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%%    Dix revisited: a formalism for rays in layered media
%%    by  Joe Dellinger
%%    and Francis Muir
%%    submitted to the CJEG special issue on anisotropy
%%    containing papers presented at the 5IWSA in Banff, Alberta, Canada
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aniso-trop-ic
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\email{jdellinger@trc.amoco.com, francis@pangea.stanford.edu, respectively}

\lefthead{Dellinger \& Muir}
\righthead{Dix revisited}
\footer{}
\thispagestyle{sep}

\title{DIX REVISITED: A FORMALISM FOR RAYS IN LAYERED MEDIA}
\author{Joe Dellinger and Francis Muir}

\noindent
(Presented at the 5th International Workshop on Seismic Anisotropy
in Banff, Alberta, Canada.
Appeared in the 1993 Journal of the Canadian Society of Exploration
Geophysics special 5IWSA issue on pages 93 to 97.)


\mhead{ABSTRACT}
Dix shows us how to calculate the moveout velocity of
a stack of {\em isotropic\/} layers,
but what about
{\em anisotropic\/} layers requiring higher-order paraxial
approximations?
The usual derivation requires a great deal of algebra
even for the standard hyperbolic-moveout case.
The key is to realize that the Dix equations are
an equivalent-medium theory: they provide a formula for replacing
a heterogeneous layer stack with an equivalent homogeneous block.
Another equivalent medium theory, the Schoenberg-Muir calculus,
suggests a cleaner way of deriving Dix's result.
Identify layer variables that are {\em constant\/}
through the entire stack; these are the ``knowns''.
Identify layer variables that {\em add\/} through the stack,
and express these additive parameters in terms of
the known stack constants and elastic parameters in each layer.
The coefficients multiplying the stack constants in this formula
are the {\em layer-group elements\/}.
Map from layer parameters to layer-group elements, sum over all layers, and
map back to find the {\em equivalent medium\/}.

For the Dix case,
the stack constant is the ray parameter $p = d T / d x$,
the additive variable is the traveltime through each layer $T$, and
the ``layer elastic constants'' are given by a moveout equation $T(x)$.
We are interested in paraxial (near-offset) traveltime behavior;
this suggests finding a power series for $T$ in terms of $x$, but
$T$ in terms of $p$ is also paraxial and a far better choice
because $p$ is {\em constant through all layers\/}.
To get this series,
expand $p = dT(x)/dx$ as a power series in $x$,
revert to obtain $x$ as a series in $p$, and then
substitute into the series for $T$ in terms of $x$.
Because the $T$'s for each layer sum for any given $p$,
{\em the coefficients of each power of $p$ are thus
the additive ``layer-group'' parameters\/}.
For the standard case the first layer-group parameter is
``vertical traveltime'' and the second is ``moveout velocity squared''.
The equivalent-medium algorithm similarly provides
a direct method for calculating the analogous
Dix layer-group parameters for arbitrary anisotropic systems.

\mhead{INTRODUCTION}
The Dix equations are usually written in the form
\begin{equation}
T(0)_{\mbox{\rm\scriptsize total}} = \sum_i T(0)_i
\label{Dixcan1}
\end{equation}
and
\begin{equation}
V_{\mbox{\rm\scriptsize RMS}} =
\sqrt{
{
\sum_i T(0)_i V^2_i
\over
\sum_i T(0)_i
}
}
,
\label{Dixcan2}
\end{equation}
where $T(0)_i$ is the vertical traveltime and
$V_i$ is the moveout velocity for the $i$th layer,
$T(0)_{\mbox{\rm\scriptsize total}}$ is the total traveltime through
the stack, and $V_{\mbox{\rm\scriptsize RMS}}$ is the near-offset
moveout velocity for the stack.
This form for the equations is favored because it corresponds directly
to surface data measurements.
It is also {\em consistent\/}: the stack as a whole is parameterized in the
same way as the individual layers.
This would not have been the case if we had used, for example,
layer thickness and isotropic velocity;
the vertical velocity for the stack as found from
$T(0)_{\mbox{\rm\scriptsize total}}$
is not generally the same as the near-offset moveout velocity
$V_{\mbox{\rm\scriptsize RMS}}$,
so an isotropic parameterization for the stack as a whole is not possible.

