Adding figures and most of chapter 8.

chapters/ch6.tex

@@ -11,7 +11,15 @@ \section{Introduction}%
 with stationary nuclei and for nuclei in an electron cloud. The interatomic
 interactions determined by the electron cloud often is difficult to solve so
 empirical and semi-empirical potentials like \textbf{Moore} or
-\textbf{Leonard-Jones} are often used.
+\textbf{Leonard-Jones} are often used. Figure~\ref{fig:diatomicpotential} shows
+an experimentally determined interatomic potential with a dashed line
+representing the Moore's potential.
+\begin{figure}[htpb]
+	\centering
+	\includegraphics[width=0.6\linewidth]{interatomic_potential.png}
+	\caption{Interatomic potential of a diatomic species.}
+	\label{fig:diatomicpotential}
+\end{figure}
 
 \section{Rigid Rotor Harmonic Oscillator}%
 \label{sec:RR}
@@ -81,7 +89,17 @@ \subsection{Energy Levels and Spectra}
 found for different diatomic molecules. \textit{Note,} the vibrational spectra
 consists of one line. One other quick fact is that for a given electronic state
 there is an vibrational energy $D_o$ that corresponds to a disassociating
-molecule.
+molecule. Figure~\ref{fig:diatomicelectronic} shows the potential the
+vibrational mode operates in with some common quantities: $D_o$ represents the
+dissociation energy for the molecule, $D_e$ represents the depth of the
+potential well, the top function represents the first excited electronic state.
+\begin{figure}[htpb]
+	\centering
+	\includegraphics[width=0.7\linewidth]{electronic_states.png}
+	\caption{Electronic potentials for the ground and first excited state with
+	relevant quantities marked.}
+	\label{fig:elecronic_states}
+\end{figure}
 
 \subsection{Deriving the Partition Function}
 In order to derive the partition function, a zero state energy must be assigned.

chapters/ch7.tex

@@ -242,7 +242,7 @@ \subsection{Deriving the Louiville Equation}
 thermodynamical observed variables.
 
 We now turn our attention to the flux of phase points into and out of an
-arbitrary volume about a point $X\inP$. We will prefix quantities in this volume
+arbitrary volume about a point $X\in P$. We will prefix quantities in this volume
 with $\delta$. The instantaneous number of phase points in $\delta V$ is given
 by,
 \begin{equation*}

