Among the given layered models, labelled as P, Q, R and S, which of the following pairs is/are NOT possible to distinguish according to the principle of equivalence? Model parameters read off the figure -- (P): \(\rho_1 = 1\ \Omega\text{m}, h_1 = 10\ \text{m}\); \(\rho_2 = 100\ \Omega\text{m}, h_2 = 20\ \text{m}\); \(\rho_3 = 0\ \Omega\text{m}\). (Q): \(\rho_1 = 5\ \Omega\text{m}, h_1 = 10\ \text{m}\); \(\rho_2 = 400\ \Omega\text{m}, h_2 = 5\ \text{m}\); \(\rho_3 = 0\ \Omega\text{m}\). (R): \(\rho_1 = 40\ \Omega\text{m}, h_1 = 20\ \text{m}\); \(\rho_2 = 2\ \Omega\text{m}, h_2 = 10\ \text{m}\); \(\rho_3 = 100\ \Omega\text{m}\). (S): \(\rho_1 = 30\ \Omega\text{m}\); \(\rho_2 = 3\ \Omega\text{m}, h_2 = 10\ \text{m}\), with \(h_1 = 5\ \text{m}\); \(\rho_3 = 100\ \Omega\text{m}\).
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For a resistive middle layer (K-type), curves are equivalent when rho2*h2 (transverse resistance) matches; for a conductive middle layer (H-type), they are equivalent when h2/rho2 (conductance) matches. Compute both for all four models.
A different way to see this is to directly compare which single lumped parameter each curve type is sensitive to, rather than computing T and S from scratch for every pair.
For a three-layer sounding curve, in the limit of a thin, non-outcropping middle layer, the far-field apparent resistivity behaviour depends on \(\rho_1\), \(\rho_3\), and only ONE composite number describing the middle layer -- never on \(\rho_2\) and \(h_2\) separately. Which composite number matters depends on the sequence type:
Model
Sequence
Type
Equivalence parameter
Value
P
1, 100, 0
K (resistive middle)
\(T=\rho_2h_2\)
2000
Q
5, 400, 0
K (resistive middle)
\(T=\rho_2h_2\)
2000
R
40, 2, 100
H (conductive middle)
\(S=h_2/\rho_2\)
5
S
30, 3, 100
H (conductive middle)
\(S=h_2/\rho_2\)
3.33
Reading straight down the last column: P and Q share the same T = 2000, so no VES inversion can uniquely separate their \(\rho_2\) and \(h_2\) -- they map to the same curve, i.e. they are NOT distinguishable. R and S have different S values (5 vs 3.33), so despite both being H-type, they are physically distinguishable curves. This immediately eliminates (A), (B) and (D), leaving (C) P, Q as the only equivalent, hence non-distinguishable, pair.
