Radio Propagation and Remote Sensing of the Environment - Chapter 8 pdf

The fluctuation field itself can be described as the field radiated by random currents with density jr.. The mean value of this density is equal to zero, and the spatial correlation func

Trang 1

Radiation

8.1 EXTENDED KIRCHHOFF’S LAW

The background of the thermal radiation theory of heated bodies will be discussed

in this chapter This radiation appears as the result of random motion of charged particles inside the body The velocities of this movement are stochastic and, in particular, they change their value and direction occasionally as a result of the interaction (collisions) of particles with each other The radiated field strength is random, and its intensity depends on the particle energy and, consequently, on the temperature of the body In this connection, the radiation under discussion is referred

to as thermal

We have to imagine that these bodies, and the fluctuation field generated by them, are in a giant thermostat that maintains the thermodynamic balance This means that the charged particles of the body interact with the given fluctuation field, derive energy from it, reradiate it afresh, and then absorb, reradiate, and so on In

a word, the radiating and absorbing energies are balanced in an equilibrium state for the fluctuation field The fluctuation field itself can be described as the field radiated by random currents with density j(r) The mean value of this density is equal to zero, and the spatial correlation function of its frequency spectrum is defined

on the basis of the fluctuation–dissipation theorem (FDT):24,56

(8.1)

represent corresponding coordinate components of the current vector The FDT, described in such a way, is correct over practically the entire electromagnetic spec-trum (at least, for wavelengths that exceed interatomic or intermolecular distances)

true) Therefore, we will use this approach when the averaged energy of quantum oscillator:

(8.2)

b

′





 ′′

2

8 2k T ( )ωω δ(r′ − ′′r )δαβ

=h 2π=1 05 10 ⋅ − 27erg⋅sec

ω << k Tb

Θ( , )ω T ω ω

k T







TF1710_book.fm Page 221 Thursday, September 30, 2004 1:43 PM

Trang 2

222 Radio Propagation and Remote Sensing of the Environment

(8.3)

is valid in the radiofrequency band It is characteristic that orthogonal components

of the fluctuation current are not correlated For the current components themselves, the spatial correlation radius in this case is equal to zero In fact, it has the scale of particles interaction — for example, interatomic distances in a solid body or free path length in a gas Because the wavelengths considered here exceed these distances,

it is possible to neglect their variation from zero The most important fact is that the spectral density of the fluctuation current is defined by the imaginary part of the body permittivity (i.e., its ability to absorb electromagnetic waves) Kirchhoff’s law applies here implicitly, as it connects radiating and absorbing properties of the body Let us point out in this connection, that we are dealing with nonmagnetic materials,

so the magnetic losses default, and we do not need to represent the magnetic fluctuation currents

In order to calculate the fields generated by fluctuation currents, we need to know Green’s function of the considered body — that is, the diffraction field excited inside the body by a single current source:

(8.4)

where e is a single vector, generally speaking, of arbitrary direction The field , excited by this current, is the diffraction field To determine the fluctuation field, we can use the mutuality theorem in the form of Equation (1.64) The

(Equation (8.4)) and the diffraction field generated by it Omitting unnecessary subscripts, we now have the following equation for the fluctuation field:

(8.5)

Also, we have the expression for the calculation of the fluctuation field component, oriented in the direction of vector e Its average value is equal to zero, as the average value of the fluctuation current is also equal to zero In this connection, let us point out that the diffraction field is the determining value Let us also emphasize another very important fact The imaginary part of the dielectric permittivity in Equations (8.1) and (8.3) can be a function of the coordinates Particularly, it can be equal to zero if, for example, part of the considered volume V occupies a vacuum So, the volume can include as the actual heated body, which serves as a fluctuation field source, any part of space up to infinity It is important that point r of the dipole, existing in the diffraction field, is situated inside the volume, but it can be outside

Θ = k Tb ,

( )r ( )r′′ = ′′( ) (r′ − ′′r )

jG( )r′ =eδ(r− ′r),

E Hd, d

j E1, 1 j E2, 2

e E r⋅ ( )=∫j r( )′ ⋅Ed(r r e′ ) r′

V

d

Trang 3

Radio Thermal Radiation 223

of the material body In particular, it can be removed from the radiating body so far that the incident on the body wave is practically a plane wave

Due to the isotropy of the fluctuation currents and statistical properties and based

on the FDT, we now have the following for the mean intensity of the field component:

(8.6)

If we now recall Equation (1.20), describing the density of losses of electromagnetic energy in a substance, then the last result can be rewritten as:

(8.7)