chapters/ch8.tex

@@ -0,0 +1,288 @@
+\section{Form of Polyatomic Partition Function}%
+\label{sec:polyppf}
+With the knowledge of appropriateness to treat particular degrees of freedom
+classically, we can approach the topic of polyatomic molecule using an
+suitable combination of classical and quantum mechanics. Of course the full
+quantum mechanical solution will be strictly more accurate, but not necessarily
+usefully more accurate.
+\subsection{Approximations}
+In order to derive the vibrational and electronic modes, the Born-Oppenheimer
+approximation is necessary. In doing so we wish to arrive at a description of
+the electronic potential in as a function of a currently unknown number of
+variables. To fully describe a polyatomic molecule's position $3n$ coordinates
+are needed, where $n$ is the number of atoms in the molecule. 3 coordinates can
+describe the center of mass of the particle. 2 $(\theta \text{ and }\phi)$ or 3
+$(\theta,~\phi,\text{ and }\psi)$ coordinates uniquely specify the rotation of a
+rigid linear and non-linear body respectively. This leaves $3n - 5$ or $3n - 6$
+degrees of freedom internal to the molecule. Thus, the internal motion of the
+molecule occurs in a potential energy landscape of $3n - 5$ or $3n - 6$
+dimensions.
+
+Furthermore, in order to deal with vibrational and rotational energy modes
+separately, we will make the rigid-rotor harmonic oscillator approximation
+again.
+
+\subsection{Translational, Electric, and Nuclear Partition Functions}
+The translational partition function is identical to the previous cases except
+that we must be careful to define translation as the movement of the center of
+mass of the particle. Therefore, the translational partition function is
+\begin{equation}
+	q_{trans} = {\left(\frac{2\pi mkT}{h^3}\right)}V.
+\end{equation}
+The electronic partition function is also identical. We set the energy zero
+to the energy where all atoms are separated in their ground state, in
+other words the ground state is at an energy, $-D_e$. The partition function is
+then,
+\begin{equation*}
+	q_{elect} = \omega_{e1} e^{D_{e}/kT}.
+\end{equation*}
+We ignore any state other than the ground state for nuclei and set the nuclear
+partition function, $q_{nucl} = 1$.
+
+\section{Vibrational Partition Function}%
+\label{sec:polyvpf}
+Given the previous analysis of the potential energy field of the nuclei of the
+molecule, the nuclei vibrate in $(3n - 5)$ or $(3n - 6)$ dimensions. These
+relative dimensions will involve combinations of the nuclei's positions such
+that in the squaring of terms leave cross terms. An example would be the
+combination,
+\begin{equation*}
+	(x_2 - x_{1})^{2} = x_{2}^2 - 2 x_1 x_2 + x_1^2.
+\end{equation*}
+The square comes from the assumption that force is roughly linear with small
+deviations from the equilibrium positions this is the harmonic oscillator
+assumption. Luckily using a method called \textit{normal coordinate analysis}
+can determine a set of coordinates $\{Q_{i}\}$ such that each harmonic
+oscillator is devoid of cross terms. The Hamiltonian is thus
+\begin{align*}
+	\H &= K + U\\
+	   &= - \sum_{j=1}^{\alpha}{\frac{\hbar^2}{2 \mu_j}
+	\frac{\partial^2}{\partial Q_j^2}} +
+	\sum_{j=1}^{\alpha}{\frac{k_j}{2}Q_{j^2}} \quad \alpha = 3n - 5
+	\text{ or } 3n - 6
+\end{align*}
