1

5.3.3 The general solution for plane waves incident on a layered half-

space

The general solution to the Helmholz equation in rectangular

coordinates

The vector propagation constant

Vector relationships between vectors k, E, and H

Properties of uniform plane waves, characteristic impedance

Reflection and transmission of plane waves at a plane interface

TE mode reflection and transmission at a plane interface

TM mode reflection and transmission at a plane interface

Reflection and transmission of plane waves incident at real angles of

incidence

Surface impedance formulation of reflection and transmission

The general impedance of an n-layered half-space

The general solution to the Helmholz equation in rectangular coordinates

In section 5.2 we derived the Helmholtz equation in either magnetic or

electric fields. In the simple description of the MT method in section 5.3.2

we looked at a simple solution for the Helmholtz equation in H when the

horizontal derivatives were zero and there was only one horizontal

component of the field Hy. Here we will look at the general solution to the

vector Helmholz equation where there are no such simplifying assumptions.

The behavior of magnetic and electric fields at the surface of a layered half-

2

space of arbitrary conductivity and dielectric in the layers as a function of

frequency and the angle of incidence of the field on the surface. This general

treatment will consider the orientation of the electric and magnetic fields

with respect to the interface and the impedance and reflection coefficient for

such fields. We will use a right handed rectangular coordinate system with z

positive down.

For any rectangular component of the field, say Hx, the Helmholz equation

is:

2 2 22

2 2 20

x x xx

H H Hk H

x y z

where we have used an i te time dependence. [from here on the

dependence is implied and will not be denoted explicitly].

A classic method of solution is separation of variables: assume that the

solution can be written as the product of three independent solutions in each

coordinate viz:

xH X x Y y Z z

Substituting this solution in the Helmholz equation yields three

independent ordinary equations:

22

2

22

2

22

2

1

1

1

x

y

z

d Xk

X dx

d Xk

Y dy

d Xk

Z dz

3

and the condition that

2 2 2 2 0 x y zk k k k

implies that there are only two independent equations since

2

2 2 2 2

2

1

z x y

Xk k k k

Z z

Particular solutions to these ordinary equations are:

2 2 2

, , x yyxi k k k zik yik x

X Ae Y Be and Z Ce

so the general solution is:

2 2 2

', , ,

x yyx

i k k k zik yik xx x y x yH x y z H k k e e e k k

This is a Fourier transform and in fact we get to this solution more elegantly

by simply starting all over and taking the Fourier transform of the Helmholz

equation.

Using the following Fourier transform pair:

'

'

1

2

x

x

ik xx x

ik x

H x H k e dk

H x H x e dx

and applying the first directly to the Helmholz equation, in other words

taking the spatial transform, yields an algebraic equation in transform space:

2 ' 2 ' 2 ' 2, , , , , , 0 x x x y z y x x y z z x x y zk H k k k k H k k k k H k k k k

or 2 2 2 2 0 x y zk k k k .

The ki are not independent so the inverse transform back to x,y,z space

becomes:

4

2 2 2

'1, , ,

4

x yyx

i k k k zik yik xx x y x yH x y z H k k e e e k k

Before going on to apply the general solution to particular problems, we will

examine the particular solutions and find out the physical significance of the

ki parameters.

The Vector propagation constant

First, to simplify the ensuing algebra, we will consider a two

dimensional model of the Earth-air interface and a field above the interface

that has no variation in y, so yk is zero. [Over a uniform half-space the

coordinates can always be rotated such that y axis coincides with the

direction in which the field is invariant]. Consider the particular solution for

a component of the electric field in the y direction:

0 0

x zx z

i k x k zik x ik zy y yE E e e E e

The bracketed term in the exponent has the form of a vector dot product of a

position vector r ix z and a vector propagation constant

x zk ik k where ,i j and are unit vectors in the x, y and z axes

respectively; this solution has the form

0 ik ry yE E e

The following sketch illustrates these vectors in two dimensions:

5

Figure 5.3.3.1

Here we can identify and x zk k as the rectangular components of a vector k

making an angle with the vertical, z, axis. In this illustration and x zk k can

be interpreted as the rectangular components of k ,

sin and cos x zk k k k implying values limited to k. But be careful

because in the general solution or x zk k can have values far greater than k

suggesting meaning for that is not intuitive. We return to these solutions

later.

