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\begin{document}
\title{\Huge{Stars and {\bf Galaxies}}}
\author{Tom Theuns\\
			Office 307\\
        Institute for Computational Cosmology\\
			Ogden Centre for Fundamental Physics\\
			Durham University\\
         and\\
         University of Antwerp\\
	      Campus Drie Eiken\\
			tom.theuns@durham.ac.uk}
\date{2006-2007}
\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 galaxy properties and of some of the physical processes
that we think shape galaxies. 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. Suggestions on material included in this course, its
presentation, and lists of errors in the notes, 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. Given five differences between
football and basketball. 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 quite
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
there.
\newpage
\chapter{Introduction}
\section{Historical perspective}
Galaxies were discovered in the 20th century. This is surprising, since
you can see many of them on a clear night with the naked eye. They
differ from stars in that they are {\em extended}. For example, the
nearby {\it Andromeda galaxy} has an extent of several degrees
(diameter of full moon is 0.5 degrees). And the {\it Large} and {\it
Small Magellanic clouds} in the Southern hemisphere, are even
bigger. 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 small increase in surface brightness over the night sky.

They were called {\it nebulae} to distinguish them from stars. However,
there are other extra-solar objects which are also extended on the sky:
Planetary Nebulae, star clusters, 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 were the
relations between them. And so many of the catalogues of nebulae, for
example 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 unrelated objects.

Immanual Kant, the German philosopher, suggested that the reason we see
a faint band of stars across the night sky -- usually called the {\it
Milky Way} (MW for short) -- is because most MW stars, as well as the
Sun, lie in a disk. And so, when you look out at night in the plane of
the disk, you see a faint band of light. Galileo already had resolved
this band of light into stars. Kant claimed that most of the other
nebulae were just other galaxies. It took until the 1920s before this
hypothesis was proved correct. The historical introduction in BM is
well worth reading.

As a final point: why 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) humans can {\em see} stars, we are
naturally inclined to study them. But, for example, although a {\it
cluster of galaxies} contains thousands of galaxies, they are by no
means its main constituent. Most of the {\em mass} is in dark matter --
which we cannot see at all. But most of the baryons are 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 you are an amateur
astronomer. http://www.aao.gov.au/images.html/ shows professional
images. Such professional pages (for example the {\it Hubble Space
Telescope} (HST) archive) tend to contain a large fraction of unusual
galaxies -- simply because those are the ones that astronomers tend to
study.


\subsection{Vocabulary}
{\bf Colours.} These images are taken through {\it broad-band filters},
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 and \R,
and so each individual exposure might as well be in black and white,
the relative 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. Images taken trough different filters are then combined
to produce a {\it colour image}, scaling the different bands trying to
mimic the colour your eye would see.

Sometimes though, one of the filters used is a {\it narrow-band
filter}, for example H$\alpha$ or O[III]. These filters block all light
which is not in a narrow range around the hydrogen H$\alpha$
transition, or a transition in doubly ionised oxygen. Combining such a
narrow-band image with images made with others filters produces a {\it
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 -- for example H$\alpha$ emission. The really
pretty colour pictures of Planetary Nebulae are false colour images.\\

{\bf Luminosity, flux and surface brightness}

A star emits a certain amount of energy per unit time. For example the
Sun has a {\em luminosity} $L=1L\odot=3.4\times 10^{26}$W. This
quantity is not observable however: what we can measure is how much of
this energy we receive, per unit time, per unit surface area (assuming
you put the surface perpendicular to the radiation!), at a given
distance from the source. The observable quantity, energy received per
unit time per unit of surface area, is called {\rm flux}. Clearly the
flux will depend both on the luminosity of the star, and on its
distance. How?

Consider a sphere of radius $r$, with the star at the centre. Since the
star will distribute all of its energy equally over the surface of the
sphere, the flux $F=L/(4\pi r^2)$. (As an exercise, find the unit of
flux).

It is the {\em flux} of the star that determines how bright it appears
to be. Indeed, star A may be brighter than star B, either because it is
more luminous, $L_A\gg L_B$, or because it is nearer to us, $r_A\ll
r_B$.

Although $F$ is in principle observable, in practise astronomers
express brightness in terms of magnitudes $m$, where the apparent
magnitude $m=-2.5\log_{10} (F)+C$, where $C$ is some constant that
depends on the magnitude system used. Since the $F$ depends on
distance, so does $m$, it is not an intrinsic quantity of the star --
hence \lq apparent magnitude\rq. The absolute magnitude, $M$, is the
apparent magnitude when $r=10$pc.

Since a galaxy contains many stars, we can also talk about its
luminosity, and the flux we receive from it. However, because galaxies
are spatially extended\footnote{{\it Very} nearby stars can be resolved
with special techniques.}, we can also try to measure what fraction of
the flux comes from the centre of the galaxy and what fraction comes
from the outer parts, say. This gives rise to the term {\em surface
brightness}.

