\title{Why Stationary the Stationary Phase analysis and the Fresnel Zone method are equivalent
to evaluate a surface integral?}
\author{
Herman Jaramillo \\
}
\date{\today}
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\begin{document}
\maketitle
\section{Introduction}
This document was motivated by a discussion with my friend Andreas Rueger
about the meaning of the Huygens principle and the work of Fresnel.
I evaluated the Fresnel--Kirchhoff integral using the method of stationary
phase and then rewrote Born and Wolf~\cite{born} notes about Fresnel method to
compute the wavefield as an integration over a wavefront following
Huygens principle. Both results agree. Is this a coincidence?
I found an article by Kampen~\cite{kampen}
\href{http://igitur-archive.library.uu.nl/phys/2006-1207-201356/kampen_58_method.pdf}{in
this}~\footnote{
http://igitur-archive.library.uu.nl/phys/2006-1207-201356/kampen\_58\_method.pdf}
website, which explains the fact that those evaluations are equivalent.
I use Kampen's work as a reference for these notes. Loosly speaking, the
Huygens principle state that the wavefield at some point is an integration
over any wavefront before the point was reached. In integral terms this is
\section{The equivalence}
\begin{figure}[ht]
\caption{A wave starts at a point source $P$. After sometime, a wavefront
is located in the blue sphere (only a main cross--section in the $xy$
plane is shown). A disturbance at point $P$, according to Huygens principle
is understood as the superposition of all disturbances starting at the
wavefront.}
\begin{tikzpicture}[>=latex]
\coordinate [label=150:](a) at (0,0) node[below] {$P_0$};
\coordinate (p) at (-1: 12 cm);
\draw[line width=2.0] (a) -- (p) node[below] {$\quad P$};
%arc above
\draw[color=blue, line width=2] (0:4cm) arc (0:90:4cm) node[above] {$S$};
%arc below
\draw (0:4cm)[color=blue, line width=2] arc (0:-90:4cm) ;
%arcs
\draw (0cm:4cm) node[below] {$\quad C$} arc (180:188:8cm)
node[below] {\scriptsize $\; Z_1$};
\draw (0cm:4cm) arc (180:172:8cm) ;
\draw (0cm:3.8cm) arc (180:190:8.2cm)
node[below] {\scriptsize $\; Z_2$};
\draw (0cm:3.8cm) arc (180:170:8.2cm);
\draw (0cm:3.6cm) arc (180:192:8.4cm)
node[below] {\scriptsize $\; Z_3$};
\draw (0cm:3.6cm) arc (180:168:8.4cm);
\draw (0cm:3.4cm) arc (180:194:8.6cm)
node[below] {\scriptsize $\; Z_4$};
\draw (0cm:3.4cm) arc (180:166:8.6cm);
\draw (0cm:3.2cm) arc (180:195:8.8cm) ;
\draw (0cm:3.2cm) arc (180:165:8.8cm);
\draw (0cm:3.cm) arc (180:196:9.0cm) ;
\draw (0cm:3.cm) arc (180:164:9.0cm);
\draw (0cm:2.8cm) arc (180:197:9.2cm) ;
\draw (0cm:2.8cm) arc (180:163:9.2cm);
\draw (0cm:2.6cm) arc (180:198:9.4cm);
\draw (0cm:2.6cm) arc (180:162:9.4cm);
\draw (0cm:2.4cm) arc (180:198.5:9.6cm);
\draw (0cm:2.4cm) arc (180:161.5:9.6cm);
\draw (0cm:2.2cm) arc (180:199:9.8cm);
\draw (0cm:2.2cm) arc (180:161:9.8cm);
\draw (0cm:2.0cm) arc (180:199:10.0cm);
\draw (0cm:2.0cm) arc (180:161:10.0cm);
\draw (0cm:1.8cm) arc (180:199.5:10.2cm);
\draw (0cm:1.8cm) arc (180:160.5:10.2cm);
\draw (0cm:1.6cm) arc (180:200:10.4cm);
\draw (0cm:1.6cm) arc (180:160:10.4cm);
\coordinate [label=150:](a) at (0,0) node[below] {$P_0$};
\coordinate (q) at (3,2.64575131106459059050);
\draw (q) node[anchor=south] {$Q$};
\fill [black] (q) circle (2pt);
%rays
\coordinate (q2) at (4.2,3.70405183549042682670);
\coordinate (q3) at (3.6,3.17490157327750870860);
\draw[-latex] (a) -- (q2);
\draw[-latex] (a) -- node[anchor=south] {$r_0$} (q);
\draw[-latex] (q) -- node[anchor=south] {$s$} (p);
% angle at P_0$
\draw[->] (1,0) node [anchor=south east] {$\theta$} arc (0:43:1.0cm);
\coordinate (q4) at (3.8,3.1);
\draw[<-] (q3) arc (30:-15:1.0cm);
\draw (q4) node[anchor=north east] {$\chi$};
%b
\draw[|-|] (4,-0.5)--(8,-0.5) node {$\quad b$};
\draw[|-|] (8.4,-0.5)--(12,-0.5);
\end{tikzpicture}
\label{FH}
\end{figure}
The Huygens principle is writen as certain integral
\bea
\label{int}
I(P) = \int_{S} A(x,y) \exp{[-\mathrm{i} k \phi(x,y)]} dx dy.