A small digression at this point will help explain why, in later sections,

we will study things like reflection coefficient and surface impedance in

terms of xk . If we consider the full solution for a field, xiwt ik xe , and consider

where a field of constant phase moves in space and time we have:

a constant xt k x . And so x

x

dxV

dt k

, the phase velocity is in the x

direction. From the relation that wavelength, ,V

f , we find that such a

6

wave has a spatial wavelength of 2

x

x xk f k

. So a wave propagating in

the x direction has an implicit horizontal wavelength. The general solution,

which is an integral over all xk is made up of the superposition of waves of

different horizontal wavelength. We will see how the reflection and

refraction, and surface impedance, of such a wave depends strongly on x .

Vector relationships between , and Hk E

We can now show the vector relationships between the fields and the

vector propagation constant by taking the Fourier transform of the individual

Maxwell equations. For example the E equation can be written out in

full as:

yzx

x zy

y xz

EEi H

y z

E Ei H

z x

E Ei H

x y

Taking the Fourier transform of each of these yields:

, , , , , ,

, , , , , ,

, , , , , ,

y z x y z z y x y z x x y z

z x x y z x z x y z y x y z

x y x y z y x x y z z x y z

ik E k k k ik E k k k i H k k k

ik E k k k ik E k k k i H k k k

ik E k k k ik E k k k i H k k k

7

Since , and x y zk k k and kz are now known as components of a vector k this

set of equations can be written in a much more compact form:

k E H

and the same process yields the second curl equation

k H i E

Finally the divergence equations 0 and 0 E H and Ñ×H = 0

simply transform to:

0

0

k E

k H

So for this propagating plane wave we can see that E and H must always

be perpendicular to one another and that both are perpendicular to the vector

direction of the propagation constant. These simple rules show that there can

never be component of the field in the direction of propagation of a uniform

plane wave field.

Properties of uniform plane waves

To simplify the algebra involved let’s consider the nature of a wave

propagating in the z direction. Since solutions for each component are

simply additive we can illustrate properties by choosing any component.

First lets consider the solution for Ey i.e. a solution to

2

2

20

y

y

Ek E

z

8

is 0 ikz i t

y yE E e and from Faraday’s law

y

x y

Ei H ikE

z so the

ratio of the electric to magnetic field for a plane wave propagating in the

positive z direction is the characteristic or intrinsic impedance, denoted here

by and is equal to:

y

y

E

x

E

H k

Note a couple of things about the characteristic impedance. We could have

started this process from the Helmholz equation in xE viz.

22

20

xx

Ek E

z

The solution is for a wave travelling in the +z direction is 0 ikz i t

x xE E e

and now from Faraday’s law

xy x

Ei H ikE

z , so

x

xE

y

E

H k

.

The component chosen affects the sign of the intrinsic impedance.

If we had chosen a wave travelling in the negative z direction the signs of

the z derivative would have reversed and so

y yE E and x xE E

Finally some sharp eyed reader may have tried to get to these relationships

by starting with the Helmholz equation in, say, xH viz.

22

20

xx

Hk H

z

The solution is 0 ikz i t

x xH H e and now from Faraday’s law

9

xy x

Hi E ikH

z

And so y

y

E

x

E ik

H i

which doesn’t look much like

k

.

But wait, with a little multiplying top and bottom and factoring watch what

happens:

2 2

22

i i ik

k i iii

Either definition of the intrinsic impedance is ok.

Now let’s look at the propagation characteristics for various values of

, , and .

In general k is complex, that is 2Re Im k k ik i .

Squaring both sides and equating real and imaginary terms yields the

following relations:

2 2 2Re Im Re Im and 2k k k k .

After a tedious bit of algebra we solve for Re Im and k k :

After a tedious bit of algebra we find the following equations for kR and kI :

12 2

1 1

2

Rek

10

and

12 2

Im

1 1

2

k

There are three limiting cases to be considered.

1) If 0 (free space) then Im 0k and Re k

2) If then Re Im2

k k

3) If , then is small and

,

12 22 1

1 12

[based on 1

21 12

when 1 ]

In this high frequency limit Re k , the same value as in a non-

conductive medium and Im2

k

.

The skin depth is therefore 2

2 .

