APPENDIX

The proposition I want to prove is that the vector perpendicular to the constant-offset isochron bisects the angle between the ray coming from the source and the one returning to the receiver. This is the direction of the zero-offset ray, as the incident angle is equal to the reflected angle and the zero-offset ray bisects both angles. The gradient to a constant-offset isochron is a vector

$$\bar{R}=T_x\bar{i}+T_z\bar{j}$$

where

$$\left \{ \begin{array} {l} T=\mbox {constant-offset travel-t... ...or of $x$} \\ \bar{j}= \mbox{versor of $z$} \end{array} \right.$$

The constant-offset travel-time field can be decomposed into the sum of the travel-time from the source (T_s) and the travel-time to the receiver (T_r). The vector $\bar{R}$ can be therefore as the sum of vectors

$$\bar{R}= {(T_{sx}+T_{rx})\bar{i}+(T_{sz}+T_{rz})\bar{j}}$$

where T_sx, T_sz, T_rx, T_rz are the partial derivatives of the travel-time from source and receiver, respectively. If we note

$$\left \{ \begin{array} {l} \bar{R_s}=T_{sx}\bar{i}+T_{sz}\bar{j} \\ \bar{R_r}=T_{rx}\bar{i}+T_{rz}\bar{j}\end{array} \right.$$

then

$$\bar{R}=\bar{R_s}+\bar{R_r}$$

which are two vectors: $\bar{R_s}$ along the ray from the source and $\bar{R_r}$ along the ray from the receiver. The gradient of the constant-offset travel-time field is a sum of the two vectors $\bar{R_s}$ and $\bar{R_r}$, and because the two vectors have equal length $\Vert\bar{R_s}\Vert=\Vert\bar{R_r}\Vert=slowness$ (from the eikonal equation), the vector $\bar{R}$ bisects the angle between the vectors $\bar{R_s}$ and $\bar{R_r}$.

 

Figure 7: The gradient of the constant-offset travel-time field bisects the angle between the ray from the source and the ray to the receiver.