Diffraction from a half-plane

**Problems in Exploration Seismology and their Solutions**
Series	Geophysical References Series
Title	Problems in Exploration Seismology and their Solutions
Author	Lloyd P. Geldart and Robert E. Sheriff
Chapter	2
Pages	7 - 46
DOI	http://dx.doi.org/10.1190/1.9781560801733
ISBN	ISBN 9781560801153
Store	SEG Online Store

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Problem 2.19

The general equation for determining the diffraction effect of a plane surface is given by equation (2.19b). Show that the diffraction effect of the half-plane in Figure 2.19a is given by the integral $\int \nolimits _{t=tr}^{+\infty }\phi \left(t\right)\mathrm {d} t$ where

{\begin{aligned}\phi \left(t\right)&={\frac {\left(4chy_{0}/\pi V^{3}t\right)}{\left(t^{2}+t_{y}^{2}-t_{r}^{2}\right)(t^{2}-t_{r}^{2})^{1/2}}},\quad t>t_{r}.\\&=0,\qquad \qquad \qquad \qquad \qquad \quad \ \ t<t_{r},\end{aligned}}

(2.19a)

where $c$ is a constant, $V$ is the velocity (assumed to be constant), while $h,y_{0},$ , $r$ , and $\xi$ are defined in Figure 2.19a, and $t$ , $t_{y}$ , and $t_{r}$ are two-way traveltimes along $\xi ,y_{0}$ , and $r$ .

Figure 2.19a Calculating diffraction effect of a plane surface

S

.

Background

The use of rays to describe wave propagation simplifies the phenomenon by ignoring diffraction (spreading of energy radiating from a virtual point source). Since a wave is reflected by all parts ofa surface, we can consider each point on the surface as a point source (Huygens's principle, see problem 3.1) and integrate over the surface to get the correct total effect. For a coincident source and receiver, the integral of the effects of point sources over the surface $S$ can be transformed into a line integral around the boundary of the surface (see Sheriff and Geldart, 1995, Section 2.8.2). When the origin is over the surface, the integration gives two terms, one term representing the reflection given by ray theory, the other the diffraction. When the origin is not over the surface, the reflection term is zero leaving only the diffraction term.

The diffraction response of a plane area to a unit impulse (see Sheriff and Geldart, 1955, Section 15.2.5) emitted by a source at the origin and recorded at the origin is obtained by integrating the quantity $\phi \left(t\right)$ around the entire boundary of the area, where $\phi \left(t\right)$ is given by equation (2.19b) [see Sheriff and Geldart, 1995, equation (2.131)]:

{\begin{aligned}\phi \left(t\right)=\left(ch/\pi V^{2}t^{2}\right)\left({\frac {\mathrm {d} \theta }{\mathrm {d} t}}\right),\end{aligned}}

(2.19b)

where $\phi \left(t\right)$ is the response of a unit element of the boundary, $t$ is the two-way traveltime from the origin to the element of the boundry, and $\delta$ is the angle shown in Figure 2.19a.

Solution

In Figure 2.19a

${\begin{aligned}OA&=\xi ,\\\xi ^{2}&=r^{2}+x^{2}=r^{2}+y_{0}^{2}\tan ^{2}\theta ,\end{aligned}}$

where $y_{0}$ is normal to $BD$ and $x=O'A$ . The points $B$ and $D$ are in fact at $\pm \infty$ ; thus, on integrating, $x$ goes from $-\infty$ to $+\infty$ while 6 goes from $-\pi /2$ to $+\pi /2$ . Dividing by $\left(V/2\right)$ , we get the following relation between the traveltimes:

{\begin{aligned}(2\xi /V)^{2}=t^{2}=t_{r}^{2}+t_{y}^{2}\tan ^{2}\theta .\end{aligned}}

(2.19c)

Then

{\begin{aligned}{\frac {\mathrm {d} t}{\mathrm {d} \theta }}=\left(t_{y}^{2}/t\right)\tan \theta \sec ^{2}\theta \end{aligned}}

(2.19d)

and

{\begin{aligned}{\frac {\mathrm {d} \theta }{\mathrm {d} t}}=\left(t/t_{y}^{2}\right)\cot \theta \cos ^{2}\theta .\end{aligned}}

(2.19e)

From equation (2.19c) we have

${\begin{aligned}{\tan }^{2}\theta &=\left(t^{2}-t_{r}^{2}\right)/t_{y}^{2},\\{\cos }^{2}\theta &=1/\left(1+{\tan }^{2}\theta \right)\\&=1/\left[1+\left(t^{2}-t_{r}^{2}\right)/t_{y}^{2}\right]\\&=t_{y}^{2}/\left(l^{2}+l_{y}^{2}-t_{2}^{r}\right),\end{aligned}}$

so

${\begin{aligned}{\frac {\mathrm {d} \theta }{\mathrm {d} t}}&=\left({\frac {t}{t_{y}^{2}}}\right){\frac {t_{y}}{(t^{2}-t_{r}^{2})^{1/2}}}{\frac {t_{y}^{2}}{\left(t^{2}+t_{y}^{2}-t_{r}^{2}\right)}}\\&={\frac {t_{y}t}{(t^{2}+t_{y}^{2}-t_{r}^{2})(t^{2}-t_{r}^{2})^{1/2}}}.\end{aligned}}$

Substituting this expression in equation (2.19b), we obtain the following result for the diffraction effect of a unit length of the boundary in Figure 2.19a:

${\begin{aligned}\phi \left(t\right)={\frac {\left(ch/\pi V^{2}t^{2}\right)\left(t_{y}t\right)}{\left(t^{2}+t_{y}^{2}-t_{r}^{2}\right)(t^{2}-t_{r}^{2})^{1/2}}}.\end{aligned}}$

Substituting $t_{y}=2y_{0}/V$ in the numerator, we get

${\begin{aligned}\phi \left(t\right)={\frac {\left(2chyo/\pi V^{3}t\right)}{\left(t^{2}+t_{y}^{2}-t_{r}^{2}\right)(t^{2}-t_{r}^{2})^{1/2}}}.\end{aligned}}$

To get the total diffraction effect of the half-plane, we integrate this expression around the four sides of the half-plane. Three of the four sides are at infinity so that $t$ is infinite and the fraction, being proportional to $t^{-2}$ , vanishes. Therefore the effect reduces to the integral along $BD$ . Because of symmetry, we can integrate along $OD$ and double the result. Thus, the diffraction function for a half-plane is

${\begin{aligned}\phi \left(t\right)&={\frac {\left(4chyo/\pi V^{3}t\right)}{\left(t^{2}+t_{y}^{2}-t_{r}^{2}\right)(t^{2}-t_{r}^{2})^{1/2}}},\quad t\geq t_{r},\\&=0,\qquad \qquad \qquad \qquad \qquad \quad \ \,t\leq t_{r}.\end{aligned}}$

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