The usual method of deriving equations (\ref{Dixcan1}) and~(\ref{Dixcan2})
involves finding a power series for the stack traveltime
in the form
\begin{equation}
T(x)^2 = C_0 + C_1 x^2 + C_2 x^4 + C_3 x^6 + \ldots
\label{powermess}
\end{equation}
and comparing this stack equation with
the corresponding one for a single layer,
\begin{equation}
T(x)^2 = T(0)^2 + {x^2 \over V^2}
;
\end{equation}
see for example section 4.1 in Hubral and Krey (1980).
This method is not particularly well suited
to the task of finding anisotropic extensions
of the Dix equations.
The algebra involved is tedious even for the isotropic case,
and a general expression for the $C_i$ in equation~(\ref{powermess})
provides more information than we need.
We propose instead a direct method that makes use of key
concepts from the Schoenberg-Muir calculus (Schoenberg and Muir, 1989).
Although the Dix equations and the Schoenberg-Muir calculus may appear
to be unrelated, both are {\em equivalent-medium theories\/};
both show how to replace a stack of layers with a bulk
homogeneous equivalent that is (in some sense) indistinguishable from
the heterogeneous stack.
The concepts underlying the Schoenberg-Muir derivation apply equally well
to rays in layered media
and provide a framework for an alternative derivation of the Dix equations
more amenable to anisotropic extension.

We begin with the canonical problem of the type, ``springs in series''.
We then show how the same concepts apply to the Schoenberg-Muir calculus,
and finally recast Dix's model of paraxial rays in layered media.

\mhead{SPRINGS IN SERIES}
A standard high-school physics problem asks the student to find the
effective ``spring constant'' of several springs in series.
Although this problem is trivial it serves to illustrate
in canonical form the fundamental features of all equivalent
layered-medium problems.

Each spring obeys Hooke's law,
\begin{equation}
F_i = k_i  \; \Delta x_i
,
\label{hook}
\end{equation}
where $i$ is the spring number,
$F_i$ is the tension, $k_i$ is the ``spring constant'' (stiffness),
and $\Delta x_i$ is
the displacement from the equilibrium position.

We get two more equations from the way the springs are connected.
First, the tension is the same in all the springs:
\begin{equation}
F_i \equiv F
.
\label{constant}
\end{equation}
Second, the individual displacements of all the springs add to give the
total displacement:
\begin{equation}
\Delta x_{\mbox{\rm\scriptsize total}} = \sum \Delta x_i
.
\label{adds}
\end{equation}

To solve the problem we write the additive term $\Delta x$
as a function of the globally constant tension $F$,
\begin{equation}
\Delta x_i = {1 \over k_i} \;\; F
,
\label{springone}
\end{equation}
and sum over all the springs to find the total displacement
\begin{equation}
\Delta x_{\mbox{\rm\scriptsize total}} =
\sum \Delta x_i =
\sum \Biggl( {1 \over k_i} \;\; F \Biggr) =
\Biggl( \sum {1 \over k_i} \Biggr) \;\; F
.
\label{springtotal}
\end{equation}
Comparing equation (\ref{springtotal}) with~(\ref{springone})
we see that springs in series behave like a single spring with
a stiffness $k_{\mbox{\rm\scriptsize total}}$ determined by the equation
\begin{equation}
{1 \over k_{\mbox{\rm\scriptsize total}}} = \sum {1 \over k_i}
.
\end{equation}

The equivalence of form between
equations (\ref{springone}) and~(\ref{springtotal})
is clearly the key to this problem.
Written this way it is clear that we can add springs in series
by summing their {\em compliances\/} $1/k_i$.
The coefficient on the constant term in equation~(\ref{springone}),
$1/k_i$, provides an alternative way of representing
the spring properties, the ``{\em spring-group\/}'' representation.
There is really no reason except convention which prevented
us from starting this derivation by writing Hooke's law as
\begin{equation}
\Delta x = c \; F
,
\label{springSM}
\end{equation}
with the compliance $c = 1/k$ called the ``spring constant''.
If Hooke's law were normally written this way,
generations of beginning physics students could have answered
``when springs are connected in series the spring constants add''
and actually have been correct.

