APPENDIX A

In this appendix, I derive the formula for calculating the Jacobian function J(x,z) and the partial differential equation (14).

From equation (12), we have

$$J(x,z) = {\pmatrix{\displaystyle{\partial x \over \partial s... ...} \displaystyle{\partial \gamma \over \partial x}}. \eqno(A.1)$$

The expression in the denominator of the last line is a cross-product of two gradient vectors:

$${\partial s \over \partial x}{\partial \gamma \over \partial... ... = {1 \over m(x,z)}\nabla \tau \times \nabla \gamma. \eqno(A.2)$$

$\gamma(x,z)$ is the take-off angle of the ray that reaches point(x,z), and is constant along each ray. Thus, the gradient directions of $\gamma$ are always orthogonal to rays. On the other hand, the gradient directions of traveltimes are always tangential to rays. Consequently, $\nabla \tau$ and $\nabla \gamma$ are orthogonal, which yields

$$\nabla \tau \cdot \nabla \gamma=0 \eqno(A.3)$$

and

$$\vert\nabla \tau \times \nabla \gamma\vert = \vert\nabla \tau\vert\vert\nabla \gamma\vert. \eqno(A.4)$$

In 2-D, equation (A.3) can be rewritten as

$$\tau_x\gamma_x+\tau_z\gamma_z=0 \eqno(A.5)$$

which is equation (14). From equations (A.1), (A.2) and (A.4), we can derive

$$\vert J(x,z)\vert= {m(x,z) \over \vert\nabla \tau\vert\vert\nabla \gamma\vert}. \eqno(A.6)$$

The eikonal equation states that $\vert\nabla \tau\vert^2=m^2(x,z)$ or $\vert\nabla \tau\vert=m(x,z)$. Applying this relation to equation (A.6), we get equation (12):

$$\vert J(x,z)\vert= {1 \over \vert\nabla \gamma\vert}= {1 \o... ...\left({\partial \gamma \over \partial z}\right)^2}}. \eqno(A.7)$$