424d1-fall2014.pdf

Geophysics 424 September 2014

Martyn Unsworth, University of Alberta, 2014 1

D1 Theory of magnetotellurics over a 1-D Earth D1.1: Magnetotelluric response of a halfspace at normal incidence

Consider a plane, polarized, electromagnetic wave that is normally incident on the Earth-air interface. A transmitted wave will enter the Earth and diffuse downwards, while a reflected wave will travel upwards. Incident wave The incident wave is polarized in the x-direction has an amplitude Ei , frequency and wave number k0 and since the conductivity of the air is zero, this wave will not be damped and the wavenumber is real and defined as

000 k (1)

tizikiix eeEtzE

0),( (2)

Reflected wave This will have an amplitude Er and can be described by a very similar equation to the incident wave. Since it travels in the opposite direction (upwards), the sign of the spatial term will be the opposite to the incident wave.

tizikrrx eeEtzE

0),( (3) Transmitted signal This will have an amplitude Et and since the Earth conductivity is relatively large, it will propagate in the Earth by diffusion. This assumes that displacement current is negligible compared to conduction current, as discussed previously.

2)1( 11

ik (4)

Thus the transmitted signal can be described by:



tizkttx eeEtzE

1),( (5) Note that the wavenumber (k1) is a complex number. The real part will cause exponential decay with depth, while the imaginary part causes an oscillation.

The signal will decay with a characteristic skin depth, where f

503 (m)

Assuming that Ei is known, we now need to derive values for Et and Er to describe the energy partition at the Earths surface. This requires two equations which are obtained from imposing boundary conditions at z = 0 Boundary condition 1

Continuity of the horizontal electric field (Ex ) requires that

),(),(),( tzEtzEtzE txrx

ix (6)

at z = 0 for all values of t. It is simple to show that

tri EEE (7) Boundary condition 2

ztzE

itzH xy

),(1),(

(8)

is also continuous at z = 0

Incident wave tizikiix eeikE

ztzE

0)(),( 0 (9)

Reflected wave tizikrrx eeikE

ztzE

0)(),( 0 (10)

Transmitted wave tizkttx eekE

ztzE

1)(),( 1 (11)

Thus continuity of Hy requires that

tri EkEikEik 100 (12)

This can be rearranged to give

0

1

ikEkEE

tri (13)

Adding (7) and (13) gives



0

12ikEkEE

tti (14)

)2

(0

10

ikkikEE ti (15)

Now re-arrange to express the transmitted amplitude as a ratio of the amplitude of the incident wave.

)2(10

0

kikik

EE

i

t

(16)

Consider some typical numbers for a magnetotelluric survey:

= 1 rad / s, 1 = 0.01 S/m, 0 = 4 10-7 H/m, 0 = 8.85 10-12 F/m

k0 = 3.33 10-8 m-1 and k1= (1-i) 7.93 10-5 m-1

Thus it is clear that |k0|



Consider the value of impedance at z = 0 for the case of a uniform halfspace Earth.

1

0

1

0

1

0

10

0

1

0

2)1()(

iii

ii

kiZ xy

Note that Zxy )( depends only the properties of the Earth, and not the air. The impedance is a complex number with magnitude:

1

0

22

)()()(

y

xxy H

EZ (21)

Re-arranging (21) we can show that

20

20

1

y

xxy

HEZ

(22)

This expression can also be written in terms of the resistivity of the Earth (1) as:

2

0

2

01

11y

xxy H

EZ

(23)

Note that all terms on the right hand side of (23) can be measured, and this shows how surface measurements of electric and magnetic fields can be used to measure the resistivity of the Earth. It is important to consider which part of the Earth is being sampled in such a measurement. Since the EM fields attenuate in the Earth with a length scale of a skin depth (), this measurement samples a hemisphere around the observation site, radius . Apparent resistivity In reality, the resistivity of the Earth will not be constant over the hemisphere. In this case it is usual to define the apparent resistivity as a function of frequency )( as:

2

0 )()(1)(

y

xa H

E (24)

If the Earth has a uniform resistivity, then the analysis above shows that a = 1 In general, the resistivity will not be constant with depth. In this case, the apparent resistivity can be considered as the average resistivity over a hemisphere with radius equal to the skin depth.