\shead{The spring group}
Does the ``spring group'' form a group in the formal mathematical sense?
A group is a set of elements and
a binary operator ``$+$'' that satisfies
four conditions (Herstein, 1975):
\begin{list}{}{}{}
\item[1)] Closure: $A$ and $B$ in the group implies $A+B$ in the group.
\item[2)] Associativity: $A$, $B$, $C$ in the group implies that
$A + (B+C) = (A+B) +C$.
\item[3)] Identity: There exists an element $E$ in the group such that
$A + E = E + A = A$ for all $A$ in the group.
\item[4)] Inverse: For every element $A$ in the group there exists another
element $A_{\mbox{\rm\scriptsize inverse}}$ in the group such that
$A + A_{\mbox{\rm\scriptsize inverse}} = A_{\mbox{\rm\scriptsize inverse}} + A = E$.
\end{list}
If in addition a group also satisfies
\begin{list}{}{}{}
\item[5)] Commutativity: For all $A$ and $B$ in the group $A+B = B+A$.
\end{list}
it is said to be an {\em Abelian\/} Group.

For our spring example the elements in the group are springs parameterized
by their compliance $1/k$.
The binary operator ``$+$'' represents connecting two springs in series
and replacing the result with an equivalent single spring.
Note that in terms of compliance (the ``spring-group'' representation)
``$+$'' behaves exactly like standard addition.
Given this definition,
closure, associativity, and commutativity are obvious enough.
The identity element is just the infinitely stiff spring with $1/k = 0$.
In order to have inverses as required we must allow $1/k < 0$;
to get the inverse of a given spring just change the sign
on the spring compliance $1/k$.

It appears that if we allow springs with negative $k$ into our set
the ``spring group'' is indeed a formal Abelian group.
Is there any practical reason to do so?
The group notation does not seem particularly enlightening for the
trivial case of springs in series,
but the group concept of inverse elements is quite a useful one
for more complex equivalent-medium systems such as the Schoenberg-Muir
calculus.
Inverse elements form the theoretical basis for layer stripping and
more generally allow us to decompose a bulk equivalent
(what we can actually observe) back into candidate heterogeneous models
that may represent the real earth.

\mhead{THE CANONICAL METHOD}
The method we have used to solve the simple spring problem is quite general
and applies equally well to other layered-medium problems, for
example the calculus of Schoenberg-Muir and the Dix equations.
For this reason it is worthwhile to pause here and enumerate
the steps we used explicitly:
\begin{list}{}{}{}
\item[1)] Find the general equation describing layer behavior
(e.g., equation~(\ref{hook})).
\item[2)] Divide the variables that occur in the general layer equation
into three classes:
\begin{list}{}{}{}
\item[2a)] The {\bf layer} parameters are assumed given (e.g., the $k_i$).
\item[2b)] The {\bf constant} parameters have the same value for all layers
in the stack (e.g., $F$).
\item[2c)] The {\bf additive} parameters sum through the stack
(e.g., the ${\Delta x}_i$).
\end{list}
\item[3)] Rewrite the equation from step~1 so it has the form
\begin{equation}
{\mbox{\rm\bf additive}\,}_i = \mbox{\rm Function}({\mbox{\rm\bf layer}\,}_i) \cdot
\mbox{\rm\bf constant}
\end{equation}
(e.g., equation~(\ref{springone})).
\item[4)] Identify the term ``$\mbox{\rm Function}({\mbox{\rm\bf layer}\,}_i)$''
as an alternative {\em layer-group\/} representation of the layer parameters
(e.g., the $1/k_i$). To find the homogeneous equivalent of a stack, convert
from the standard representation to the layer group, sum, and convert back.
\end{list}

\mhead{THE SCHOENBERG-MUIR CALCULUS}
We now apply these steps to the the layer-averaging techniques
of Backus (1962), Helbig and Schoenberg (1987) and
Schoenberg and Muir (1989). These methods consider the effect of
a static stress on a welded stack of infinite elastic layers.
Although derived for the static case, the results are also applicable
to the case of propagating waves so long as the wavelength
is much longer than the scale of the layering.

