One of the major difficulties with depth tomography is that our reflectors may be significantly mispositioned. To illustrate this fact I constructed a synthetic model (Figure~\ref{fig:synth-model}). The model is an anticline with six reflectors, one above the anticline, four within the anticline, and one flat reflector representing basement rock. For added difficulty there is a low velocity layer between the second and third reflector. The model was used to do acoustic wave modeling, with the resulting dataset having 32 meter common midpoint (CMP) spacing and 80 offsets spaced 64 meter apart. If we use as our initial estimation of the slowness $s(z)$ function from outside the anticline, the migrated reflector positions are pulled up due to using too low a velocity within the anticline (Figure~\ref{fig:mig0}). Our reflector mispositioning causes significant errors in our linearized tomography problem. For example, if we compare a set of true raypaths (Figure~\ref{fig:rays}) versus our estimated raypaths, we can see that we are back projecting our moveout errors to the wrong positions in the slowness model. Another way to think about it is that our linear approximation, equation (\ref{eq:linear-approx}) is poor, causing us to have to take smaller step sizes per non-linear iteration, and increasing the chance that we will get stuck in local minima or maxima. \plot{synth-model}{height=3.3in,width=6.0in}{Left panel is the synthetic velocity model with six reflectors spanning the anticline. The right panel shows the data generated from this model.} \plot{mig0}{height=3.0in, width=6.0in}{Left panel is our initial guess at the velocity function, the right panel shows the zero offset ray parameter reflector position using this migration velocity. The correct reflector positions are shown as `*'. Note that reflectors are significantly mispositioned.} \plot{rays}{height=3.0in, width=6.0in} {The solid lines show raypath through the correct velocity and the correct reflector position. The dashed lines are raypaths through our initial model and our initial reflector positions. Note that our estimated raypaths have significant error, therefore our tomography operator will have significant error.} \par When doing time domain velocity analysis, reflector mispositioning is not as much of a problem. Reflector positioning in depth is done in two stages \cite{GPR25.04.07380745,GEO46.05.07340750}. The first step is to find the velocity, $v_{\rm rms}$, that best focuses the data. This can be relatively easier in time because reflector position is much less effected by velocity. After $v_{\rm rms}$ has been estimated, the reflectors are mapped in depth using a velocity function, sometimes different than $v_{\rm rms}$. The disadvantage of time domain methods is that they cannot accurately define wave behavior in complex media. This is clearly indicated in Figure~\ref{fig:time}. A single hyperbola can not accurately describe the wave behavior even in this relatively simple model. \sideplot{time}{width=3.0in,height=3.0in}{The solid curve represents the correct travel times from $x=8$ along the bottom reflector to the surface. The dashed lines show the arrival times that time migration would use for three different $v_{\rm rms}s$. Note that both curves differ significantly from the correct curve.} \par The tau domain offers a good alternative that combines the best of both depth and time velocity analysis. The tau, or vertical travel time, domain is defined by the transform \beqa \tau(z,x) &=& \int_0^z 2 s(z',x) dz' \label{eq:transform1} \\ x'(z,x) &=& x \label{eq:transform2} . \eeqa In this domain, flat reflectors are independent of the slowness field and dipping reflector movement is much more limited (Figure~\ref{fig:tau-initial}). Additionally the coordinate transform has not diminished our ability to handle complex wave behavior. \plot{tau-initial}{height=3.0in,width=6.0in}{Left panel is the same model as in Figure~\ref{fig:synth-model} in tau, the right is our initial guess in tau. Note how our reflectors positions are much more accurate in the tau case compared to the depth case (the right panel of Figure~\ref{fig:mig0}).