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Root-mean-square velocity

When a ray travels in a depth-stratified medium, Snell's parameter $p=v^{-1}\sin\theta$ is constant along the ray. If the ray emerges at the surface, we can measure the distance $x$ that it has traveled, the time $t$ it took, and its apparent speed $dx/dt=1/p$. A well-known estimate $\hat v$ for the earth velocity contains this apparent speed.
\begin{displaymath}
\hat v \eq \sqrt{ {x\over t}  {dx\over dt} }
\end{displaymath} (18)

To see where this velocity estimate comes from, first notice that the stratified velocity $v(z)$ can be parameterized as a function of time and take-off angle of a ray from the surface.
\begin{displaymath}
v(z) \eq v(x,z) \eq v'(p,t)
\end{displaymath} (19)

The $x$ coordinate of the tip of a ray with Snell parameter $p$ is the horizontal component of velocity integrated over time.
\begin{displaymath}
x(p,t) \eq \int_0^t  v'(p,t)  \sin\theta(p,t) dt
\eq p \int_0^t v'(p,t)^2 dt \
\end{displaymath} (20)

Inserting this into equation (3.18) and canceling $p=dt/dx$ we have
\begin{displaymath}
\hat v \eq
v_{\rm RMS}\eq \sqrt{ {1\over t}  \int_0^t v'(p,t)^2 dt  }
\end{displaymath} (21)

which shows that the observed velocity is the ``root-mean-square'' velocity.

When velocity varies with depth, the traveltime curve is only roughly a hyperbola. If we break the event into many short line segments where the $i$-th segment has a slope $p_i$ and a midpoint $(t_i,x_i)$ each segment gives a different $\hat v(p_i,t_i)$ and we have the unwelcome chore of assembling the best model. Instead, we can fit the observational data to the best fitting hyperbola using a different velocity hyperbola for each apex, in other words, find $V(\tau )$ so this equation will best flatten the data in $(\tau,x)$-space.

\begin{displaymath}
t^2 \eq \tau^2 + x^2/V(\tau)^2
\end{displaymath} (22)

Differentiate with respect to $x$ at constant $ \tau $ getting
\begin{displaymath}
2t  dt/dx \eq 2x/V(\tau)^2
\end{displaymath} (23)

which confirms that the observed velocity $\hat v$ in equation (3.18), is the same as $V(\tau )$ no matter where you measure $\hat v$ on a hyperbola.


next up previous [pdf]

Next: Layered media Up: CURVED WAVEFRONTS Previous: CURVED WAVEFRONTS

2009-03-16