Electromiagnetic theory

8/11/2019 Electromiagnetic theory

1/16

~

ELECTROMAGNETIC THEORY

A. B. Lahanas

University of Athens, Physics Department, Nuclear and Particle Physics Section, Athens 157 71, Greece

Abstract

An introduction to Electromagnetic Theory is given with emphasis on wave

propagation phenomena in free space and inside wave guides. We also discuss

the radiation emitted by moving electric charges, an issue which is particularlyimportant in accelerator physics.

1. INTRODUCTION

The topics that will be covered in this lecture are the following:

Maxwell Equations

Conservation of Energy - The Poynting vector

Propagation of Electromagnetic (EM) Waves

Power Absorption by Conducting Surfaces

Propagation of EM Waves in Wave Guides

Energy Flow and Power Losses in Wave GuidesPotentials - Radiation by Moving Charges

All material covered in this lecture and details of the calculations involved can be found in standard

textbooks (see for instance [1, 2, 3] ). In writing this lecture I have benefited from lectures given in

previous CERN schools ( see [4] ).

2. MAXWELL EQUATIONS

We start by reviewing Maxwellsequations :

Electric charges whose density is are the sources of the electric fieldE~ . In theMKSA system this is

expressed by Gausss law

r~ E~ = : (1)0Electric currents with densityJ~

expressed by Amperes law

= ~u are the sources of the magnetic induction fieldB~ . This is

r~ B~ = 0J~ : (2)Field lines ofB~ are closed. This is equivalent to the statement that there are no magnetic monopoles.

Mathematically this is expressed by the equation

r~ B~ = 0 : (3)The electromotive force around a closed circuit is proportional to the rate of change of flux of the field

Bthrough the circuit (Faradayslaw ). In differential form this law is expressed by the following formula

r~ E~ = @B~

@t: (4)

Within material media having polarizationP~ and magnetizationM~

the following replacements

the above laws still hold with

) r~ P~ ; J~)J~ + r~ M~@P~

+@t

+ 0@E~

:@t


2/16

That is to the true charge density we have to add the polarization charge density and to the true current

density we have to add the contributions of the magnetization current, the polarization current and the

displacement current introduced by Maxwell. In terms of the electric displacement and magnetic fields,defined by D~ 0 E~ + P~ andH~

1 B M respectively, Maxwell equations can be brought

into the following form 0~ ~

r~ D~

r~ E~

=

= @B~@t

r~ H~

r~ B~

= J~

= 0

@D~+

@t

In some materials ( Linear media ) it happens thatD~ = E~ ; B~ = H~ ; where the quantities

; are called the dielectric constant and magnetic permeability of the medium respectively.

2.1 The Continuity Equation

The electric charge is conserved. Actually we have never observed in the laboratory a violation of this

conservation law. This conservation law is expressed by the following Continuity Equation

r~ J~ + @@t

= 0 (5)

where is the charg

Re density andJ~ = ~u is the current density. This equation follows from Maxwell

equations and it is not an independent hypothesis.

The quantity J~ d~S represents the charge flowing out of surface S per unit time ( this is

continuity equation it follows that r~H J~ = 0 . In this case we say that we have steady currents. Formeasured in Amperes

S

in the system MKSA ). If the charge density is time independent then from the

the steady current case the integral J~

d~S over any closed surface S vanishes. In order to see itsconsequence, consider the case of a s

Surface crosing N wires carrying currents I1; I2; :::;IN flowing into

(or out) a node surrounded by the surface. Then the vanishing of this closed surface integral results to

I1 + I2 + ::: + IN = 0 , which is the well known Kirchoffscurrent conservation law.

2.2 The Lorentz force

The force acting on a charge e , which is at rest within an electric field, isF~ = e E~ . Also the force

acting on a small wire element d~l, carrying electric current I, which is placed in a magnetic field, is

F~ = I d~l B~ . These two suggest that for a charge e moving with velocity u , the total force acting

on it is

F~ = e ( E~ + ~u B~ ): (6)

This is the wellknownLorentz force. For a continuous charge and current distribution, andJ~ = ~u ,it is convenient to define the force densityf~ , that is force per unit volume. On account of eq. (6) this is

given by

f~ = E~ + J~ B~ : (7)

3. CONSERVATION OF ENERGY - The Poynting vector

From the previous section it becomes evident that the rate of doing work on a unit volume of the distri-

bution is given by

f~ ~u = (E~ + ~u B~ ) ~u = J~ E~ : (8)


3/16

Using Maxwell equations, and after some trivial mathematical manipulations, the right hand side of this

equation can be written as

J~ E~ = E~ (

@D~

@tr~ H~ )

= E~@D~

@t

H~ r~ E~ + r~ (E~ H~)

