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\centerline{\bf LECTURE 2} 

\centerline{BASIC CONCEPTS}
\centerline{States of a System}
Let's consider how we specify the state of a system with N particles 
in both classical mechanics and quantum mechanics. 

In classical mechanics if we have a single particle in one dimension
then we can describe the system completely by specifying the position
coordinate $q$ and the momentum coordinate $p$. We can represent this
graphically by labeling one axis with $q$ and one axis with $p$:

\begin{center}
\leavevmode
\epsfxsize=2.0 true in
\epsfbox{phasespace.eps}
       \end{center}

We call the space spanned by $p$ and $q$ ``phase space.'' It's the
space in which the point $(q_1,p_1)$ exists.
If our system has N particles and exists in 3D, then we must provide 
$\vec{p}_i$ and $\vec{q}_i$ for all N particles. Since $\vec{p}$ and
$\vec{q}$ are each 3 dimensional vectors, if we want to represent the
system as a point in phase space, we need 6N axes. In other words
phase space is 6N dimensional. The coordinates of the point representing
the system in phase space are $(q_{x1},q_{x2},...,q_{zN},p_{x1},...p_{zN})$. 

Spatial coordinates and momenta are continuous variables. To obtain a
countable number of states, we divide phase space into little boxes
or cells.  For our one particle in 1D example, the volume of one of these
cells is
\be
\delta q\;\delta p=h_o
\ee
where $h_o$ is some small constant having the dimensions of angular
momentum. The state of the system can then be specified by stating that 
its coordinate lies in some interval between $q$ and $q+\delta q$ and
between $p$ and $p+\delta p$.
For our N particle system, $f=3N$ spatial coordinates
and $f$ momentum coordinates are required to specify the system. So the 
volume of a cell in phase space is
\be
\delta q_1...\delta q_f\;\delta p_1 ... \;\delta p_f=h_o^f
\ee
Each cell in phase space corresponding to a state of the system can be
labeled with some number. The state of a system is provided by specifying
the number of the cell in phase space within which the system is located.

One microscopic state or {\it microstate} of the system 
of N particles is defined by specifying 
all the coordinates and momenta of each particle. An N particle system
at any instant of time is specified by only one point in a 6N 
dimensional phase space and the corresponding microstate by 
the numerical label of the cell in which this point is located.
As the system evolves in time, the coordinates of the particles
change, and the N particle system follows some trajectory in this 6N
dimensional phase space. 

A macroscopic state or {\it macrostate} of the system is determined
by only a few macroscopic parameters such as temperature, energy,
pressure, magnetization, etc. Note that a macrostate contains much
less information about a system than a microstate. So a given macrostate
can correspond to any one of a large number of microstates.

How would quantum mechanics be used to describe a system? Any system of
N interacting particles can be described by a wavefunction
\be
\psi_{\{n\}}(q_1, ... , q_f)
\ee
where the $q_i$ are the appropriate ``coordinates'' for the N particles.
The coordinates include both spin and space coordinates for each particle
in the system. A particular state (or a particular wavefunction) is then
specified by providing the values of a set of quantum numbers $\{n\}$.
This set of quantum numbers can be regarded as labelling this state.
Different values of the quantum numbers correspond to different states.
For simplicity let's just label the states by some index $r$, where
$r=$ 1, 2, 3, ... The index $r$ then labels the different microstates.
In quantum mechanics, $h_o$ is replaced by Planck's constant $h$.

Thus both classical and quantum mechanics lead to a countable number of
microstates for an N particle system.


\centerline{Ensemble}
Statistical mechanics is based on probability considerations, and 
averages over appropriate quantities. Thus one approach to statistical
mechanics is a consideration of a large number of identically prepared
systems, all subject to the same initial conditions and the same
set $\{X\}$ of external parameters
such as the total energy, particle number, and volume. This hypothetical
collection of identical systems is called an {\it ensemble}. The
systems in the ensemble will, in general, be in different states and will,
therefore, also be characterized by different macroscopic parameters
(e.g., by different values of pressure or magnetic moment). (These 
macroscopic parameters are not the ones in set $\{X\}$, but
they may be conjugate to them. For example pressure $p$ is
conjugate to volume $V$ because $p\;dV$ is the work done by pressure
$p$ in changing the volume by $dV$.) We can
calculate the probability of the occurrence of a particular value
of such an external parameter, i.e., we can determine the fraction of 
cases in the
ensemble when the parameter assumes this particular value. For example
we can calculate the average pressure. Another
way to say this: any variable
or property that we are attempting to calculate will then be obtained
from an averging procedure over all members of the ensemble.
The aim of theory will be to predict the probability of occurrence in the
ensemble of various values of such a parameter on the basis of some
basic postulates.

