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\centerline{\bf LECTURE 10} 
\centerline{\bf Simple Applications of Statistical Mechanics}
We have seen that if we can calculate the partition function 
\be
Z=\sum_r e^{-\beta E_r}
\ee
then we can derive just about anything we want to know from the
partition function such as the mean internal energy, the entropy, the pressure,
the Helmholtz free energy, the specific heat, etc. Let's give some
examples of how this works.

\centerline{Paramagnetism}
You may remember from E\&M that the magnetic behavior of materials usually
arises from the angular momentum of the electrons. The electrons have 2 forms
of angular momentum: spin and orbital. Classically their
orbital angular momentum is associated with their orbital motion 
around the nucleus. Quantum mechanically the orbital angular momentum
is associated with their spatial wavefunction. Even if an electron
has zero orbital angular momentum, it still has acts like a tiny magnet
and has a magnetic moment $\approx\vec{\mu}_B$. Associated with this magnetic
moment is the spin angular momentum $\hbar\vec{S}$ of the electron.
Protons and neutrons also have angular momentum $\hbar\vec{J}$ and magnetic
moments $\vec{\mu}$. This gives rise to nuclear magnetic moments which
are involved in NMR and MRI.
The magnetic moment of an atom, ion or single elementary particle such
as an electron or proton in free space is proportional to its angular
momentum:
\be
\vec{\mu}=g\mu_B\vec{J}=-\gamma \hbar \vec{J}
\ee
where the proportionality factor $g$ is the ``spectroscopic splitting''
factor or just the g factor, $\mu_B$ the Bohr magneton, $\gamma$
the gyromagnetic ratio, and $\hbar \vec{J}$ the total angular
momentum. Rather than considering the case of arbitrary $\vec{J}$
(see Reif 7.8, pages 257-262), let us begin by considering the
simple case of spin-1/2 where $J=1/2$ and $g=2$. Then the allowed
quantum energy states are
\begin{eqnarray}
E&=&-\vec{\mu}\cdot \vec{H} \nonumber\\ 
&=&-g\mu_B\vec{J}\cdot \vec{H} \nonumber\\
&=&-g\mu_B J_zH \;\;\;\;\;\;\;\;\;({\rm since}\;\;\vec{H}\|\hat{z})\nonumber\\
&=&-2\mu_B(\pm\frac{1}{2})H\nonumber\\
&=&\pm\mu_BH
\end{eqnarray}
where $H$ is the external magnetic field. Let us also assume that we have $N$
such independent ions or atoms all in contact with a heat reservoir
at temperature $T$. A physical system that approximates this model is
a metal with magnetic impurities like Mn$^{2+}$.