Step~1: Hooke's law (equation~(\ref{hook})) generalizes to
\begin{equation}
\left[ \matrix { \sigma_1 \cr \sigma_2 \cr \sigma_3
\cr \sigma_4 \cr \sigma_5 \cr \sigma_6 } \right] \ =\
\left[ \matrix { C_{11} \  C_{12} \  C_{13}
\  C_{14} \  C_{15} \  C_{16} \cr
C_{12} \  C_{22} \  C_{23} \  C_{24} \  C_{25} \  C_{26} \cr
C_{13} \  C_{23} \  C_{33} \  C_{34} \  C_{35} \  C_{36} \cr
C_{14} \  C_{24} \  C_{34} \  C_{44} \  C_{45} \  C_{46} \cr
C_{15} \  C_{25} \  C_{35} \  C_{45} \  C_{55} \  C_{56} \cr
C_{16} \  C_{26} \  C_{36} \  C_{46} \  C_{56} \  C_{66} }
\right] \
\left[ \matrix { \epsilon_1 \cr \epsilon_2 \cr \epsilon_3
\cr \epsilon_4 \cr \epsilon_5 \cr \epsilon_6 } \right]
,
\label{hookS}
\end{equation}
where
$$
\left[ \matrix { \sigma_1 \cr \sigma_2 \cr \sigma_3
\cr \sigma_4 \cr \sigma_5 \cr \sigma_6 } \right] \ \ =\ \
\left[ \matrix { \sigma_{11} \cr \sigma_{22} \cr \sigma_{33}
\cr \sigma_{23} \cr \sigma_{31} \cr \sigma_{12} } \right]
\ \ \ \ \ \ {\rm and}\ \ \ \ \ \
\left[ \matrix { \epsilon_1 \cr \epsilon_2 \cr \epsilon_3
\cr \epsilon_4 \cr \epsilon_5 \cr \epsilon_6 } \right] \ \ =\ \
\left[ \matrix { \epsilon_{11} \cr \epsilon_{22} \cr \epsilon_{33}
\cr 2 \epsilon_{23} \cr 2 \epsilon_{31} \cr 2 \epsilon_{12} }
\right] \ .
$$
Note that
equation~(\ref{hookS}) is not exactly the three-dimensional equivalent
of equation~(\ref{hook}). In equation~(\ref{hook}) the ``strain'' $\Delta x$
is measured in units of displacement, whereas in equation~(\ref{hookS}) the
strain is a dimensionless
proportionality of the form $\Delta \mbox{\rm length} / \mbox{\rm length}$.
Similarly, in equation~(\ref{hook}) the left hand side is in units of
force, whereas in equation~(\ref{hookS}) the left hand side is in units of
force per unit area.
We will have to take these length normalizations into account in step~2c.

Step~2a: The layer parameters are the stiffness matrix $\bf C$, the layer
thickness $h$, and the layer density $\rho$.

Step~2b: Instead of ``springs in series'' we have
a stack of infinite flat layers with welded interfaces.
(We will assume the layering is normal to the $z$ axis.)
Instead of a constant scalar spring tension
we have that
the stress normal to the layering must be the same in all layers.
Thus equation~(\ref{constant}) becomes
\begin{equation}
\sigma_{3_{\,i}} \ \equiv\  \sigma_3 \ ,\ \ \ \
\sigma_{4_{\,i}} \ \equiv\  \sigma_4 \ ,\ \ \ \
\sigma_{5_{\,i}} \ \equiv\  \sigma_5 \ .
\label{constantS1}
\end{equation}
Because the layers are infinite and welded together and so cannot accommodate
any differential horizontal expansion, we
also find that the strain tangential to the layering must be the same
in every layer:
\begin{equation}
\epsilon_{1_{\,i}} \ \equiv\  \epsilon_1 \ ,\ \ \ \
\epsilon_{2_{\,i}} \ \equiv\  \epsilon_2 \ ,\ \ \ \
\epsilon_{6_{\,i}} \ \equiv\  \epsilon_6 \ .
\label{constantS2}
\end{equation}

Step~2c: The first two additive parameters are obvious ones.
The thickness $h$ of each layer adds,
as does the areal density $h \, \rho$; thus the first two
layer-group elements are trivially $h$ and $h \, \rho$.

To obtain vertical displacements, which sum through the stack,
we must multiply the vertical components of strain by the layer thickness $h$.
Thus equation~(\ref{adds}) becomes
\begin{equation}
h_{\mbox{\rm\scriptsize total}} \, \epsilon_{3_{\,\mbox{\rm\scriptsize total}}} = \sum h_i \; \epsilon_{3_{\,i}} \ ,\ \ \ \
h_{\mbox{\rm\scriptsize total}} \, \epsilon_{4_{\,\mbox{\rm\scriptsize total}}} = \sum h_i \; \epsilon_{4_{\,i}} \ ,\ \ \ \
h_{\mbox{\rm\scriptsize total}} \, \epsilon_{5_{\,\mbox{\rm\scriptsize total}}} = \sum h_i \; \epsilon_{5_{\,i}} \ ,
\label{addsS1}
\end{equation}
where $h_i$ is layer thickness and $h_{\mbox{\rm\scriptsize total}} = \sum h_i$
is the total thickness of the stack of layers.