+where $\mu_j$ and $k_j$ are effective masses and force constants for these
+normal coordinates respectively. In solving the above Hamiltonian, the total
+energy becomes,
+\begin{equation*}
+	\varepsilon = \sum_{j=1}^{\alpha}{(n_j + \frac{1}{2})h \nu_j}
+	\quad\text{for}\quad n_j = 0, 1, 2, \ldots
+\end{equation*}
+where,
+\begin{equation*}
+	\nu_j = \frac{1}{2\pi} {\left(\frac{k_{j}}{\mu_{j}}\right)}^{1/2}.
+\end{equation*}
+
+The normal coordinates come from analyzing the potential modes of vibration in a
+molecular system. Each independent mode $M$ has a corresponding coordinate $Q$
+that denotes deviation from the equilibrium position inherent to the mode. For
+$CO_2$, an example of normal coordinate analysis can be found, see
+Fig~\ref{fig:polyco2}.
+\begin{figure}[htpb]
+	\centering
+	\includegraphics[width=0.8\linewidth]{co2.png}
+	\caption{$CO_2$ normal coordinate analysis}
+	\label{fig:polyco2}
+\end{figure}
+With the normal coordinates then, the partition function is
+\begin{equation*}
+	q_{vib} = \prod_{j=1}^{\alpha}{\frac{e^{-\theta_{\nu_{j}/2T}}}{(1 -
+	e^{-\theta_{\nu_{j}}/T}}}.
+\end{equation*}
+The energy and heat capacity are,
+\begin{align*}
+	E_{vib} &= Nk \sum_{j=1}^{\alpha}{\left(\frac{\theta_{\nu_j}}{2} +
+			\frac{\theta_{\nu_j} e^{-\theta_{\nu_{j}/2T}}}{1 -
+	e^{-\theta_{\nu_{j}}/T}}\right)}\\
+		C_{v,vib} &= Nk
+		\sum_{j=1}^{\alpha}{\left[\left( \frac{\theta_{\nu_{j}}}{T}\right)^2
+			\frac{e^{-\theta_{\nu_{j}/2T}}}{1 - e^{-\theta_{\nu_{j}}/T}}.
+	\right]}\\
+\end{align*}
+
+\section{Rotational Partition Function}%
+\label{sec:polyrpf}
+The rotational partition function depends a lot on the molecules inherent
+symmetries. A theorem of rigid body motion which is the mode of energy in
+consideration here, is always definable on 3 orthogonal principle axis where the
+inertia tensor becomes diagonal. Each entry in the tensor is defined by
+\begin{equation*}
+	I_{ij} = \sum_{l=1}^{n}{m_l (x_{l,i} - x_{i,cm})(x_{l,j} - x_{j,cm}).
+\end{equation*}
+So the theorem states there are orthogonal coordinate axes that make $I_{ij} =
+0 ~ \text{for} ~ i \ne j$.
+\begin{equation*}
+	\left[\begin{array}{c c c}
+		I_{xx} & 0 & 0\\
+		0 & I_{yy} & 0\\
+		0 & 0 & I_{zz}
+	\end{array}\right].
+\end{equation*}
+From the diagaonalization of the tensor $\overline{I}$ 4 cases exist: 3
+identical $I$'s, 2 equal $I$'s, 3 unique $I$'s, and 2 identical nonzero $I$'s
+and 1 zero $I$. These different cases are known as
+spherical top, symmetric top, asymmetric top, and linear.
+
+As a brief aside, a nomenclature for these 3 values have the subscripts as $I_a,
+I_b, I_c$ and a convenient rescaling of the moments is given by,
+\begin{equation*}
+	{(\overline{A},\overline{B},\overline{C})} = \frac{h}{8\pi I_{a,b,c} c}.
+\end{equation*}
+
+\subsection{Linear Molecules}
+For linear molecules, the problem is the same as the diatomic case. The moment
+of inertia is given by $I = \sum_{i=1}^{n}{m_i r_i^2}$, where $r_{i}^2$ is the
+distance from the center of mass. The final partition function is the same as
+before
+\begin{equation*}
+	q_{rot} = \frac{8 \pi^2 IkT}{\sigma h^2} = \frac{T}{\sigma \Theta_r}.
+\end{equation*}
+Once again $\sigma$ is the symmetry number which represents the number of
+indistinguishable configurations a molecule has.
+
+\subsection{Non-Linear Molecules}
+Non-linear rigid bodies have at least two different diagonal moments of