If we consider only one atom or ion, the partition function $\xi$ of
that one atom or ion becomes
\be
\xi=\sum_{\rm all\;states\;r}e^{-\beta E_r}=e^{\beta\mu_BH}
+e^{-\beta\mu_B H}
\label{eq:xi}
\ee
and
\be
Z=\xi^N=\left(e^{\beta\mu_BH} +e^{-\beta\mu_B H}\right)^N
\label{eq:Z2state}
\ee
The mean energy can be calculated from
\be
\overline{E}=-\frac{\partial \ln Z}{\partial \beta}
=-N\frac{\partial\ln\xi}{\partial \beta}=
-\frac{N}{\xi}\mu_B H\left(e^{\beta\mu_BH} +e^{-\beta\mu_B H}\right)^N
\ee
\be
\overline{E}=-N\mu_B H\;\;\frac{e^{\beta\mu_BH} -e^{-\beta\mu_B H}}
{e^{\beta\mu_BH} +e^{-\beta\mu_B H}}
\ee
\be
\overline{E}=-N\mu_B H\tanh \left(\frac{\mu_B H}{k_BT}\right)
\ee
The total magnetic moment or magnetization can be calculated from
\begin{eqnarray}
\overline{M}&=&N\overline{\mu}\nonumber\\
&=&N\frac{\mu_{\uparrow}-\mu_{\downarrow}}{\mu_{\uparrow}+\mu_{\downarrow}}
\nonumber\\
&=&N\;\;\frac{e^{\beta\mu_BH} -e^{-\beta\mu_B H}}
{e^{\beta\mu_BH} +e^{-\beta\mu_B H}}
\end{eqnarray}
or
\be
\overline{M}=N\mu_B\tanh \left(\frac{\mu_B H}{k_BT}\right)
\ee
Alternatively we can calculate the generalized force associated with
the magnetic field. Recall that
\be
dW=\overline{X}dx
\ee
and
\be
\overline{X}=\frac{1}{\beta}\frac{\partial\ln Z}{\partial x}
\ee
>From Reif 11.1 we find that magnetic work is
\be
dW=-\vec{M}\cdot d\vec{H}
\ee
where $\vec{M}$ is the magnetization. Suppose $\vec{H}$ is pointing
along the $\hat{z}$ axis. We can identify $H$ with the external parameter
$x$, and $M$, the z--component of $\vec{M}$, with the generalized force. 
Then the generalized force is
\begin{eqnarray}
M&=&\frac{1}{\beta}\frac{\partial\ln Z}{\partial H}\nonumber\\
&=&\frac{N}{\beta}\frac{\partial\ln\xi}{\partial H}\nonumber\\
&=&N\mu_B\;\;\frac{e^{\beta\mu_BH} -e^{-\beta\mu_B H}}
{e^{\beta\mu_BH} +e^{-\beta\mu_B H}}\nonumber\\
&=&N\mu_B\tanh\left(\frac{\mu_B}{k_BT}\right)\nonumber\\
&=&-\frac{\overline{E}}{H}
\end{eqnarray}
We see that
\be
\overline{E}=-\overline{M}H
\ee

We can examine $\overline{M}$ in the limit of high and low temperature.
\be
{\rm If}\;\;T\rightarrow 0\;\;\left({\rm or}\;\;
\frac{\mu_BH}{k_BT}\rightarrow\infty\right)\;\;\;\;\;\;\;
\tanh\left(\frac{\mu_B}{k_BT}\right)\rightarrow 1\;\;{\rm and}\;\;
\overline{M}\rightarrow N\mu_B
\ee
Physically at low temperatures all the magnetic moments are aligned with
the magnetic field.
\begin{center}
\leavevmode
\epsfxsize=2.0 true in
\epsfbox{tanh.eps}
       \end{center}

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

At high temperatures, $\mu_BH/k_BT\rightarrow 0$ and $T\rightarrow \infty$.
Since $\tanh(x)\rightarrow x$ as $x\rightarrow 0$, we obtain
\be
\overline{M}\cong N\mu_B\left(\frac{\mu_BH}{k_BT}\right)=
\frac{N\mu_B^2 H}{k_B}\;\frac{1}{T}
\label{eq:Curie}
\ee
We expect the magnetization to be proportional to the magnetic field $H$.
This is what characterizes a paramagnet.
How easy or hard it is to magnetize the system is reflected in the magnetic
susceptibility $\chi$:
\be
M=\chi H
\ee
>From eq. (\ref{eq:Curie}) we can read off the susceptibility:
\be
\chi=\frac{\overline{M}}{H}=\frac{N\mu_B^2}{k_B}\;\frac{1}{T}
\;\;\;\;\;\;\;\;\; T\rightarrow \infty \;\;({\rm high \; temperature\;
limit})
\ee
The fact that $\chi$ varies as $1/T$ is know as the Curie law. $\chi$ is
called the Curie susceptibility because it goes as $1/T$.
\begin{center}
\leavevmode
\epsfxsize=2.0 true in
\epsfbox{Curie.eps}
       \end{center}