Phase

)()(tan)(tan)( 11

y

xxy H

EZ (25)

The phase of Zxy )( is the phase angle between Ex )( and Hy )(

tiixx eeEtE E

|)(|),(

tiiyy eeHtH H

|)(|),(

)(|)(|)( HEixyxy eZZ

For an electromagnetic wave traveling in free space Ex and Hy will be in phase with () = 0

http://www.ccrs.nrcan.gc.ca/glossary/index_e.php?id=3104 For a half space, the phase is given by:

42)1(tan)()(

1

01

iZ xy (26)

This phase angle will be observed at all frequencies over a uniform Earth (halfspace)



Summary of D1.1

You should be able to repeat this derivation from first principles, without using notes.

Note that the absolute value of the incident EM wave amplitude is not important.

This is because the impedance depends on the ratio of electric and magnetic fields. This makes MT easy to apply as the amplitude of the incoming wave does not need to be known.

The reflection and transmission is illustrated in the MATLAB movie

MT_physics_correct_movie.m. Here the incident wave is shown in blue and the reflected wave in the air is shown in red. In the Earth, the EM energy diffuses and is shown by the black curve.

An animated version of this figure can be found at

http://www.ualberta.ca/~unsworth/UA-classes/424/notes424-2012.html

This figure is subject to some artistic license. Can you identify two things that are not quite correct?



D1.2: Magnetotelluric response of a halfspace, at non-normal incidence Derivation for apparent resistivity and phase in notes We will prove that the values of apparent resistivity and phase derived from surface EM fields do not depend on the angle of incidence of the EM wave. This simplifies MT data analysis, since we do need to know the angle of incidence for a given EM wave.



D1.3: Magnetotelluric response of a 2-layer earth Qualitative solution for apparent resistivity of a layered Earth (1) f = 10 Hz = 500 m in upper layer

At high frequency, the skin depth is much less than the thickness of the layer. Average resistivity over a hemisphere, radius = is the resistivity of the upper layer.

(2) f = 1 Hz = 1580 m in upper layer

The EM signals are now just sampling the lower layer, Average resistivity begins to increase as the 100 m layer is being sampled.

(3) f = 0.1 Hz = 5000 m in upper layer

Hemisphere dominated by lower layer, so apparent resistivity close to 100 m

(4) f = 0.01 Hz = 15800 m in upper layer Apparent resistivity approaches 100 m asymptotically



Quantitative solution for apparent resistivity of a 2-layer Earth Assume a plane EM wave with angular frequency and amplitude A1 hits the ground at normal incidence. The electric field is polarized in the x-direction, and the magnetic field is polarized in the y-direction.

In air, the EM field travels as a wave with ok Down going wave tizikx eeAtzE o

1),( Up going wave tizikx eeAtzE o

2),(

In the layer the EM field diffuses with 2

)1( 11ik

Down going signal tizkx eeBtzE 11),(

Up going signal tizkx eeBtzE 12),(

In the halfspace the EM field diffuses with 2

)1( 22ik

Down going signal tizkx eCetzE 2),(

To compute the impedance at z = 0, we can specify the incident wave amplitude (A1) and need to compute A2, B1, B2 and C. Thus we should apply four boundary conditions.

