Schmucker-Weidelt Lecture Notes, Aarhus, 1975 - MTNet

i.e. source and receiver are interchangeable. For a proof write
(3.42) for G (r 1 rt ) and a corresponding equation for G (r 1 Y:* ) .
n -a - 1 - -
Multiply these equations crosswise by G (r (r' ) and Gn (rolr' ) , subn-tract
and integrate over the full space. Then the result is ob-
tained on using Green's theorem.
For a sol.ution of (3.44) we have to put a line current at each point
r of the anomalous domain and have to compute the resulting elec-
--O
tric field at each point of this domain replacing its anomaious
structure by the normal conductivity. Assume a rectangular anomalous
don~ain wLth NY cells in. y-direction and iqZ cells in z-direction.
Because of horizontal isotropy the field depends in horizontal d.irec-
tion only on the distance between source axid receiver. Hence, we
need field values only f0r.N.Y horizontal distances. Due to the
layering there is no isotropy in vertical direction. Here we have
to put the line current into the center of each cell. Because of
reciprocity the sorresponding field v.alues have to be calculated
in and below the depth of the source. Hence, for the kernel Gn a
1
total of NY - TNZa(NZ+l) field values has to be calculated.
A second set of kernels is required which transform on using (3.44)
the electrical field in the anomalous domain into the electromagne-
tic surface field for all three components E xr H yr H z . The number of
required kernel data depends on the range where the field is to be
ev~luated.
It remains to calculate G (r ir) for a layered structure. Assume L
n -a -
layers with conductivities oo = 0, al, ...r csL and upper edges
h, ~ 2 ' "'I hL' hL+, - O0.
-
Let the source and observation point be placed in the m-th and p-
layer, respectively, and let in the m-th layer
+ +
6 A-f-(z),
omm
z - < z
0
+ + 6 F3-f (Z),
. Lmm Z > Z - 0