+inertia. The principle axis are usually located on the axis of symmetry and
+highly symmetric molecules usually have more of their moment of inertia
+components the same.
+
+\subsubsection{Spherical Tops}
+The case of the spherical top applies to molecules like $CH_4$ and $CCl_4$ and
+other highly symmetric molecules. The energy levels of this system are
+\begin{align*}
+	\varepsilon_j &= \frac{J(J+1)\hbar^2}{2I} \qquad J = 0,1,2,\ldots\\
+	\text{with degeneracy}\\
+	\omega_j &= (2J + 1)^2.
+\end{align*}
+The taking the classical approximation $q_{rot}$ is given by
+\begin{equation*}
+	q_{rot} = \frac{1}{\sigma}
+	\int_{0}^{\infty}{(2J + 1)^2 e^{-J(J+1)\hbar^{2}/2kT}\d{J}}.
+\end{equation*}
+In addition, at large values of $J$ the $(J + 1) \approx J$, so
+\begin{equation*}
+	q_{rot} &= \frac{1}{\sigma}
+	\int_{0}^{\infty}{4J^2 e^{-J^2 \hbar^{2}/2IkT}\d{J}}\\
+\end{equation*}
+Let,
+\begin{align*}
+	u &= 2J &\text{and} \quad
+	\d{v} = 2J e^{-J^2 \hbar^{2}/2kT}\\
+	\d{u} &= 2 \d{J}  &\text{and} \quad
+	v = \frac{4IkT}{\hbar^2} e^{-J^2 \hbar^{2}/2IkT},
+\end{align*}
+Here, I note that the integral is directly analogous to that used to prove the
+law of equipartition derived in Section\ref{sec:eoe}. Then, the final solution
+is
+\begin{equation*}
+	q_{rot} = \frac{1}{\sigma}
+	{\left( \frac{8\pi^{2/3} IkT}{h^2} \right)}^{3/2}.
+\end{equation*}
+Note that the $3/2$ power comes from the constant of $v$ and the
+evaluation of $\int{v \d{u}}$ which provides 1 and $1/2$ power
+respectively.
+
+\subsubsection{Symmetric Tops}
+The symmetric top also has a analytical quantum mechanical solution. A theorem
+states that any rigid body with at least $n \ge 3$ fold axis of symmetry is at
+least a symmetric top. The energy levels are
+\begin{align*}
+	\varepsilon_{JK} &= \frac{\hbar^2}{2}
+	{\left( \frac{J(J+1)}{I_A} + K^2
+	{\left[\frac{1}{I_c} - \frac{1}{I_{a}}\right]}
+	\right)}\\
+	J &= 0,1,2,\ldots
+	~ \text{and} ~ K = J, J-1, \ldots, -J\\
+	\text{with degeneracy}\\
+	\omega_{JK} &= (2J+1).
+\end{align*}
+$K$ represents the distribution of energy in the non symmetric axis. By
+definition then,
+\begin{equation*}
+	q_{rot} = \frac{1}{\sigma}
+	\sum_{J=0}^{\infty}{(2J+1) e^{-\alpha_A J(J+1)}}
+	\sum_{K=-J}^{J}{e^{-(\alpha_C - \alpha_{A})K^2}},
+\end{equation*}
+with
+\begin{equation*}
+	\alpha_j = \frac{\hbar^2}{2I_j kT}.
+\end{equation*}
+The classical limit (integral) of this equation evaluates much the same as
+previously using the results of the error function as has been repeatedly done,
+\begin{equation*}
+	q_{rot} = \frac{\pi^{1/2}}{\sigma}
+	{\left(\frac{8\pi^2 I_A kT}{h^2}\right)}
+	{\left(\frac{8\pi^2 I_C kT}{h^2}\right)}^{1/2}.
+\end{equation*}
+\subsubsection{Asymmetric Tops}
+The case of asymmetric tops is only numerically solvable in quantum mechanics;
+thus, a fully classical approach is necessary to get a closed form partition
+function. The classical solution is of the same ilk as the other as one
+might expect,
+\begin{equation*}
+	q_{rot} = \frac{\pi^{1/2}}{\sigma}
+	{\left(\frac{8\pi^2 I_A kT}{h^2}\right)}^{1/2}
+	{\left(\frac{8\pi^2 I_B kT}{h^2}\right)}^{1/2}
+	{\left(\frac{8\pi^2 I_C kT}{h^2}\right)}^{1/2}.
+\end{equation*}
+With rotational temperatures,
+\begin{equation*}
+	q_{rot} = \frac{\pi^{1/2}}{\sigma}
+	{\left(\frac{T^3}{\Theta_A \Theta_B \Theta_{C}}\right)}.
