Добавил:
Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз: Предмет: Файл:

B.Thide - Electromagnetic Field Theory

.pdf
Скачиваний:
42
Добавлен:
07.05.2013
Размер:
1.09 Mб
Скачать

7.3 THE RADIATION FIELDS

103

the total electric field:

E(t;x) = Z 1d!E!(x) e i!t

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

4 "0 ZV

 

(tret0 xx0

 

x0

 

3

0

)

 

d3x0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

;

 

 

)(x

 

 

x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

j

 

 

 

j

 

 

 

 

 

 

 

 

 

 

0}

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

d3x

 

 

 

 

 

 

|+

 

 

 

 

 

 

 

[j{zret0 0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Retarded Coulomb field

 

 

4

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

4 "0c ZV

 

(t

 

;x )

x x0

 

 

 

 

 

 

x )

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

(x

 

 

x )](x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+|

1

 

 

 

 

 

 

 

 

 

 

 

 

j

 

 

 

 

 

j

x0)]

 

(x

x0)}

3

 

 

 

 

 

 

 

[j(tret0 ;x0){z(x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(7.23)

 

 

 

 

 

 

 

 

 

 

 

 

Intermediate field

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ZV

 

 

 

 

 

 

 

 

x

 

 

x0

 

4

 

 

d x0

 

 

 

 

4 "0c

 

 

 

 

 

 

 

 

 

 

j

 

 

|

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

j

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

[j˙(tret0 ;x0)

{z(x x0)] (x x0) 3}

 

 

 

 

 

 

 

 

 

 

 

 

 

Intermediate field

 

 

 

 

 

 

 

 

 

 

+

 

ZV

 

 

 

 

 

 

 

 

x

x0

 

3

 

 

d x0

 

4 "0c2

 

 

 

 

 

 

 

 

j

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

j

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

|

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

{z

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

}

Radiation field

Here, the first term represents the retarded Coulomb field and the last term represents the radiation field which carries energy over very large distances. The other two terms represent an intermediate field which contributes only in the near zone and must be taken into account there.

With this we have achieved our goal of finding closed-form analytic expressions for the electric and magnetic fields when the sources of the fields are completely arbitrary, prescribed distributions of charges and currents. The only assumption made is that the advanced potentials have been discarded; recall the discussion following Equation (3.35) on page 45 in Chapter 3.

7.3 The Radiation Fields

In this section we study electromagnetic radiation, i.e., the part of the electric and magnetic fields fields, calculated above, which are capable of carrying energy and momentum over large distances. We shall therefore make the assumption that the observer is located in the far zone, i.e., very far away from the source region(s). The fields which are dominating in this zone are by definition the radiation fields.

From Equation (7.12) on page 100 and Equation (7.23) above, which give

Draft version released 31st October 2002 at 14:46.

Downloaded from http://www.plasma.uu.se/CED/Book

i

i

i

104

ELECTROMAGNETIC FIELDS FROM ARBITRARY SOURCE DISTRIBUTIONS

the total electric and magnetic fields, we obtain

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

˙

;x

)

 

(x

 

x

)

 

Brad(t;x) = Z 1d!Brad!(x) e i!t =

0

ZV

j(tret0

 

0

 

 

 

2

 

0

 

d3x0

4 c

 

j

x

 

x0

j

 

 

 

1

 

 

 

 

 

 

 

 

 

 

(7.24a)

Erad(t;x) = Z 1d!Erad

!(x) e i!t

1

 

=

1

ZV

4 "0c2

where

@t t=tret0

j˙(tret0 ;x0)

def

 

@j

 

˙ 0 0 0 0

[j(tret;x ) (x x )] (x x ) d3x0 jx x0j3

(7.24b)

(7.25)

Instead of studying the fields in the time domain, we can often make a spectrum analysis into the frequency domain and study each Fourier component separately. A superposition of all these components and a transformation back to the time domain will then yield the complete solution.

The Fourier representation of the radiation fields Equation (7.24a) and Equation (7.24b) above were included in Equation (7.11) on page 100 and Equation (7.22) on page 102, respectively and are explicitly given by

B! (x) =

 

2 Z 1 dt B

rad

(t;x) e

i!t

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

rad

 

1

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= i 4 0 ZV

!