This last result is very important because it means that in a conductive

medium at very high frequencies the skin depth becomes independent of

frequency. The skin depth may be very small but it depends only on the

11

conductivity, dielectric permittivity and magnetic permeability. The phase

velocity on the other hand is independent of the conductivity. We will return

to these results in section 5.11 on Radar.

There are a few details left to discuss about propagation in free space. Again

we will consider an Ey field propagating in the positive z direction the form

for which is

0 ikz i t

y yE E e

Now let’s consider this solution for fields in free space where 0 and so k

is real and equal to 0 0 . The complete solution is:

0 0

0

i z t

y yE E e

This is a true propagating wave. A point of constant phase is defined by

0 0 t z = constant

and 0 0

1 ph

dzV

dt the phase velocity ( in + z).

The constants 7 120 04 10 and 8.85 10 yield a phase velocity of

82.998 10 which is the velocity of light in free space. This remarkable

derivation is due to Maxwell who predicted it from first principles far ahead

of its accurate experimental determination.

The characteristic impedance of free space is:

0 0

0

377 Ohms. Eyk

12

For completeness the basic solution we derived in section 5.3.2 for an H y

component ‘propagating’ in a good conductor where is repeated

here.

2

2

20

y

y

Hk H

z

the solution for which is (including the time dependence):

0 i t ikz

y yH H e

The sign must be chosen so that the fields do not grow exponentially with z.

In this case we choose e-ikz so that on expanding k we get the solution:

220

i t zz

y yH H e e

And we defined the skin depth, , as

7

2 2500

2 4 10

ff

We now have all the basic definitions, formulations and propagation

characteristics to address what happens when a propagating field in one

medium is incident on another medium. Our major interest in in plane waves

incident on a layered earth.

The schematic diagram below shows an incident field approaching a

horizontal interface between medium 1 and medium 2 with a propagation

vector. The position vector lies in the interface and the unit vector is normal

to the interface and is by our coordinate convention positive down. We will

assume from experimental results that there is energy reflected from the

13

interface and some energy may be transmitted across the interface into the

medium 2.

Figure 5.3.3.2

We have drawn this diagram in the conventional way used in most texts to

show the incident field as a ray approaching the interface at an angle 1 .

Remember though that 1k is a vector composed of an x component and a z

component where xk and zk are can have values far higher than the

geometric interpretation 1 1cosk or 1 1sink . As drawn the diagram

represents a field incident at what is termed a real angle of incidence. In the

derivations that follow we will consider the general case of arbitrary xk and

zk and highlight the case of real angles of incidence as we go. We define the

plane of incidence as the plane containing 1k and the unit normal to the

surface n .

Before going on to derive the amplitude relations of the reflected and

transmitted fields we can derive some important properties of the fields at an

interface from their vector representation. We have prescribed and incident,

14

reflected and transmitted field, say vector electric field, by

'1 1 2'

1 1 2, and ik r ik r ik rE e E e E e respectively. Exactly the same form applies to

the vector magnetic field.

The boundary conditions are that the tangential components of the field must

be continuous across the interface. Thus, on the interface the sum of the

tangential incident and reflected fields must equal the tangential transmitted

field. First consider the incident field to have only a y component, i.e. only a

tangential component. Then the boundary condition yields:

'1 1 2'

1 1 2 ik r ik r ik r

y y yE e E e E e

This condition must hold for any position, r , in the interface and this can

only be assured if ; '

1 1 2 k r k r k r .

Now suppose that r locates an arbitrary point on the interface, it is on the

horizontal xy plane in the above sketch. Now choose another vector t also

lying in the interface so that r can be written as r t n . Substituting this

form for r in the above equation and noting the vector identity

k t n t n k we arrive at:

'1 1 2 n k n k n k

This is the general form of Snell’s Law.

An immediate consequence of this relationship is that all the propagation

vectors must lie in the same plane, the plane of incidence. Also, in the plane

of the interface, z is zero so

15

x z xk r k x k z k x

so again if '1 1 2 k r k r k r and this holds for any point in the interface,

any x, then we have the important result that

1 2x xk k .

To put this general form of Snell’s Law in its familiar form we return to the

representation of the incident fields in the above sketch where the

propagation vectors of the incident reflected and transmitted fields lie at

angles '

1 1 2, and with respect to the normal to the interface n . Because

'1 1 2 n k n k n k can now be written

' '1 1 1 1 2 2sin sin sin k k k

Since '

1 1k k then '

1 1 so the angle of reflection is equal to the angle of

incidence.