Consider a small patch of galaxy, namely that part contained in a
(small) solid angle $d\Omega$. Observationally we can measure the flux,
$dF$, of light, coming from the part of the galaxy contained within
$d\Omega$.  The quantity $dF/d\Omega$ is called {\em surface
brightness} (SB), and so it has dimensions
energy/time/area/steradian.\footnote{A solid angle is usually expressed
in steradians, or arc sec$^2$, see
http://whatis.techtarget.com/definition/0,,sid9\_gci528813,00.html.}

The surface brightness of a galaxy does not depend on its distance. To
see this, assume that the patch of galaxy contained within $d\Omega$
contains $N$ stars, with mean luminosity $L$, when it is at a distance
$r$. The surface brightness SB=$N\,L/(4\pi r^2)/d\Omega$. Now increase
the distance to the galaxy by a factor of 2. The flux from each
individual star will decrease by a factor $2^2$, since $F\propto
1/r^2$. However the number $N$ of stars within $d\Omega$ will increase
by a factor $2^2$, since the physical size of the patch enclosed by
$d\Omega$ will double when $r$ doubles. Hence SB is distance
independent\footnote{This is only true for relatively nearby
galaxies. On cosmological distance the SB decreases with distance
because of the redshift.}, therefore whereas with brightness there was
an intrinsic quantity (luminosity), and an observable one (flux),
there is only one surface brightness.

A galaxy's SB depends on its distribution of stars.
Assume a face-on galaxy at distance $r$ has a surface density $\Sigma$
of stars (in stars/pc$^2$, say), all with the same luminosity $L$. The
surface area of galaxy $dS$, contained within a solid angle $d\Omega$,
is $dS=d\Omega\,r^2$. Therefore the number, $dN$, of stars within
$d\Omega$ is $dN=\Sigma\,dS=\Sigma\,d\Omega\,r^2$. The flux received from
a single star is $L/(4\pi r^2$, and therefore the flux received from
all $N$ stars within $d\Omega$ is $dF=(L/(4\pi r^2))\times
(\Sigma\,d\Omega\,r^2)\,=\Sigma\,L\,d\Omega/(4\pi)$. The surface
brightness is the flux per unit solid angle, is $df/d\Omega=\Sigma\,L/(4\pi)$ is
distance independent, as we saw before.


\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, E galaxies for short)
\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 $n=10(1-b/a)$, 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). 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, S galaxies for short)
\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 inferred the {\em stellar} properties from the fact that E galaxies
tend to be redder than S. This suggested that they lack the young
massive stars that produces most of the blue light. Hence our inference
that the stellar population is likely to be older, with the massive
stars that have short life times already dead. We will see later that
the SB distribution, $SB(r)$, are quite different for these two types.

These are the archetypal properties of such systems, of course there
are exceptions. For example, some ellipticals do have dust, as
well as evidence for recent star formation. Some even have a small
disk. Given that some spirals have a very large bulge, and some
ellipticals 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 in the first place.

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.

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

There are many schemes for classifying galaxies. A good scheme should
be unambiguous, and ideally have some physics behind it. Classification
along 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
(S), along the bottom are barred spirals (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 the filter
you use to observe them in.

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.

Yet 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.

Most smaller galaxies are of the Irregular type: no disk, no nice
spherical distribution of stars either, but 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 some properties of Globular 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 Know where to find web-pages with images of galaxies.
\end{itemize}
\newpage


\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 Milky Way. 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 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} There are two types of star clusters, {\it
i.e.} groups of $10^3$s to $10^5$s of stars. {\bf Globular Clusters}
(GCs) are spherical, gravitationally bound systems of $10^5-10^6$ old
stars. Most galaxies contain 10s to 1000s of GCs, spherically
distributed around the parent galaxy. The MW has about 150 GCs. 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. So they differ from GC in that
they are typically smaller, have young stars, and lie in the galactic
plane. Star clusters 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} 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. From the fact that the number of stars
falls with magnitude much faster perpendicular to the disk, than in the
plane of the disk, one deduced that the stars around the Sun are indeed
distributed in a disk, with axes ratio about 5:1.

How does it work? Suppose you count the number of stars $dN$, in a
given solid angle $d\Omega$, that have flux fainter than $F$, but
brighter than $F-dF$ (with $dF>0$). If all stars have luminosity $L$,
then stars with flux $F$ are at distance $r$, given by $F=L/(4\pi
r^2)$. Similarly, stars of flux $F-dF$ are at distance $r+dr$, given by
$F-dF=L/(4\pi\,(r+dr)^2)$. If $dr\ll r$, then $dF=2F\,dr/r$. The volume
of the fraction of a spherical shell that falls within a solid angle
$d\Omega$, between $r$ and $r+dr$ is $dV=d\Omega\,r^2\,dr$. Combining
all this we get for the number of stars per unit solid angle, with flux
between $F$ and $F-dF$

\begin{equation}
{dN\over d\Omega}=n(r)\,r^2\,dr = n(r)\,r^2\,{r\over 2F}\, dF = {1\over
2}\,({L\over 4\pi F})^{3/2}\,n(r)\,{dF\over F}\,.
\end{equation}

The function $dN(F)/d\Omega$ in the plane of the MW drops much slower
with decreasing $F$ than perpendicular to the plane, meaning $n(r)$
drops much slower within the disk, then perpendicular to it. Which is
because the stars are distributed in a disk around the Sun, not
spherically symmetric. This is how you can determine $n(r)$ by
counting stars.\\


\noindent{\bf Standard candles} are objects with a known absolute
property, for example a known size, or known luminosity. It is then
possible to determine the distance to the standard candle by measuring
their angular size , or apparent luminosity, respectively. 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 as standard candles (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 appears to move (in degrees,
say), the length of your arm, and the distance between your eyes. If
you now one distance (e.g. between your eyes), you can determine the
other (length of arm).