 

 

0x

x0

2

 

0

)

 

eikjx x0j d3x0

 

 

 

 

 

(7.26a)

 

 

 

 

k

 

 

 

 

j

 

 

(x )

 

(x

 

x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= i 40

ZV j!x

 

j

 

j

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

x0

eikjx x0j d3x0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(x )

k

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E! (x) =

 

 

 

 

 

 

 

 

 

j

 

 

 

 

j

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2 Z 1 dt E

rad

(t;x) e

i!t

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

rad

 

1

 

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

= i 4 "0c ZV

 

[j!(x0) (x

x00

 

3

 

(x

 

0

)

eikjx x0j d3x0 (7.26b)

 

 

 

 

 

k

 

 

 

 

 

 

 

 

x

 

 

 

x )]

 

 

x

 

 

 

 

 

 

 

4 "0c ZV

 

 

 

 

 

 

j

 

 

 

j

 

 

 

 

 

j

 

 

j

 

0

 

 

 

 

 

 

 

jx x0j2

 

 

 

 

 

 

 

 

 

=

 

 

i

1

 

 

 

 

 

 

[j!(x0) k]

 

(x x0)

eik x

 

x0

 

d3x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ˆ 0 j j where we used the fact that k = kk = k(x x ) x x` .

=

If the source is located inside a volume V near x0 and has such a limited spatial extent that max jx0 x0j jx x0j, and the integration surface S , centred

Downloaded from http://www.plasma.uu.se/CED/Book

Draft version released 31st October 2002 at 14:46.

i

i

i

7.3 THE RADIATION FIELDS

105

 

S

 

dS = nˆd2x

x x0

 

kˆ

x0

x x0

 

x0 x0

 

x

x0

 

 

V

O

FIGURE 7.1: Relation between the surface normal and the k vector for radiation generated at source points x0 near the point x0 in the source volume V. At distances much larger than the extent of V, the unit vector nˆ, normal to the surface S which has its centre at x0, and the unit vector

ˆ 0

k of the radiation k vector from x are nearly coincident.

on x0, has a large enough radius jx x0j max jx0 x0j, we see from Figure 7.1 that we can approximate

 

 

 

 

k

x

x0

 

k (x0

 

x0)

(7.27)

k

x

 

x0

 

k (x

x0)

k (x x0) k (x0

x0)

 

 

 

 

 

j

 

j

 

 

 

 

Recalling from Formula (F.45) and Formula (F.46) on page 166 that dS = jx x0j2 d= jx x0j2 sin d d'

ˆ

and noting from Figure 7.1 that k and nˆ are nearly parallel, we see that we can approximate.

ˆ

 

ˆ

 

 

 

 

jx x0j2

 

jx x0j2

 

 

 

k dS

=

k nˆ

dS

 

d

(7.28)

Both these approximations will be used in the following.

Within approximation (7.27) the expressions (7.26a) and (7.26b) for the

Draft version released 31st October 2002 at 14:46.

Downloaded from http://www.plasma.uu.se/CED/Book

i

i

i

106

ELECTROMAGNETIC FIELDS FROM ARBITRARY SOURCE DISTRIBUTIONS

radiation fields can be approximated as

B!rad(x) i

4 eikjx x0j ZV

x

x0

 

e ik (x0 x0) d3x0

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

j!(x0)

 

k

 

 

 

 

 

 

 

 

 

0

 

 

ik x

x0

 

j

 

j

 

 

 

 

 

 

 

 

(7.29a)

 

i

 

e j j

ZV [j!(x0) k] e ik (x0 x0) d3x0

 

 

 

 

 

 

 

4

x x0

 

 

 

 

 

 

!

 

 

 

 

j

 

 

 

j

 

j ZV

 

 

 

jx x0j2

 

 

 

 

 

0

 

4"0c

 

j

 

 

 

 

 

 

Erad(x)

 

i

1

 

 

eik x

x0

 

[j!(x0) k] (x x0)

e

 

ik

(x0

 

x0) d3x

 

 

 

 

 

 

 

 

 

 

 

 

 

i 4"0c

 

x j

x0 j

x x0

ZV [j!(x0) k] e ik (x0 x0) d3x0

 

 

 

 

1

 

 

 

eik x

x0

(x x0)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

j

 

j

j

j

 

 

 

 

 

 

 

 

(7.29b)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

I.e., if max jx0 x0j jx x0j, then the fields can be approximated as spherical waves multiplied by dimensional and angular factors, with integrals over points in the source volume only.

7.4 Radiated Energy

Let us consider the energy that is carried in the radiation fields Brad, Equation (7.26a), and Erad, Equation (7.26b) on page 104. We have to treat signals with limited lifetime and hence finite frequency bandwidth differently from monochromatic signals.