(1) Horizontal electric field at z=0 (2) Horizontal magnetic field at z=0 (3) Horizontal electric field at z=h (4) Horizontal magnetic field at z=h

After a modest amount of algebra, you can show (see Assignment 1) that:

)])(())([()(2

2112

211

2112 1 kkkikekkkik

kkikABo

hko

o

and

)])(())([()(2

2112

211

2211

1 1

1

kkkikekkkikekkikAB

ohk

o

hko

Now we can calculate the impedance (Zxy) at the surface of the Earth (z=0)



We can write 21 BBEx and )(1

211 BB

ik

zE

iH xy

Thus )()(

21

21

1 BBBB

ki

HEZ

y

xxy

Now we need to substitute for the values of B1 and B2 that were previously defined. Note that they have the same denominator. After some algebra (also in Assignment!) can show that:

)]()[()]()[(

212

21

212

21

11

1

kkekkkkekk

ki

HEZ hk

hk

y

xxy

To check this expression, consider the high and low frequency limits. What answers would you expect? (1) High frequency limit of Zxy

As becomes large, k1 becomes large. Thus hke 12 becomes very large and

12

21

221

1 ])[(])[(

1

1

ki

ekkekk

ki

HEZ hk

hk

y

xxy

Comparing this with the result derived in D1.1, we see that this is the result for a halfspace of conductivity 1. Is this as expected? Why? The apparent resistivity can be calculated as:

111

21

2 11

k

Zxya

Thus the impedance (computed from field measurements of Ex and Hy) depends only on the properties of the upper layer.

(2) Low frequency limit of Zxy

When becomes very small, k1 becomes small. Thus 112 hke

22

1

12121

2121

1 ]2[]2[

)]()[()]()[(

ki

kk

ki

kkkkkkkk

ki

HEZ

y

xxy

Comparing this with the result derived in C4.1.1, we see that this is the result for a halfspace of conductivity 2. Is this as expected? Why?



Again, the apparent resistivity is given by:

222

22

2 11

k

Zxya

At low frequency the impedance (computed from field measurements of Ex and Hy) depends only on the properties of the lower layer.

(3) Zxy at intermediate frequencies

Zxy can be evaluated quite simply in MATLAB, as you will discover in Assignment 1. At what frequency would you expect the change from high to low frequency limit to occur? Describe the answer in terms of the physics of EM diffusion in the Earth.

Example Consider 1= 100 ohm-m, 2= 10 ohm-m and h =1000 m. Sketch the shape of the apparent resistivity curve that would be expected as a function of frequency (f).



We can also plot the phase, (Zxy), as a function of frequency (or period).

Previously we showed that at high frequency, 1k

iZxy

and 4

)( xyZ

At low frequency, 2k

iZxy

and again4

)( xyZ

What happens to the phase at intermediate frequencies? One way is to evaluate the full expression in MATLAB. An alternative is to use the fact that we can write an approximate expression for as:

)loglog1(

4 Ta

where T is the period of the EM signal (T=1/f)

In this example a decreases with T (since decreases with depth). Thus the derivative

term is negative and > 4 .

Similarly, if increases with depth, a increases with T. The derivative term will be

positive and < 4 .

It can be shown that, 1log

log1

T

a . Thus 0



D1.4 MT response of multiple layers

Apparent resistivity and phase can be computed with a similar approach to that used in D1.3 with waves traveling up and down in each layer.

Can derive a recursion relation that allows the impedance at one interface (Zn) to be derived from the impedance at the interface directly below (Zn+1). This is covered in Lab 3 in Geophysics 424.

General description of the calculations was first reported by Cagniard (1953). The

basics of the MT method was independently described by Tikhonov (1950), who also addressed the inverse problem.

Set of apparent resistivity curves for a 3-layer Earth model was described by Yungul (1961)



EXAMPLE 1 : Conductive/resistive layer and halfspace

At short periods (T < 0.01 s) the apparent resistivity equals that of the upper layer (100 m).

Between periods of T = 0.01 and T = 0.1 s the apparent resistivity increases slightly. This is a resonance phenomenon that occurs when the skin depth is approximately equal to the thickness of the layer. It is rarely seen in field MT data. Note that this is also evident as slight decrease in phase.

