PSN-L Email List Message

Subject: Re: Dissipation of Seismic Waves
From: John or Jan Lahr JohnJan@........
Date: Tue, 24 Oct 2006 23:49:40 -0700


At 06:14 PM 10/23/2006, you wrote:
>Hi All -
>
>I believe that the intensity of light drops by the square of the 
>distance i.e., double the distance and the new value is the square 
>root of the distance.  How does this apply to seismic waves which 
>are basically sound waves?  In addition, how does this affect the 
>amplitude and radiated seismic energy of the waves?
>
>Bob Hancock

Hi Bob,

The graphs on the page below of amplitude versus distance show how 
seismic waves decay with distance.

Ground motion amplitude 
and velocity as a function of distance, magnitude, and depth.  (Only 
works well with MSIE -- Not Firefox.)  Keep in mind that bacteria 
range from about 1 to 10 microns in diameter, while a human hair may 
be from 50 to 100 microns in diameter.

For waves that expand in three dimensions, ignoring conversion to 
heat, the total energy flux through a sphere at any radius will be 
constant.  The surface area of the sphere is 4 pi r^2, so the energy 
per unit area must be proportional to 1/r^2.  The energy carried by 
the wave is proportional to the amplitude squared (A^2), so

A^2 is proportional to 1/r^2, or A is proportional to 1/r.

I'm not sure how to relate this to the amplitude at the surface 
versus geocentric distance for body waves in a spherical earth with 
velocity that varies with depth!

For surface waves that are constrained to the surface of the earth, 
ignoring conversion to heat or leaking of some energy into the 
interior, the total energy flux through a circular zone or any radius 
will be constant.  The area of the outer edge of a circular zone of a 
given thickness is proportional to the radius of the zone, so the 
energy per unit area must be proportional to 1/r.  Then, since the 
energy is proportional to A^2,

A^2 is proportional to 1/r, or A is proportional to 1/sqrt(r).

 From the formula for surface wave magnitude, Ms = log (A/T) + 1.66 
log D + 3.3,
one can show that the Amplitude is proportional to 1/ (D^1.66)
This is much faster decay than 1/(D^.5), due, I suppose, to 
attenuation (conversion of energy to heat) and  leaking of energy 
into the interior of the Earth.  Actually the amplitude increases 
again near the opposite side of the earth as the surface waves 
converge from all 
directions.  See:  http://jclahr.com/science/software/magnitude/ms/msmag.html

Don't know if this helps much!

Cheers,
John 


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