There is one more additive equation to find.
Think of squeezing our layercake horizontally in a vise;
all layers are forced to undergo the same horizontal distortion.
The various layers will exert differing forces on the vise blades;
to find the total traction, sum over the horizontal forces exerted by each
of the layers.
Stress is force per area, so force is area times stress. The (infinite)
horizontal dimension tangential to the vice blades is the same for all layers
and so can be factored out,
leaving only the vertical thickness $h$ multiplied by the horizontal components
of stress.
We thus have
\begin{equation}
h_{\mbox{\rm\scriptsize total}} \, \sigma_{1_{\,\mbox{\rm\scriptsize total}}} = \sum h_i \; \sigma_{1_{\,i}} \ ,\ \ \ \
h_{\mbox{\rm\scriptsize total}} \, \sigma_{2_{\,\mbox{\rm\scriptsize total}}} = \sum h_i \; \sigma_{2_{\,i}} \ ,\ \ \ \
h_{\mbox{\rm\scriptsize total}} \, \sigma_{6_{\,\mbox{\rm\scriptsize total}}} = \sum h_i \; \sigma_{6_{\,i}} \ .
\label{addsS2}
\end{equation}

Step~3: We can now rearrange equation~(\ref{hookS}) to
segregate the additive and constant elastic terms, obtaining
\begin{equation}
\left[ \matrix {
\sigma_1 \cr \sigma_2 \cr \sigma_6 \cr \cr \sigma_3 \cr \sigma_4 \cr \sigma_5
} \right] \ =\
\left[ \matrix {
C_{11} \  C_{12} \  C_{16} \ \ \ \ \ C_{13} \  C_{14} \  C_{15} \cr
C_{12} \  C_{22} \  C_{26} \ \ \ \ \ C_{23} \  C_{24} \  C_{25} \cr
C_{16} \  C_{26} \  C_{66} \ \ \ \ \ C_{36} \  C_{46} \  C_{56} \cr
\cr
C_{13} \  C_{23} \  C_{36} \ \ \ \ \ C_{33} \  C_{34} \  C_{35} \cr
C_{14} \  C_{24} \  C_{46} \ \ \ \ \ C_{34} \  C_{44} \  C_{45} \cr
C_{15} \  C_{25} \  C_{56} \ \ \ \ \ C_{35} \  C_{45} \  C_{55}
} \right] \
\left[ \matrix {
\epsilon_1 \cr \epsilon_2 \cr \epsilon_6 \cr \cr \epsilon_3 \cr \epsilon_4 \cr \epsilon_5
} \right]
\label{hookS2}
.
\end{equation}
Written explicitly as a block matrix equation
in terms of the additive and constant parameters,
\begin{equation}
\sigma_{\mbox{\rm\scriptsize add}} = 
h
\left[ \matrix {\sigma_1 \cr \sigma_2 \cr \sigma_6 } \right],
\ \ \ \ 
\epsilon_{\mbox{\rm\scriptsize add}} = 
h
\left[ \matrix {\epsilon_3 \cr \epsilon_4 \cr \epsilon_5 } \right],
\ \ \ \ 
\sigma_{\mbox{\rm\scriptsize const}} = 
\left[ \matrix {\sigma_3 \cr \sigma_4 \cr \sigma_5 } \right],
\ \ \ \ 
\epsilon_{\mbox{\rm\scriptsize const}} = 
\left[ \matrix {\epsilon_1 \cr \epsilon_2 \cr \epsilon_6 } \right],
\end{equation}
equation~(\ref{hookS2}) becomes
\begin{equation}
\left[ \matrix {
{1 \over h} \sigma_{\mbox{\rm\scriptsize add}}  \cr \cr \sigma_{\mbox{\rm\scriptsize const}}
} \right] \ =\
\left[ \matrix {
{\mbox{\bf C}}_{TT} \ \ \ \ \ \  {\mbox{\bf C}}_{TN} \cr
\cr
\cr
{\mbox{\bf C}}_{TN}^T \ \ \ \ \ \ {\mbox{\bf C}}_{NN}
} \right] \
\left[ \matrix {
\epsilon_{\mbox{\rm\scriptsize const}}  \cr \cr {1 \over h} \epsilon_{\mbox{\rm\scriptsize add}}
} \right]
\label{hookS3}
.
\end{equation}