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.

The parallax and Cepheid variables are the first two steps in the {\em
distance ladder}, which use one method (e.g. parallax) to calibrate
another distance measure (e.g. Cepheids), that then can be used to
calibrate another distance indicator, and so go to go to greater and
greater distances. The Hubble space telescope recently measured
Cepheids in the Virgo cluster (at a distance of $\sim\,17\Mpc$),
thereby providing the first accurate distance to that cluster of
galaxies, and getting an accurate measurement of Hubble's constant in
the process.


\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. He
counted stars on photographic plates as function of apparent magnitude
(brightness), 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. This is in fact wrong but for an
interesting reason.

Kapteyn assumed that the flux received drops as $1/r^2$. One reason
this could be wrong is because of absorption or scattering of the
stellar light, in which case the flux will drop faster than $1/r^2$.
Of course, not taking absorption into account would give you the wrong
answer, and Kapteyn was fully aware of this. How can we test whether
absorption is important? For {\it Rayleigh scattering} off atoms, there
is a colour dependence: blue light is scattered more than red light. As
a consequence fainter and hence more distant stars would be expected to
be {\it redder} on average. Now Kapteyn found this indeed to be the
case, yet the {\it amount} of reddening was too small for absorption to
be important. The mistake in the reasoning 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. Consequently the size of the MW as well as the position of the
Sun in it came out wrong.

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
for some time.

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.

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.  In one great swoop,
the universe because much bigger, and in addition it was found to be
expanding! It is difficult to underestimate the significance of this
discovery.

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, the MW was
recognised to be just one of millions of other galaxies, the Universe
was found to be big and expanding

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\,. This
involves predicting the luminosity of a star from its spectrum.  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 these 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 the 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 we 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, but you get the idea.)

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 Derive the relation between the stellar density and star counts
\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}

\newpage

\chapter{The modern view of the Milky Way}
Our observational picture of the MW has been clarified due to steady
hard work and occasionally in great leaps with the advent of new
technologies.


\section{New technologies}
\subsection{Radio-astronomy}
Radio astronomy used the new technology developed for radar and opened
a new window for astronomers to gaze
through. 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 the
energy levels are quantised. And transitions between them correspond to
rotational or vibrational transitions. Since the energy-levels are
lower they 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 in good \lq mirror\rq\ the 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 500nm, say, but for radio-waves, 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 $D^2$.}

An electron moving in a magnetic field will feel an acceleration,
causing it to circle around the field lines. Such {\em synchrotron
radiation} can also be observed in the radio.

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 water
vapour in the atmosphere will otherwise absorb a large fraction of the
IR light. 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 strongly
absorbed by 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 may seem 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 determine
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 the \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. So getting these
parallax measurements accurate is of major importance.

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 \lq of almost inconceivable extent and
implication\rq\, (according to their 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. The three components are 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
{\em solar galacto-centric distance}, and there is some debate about
its value. $R_0=8.5\kpc$ is usually quoted, but some studies find
$R_0=8.0\kpc$. $R_0$ is usually determined from finding the centre of a
set of objects which are presumed to be randomly distributed about the
galactic centre, for example globular clusters, or variable stars such as
RR Lyrae or Mira variables, or O and B stars. Another method uses the
fact that there is 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.

Besides 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 $\approx 220\kms$.

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
itself (i.e., they may have formed somewhere else). One neat age
estimator are White Dwarfs. As you recall, these do not undergo nuclear
fusion anymore, but are presently just cooling down. 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 rotation
axis, $z$ 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 $z=0$, hence the
scale height of young stars is 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 disk!

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 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 $z_h\approx 1\kpc$. Its existence has been discovered
only recently, in the 1980s. The difference with the thin disk is not
just the kinematics of its stars: it turns out that thick disk stars
are on average 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 the notation [Fe/H] 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, its 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}-1]\,.
\label{eq:r4}
\end{equation}

This is a famous fit to the intensity 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 {\it 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
about 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 stellar 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, {\em 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. For some elements, like e.g. Carbon, we don't
really know which stars are the dominant source.

Super Novae (SNe) are a major source of metals. There are two basic
types: 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\footnote{And they produced the
stuff we are made of}. 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,
for [Fe/H]=-1, it has one tenth the solar value.

One might expect the metallicity of a star to be related to when it
formed. 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 quite 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 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 the lowest metallicity
currently known, has [Fe/H]$\approx -5$, and might well be one of the
first stars to have formed 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. Use spherical trigonometry to convert between $(l,b)$ and right
ascension $\alpha$ and declination $\delta$. The celestial equator is
inclined wrt the plane of the MW, hence 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 know
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 {\em stuff} between the stars:
gas, dust, magnetic fields and cosmic rays. Stars form in molecular
clouds, regions of the ISM which are dense and cold. During their
lifetimes, they return material back to the ISM in the form of stellar
winds and Planetary Nebulae, and at their deaths during super nova
explosions. The detailed interplay between stars and the ISM is
complicated and 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.