At period of T = 0.1 s the apparent resistivity starts to decrease rapidly as the conductive layer (10 m) is detected by the EM signals.

For the 2 layer model, the apparent resistivity then asymptotes at 10 m at long periods.

For the 3 layer model the apparent resistivity increases at T > 1 s as the electromagnetic signals enter the third layer (100 m ) below 2 km depth.

Can you confirm that decreasing apparent resistivity with increasing period corresponds to > 45 and vice versa?



This is identical to the previous example, except that the second layer is a resistor compared the first layer.

Resonance phenomena is observed again, but in opposite sense. Can you confirm that decreasing apparent resistivity with increasing period corresponds to

> 45 and vice versa?

Compare a second layer with resistivity contrast of 10 compared to the 1st and 3rd layers. Effect of second layer is observed at the same period in each case (T = 0.01 s) Apparent resistivity at long period is same in each case (T ~ 10000 s) Model with conductive second layer has greater effect on apparent resistivity. This is

because MT signals are strongly attenuated by the conductor and this makes a significant change at the surface, where the impedance is computed. In contrast, the EM signals travel through the resistive layer with minimal attenuation. Thus the resistive layer does not significantly change the surface impedance.



EXAMPLE 2 : Varying depth to a conductive layer

As depth to the layer increases, the period at which it is detected increases. This is as expected from the skin-depth equation. Can you verify that the graphs above are correct for the 1 km thick layer that that resistivity = 100 m

The magnitude of the response decreases as the layer becomes deeper. This is because the apparent resistivity represents an average resistivity from the surface to the maximum depth of exploration.

MT is good at determining the depth of a conductive layer. With typical quality field MT data (errors of 1-2 % in apparent resistivity), this can be determined within 10%



EXAMPLE 3 : Layer of constant conductance

Conductance of a layer is the product of conductivity and thickness Each model has a layer with conductance 100 Siemens. Thus to maintain constant

conductance, as the layer becomes thinner, the conductivity increases Once the layer become thin (compared to its depth), the MT curves for different

combinations of thickness and conductivity cannot be distinguished. This is an example of non-uniqueness. MT cannot separately determine the thickness and

resistivity of the layer. Only the conductance can be determined. Addition information, such as constraints from other geophysical methods, rock properties

or well logs must be used to overcome this non-uniqueness. Note that for the surface layer, resistivity and thickness can be individually determined.

The apparent resistivity at the highest frequency will be the true resistivity (provided that the highest frequency is high enough).



EXAMPLE 4 : Two conductive layers



Upper layer has a conductance of 100 S, while the lower layer has a variable conductance.

The lower layer can only be detected with MT when its conductance is greater than that of the upper layer.



Examples of this effect from field data

Mareschal et al., (1991) described an MT survey that was designed to image the Grenville Front in Ontario. This is a major crustal scale, suture zone of Proterozoic age that has unusual seismic reflections that were thought to be due to fluids. However a shallow layer of black shale (low resistivity) that was close to the surface in the Phanerozoic sedimentary section made it impossible to image any conductive features in the crystalline basement rocks.

Zhang and Pedersen (1991) showed that the presence of a sedimentary basin (low resistivity) could make it difficult to detect the lower continental crust (also low resistivity). See also Geophysics 424 Mid-term exam, Question 3, 2010 for a discussion of this effect.

D1.5 Non-uniqueness and the MT inverse problem

So far we have only considered the MT forward problem. This involves making a prediction of the MT data that would be observed over a given (1-D) resistivity model.

Remember that forward problems have a unique solution.

In general, geophysical data analysis requires that we also solve the inverse problem. This requires that a resistivity model is found that will fit a given MT dataset.

The inverse problem is non-unique with the non-uniqueness arising from two

distinct issues. (a) Inherent non-uniqueness.

