In a transversely isotropic medium,
the phase velocity for *P* and *SV* waves
is expressed as follows (Auld, 1990):

(7) | ||

An equation for the transmitted phase angle as a function of
the incident *p* can be obtained after substituting equation (7) into
equation (5), resulting the following fourth order polynomial in
:

(8) |

As shown in Appendix A, the coefficients *a*_{i} are the
same for both *P* and *SV* waves (the in
front of the square root in equation (7) is lost in the derivation of (8))
and they are independent
on the sign of *p*.
In principle, by solving this equation at each interface,
it should be possible to find the
angle of the transmitted phase by computing the arctangent of one
of its roots. However, the procedure is not straight forward. In one hand,
we have to identify the solutions that correspond to *p* or -*p* and
*P* or *SV* waves. On the other hand, finding an angle from its
tangent is not a one to one transformation because and therefore, infinite angles satisfy equation (8).
Besides these problems, we have to deal usually
with the difficulties of solving
equation (8) numerically.

Let's examine first the numerical problems. It is well known that
most algorithms used to find roots of polynomials fail in the presence
of multiple (or close to multiple) roots. This is the case of equation (8)
when *p* = 0. From equation (5), we obtained that in this case
the transmitted phase angle is
and therefore, it is not necessary to solve equation (8) when
*p*=0. Problems may
also arise when because the roots are close of being
multiple. The way I found around this problem was based on the assumption
that the slowness
surface is an smooth function of and then, when
, .If this is true, can
be used as ``starting'' point to find by successive applications
of equation (5) as follows:

(9) |

Figure shows how to solve equation (8) graphically.
From the construction
with the slowness surfaces, we see that the infinite roots of equation
(8) that I mention before correspond to only eight different
points in the slowness surfaces for both *P* and *SV* waves.
In this figure
I assume that the incident phase (not shown) is crossing the interface
from medium 1 to medium 2. For simplicity, I assume that both media
are isotropic. The slowness surfaces in medium 2 are represented one half
by continuous circles and the other half by dotted circles. The slowness
surfaces in medium 1 are not shown. For a given value
at the interface, this figure shows the position of all the
possible points in the *P* and *SV*
slowness surfaces whose angle measured from
the symmetry axis satisfy equation (8). These points are obtained by
simple application of the kinematic boundary condition (3). The points
*A, B, C,* and *D*
are the solutions for negative *p* and the points *H*,
*G, F* and *E*
are obtained by adding to the previous ones
and they are the solutions for positive *p*.
If two of the roots of (8) are complex and the
and there are no transmitted *P* waves. If
the four foots are complex and
there are no transmitted *SV* waves either. I will assume that when
the transmitted ray travels parallel to the interface, it doesn't go
back to the previous medium, which is a simplification of real the behavior of
head waves.

Figure 4

When the axis of symmetry and the normal to the interface
are parallel (axis of symmetry 1 in Figure ),
the arctangent of the roots
of equation (8) give the angles of the points *A*, *B*, *E* and *F*. This
is because the FORTRAN function ATAN gives its results in the
interval and this is exactly the domain of the phase
angle measured from the axis of symmetry. If we take 2 as the axis of symmetry
(which is true here because the medium is isotropic), the roots
of equation (8) give the points *E*, *F*, *G* and *H*.
None of this points is the
correct one when, for example, a *P* wave impinges on the interface
with negative *p*. The point *B* is the solution in this case.

The previous example tells us that when the axis of symmetry and the
normal to the interface are not parallel, the roots of equation
(8) may not give the expected direction for the refracted phase.
For this reason, it is necessary to examine all the eight angles
and pick the one that verifies equation (5) for the desired wave type
(with this, the number of possible angles is reduced from eight to two)
and produces a ray pointing towards the expected direction (which further
reduces the number of angles from two to one). This will be explained in more
detail in the next section, for *SH* waves.

11/18/1997