The colour of the sky during the day is blue, because it is
(predominantly) the blue light from the Sun that is scattered toward
you. At sunset or sunrise, the sun light passes through a lot more
atmosphere than at midday. The increased path length causes the
enhanced reddening of the sun. This is partly due to scattering, partly
due to absorption on little dust grains. The same processes of
scattering and absorption also occur in the ISM.

{\em Scattering} changes the direction of the incoming photon, but not
its 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 decreasing 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 is 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}. Again, the 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 longer wavelengths, typically in
the IR. For example in star forming regions, young stars produce large
amounts of dust. Even though these stars may be very bright, we cannot
see them at visual wavelengths because of the dust absorbs their
light. The absorbed light is re-radiated in the IR, making the region
very bright in the IR. And so you understand that the newly planned
space-telescope which will work in the IR, 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. Some IR observatories have moved to the South pole,
where there is much less water vapour.}\\

The scattering and absorption by dust grains may seem to be just a
nuisance for astronomers, given that they decrease the amount of
starlight 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). This is a slightly confusing
convention: you would think HII to be double ionised hydrogen --
impossible of course -- it means H$^+$. And when talking, you cannot
distinguish HII from H$_2$ -- both are pronounced as H-two. So, CIV is
{\em triply} ionised carbon.

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 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 to be more energetic 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}.\\

How about the reverse process? 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, HII + e $\rightarrow$ HI, 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. This process is called {\em radiative cooling}
and it is very important for 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 and HII regions}
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 its surroundings.

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 its HII form. And so massive bright
stars are often surrounded by {\em HII regions}. These regions of
ionised Hydrogen gas delineate spiral arms, because star formation
occurs predominantly in spiral arms and massive stars do not live for
very long, and so are found close to where they form.

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
its temperature, density, and the relative abundance of the elements.

You should remember two things about this, (1) understand the physics
that determines the properties of the {\em spectrum} that you get from
such a nebula, and (2) understand {\em what sets their size} (the
so-called Str\"omgren radius)

{\noindent\em Nebular spectra} 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. 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.\\

{\noindent\em Str\"omgren spheres.} 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 $n$ (in HI
atoms per cm$^{3}$ say). 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, $v$, 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 $\propto 1/R^2$.\\

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 start to decrease even more. Eventually, an
equilibrium is reached, in which the number of photo-ionisations
within $R$ equals the number of recombinations. The stalling radius is
called 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 {\em rate} at which HII ions
recombine is proportional to product of electron and 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$^{-1}$,
$\alpha$ has 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, hence

\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. 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, its low probability becomes an advantage, since
the 21-cm radiation can pass through a lot of HI gas, without it being
absorbed again. Given the long wavelength, 21-cm radiation is not
absorbed by dust.

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 transition is from 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 with 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 equilibria.

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.\\

This derivation 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\footnote{An even better way would be to find the
dispersion relation that relates the speed of a wave to its
wavelength.}

\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 3}\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.

\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.
\end{itemize}


\chapter{Dynamics of galactic disks}
The stars in the Milky Way disk are on (almost) circular orbits, with
gravity balancing centripetal acceleration. Given that most of the
light of the disk comes from its central parts, we would expect the
circular velocity in the outer parts of the disk to fall with distance
as appropriate for Keplerian motion. We should also be able to compute
how velocities of stars in the solar neighbourhood depend on
direction. Observations do not follow these expectations at all, which
leads to the startling conclusion that most of the mass in the Milky
Way is invisible.

\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 could expect that most of the mass would be interior to the orbit
as well. Newton's law then 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, {\em i.e. } 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 {\em differential
rotation}\footnote{The inner parts go around faster than the outer
parts, so differential rotation, as opposed to say solid body
rotation.}, with $V_c$ decreasing with increasing $R$. Since the period
$P$ of the orbit $P=2\pi R/V_c$, we expect that $P\propto R^{3/2}$,
just as is the case for the motion of planets around the Sun.

If the stars around the Sun are on circular orbit, then there are
relations between the relative velocity of the star, its distance, and
its direction in the disk. These are described by Oort's constants.

\subsection{Oort's constants}
\begin{figure*}[!ht]
\center\resizebox{.9\textwidth}{!}{\includegraphics{oort.ps}}
\caption{(Taken from http://www.csupomona.edu/\~{}jis/1999/kong.ps)
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 right-angled 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. Combining these
equations gives

\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 velocity 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 a 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 $R_0\approx 8.5\kpc$ hence

\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 either $V_c$ 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 $V_c(R)$: they will have (small) {\em peculiar}
velocities 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
motion}. 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
{\em 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 \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.} $d=R_0\cos(l)$.
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. {\bf the Milky Way's 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, $V_c(R)=constant$, it means there must
be more mass in the outskirts of spirals than we've assumed so
far. Let's compute how much more.

Assume the density $\rho$ of the MW is $\rho=\rho_0(R_0/R)^{\alpha}$,
where $\rho_0$ and $R_0$ are constants. 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. Note that this dark matter needs to be (1) massive, (2)
invisible, {\em i.e.} it does not {\em absorb} light, nor does it {\em
emit} light, it simply does not interact with photons.

