Computing the traveltime: the theory

Let us denote by n a value smaller 1 less than the total number of bounces of the wave in the earth. Let us pretend ``not to know'' that the horizontal coordinates of S and G are -h and h, respectively, and denote them with s and g instead, since we will later need a more general expression that can be differentiated with respect to these variables. We start by generating the sequences $\phi_i$ and l_i, according to (5) and (6), and keeping in mind the nonphysical prependix $\phi_0=-\frac{\pi}{2}$.

Using the fact that n is always even because the total number of bounces inside the earth is always odd, and substituting into (25), we find the image cascaded through n reflection operations from the source to be  

$${\bf q}_n^S = s \left[ {\begin{array} {*{20}c} {-\sin\beta_n... ..._j} \\ {\left(-1\right)^j \cos\beta_j} \\ \end{array}} \right].$$ (27)

The receiver image is obtained from a single reflection operation, through the last reflecting interface:  

$${\bf q}_1^G=g\left[ {\begin{array} {*{20}c}{\cos2\phi_{n+1}}... ...{-\sin\phi_{n+1}} \\ {\cos\phi_{n+1}} \\ \end{array}} \right].$$ (28)

The traveltime is the distance between ${\bf q}_n^S$ and ${\bf q}_1^G$divided by the velocity. This distance will be computed as the magnitude of the vector ${\bf q}_n^S-{\bf q}_1^G$. By making the notations  

$${\bf u}_1 = \frac{2}{v}\left\{\sum\limits_{j=0}^{n-1} l_{n-j... ...i_{n+1}} \\ {-\cos\phi_{n+1}} \\ \end{array}} \right]\right\},$$ (29)

 

$${\bf u}_2=\frac{1}{v}\left[ {\begin{array} {*{20}c} {-\sin\beta_n} \\ { \cos\beta_n} \\ \end{array}} \right],$$ (30)

 

$${\bf u}_3=-\frac{1}{v}\left[ {\begin{array} {*{20}c}{\cos2\phi_{n+1}} \\ {\sin2\phi_{n+1}} \\ \end{array}} \right],$$ (31)

we can write:  

$$t = \left\vert{\bf u}_1+s{\bf u}_2+g{\bf u}_3\right\vert.$$ (32)

In particular, for s=-h and g=h and  

$${\bf u}_4={\bf u}_2-{\bf u}_3=\frac{1}{v}\left[ {\begin{arr... ...left(-\beta_n\right)+\sin2\phi_{n+1}} \\ \end{array}} \right],$$ (33)

the traveltime can be written as  

$$t = \left\vert{\bf u}_1-h{\bf u}_4\right\vert.$$ (34)

This vector magnitude can be computed using scalar products:  

$$t=\sqrt{\left[{\bf u}_1-h{\bf u}_4\right]\cdot\left[{\bf u}_1-h{\bf u}_4\right]}$$ (35)

or it can be written as  

$$t^2 = {\bf u}_4\cdot{\bf u}_4\left(h-\frac{{\bf u}_1\cdot{\b... ...eft({\bf u}_1\cdot{\bf u}_4\right)^2}{{\bf u}_4\cdot{\bf u}_4},$$ (36)

which is the equation of a hyperbola with the apex at  

$$h_{apex}=\frac{{\bf u}_1\cdot{\bf u}_4}{{\bf u}_4\cdot{\bf u}_4},$$ (37)

 

$$t_{apex}= \sqrt{{\bf u}_1\cdot{\bf u}_1-\frac{\left({\bf u}_1\cdot{\bf u}_4\right)^2}{{\bf u}_4\cdot{\bf u}_4}}.$$ (38)