Further in non conducting media k and the phase velocity ,V, is

k

so Snell’s Law can be written in the familiar form:

1 2

1 2

sin sin

V V

Another important result for plane waves in free space incident at a real

angle to a conducting ground can be found from

1 11 1 2 2 2

2

sinsin sin or sin

kk k

k

If k2 >> k1 then 2 2sin 0 and 90 no matter what the incident real

angle. The refracted wave propagates vertically even for incident fields

propagating horizontally. This condition applies for a range of k1x beyond k1.

16

For the wave in medium 2 to propagate vertically 2 2x zk k and since

1 2x xk k then in terms of horizontal wavelength in the incident medium

1

2

2x

k

. In a practical case where the ground has a resistivity of 100 Ohm-

m and at 1,000 Hz we find that the horizontal wavelength must only be

greater than ~700 m for the wave to be refracted vertically. This implies that

if the incident field is actually produced by a finite nearby source then it is

simply necessary to ensure that the bulk of the integral for the fields comes

from horizontal wave lengths greater than 700 m. This is a fundamental

statement about using controlled sources to simulate MT results in systems

like CSAMT and Stratagem.

Reflection and refraction of plane waves at a plane boundary

We now have some basic properties of incident, reflected and refracted

fields and we will now derive the amplitudes of the reflected and refracted

fields, and in the process determine the surface impedance of a layered

media.

Any incident plane wave field has the electric and magnetic field vectors

orthogonal in the plane that is perpendicular to the direction of propagation.

The field vectors can always be broken into two component one of which is

perpendicular to the plane of incidence and consequently parallel or

tangential to the interface. Since the boundary conditions are on the

tangential components it is convenient to set up the reflection refraction

17

problem in terms of electric or magnetic field components that are

perpendicular to the plane of incidence. This field decomposition has two

important definitions:

(1) An incident wave with electric field normal to the plane of incidence is

called the TE mode. The incident wave has an electric field transverse to the

plane of incidence .

(2) An incident wave with magnetic field normal to the plane of incidence is

called the TM mode. The incident wave has an magnetic field transverse to

the plane of incidence .

We will see that the amount of incident field that is reflected or transmitted

depends on the angle of incidence and on the mode.

TE mode reflection and refraction at a plane boundary

We will assume that the coordinate system can always be rotated such that

the xz plane is the plane of incidence. The TE mode then has only a y

component of E and the TM mode has only a y component of H. [Note that

an incident field with E and H in arbitrary directions can always be broken

down into a field with an E component and an H component perpendicular

to the plane of incidence, solved for each mode separately, and combined for

the total field in the reflected and transmitted fields.]

We will set up the problem for the y component of E, Ey,

The incident, reflected and transmitted fields are

'1 1 2'

1 1 2, and ik r ik r ik ry y yE e E e E e

The tangential E continuous boundary condition simply yields:

18

'1 1 2 y y yE E E

From

i

BE k E i H

t and that the tangential component of

H n

we find that (dropping the y subscript):

' ' '1 1 1 1 2 2

1 2

k E n k E n k E n

Using the vector identity A B C B C A A C B and noting that

'1 1 2 0 n E n E n E because E is perpendicular to the plane of incidence

and therefore perpendicular to n , this boundary condition reduces to:

' '1 1 1 1 2 2

1 2

E n k E n k E n k

Using both boundary condition equations and solving for '1 2 and E E (and

noting that '1 in k n k ), we find that:

'2 1 1 2'

1 1 '2 1 1 2

n k n kE E

n k n k

and

1 2

2 1

1 2 2 1

2

n kE E

n k n k

These are the Fresnel equations in vector form for the reflected and

transmitted E field amplitudes for the TE Mode. The transmitted field is

19

called the refracted field in studies of light. The ratios of '1 2

1 1

and E E

E E are the

reflection coefficient and transmission coefficient and t respectively.