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 accounted 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: 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. The {\em density
contrast}\footnote{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 going round and round) -- 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 proportional to its radius, $P\propto R$. Now concentrate on two
green stars, one at distance $R$ (star 1), the other at distance $2R$
(star 2), having periods $P_1=P$ and $P_2=2\,P$. 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 wound some spiral arms
are, this would imply they are all very recently formed, placing us at
a peculiar instant in time where spiral structure forms.

In cases where we were are 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.ps hscale=80 vscale=80}}
\put( 3.5, 0){\special{psfile=wind3.ps 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. The more messy pattern of flocculent spirals is probably
due to a gas-dynamical instability in the disk. And once excited, a
spiral pattern may live for a long time due to the spiral-density
wave.


\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{The Dark Halo}

Measurements of the rotation curve from HI 21-cm emission, 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 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 -- suggesting once more the
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 failure of the point mass 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}

For the Sun, $R\approx 8.5\kpc$, $V_c=220\kms$, $v_e\ge 500\kms$,
implying $R_\star\ge 41\kpc$, 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, our reasoning again suggests 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 therefore
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 {\em 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 $\theta=\pi$), 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. I }
Elliptical galaxies are spheroidal stellar systems with smooth
luminosity profiles which contain old, metal rich stars. They have a
large range in sizes. Usually, rotation is unimportant in ellipticals,
unlike for spirals. They often contain low density, very hot gas that
emits X-rays. The properties of this hot gas, and also the stellar
dynamics, suggest that also in ellipticals most of the mass is in dark
matter.

\section{Luminosity profile}
In contrast to spirals galaxies, elliptical have smooth surface
brightness (SB) profiles, ellipsoidal in shape. The SB profile of some
elliptical are shown in Figure~(\ref{fig:2D}). Note how in the center
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 foreground MW stars.

\begin{figure*}
\center\resizebox{.9\textwidth}{!}{\includegraphics{ellipticals_2D.ps}}
\caption{(Taken from astro-ph/0206097) Lines of constant surface
brightness (isophotes) in the \KB\ band for 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.  A more
general fit which sometimes used, is $I(r)\propto
\exp[-(r/r_e)^{1/n}]$, where $n=4$ corresponds to the de Vaucouleurs
$r^{1/4}$ fit.

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

\begin{figure*}
\center\resizebox{.8\textwidth}{!}{\includegraphics{3379.ps}}
\caption{Luminosity profile for galaxy NGC 3379 (open symbols) and
$r^{1/4}$ fit (drawn line). Taken from de Vaucoleurs \& Capaccioli,
ApJS 40, 1979.}
\label{fig:3379}
\end{figure*}

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

\begin{figure*}
\center\resizebox{.8\textwidth}{!}{\includegraphics{1399_GCS.ps}}
\caption{Image of NGC~1399 with smooth, best fit $r^{1/4}$ profile
subtracted (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
centers of big clusters of galaxies have an extended halo that is not
fit well by the de Vaucouleurs profile. Such galaxies are called cD
galaxies. Fig.~(\ref{fig:1399}) show the profile of NGC~1399, which is
the cD galaxy in the nearby Fornax cluster of galaxies. 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 center!\\

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.\\

{\em 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. We discussed this before: if the atmosphere
of a star contains more metals, then the dust and the gas will scatter
preferentially the {\em blue light} hence the stars appears
redder. 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 appears blue either 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 th
metallicity $Z$, for example from stellar spectra, or, the astronomer's
favourite, from HII regions.\\

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

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

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

\section{X-rays}
Elliptical galaxies emit X-rays. I want to briefly discuss {\em why},
{\em how} X-rays are observed, and {\em 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 extra-solar X-rays.\\

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 {\em 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} from the galaxy. The process that
produces them is called {\em thermal bremsstrahlung}.

\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 and this process is therefore also called
free-free radiation. {\bf This is all you need to know about this}, but
if you want more details, here they come.\\

\subsubsection{radiated power}
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 one way to determine the temperature $T$ of the hot gas.

\section{Evidence for dark matter from X-rays}
An important application of X-ray observations is to probe the
gravitational potential of the galaxy, and hence infer its density
distribution. Consider a thin shell of hot gas. The gravitational force
due to mass enclosed by the shell (consisting of stars, gas and dark
matter) will pull the shell inward. At the same time, if the pressure
decreases outward, then the pressure gradient will try to push the
shell outward. In {\em hydrostatic equilibrium}, these forces
balance. By estimating the pressure gradient from X-rays, we can infer
the gravitational force and hence look for evidence for dark
matter. Let's assume for simplicity that the elliptical galaxy is
spherically symmetric.

Let $M(<r)$ be the mass inside the shell of radius $r$ (consisting of
stars, gas and dark matter: $M=M_\star+M_{\rm gas}+M_{\rm dm}$), and
$dr$ its thickness The gravitational force on the shell is
\begin{equation}
F_g(r)= {GM(<r)M_s\over r^2}\,,
\label{he:1}
\end{equation}
where $M_s =4\pi r^2\rho_{\rm gas}(r) dr$ is the mass of the
shell.

