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We want to prove the following theorem:

If $\vec u(r,\theta,\phi)$ is a displacement field whose divergence $q(r,\theta,\phi) = \nabla \vec u(r,\theta,\phi)$ has the following properties:
$\bullet \quad q$ fulfills the Laplace equation $\Delta q = 0$ outside a finite source region
$\bullet \quad q$ tends uniformly to zero for $r \rightarrow \infty$after multiplication with r2
$\vec u$ displaces the same volume through any spherical surface enclosing the source region.

We note that the static displacement field of a moment-tensor source in a homogeneous isotropic full space has the required properties, as can be analytically derived from the explicit solution (the use of computer algebra is recommended). For a LVD source q is given by eq. (6). It is a multiple of the classical quadrupole potential, $P_{2}(\cos \theta)/r^3$, and therefore a solution of $\Delta q = 0$. For our present purpose we need the theorem only for point sources, but may as well prove it for distributed sources at the same time.

The proof of the theorem is as follows:

Under the above conditions, q can be expanded into a series of spherical harmonics associated with negative powers of the radius:

\begin{displaymath}q(r,\theta,\phi) = \sum_{\ell=2}^\infty \ \sum_{m=-\ell}^\ell...
...\ r^{-\ell - 1} \ P_{\ell}^{\,m} (\cos \theta) \ e^{i m \phi}
\end{displaymath} (7)

The origin may be chosen arbitrarily. The expansion will converge outside the smallest sphere around the origin that encloses the source. The condition that spherical surfaces enclose the source region will from now on be understood. The $\ell = 0$ and $\ell=1$ terms are absent for any choice of the origin due to our assumption on the asymptotic behaviour of q for large radii. The integral or average of q over any spherical surface, which is related to the $\ell = 0$term of the expansion, is therefore zero. Generally, any solution of the Laplace equation that decays faster than r-1 for $r \rightarrow \infty$ must have zero average over any spherical surface enclosing the source. The same result can be derived without invoking the series expansion by setting up Green's formula

\begin{displaymath}\int\!\!\!\int\!\!\!\int_{Volume}{(u \Delta v - v \Delta u) \...\!\!\!\int_{Surface}{(u \nabla v - v \nabla u) \, d \vec S}
\end{displaymath} (8)

with u=q and v=1/r, and integrating over a spherical shell whose outer radius is then made infinite.

Consequently, if we integrate q over a spherical shell of any thickness, the result must be zero.

Since q is the divergence of $\vec u$, this implies that $\vec u$displaces the same total volume through the inner and outer surfaces of the shell. The volume displacement through all spherical surfaces having a common center is thus the same.

We now consider two such sets of surfaces. We select surface S1from the first set and S2 from the second set so that they have the same radius r. The surfaces intersect because they both enclose the source. Let the volumes displaced by $\vec u$ through these surfaces be D1 and D2. The difference D2 - D1may be calculated by integrating q over the differential volume enclosed between S1 and S2 (one of the two bowl-shaped halves of this volume enters with a negative sign). By making r sufficiently large, D2 - D1 can be made arbitrarily small because the volume of integration increases with r2 while the integrand tends uniformly to zero faster than r-2. On the other hand, the displaced volume is constant within each set, so D1 must equal D2 for any value of r. The displaced volume is thus the same for all surfaces of both sets. This completes our proof.

next up previous contents
Next: About this document ... Up: Draft On the relationship Previous: References
Erhard Wielandt