxy cluster), then lensing may distort the image,
(4) multiple images: in general a single source will be imaged in
multiple images. All these effects have been observed.\\

In general, the deflection and amplification in real systems is usually
very small, therefore our view of the Universe is not greatly distorted by
gravitational lensing. But occasionally, some objects are lensed quite
strongly. We'll start by deriving a very simple -- but unfortunately
not quite correct -- expression for the deflection angle, but also
state the correct result. Then we'll be armed with enough understanding
to look at some applications of lensing, and you'll realise the
tremendous potential of this technique, on a really wide variety of
fronts, from constraining the nature of dark matter, via measuring the
masses of galaxy clusters, to detecting planets and constraining models
of stars.

\section{The lens equation}
\subsection{Bending of light}
Suppose a particle with mass $m$ and velocity $v$ flies with impact
parameter $b$ past an object with mass $M$. The gravitational pull of
$M$ will deflect $m$, and give it a component $v_\perp$ perpendicular
to its initial $v$. For the case $m=M$, this is in fact the situation
we envisaged when deriving the relaxation time in
Section~\ref{sect:relax}. There we found that (see Eq.~\ref{eq:dv})
when the deflection angle $\alpha$ is small,

\begin{equation}
v_\perp = {2GM\over b\,v}\,.
\end{equation}

Convince yourself that this is the correct result also in case $m\ne
M$. Finally, since $\tan(\alpha)=v_\perp/v$ and when $\alpha$ is small
so that $\tan(\alpha)\approx \alpha$,

\begin{equation}
\alpha = {2GM\over b\,v^2}\,.
\label{eq:lens0}
\end{equation}

Now here's the trick: the expression for $\alpha$ does not depend on
the mass $m$ of the particle being reflected, only on the mass $M$ of
the deflector. So we might be tempted to put $m=0$, $v=c$ and claim
this is the angle by which light will be deflected as well.\\

Applying this to starlight passing close to the sun, hence using
$M=\msun$, $R=\rsun=7\times 10^5\km$, we get $\alpha=0.83\arcsec$ --
so $\alpha\ll 1$ is certainly satisfied. This was actually the value
Einstein found in 1911, though one Johann Soldner already obtained it
almost a century earlier.\\

Curiously, the answer is wrong! Once Einstein formulated his theory of
General Relativity, he did the problem again, and found that the
deflection angle is {\em twice} our \lq Newtonian\rq\ value. So the
correct deflection angle for light is

\begin{equation}
\alpha = {4GM\over b\,c^2}\,.
\label{eq:lens1}
\end{equation}

and it was of course a great vindication of the theory when Eddington
verified this during a solar eclipse in 1920. Note that the deflection
angle increases with $M$ and decreases with $b$ -- the closer you pass
to the more massive an object, the bigger is the deflection.\\

In the case of the Sun, the deflection is so small that you might think
it if of little use in Astronomy. This was actually true for a long
time, but recently it has really taken off as a new way of looking at
the Universe. Before I show you some applications, we need to do a
little bit more maths first.


\subsection{Point-like lens and source}
\begin{figure*}
\center\resizebox{.8\textwidth}{!}{\includegraphics{evafig1.pdf}}
\caption{The source (S), lens (L) and observer (O) are all
aligned. Light from S is deflected by an angle $\alpha$ toward
O. Because of symmetry, O sees the source S as ring of radius $R_E$
centred around L.}
\label{fig:er}
\end{figure*}

\begin{figure*}
\center\resizebox{.8\textwidth}{!}{\includegraphics{evafig2.pdf}}
\caption{The source (S), lens (L) and observer (O) are now
misaligned, and O sees two images of S, one of which is depicted here while
the other one lies on the other side of L.}
\label{fig:er2}
\end{figure*}

Suppose the light source (S), lensing mass $M$ (L) and observer (O) are
exactly aligned, as in Fig.\ref{fig:er}. Also assume they are all
point like (i.e. not extended like for example a galaxy). Light rays
from S are deflected by L over an angle $\alpha$ toward O. Because of
symmetry of the situation, O sees a ring of light around L: an Einstein
ring. Of course the figure is not to scale: $\alpha$ should be small if
we want to apply the lensing equation.\\

Introducing the distances between observer and lens $D_{OL}$, observer
and source $D_{OS}$, and lens and source $D_{LS}$, we can compute $R_E$
and the angle $\theta_E$ under which O sees the ring. Since
$\psi+\theta_E=\alpha$ we get, using the lensing equation and the figure,

\begin{eqnarray}
\theta_E &=& {R_E\over D_{OL}}\nonumber\\
\psi     &=& {R_E\over D_{SL}}\nonumber\\
\alpha   &=& {4GM\over R_E\,c^2}\,,
\end{eqnarray}

and hence
\begin{equation}
R_E^2 = {4GM\over c^2}\,{D_{OL} (D_{OS}-D_{OL})\over D_{OS}}\,.
\end{equation}