This concept, and the term enemble, were introduced by J. W. Gibbs, an
American physicist around the turn of the 20th century.

Another basic approach to statistical mechanics, proposed by Boltzmann
and Maxwell, is known as the {\it ergodic hypothesis}. According to
this view, the macroscopic properties of a system represent averages
taken over the microstates traversed by a single system in the course
of time. It is supposed that systems traverse {\it all} the possible
microstates fast enough that the time averages are identical with
the averages taken over a large collection of identical and independent
systems, i.e., an ensemble. This is the idea behind Monte Carlo simulations.

\centerline{Basic Postulates of Statistical Mechanics}
To make any progress, we need some basic postulates about the relative
probability of finding a system in any of its accessible states. Usually
one only has partial information about a system. We don't know every
single thing about every particle. The states which are compatible with
the information that we have about the system are called ``accessible
states.'' The accessible states don't violate or contradict any
information that we have about the system. Now consider a thermally isolated
system. It cannot exchange energy with the system so its total energy 
is fixed or conserved. We would like to make some statements about the
system in equilibrium. When the system is in equilibrium, things are not
changing in time. The macroscopic parameters are time--independent.
The probability of finding the system in any one state is independent
of time, i.e., the representative ensemble is the same irrespective of time.
This leads to the fundamental postulate of statistical mechanics:

{\it An isolated system in equilibrium is equally likely to be in any of
its accessible states.} In other words if phase space is subdivided into
small cells of equal size, then an isolated system in equilibrium is
equally likely to be in any of its accessible cells. 

Certainly this seems reasonable. There is no reason for one microstate to
be preferred over another, as long as each microstate is consistent with
the macroscopic parameters. There are also more rigorous reasons to accept
this postulate. It's a consequence of Liouville's theorem (see Appendix
13 of Reif) that if a representative ensemble of such isolated systems
are distributed uniformly over their accessible states at any one time,
then they will remain uniformly distributed over these states forever.

One can think of the accessible states as the ``options'' that a system
has available to it. Lots of accessible states means lots of possible
microstates that the system can be in.

What if an isolated system is not equally likely to be found in any of
the states accessible to it? Then it is not in equilibrium. But it will
approach equilibrium. The system will make transitions between
all its various accessible states as a result of interactions
between its constituent particles. Once the system is equally likely to
be in any of its accessible states, it will be in equilibrium, and it
will stay that way forever (at least as long as it is isolated). The
idea that a nonequilibrium system will approach equilibrium is a 
consequence of the H theorem (Appendix 12 in Reif). If we think in
terms of an ensemble of systems distributed over the points in phase space
in some arbitrary way, then the ensemble will evolve slowly in time
until phase space is uniformly occupied. The characteristic time 
associated with attaining equilibrium is called the ``relaxation time.''
The magnitude of the relaxation time depends on the details of the system.  
The relaxation time can range from less than a microsecond to longer
than the age of the universe (e.g., glass). Indeed the glass transition is
a good example of a system falling out of equilibrium because
the experimenter cannot wait long enough for the system to equilibrate. 
Calculating the rate of
relaxation toward equilibrium is quite difficult, but once equilibrium
is reached and things become time--independent, the calculations become
quite straightforward. For example, many of the properties of the
early universe have been calculated using the assumption that things were
in equilibrium.

\centerline{Probability calculations}
From this basic postulate, how do we calculate various quantities of
interest? Let us consider a system of total energy between $E$ and
$E+\delta E$. Let $\Omega(E)$ be the total number of microstates
that satisfy this condition. Suppose that $\Omega(E;y_k)$ is the number of
states contained within $\Omega(E)$ that are characterized by the parameter
$y$ having the value $y_k$. For example, $y$ might be the magnetic moment
of the system or the pressure exerted by the system. Since all states 
are equally likely, we have for the probability $P(y_k)$ that the 
parameter $y$ of the system assumes the value $y_k$
\be
P(y_k)=\frac{\Omega(E;y_k)}{\Omega(E)}
\ee
To calculate the mean value of the parameter $y$ of the system, we
simply take the average over the systems in the ensemble; i.e.,
\begin{eqnarray}
\overline{y}&=&\sum_k P(y_k)\;y_k \nonumber\\
&=& \sum_k\frac{\Omega(E;y_k)}{\Omega(E)} \;y_k \nonumber\\
&=& \frac{\sum_k \Omega(E;y_k)\;y_k}{\Omega(E)} 
\end{eqnarray}
Here the sum over $k$ denotes a sum over all possible values which the
parameter $y$ can assume. Note that to calculate the average value of
the parameter $y$, we simply need to count states. However, this
may be highly nontrivial.

