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• Source deconvolution (like filtering methods) is done in the frequency domain (where it is a simple division)
• Migration redistributes reflection energy from dipping structures to its “true” location in two-way travel-time, using redundant sampling by multiple sources… Depth migration also removes velocity effects
Read for Mon 13 Apr: S&W 157-176 (§3.4–3.5)
€
s t( ) = w t( )⊗ r t( ) = w t − τ( )r τ( )dτ−∞
∞
∫ ⇔ S ω( ) = W ω( )R ω( )
Ray Theory in a Spherical EarthMost of us (with a few exceptionsfound mostly in legislative bodies)would agree that the Earth is betterapproximated by a sphere than aflat surface.
At distances >20°, we have to takethis into account somehow…
Two common approaches are
• Build a spherical geometry into the equations used, or
• Assume a Cartesian geometry but then change the layer velocities and thicknesses to give travel-times equivalent to those of a sphere (called an “Earth flattening transformation”)
Let’s consider Snell’s Law in a spherical geometry:
Here the ray path is in red;velocity is constant withinspherical shells and increasesat each layer boundary.
At point A, locally the boundaryis ~flat & we can use regularSnell’s Law at the interface:
But '1 does not equal !Consider the right triangles formed by d, r1 and r2:
2
1
1
1 sinsin
υ
υ ′
=
Since sine = opposite/hypotenuse, sin'1 = d/r1 andsin2 = d/r2, or d = r1sin'1 = r2sin2.
2
1
1
1 sinsin
υ
υ ′
=
€
r1 sinθ1
υ1
=r2 sinθ2
υ 2
Snell’s Law in a spherical geometry:
Now, multiply both sides ofSnell’s Law:
by r1:
Since:
we can write Snell’s Lawin spherical coordinatesas:
€
r1 sinθ1
υ1
=r1 sin ′ θ 1
υ 2
€
d = r1 sinθ1' = r2 sinθ2
Snell’s Law in a spherical geometry:
We can alternatively write
in terms of slowness as:
which for an arbitrary ith layeris simply:
Thus the r multiplier correctsfor the change in orientation ofthe interface normal as the rayapproaches the center of the Earth.
€
r1 sinθ1
υ1
=r2 sinθ2
υ 2
€
u1r1 sinθ1 = u2r2 sinθ2 = p
€
p = uiri sinθ i
Δ=
d
dTp
Snell’s Law in a spherical geometry:
But haven’t we changed theunits of the ray parameter?
Short answer: yes! p has unitsof [time/distance]*radius, but recall that arc length on a circleis equal to radius * angle inradians. Thus p actually can bethought of as having units of[time]/[ in radians].
In other words (and thisshouldn’t come as a surprise):
Δ
r
rΔ
More rigorously, let’s consider two rays with slightly different ray parameters:Ray 1
p1 = pT1 = TΔ1 = Δ
Ray 2p2 = p + dpT2 = T+dTΔ2=Δ+dΔ
In the limit as dΔ goes to zero,the “triangle” at right satisfies:
and thus:
Similar to the Cartesian casewhere the ray parameter is theinverse of the apparent velocity cx along the surface, we have:
€
sin i =v0dT
r0dΔ
€
dT
dΔ=
r0 sin i
v0
= p
€
1
cx
=1
dΔdT
=dT
dΔ= p
With some easy math, which we’ll skip (but you can see p159eqns 9-15), we can show that travel-time is:
where: r0 = surface radiusrp = deepest point on ray path (“turning radius”) = r/v;
Angular distance is:
And in this format,
€
T p( ) = uds∫ =ds
v∫ = 2
ζ 2dr
r ζ 2 − p2rp
r0
∫
€
Δ p( ) = 2pdr
r ζ 2 − p2rp
r0
∫
€
τ p( ) = T p( )− pΔ p( ) = 2ζ 2 − p2
rdr
rp
r0
∫
€
dp
dΔ=
d 2T
dΔ2< 0
For a slowly increasingvelocity with depth,slope of the travel-timecurve (= p) decreaseswith Δ and the intercept( τ) increases with Δ(so decreases with p).
With a sufficiently steepgradient in velocity overa small distance, canget triplication (as in theCartesian case).