If O, L and S are not aligned, as in Fig.\ref{fig:er2}, then O does not
see a ring. Denote the angular position of S from the line OL by
$\theta_S$, and the angle between S and its image I by $\theta_I$. Then
from the figure you see that
$\alpha(D_{OS}-D_{OL})=D_{OS}(\theta_I-\theta_S)$. A bit of juggling
and using the lensing equation gets you to:

\begin{equation}
\theta_I^2-\theta_I\theta_S = \theta_E^2\,.
\end{equation}

This is a quadratic equation for $\theta_I$ for a given $\theta_S$ and
so there will be {\em two} images, at positions $\theta_{I,\pm}$.\\

Often, the change in position will be very small and not
observable. But besides being lensed, an image will also be {\em magnified}
(or de-magnified). So when the displacement is too small to see two
images, you'll only see one but magnified by total magnification
$A=A_++A_-$ of each image separately. Interestingly, for a small
fraction of peculiar alignments, $A$ may diverge, and so potentially
you can get very large amplifications.

\subsection{More realistic lensing}
In the previous section we assumed that both lens and source where just
point masses. When the lens is extended, you can get multiple images
(3,5,7,...), depending on the impact parameter and the mass
distribution of the lens. If the {\em source} is extended as well, then
different parts of the source may be deflected and amplified in various
ways, and so the image may be highly deformed. We are now in a position
to look for applications. To wet your appetite, let me show you two
beautiful examples.\\

Figure \ref{fig:EinsteinRing} shows an Einstein ring. The lens is
probably an elliptical galaxy, and the source a more distant
spiral. Because of the near perfect alignment, the spiral has been
imaged into a ring.\\

Figure \ref{fig:EinsteinCross} shows how a nearby spiral galaxy has
lensed a very distant background quasar and produced multiple images of
it, in this case five.

\begin{figure*}
\center\resizebox{\textwidth}{!}{\includegraphics{gravring.pdf}}
\caption{Hubble Space Telescope image of an Einstein ring, where a
fortuitous alignment has caused an intervening elliptical galaxy to
lense a background spiral galaxy into a ring.}
\label{fig:EinsteinRing}
\end{figure*}

\begin{figure*}
\center\resizebox{\textwidth}{!}{\includegraphics{EinsteinCross.pdf}}
\caption{Hubble Space Telescope image of the \lq Einstein cross\rq,
where a distant quasar has been lensed into multiple images by an
intervening spiral gaalxy.}
\label{fig:EinsteinCross}
\end{figure*}

\section{Micro-lensing}
\begin{figure*}
\center\resizebox{.8\textwidth}{!}{\includegraphics{microlensing.pdf}}
\caption{Luminosity of a star in the LMC in the blue (top) and red
(bottom) as a function of time. Lensing causes it to brighten around
day 420.}
\label{fig:lmc_lensing}
\end{figure*}
If\footnote{I took this part, including the figures, from W Evans' 2003
review, see astro-ph/0304252} both source and lens are stars, then we
can use our results from the point-like models. We expect very small
deflections, but potentially large magnifications. We only get large
magnifications for chance alignments.

So as foreground stars move in front of background stars -- for example
a star in the Milky Way disk moves in front of a star in the bulge --
it can occasionally amplify the bulge star by a large amount. However,
since the alignment has to be very precise to get a large
amplification, the bulge star will not be amplified for a long time,
before the disk star moves away from the ideal lensing
alignment.

Because the alignment has to be so good, to have a chance of seeing
significant amplification, you need to look at millions of stars. So
people observed millions of stars toward the LMC and toward the bulge,
to find stars that brightened by a lot over a few days. This phenomena
is called {\em micro-lensing} (because the deflection angle is so
small, $\sim 10^{-6}\arcsec$).

There is one crucial aspect of lensing that we haven't paid attention
to yet: the deflection and amplification is {\em independent of the
wavelength of the light}, as you can see from Eq.~(\ref{eq:lens1}).
This is important if you want to detect micro-lensing, because the
brightness of many stars varies anyway, because they are variable
stars. But in general, the change in brightness for a variable star
depends on the colour you observe it in. But the change in brightness
due to micro-lensing is colour independent. This makes it possible to
distinguish variable stars from lensed stars. In addition, the light
curve -- luminosity as function of time -- is very easily computed in
terms of $M$ and the relative velocity of L and S. Combined, it allows
one to confidently detect micro-lensing. Figure~\ref{fig:lmc_lensing}
shows an example of micro-lensing of a star in the LMC, which brightens
considerably around day 420, equally much in the blue as in the
red.\\

Although we so far only considered lensing by stars, in fact any
massive object will produce lensing. This has been exploited by the
{\bf MACHO collaboration}\footnote{http://wwwmacho.mcmaster.ca/} to
constrain the nature of the dark matter in the halo of the Milky
Way. Indeed, suppose dark matter consisted of some type of massive
objects, like black holes, or Jupiter-like objects, or some type of
{\bf MA}ssive {\bf C}ompact {\bf H}alo {\bf O}bject in general. Since
we know that the outer parts of the Milky Way halo is dominated by dark
matter, there should be many such objects between the Sun and the
LMC. If they exist, then we can detect them by their lensing signal.\\