\centerline{Density of States}
Density of states is a useful concept. A macroscopic system, like a cup
of coffee or a block of copper, has a great many degrees of freedom.
Let $E$ be the energy of the system. Suppose we divide up the energy
scale into small regions, each of size $\delta E$, where $\delta E$
is much larger than the spacing between energy levels but macroscopically
small. Let $\Omega(E)$ be the number of states whose energy lies between
$E$ and $E+\delta E$. Then $\Omega(E)$ must be proportional to
$\delta E$ and we can write
\be
\Omega(E)=\rho(E)\delta E
\ee
where $\rho(E)$ is the ``density of states''. (Your book writes it as
$w(E)$.) The density of states is a characteristic property of the system
which measures the number of states per unit energy range. For example
one could have the number of states per eV. The density of states is
an important concept in systems with many particles. For example in a simple
metal where electrons conduct electric current, the density of electron states
near the Fermi energy determines how good a conductor the metal is.
If the density of states is high, then the metal is a good conductor because
the electrons near the Fermi energy will have a lot of empty states to choose
from when they hop. If $\rho(E)$ is small, then the metal is a poor conductor
because the electrons will not have many empty states to hop to. The density
of states is often useful in converting sums into integrals over energy:
\be
\sum_{i}f_i\rightarrow\int dE\rho(E)f(E)
\ee

\centerline{Interaction Between Macroscopic Systems}
Macroscopic systems are described by specifying some macroscopically
measurable independent parameters like the volume V or the applied
external electric and magnetic field. Now
consider two macroscopic systems A and A$^{\prime}$ which can interact with one
another so that they can exchange energy. Their total energy
$E+E^{\prime}$ remains constant since the combined system A$^o$ 
consisting of A and A$^{\prime}$ is isolated. They can interact in two ways:
mechanically and/or thermally. If they interact thermally, they exchange
heat but the energy levels of the systems do not change though
their occupation might. If they interact mechanically, the external
parameters (like volume) are changed and some of the energy levels are
shifted. 

\centerline{Thermal Interaction}
Let's consider the thermal interaction. In a purely thermal
interaction, energy is transferred from one system to the other.
If we have an ensemble of interacting systems (A+A$^{\prime}$), the
amount of energy transferred to each system A is not exactly the same 
for the different members of the ensemble. We can
however talk in terms of the change in the {\it mean} energy of each
of the systems. This is called ``heat.'' More precisely, the change
$\Delta \overline{E}$ of the mean energy of system A is called the 
``heat $Q$ absorbed'' by this system; i.e., 
\be
Q=\Delta \overline{E}
\ee
Heat is energy transfer.
The heat can be positive or negative. $-Q$ is the heat given off by
a system; $Q$ is the heat absorbed by the system. Since the total energy
is unchanged
\be
\Delta \overline{E}+\Delta \overline{E}^{\prime}=0
\ee
or
\be
Q+Q^{\prime}=0
\ee
or
\be
Q=-Q^{\prime}
\ee
This is just conservation of energy. 

\centerline{Mechanical Interaction}
Now suppose that the systems A and A$^{\prime}$ cannot interact thermally, i.e.,
they are themally isolated. However they can interact mechanically.
For example A$^{\prime}$ could do work on A. Consider a cylinder separated into two
parts by a movable piston. Let A be the gas in one part and A$^{\prime}$ be the
gas in the other part. Suppose A$^{\prime}$ expands, moves the piston and
compresses the gas in A.
%If A is the gas in a cylinder, then
%A$^{\prime}$ could be a piston that changes the 
%volume of A$^{\prime}$ and compresses the gas.
This changes the energy of A by heating up the gas. 
As before we think of an ensemble of identical systems and speak in
terms of the change in the mean energy. If the change in the mean 
energy due to the change of the external parameters is denoted by  
$\Delta_{x}\overline{E}$, then the macroscopic work done {\it on} the
system is defined as 
\be 
{\cal W}=\Delta_{x}\overline{E}
\ee
The macroscopic work $W$ done {\it by} the system is the negative of
${\cal W}$:
\be
W=-{\cal W}=-\Delta_{x}\overline{E}
\ee
Conservation of energy dictates that 
\be
W+W^{\prime}=0
\ee
or
\be
W=-W^{\prime}
\ee
Doing work on a system changes the positions of the energy levels
and the occupation of different states.