We can also calculate the Helmholtz free energy using eqns. (\ref{eq:Z2state})
and (\ref{eq:xi}):
\be
F=-k_BT\ln Z=-k_BT\ln(\xi^N)=-Nk_BT\ln \xi
\ee
where 
\be
\xi=e^{\beta\mu_BH} +e^{-\beta\mu_B H}=2\cosh(\beta \mu_BH)
=2\cosh\left(\frac{\mu_BH}{k_B T}\right)
\ee
So
\be
F=-Nk_BT\ln\left(2\cosh\left(\frac{\mu_BH}{k_B T}\right)\right)
\label{eq:F2state}
\ee
We can also calculate the heat capacity. Recall from lecture 9 that
\be
C_V=-T\frac{\partial^2 F}{\partial T^2}
\ee
Taking two derivatives of (\ref{eq:F2state}) yields
\be
C_V=Nk_B\left(\frac{\mu_BH}{k_B T}\right)^2
{\rm sech}^2\left(\frac{\mu_BH}{k_B T}\right)
\ee
The specific heat per spin is
\be
c_V=\frac{C_V}{N}=k_B\left(\frac{\mu_BH}{k_B T}\right)^2
{\rm sech}^2\left(\frac{\mu_BH}{k_B T}\right)
\ee
This is called the Schottky specific heat. It is the specific heat
characteristic of two state systems. A two state system has only 2 states
available to it. For example a spin--1/2 object has spin--up and spin--down
states. Another example is an object that can only access
the lowest states in a double well potential. In considering such discrete
states, we are thinking quantum mechanically.
\begin{center}
\leavevmode
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\centerline{Classical Harmonic Oscillator}
Consider a classical harmonic oscillator with a spring constant $\kappa$.
The typical example is a mass on a spring. The energy is given by
\be
E=\frac{p^2}{2m}+\frac{1}{2}\kappa x^2
\ee
Assume the harmonic oscillator is in contact with a thermal reservoir at 
temperature $T$. The partition function is
\begin{eqnarray}
Z&=\& \frac{1}{h_o}\int e^{-\beta E}dxdp\nonumber \\
&=& \frac{1}{h_o}\int^\infty_{-\infty}dp\; 
\exp\left(-\frac{\beta p^2}{2m}\right) 
\int^{\infty}_{-\infty}dx\; \exp\left(-\frac{\beta\kappa x^2}{2}\right) 
\label{eq:ZSHO}
\end{eqnarray}
Recall from Reif Appendix A4 that
\be
\int_{-\infty}^{\infty}e^{-\alpha x^2}dx=\left(\frac{\pi}{\alpha}\right)^{1/2}
\ee
Using this (\ref{eq:ZSHO}) becomes
\be
Z=\left(\frac{2\pi}{h_o\beta \kappa}\right)^{1/2}
\left(\frac{2\pi m}{\beta h_o}\right)^{1/2}=\frac{1}{\beta}
\frac{2\pi}{h_o}\left(\frac{m}{\kappa}\right)^{1/2}
\ee
Or
\be
\ln Z=-\ln \beta+\ln\left(
\frac{2\pi}{h_o}\left(\frac{m}{\kappa}\right)^{1/2}\right)
\ee
and 
\be
\overline{E}=-\frac{\partial\ln Z}{\partial\beta}=\frac{\partial\ln\beta}
{\partial \beta}=\frac{1}{\beta}=k_B T
\ee
This makes sense; we expect the mean energy to be of order $k_BT$.
It turns out that $\overline{E}-k_BT$ has equal contributions from
the kinetic energy and from the potential energy. Each contributes 
$k_BT/2$. One can show this explicitly. The mean kinetic energy is
\be
\overline{\rm KE}=\frac{\int_{-\infty}^{\infty}\frac{p^2}{2m}
e^{-\beta p^2/2m}dp}{\int^{\infty}_{-\infty}dp e^{-\beta p^2}{2m}
\ee
Use Reif Appendix A4 to evaluate the numerator and denominator:
\be
\int_{-\infty}{\infty}e^{-\alpha x^2}dx=\left(\frac{\pi}{\alpha}\right)^{1/2}
\ee
and
\be
\int_{-\infty}{\infty}e^{-\alpha x^2}\;x^2\;dx=\frac{\sqrt{\pi}}{2}
\alpha^{-3/2}
\ee
So
\begin{eqnarray}
\overline{KE}&=&\frac{\frac{1}{2m}\frac{\sqrt{\pi}}{2}\left(\frac{2m}{\beta}
\right)^{3/2}}{\sqrt{\pi}\left(\frac{2m}{\beta}\right)^{1/2}}
\nonumber\\
&=&\frac{k_BT}{2}
\end{eqnarray}
Similarly the mean potential energy is given by 
\begin{eqnarray}
\overline{PE}&=&\frac{\int^{\infty}_{-\infty}dx\frac{1}{2}\kappa x^2
e^{-\beta\kappa x^2/2}}{\int_{-\infty}^{\infty}dx e^{-\beta\kappa x^2/2}
\nonumber\\
&=&\frac{\frac{\kappa}{2}\frac{\sqrt{\pi}}{2}\left(\frac{2}{\beta\kappa}
\right)^{3/2}}{\sqrt{\pi}\left(\frac{2}{\beta\kappa}\right)^{1/2}}\nonumber\\
&=&\frac{k_B T}{2}
\end{eqnarray}
The fact that the mean kinetic and potential energies of a harmonic
oscillator equals $k_BT/2$ is an example of the classical equipartition theorem.