After a little matrix algebra we finally obtain Hooke's law in the
desired form,
the elastic equivalent of equation~(\ref{springSM}):
\begin{equation}
\left[ \matrix {
\sigma_{\mbox{\rm\scriptsize add}}  \cr \cr \epsilon_{\mbox{\rm\scriptsize add}}  
} \right] \ =\
h \;
\left[ \matrix {
{\Bigl({\mbox{\bf C}}_{TN} {\mbox{\bf C}}_{NN}^{-1}\Bigr)}
& \ \ \ \ \ &
{\mbox{\bf C}}_{TT}  - {\mbox{\bf C}}_{TN} {\mbox{\bf C}}_{NN}^{-1} {\mbox{\bf C}}_{TN}^T
\cr
\cr
\cr
{\mbox{\bf C}}_{NN}^{-1}
& \ \ \ \ \ &
- {\Bigl({\mbox{\bf C}}_{TN} {\mbox{\bf C}}_{NN}^{-1}\Bigr)}^T
} \right] \
\left[ \matrix {
\sigma_{\mbox{\rm\scriptsize const}}
\cr \cr
\epsilon_{\mbox{\rm\scriptsize const}}
} \right]
\label{hookS4}
.
\end{equation}
Equation~(\ref{hookS4}) is really nothing more
than another form of Hooke's law, one better suited to the geometry of
layered media.

Step~4: The somewhat complicated matrix coefficient
on the constant vector parameter in equation~(\ref{hookS4}),
\begin{equation}
h \;
\left[ \matrix {
{\Bigl({\mbox{\bf C}}_{TN} {\mbox{\bf C}}_{NN}^{-1}\Bigr)}
& \ \ \ \ \ &
{\mbox{\bf C}}_{TT}  - {\mbox{\bf C}}_{TN} {\mbox{\bf C}}_{NN}^{-1} {\mbox{\bf C}}_{TN}^T
\cr
\cr
\cr
{\mbox{\bf C}}_{NN}^{-1}
& \ \ \ \ \ &
- {\Bigl({\mbox{\bf C}}_{TN} {\mbox{\bf C}}_{NN}^{-1}\Bigr)}^T
} \right]
,
\label{SMfinal}
\end{equation}
defines the Schoenberg-Muir layer-group representation of the elastic
stiffness matrix $\bf C$.
Note this $6$ by~$6$ block matrix has $21$ independent elements,
the same number as $\bf C$;
the complete layer group also includes two more independent
parameters, $h$ and $h \, \rho$, for a total of $23$ group elements.

Does the Schoenberg-Muir system form an Abelian group? Yes.
Association and commutativity are obviously satisfied.
The inverse can be formed simply by changing the sign of $h$,
and the identity element is the layer of zero thickness.
The group is closed in the layer-group domain, although
it is possible to create representations in the layer-group domain
by layer subtraction that correspond to physically disallowed elastic media
(Schoenberg and Muir, 1989).
The group concept of subtraction coupled with the Schoenberg-Muir calculus
does provide a useful theoretical framework for decomposing observed elastic
properties into geophysically meaningful components;
for example observed orthorhombic anisotropy can be decomposed into
a transversely isotropic background with fractures
(Hood, 1991) and (Hood and Schoenberg, 1992).

\mhead{THE DIX EQUATIONS}
The familiar Dix equations, equations (\ref{Dixcan1}) and~(\ref{Dixcan2}),
form a paraxial (near-offset) equivalent-medium system
accurate to second order in offset.
Although the moveout recorded over
a stack of even two isotropic layers is not exactly hyperbolic,
it is well known that for near-vertical
propagation a stack of isotropic layers produces a moveout that is close
enough to hyperbolic for most practical purposes.
We next show how to extend the Dix equations to encompass arbitrary
anisotropic moveout using the same canonical equivalent-medium method.
We base our extension here
on the ``first anelliptic anisotropic approximation''
of Dellinger and Muir (1992), although the
solution method itself is completely general.

Step~1:
The first-anelliptic moveout equation is
\begin{equation}
T(x)^2 =
{{
T(0)^4
+
( F_W + 1 )
T(0)^2 V_{\mbox{\rm\scriptsize NMO}}^{-2} x^2
+
F_W^2
V_{\mbox{\rm\scriptsize NMO}}^{-4} x^4
\over
T(0)^2 + F_W V_{\mbox{\rm\scriptsize NMO}}^{-2} x^2
}}
,
\label{An1NMO}
\end{equation}
where
$x$ is the offset,
$T(0)$ is the vertical traveltime,
$V_{\mbox{\rm\scriptsize NMO}}$ is the near-offset NMO velocity,
and $F_W$ is a dimensionless anisotropy parameter.
If $F_W \equiv 1$ the moveout described by equation~(\ref{An1NMO})
becomes exactly hyperbolic, reproducing the standard Dix moveout equation;
the more $F_W$ departs from unity the more nonhyperbolic the moveout becomes.

