\def\figdir{Figs}
%Velocity estimation is fundamentally an inverse problem.
%The correct solution is to do
%full wave-form inversion \cite{GEO51-10-18931903,Mora.sepphd.52}, but
%this is generally impractical. 
Tomography, or at least the definition of tomography I will
be using in thesis, starts from
the idea that there is an operator
$\tomo{nl}$ that relates  slowness $\bf s$
to  travel times  $\bf t$,
\begin{equation}
\bf t = \tomo{nl} \bf s\end{equation}
One way to think of this operator is in terms of rays.  
Figure~\ref{fig:cor.overlay}  shows a simple two layered 
model with sources along
top and left edges and receivers along all four edges.
% every $.2~km$  of the model space every $.2~km$ in depth and every $.5~km$ in
The simplest way to think of
of the operator $\tomo{nl}$ is
as simply  integrating the slowness along the raypath.

\activesideplot{cor.overlay}{width=3.0in,height=3.0in}{ER}
{A simple example of tomography with the sources and receivers every
$.2~km$ in depth and every $.5~km$ in x.  Overlaid are
the raypaths connecting the sources to the receivers.}

This simplistic view of the tomography has a number of inconsistencies
with the {\it reflection tomography problem}.
The first is that we  wouldn't be doing tomography
if we actually knew the slowness field before hand.
What we actually have is some initial guess, $\bf s_0$, of the slowness field.
If we find the raypaths connecting the same source and receivers
in a constant slowness field, 
Figure~\ref{fig:init.overlay}, we encounter
another problem: our raypaths are significantly different.   An alternate
way to state this observation is that our operator is model dependent 
and therefore non-linear.

\activesideplot{init.overlay}{width=3.0in,height=3.0in}{ER}
{The same recording geometry as Figure~\ref{fig:cor.overlay}, with
a constant velocity field.  Note that the ray paths are significantly
different.}