7.4.1 Monochromatic signals

If the source is strictly monochromatic, we can obtain the temporal average of the radiated power P directly, simply by averaging over one period so that

hSi h

1

 

i

1

 

E B

1

 

E!e

i!t

(B!e

i!t

)

(7.30)

2 0 Re

21 0 Re

= E

 

H =

 

 

 

=

 

 

 

 

 

 

 

 

= Re E! B e i!tei!t = Re E! B

2 0 ! 2 0 !

Using the far-field approximations (7.29a) and (7.29b) and the fact that 1=c = p"0 0 and R0 = p 0="0 according to the definition (2.18) on page 28, we obtain

hSi = 32 2 R0

 

x

x0

2

ZV

(j! k)e ik (x0 x0) d3x0

 

2

 

x

x0

(7.31)

1

 

j

 

1

j

 

 

 

j

x

x0

j

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Downloaded from http://www.plasma.uu.se/CED/Book

Draft version released 31st October 2002 at 14:46.

i

i

i

7.4 RADIATED ENERGY

107

or, making use of (7.28) on page 105,

 

 

d = 32 2 R0

ZV (j! k)e ik (x0 x0) d3x0

(7.32)

 

dP

1

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

which is the radiated power per unit solid angle.

7.4.2 Finite bandwidth signals

A signal with finite pulse width in time (t) domain has a certain spread in frequency (!) domain. To calculate the total radiated energy we need to integrate over the whole bandwidth. The total energy transmitted through a unit area is the time integral of the Poynting vector:

Z1 S(t) dt = Z1(E H) dt

 

 

 

1

1

 

 

 

 

 

= Z1 d!Z1 d!0 Z1

(E! H!0)e i(!

!0)t dt

 

1

1

1

+

 

If we carry out the temporal integration first and use the fact that

Z 1

e i(!+!0)t dt = 2 (!+!0)

1

Equation (7.33) can be written [cf. Parseval's identity]

(7.33)

(7.34)

Z1 S(t) dt = 2 Z1(E! H !) d!

 

 

 

1

1

 

 

 

= 2

Z01(E! H !) d!+Z1(E! H !) d!

 

 

 

0

 

 

= 2 Z01(E! H !) d! Z01(E! H !) d!

(7.35)

 

Z 1

Z 1

 

= 2

2 Z

= 0 0

2 Z

= 0 0

0

(E! H !) d!+ 0

(E ! H!) d!

1

(E! B ! +E ! B!) d!

1

(E! B! +E! B!) d!

where the last step follows from the real-valuedness of E! and B!. We insert the Fourier transforms of the field components which dominate at large

Draft version released 31st October 2002 at 14:46.

Downloaded from http://www.plasma.uu.se/CED/Book

i

i

i

108

ELECTROMAGNETIC FIELDS FROM ARBITRARY SOURCE DISTRIBUTIONS

distances, i.e., the radiation fields (7.26a) and (7.26b). The result, after integration over the area S of a large sphere which encloses the source, is

U =

4 r "0

ZS Z0

1 ZV

x x0

eikjx x0j d3x0

 

kˆ

nˆ dS d!

(7.36)

 

1 0

 

 

j!

k

 

 

2

 

 

 

 

 

 

 

 

 

j

j

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Inserting the approximations (7.27) and (7.28) into Equation (7.36) above and also introducing

Z 1

U = U! d! (7.37)

0

and recalling the definition (2.18) on page 28 for the vacuum resistance R0 we obtain

dU! d! d

Z

1

R0

4 V

2

(j k)e ik (x0 x0) d3x0 d! (7.38)

!

which, at large distances, is a good approximation to the energy that is radiated per unit solid angle d in a frequency band d!. It is important to notice that Formula (7.38) includes only source coordinates. This means that the amount of energy that is being radiated is independent on the distance to the source (as long as it is large).

Bibliography

[1]F. HOYLE, SIR AND J. V. NARLIKAR, Lectures on Cosmology and Action at a Distance Electrodynamics, World Scientific Publishing Co. Pte. Ltd, Singapore, New Jersey, London and Hong Kong, 1996, ISBN 9810-02-2573-3(pbk).

[2]J. D. JACKSON, Classical Electrodynamics, third ed., John Wiley & Sons, Inc., New York, NY . . . , 1999, ISBN 0-471-30932-X.

