\input{../appendix/macros_act}
\chapter{Tau operator in 3-D}
\label{chap:op3d}

Starting with mapping slowness $s_m$ in terms of $x$,$y$, and $z$, we
can transform into tau space through
\beqa
\tau(z,x,y) &=& \int_0^z 2 s_m(z',x,y) dz' \\
x'  &=& x \\
y'  &=& y,  \label{eq:trans3d}
\eeqa
where $\tau$ is the two traveltime, $x'$ is our new $x$ coordinate, and
$y'$ is our new $y$ coordinate.

Using the chain rule we can derive the relationship between
the derivatives of our coordinates,

\beqa
\frac{\partial t}{\partial z} &=&
\frac{\partial t}{\partial \tau} 
\frac{\partial \tau}{\partial z}  +
\frac{\partial t}{\partial x'} 
\frac{\partial y'}{\partial z}  +
\frac{\partial t}{\partial y'} 
\frac{\partial y'}{\partial z}  = 
\frac{\partial t}{\partial \tau} 2\smf \\  
\frac{\partial t}{\partial x} &=&
\frac{\partial t}{\partial \tau} 
\frac{\partial \tau}{\partial x}  +
\frac{\partial t}{\partial x'} 
\frac{\partial x'}{\partial x}  +
\frac{\partial t}{\partial y'} 
\frac{\partial y'}{\partial x}  = 
\frac{\partial t}{\partial x'} + 
\frac{\partial t}{\partial y'} + 
\frac{\partial t}{\partial \tau} \int_0^z \frac{\partial}{\partial x} 2s_m(z',x,y) dz' \\
\frac{\partial t}{\partial y} &=&
\frac{\partial t}{\partial \tau} 
\frac{\partial \tau}{\partial y}  +
\frac{\partial x'}{\partial y}  +
\frac{\partial t}{\partial y'} 
\frac{\partial y'}{\partial y}  = 
\frac{\partial t}{\partial y'} + 
\frac{\partial t}{\partial \tau} \int_0^z \frac{\partial}{\partial y} 2s_m(z',x,y) dz' .
\eeqa

We can simplify the above relations by defining two $\sigma$ quantities,
one in the $x$-direction, $\sigma_x$ and one in the $y$-direction,
$\sigma_y$,

\beqa
\sigma_x(z,x,y) &=& \int_0^z \frac{\partial}{\partial x} 2s_m(z',x,y) dz' \\
\sigma_y(z,x,y) &=& \int_0^z \frac{\partial}{\partial y} 2s_m(z',x,y) dz'  .
\eeqa

Taking the derivative of both sides of the transform of (\ref{eq:trans3d})
we  can obtain a relation for $dz$, $dx$, and $dy$,

\beqa
dz &=& \frac{d \tau}{2 s_m } -\frac{ \sigma_x dx'}{2 s_m} - \frac{\sigma_y dy
'}{2 s_m} \label{eq:der3d1}\\
dx &=& dx' \label{eq:der3d2}\\
dy &=& dy' \label{eq:der3d3}  .
\eeqa

To obtain our tomography operator in 3-D we begin by defining the
the traveltime along a single ray segment in tau space using equations
(\ref{eq:der3d1}-\ref{eq:der3d3}),

\begin{equation}
\widetilde{dt} = \sqrt{ \left(\widetilde{S_f} \widetilde{dx'} \right)^2 + \left( \widetilde{S_f} \widetilde{dy'}\right)^2 +
\left(\frac{\widetilde{ d \tau}- \widetilde{\sigma_x} \widetilde{dx'}- 
\widetilde{\sigma_y} \widetilde{dy'} }{2}\right)^2}\end{equation}
We can then take the derivative with respect to the focusing slowness $s_f$,

\beqa
\frac{ d (\widetilde{dt})} {d S_f} = 
\frac{ \widetilde{S_f} \left( \widetilde{dx'}^2 + \widetilde{dy'}^2\right)}
{\widetilde{dt}} 
{d\widetilde{S_f} \over ds_f} - 
\frac{\left(\widetilde{d\tau} - \widetilde{dx'}
\widetilde{\sigma_x} -\widetilde{dy'}\widetilde{\sigma_y} \right)
}{ 4 \widetilde{dt}} \left( \widetilde{dx'} {d\widetilde{\sigma_x} \over ds_f}
+
\widetilde{dy'} {d\widetilde{\sigma_y} \over ds_f} .
\right)
\eeqa

We new have an expression in terms of  ${d\widetilde{S_f} \over ds_f}$, 
${d\widetilde{\sigma_x} \over ds_f}$, and ${d\widetilde{\sigma_y} \over ds_f}$.
To get an expresion in terms of just $\frac{d\widetilde{S_f}} {ds_f}$
we start by taking the parital derivative of $\tau$ with respect to $x'$,

\beqa
\frac{\partial \tau}{\partial x'} &=& 
\frac{\partial \tau}{\partial x}
\frac{\partial x}{\partial x'} +
 \frac{\partial \tau}{\partial y}
 \frac{\partial y}{\partial x'}+
 \frac{\partial \tau}{\partial z}
 \frac{\partial z}{\partial x'}\\
\frac{\partial \tau}{\partial x}
\frac{\partial x}{\partial x'} &=& 
\frac {\partial}{\partial z}
\int_0^z 2 s(z',x,y) dz'  \nonumber \\
\sigma_x &=& - 2 s_f(\tau, x')
\frac{\partial z}{\partial x}  \nonumber \\
\sigma_x(z,x,y) &=& -  s_f(\tau, x',y)
\int_0^\tau  \frac{\partial}{\partial x'} \frac{1}{ s_f(\tau',x',y')}  d \tau'  \nonumber 
\eeqa

and similarly
\begin{equation}
\sigma_y(z,x,y) =-  s_f(\tau, x,y')
\int_0^\tau  \frac{\partial}{\partial y'} \frac{1}{ s_f(\tau',x',y')}  d \tau' .\end{equation}
We then the derivative with respect to $s_f$ and evalualte at the
ray segment,
\beqa
\frac{\partial  \widetilde{\sigma_x}}{\partial s_f}
&=& {\widetilde{\sigma_x} \over \widetilde{S_f}}
\frac{\partial \widetilde{S_f}} { \partial s_f} -
\widetilde{S_f} \int_0^\tau \frac{\partial^2 s_f(\tau',x',y') }{\partial x'
\partial s_f}
\partial \tau' \\
\frac{\partial  \widetilde{\sigma_y}}{\partial s_f}
&=& {\widetilde{\sigma_y} \over \widetilde{S_f}}
\frac{\partial \widetilde{S_f}} { \partial s_f} -
\widetilde{S_f} \int_0^\tau \frac{\partial^2 s_f(\tau',x',y') }{\partial y'
\partial s_f}
\partial \tau' .  \label{eq:dtau3dop}
\eeqa

%\beq
%\frac{d(dt)}{ds_f} = \fract{\widetilde{S_f}dx'
%\eeq
