Analytical evaluation of the tangent plane to the impulse response

In this appendix I derive the expressions for evaluating the derivatives of image depth $z_\xi$ with respect to the subsurface offset $h_\xi$ and the midpoint $m_\xi$;these derivatives are computed along the tangent plane to the impulse response of the generalized migration operator, which is defined in equations 18-24.

I start by deriving the equation for the vector normal to the impulse-response surface, ${\vec n}$: where $\vec z_\xi$, $\vec m_\xi$, and $\vec h_\xi$ are respectively the unit vectors along the three dimensions $z_\xi$, $m_\xi$, and $h_\xi$.

The equation of the tangent plane at the image point with coordinates $\left({\widebar z_\xi},{\widebar m_\xi},{\widebar h_\xi}\right)$is given by:

$$\begin{eqnarray} T\left(z_\xi,m_\xi,h_\xi\right) &=& \left( \frac{\partial m_\xi... ...tial \alpha_x} \right) \left(h_\xi- {\widebar h_\xi} \right) =0.\end{eqnarray}$$ (32)

The derivative of the depth with respect o the subsurface offset, at constant midpoint, is given by:  

$$\left. \frac{\partial z_\xi}{\partial h_\xi} \right\vert _{m... ...{\partial \gamma} \frac{\partial h_\xi}{\partial \alpha_x} }.$$ (33)

and similarly the derivative of the depth with respect to the midpoint, at constant subsurface offset, is given by:  

$$\left. \frac{\partial z_\xi}{\partial m_\xi} \right\vert _{h... ...{\partial \gamma} \frac{\partial h_\xi}{\partial \alpha_x} }.$$ (34)

To evaluate equations 33-34. we need to evaluate the following partial derivatives, obtained by differentiating the expressions in equations 18-20:

$$\begin{eqnarray} \frac{\partial z_\xi}{\partial \alpha_x} &=& -L\left(\alpha_x,\... ...\gamma\right)}{\partial \gamma} \frac{\sin \gamma}{\cos \alpha_x}.\end{eqnarray}$$ (35)

The derivative of path length are evaluated as follows:

$$\begin{eqnarray} & \frac{\partial L}{\partial \alpha_x} = \frac{-t_{D}} {\left[\... ...frac{\left(S_r-S_s\right)\tan \gamma}{\cos ^2 \alpha_x} \right], &\end{eqnarray}$$ (36)

and

$$\begin{eqnarray} & \frac{\partial L}{\partial \gamma} = \frac{-t_{D}} {\left[\le... ...frac{\left(S_r-S_s\right)\tan \alpha_x}{\cos ^2 \gamma} \right]. &\end{eqnarray}$$ (37)

 

• Application to the isotropic case