So, the intensity of the fluctuation field is determined by the value of the thermal losses of the diffraction field excited in the body by a unit current applied at the point where the fluctuation field is being studied and directed according to the vector

of the fluctuation field polarization

The result obtained is sometimes referred to as the extended Kirchhoff’s law It

is called extended because the law initially formulated by Kirchhoff was restricted

to the case of a body large in size compared to the wavelength (i.e., the geometrical optics problem) No such limit is stated in the relations being considered here, so the extended Kirchhoff’s law applies

We should point out the dependence of the integrands in the previous formulae

on vector e and the diffraction field dependence on the auxiliary dipole polarization; thus, the polarization is dependent on the radiation of the heated bodies Let us suppose now that the radiation is detected by a receiver responding to only one linear polarization We can assume that the receiver is rather distant and detects the waves with polarization orthogonal to the line connecting the receiver and the center

of gravity of the radiating body We can direct the z-axis along this line The discussion in this case is about the reception of waves, the polarization of which is oriented in the plane {x, y} Let us consider the case of a receiver detecting the x-polarization In this case, it will react not only to the x-polarization waves but also

to waves polarized in the plane The difference between these waves and the x-polarized waves is that the power of their fluctuations will be detected by the receiver with the weight , where η is the angle between vectors e and ex In the case

of statistical independence of waves of different polarization, the fluctuation intensity

of the detected x-polarized field will be equal to the weighted sum of intensities of all the fields polarized in the {x, y} plane In other words,

(8.8)

e E r⋅ ( )2 = 2 ∫ ′′( ),r′ E (r r e, ,′ )2 3r′

4

ω

π ε ω

k T

d

V

b

d

e E r⋅ ( )2 = 2 k T ∫Q(r r e, ,′ )d3r′

V

b

π

cos2η

0

2

= k T ∫ ∫ ( ′ ) ′

V

π

r r, , cos r

Trang 4

In the case of the receiver responding only to the y-component of waves, we have:

(8.9)

If the receiver detects both orthogonal polarizations, then the power of the resulting field will be equal to the sum of Equations (8.8) and (8.9)

Let us now compute the spectral density of Poynting’s vector z-component for different polarizations To do so, we must take into account that the spectral densities given by Equation (8.7) and others are two-way (i.e., they are applicable over the entire real axis of frequencies from – to + We, however, are interested in a one-way spectral density that is associated with the positive half-axis of frequencies This means that the previous expressions should be multiplied by 2:

(8.10)

8.2 RADIO RADIATION OF SEMISPACE

Let us now consider the radiation of semispace We will first consider the case when the receiving antenna is situated in a vacuum and directed perpendicularly to the plane boundary of the radiating medium In this case, the diffraction field and, consequently, the volume density of absorption do not depend on the polarization The integration in Equation (8.10) can be performed over the angle to give the factor

π Further, we are interested in the spectral density of the power flow that is detected

by the receiver for any linear polarization:

Here, is the antenna area, and integration with respect to s represents integration over the plane that is perpendicular to the z-axis Writing the expression in such a form shows that we have implicitly used the geometrical optics approach Equation (8.11) assumes that the radiation field in the view of plane waves coming from different directions reaches the antenna and summarizes their intensities with a weight given by the antenna area value One can point out, in this connection, that this approximation requires the position of the point in the field being searched to

be at a distance from the interface much greater than the wavelength Equation (8.11)

0

2

= k T ∫ ∫ ( ′ ) ′

V

π

r r, , sin r

S

E E

x y

x

y

ω

( ) ( )

















=

















c

2





= ( ′ )











ck T

b

η

η η

2

2 2

r r, , cos

sin

3 0

2

′

∫

V

π

b

d

π

∞

∫

0

∫∫

A e

Trang 5

implies at the same time that the antenna reacts only to polarization orthogonal to the z-axis, which occurs only in the case of a highly directional antenna Therefore, the main area of integration with respect to s in Equation (8.11) is concentrated close to the coordinate origin From here, using the geometrical optics approxima-tion, the diffraction field can be expressed as:

where is the field of auxiliary dipole on the interface and F(0) is the Fresnel reflective coefficient at zero incident angle After integration over z and simple transforms, we obtain:

The field on the surface is easily calculated if we take into account that it is generated

the first term of Equation (1.40)) The dipole field on the surface being considered close to point s = 0 is:

(8.12)

solid angle element) and use Equation (1.122) As a result,

Just the same result will apply to a y-field component, so the spectral density of the total power flow can be written as:

(8.14)

This result is obvious and reflects the detailed balance between emitted and reflected energy flows