[3]L. D. LANDAU AND E. M. LIFSHITZ, The Classical Theory of Fields, fourth revised English ed., vol. 2 of Course of Theoretical Physics, Pergamon Press, Ltd., Oxford . . . , 1975, ISBN 0-08-025072-6.

[4]W. K. H. PANOFSKY AND M. PHILLIPS, Classical Electricity and Magnetism, second ed., Addison-Wesley Publishing Company, Inc., Reading, MA . . . , 1962, ISBN 0-201-05702-6.

[5]J. A. STRATTON, Electromagnetic Theory, McGraw-Hill Book Company, Inc., New York, NY and London, 1953, ISBN 07-062150-0.

[6]J. A. WHEELER AND R. P. FEYNMAN, Interaction with the absorber as a mechanism for radiation, Reviews of Modern Physics, 17 (1945), pp. 157–.

Downloaded from http://www.plasma.uu.se/CED/Book

Draft version released 31st October 2002 at 14:46.

i

i

i

8

Electromagnetic Radiation and Radiating Systems

In Chapter 3 we were able to derive general expressions for the scalar and vector potentials from which we then, in Chapter 7, calculated the total electric and magnetic fields from arbitrary distributions of charge and current sources. The only limitation in the calculation of the fields was that the advanced potentials were discarded.

Thus, one can, at least in principle, calculate the radiated fields, Poynting flux and energy for an arbitrary current density Fourier component and then add these Fourier components together to construct the complete electromagnetic field at any time at any point in space. However, in practice, it is often difficult to evaluate the source integrals unless the current has a simple distribution in space. In the general case, one has to resort to approximations. We shall consider both these situations.

8.1 Radiation from Extended Sources

Certain radiation systems have a geometry which is one-dimensional, symmetric or in any other way simple enough that a direct calculation of the radiated fields and energy is possible. This is for instance the case when the current flows in one direction in space only and is limited in extent. An example of this is a linear antenna.

109

i

i

i

110

ELECTROMAGNETIC RADIATION AND RADIATING SYSTEMS

 

 

 

 

 

 

L

 

L

2

 

2

 

 

 

 

 

 

 

 

 

 

FIGURE 8.1: A linear antenna used for transmission. The current in the feeder and the antenna wire is set up by the EMF of the generator (the transmitter). At the ends of the wire, the current is reflected back with a 180 phase shift to produce a antenna current in the form of a standing wave.

8.1.1Radiation from a one-dimensional current distribution

Let us apply Equation (7.32) on page 107 to calculate the power from a linear, transmitting antenna, fed across a small gap at its centre with a monochromatic source. The antenna is a straight, thin conductor of length L which carries a one-dimensional time-varying current so that it produces electromagnetic radiation.

We assume that the conductor resistance and the energy loss due to the electromagnetic radiation are negligible. The charges in this thin wire are set in motion due to the EMF of the generator (transmitter) to produce an antenna current which is the source of the EM radiation. Since we can assume that the antenna wire is infinitely thin, the current must vanish at the end pointsL=2 and L=2 and. Furthermore, for a monochromatic signal, the current is sinusoidal and is reflected at the ends of the antenna wire and undergoes there a phase shift of radians. The combined effect of this is that the antenna current forms a standing wave as indicated in Figure 8.1

For a Fourier component !0 the standing wave current density can be written as j(t0;x0) = j0(x0)expfi!0t0g [cf. Equations (7.7) on page 99] where

j0(x0) = I0 (x0) (x0) sin[k(L=2

 

x30

)] xˆ

3

(8.1)

1

2

 

 

 

 

 

 

 

sin(kL

=2)

 

 

 

 

 

 

 

 

 

 

Downloaded from http://www.plasma.uu.se/CED/Book

Draft version released 31st October 2002 at 14:46.

i

i

i

8.1 RADIATION FROM EXTENDED SOURCES

111

 

rˆ

x3 =z

'ˆ

L

x

 

2

 

 

ˆ

 

 

 

ˆ

j!(x0)

k

x2

 

 

'

x1

 

L2

 

FIGURE 8.2: We choose a spherical polar coordinate system (r =jxj; ;') and orient it so that the linear antenna axis (and thus the antenna current density j!) is along the polar axis with the feed point at the origin.

where the current amplitude I0 is a constant (measured in A).