Step~2a: The layer parameters are vertical traveltime through the layer,
$T_i(0)$, layer moveout velocity, $V_{{\mbox{\rm\scriptsize NMO}}_{\,i}}$,
and layer anisotropy factor, $F_{W_{\,i}}$.

Step~2b: The constant parameter is the {\em ray parameter\/}~$p$.
Although $p$ does not occur directly in equation~(\ref{An1NMO})
we can find it as a function of the other parameters by using
the formula $p(x) = {dT \over dx}(x)$.

Step~2c: The additive parameters are traveltime $T$ and offset $x$.

Step~3: $T$ and $x$ cannot be written as linear functions of $p$
because the Dix model is a paraxial equivalent-medium system
good for ``small'' offsets.
Instead we must express $T$ and $x$ as power series in $p$;
each distinct power of $p$ ($p^0$, $p^2$, $p^4$, etc) then functions
as an independent constant parameter.

The required power series $T(p)$ can be conveniently calculated in
the following way:
\begin{list}{}{}{}
\item[1.] Expand $T(x)$ as a power series about $x=0$.
\item[2.] Differentiate the power series $T(x)$ with respect to $x$, giving
the power series $p(x) = {dT \over dx}(x)$.
\item[3.] Revert this power series for $p(x)$ to obtain a series
for $x(p)$ in terms of powers of $p$. (See Knuth (1981) for details.)
\item[4.] Compose the power series for $T(x)$ with the power series for $x(p)$,
obtaining a series for $T(p)$ in powers of $p$.
\item[5.] The coefficients of the lowest powers of $p$ define the layer-group
representation.
\end{list}
Note that because
\begin{equation}
T(p) = T(0) + p\, x(p) - \int_0^p x(p^\prime) d{p^\prime}
\end{equation}
the series $x(p)$ provides a redundant subset
of the group elements in $T(p)$ and can safely be ignored.

Although this algorithm does require somewhat less algebra than
the standard method described in the introduction, the amount
required can still be formidable.
Fortunately, each step corresponds
to a basic command in the program Mathematica (TM) (Wolfram, 1988),
and the desired result is given {\em directly\/} in the form of
coefficients in the power series $T(p)$.
If we use this algorithm with equation~(\ref{An1NMO}) we obtain
%                           2     2                   2         2     4
%               2    Sqrt[t0 ] lam    3 (1 + 4 f - 4 f ) Sqrt[t0 ] lam
%Out[6]= Sqrt[t0 ] + -------------- + --------------------------------- + 
%                        2 mnmo                          2
%                                                  8 mnmo
% 
%                       2       3       4         2     6
%     5 (1 + 12 f + 12 f  - 56 f  + 32 f ) Sqrt[t0 ] lam          8
%>    --------------------------------------------------- + O[lam]
%                                 3
%                          16 mnmo
\begin{equation}
T(p) = \biggl[ T(0) \biggr] \, p^0 +
\biggl[ T(0) V_{{\mbox{\rm\protect\scriptsize NMO}}}^2 \biggr] \, {1\over 2} p^2 +
\label{dix-group}
\end{equation}
$$
\biggl[ T(0) (1 + 4 F_W - 4 F_W^2) V_{{\mbox{\rm\protect\scriptsize NMO}}}^4 \biggr] \, {3\over 8} p^4 +
$$
$$
\biggl[ T(0) (1 + 12 F_W + 12 F_W^2 - 56 F_W^3 + 32 F_W^4) V_{{\mbox{\rm\protect\scriptsize NMO}}}^6 \biggr] \, {5\over 16} p^6 +
\ldots
.
$$

Step~4:
From the $p^0$ term we find the first Dix layer-group element, ${T(0)}$.
This group element corresponds to equation~(\ref{Dixcan1}),
``vertical traveltimes add''.
From the $p^2$ term we find the second Dix layer-group element,
${{T(0)} {V_{{\mbox{\rm\protect\scriptsize NMO}}}}^2}$.
This group element corresponds to Dix's familiar RMS-velocity equation,
equation~(\ref{Dixcan2}).