This is due to the physics of the geophysical method. e.g. cannot separately determine the thickness and conductivity of a conductive layer. No amount of data or expensive software can overcome this type of non-uniqueness.

(b) Non-uniqueness due to noisy data

With large error bars in the data, a range of resistivity models can be derived that all fit the measured MT data. This type of non-uniqueness can be reduced through the collection of higher quality data with smaller error bars.

Parker and Whaler (1981) showed that the conductivity model that would give the best possible fit to an MT sounding curve would be a set of functions.



D1.6 Measuring the misfit between data and a model response To interpret MT data, we need a quantitative way to measure how well a model response fits a set of measured MT data.

Consider a simple MT curve with apparent resistivity data defined at three periods as (1, 2, 3) and shown by the blue circles. A resistivity model has been found that fits the data and the forward response has been computed as (m1 , m2 , m3) and is plotted as the red line. This figure was generated with the MATLAB function rms_defm.m Now consider the misfit function

233

222

211 )()()( mmm

This provides a measure of the misfit. Note that the squared terms are needed since the response (m) can be greater or less than the data point (d). This misfit measure does not allow for the fact that some data points will have larger uncertainties (error bars) than others. Clearly, we should pay more attention to the points with smaller error bars. If the three data points have uncertainties of (e1 e2 e3) then the misfit can be written in terms of the residual

)(

1

111 e

mr

which is a normalized misfit. In an ideal case the residual should be in the range -1 to +1, with the model response passing through the error bar.



In the example above the data point at T = 10 s has an unacceptable misfit. The other two data points (with larger error bars) have acceptable misfits. The misfit equation to be written in terms of residuals as:

2

3

332

2

222

1

11 )()()(e

me

me

m

It is useful to normalize this equation by the number of points, and take the square root so for a statistically acceptable fit (within 1 residual), the misfit is 1. This function is the root-mean-square misfit, usually abbreviated to the r.m.s. misfit , and can be written as:

2

3

332

2

222

1

11 )()()(31

em

em

em

rms

In the general case of N data points we can write

Ni

i i

irms e

mN 1

2i )(1

This can be evaluated with the following MATLAB function

function [misfit,res] = rms(d,err,m,n) misfit = 0.; for i = 1:n

res(i) = (d(i)-m(i))/err(i); misfit = misfit +(res(i))^2;

end misfit = sqrt(misfit/n);

end Notes

(1) The figure above illustrates the danger of judging the fit by a single number! Ideally the residuals should have a Gaussian distribution and not have any system variation. In the example above, the residuals show a systematic variation with period.

(2) There are other ways of estimating misfits Could add some figures generated with MATLAB script rms_plot_v1.m to demonstrate this and other aspects of data fitting.



D1.7 Examples of field MT data Example 1 : TBT180 This MT station is located in Southern Tibet and data was collected in 1995

Fit data with a trial-and-error approach using the MATLAB code MT1Dfit.m

Sketch the final model on the figure.

Before we start, think how many layers are needed.

How can the resistivity structure be interpreted in this tectonic environment?

Details in paper by Wei et al., (2001).

We will apply this code to some other MT data in Assignment 2

Fitting also presented in 2008 Cornell University MT short course

Note that the error bars are plotting as the dashed lines above, and below, the actual data points.



Example 2 : CXB02

Before starting to fit the data, decide what is the minimum number of layers needed to fit the data.

Fitting also presented in 2008 Cornell University MT short course



Example 3 : Historical MT data from Central Alberta

Figure 6.27 from Telford shows a number of MT apparent resistivity curves collected in Central Alberta. Instruments used at the University of Alberta in the 1970s had a limited bandwith (signal period recorded was in the range 10 to 1000 s). The increasing resistivity in this period range corresponds to the transition from the sedimentary rocks of the Western Canadian Sedimentary Basin (low resistivity) to the crystalline Pre-Cambrian basement rocks (high resistivity).