In order to evaluate Formula (7.32) on page 107 with the explicit monochromatic current (8.1) inserted, we use a spherical polar coordinate system as in Figure 8.2 to evaluate the source integral

ZV0d3x0

j0 k)e ik (x0 x0)

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

L=2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

sin[k(L=2

 

 

 

)]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

 

 

 

 

 

 

x30

 

 

ikx30 cos

 

 

ikx0 cos

 

 

 

 

 

 

 

 

 

 

 

L=2

I0

 

sin(kL=2)

 

 

 

 

k sin e

 

 

e

 

 

dx30

 

 

 

 

 

 

 

Z

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

2

 

 

 

 

 

 

 

 

L=2

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

k2 sin

 

 

ikx0 cos

2

 

 

 

 

 

 

 

0

 

 

 

0

 

0

 

 

 

 

0 sin2(kL=2)

 

 

 

 

 

 

Z0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

 

 

 

 

= I

 

 

 

 

 

 

 

e

 

 

 

 

2

sin[k(L=2

 

 

x )]cos(kx

 

cos ) dx

 

 

 

= 4I2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos[(kL=2)cos ] cos(kL=2)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin sin(kL=2)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(8.2)

Inserting this expression and d = 2 sin d into Formula (7.32) on page 107

Draft version released 31st October 2002 at 14:46.

Downloaded from http://www.plasma.uu.se/CED/Book

i

i

i

112

ELECTROMAGNETIC RADIATION AND RADIATING SYSTEMS

and integrating over , we find that the total radiated power from the antenna is

0

0

4 Z0

 

 

 

 

sin sin(kL=2)

 

 

 

 

 

2 sin d

 

P(L) = R

I2

1

 

 

 

cos[(kL=2)cos ] cos(kL=2)

 

 

(8.3)

 

 

 

 

 

 

 

One can show that

12

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

kL!0

 

 

R0I0

 

 

 

 

 

 

 

 

 

 

lim P(L)

=

 

 

L

 

2

2

 

 

 

 

 

 

 

 

 

 

(8.4)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where is the vacuum wavelength.

 

 

 

 

 

 

 

 

 

 

The quantity

 

 

 

 

 

21 I02

= R0 6

 

197

 

 

 

 

 

 

 

 

Rrad(L) =

Ieff2

=

 

 

 

 

 

 

(8.5)

 

P(L)

 

P(L)

 

 

L

2

 

L

 

2

 

 

 

is called the radiation resistance. For the technologically important case of a half-wave antenna, i.e., for L = =2 or kL = , Formula (8.3) above reduces to

 

 

4

 

Z0

cos2

 

cos

 

 

= R0I02

1

 

 

 

 

P(=2)

 

 

 

 

 

 

2

 

d

 

 

 

 

 

sin

 

 

 

 

 

 

 

 

 

The integral in (8.6) can always be evaluated numerically. fact also be evaluated analytically as follows:

(8.6)

But, it can in

Z0

cos2

 

 

 

 

cos

 

 

 

 

2

sin

d = [cos ! u] = Z

 

1

1

u2

du =

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

cos2

 

u

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2

 

 

 

 

 

 

 

 

cos2

 

2 u =

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

+cos( u)

 

 

 

 

 

 

 

=

2

Z 1

(1 +u)(1 u) du

 

 

 

 

 

 

 

 

 

 

 

 

1

1

 

1

+cos( u)

 

 

 

 

 

 

 

 

 

 

 

 

= 4

Z

1

1

 

(1 +u)

 

 

 

du + 4 Z

 

1

(1 u)

du

 

1

1

 

 

+cos( u)

 

 

 

1

 

1

1 +cos( u)

(8.7)

= 2

 

1

 

 

 

(1 +u)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Z

 

 

 

 

 

 

du = 1 +u !

 

 

 

1

1

1 +cos( u)

 

 

 

 

h

 

 

 

v

i

 

=

2

2

1

v

dv =

2

 

 

 

 

 

 

 

Z0

 

 

[ +ln2 Ci(2 )]

 

1

 

 

 

 

 

 

cosv

 

 

 

 

 

1

 

 

 

 

 

 

 

 

 

 

1:22

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

where in the last step the Euler-Mascheroni constant = 0:5772 ::: and the cosine integral Ci(x) were introduced. Inserting this into the expression Equation (8.6) we obtain the value Rrad(=2) 73 .

Downloaded from http://www.plasma.uu.se/CED/Book

Draft version released 31st October 2002 at 14:46.

i

i

i

Соседние файлы в предмете Физика