# Amplitude decrease with geometrical spreading

1. ### AlecYates

12
Hey,

I'm looking at amplitude decrease of a seismic pulse as a result of geometrical spreading.

Starting with I = E / (4 * pi * r2) where E = original energy from source, we know that energy falls off as 1/r2, thus amplitude falls off as 1/r.

From wikipedia: "The energy or intensity decreases (divided by 4) as the distance r is doubled;"

This makes sense to me, as when r is doubled we have the energy divided by (2r)2 = 4r2 (which is 4x r2).

From this same principle, I would expect that the amplitude is divided by 2 when the distance is doubled as we have 1/2r instead of 1/r.

However from a Louisiana State University website:

"Geomteric spreading makes the amplitude of a signal falls off in proportion to the distance traveled by the ray. So that if the path of flight is doubled the amplitude will decrease by a factor of: square root of 2."

I can't see how they got their factor of √2 instead of 2.

Is one a mistake or am I missing something?

Cheers

2. ### Andrew Mason

6,875
Think of a compression/sound wave spreading out from a source equally in all directions as concentric spherical shells of energy. The energy in a given shell is constant. So the energy density is inversely proportional to the area of that sphere: i.e. energy per unit area varies as 1/r^2 where r is the radius of the shell.

The question you are asking has to do with the relationship between amplitude and energy density of a wave front. Think of the vibration of a spring: the energy contained in the spring is proportional to the square of the maximum amplitude. PE = kx^2/2 .

Since the energy contained in the wave front is proportional to the square of the amplitude and the energy density is inversely proportional to r^2, how would amplitude vary with r?

AM