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Seismic imaging · A course with experiments

L2 · Chapter 6

Travel-time tomography and what the resolution matrix shows

Each ray crosses a few cells of a grid, and its delay is the sum of the slowness changes it meets, each weighted by the length of the ray in that cell. Many rays give a linear system. It usually has fewer independent equations than unknowns, so some patterns leave no trace in the data at all. A tomographic image is one chosen answer among many, and the resolution matrix records how that choice blurs the true structure.

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Each delay is a sum over the cells a ray crosses

Split the region into cells with a constant slowness perturbation each, the change in inverse speed relative to a background. If rays are straight and the perturbations are small, a ray’s delay is the sum over cells of the length of the ray in each cell times that cell’s perturbation. Section 4.19 of the MIT 12.201 notes writes the system this way, with the matrix holding path lengths.

Rawlinson and Spakman write the same system as their Eq. 6 and list what the error term holds: picking error, error of the forward theory, linearization error and the error of describing a smooth Earth with cells. The tomography Lab computes its delays by integrating smooth anomalies along each ray, so its data carry that last error too.

di=∑jGij mj+ϵi,Gij=length of ray i in cell jd_i=\sum_j G_{ij}\,m_j+\epsilon_i,\qquad G_{ij}=\text{length of ray } i \text{ in cell } j

Damping and smoothing choose one answer

Patterns that no ray can see form the null space of the matrix. Adding any of them to a model leaves every delay unchanged, so the data alone cannot pick a model. The usual cure adds penalties to the misfit: one on the size of the model, called damping, and one on its roughness, called smoothing. Rawlinson and Spakman write this cost as their Eq. 7 and the resulting generalized inverse as Eq. 9.

The tomography Lab uses a damping weight and a smoothing weight, each set on a log scale and multiplied by the mean diagonal of the data term so that the sliders mean the same thing for every ray geometry.

S(m)=(d−Gm)TCd−1(d−Gm)+ε2mTm+η2mTDTDm,m^=G−gdS(\mathbf m)=(\mathbf d-\mathbf G\mathbf m)^{\mathsf T}\mathbf C_d^{-1}(\mathbf d-\mathbf G\mathbf m)+\varepsilon^2\mathbf m^{\mathsf T}\mathbf m+\eta^2\mathbf m^{\mathsf T}\mathbf D^{\mathsf T}\mathbf D\mathbf m,\qquad \hat{\mathbf m}=\mathbf G^{-g}\mathbf d

The resolution matrix maps truth into the image

Insert the true model into the solution. The image equals the resolution matrix times the truth plus the inverse applied to the errors (Rawlinson and Spakman, Eqs. 10 to 13). If the resolution matrix were the identity, every cell would be recovered exactly. In practice each row is a blur: it shows which true cells leak into one image cell.

The noise term is a separate matter. Its spread is the model covariance of Eq. 14, which depends only on the assumed data errors and the inverse, not on the data. A model can have a resolution matrix close to the identity and still be dominated by noise.

A synthetic test inverts noise-free delays from a made-up model through the same matrix and inverse. By Eq. 16 its output is the resolution matrix times the made-up model, which is why checkerboard and spike tests reveal properties of the resolution matrix.

m^=R mtrue+G−gϵ,R=G−gG,CM=G−gCd(G−g)T\hat{\mathbf m}=\mathbf R\,\mathbf m_{\rm true}+\mathbf G^{-g}\boldsymbol\epsilon,\qquad \mathbf R=\mathbf G^{-g}\mathbf G,\qquad \mathbf C_M=\mathbf G^{-g}\mathbf C_d\big(\mathbf G^{-g}\big)^{\mathsf T}

One test pattern cannot show all the blur

A checkerboard test can look perfect while larger structures are lost. Rawlinson and Spakman cite an idealized experiment by Lévêque and co-authors in which a fine checkerboard was recovered perfectly and a coarse one sat almost entirely in the null space, with the same rays. A recovered checkerboard therefore shows only that this pattern is resolved; a failed one shows a lack of resolution.

In the tomography Lab the checkerboard test gives a pattern match of 0.899 with the full ring, higher than the 0.724 for the two smooth anomalies. The test pattern looks better resolved than the real target.

