Hyperbolic Moveout

Claim: To good approximation, for ``small'' offsets,

$$T(t_0,x)=\sqrt{t_0^2+ \frac{x^2}{v_{\rm RMS}^2(t_0)}},\,\,T_0(T(t_0,x),x) \equiv t_0$$

where the RMS velocity is

$$v_{\rm RMS}(t_0)=\sqrt{\frac{1}{t_0}\int_0^{t_0}\,v^2}$$

Justification [Continuum derivation of the hyperbolic moveout approximation]: The 2-way traveltime $T(t_0,x) =\tau_2(z,x)$ from the surface at z=0 to depth z and back at offset x is related to the solution of the eikonal equation $\tau(z,x)$ with point source at z=x=0 by

$$\tau_2(z,x)=2\tau\left(z,\frac{x}{2}\right)$$

Thus

$$\frac{1}{4}\left(\frac{\partial \tau_2}{\partial z}\right)^2+ \left(\frac{\partial \tau_2}{\partial x}\right)^2 =\frac{1}{v}^2$$

Differentiate this twice with respect to x and use the vanishing of odd-order x derivatives at x=0 (implied by symmetry) to conclude that the second x derivative

$$q(z)=\frac{\partial^2 \tau}{\partial x^2}(z,0)$$

satisfies

$$\frac{1}{2v}\frac{dq}{dz}+q^2=0$$

Introduce temporarily a new depth coordinate

$$\sigma(z)=2\int_0^z \, v$$

Then in terms of $\sigma$, q satisfies the Ricatti equation

$$\frac{dq}{d\sigma}+q^2=0$$

The solution which is singular at $\sigma=0$, i.e. z=0, is

$$q(\sigma)=\frac{1}{\sigma}=\frac{1}{2\int_0^z \,v}$$

Since $dz = \frac{1}{2}v dt_0$, you can also write this as

$$q(\sigma)=\frac{1}{\int_0^{t_0}\,v^2}$$

Thus

$$T(t_0,x)=t_0 + \frac{x^2}{2}\frac{\partial^2 T(t_0,0)}{\partial x^2} +...$$

$$= t_0 + \frac{x^2}{2\int_0^{t_0}\,v^2}+...$$

Since

$$\frac{\partial^2}{\partial x^2}(T(t_0,x))^2_{x=0}=2 t_0\frac{\partial^2 T} {\partial x^2}(t_0,0)$$

the above can be rewritten

$$T(t_0,x)^2 = t_0^2 + \frac{x^2}{\frac{1}{t_0}\int_0^{t_0}\,v^2}+...$$

$$=t_0^2+\frac{x^2}{v_{\rm RMS}^2(t_0)} + ...$$

which reveals that the hyperbolic moveout approximation is just the second order Taylor expansion of T² in x, which should be good for ``small'' x.

This report adopts the hyperbolic moveout approximation, i.e. truncate the Taylor expansion above and take

$$T(t_0,x)^2 = t_0^2 + \frac{x^2}{v_{\rm RMS}^2(t_0)}$$

This amounts to assuming that all events in the data have precisely hyperbolic moveout. Of course this assumption is not entirely consistent with geometric optics. It has been suggested that the deviation of actual two-way time from the hyperbolic moveout approximation may be mistaken for evidence of anisotropy in some cases. In any case the error caused by replacing actual two way time by its hyperbolic moveout approximation is not an asymptotic error in the sense of the last section, so I will treat it as a component of data noise.

The reciprocal square RMS velocity, or RMS square slowness is the primary expression of velocity in the above formula. It occurs so often as to warrant its own notation:

$$u(t_0) \equiv (v_{\rm RMS}(t_0))^{-2}$$

The conditions defining the mute can be restated: since

$$\frac{\partial T}{\partial t_0}(t_0,x) = \frac{t_0+\frac{x^2}{2} \frac{\partial u}{\partial t_0}(t_0)}{T(t_0,x)}$$

the quantity on the right hand side of this equation must be bounded away from zero. Since v generally increases with depth, hence u decreases, such a lower bound will only be possible for t₀ exceeding a threshold for each x, which is the mute boundary mentioned before. In the data, i.e. (t,x), coordinates, the stretch factor condition becomes

$$s(t,x)= \frac{\partial T_0}{\partial t}(t_0,x)= \left(\frac{... ...{2} \frac{\partial u}{\partial t_0}(T_0(t,x))} <C_{\rm stretch}$$

and as before the mute $\phi$ must be supported in the set specified by this condition.

The upper and lower velocity envelopes implied by membership of the velocity in $\cal{A}$ imply corresponding envelope mean square slownesses ($u_{\rm min}$ corresponding to $v_{\rm max}$ and vis-versa) so that $u_{\rm min}(t_0) \le u(t_0) \le u_{\rm max}(t_0)$.

It is usually reasonable to assume the lower velocity bound to be constant (independent of t₀) - for example, equal to sound velocity in water, or close to it. Then $u_{\rm max}$ is also constant, so you can explicitly estimate a lower bound for T₀:

$$T^2(t_0,x)=t_0^2+x^2 u(t_0) \le t_0^2 + x^2 u_{\rm max}$$

so

$$T_0(t,x) \ge \sqrt{t^2 - x^2 u_{\rm max}}$$

The velocity bounds also imply a bound on the derivative of u:

$$\frac{du}{dt_0}= -\frac{u^2}{t_0}\left(v^2-\frac{1}{u}\right... ...\left(v^2-v^2(0) - \frac{1}{t_0}\int_0^{t_0}(v^2-v^2(0))\right)$$

The bounds on v, the known value of v at the surface, and the maximum two way time imply bounds on the slope

$$\frac{v^2(t_0)-v^2(0)}{t_0}$$

whence a bound $u'_{\rm max}$ on the derivative of u follows immediately.

Since both the lower bound on T₀ and the uppoer bound on the derivative of u are uniform over $\cal{A}$, a $\cal{A}$-uniform bound on the stretch factor follows:

$$s(t,x)\le \frac{t}{\sqrt{t^2-x^2 u_{\rm max}} -\frac{x^2}{2}u'_{\rm max}}$$

From this you can derive a $\cal{A}$-uniform mute boundary. Therefore assume henceforth that $\phi$ is a $\cal{A}$-uniform mute.