Reduction to the 2D case

The 2D case is identified here as the waveguide with cylindrical symmetry discussed in section 2.2.3, so the coordinates $r$, $z$ and $\theta$ stand for horizontal distance, depth and ray slope related to the horizontal, respectively. The Gaussian beam expression for this case follows readily from the expression for the 2.5 case, by taking $n_2 = 0$ and $n = n_1$, which provides the expression
P(s,n) = \frac {1}{4\pi}\sqrt{ \frac {c(s)}{c(0)} \frac {\co... + \frac{1}{2}\frac {p(s)}{q(s)} n^2 \right) \right]  ;
\end{displaymath} (3.7)

in this expression $p(s)$ and $q(s)$ are related through the so-called dynamic equations:
\frac{dq}{ds} = c(s) p(s) \hskip5mm  , \hskip5mm \frac{dp}{ds} = -\frac {c_{nn}}{c^2} q(s)  .
\end{displaymath} (3.8)

The second-order derivative along the normal can be written in terms of derivatives along $r$ and $z$ as[1]
c_{nn} = \left( \frac{dr}{dn} \right) ^2c_{rr} + 2 \left( \f...
...} \right) c_{rz} + \left( \frac{dz}{dn} \right) ^2 c_{zz}  ,
\end{displaymath} (3.9)


c_{rr} = \frac{\partial{^2c}}{\partial{r^2}} \hskip5mm , \hs...
c_{rz} = \frac{\partial{^2c}}{\partial{r \partial z}}  ,


\frac{dr}{dn} = -\sin\theta\hskip5mm , \hskip5mm
\frac{dz}{dn} = \cos\theta\hskip5mm .

Those derivatives can be identified as the components of the polarization vector $\mbox{$\mathbf{e}$}_1(s)$, which in the 2D case correspond to:

\mbox{$\mathbf{e}$}_s(s) =
-\sin\theta(s) \\
\end{array} \right]

(see Fig.3.3).

Figure 3.3: Polarization vectors for the 2D case.

The beam width and curvature, $L(s)$ and $K(s)$, respectively[1,9], can be calculated from $p(s)$ and $q(s)$ by comparing the following expressions:

i\omega \frac{1}{2} \frac {p}{q}n^2 = \frac {i\omega K(s)}{2c(s)}n^2 + \frac {1}{L^2}n^2  ,

which yields that
K(s) = c(s) \makebox{Re}\left[ \frac {p(s)}{q(s)} \right]  ,
\end{displaymath} (3.10)

L(s) = \sqrt{ \frac {-2}{\omega \displaystyle { \makebox{Re}\left[ \frac {p(s)}{q(s)} \right] }} }  .
\end{displaymath} (3.11)

As shown by Eq.(3.11) the Gaussian beam approximation requires that $p(s)/q(s) > 0$. To this end it would be sufficient to select a non-zero real value for $p(0)$, plus the choice $q(0) = i\varepsilon$, being $\varepsilon$ a real number, as small as possible.
Orlando Camargo Rodríguez 2012-06-21