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\title{Answers to Examples Class 2}
\author{}
\begin{document}
\maketitle

\section*{Calcium Chloride}

\begin{figure}[!htp]
\begin{center}
\includegraphics[width=0.65\linewidth]{EA7.eps}
\caption{Structure projected along [001]. The Bravais lattice is $P$ (the Ca at the origin has a different environment to that at the centre of the unit cell). Diads parallel to [001] go through each Ca and down the centres of the four sides of the cell containing [001]. (002) planes at $z=0$ and $\frac{1}{2}$ are mirrors. (200) diagonal glide $(n)$ planes cut at $x=\frac{1}{4}$ and $x=\frac{3}{4}$. (020) diagonal glide $(n)$ planes cut at $y=\frac{1}{4}$ and $y=\frac{3}{4}$. $2_1$ screw axes lie in the glide planes at heights $z=\frac{1}{4}$ and $z=\frac{3}{4}$ (only the former is illustrated). Point group $mmm$.}
\label{fig:EA7}
\end{center}
\end{figure} 


\begin{figure}[!htp]
\begin{center}
\includegraphics[width=0.7\linewidth]{EA8.eps}
\caption{Translational symmetry neglected in determining point group, which is $mmm$. Point symmetry of Ca is $2/m$ and that of Cl  is $m$.}
\label{fig:EA8}
\end{center}
\end{figure} 


\section*{Martensite}

The strain which transforms the structure of austenite into martensite is the Bain strain, which involves a compression along one edge and uniform expansion along the other two edges of the unit cell of austenite.


Suppose that the austenite is represented as a sphere with its unit cell
edges denoted by the vectors ${\bf a_i}$ with $i=1,2,3$, as illustrated
in Fig.~\ref{fig:ellipse}a,b. The Bain strain changes the sphere into an ellipsoid of
revolution about ${\bf a_1}$. There are no lines in the
$(0~0~1)_\gamma$ plane which are undistorted. However, it is possible to
find  lines such as $wx$ and $yz$ are undistorted by the deformation, but
are rotated to the new positions $w'x'$ and $y'z'$. Since they are
rotated by the Bain deformation they are not invariant--lines. In fact,
the Bain strain does not produced an invariant--line strain. It can be
converted into an invariant--line strain by adding a rigid body rotation
as illustrated in Fig.~\ref{fig:ellipse}c. 
\begin{figure}[!htp]
\begin{center}
\includegraphics[width=0.9\linewidth]{ellipse.eps}
\caption{Bain strain and rigid body rotation.}\label{fig:ellipse}
\end{center}
\end{figure} 

It is also apparent from Fig.~\ref{fig:ellipse}c that there is no possible rotation
which would convert {\bf B} into an invariant--plane strain because
there is no rotation capable of making two of the non--parallel
undistorted lines into invariant--lines. Thus,  it is impossible to
convert austenite into $\alpha'$ martensite by a strain which is an
invariant--plane strain. A corollary to this statement is that the two
crystals cannot ever be joined at an interface which is fully coherent
and stress--free. 



\section*{Orientation Relationships}
Inspection of the matrix reveals that
\begin{equation}J_{11} + J_{22} + J_{33} =  1 + 2\cos\theta\end{equation}
and the components of the unit vector {\bf u} along the axis of rotation are given by
\begin{equation}\begin{array}{ccc}u_1  & = & {{(J_{23}-J_{32})}/{2\sin\theta}} \cr
u_2 & = & {{(J_{31}-J_{13})}/{2\sin\theta}} \cr
$$u_3 & = & {{(J_{12}-J_{21})}/{2\sin\theta}} \end{array}\end{equation}
It follows that the matrix describes a rotation of 90$^\circ$ about [001]$_X$, and hence represents a symmetry operation so that $\Sigma=1$.

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