} %Before deriving the a ray based tau tomography it is useful to derive %tau domain Eikonal equation %in this domain. %First, write the Eikonal equation in terms of both a mapping $\vmap$ %and a focusing $\sfoc$ slowness %\beq %{ 1 \over \smap(z,x)^2 }\left[{\partial t (z,x) \over \partial z}\right]^2 + %%{ 1 \over \sfoc(z,x)^2} \left[{\partial t (z,x) \over \partial x}\right]^2 = 1 \label{eq:eik-depth} . %\eeq %We can substitute $\sigma_m$ into %(\ref{eq:eik-depth}) and %end up with a new Eikonal equation in tau, %\beq %4\left[ { \part(\tau,x') \over \partau } \right] + %{1 \over \sfoc(\tau,x')^2 }\left[ {\part(\tau,x') \over \parxp} | \sigma_m(\tau,x') %{\part (\tau,x') \over \partau} \right]^2 =1 . %\eeq %The advantage of the %new Eikonal equation is that while it depends directly on our focusing velocity, %but indirectly on the mapping velocity through $\sigma$. \subsection{Tau operator} In order to perform tomography in the tau domain, we must find an operator equivalent to our depth domain operator (\ref{eq:tomo-base}) in the tau domain. This amounts to applying some simple coordinate transforms to our already derived depth equations. We will start from our relationship for a single ray segment, equation (\ref{eq:segment}). We will begin by rewriting it in terms of focusing and mapping slowness ($s_f$ and $s_m$): \beq \widetilde{dt} = \sqrt{\left(\widetilde{dz}\widetilde{S_m}\right)^2+ \left(\widetilde{dx}\widetilde{S_f}\right)^2} \label{eq:tau-segment} \eeq where $\widetilde{}$ represents a ray segment quantity. By applying the chain rule to (\ref{eq:transform1}) and (\ref{eq:transform2}) we find a relationship between our derivatives: \beqa {\part \over \parz} &=& {\part \over \partau}{\partau \over \parz} + {\part \over \parxp}{\parxp \over \parz} = {\part \over \partau} {2 \smap(z,x)} \\ {\part \over \parx} &=& {\part \over \parxp}{\parxp \over \parx} + {\part \over \partau}{\partau \over \parx} = {\part \over \parxp} + {\part \over \partau} \int_0^z {\partial \over \parx} \left[{ 2 \smap(z',x) } \right] dz' . \eeqa Rather than carrying around the integral we can define a differential mapping factor $\sigma$, \beq \sigma(z,x)= \int_0^z {\partial \over \parx} \left[{ 2 \smap(z',x) } \right] dz' . \eeq The differential mapping factor has non-zero values only in areas at and below where the medium deviates from $s(z)$. For the synthetic model it is only inside the anticline (Figure~\ref{fig:sigma}). \sideplot[t]{sigma}{width=3.0in,height=3.0in}{Sigma values for the synthetic model in Figure~\ref{fig:tau-initial}} We can use equations (\ref{eq:transform1}) and (\ref{eq:transform2}) to write a relationship between change in $(z,x)$ and ($\tau,x')$ space, \beqa \widetilde{dz}&=&{\widetilde{d\tau} \over 2 \widetilde{S_m}} - {\widetilde{\sigma} \widetilde{dx'} \over 2 \widetilde{S_m}} \\ \widetilde{dx}&=&\widetilde{dx} . \eeqa We apply these definitions to equation (\ref{eq:tau-segment}) and come up with a new definition for $dt$ in tau space, \beq \widetilde{dt} = \sqrt{\left( \left(\widetilde{S_f} \widetilde{dx'}\right)^2 + { \widetilde{d\tau} - \widetilde{\sigma} \widetilde{dx'} \over 2} \right)^2} . \eeq We can then take the derivative with respect to $s_f$, \beq {\widetilde{dt} \over ds_f} ={ \widetilde{S_f} \widetilde{dx'}^2 \over \widetilde{dt}} {d\widetilde{S_f} \over ds_f} - {(\widetilde{d\tau} - \widetilde{dx'} \widetilde{\sigma} ) \widetilde{dx'} \over 4 \widetilde{dt}} {d\widetilde{\sigma} \over ds_f} . \eeq Unfortunately, our operator isn't quite as simple as its depth counterpart. In this case we do not have a linear relationship between $d\widetilde{S_f} \over ds_f$ and ${\widetilde{dt} \over ds_f}$ because of an intermediate ${d\widetilde{\sigma}\over ds_f}$ term. We can find a linear relationship between ${d\widetilde{\sigma}\over ds_f}$ and $d\widetilde{S_f} \over ds_f$ by first writing \beqa \frac {\partial \tau }{ \partial x'}&=& \frac{\partial \tau}{\partial x} \frac{\partial x}{\partial x'} + \frac{\partial \tau}{\partial z} \frac{\partial z}{\partial x} \\ \frac{\partial \tau}{\partial x} &=& \frac {\partial \tau }{ \partial x'} - \frac {\partial}{\partial z} \int_0^z 2 s(z',x) dz' \nonumber \\ \sigma(z,x) &=& - 2 s_f(\tau, x') \frac{\partial z}{\partial x} \nonumber \\ \sigma(z,x) &=& - s_f(\tau, x') \int_0^\tau \frac{\partial}{\partial x'} \frac{1}{ s_f(\tau',x')} d \tau' \nonumber . \eeqa We then take the partial derivative with the respect to $s_f$ and evaluate along the ray segment \beq \frac{\partial \widetilde{\sigma}}{\partial s_f} = {\widetilde{\sigma}\over \widetilde{s_f}} \frac{\partial \widetilde{S_f}} { \partial s_f} - s_f(\tau,x') \int_0^\tau \frac{\partial^2 s_f(\tau',x') }{\partial x' \partial s_f} \partial \tau' . \label{eq:dtauop} \eeq With this relationship (\ref{eq:dtauop}) we can then create a new complex tomography operator in tau space: \beq \dt \approx \tautomo \ds \label{eq:tomo-base-tau} . \eeq