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Geology 5640/6640Seismology

Last time: Surface Waves• Surface waves (evanescent waves trapped at a surface) include Stoneley, Lamb as well as Rayleigh, Love waves

• Stoneley waves (“tube” waves) occur in boreholes• Lamb waves occur in plates or low-velocity waveguides• Rayleigh waves are P-SV wave interference patterns• Love waves consist of only SH-wave motion• Rayleigh waves occur at a free surface (i.e. i3 = 0). For propagation, require the same apparent velocity cx for both P and SV. Imposing these conditions we can solve for amplitudes A, B, leading to the Rayleigh function:

02 Mar 2011

Reading for Fri 4 Mar: S&W 89–100 Assignment 3 due 4 Mar!€

2cx

2

α 2−1

⎝ ⎜ ⎜

⎠ ⎟ ⎟A + 2 −

cx2

β 2

⎝ ⎜

⎠ ⎟B = 0

cx2

β 2−2

⎝ ⎜

⎠ ⎟A + 2

cx2

β 2−1

⎝ ⎜

⎠ ⎟B = 0

⇒ 2 −cx

2

β 2

⎝ ⎜

⎠ ⎟

2

+ 4cx

2

β 2−1

⎝ ⎜

⎠ ⎟cx

2

α 2−1

⎝ ⎜

⎠ ⎟= 0

Recall from our tensor review that these have a non-trivial (non-zero) solution only if the determinant is 0. That means we have to solve the roots of a polynomial equation, which we call the Rayleigh function:

This can also be written (e.g., p150 in Shearer):

To simplify this equation we can assume a Poisson’s solid ( = 0.25 => = ), in which case:

This has three roots but only one satisfies cx < < , so:

2 −cx

2

β 2

⎝ ⎜

⎠ ⎟

2

+ 4cx

2

β 2−1

⎝ ⎜

⎠ ⎟cx

2

α 2−1

⎝ ⎜

⎠ ⎟= 0

2 p2 −1

β 2

⎝ ⎜

⎠ ⎟+ 4 p2 p2 −

1

α 2

⎝ ⎜

⎠ ⎟ p2 −

1

β 2

⎝ ⎜

⎠ ⎟= 0

cx2

β 2

⎝ ⎜

⎠ ⎟cx

6

β 6−8

cx4

β 4+

56

3

cx2

β 2−

32

3

⎣ ⎢

⎦ ⎥= 0

cx = 2 −2

3

⎝ ⎜

⎠ ⎟β = 0.92β

Once we know the propagation velocity cx, we can solve for the amplitudes A, B, from the system of equations:

These are amplitudes in terms of displacement potentials , … We’d like to know the amplitudes in terms of displacements. For P- and SV-waves only:

If we evaluate these and take only the real component, we get (for the Poisson solid case):

2cx

2

α 2−1

⎝ ⎜ ⎜

⎠ ⎟ ⎟A + 2 −

cx2

β 2

⎝ ⎜

⎠ ⎟B = 0

cx2

β 2−2

⎝ ⎜

⎠ ⎟A + 2

cx2

β 2−1

⎝ ⎜

⎠ ⎟B = 0

⇒ B =

A 2 −cx

2

β 2

⎝ ⎜

⎠ ⎟

2rβ

ux =∂φ

∂x−

∂ψ

∂z

uz =∂φ

∂z−

∂ψ

∂x

ux = Akx sin ωt − kx x( ) e−0.85kz z −0.58e−0.39kz z[ ]

uz = Akx cos ωt − kx x( ) −0.85e−0.85kz z −1.47e−0.39kz z[ ]

Here’s what this looks like:

But that’s forconstant velocity& = only!

More generally velocity (and density) vary with depth in the Earth…

Here’s how velocity cx depends on VP

and VS (& density ) for two different periods T (= x/cx)

Now let’s consider particle motion in a Rayleigh wave, i.e., displacement of a specific point in the Earth. At z = 0:

cos and sin are 90° out of phase, so a particle on the surface traces an elliptical motion. At depth, the sense of rotation changes from retrograde (CCW) to prograde (clockwise).

ux = 0.42Akx sin ωt − kx x( )

uz = 0.62Akx cos ωt − kx x( )

Love Waves• Consist of SH motion only

• Need a positive velocity jump or gradient to exist

• Are inherently dispersive

(Note Rayleigh waves are also dispersive on Earth, but they don’t have to be: They are because different wavelengths sample different depths, and velocity increases with depth).

For a simple Love wave, let’s consider a (lower-velocity) layer over a (higher-velocity) infinite half-space:

Consider a SH wave which is a simple harmonic (sin). As it bounces around in the layer, there are multiple places on the wavepath where the phase is the same. Let’s look for places where the wave might experience constructive reinforcement at the surface, i.e. conditions for which the phase at point A is the same as at point Q.

First we need to remind ourselves what phase is! A complex number can be represented in the complex plane as z = x + iy = r ei, where is the phase angle:

So for a typical SH wave represented as

the phase is = t – kxx – kzz. To simplify, we can rotate our spatial axes into the propagation direction (or ray) to get:

where k is the modulus of the wavenumbers and d is distance along the raypath.

(real)

(imag)

x

yr

uy = Bei ωt−kx x−kz z( )

uy = Bei ωt−kβ d( )

Some useful quantities in this way of representing things are the angle which we rotated the coordinate axes (also angle of incidence!) j:

and by the relationship to V = f,

Phase is now = t – kd. If the phase difference between point A and point Q is some multiple of 2, then the waves will constructively interfere! We’ll arbitrarily choose time t = 0:

cos j =kz

kβ =kz

cos j=

ω

β

φ A( ) =ωt − kβ d = 0

φ Q( ) =ωt − kβ d = −kβ AB+ BQ( )

But now we need to consider a complicating factor: Recall that for an evanescent wave, a phase change will occur at the interface between the layer and the halfspace if the ray is post-critical! The phase change is given by (S&W 2.6.23):

So we’ll get constructive interference IF

After some trig substitutions this turns out to be:

Δφ=2tan−1 μ 2rβ 2

*

μ1rβ1

⎝ ⎜ ⎜

⎠ ⎟ ⎟

kβ AB+ BQ( )+2tan−1 μ 2rβ 2

*

μ1rβ1

⎝ ⎜ ⎜

⎠ ⎟ ⎟= 2nπ

kβ −2h cos j( )+2tan−1 μ 2rβ 2

*

μ1rβ 1

⎝ ⎜ ⎜

⎠ ⎟ ⎟= 2nπ

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