\centerline{Generalized Force}
In introductory physics we defined work as the force on an object times
the distance it moves (``force times distance''). Now we have a system
with $10^{23}$ particles. How do we define work? ``Pressure times volume.''
We mentioned earlier that when something changes the volume by applying
pressure, the mean energy changes and work has been done on the system.
Pressure has units of force per unit area. The gas pushes on a wall and
produces pressure which is the force per unit area on the wall. Notice
that $F/A$ has the same units as energy/volume. In fact the definition
of the (mean) pressure is
\be
\overline{p}=-\frac{\partial \overline{E}}{\partial V}
\ee
(We should keep the entropy fixed in this derivative.)

We can make this more formal. When we say ``macroscopic work,'' we mean more
than $pdV$ or $Fdx$ where $F$ is a force. Let the energy of some 
microstate $r$ depend on external parameters $x_1, ... ,x_n$.
\be
E_r(x_1, ... ,x_n)
\ee
Then when the parameters are changed by infinitesimal amounts, the
corresponding change in energy is
\be
dE_r=\sum_{\alpha=1}^{n}\frac{\partial E_r}{\partial x_{\alpha}}\; dx_{\alpha}
\ee
The work $dW$ done {\it by} the system when it remains in this
particular state $r$ is
then defined as 
\be
dW_r\equiv -dE_r=\sum_{\alpha} X_{\alpha,r}\;dx_{\alpha}
\ee
where 
\be
X_{\alpha,r}\equiv -\frac{\partial E_r}{\partial x_{\alpha}}
\ee
This is called the ``generalized force'' conjugate to the
external parameter $x_{\alpha}$ in the state $r$. Note that if $x_{\alpha}$
denotes a distance, then $X_{\alpha}$ simply is an ordinary force.

Consider an ensemble of similar systems. If the external parameters 
are changed quasi--statically so that the system remains in equilibrium
at all times, then we can calculate the mean value averaged over all
accessible states $r$
\be
dW=\sum_{\alpha=1}^{n}\overline{X}_{\alpha}\;dx_{\alpha}
\ee
where
\be
\overline{X}_{\alpha}\equiv -\overline
{\frac{\partial E_r}{\partial x_{\alpha}}}
\ee
is the mean generalized force conjugate to $x_{\alpha}$. 
Note that
\begin{eqnarray}
\overline{X}_{\alpha}&\equiv& -\overline
{\frac{\partial E_r}{\partial x_{\alpha}}}\nonumber \\
&=&\sum_{r=1}^{N} P_r\left(-\frac{\partial E_r}{\partial x_{\alpha}}\right)
\nonumber \\
&=&-\frac{\partial}{\partial x_{\alpha}}\sum_{r=1}^{N} P_r E_r\nonumber \\
&=&-\frac{\partial}{\partial x_{\alpha}}\overline{E}
\end{eqnarray}
The macroscopic work $W$ resulting from a finite quasi--static change of 
external parameters can then be obtained by integration.

Examples
\begin{enumerate}
\item  Force times distance
\be
dW=F_x\;dx
\ee
where $x$ is the linear dimension and $F_x$ is the ordinary force in the
$x$ direction.
\item Pressure times volume
\be
dW=\overline{p}\; dV
\ee
where $\overline{p}$ is the average pressure and $V$ is volume. We wrote 
down the expression for
pressure before, but now we can be more precise. The pressure is the
generalized force associated with changes in volume.
\be
\overline{p}=-\overline{\frac{\partial E_r}{\partial V}}=
\sum_{r=1}^{N} P_r\left(-\frac{\partial E_r}{\partial V}\right)
=-\frac{\partial}{\partial V}\sum_{r=1}^{N} P_r E_r=
-\frac{\partial}{\partial V}\overline{E}
\ee
or
\be
\overline{p}=-\frac{\partial \overline{E}}{\partial V}
\ee
where $E$ is the macroscopic energy and $V$ is the volume.
\end{enumerate}
Your book talks about quasi--static processes in which the process occurs
so slowly that the system can be regarded as being in equilibrium throughout.
For example, the piston can be moved so slowly that the gas is always
arbitrarily close to equilibrium as its volume is being changed. 
In this case the mean pressure has a well--defined meaning. If the 
volume is changed by an infinitesimal amount $dV$, then the work done
is 
\be
dW=\overline{p}dV
\ee
If the volume is changed from an initial volume $V_i$ to a final volume
$V_f$, then the macroscopic amount of work done is given by
\be
W_{if}=\int_{V_i}^{V_f}dW=\int_{V_i}^{V_f}\overline{p}dV
\ee
The work done is the area under the pressure curve between $V_i$
and $V_f$ in the P-V diagram.
\begin{center}
\leavevmode
\epsfxsize=2.0 true in
\epsfbox{onepath.eps}
\end{center}