In the standard case the first two powers of $p$ exhaust the available
free layer parameters $T(0)$ and
${V_{{\mbox{\rm\protect\scriptsize NMO}}}}$, and
the coefficient of $p^4$ cannot also be made consistent.
The Dix equations thus form a paraxial equivalent-medium theory exact
up to second order in $p$.
(Note that the speed of convergence of the series in equation~(\ref{dix-group})
actually depends on the magnitude of a {\em dimensionless\/} stack parameter,
$p\, {V_{{\mbox{\rm\protect\scriptsize NMO}}}}$.)

For the first-anelliptic approximation the $F_W$ layer parameter yet remains,
and so the $p^4$ coefficient in equation~(\ref{dix-group})
defines a third Dix layer-group element,
\begin{equation}
T(0)
V^4_{{\mbox{\rm\protect\scriptsize NMO}}}
(1 + 4 {F_W} - 4 {F_W}^2)
\ \ .
\label{dix-group3}
\end{equation}
If $F_W \neq 1$ the calculated stack moveout is {\em non\/}hyperbolic.
Nonhyperbolic moveout can be caused by intrinsically anisotropic
layers within the stack ($F_{W_{\,i}} \neq 1$), but
more generally it will also be caused by
ray bending at layer boundaries not accounted for by the first two
layer-group elements.
The three-term first-anelliptic extension of the Dix
equivalent-medium system is exact up to order $p^4$ and so
should do a better job of accounting for ray bending than the
first two terms alone can, at least paraxially.
(Note we have been discussing the power series $T(p)$ here,
not the series $T^2(x^2)$ which is more commonly found
in the literature; see for example Hake et. al. (1984).)

Our Dix layer group also forms a true Abelian group.
In terms of the layer-group parameters there is no difficulty satisfying
closure, associativity, and commutativity.
The identity is the zero-traveltime layer, and to obtain the inverse
just change the sign of $T(0)$.
As was the case for the Schoenberg-Muir layer group,
it is possible to create
representations of physically disallowed media by subtraction
in the Dix layer-group domain.
For example, the infamous layers with imaginary interval
velocities that are the bane of layer strippers everywhere
are the result of subtraction in the Dix layer-group domain without
due regard for physical constraints.
For our first-anelliptic example uncareful subtraction will similarly
create layers with complex $F_W$.

\mhead{CONCLUSIONS}
We have shown that the Schoenberg-Muir calculus and the Dix equations
share a common theoretical underpinning: both are equivalent-medium
systems, and both can be derived in a similar way.
The equivalent-medium formulation of the Dix model provides a
convenient direct algorithm for finding the extension
of the Dix equations to anisotropic layered-medium systems.

\ACK
Francis Muir wishes to acknowledge the Stanford Exploration Project
for support.
This is University of Hawai`i at M\=anoa
%
% Note the glottal stop (backwards single quote) between the two i's in Hawaii
% and the macron (long bar) over the first a in Manoa...
% If the Quebeckers can insist on their accents and cedillas, I might as well
% ask you to try to spell Hawaii and Manoa correctly! :-)
%
School of Ocean and Earth Science and Technology publication
number~3136.

\clearpage
\mhead{REFERENCES}
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by horizontal layering: J. Geophys. Res., {\bf 67}, 4427--4440.}
\reference{Dellinger, J., and Muir, F., 1993, Anelliptic approximations for TI
media: Journal of Seismic Exploration, {\bf 2}, 23--40.}
\reference{Hake, H., Helbig, K., and Mesdag, C. S., 1984,
Three-term Taylor series for $t^2$-$x^2$ curves of P- and S-waves over layered
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\reference{Hood, J.~A., 1991, A simple method for decomposing
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\reference{Hood, J.~A. and Schoenberg, M., 1992,
NDE of fracture-induced anisotropy,
{Rev.\ Prog.\ Quantitative NDE,} {\bf 11B},
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\reference{Hubral, P., and Krey, T., 1980,
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\reference{Knuth, D. E., 1981,
The art of computer programming, volume 2: seminumerical algorithms:
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\reference{Schoenberg, M., and Muir, F., 1989, A calculus for finely
layered anisotropic media: Geophysics, {\bf 54}, 581--589.}
\reference{Wolfram, S., 1988, Mathematica: A System for Doing Mathematics
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\end{document}