Let $p(r)$ be the pressure in the hot gas. The pressure force on a
surface is $p(r)$ times the surface area, $4\pi r^2$. The net outward
force is the difference between the outward and inward forces:
\begin{equation}
F_p(r) = 4\pi r^2 (p(r)-p(r+dr)) = -4\pi r^2 {dp\over dr}\,dr\,
\label{he:3}
\end{equation}
where the pressure was expanded in a Taylor series. Combining the last
three equations gives

\begin{equation}
{GM(<r)\over r^2} 4\pi r^2 \rho_{\rm gas}(r) dr = -4\pi r^2 {dp\over
dr} dr\,,
\end{equation}
hence
\begin{equation}
{GM(<r)\over r^2} = -{1\over \rho_{\rm gas}(r)} {dp\over dr}\,,
\label{eq:HE0}
\end{equation}
or alternatively in terms of the gravitational potential $\Phi$

\begin{equation}
\nabla\Phi=-\nabla p/\rho_{\rm gas}\,.
\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
$T(r)$, and since the X-ray intensity $\propto \rho_{\rm gas}^2
T^{1/2}$, we can also determine $\rho_{\rm gas}(r)$. Hence $p\propto
rho_{\rm gas}T$, and so we have determined 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_{\rm gas}\,.
\end{equation}

If we take $T$ {\em constant} then the rhs of Eq.~(\ref{eq:HE0}) becomes
$-{\boltz\,T\over \mu\,\mp}{d\ln(\rho_{\rm gas})/dr}$, and hence
\begin{equation}
G M(<r) = -{\boltz T\over \mu\,\mp} r^2 {d\ln\rho_{\rm gas}\over dr}\,.
\label{eq:HE3}
\end{equation}

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

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

since $dM(<r)/dr=4\pi\,r^2\,\rho_{\rm tot}$. $\rho_{\rm tot}$ is the
{\em total mass density}, {\em i.e.} due to stars, gas and dark matter.\\

Now assume that the gas density and total mass density are
proportional, $\rho_{\rm gas}\propto \rho_{\rm total}$, and try a
solution of the form $\rho_{\rm
totcl}=\rho_0\,(r_0/r)^\beta$. Substitution shows that this is indeed a
solution when $\beta=2$, in which case
\begin{equation}
\rho_{\rm tot}(r) = {\boltz\,T\over 2\pi\,G\mu\mp}\,r^{-2}\,.
\end{equation}

So a measurement of $T$ can constrain the {\bf total} mass density
$\rho_{\rm tot}$. Estimating the stellar mass density from the
luminosity, the gas density from the X-ray emissivity, we find
$\rho_{\rm dm}=\rho_{\rm total}-\rho_\star-\rho_{\rm gas}$. More
detailed modelling of this type confirms that also elliptical galaxies
have most of their mass in the form of dark matter.


\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{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. The aim of this
chapter is twofold: ({\it i}) explain why you can derive fluid-like
equations to describe the structure of ellipticals, and ({\it ii}) use
these fluid equations to explain why elliptical are elliptical in
shape.

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 Jeans'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 due to near stars?

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 $r=|\bfr|$. In such a time-independent potential both the angular
momentum, ${\bf L}=m\,{\bf r}\times{\bf v}$, and energy,
$E=v^2/2+\Phi$, 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), and then sum over encounters. This
derivation was first done by Chandrasekhar, the derivation here is
slightly different from BT (p.~188).\\

Let's compute the change in velocity $\delta\bfv$ of this star. Because
we want to treat the case where the effect of encounters is small,
we'll approximate the orbit as a straight line, traversed with constant
velocity $v$. 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. Combined with Newton's law


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

yields the net change $|\delta\bfv_\perp|$ over a single 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 time this force acts, $\Delta t\sim b/v$. Because
of symmetry, encounters are equally likely to change the velocity in
the positive direction, as in the negative direction. Therefore the
mean change $\langle \delta\bfv_\perp\rangle=0$. We will therefore
consider the rate of change of $\delta\bfv_\perp^2$.\\

To compute the rate of change of $\delta\bfv_\perp^2$, due to many
encounters, proceed as follows. Suppose the test star moves with
velocity $v$ through a stellar system with radius $R$, and (uniform)
number density of stars, $n$.  When travelling for a time $\delta t$,
the number of encounters with impact parameter between $b$ and $b+db$,
is simply the volume of the cylindrical shell, $dV=2\pi\,b\,db\,\times
v\delta t$, times the number density of stars, $n$. Here, $v\delta t$
is the length of the cylinder, $2\pi\,b\,db$ is the surface area of the
shell. Since each of these encounters leads to a change in velocity
squared by an amount $\delta\bfv_\perp^2$, we find for the rate of
change

\begin{equation}
{d^2 \delta\bfv_\perp^2\over dt\,db} = n\,v\,2\pi\,b\,\times ({2Gm\over
b\,v})^2\,.
\end{equation}

