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Various meanings of H = 0

Recall the various forms of the stepout operator:

3|c|    
3|c| Forms of stepout operator 2H/v    
3|c|    
     
ray trace Fourier PDE
     
     
$\displaystyle {\strut dt\over dh}$ $\displaystyle {\strut k_h\over\omega}$ $\partial_h^t\ =\ \int\limits_{-\infty}^t
 dt \displaystyle {\strut \partial\over\partial h}$
     

Reciprocity suggests that travel time is a symmetrical function of offset; thus dt/dh vanishes at h = 0. In that sense it seems appropriate to apply equation (56) to zero-offset sections. More precisely, the ray-trace expression dt/dh strictly applies only when a single plane wave is present. Spherical wavefronts are made from the superposition of plane waves. Then the Fourier interpretation of H is slightly different and more appropriate. To set $ \omega = 0 $ would be to select a zero frequency component, a simple integral of a seismic trace. To set kh = 0 would be to select a zero spatial-frequency component, that is, an integration over offset. Conventional stacking may be defined as integration (or summation) over offset along a hyperbolic trajectory. Simply setting kh = 0 is selecting a hyperbolic trajectory that is flat, namely, the hyperbola of infinite velocity. Such an integration will receive its major contribution from the top of the data hyperboloid, where the data events come tangent to the horizontal line of integration. (For some historical reason, such a data summation is often called vertical stack). Of the total contribution to the integral, most comes from a zone near the top, before the stepout equals a half-wavelength. The width of this zone, which is called a Fresnel zone, is the major factor contributing to the integral. The Fresnel zone has been extracted from a field profile in Figure 29.

 
denmark
denmark
Figure 29
(left) A land profile from Denmark (Western Geophysical) with the Fresnel zone extracted and redisplayed (right).


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The definition of the Fresnel zone involves a frequency. For practical purposes we may just look at zero crossings. Examining Figure 29 near one second we see a variety of frequencies. In the interval between t=1.0 and t=1.1 I see about two wavelengths of low frequencies and about 5 wavelengths of high frequencies. The highest frequencies are the main concern, because they define the limit of seismic resolution. The higher frequency has about 100 half wavelengths between time zero and a time of one second. As a rough generality, this observed value of 100 applies to all travel times. That is, at any travel time, the highest frequency that has meaningful spatial correlations is often observed to have a half period of about 1/100 of the total travel time. We may say that the quality factor Q of the earth's sedimentary crust is often about 100. So the angle that we are typically thinking about is $\cos \,$ 8$^\circ$ $=\ .99$.

Theoretically, the main differences between a zero-offset section and a vertical stack are the amplitude and a small phase shift. In practical cases they are unlikely to migrate in a significantly different way. It would be nice if we could find an equation to downward continue data that is stacked at velocities other than infinite velocity.

The partial-differential-equation point of view of setting H = 0 is identical with the Fourier view when the velocity is a constant function of the horizontal coordinate; but otherwise the PDE viewpoint is a slightly more general one. To be specific, but not cluttered, equations (54) and(55) can be expressed in 15$^\circ$retarded, space-domain form. Thus,  
 \begin{displaymath}
\left[ {\partial\ \over \partial z}\ +\ 
{v \over - \, i \om...
 ...artial^2\ \over \partial h^2} \ \,
\right) \ \right] \ U' \eq 0\end{displaymath} (62)
Integrate this equation over offset h. The integral commutes with the differential operators. Recall that the integral of a derivative is the difference between the function evaluated at the upper limit and the function evaluated at the lower limit. Thus,  
 \begin{displaymath}
\left. \left( \ {\partial\ \over \partial z} \ +\ {v
\over -...
 ...
 { h \ =\ +\,\infty}
\cr \cr
 { h \ =\ -\,\infty}
}
\ \ =\ \ 0\end{displaymath} (63)
The wave should vanish at infinite offset and so should its horizontal offset derivative. Thus the last term in (63) should vanish. So, setting H = 0 has the meaning  
 \begin{displaymath}
(\hbox{Paraxial operator}\ \ (\hbox{vertical stack} \eq 0\end{displaymath} (64)

A problem in the development of (64) was that, twice, it was assumed that velocity is independent of offset: first, when the thin-lens term was omitted from (62), and second, when the offset integration operator was interchanged with multiplication by velocity. If the velocity depends on the horizontal x-axis, then it certainly depends on both midpoint and offset. In conclusion: If velocity changes slowly across a Fresnel zone, then setting H = 0 provides a valid equation for downward continuation of vertically stacked data.


previous up next print clean
Next: Clayton's cosine corrections Up: THE MEANING OF THE Previous: Conventional processing separable approximation
Stanford Exploration Project
10/31/1997