Noting that zn k k and that 2 2 2 x zk k k and 1 2x xk k we can expand the

above Fresnel equations into:

2 2 2 2' 2 1 1 1 2 12 1 1 21 1 1

2 2 2 22 1 1 2 2 1 1 1 2 1

x xz z

z z x x

k k k kk kE E E

k k k k k k

and similarly

2 22 1 1

2 12 2 2 2

2 1 1 1 2 1

2

x

x x

k kE E

k k k k

TM mode reflection and refraction at a plane boundary

Following the same derivation used above for the TE mode we can obtain

the Fresnel equations for the TM mode: now H is perpendicular the plane of

incidence and E is in the plane of incidence.

2 2 2 2' 1 2 2 1 2 1 2 11 1

2 2 2 21 2 2 1 2 1 2 1

x x

x x

k k k k k kH H

k k k k k k

and

2 21 2 2 1

2 12 2 2 2

1 2 2 1 2 1 2 1

2

x

x x

k k kH H

k k k k k k

20

The amplitude reflection coefficients are defined for each mode as

'1

1

y

TE

y

Er

E and

'1

1

y

TM

y

Hr

H.

It is quite clear that these reflection coefficients are different for each mode,

and that the reflected (and for that matter the transmitted) fields can have a

different phase than the incident field. This can be simply illustrated with the

limiting case of normally incident fields, i.e. when 1 2 0 x xk k . For added

simplicity let 1 2 .

Then

1 2

1 2

TE

k kr

k k and 2 1

1 2

TM

k kr

k k

So in this simple situation we see that the sign of the coefficient is different

for each mode. This has particular significance when the incident field is in

free space and 1 k and the field is incident on a conducting half-

space for which 2 k i . In this situation, which is typical for most

earth conductivities and frequencies less than a megahertz, 2 1k k and so

'1

1

1 y

y

H

H. The reflected field is reflected in phase and is essentially the

same amplitude as the incident field. This means that the total magnetic field

on the ground surface is very close to twice the amplitude of the incident

field and is independent of the ground conductivity.

On the other hand

'1

1

1 y

y

E

E so the electric field is reduced almost to zero.

Just how small the net surface electric field is can be found from the

transmission coefficient:

21

2 22 1 12

2 2 2 21 2 1 1 1 2 1

2

x

x x

k kE

E k k k k

For simplicity suppose 2 1 and assume normal incidence, i.e. 1 0xk ,

then

2 1

1 1 2

2

E k

E k k

and if 2 1k k then

2 4

1

22

iEe

E i

The net surface field has a phase shift of 45 degrees and for typical ground

conductivity of 0.01 S/m and for an angular frequency of 1,000 (f = 169 Hz)

the net field is reduced by 1.88 *10-3

.

Neither result should be surprising since these are exactly the relationships

that hold in the limit at the boundary of a perfect conductor. What is

important to note about this result is that the small residual electric field on

the boundary is very sensitive to conductivity. In all geophysical

measurements employing incident plane wave fields it is the electric field

that has information about the conductivity of the ground.

The graphical window associated with this section has an option to calculate

the TE and TM reflection coefficients for arbitrary kx for any values of

.

22

Reflection and refraction of plane waves at incident on the ground at real

angles of incidence

When the plane wave is incident at a real angle of incidence the Fresnel

equations can all be rewritten in terms of sines and cosines of the angle of

incidence 1 . For the TE mode the reflection and refraction coefficients

become:

TEr

2 2 2' 2 1 1 1 2 1 11

2 2 21

2 1 1 1 2 1 1

cos sin

cos sin

k k kE

E k k k

2 2 1 1

2 2 21

2 1 1 1 2 1 1

2 cos

cos sin

TE

E kt

E k k k

and for the TM mode they are:

2 2 2 2' 1 2 1 2 1 2 1 11

2 2 2 21

1 2 1 2 1 2 1 1

cos sin

cos sin

TM

k k k kH

rH k k k k

and

2

2 1 2 1

2 2 2 21

1 2 1 2 1 2 1 1

2 cos

cos sin

TM

H kt

H k k k k

The graphical window associated with this section has an option to calculate

the TE and TM reflection coefficients for arbitrary 1 for any values of

or.