It is convenient to express the power detected by an ideal receiver (i.e., by a receiver with perfectly matched circuits) in the temperature scale Such temperature

is called brightness and is equal to:

Ed( )s,z ≅Ed( )0,z ≅E0i 1+ ( )0  z,

ik

Ei

0

W

c k T F

i

x

2 b

e

E

= (1− 0 ) ∫

8

2 2

( )

p= −1 iω

ik

k i

e

0 2

= ω

r p

r

d2s r= Ωd

Wx( )ω = k Tb  − F( )

 

2

W( )ω = − F( ) k T

1 0 

2 b

Trang 6

(8.15)

Note that our method of calculation leads to the-so called antenna temperature; however, there is no difference between brightness and antenna temperature in the considered case of semispace

The black-body reflective coefficient is equal to zero and, in this case, the brightness temperature is simply equal to the temperature of the black body So, the brightness temperature is the temperature of a black body at which it radiates with the same intensity as the heated body at a given polarization and frequency In the example discussed here, the following value is the emissivity (coefficient of emis-sion):

(8.16) and it is more convenient to write:

(8.17)

This last expression is considerably more widely used than the particular case from which it was obtained At the inclined observation of a plane-stratified medium, the emissivity is equal to:

Here, the second summand is equal to half the sum of the reflective coefficients for the horizontal and vertical polarized waves These coefficients for a plane-layered medium differ, in general, from the Fresnel ones The angle θ is the zenith angle of observation in this case Let us emphasize the fact that Equation (8.18) concerns media inside which radiation that has penetrated is fully absorbed So, for example, Equation (8.18) has to be modified by adding the transmission coefficient in the case

of a limited-thickness layer In the general case,

(8.19)

Let us point out two circumstances regarding the radiation of a semispace filled by

a transparent, or more exactly, a weakly absorbing medium First, the brightness temperature is equal to the temperature of the body observed on vertical polarization

at the Brewster angle This means that the medium is like a black body (the emissivity

is equal to unity) under the specified conditions If we were to perform the vertical

T( )ω = − F( ) T

1 0 

2

κ ω( )= −1 F( )0 2

T( )ω =κ ω( ) T

κ ω θ( , )= −  ( )θ + ( )θ

2

Fh Fv

κ ω θ( ), i = −1 1F ( )θi +F ( )θi +T ( )θi +T ( )θi 

Trang 7

polarization emission measurement and change, in the process, the angle of obser-vation, then the body temperature is determined at the angle of maximum radiation, and the dielectric constant is calculated by this angle value

The second circumstance is connected with Equation (3.25), from which it follows that the temperature of an emitting semispace can be easily determined by observing both polarizations at θ = 45° It is computed by the formula:

(8.20)

The emissivity is a function of frequency This frequency dependence is twofold due to the permittivity frequency dependence and to the interference and resonance phenomena that are described by the diffraction field

The temperature of natural media cannot be constant over the space This situation is typical, for example, for soil, which is not uniformly heated by solar radiation In these cases, the formulae defining the brightness temperature demand elaboration One should take into account when performing the corresponding cal-culations that, strictly speaking, a medium that is not uniformly heated is not in equilibrium, even with the heat transfer process; however, if spatial temperature gradients are rather small, then the medium can be assumed to be locally in equi-librium The definition of small gradients is not formulated in the general case and always requires elaboration, taking into account the peculiarities of the problem being studied One can assume, in our cases, that the demand of small spatial gradients of temperature is always fulfilled

The temperature spatial variations are followed by a spatial change of the heated medium permittivity, as the permittivity is a temperature function The spatial changes, in this case, should also be taken into account Further, we will summarize the problem and take into consideration permittivity spatial changes caused by various factors, not only temperature Among these are spatial variation of the medium density or concentration changes in impurities

Let us examine, for instance, the case of a semispace that is not uniformly heated (e.g., soil in the morning) The permittivity of the medium will be assumed to be a function only of depth, and we will concentrate on observation at the nadir Instead

of Equation (8.11), we now have:

(8.21)

Let us assume that the temperature and permittivity change slightly on a scale of the order of the wavelength We can use, in this case, the Wentzel–Kramers–Brillouin (WKB) approximation for the diffraction field According to this, the reflection inside the medium cannot be taken into consideration, and we should analyze only the

= ( )

−

(h) (h) (v) 2

∞

∫

ω

0

Trang 8

wave reflection on the medium–vacuum interface, which is characterized by the reflective coefficient:

As a result,

(8.22)

where the coefficient of absorption is:

(8.23)

After rather simple transforms, we obtain:

(8.24)

