Can I abandon the material in this section?

We could work out the mathematical problem of finding an analytic solution for the travel time as a function of distance in an earth with stratified v(z), but the more difficult problem is the practical one which is the reverse, finding $v(\tau)$ from the travel time curves. Mathematically we can express the travel time (squared) as a power series in distance h. Since everything is symmetric in h, we have only even powers. The practitioner's approach is to look at small offsets and thus ignore h⁴ and higher powers. Velocity then enters only as the coefficient of h². Let us why it is the RMS velocity, equation (25), that enters this coefficient.

The hyperbolic form of equation (24) will generally not be exact when h is very large. For ``sufficiently'' small h, the derivation of the hyperbolic shape follows from application of Snell's law at each interface. Snell's law implies that the Snell parameter p, defined by  

$$p \eq \frac{\sin\theta_i}{v_i}$$ (36)

is a constant along both raypaths in Figure 10. Inspection of Figure 10 shows that in the ith layer the raypath horizontal distance $\Delta x_i$ and travel time $\Delta t_i$are given on the left below by

$$\begin{eqnarray} \Delta x_i &=& \Delta z_i \tan\theta_i \eq \frac{v_i\Delta\ta... ...tau_i \left(1+ {\tiny 1 \over 2} p^2 v_i^2 \right) + O(p^4) \ \ .\end{eqnarray}$$ (37) (38)

The center terms above arise by using equation () to represent $\tan\theta$ and $\cos\theta$as a function of $\sin\theta$ hence p, and the right sides above come from expanding in powers of p. Any terms of order p³ or higher will be discarded, since these become important only at large values of h. Summing equation () and () over all layers yields the half-offset h separating the midpoint from the geophone location and the total travel time t.

$$\begin{eqnarray} h &=&\frac{p}{2}\ \tau \ V^2(\tau) + O(p^3) \\ t &=&\tau \left( 1 + \frac{1}{2} p^2 V^2(\tau) \right) + O(p^4) \ \ \ .\end{eqnarray}$$ (39) (40)

Solving equation () for p gives $p=2h/(\tau V^2)$,justifying the neglect of the O(p³) terms when h is small. Substituting this value of p into equation () yields

$$t \eq \tau \left( 1 + \frac{2h^2}{\tau^2 V^2(\tau)} \right) + O(p^4) \ \ \ .$$ (41)

Squaring both sides and discarding terms of order h⁴ and p⁴ yields the advertised result, equation (24).

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