Worked example

A grid of four cells with a hidden pattern

Take four square cells of side 1 km, numbered across the top row and then the bottom row. Two horizontal rays run along the rows and two vertical rays along the columns, so each ray spends 1 km in each of two cells.

The four delays add the top row, the bottom row, the left column and the right column. The pattern plus one, minus one, minus one, plus one adds to zero along every ray. It is in the null space, so the matrix has rank three, not four.

The minimum-length solution projects out that pattern, and its resolution matrix is the identity minus the outer product of the normalized null vector. Put a slowness anomaly of 0.0100 s/km in the top-left cell only. The data are 0.0100 s on the top-row ray and the left-column ray, and zero on the other two. The recovered model is 0.0075 in the top-left cell, 0.0025 in its two neighbours and minus 0.0025 in the opposite corner: three quarters of the spike, smeared into a cross.

Add one diagonal ray from the top-left corner to the bottom-right corner. It spends 1.41 km in each of the two diagonal cells, its delay changes when the hidden pattern is added, the null space disappears and the spike is recovered exactly.

n=12(1,−1,−1,1),R=I−nnT,R (0.0100,0,0,0)=(0.0075, 0.0025, 0.0025, −0.0025) s/km\mathbf n=\tfrac12(1,-1,-1,1),\quad \mathbf R=\mathbf I-\mathbf n\mathbf n^{\mathsf T},\quad \mathbf R\,(0.0100,0,0,0)=(0.0075,\ 0.0025,\ 0.0025,\ -0.0025)\ \mathrm{s/km}

An experiment to think through

Predict, change, explain

  1. In the tomography Lab, compare Stations all around with Two wells: sideways smear. Predict the direction in which the two-well image smears an anomaly.
  2. Choose Two wells: sideways smear and drag the green probe cell across the region.
  3. Explain the direction of the smear from the directions of the rays between the wells.
Invert ray delays and probe the resolution

The Labs use flat, isotropic, elastic or acoustic rock with made-up but realistic speeds; none of them reproduces a real survey. Each Lab states its model and limits in its notes and can export the data behind its readouts.

Try explaining it yourself

1. In the four-cell example, which data combination is independent of the model?

Read an explained answer

The sum of the two row delays equals the sum of the two column delays, because both equal the sum over all four cells. That is the one redundancy that leaves the rank at three.

2. The Lab’s preset No damping: noise takes over has a mean resolution diagonal of 0.586, higher than 0.422 for the default, yet its pattern match is 0.031. How can both be true?

Read an explained answer

The resolution matrix describes only the first term of the image, the blurred truth. Cutting damping and smoothing to a thousandth makes it closer to the identity, but the inverse then amplifies the 80 ms pick noise. The noise term, measured by the model covariance, swamps the image.

3. Why must a synthetic test use the same matrix and the same inverse as the real inversion?

Read an explained answer

Only then is its output the same resolution matrix times the test model, as in Eq. 16. A different ray set or a different regularization tests a different resolution matrix.

Sources to read alongside

  1. MIT OpenCourseWare, 12.201 Essentials of Geophysics (Fall 2004), lecture notes, Chapter 4: Seismology

    Section 4.6 separates the elastic wave equation into P and S waves and gives their speeds (Eqs. 4.34 and 4.37). Sections 4.14 to 4.16 derive Snell’s law from Fermat’s principle (Eqs. 4.91 to 4.96) and show that the ray parameter is the slope of the travel-time curve (Eq. 4.98). Section 4.19 notes that in travel-time tomography the matrix entries are the ray lengths in each block.

  2. N. Rawlinson and W. Spakman, On the use of sensitivity tests in seismic tomography, Geophysical Journal International 205, 1221–1243 (2016)

    Section 2 writes the damped least-squares solution and its resolution matrix (Eqs. 7 to 14) and shows that a noise-free synthetic test samples the same resolution matrix (Eq. 16). Section 2.5 lists eight consequences. Section 3 compares checkerboard and spike tests on a 3844-unknown surface-wave example and ends with six recommendations.

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