# Kirchhoff migration

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 9 295 - 366 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 9.27a

Prove that the equation of a diffraction curve (see Figure 9.27a) is

{\begin{aligned}z^{2}-x^{2}=z_{0}^{2},\end{aligned}} where $z_{0}$ is the depth of the diffracting point (assuming constant velocity).

### Background

Migration is the process of repositioning reflection data so that reflection events are located where the reflector in the subsurface is, as shown in Figure 9.27a. $S_{0}$ , $S_{1}$ , $S_{2}$ , ..., are a series of coincident sources and receivers and $P$ is a diffracting point on a plane reflector $PQ$ . The data recorded at $S_{0}$ are ordinarily plotted directly below $S_{0}$ before migration is performed. If the velocity is constant, the reflection from $P$ that is observed at $S_{0}$ appears on the $S_{0}$ trace at $M$ , and the reflection point must lie on a circle through $M$ centered at $S_{0}$ . The circles such as $PM$ are wavefronts and radii such as $S_{0}P$ are raypaths which are orthogonal to the respective wavefronts.

### Solution

The locus of point M as the source and coincident receiver are moved is a diffraction curve (see Figure 6.5a). If O is the origin and the coordinates of $M$ are ($x,z$ ), then, because $S_{0}P=S_{0}M$ .

 {\begin{aligned}z^{2}=x^{2}+z_{0}^{2},\quad \mathrm {so} \quad z^{2}-x^{2}=z_{0}^{2}.\end{aligned}} (9.27a)

## Problem 9.27b

Show that an unmigrated reflection is tangent to the diffraction curve.

### Solution

We find the equation of an unmigrated reflection $MN$ (Figure 9.27a) and determine its slope, then find the slope of the diffraction curve and show that the point where the slope of the diffraction curve is the same as that of $MN$ is also on $MN$ . To get the equation of $MN$ , we need the equation of the reflector $PQ$ because it meets $MN$ at the outcrop on the $x$ -axis.

The line $PQ$ passes through $P(0,\;z_{0})$ with slope ${\rm {\;tan\;}}\xi$ , where $\xi$ is the dip, so its equation is

 {\begin{aligned}z=x{\rm {\;tan\;}}\xi +z_{0}.\end{aligned}} (9.27b)

This line outcrops (that is, at $z=0$ ) at $x=-z_{0}{\rm {\;cot\;}}\xi$ , and $MN$ passes through the point $(-z_{0}{\rm {\;cot\;}}\xi ,\;0)$ as well as the point $M$ , whose coordinates are $x=z_{0}{\rm {\;tan\;}}\xi$ , $z=S_{0}M=S_{0}P=z_{0}/{\rm {\;cos\;}}\xi$ . The equation of $MN$ now becomes

 {\begin{aligned}{\frac {z-z_{0}/\cos \xi }{x-z_{0}{\tan \xi }}}={\frac {z_{0}/\cos \xi }{z_{0}\tan \xi +z_{0}\cot \xi }}={\frac {1}{\sin \xi +\cos ^{2}\xi /\sin \xi }}=\sin \xi .\\{\mbox{Thus}}\qquad \qquad z&=z_{0}/\cos \xi +x\sin \xi -z_{0}\tan \xi \sin \xi \\&=x\sin \xi +z_{0}\cos \xi ,\end{aligned}} (9.27c)

and the slope of $MN$ is ${\hbox{d}}z/{\hbox{d}}x={\rm {\;sin\;}}\xi$ .

The slope of the diffraction curve in equation (9.27a) is ${\hbox{d}}z/{\hbox{d}}x=x/z$ , so we need to find a point on the curve in equation (9.27a) where this slope equals that of $MN$ , that is, where $x/z=\sin \xi$ . Substituting this value of $x$ in equation (9.27a) gives

{\begin{aligned}z^{2}-(z\sin \xi )^{2}=z_{0}^{2},\quad \mathrm {so} \quad z=z_{0}/\cos \xi ,x=z_{0}\tan \xi .\end{aligned}} We now substitute these coordinates in equation (9.27c) to show that they satisfy the equation of $MN$ . Thus,

{\begin{aligned}z_{0}/\cos \xi =(z_{0}\tan \xi )\sin \xi +z_{0}\cos \xi \;,\;1=\sin ^{2}\xi +\cos ^{2}\xi ;\end{aligned}} therefore, the diffraction curve and $MN$ both pass through this point and, since they have only one common point, $MN$ must be a tangent.

## Problem 9.27c

Show that the coordinates of $P$ and the slope of the wavefront at $P$ (hence also the dip $\xi$ ) can be obtained from the recorded data.

### Solution

Assuming that the seismic line is normal to strike, we obtain $z_{0}$ from the minimum value of the traveltime $t_{0}$ , which occurs when the source is directly over $P$ ; this fixes the $x$ -coordinate of $P$ . If the velocity is constant, by swinging an arc with center at any source and radius ${\frac {1}{2}}Vt_{0}$ and then drawing the tangent from $P$ to the arc, we get the dip $\xi$ .

## Problem 9.27d

How will parts (a), (b), and (c) change if the velocity changes in the vertical direction (e.g., if it increases with depth)?

### Solution

In this case the raypath will curve, a wavefront may not be the arc of a circle, the diffraction curve will no longer be a hyperbola, an unmigrated reflection will still be tangent to the diffraction curve, and locating the reflecting point will be more complicated. However, assuming that a diffraction curve is hyperbolic often results in solutions that are close enough to be useful and most data processing makes this assumption.