This integral depends on the path taken from the initial to the final 
volume. It is not path independent. For example,
in the figure, the area under the path $i\rightarrow b\rightarrow f$
is twice the area under the path $i\rightarrow a\rightarrow f$, even
though the endpoints of both paths are the same.
\begin{center}
\leavevmode
\epsfxsize=2.0 true in
\epsfbox{workpath.eps}
\end{center}

So $dW$ is not an exact differential.
(Recall in electromagnetism, the potential difference is path
independent.) $dW$ is not the difference of 2 numbers referring to 2
neighboring macrostates; rather it is characteristic of the process
of going from state $i$ to state $f$. Similarly the infinitesimal
amount of heat $dQ$ absorbed by the system in some process is also
not an exact differential and in general, will depend on how the process
occurs.

\centerline{General Interaction between 2 Systems}
In general two systems interact both thermally and mechanically.
Let $Q$ be the heat absorbed {\it by} the system and let $W$ be
the work done {\it by} the system. Then the change in the mean
energy $\Delta\overline{E}$ is given by
\be
\Delta\overline{E}=Q-W
\ee
This is the first law of thermodynamics.
If we write
\be
Q=\Delta\overline{E}+W
\ee
then we can view the heat $Q$ as the mean energy change not due to a change in
the external parameters. For infinitesimal changes, we can write
\be
dQ=d\overline{E}+dW
\ee
Note that $d\overline{E}$ is an exact differential. The change in
the mean energy is independent of the path taken between the initial
and final states. The energy is characteristic of the state, not of
the process in getting to that state. 

For example, suppose we push a cart over a bumpy road to the top of
a hill. Let us suppose there are 2 roads to the top of the hill.
How much work we do and how much is lost to friction and heat depends
on which road we take and how long the road is. However, at the end
of our journey at the top of the hill, the (potential) energy is 
independent of the road we chose. This is why $dQ$ and $dW$ are inexact
differentials but $dE$ is an exact differential.

Note that if $dQ=0$, $d\overline{E}=-dW$ is an exact differential. So if
$Q=0$, then $\Delta\overline{E}_{if}=-W_{if}$. On the other hand, if $dW=0$,
$d\overline{E}=dQ$ and $dQ$ is an exact differential. So if $W=0$,
$\Delta\overline{E}_{if}=Q_{if}$.

\centerline{Exact and Inexact Differentials}
We must now make a small digression to remind ourselves of the difference
between exact and inexact differentials. Consider any function of
two independent variables $F(x,y)$. Then the differential $dF$ is 
defined by
\begin{eqnarray}
dF&=&F(x+dx,y+dy)-F(x,y)=\frac{\partial F}{\partial x}\;dx+
\frac{\partial F}{\partial y}\;dy\nonumber\\
&=&A(x,y)dx+B(x,y)dy\\
\end{eqnarray}
and
\be
\Delta F=F_f-F_i=\int_i^f dF=\int_i^f(A\;dx+B\;dy)
\ee
Note that the integral of an exact differential depends only on the endpoints
(initial and final points) and not on the path of integration. 

However, not every function is an exact differential. Consider
\be
dG=A^{\prime}(x,y)\;dx+B^{\prime}(x,y)\;dy
\ee
It is {\it not} guaranteed that there will exist a function $G(x,y)$ such
that
\be
dG=G(x+dx,y+dy)-G(x,y)
\ee
That is, it is not always true that
\be
\int_i^f dG
\ee
is independent of the path between the endpoints. The integral may depend
on the path of integration. As an example, consider
\be
dG=\alpha dx+\beta \frac{x}{y}\;dy=\alpha dx+\beta x d(\ln y)
\ee
It is easy to show that
\be
\int_{i\rightarrow a\rightarrow f} dG=\alpha + 2\beta\ln 2
\ee
and
\be
\int_{i\rightarrow b\rightarrow f} dG=\alpha + \beta\ln 2
\ee
\begin{center}
\leavevmode
\epsfxsize=2.0 true in
\epsfbox{path.eps}
       \end{center}
Note however that if 
\be
dF\equiv \frac{dG}{x}=\frac{\alpha}{x}dx+\frac{\beta}{y}dy
\ee
then $dF$ is an exact differential with 
\be
F=\alpha \ln x + \beta\ln y
\ee
and 
\be
\int^{f}_{i}dF=\int^{f}_{i}\frac{dG}{x}=(\alpha + \beta)\ln 2
\ee
independent of path. The factor $1/x$ is called an integrating factor for
$dG$.


\end{document}