Let us now regard two extreme cases The first one refers to weak absorption

in the sense that the imaginary permittivity part is much smaller than its real one;

it is reasonable to do so to obtain:

(8.25)

In the case of strong absorption, when the real and imaginary parts of the permittivity are comparable, the integral from the absorbing coefficient changes quickly on the wavelength scale Other cofactors in Equation (8.24) can be assumed to be “slow” functions, which gives us the opportunity, using integration by parts, to obtain an expansion with respect to the reverse degree of ΛΓ(0), where Λ is the scale of the temperature or permittivity change Because ΛΓ(0) ≈ l/λ in this case, the series terms decrease quickly The sum of the first two is represented in the form:

F= −

1

0 0 0

ε

ε , ε ε( ).

E

z

2 2

0

1













∫

d

ω

ε

ε exp Γ ζ ζ ,

Γ(z)= =  (z)− ∗(z)= ′′( z)

∗

1

21

2

0

z

Γ exp ∫∫Γ









∞

0

′′ << ′

ε ε

T= −( F ) T − ( )d d











∫

∞

0 0

z

∫∫

Trang 9

(8.26)

It follows from Equation (8.26), that, at strong absorption, the temperature spatial variability is not appreciably displayed in the intensity of thermal radiation It is understandable, because in this case the essential contribution to the radiation is brought by the fluctuation currents that are situated close to the medium interface Let us now consider the case of an eroded boundary of radiating medium to see what corrections the unsharpness of border introduces into the emissivity We will assume that thickness d of the transient layer is small in comparison with the wavelength In order to calculate the emissivity at the zero incident angle, we will use Equation (3.119) to obtain:

(8.27)

We also must use the model for the permittivity depth distribution:

(8.28)

The advantage of this model is that the function describing the permittivity depth profile is integrated with the points z = 0, d and has zero derivatives there For this reason, it “is smoothly connected” with the permittivity values on the boundaries of the transient layer Integration over the equations provided in Section 3.8 gives us:

(8.29)

It is easy to see that, as was expected, the correction is small at the accepted approaches; however, we can also see that it increases the emissivity It takes place because of better matching between the vacuum and the emitting medium The transient layers improve the matching rather appreciably in some cases, as can be seen by analyzing Equation (3.41) Without going into the details, let us

Let us now set the conditions at which the reflective coefficient of

in order to change the cosine in the numerator of Equation (3.41) to –1; in fact, this

d

T

+













∗

0

2

ε 





















=

z 0

κ ω( )= −( ) −  + −( ) ( )

1 F f2 1 k2 2F I f 1 F f 2J12 0

ε(z)= + (εd− )z εd z





 − ( − )







κ ω( )= −( ) +( ε − )











1 20

2

2 2

F

k d

d

1<ε2<ε3

ψ12=ψ23= π

2ψ π=

Trang 10

will take place at any odd numbers of π Here, however, we will restrict ourselves

to a very simple case Under these conditions, we are dealing with a quarter-wave

we are discussing layers for which the thickness is on the order of the wavelength

as well as weakly absorbing media, then we should assume that τ = 0 in Equation (3.41) Then, the second requirement of converting the reflective coefficient to zero and, correspondingly, the emissivity to unity will be satisfied by the equality

which leads to the necessary validity of the relation

We should emphasize the resonant character of the effect described, as the monochromatic emission is under discussion Because reception of thermal radiation takes place in the frequency bandwidth, which is often a rather wide one, then the mentioned resonance can be eroded and full conversion of the emissivity to unity does not occur This erosion is found to be weak in the case of a narrow bandwidth The corresponding analysis should be carried out using, for example, Equation

the following is obtained:

(8.30)

Here, f0 is the central frequency and ∆f is the bandwidth of the receiving frequencies

It would appear that the difference between the quarter-wave layer emissivity and unity seems to be small

Let us consider, finally, the radiation of a weakly reflected layer The diffraction field can be described by the WKB approximation in this case; therefore, the expres-sion for the brightness temperature is obtained from Equation (8.25), where the reflective coefficient is set equal to zero (i.e., no sharp jump of permittivity on the border) So,

(8.31)

Here, d is the layer thickness In the case of constant temperature inside the layer,

(8.32)

d= λ 4 ε

2

=

β2−β1<<1

ε

= − ≅ − ( − ) 







192

2

2 2 2

2

f

∆

z d











∫

∫ z Γ z exp Γ ζ ζ z

0 0

d

= (1− − ) = ∫ ( )

0

τ , τ Γ ζ ζ

Định dạng
Số trang	20
Dung lượng	455,01 KB