To detect such MACHO's, the collaboration imaged millions of stars in the
LMC for four years, and indeed found several lensing events like the
one pictured in Fig.\ref{fig:lmc_lensing}. However, it is now believed
that in each case, the lens is actually a normal star in the LMC, and
not some exotic kind of matter. As a consequence, this experiment has
now ruled-out a wide range of dark matter candidates. If the dark
matter is some type of elementary particle, it will not lens LMC stars
(well, it will, but $M\rightarrow 0$!), so that is still an option.\\

The future of micro-lensing is very bright. Suppose for example that
the lens is a binary star -- then the light curve of the lensed object
may be much more complicated, as the source may be lensed by each
component of the binary in turn. Such events have already been detected
toward the bulge. This is also a very promising way for detecting {\em
planets} around distant stars, and planned satellite missions will
undoubtedly detect planets in this way.

\section{Strong lensing}
\begin{figure*}
\center\resizebox{\textwidth}{!}{\includegraphics{abell2218rv.pdf}}
\caption{Hubble Space Telescope image of the rich cluster of galaxies
Abell 2218. The large galaxies in the image are all massive elliptical
galaxies in this galaxy cluster. Tens of radial arcs are images of
background galaxies that have been lensed and distorted by the dark
matter in the cluster.}
\label{fig:abel2218}
\end{figure*}
The effect of lensing increases with the mass of the lens. So can we
use galaxy clusters with masses $\sim 10^{15}\msun$ as lens? Yes we
can, and we do get large deflections of several tens of arc seconds up
to several arc minutes.\\

Figure~\ref{fig:abel2218} is an HST picture of a rich cluster of
galaxies -- you can see many massive elliptical galaxies in the
cluster, as you'd expect. Because of the superb image quality of HST,
you also see tens of radial arcs, centred on the cluster centre. These
are images of galaxies that lie far behind the cluster, that have been
lensed by the dark matter in the cluster. Clearly, their image have been
distorted a lot by the lensing process.\\

Several of these lensed galaxies have multiple images, i.e. you see the
same galaxy at various positions in the cluster. In addition, some
images of galaxies may have been magnified by a lot, sometimes as much
as a factor of 30. Because of this, the cluster really acts as a
telescope, allowing you to study faint and distant galaxies, using the
dark matter in the cluster to increase the brightness of the galaxy
light. Astronomers are now deliberately targeting foreground clusters
in order to study better the faint galaxies behind it, that would not
be bright enough to observe in the absence of a lensing.\\

The positions and magnifications of these lensed galaxies can also be
used to infer the distribution of matter in the cluster, since it is
that which causes the lensing. This is a very important application of
lensing. Recall how we inferred the presence of dark matter in clusters
so far. Assuming virial equilibrium, we estimated the mass in the
cluster from the velocities of the galaxies. Another way was to assume
that the hot gas in the cluster was in hydrostatic equilibrium, and
obtain the mass from the X-ray observations of the hot gas. In either
case, the computed masses were far higher than what we could account
for in either stars or gas -- hence our claim that galaxy clusters
contained a lot of dark matter.\\

But we had to assume something -- virial equilibrium for the stars,
hydrostatic equilibrium for the gas. The great thing about
gravitational lensing is that all we care about is the amount of mass
present: we do not need to assume any type of equilibrium, since the
velocity of the mass is irrelevant. The lensing mass and X-ray masses
are in reasonably good agreement for the still relatively few clusters
studied so far, and so this is again very convincing evidence for the
presence of dark matter.\\


\section{Weak lensing}
The arc lets seen in clusters arise because the image of the background
galaxy is strongly deformed. Equation~(\ref{eq:lens1}) shows that as we
look at galaxies further from the projected centre of the cluster --
and hence as the impact parameter $b$ increases -- the distortion will
become smaller and smaller. Eventually, we won't be able to detect the
deformation for a single galaxy any more, but we could look at many
galaxies in a given small patch of sky, and try to determine whether
perhaps they are all slightly elongated in the same direction, as would
be the case if there were a lot of mass nearby. This is called {\em
weak lensing}: you try to identify the presence of a large mass in a
given direction, from the fact that galaxies nearby in projection, tend
to be elongated in a same direction. Astronomers are now searching
through stacks of images trying to find whether may be there are some
really massive objects that have no galaxies associated with
them. Another use of weak lensing is just try to map directly the
distribution of matter in the Universe, irrespective of whether it
emits light as well.

\pagebreak
\section{Summary}
After having studied this lecture, you should be able to 

\begin{itemize}
\item derive the Newtonian lensing equation (\ref{eq:lens0})
\item explain why alignment produces an Einstein ring, and derive its
radius.
\item explain how micro-lensing works and has been used to search for
massive compact halo objects in the Milky Way halo
\item explain how gravitational lensing has been used to estimate the
mass of galaxy clusters, and hence infer that they contain dark matter.
\end{itemize}

\tableofcontents
\end{document}
