Two point raytracing for reflection off a 3D plane

Reflection gradient

If we are interested in map migration, the information we have is not the reflector normal, but the normal to the arrival time surface. To calculate this slope, we can conflate distance and time by choosing an arbitrary temporal unit, say a glorp equated to $1/V$ seconds. This makes a traveltime of 1 glorp correspond to 1 meter of travel distance.

I make life simpler by observing that the traveltime gradient has the same azimuth as the reflector's dip azimuth. This must be so because translating the source-receiver pair along strike does not change the reflection arrival time. I note that this does not say that the reflection point moves along the dip azimuth when the surface arrival point moves along the dip azimuth.

The next twist is that instead of translating the source-receiver pair along the dip azimuth, I'll translate the reflector plane along its normal direction. This implies that derivatives with respect to the reflector normal direction need to be scaled by the sine of the reflector dip, i.e. $\sin \theta = \sqrt{1 - ({\bf n} \cdot {\bf z})^2 }$ , as the surface intercept of the reflector moves a distance inversely related to the sine of the dip. Fortunately, even the zero dip case, where the reflector does not intersect the surface, is handled properly because the sine is zero in that case.

If we translate the initial reflection point $P$ by $-\epsilon {\bf n}$ , where my convention for $\theta$ implies $\epsilon \ge 0$ corresponds to a positive time slope, we obtain a point on the displaced reflection plane, though generally not the new reflection point $\hat{P}$ . The relation of $\hat{P}$ to $P$ can be ascertained as before by dotting with ${\bf n}$ :

$$\begin{array}{rcccl} {\bf n} \cdot ( R - \hat{P} ) & = & {\b... ...n {\bf n} ) ) & = & \beta + \epsilon \end{array}\mbox{\ \ \ .}$$

Continuing as before,

$$\begin{array}{rcrcr} R - S & = & ( \alpha - \beta ) {\bf n} ... ... \beta + \epsilon ) ) {\hat{\bf w}} \end{array}\mbox{\ \ \ ,}$$

yielding

$${\hat{\bf w}} = \left( 1 - \frac{2 \epsilon}{\alpha + \beta + 2 \epsilon } \right ) {\bf w}$$

which says that ${\bf w}$ does not rotate.

To compute changes in lengths (traveltimes), we have the relations

$$\begin{array}{rcl} R - P & = & \alpha ( {\bf n} + {\bf w} ) ... ... w}} ) + (\alpha + \epsilon ) ( {\bf n} + {\bf w} ) \end{array}$$

and

$$\begin{array}{rcl} S - P & = & \beta ( {\bf n} - {\bf w} ) \... ...eta + \epsilon ) ( {\bf n} - {\bf w} ) \end{array}\mbox{\ \ ,}$$

whence

$$\begin{array}{rcl} \hat{P} - P & = & - ( \alpha + \epsilon )... ... w} ) - \epsilon ( {\bf n} - {\bf w} ) \end{array}\mbox{\ \ .}$$

Taking first differences, we have

$$\frac{{\hat{\bf w}} - {\bf w}}{\epsilon} = -\frac{2}{\alpha + \beta + 2 \epsilon} {\bf w}$$

and

$$\frac{\hat{P} - P}{\epsilon} = - ( \alpha + \epsilon ) \frac{{\hat{\bf w}} - {\bf w}}{\epsilon} - ({\bf n}+{\bf w})$$

$$= ( \beta + \epsilon ) \frac{{\hat{\bf w}} - {\bf w}}{\epsilon} - ( {\bf n} - {\bf w} )$$

whence

$$\frac{dP}{d\epsilon} = \left( -1 + \frac{2 \alpha}{\alpha+\beta} \right) {\bf w} - {\bf n}$$

$$= \left( 1 - \frac{2\beta}{\alpha+\beta} \right) {\bf w} - {\bf n}$$

$$\makebox[0pt][r]{or, averaging the two,\ \ \ \ \ \ \ \ \ \ \ \/}$$

$$\phantom{\frac{dP}{d\epsilon}} = \frac{\alpha - \beta}{\alpha + \beta} {\bf w} - {\bf n}$$

With these in hand, we may differentiate the traveltime

$$T = T_R + T_S = \vert P - R \vert + \vert P - S \vert$$

to get

$$\frac{dT}{d\epsilon} = \left( \frac{P-R}{\vert P-R\vert} + \frac{P-S}{\vert P-S\vert} \right) \cdot \frac{dP}{d\epsilon}$$

$$= - \left( \frac{\alpha ( {\bf n} + {\bf w} ) }{T_R} + \frac{\beta ... ...ght) \cdot \left( \frac{\alpha - \beta}{\alpha+\beta} {\bf w} - {\bf n} \right)$$

$$= \frac{\alpha}{T_R} \left( \frac{ ( 1 - \vert{\bf w}\vert^2 ) \alpha + ( 1 + \vert{\bf w}\vert^2 ) \beta }% {\alpha + \beta } \right) +$$

$$\frac{\beta}{T_S} \left( \frac{ ( 1 + \vert{\bf w}\vert^2 ) \alpha + ( 1 - \vert{\bf w}\vert^2 ) \beta }% {\alpha + \beta } \right)$$

which, as remarked earlier, is then multiplied by $\sin \theta$ to obtain the surface time slope.

This last expression has a simple geometric meaning. As

$$\frac{P-R}{\vert P-R\vert}$$    and $$\frac{P-S}{\vert P-S\vert}$$

are unit vectors pointing towards the reflection point from the receiver and source respectively, their sum is necessarily parallel to their angle bisector, the normal. In particular, they sum to $-2 \cos$   $\xi$ $\,\, {\bf n}$ where $\xi$ is the angle of incidence or reflection. Dotting this with $dP/d\epsilon$ and multiplying by $\sin \theta$ we have that the time slope is simply $2 \cos \xi \sin \theta$ . Changing units from glorps back to seconds, this agrees with the well-known zero-offset result $2 \sin \theta / V$ .

Two point raytracing for reflection off a 3D plane