# 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

{\displaystyle {\begin{aligned}z^{2}-x^{2}=z_{0}^{2},\end{aligned}}}

where ${\displaystyle z_{0}}$ 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. ${\displaystyle S_{0}}$, ${\displaystyle S_{1}}$, ${\displaystyle S_{2}}$, ..., are a series of coincident sources and receivers and ${\displaystyle P}$ is a diffracting point on a plane reflector ${\displaystyle PQ}$. The data recorded at ${\displaystyle S_{0}}$ are ordinarily plotted directly below ${\displaystyle S_{0}}$ before migration is performed. If the velocity is constant, the reflection from ${\displaystyle P}$ that is observed at ${\displaystyle S_{0}}$ appears on the ${\displaystyle S_{0}}$ trace at ${\displaystyle M}$, and the reflection point must lie on a circle through ${\displaystyle M}$ centered at ${\displaystyle S_{0}}$. The circles such as ${\displaystyle PM}$ are wavefronts and radii such as ${\displaystyle S_{0}P}$ 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 ${\displaystyle M}$ are (${\displaystyle x,z}$), then, because ${\displaystyle S_{0}P=S_{0}M}$.

 {\displaystyle {\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 ${\displaystyle MN}$ (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 ${\displaystyle MN}$ is also on ${\displaystyle MN}$. To get the equation of ${\displaystyle MN}$, we need the equation of the reflector ${\displaystyle PQ}$ because it meets ${\displaystyle MN}$ at the outcrop on the ${\displaystyle x}$-axis.

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

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

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

 {\displaystyle {\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 ${\displaystyle MN}$ is ${\displaystyle {\hbox{d}}z/{\hbox{d}}x={\rm {\;sin\;}}\xi }$.

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

{\displaystyle {\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 ${\displaystyle MN}$. Thus,

{\displaystyle {\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 ${\displaystyle MN}$ both pass through this point and, since they have only one common point, ${\displaystyle MN}$ must be a tangent.

## Problem 9.27c

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

### Solution

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

## 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.