23

Surface impedance of a layered ground

The practical problem with plane waves from distant sources is that we have

no information about the incident electric and magnetic fields themselves-we

only have the total fields on the interface. We have seen that the observed

magnetic field is twice the incident field and that it is the electric field, while

much reduced, that is sensitive to the ground conductivity. It would seem

that measuring the electric field would provide the information needed. But

the incident fields are essentially random in amplitude so the measured

amplitudes vary widely in time and cannot be related directly to ground

conductivity. The solution is to reference the electric field to the magnetic

field by measuring the ratio of orthogonal tangential E and H fields on the

ground surface. We defined this ratio earlier as the surface impedance and

we will briefly re-derive the reflection and transmission coefficients in terms

of the characteristic and surface impedances, and then derive expressions for

the surface impedance of a layered medium. We will adopt a revised

notation to be compatible with the n-layered programs in the graphical

windows of this section.

First consider again a TE mode solution for an arbitrary medium i:

iz ixik z ik xyi yiE E e e

for a wave travelling in the positive z and x direction.

For a wave travelling in the negative z direction a solution is:

iz ixik z ik xyi yiE E e e

The x component of the accompanying magnetic field is obtained from

E i H so

24

1

yi izxi

i i

E kH

i z

The ratio ,

yi

xi

E

H is the characteristic impedance,

iTE = i

izk

. For a wave

travelling in the negative z direction iTE iTE .

For the TM mode we have the solution in Hy;

iz ixik z ik xyi yiH H e e

and from iH i E we derive the characteristic impedance for

the TM mode to be;

iziTM

i

ik

i

.

We can now set up the interface problem in terms of impedance. Let the free

space above the interface be denoted now by the index 0 and the ground by

index 1. At z=0, our solutions for E and H have to satisfy the boundary

condition that each is continuous at the interface. This leaves the two

equations:

0 0 1

0 0 1 1

10 0 1

y y y

y y y y

TETE TE TE

E E E

E E E E

Z

We have introduced an important concept in recognizing that on the

interface the total E and H fields define the surface impedance. For this

simple model of two media the surface impedance is the characteristic

impedance of the second medium. In the multilayer model to follow the

boundary conditions are met with the surface impedance of a layered model

25

so 1TEZ will be replaced with the general expression for the impedance of a

layered model.

Combining these two equations we arrive at a ‘new’ expression for the

reflection coefficient:

0 1 0

0 1 1

y TE TE

y TE TE

E Z

E Z

Similarly the TM reflection coefficient becomes:

0 0 1

0 0 1

y TM TM

y TM TM

H Z

H Z

These are the Fresnel equations rewritten in terms of the characteristic and

surface impedances. The program behind the graphic window uses these

expressions with either a general kx or a real angle of incidence and uses the

general n-layer impedance formula of the next section of Z. Note, as we saw

before in the general derivation of the reflection and transmission

coefficients, these equations are functions of the spatial wave number kz. as

are the characteristic impedances themselves and the surface impedance.

To bring us back to the practical matters of magnetotellurics, it is the surface

impedance that we actually compute from measurements of electric and

magnetic fields on the ground surface. Then, as we saw in the simple

introduction to MT, we assume for purposes of representing the data that the

computed impedance at a given frequency is for a simple half-space model.

This allows us to calculate an apparent resistivity via

21A Z

26

The practical impact of a finite horizontal wave number, 0xk on the TE and

TM surface impedances and hence on the apparent resistivities can be

appreciated in the following little analysis. We will assume that the earth is a

good conductor so .

The TM surface impedance of a half-space, medium 1, is:

2 2 21 11 1 1

21 1 1 1

1

xz xTM

i k kik ik kZ

k

but since 1 x oxk k and 1

1

ik

is the surface impedance for normal incidence we

have the interesting result that as 0xk increases beyond k1 the ‘normal’

impedance, and the apparent resistivity calculated from it, goes up.

For incident waves in the TE mode

11

2 2 21 1 0 0

21

1

TE

z x x

kZ

k k k k

k

So now as the horizontal wave number goes up the impedance and the

apparent resistivity goes down.

We have spent a lot of effort on the topic of arbitrary horizontal wave

number because the solutions for the fields from finite sources are integrals

over such wavenumbers and the behavior of the solutions can often be

deduced from the behavior at a single wavenumber that is representative of

the main contribution from the wavenumber spectra in the integral for the

27

full solution. This concept is dealt with fully in section 5.4 where, for

example, we derive the solution for a line source in the y direction.

