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Geology 5640/6640 Introduction to Seismology 17 Feb 2015 © A.R. Lowry 2015 Last time: Seismology as Investigative Tool Deep-Earth investigations use earthquakes as sources, three-component broadband seismometers as receivers, and analysis tools include: Normal modes Precursors Waveforms Receiver Functions Tomography Anisotropy Ambient Noise Receiver functions use transmitted and converted phase arrival times to image impedance structure Tomography uses travel-times for large numbers of criss-crossing rays to image velocity structure Anisotropy measures direction-dependence of velocity Ambient noise uses environmental Rayleigh waves Read for Wed 17 Feb: S&W 53-85 (§2.4-2.6)
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Assignment I is now posted on the course website… Due Monday, March 2 at the beginning of class Note: Not all of the relationships you’ll need have been covered in course notes, but if you’ve been reading your text you’ll know where to look!
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Waves in Spherical Coordinates: Up to now we’ve used to denote the scalar displacement potential, but shows up as a coordinate in spherical coordinates… So for now we’ll use for potential. Then our wave equation becomes Spherical coordinates relate to Cartesian as: x = r sin cos y = r sin sin z = r cos
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Using these relations we can transform the Laplacian from Cartesian to spherical coordinates: To convert to spherical coords we’ll need to use the chain rule, e.g.: If we take derivatives of the coordinate relations (last slide) we get, e.g. for ∂ ∂ x : (1)
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We now have three equations in the three unknown derivatives; solving, we get: Then substituting (2) into (1) we get: Clearly by the time we do all three derivatives and evaluate this will get into some heavy trig… But suffice to say, with the help of some identities this eventually leads us to: (2)
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For a spherically symmetric solution (i.e., a point source in a constant-velocity medium), the ∂ ∂ terms are 0. Then the Laplacian is simply and our scalar potential wave equation becomes: with solutions of the form: (Note that if there is , -dependence, this will be (r, , ,t) ). Recall our earlier solutions had a f(t±r/ ) dependence, so the spherical solution has an additional 1/r dependence…
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Ray Theory represents the full wavefield in skeletal form, in terms of travel-times along propagation paths. Pros: It’s relatively simple (analogous to optics) Can be used for many problems, including: – Determination of earth structure – Locating earthquakes – Tomographic inversion for earth structure (We saw some of these Friday) Cons: Accurate only for infinite frequency waves ( = 0 ). So, It images structures on scales of order ≤ poorly (i.e., it can’t reproduce “wavefront healing” or other diffraction-related phenomena; poorly represents steep velocity gradients).
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