+\end{equation*}
+
+\subsubsection{Hindered Rotation}
+If rotation around a particular bond --- such as the carbon-carbon bond in
+ethane --- causes a meaningful change in potential the rotation is hindered.
+Previously in this chapter, we have only considered molecules whose internal
+rotation occurs in a equa-potential field or internal rotation is unhindered
+(free). When internal rotation is free, such motion does not contribute to the
+partition function. Taking the ethane example, such internal rotation must be
+taken into consideration when deriving the partition function. Let's use the
+potential energy of the rotation of the hydrogen in ethane as shown in
+Figure~\ref{fig:polyhinderedrotation}.
+\begin{figure}[htpb]
+	\centering
+	\includegraphics[width=0.8\linewidth]{hindered_rotation.png}
+	\caption{Generic hindered rotation potential field}%
+	\label{fig:polyhinderedrotation}
+\end{figure}
+
+At low temperatures where $kT \ll V_0$ the energy mode can be treated as a
+harmonic oscillator around a energy minimum. If $kT \gg V_0$ the problem
+rotation is free and a rigid rotor approximation is valid. However, when $kT
+\approx V_0$ a different approach must be taken. In this case, we need an
+analytical form of the potential, in this case $\frac{1}{2} V_0 (1 -
+\cos{3\phi})$. The Schrödinger equation is difficult or impossible to solve
+analytically, but the eigenvalues of the function are known.
+
+\section{Thermodynamic Relations}%
+\label{sec:polytr}
+Here I copy some general thermodynamic relations,
+\begin{align*}
+	q &= q_{trans} q_{rot} q_{vib} q_{elec}\\
+	  &= {\left(\frac{2\pi mkT}{h^3}\right)}^{3/2}V \cdot
+		 \frac{\pi^{1/2}}{\sigma}
+		 {\left(\frac{T^3}{\Theta_A \Theta_B \Theta_{C}}\right)}^{1/2}
+		 {\left(\prod_{j=1}^{3n-6}{\frac{e^{-\theta_{\nu_{j}/2T}}}{(1 -
+		 e^{-\theta_{\nu_{j}}/T})}}\right)}
+		 \omega_{e1} e^{D_o /kT}\\
+	\frac{-A}{NkT} &= \ln{\left(\frac{2\pi MkT}{h^2}\right)}
+		 \frac{Ve}{N} + \ln{\left[\frac{\pi^{1/2}}{\sigma}
+		 {\left(\frac{T^3}{\Theta_A \Theta_B \Theta_{C}}\right)}^{1/2}
+		 \right]}
+		 \sum_{j=1}^{3n-6}{\left(\frac{\Theta_{\nuj}}{2T} +
+		 \ln{(1 - e^{-\Theta_{\nuj}/T})}\right)} +
+		 \frac{D_e}{kT} + \ln{\omega_{e1}}\\
+	\frac{E}{NkT} &= \frac{3}{2} + \frac{3}{2} +
+		\sum_{j=1}^{3n-6}{\left(\frac{\Theta_{\nuj}}{2T} +
+		\frac{\Theta_{\nuj}/T}{e^{\Theta_{\nuj}/T} - 1}\right)}-
+		\frac{D_e}{kT}\\
+	\frac{C_v}{Nk} &= \frac{3}{2} + \frac{3}{2} + 
+		\sum_{j=1}^{3n-6}{\left({\left(\frac{\Theta_{\nuj}}{T}\right)} +
+		\frac{e^{\Theta_{\nuj}/T}}{e^{\Theta_{\nuj}/T} - 1}\right)}\\
+	\frac{S}{Nk} &= \ln{\left[\frac{2\pi MkT}{h^{2}}\right]}^{3/2}
+	\frac{Ve^{5/2}}{N} + \ln{\left[\frac{\pi^{1/2} e^{3/2}}{\sigma}
+	{\left(\frac{T^3}{\Theta_A \Theta_B \Theta_{C}}\right)}^{1/2}\right]}
+	+
+\end{align*}

notes.tex

@@ -5,11 +5,13 @@
 \usepackage{amsmath}
 \usepackage{amssymb}
 \usepackage{enumerate}
+\usepackage{graphicx}
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 
 \title{McQuarrie Stat Mech Notes}
 \author{Brandon Butler}
 \date{2019--04--02}
+\graphicspath{{./media/}}
 
 % User defined commands
 \renewcommand{\d}{\text{d}}
@@ -32,5 +34,7 @@ \chapter{Ideal Diatomic Gas}
 \include{\chapterpath{ch6}}
 \chapter{Classical Mechanics}
 \include{\chapterpath{ch7}}
+\chapter{Polyatomic Partition Function}
+\include{\chapterpath{ch8}}
 
 \end{document}