\eea
Referring to figure~\ref{FH} the amplitude $A(x,y)$ is the stength of each
secondary source in the wavefront $Q(x,y)$ and $\phi(x,y)$ is the distance
between each point along the wavefront $[s_x(x,y), s_y(x,y), s_z(x,y)]$
and the recording (output) point $P$.
The way Fresnel evaluated the integral was by slicing it through zones $Z_j$ as shown
in the figure. We will evaluate the integral along the first Fresnel zone
(that is for all the wavefront points $Q=(s_x,s_y,s_z)$ such that their distance $d$ to
$P$ satisfies $b \le d \le j \lambda/2$, $j=1$. Then we observe that this integral
corresponds to twice the stationary phase formula corresponding to integral~\ref{int}.
The method is general and does not assume constant velocity, so the wavefront does not
have to be spherical. The idea (Fresnel's idea) is to partition the wavefront
according to the distance from recording point $P$ to the wavefront, along the stationary phase axis
(this is the axis of the normal ray. In the figure is the line that connects $P$ with $P_0$, we
will discuss this below).
The stationary phase point satisfies:
\be
\nabla s(x,y) = \phi(x,y) = 0.
\ee
Let us assume that there is only one of these points $(x_0,y_0)$. Note that each
stationary phase point is a foot of a perpendicular from $P$ to the wavefront $S$
(just as the gradient is normal to a contour). Several stationary phase pointts will mean a
non--convex wavefront and the problem is much more complicated. We are only interested
on the case of a single stationary phase point.
Figure~\ref{FH} shows the stationary phase point $C=(x_0,y_0)$.
We want to evaluate integral
\bea
\label{izp}
I_{Z_1}(P) = \int_{Z_1} A(x,y) \exp{-[\mathrm{i} k \phi(x,y)]} dx dy.
\eea
The first approximation is a high frequency approximation. We assume that $\lambda \ll 1$, and
so if the amplitude is slow variant we might say $A(x,y) \approx A(x_0, y_0)$. Also we can
expand the phase according to a Taylor series expansion around $(x_0, y_0)$ as
\be
\phi(x,y) = \phi(x_0, y_0) + \phi_{xx} (x -x_0)^2 + 2 \phi_{xy} (x -x_0)(y-y_0)
+ \phi_{yy}(y - y_0)^2.
\ee
with the notation $\phi_{uv} = \displaystyle \frac{\partial^2 \phi}{\partial u \partial v}$
evaluated at $(x_0, y_0)$.
We then write~\ref{izp} as
\be
I_{Z_1}(P) \approx A(x_0, y_0) \exp{[-\mathrm{i} k \phi(x_0, y_0)] }
\int_{Z_1}\exp{[- \mathrm{i} H(x,y)]} dx dy.
\ee
with
\be
H(x,y) = \phi_{xx} (x -x_0)^2 + 2 \phi_{xy} (x -x_0)(y-y_0)
+ \phi_{yy}(y - y_0)^2.
\ee
%\section{References}{\label{references}}
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