These fields are TE and as we get closer to the source the solution requires

shorter and shorter horizontal wavelengths to represent the large spatial

variations in the field. In that case apparent resistivities calculated in this

zone will be lower than those expected from a normally incident plane wave.

Other source would cause the apparent resistivities to go up. This issue is

important in assessing the actual sources for magnetotelluric fields and

whether apparent resistivities calculated from field data really are source

independent.

The general impedance of an n-layered half-space

We now derive the general form for the impedance of a layered half-

space. We will use the notation shown in Figure 5.3.3.3 below.

The solution for the TM mode in any layer is:

iz iz ix

iz iz ix

ik z ik z ik xyi yi yi

ik z ik z ik xxi iTM yi yi

H H e H e e

E H e H e e

From the general form of Snell’s Law 0 0 1 ...... x x x ixk k k k so in all ratios

of E to H these terms cancel and so will be dropped in the following

derivation.

28

Figure 5.3.3.3

At the surface of layer i, iz z and the above equations can be written in

matrix form viz:

iz i iz i

iz i iz i

i

ik z ik zyi yiiTM iTMxi

ik z ik zyi yi yiz z

H He eEA

H H He e

and at the bottom of the layer, 1 iz z

1 1

1 1

1

iz i iz i

iz i iz i

i

ik z ik zyi yiiTM iTMxi

ik z ik zyi yi yiz z

H He eEB

H H He e

These two equations can be combined to yield an equation for the fields at

iz in terms of the fields at 1iz both within layer i viz:

1

1

i i

xi xi

yi yiz z z z

E EA B

H H

29

After a horrible algebraic exercise of multiplying this out and noting that

1 i i iz z h we obtain:

1

2

1

2

iz i iz i iz i iz i

iz i iz i iz i iz i

i i

ik h ik h ik h ik hiTM iTMix ix

ik h ik h ik h ik hiy iyiTMz z z ziTM

e e e eE E

H He e e e

Now multiply this out into two equations and divide one by the other noting

that

1 1

1

1

i i

xi xi

yi yiz z z z

E E

H H

because E and H are continuous at the 1iz interface:

1 1

1 1

21 1

1 1

iz i iz i iz i iz i

ii i

i iz i iz i iz i iz i

ii i

ik h ik h ik h ik hiTM i x iTM i yix zz z

iTM z z ik h ik h ik h ik hiy i x iTM i yzz z

e e E e e HEZ

H e e E e e H

Finally if we divide top and bottom of the right hand side by

1

1

iz i iz i

i

ik h ik hi y z

e e H we get the disarmingly simple form for the

impedance at the top of layer i in terms of the impedance at the top of the

layer, i+1,below:

1

1 1

tanh( )

tanh( )

i TM iTM iz iiTM iTM

i TM i TM iz i

Z ik hZ

Z ik h

To find the surface impedance of an n-layer medium start by calculating the

surface impedance at the top of the layer that is on the basement, at depth

1nz , and use the characteristic impedance of the basement nTM for nTMZ .

Then by a recurring operation calculate the surface impedance at the top of

each successive layer all the way to the ground surface.

30

The TE mode impedance is derived the same way. We now start with an

incident xH field:

iz iz ixik z ik z ik xxi xi xiH H e H e e

and the accompanying yE field is now written via the TE characteristic

impedance:

iz iz ixik z ik z ik xyi iTE xi xiE H e H e e

The same procedure used above leads to the same layer to layer impedance

formula except with the TE characteristic impedances and surface

impedances.

1

1 1

tanh( )

tanh( )

i TE iTE iz iiTE iTE

i TE i TE iz i

Z ik hZ

Z ik h

These expressions are used recursively to find the surface impedance for an

arbitrary horizontal wave number 0xk .[ 2 2 2 2 2

0 0 1( sin iz i x ik k k k k

for a finite angle of incidence)].

The graphical window shows plots of the apparent resistivity derived from

this impedance, and the phase of the impedance, for multilayered models.

The resistivity 1, the permeability i and the susceptibility i for each layer

are entered as input as are the layer thicknesses and desired frequency range.

The default settings assume that the permeability and susceptibility are free

space values. Finally the impedances are calculated for either specified

horizontal wavelength ( = 2/k0x) or real angle of incidence 0. The default

is normal incidence, 0 = 0. For any model the mode, TE or TM, must be

specified on opening the window.