University of Alberta field crew at MT station FH026 near Rocky Mountain House in October 2002 (Wen Xiao and Wolfgang Soyer). Data was collected with the Phoenix V5-2000 system purchased by the University of Alberta.

An MT sounding made near Rocky Mountain House at station FH025 in 2002 shows the apparent resistivity curve at shorter period (higher frequency). This is quite close to Station 1 from Reddy and Rankin (1973) shown above. The flat section of the apparent resistivity curve (100 0.3 Hz) corresponds to the low resistivity sedimentary rocks of the WCSB. More details of the survey can be found in Xiao (2004) and Xiao and Unsworth (2006). The dip in resistivity between 0.3 and 0.1 Hz is likely due to low resistivity in the Cretaceous strata in the deeper part of the basin, where saline aquifers occur. It is probably not a resonance phenomena (D1.4). Compare Station 1 and FH025. They both pass through the point 100 m, T = 100 s. Considering the 30 years between the two measurements, and differences in technology, this is quite impressive. Note : Interpretation of MT data from Station FH025 has featured on several mid-term and final exams in Geophysics 424 in recent years .



D1.8 MT impedance tensor for a 1-D Earth

MT station 1 y

xxy H

EZ MT station 2 x

yyx H

EZ

We can define the magnetotelluric impedance tensor, Z, as:

y

x

yyyx

xyxx

y

x

HH

ZZZZ

EE

We can visualize this as the magnetic field (input) being applied to the Earth and producing an electric field (output) through electromagnetic induction. In a 1-D case, no electric field is induced parallel to the inducing field, and thus

0 xxxx HZE and 0 yyyy HZE which requires that Zxx=0 and Zyy=0

y

x

yx

xy

y

x

HH

ZZ

EE

00

Over a 1-D Earth, we will measure the same impedance, regardless of the orientation of the x and y axes. Thus in the example above we can write that

ZZZ yxxy with the apparent resistivity defined as

21 Za

In this case the impedance tensor simplifies further to:

y

x

y

x

HH

ZZ

EE

00



D1.9 References Cagniard, L., Basic theory of the magnetotelluric method of geophysical prospecting,

Geophysics, 18, 605-635, 1953. Mareschal M, RD Kurtz, M Chouteau, R Chakridi, A magnetotelluric survey on

Mantoulin Island and Bruce Peninsula along GLIMPCE seismic line J : black shales mask the Grenville Front, Geophysical Journal International, 105, 173-183, 1991.

Parker RL, KA Whaler, Numerical Methods for Establishing Solutions to the Inverse Problem of Electromagnetic Induction, Journal of Geophysical Research, 86, 9574-9584, 1981.

Reddy IK and D Rankin, Magnetotelluric measurements in Central Alberta, Geophysics, 36, 739-753, 1971.

Tikhonov, A. N., Determination of the electrical characteristics of the deep strata of the Earths crust, Dok. Akad. Nauk., USSR, 73:2, 295-297, 1950.

Wei W, MJ Unsworth, AG Jones, JR Booker, H Tan, KD Nelson, L Chen, S Li, K Solon, PA Bedrosian, S Jin, M Deng, J Ledo, D Kay, B Roberts, Detection of widespread fluids in the Tibetan crust by magnetotelluric studies, Science, 292, 716-718, 2001

Xiao W, Magnetotelluric exploration in the Rocky Mountain Foothills, MSc thesis, Department of Physics, University of Alberta, 2004.

Xiao W and MJ Unsworth, Structural imaging in the Rocky Mountain Foothills (Alberta) using magnetotelluric exploration, AAPG Bulletin, 90, 321-333, 2006.

Yungul, S.H., Magnetotelluric sounding three-layer interpretation curves, Geophysics, 26, 465-473, 1961.

Zhang P and Pedersen LB, Can MT data resolve lower crustal conductors? Physics of the Earth and Planetary Interiors, 65, 248, 1991.

424d1-fall2014.pdf

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