This is the product of the {\em rate} of encounters, $n\,v\,2\pi\,b db$, with the
velocity change {\rm per} encounter. Now we can integrate over impact
parameter to obtain for the rate of change in velocity

\begin{equation}
{d\delta\bfv_\perp^2\over dt} = \int_0^\infty  n\,v\,2\pi\,b\,({2Gm\over
b\,v})^2\,db=2\pi\,nv\,({2Gm\over v})^2\,\int_0^\infty {db\over b}\,.
\end{equation}

Unfortunately this integral diverges, both at small impact parameters
(which are few in number, but cause large velocity changes), and at
large impact parameters (which cause small velocity changes yet are
very numerous). How to overcome this divergence?\\

For a real stellar system, there cannot be encounters with impact
parameter larger than the size $R$ of the system. For the smallest
impact parameter $b$, 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 in mind we replace the previous equation by\,

\begin{equation}
{d\delta\bfv_\perp^2\over dt} \approx 2\pi\,nv\,({2Gm\over
v})^2\,\int_{b_{\rm min}}^R {db\over b}=2\pi\,nv\,({2Gm\over
v})^2\,\ln({R\over b_{\rm min}})\,.
\label{eq:TE1}
\end{equation}

We can simplify this by assuming that the stellar system is in virial
equilibrium, in which case the kinetic energy of all stars combined,
$K\sim N\,m\,v^2/2$ is half of the potential energy, $U\sim
G\,(N\,m)^2/R$, where $m$ is the mass of a star. Therefore

\begin{equation}
v^2\approx {GNm\over R}\,.
\label{Eq:VE}
\end{equation}

Combining this with the expression for $b_{\rm min}$ it is easy to show
that $R/b_{\rm min}=N$, the total number of stars.\\

So far we have computed the {\em rate} of change of the velocity. To
judge whether the change of velocity is large or small, we can multiply
the rate with the {\em crossing time} of the system, $t_{\rm cr}\equiv
R/v$. It is a measure of how long it takes to cross the stellar system,
on average. The velocity change during a single crossing of the stellar
system is therefore

\begin{eqnarray}
\delta\bfv_\perp^2\sim {d\delta\bfv_\perp^2\over dt}\,t_{\rm cr} &=& 2\pi\,nv\,({2Gm\over
v})^2\,\ln(N)\times {R\over v}\nonumber\\
 &=& {6\ln(N)\over N}\, v^2\,.
\end{eqnarray}

The first line combines Eq.(\ref{eq:TE1}) with our finding $R/b_{\rm
min}=N$ and the definition of the crossing time. The second step
assumes virial equilibrium, Eq.~(\ref{Eq:VE}). We are finally in a
position to define the {\em relaxation time} $T_{\rm relax}$, as the
time it takes for encounters to change the velocity by order of itself,
hence

\begin{equation}
{v^2\over T_{\rm relax}}\equiv {d\delta\bfv_\perp^2\over dt} =
{6\ln(N)\over N}\, {v^2\over t_{\rm cr}}\,
\end{equation}

hence we get as our final result
\begin{equation}
T_{\rm relax} = {N\over 6\ln(N)}\,t_{\rm cr}\,.
\end{equation}


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 with $N=10^5$ say, $r=10\pc$, and $v=10\kms$,
$T_{\rm cross}\approx 1\Myr$. 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, and derive fluid-like equations for the behaviour
of a galaxy. The stars will be like the gas particles in the usual gas
equations.  This is what we'll do next.

\section{Jeans equations}

\subsection{A continuity equation}
We can derive fluid-like equation for the gas of stars in an elliptical
galaxy, using the concept of a {\em 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
you count the number of cars entering the tunnel on one side and the
number that leaves 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 once.} 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.ps}}
\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.ps}}
\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 statistics of the
different types. The two degree-field galaxy redshift survey (2dF), and
the Sloan Digital Sky survey (SDSS), are two very recent {\em galaxy
redshift} surveys. In a galaxy redshift survey, one first identifies
galaxies in images of the sky, and then takes a spectrum for each of
them. Because the Universe is expanding, spectral lines of stars in a
distant galaxy are shifted from the laboratory toward longer
wavelengths: they are \lq redshifted\rq\,. There is a relation between
the distance to the galaxy and its redshift, and so a galaxy redshift
survey provides us with the 3D position of a set of galaxies, about
$2\times 10^5$ for the 2dF, about $10^6$ for the SDSS. So what kind of
general trends do they uncover? We will discuss some of these here.\\

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 a good idea of how the dark matter is distributed on a
cosmological scale, from some general considerations, backed-up by
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 how dark matter is clustered
on large scales. 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: a filamentary pattern
delineating large regions of space devoid of galaxies, called \lq
voids\rq\,, with regions of high galaxy density, clusters of galaxies,
at the intersection of the filaments.\\

\section{The density-morphology relation}
In the previous chapter we saw how some galaxies are in low density
environments, some are in higher density {\em groups}, and others in
dense {\em clusters} containing several 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.ps}}
\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 regions 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 {\bf 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 colour is
used (i.e., does $L$ refer to a \V\ -band or \I\ -band luminosity).\\


\begin{figure*}
\center\resizebox{.9\textwidth}{!}{\includegraphics{tfr.ps}}
\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\footnote{It is
unfortunate that absolute magnitude and mass are both denoted by $M$.}
that the circular velocity $V_c$ is determined by the enclosed mass
$M(<R)$,

