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Next: The Mogi model Up: Basics of the volume-source Previous: The source volume and

Relation with other source models

Since the primary physical cause of the seismic signal is the pressure P(t) in the source, we may want to express the source volume by the pressure. The result is

\begin{displaymath}V(t) = \frac{\pi a^3 P(t)}{\mu}
\end{displaymath} (13)

where $\mu$ is the shear modulus of the medium. Thus, we would need to know the three quantities a, P, and $\mu$ to calculate V and the seismic wavefield, and cannot determine these parameters individually from seismic observations.

Seismic sources are often idealized as point forces or couples. Both concepts are relevant to spherical explosive sources. Elementary theory gives the radial far-field displacement from a vertical point force F(t) in an elastic full space as

 \begin{displaymath}u_f=\frac{\cos \theta}{4 \pi r (\lambda + 2 \mu )} \, F(t-\frac{r}{c})
\end{displaymath} (14)

where $\theta$ is the angle of radiation measured against the z axis. When the same force is acting on the surface of a half-space, the P-wave displacement radiated vertically downward is just twice as large.

Another source model relevant to explosions is a couple of colinear forces $\pm F(t)$ acting at a small distance d, also called a linear vector dipole. Altough such forces do not have a moment in the sense of a torque, they define a generalized moment $M(t)=d \cdot F(t)$ and produce the radial far-field displacement

 \begin{displaymath}u_f=\frac{\cos^2 \theta}{4 \pi r c (\lambda + 2 \mu )} \, \dot{M}(t-\frac{r}{c})
\end{displaymath} (15)

G. Müller (1973) has, in analogy to the well-known definition of the seismic moment of a dislocation, defined the seismic moment of an underground nuclear explosion as

 \begin{displaymath}M_e(t)=(\lambda+2 \mu) \, S \, D(t)
\end{displaymath} (16)

where S is the area of a spherical surface around the explosion, D(t) is the radial displacement at (or of) this surface, and $\lambda$ and $\mu$ are the Lamé moduli. Obviously, $V(t)=S\/D(t)$ is the source volume, so we have

\begin{displaymath}M_e(t)=(\lambda+2 \mu) \, V(t)
\end{displaymath} (17)

Inserting this into (15) with $\theta=0$, we get the same result as from eq. (11). However, for spherical symmetry we must describe the explosion as the superposition of three mutually orthogonal linear dipoles, or equivalently by a moment tensor that is Me(t) times the unit tensor. The contributions from the three diagonal elements then add up to the spherical far-field term (11) in any direction.

Underwater explosions as seismic sources can, with respect to the downward radiated seismic signal, be modelled by a force F(t) acting vertically on the solid halfspace. Wielandt (1975) gives the equivalent force as

\begin{displaymath}F(t)=\frac{Z_1 Z_2}{Z_1+Z_2}\,\dot{V}(t)
\end{displaymath} (18)

Interestingly, the main part of the vertical force does not result from the direct pressure of the explosion products on the seabottom; it rather results from the outward travelling pressure wave in the water. The force persists as long as the gas bubble expands, even when the gas pressure has already dropped below the hydrostatic equilibrium. Although the force does not act in the bubble, it can be calculated from its volume. The relationship between the bubble volume, the force, and the seismic signal is linear, even in the presence of a shock wave.

The formula may as well be applied to explosions in a homogeneous medium, by simply taking Z1=Z2=Z. We then have

\begin{displaymath}F(t)=\frac{Z}{2} \,\dot{V}(t)
\end{displaymath} (19)

Combining this with the half-space version of eq. (14), we are back at the basic formula (11).


next up previous contents
Next: The Mogi model Up: Basics of the volume-source Previous: The source volume and
Erhard Wielandt
2001-09-21