\centerline{Classical Equipartition Theorem}
Let the energy of a system with $f$ degrees of freedom
be $E=E(q_1...q_f,p_1...p_f)$ and assume 
\begin{enumerate}
\item the total energy splits additively into the form
\be
E=\epsilon_i(p_i)+E^{\prime}(q_1,...,p_f)
\ee
where $\epsilon_i$ involves only $p_i$ and $E^{\prime}$ is independent of
$p_i$.
\item 
\be
\epsilon_i(p_i)=bp_i^2
\ee
where $b$ is a constant.
\end{enumerate}
If we replaced $p_i$ by $q_i$, the theorem is still true. We just want
$\epsilon_i$ to be a quadratic function of one component of $q$ or $p$.

Assume that the system is in equilibrium at temperature $T$. Then
\begin{eqnarray}
\overline{\epsilon}_i&=&\frac{\int_{-\infty}^{\infty}\epsilon_i e^{-\beta E}
dq_1...dp_f}{\int_{-\infty}^{\infty}e^{-\beta E} dq_1...dp_f}\nonumber\\
&=&\frac{\int_{-\infty}^{\infty}\epsilon_i e^{-\beta (\epsilon_i+E^{\prime})}
dq_1...dp_f}{\int_{-\infty}^{\infty}e^{-\beta (\epsilon_i+E^{\prime})} 
dq_1...dp_f}\nonumber\\
&=&\frac{\int_{-\infty}^{\infty}\epsilon_i e^{-\beta \epsilon_i}
dp_i\int_{-\infty}^{\infty}e^{-\beta E^{\prime}}
dq_1...dp_f}
{\int_{-\infty}^{\infty}e^{-\beta \epsilon_i}dp_i 
\int_{-\infty}^{\infty}e^{-\beta E^{\prime}} 
dq_1...dp_f}
\end{eqnarray}
where the last integrals in the numerator and denominator do not involve  
$dp_i$. These integrals cancel leaving
\begin{eqnarray}
\overline{\epsilon}_i&=&\frac{\int^{\infty}_{-\infty}\epsilon_i
e^{-\beta\epsilon_i}dp_i}{\int^{\infty}_{-\infty}e^{-\beta\epsilon_i}dp_i}
\nonumber\\
&=&\frac{-\frac{\partial}{\partial\beta}\int_{-\infty}^{\infty}
e^{-\beta\epsilon_i}dp_i}{\int^{\infty}_{-\infty}e^{-\beta\epsilon_i}dp_i}
\nonumber\\
&=&-\frac{\partial}{\partial\beta}\ln\left(
\int^{\infty}_{-\infty}e^{-\beta\epsilon_i}dp_i\right)
\label{eq:756}
\end{eqnarray}
Now use the second assumption:
\be
\int^{\infty}_{-\infty}e^{-\beta\epsilon_i}dp_i=
\int^{\infty}_{-\infty}e^{-\beta bp^2_i}dp_i=
\beta^{-1/2}\int^{\infty}_{-\infty}e^{-by^2}dy
\ee
where $y\equiv\beta^{1/2}p_i$. Thus
\be
\ln\left(\int^{\infty}_{-\infty}e^{-\beta\epsilon_i}dp_i\right)=
-\frac{1}{2}\ln\beta+\ln\int^{\infty}_{-infty}e^{-by^2}dy
\ee
Notice that the integral does not involve $\beta$. So when we take the 
derivative in (\ref{eq:756}), only the first term is involved.
\be
\overline{\epsilon}_i=-\frac{\partial}{\partial\beta}\left(-\frac{1}{2}
\ln\beta\right)=\frac{1}{2\beta}
\ee
or
\be
\overline{\epsilon}_i=\frac{1}{2}k_BT
\label{eq:equipartition}
\ee
This is the classical equipartition theorem. It says that each additive
quadratic term in the energy  (i.e., each degree of freedom) contributes
$k_BT/2$ to the mean energy of the system. 