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

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

\begin{equation}
\mol \equiv {M(<R)\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 intensity $I$ 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^2\,I}\,.
\end{equation}

So the Tully-Fisher relation is reproduced, with $\alpha=4$, if the
intensity $I$ and mass-to-light ratio \mol\ are {\em independent} of
the luminosity of the galaxy. Notice that this is not a {\em
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 $L$), 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
as tight as the corresponding $V_c(L)$ relation in spirals.


\begin{figure*}
\center\resizebox{.9\textwidth}{!}{\includegraphics{faberjackson.ps}}
\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 intensity $I_e$ (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 $R_e$. This is how it goes: suppose that
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 assumed \mol\ to depend on $L$ as
\begin{equation}
\mol\propto L^\alpha\,.
\label{eq:mol}
\end{equation}

The intensity $I_e\propto L/R_e^2$, hence
\begin{equation}
R_e\propto \sigma^{2/(1+2\alpha)}\, I_e^{-(1+\alpha)/(1+2\alpha)}\,.
\label{eq:fp1}
\end{equation}

\begin{figure*}
\center\resizebox{.9\textwidth}{!}{\includegraphics{fp.ps}}
\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 much better, 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 which was 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 as 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}), hence
the name fundamental plane.

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

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

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

Therefore, the Tully-Fisher relation can be used as a standard candle,
since it allows us to determine the luminosity from the distance
independent quantity $V_c$. Note also that $V_c$ is relatively easy to
determine from spectra, and so we have found a good way of measuring
distances to distant galaxies!\\

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
intensity $I_e$ (recall that the surface brightness and hence also
$I_e$ 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.ps}}
\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.ps}}
\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*}

\section{Epilogue}
The classification of galaxies that we have used so far, in terms of
spirals and ellipticals, is based largely on how these galaxies appear
in pictures. With several $10^5$ galaxies in modern surveys, it is not
really possible to eye-ball all of them in order to classify them. In
addition, it is not even necessarily the best way, and certainly not
the only way, of classifying galaxies and identifying different
types. With so many spectra and images available, new statistical
techniques are presently being developed, where computer programs which
do not suffer from human prejudices, such as principle component
analysis and neural networks, decide how galaxies should be grouped and
classified.

\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 how Tully-Fisher and Fundamental Plane relations are used
as standard candles.
\item explain what is the luminosity function of galaxies and sketch
it.
\end{itemize}

\chapter{Active Galactic Nuclei}
Some galaxies harbour immensely powerful sources of radiation in their
centres, detected from the longest (radio) to the shortest wavelengths
(X-rays): these are called Active Galactic Nuclei (or AGN for
short). We think they are powered by the accretion of matter onto super
massive black holes (SMBH), the same process that powers some galactic
X-ray binaries. However, the SMBH have masses ranging from $10^6$ to
$10^9\msun$, and so, unlike galactic black holes, cannot be stellar
remnants.  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\, at any given time. 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. 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{Discovery and observational properties}
Carl Seyfert reported in 1943 that a small fraction of galaxies have a
very bright nucleus, which is a source of broad emission lines produced
by atoms in a wide range of ionisation stages. These nuclei appear
unresolved, i.e. they 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 This thing is 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.

The shear luminosity of QSOs alone already suggests we are looking at
something unusual. In addition, some of these QSOs have radio-lobes
which are tens of times the size of a galaxy. What is the engine that
powers them?

\section{The central engine, and unification 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 (special)
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, 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: its mass, and its spin (i.e. its 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 its 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 its peak luminosity is about as bright as a whole galaxy, so we need
about 100 SNe and we're done. Is this a reasonable model for AGN?

The model has withstood for several decades, but it does have some
pitfalls. For example, we don't know how such SNe would generate the
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 \rho\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
approaching 1000\kms.

So first of all, they demonstrated that there is indeed a very massive
\lq dark\rq\, object in the centre of the MW.  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 {\em almost} no room to
escape the conclusion that the MDO is indeed a SMBH, at least in the MW.

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 acting
on the gas, 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\footnote{A {\em
maser} is a laser that operates in the mm regime.} 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 vary as well, 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 deep 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 Schwarzschild 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 Schwardschild radius to clinch the matter.

So let me summarise. We have evidence that probably most, if not all,
galaxies with masses of order the MW mass or higher, 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 and produces AGN activity 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 is 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.ps}}
\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.ps}}
\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.ps}}
\caption{Hubble Space Telescope image of an Einstein ring, where a
fortuitous alignment has caused an intervening elliptical galaxy (the
central spot) to lens a background spiral galaxy into a ring.}
\label{fig:EinsteinRing}
\end{figure*}

\begin{figure*}
\center\resizebox{\textwidth}{!}{\includegraphics{EinsteinCross.ps}}
\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 galaxy.}
\label{fig:EinsteinCross}
\end{figure*}

\section{Micro-lensing}
\begin{figure*}
\center\resizebox{.8\textwidth}{!}{\includegraphics{microlensing.ps}}
\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, you need to look at millions
of stars to have a chance of seeing significant amplification. 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, and distinguish it from the
case where the star is intrinsically
varying. 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.ps}}
\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 images 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 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 galaxies,
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}
