Waves in strata

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 $h^4$ and higher powers. Velocity then enters only as the coefficient of $h^2$. Let us why it is the RMS velocity, equation (3.25), that enters this coefficient.

The hyperbolic form of equation (3.24) 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 3.10. Inspection of Figure 3.10 shows that in the $i$th layer the raypath horizontal distance $\Delta x_i$ and travel time $\Delta t_i$ are given on the left below by

$$\Delta x_i = \Delta z_i \tan\theta_i \eq \frac{v_i\Delta\tau}{2} \ \frac{p v_i} {\sqrt{1-p^2v_i^2}} \eq \frac{p}{2} \Delta\tau_i v_i^2 + O(p^3)$$ (37)

$$\Delta t_i = \frac{2 \Delta z_i}{v_i \cos\theta_i} \eq \frac{\Delta\tau_i}{\s... ...}} \eq \Delta\tau_i \left(1+ {\tiny 1 \over 2} p^2 v_i^2 \right) + O(p^4) .$$ (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 $p^3$ 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 $t$.

$$h = \frac{p}{2} \tau V^2(\tau) + O(p^3)$$ (39)

$$t = \tau \left( 1 + \frac{1}{2} p^2 V^2(\tau) \right) + O(p^4) .$$ (40)

Solving equation () for $p$ gives $p=2h/(\tau V^2)$, justifying the neglect of the $O(p^3)$ 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 $h^4$ and $p^4$ yields the advertised result, equation (3.24).

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Waves in strata