Note that this theorem is true only in classical statistical mechanics as 
opposed to quantum mechanics. In quantum mechanics the energy of a system
is discretized into energy levels. At low energies the levels are rather far
apart. As the energy increases, the energy levels become more closely 
spaced. The equipartition theorem holds when mean energy is such that the
levels near it have an energy level spacing $\Delta E\ll k_BT$.

\centerline{Mean kinetic energy of a gas molecule}
Let's consider the simple example of a molecule in a gas (not necessarily
an ideal gas) at temperature $T$. Its kinetic energy is given by
\be
K=\frac{1}{2m}\left(p^2_x+p^2_y+p^2_z\right)
\ee
The kinetic energy of the other molecules do not involve the momentum $\vec{p}$
of this particular molecule. The potential energy of interaction between
the gas molecules also is independent of $\vec{p}$. So the equipartition
theorem tells us that 
\be
\overline{K}=\frac{3}{2}k_BT
\ee
For an ideal monatomic gas the energy is solely kinetic energy, so
\be
\overline{E}=N\left(\frac{3}{2}k_BT\right)=\frac{3}{2}\nu RT
\ee
where $N$ is the number of gas particles, $R=N_a k_B$ is the gas constant,
and $\nu$ is the number of moles. The molar specific heat at constant volume
is
\be
c_V=\frac{C_V}{\nu}=\left(\frac{\partial \overline{E}}{\partial T}\right)_V=
\frac{3}{2}R
\ee

\centerline{Brownian Motion}
Brownian motion was discovered by Brown, a botanist, in the 1800's and 
was explained by Einstein in 1905.
If you put a small macroscopic particle in a liquid and 
watch it in a microscope,
it jiggles around because the molecules in the liquid keep bumping into it.
This is called Brownian motion.
The molecules in the liquid are moving around because of thermal fluctuations.
To see this, let the small macroscopic particle have a mass $m$ and be
immersed in a liquid at temperature $T$. Consider the $x-$component of the
velocity $v_x$. 
\be
\overline{v}_x=0
\ee
Even though the mean value of the velocity vanishes, this does not mean
that $v_x=0$. There are velocity fluctuations so that $\overline{v^2_x}\neq 0$.
>From the equipartition theorem we have
\be
\overline{\frac{1}{2}mv^2_x}=\frac{1}{2}k_BT
\ee
or
\be
\overline{v^2_x}=\frac{k_BT}{m}
\ee
The factor of $1/m$ means that the fluctuations are negligible for 
large $m$ like a golf ball. But when $m$ is small (e.g., when the
particle is micron sized), then the velocity fluctuations become
appreciable and can be observed under a microscope. Notice that the
size of the fluctuations are proportional to temperature. The higher the
temperature, the larger the fluctuations. This is what we expect of
thermal fluctuations.
\end{document}