=

InLinear Media this takes the form

E~@ D~

@t

+ H~@B~

@t

+ r~ (E~ H~)

@W+ r~@t N

~ = J~ E~ : (9)

In (9) the quantities W ; N~ are the Energy density and the Poynting vector respectively defined by

W1

( E~ D~2

+ B~ H~) (10)

N~ E~ H~ : (11)

Integrating eq. (9) over an arbitrary v

Zolume V , whose bo

Zundary is S(V ) , we getdEV

+dt S(V )

N~ d~S = J~ E~ dV : (12)V

In this equationThe first term is the rate of change of the Electromagnetic Energy (EM) energy

EV = 1 R(E~ D~ + B~ H~) dV within the volume V .TR hequantityS(V )N

~ d~S gives the flux of EM energy across the boundary S(V) .

2 V

~

V J

R~ E dV is the power d

Rissipated, or generated, within the volume V.

As an example, within a conductor of given conductivity the current density isJ~ = E~ andtherefore J~ E~ dV becomes J . Furthermore if the conductor is a wire of constant cross

sectionV

Sand length L , the V

dV

is

electric current I = J S and the previous integral receives the

wellknown form I2R , where R is the resistance of the wire element given by R = L . One can

easily verify that except its sign, this is indeed the l.h.s. of eq. (12).S

4. PROPAGATION OF ELECTROMAGNETIC WAVES

4.1 Propagation in nonconducting media ( )= 0

In a medium with values ; , for the dielectric constant and the magnetic permeability respectively,

we can derive from Maxwell laws the following equations

f r2 @ gE~ @J~

= + r~@t2

f r2 @@t2

gB~@t

= r~ J~ : (13)

In regions where there are no charge and current distributions, their right hand sides are absent and the

electric and magnetic fields,E~ ;

given by B~ , satisfy the fr

ree wave equations. The waves travel with velocity u

1u = : (14)


4/16

qIn vacuo this is usually denoted by the symbol c and has the value c =

1

0 0' 300; 000 Km=sec :

In regions where there are nonvanishing charge and current distributions the right hand sides of eqs. (13)

are non-vanishing too and are the sources of the electromagnetic waves.

The plane waves are particular solutions of (13) in regions where sources are absent. In the

following we shall use complex notation and write the electric component of a plane wave as

E~ =E~0 exp i(~k ~x !t) :

The physical electric field measured in the laboratory is meant to be the real part of this expression. This

is the convention that we will use throughout. A similar expression holds for the magnetic field too with

E~ ; E~0 replaced byB~ ; B~0 respectively. In this expressionE~0 is the amplitude of the electric field, ~k its

wave vector and ! itsfrequency. This monochromatic pulse is a solution when the frequency is linearly

related to the magnitude k j~kj of the wave vector ~k ,! = u k :

k is called the wave number and is related to the wave length by the relation

2k = :

Using Gausss, r~ E~ = 0 , and Faradayslaw, r~ E~ + @B~ = 0 , one can immediately arrive atthe following relations for the wave number and the amplitudes @otf the electric and magnetic components:

~kE~0 = 0 ; B~

0 =1~k E~0: (15)!

Eqs. (15) state that the electric and magnetic fields of a plane wave are perpendicular to each other and

both perpendicular to the direction of the propagation ~n =~k in the sense shown in figure 1 .

E0

k

B0

=_1

k X E

0

Fig. 1: A plane wave propagating along~k. The wave front is the plane formed by the amplitudesE~0 ; B~ 0

of the electric and magnetic fields respectively .

4.2 Propagation within a conductor ( )= 0

Within a conductor the electric current density and the electric field are related by J~ = E~ , from

which it follows that r~ J~ = r~ E~ Then= : from the continuity equation one has@

+@t

= 0


5/16

is

~

which is immediately solved to yield

(~x; t) = (~x; 0) exp ( t ) : (16)

For good conductors 1014 sec1 so that from theeq. (16)weconclude that chargesmove

almost instantly to the surface of the conductor. The ratio = is called the relaxation time of the

conducting medium. For perfect conductors, = 1, so that t he relaxation time is vanishing. Forgood, but not perfect, conductors is small of the order of 10 14sec or so. For times much larger than

the relaxation time there are practically no charges inside the conductor. All of them have moved to its

surface where they form a charge density .Within a conductor the wave equation for the vector fieldE~ , see eq. (13), becomes

2

f r2 @@t2

@

@tgE~ = 0 :

Notice the appearance of a friction term @@t

which was absent in the free wave equation. If we seek

for monochromatic solutions of the formE~

the form

= E~ (~x) exp ( i ! t) , then the equation above takes on

f r2 + K2gE~ (~x) = 0where K2 = ! ( ! + i ) . This can be immediately solved to yield, for a plane wave solution

travelling along an arbitrary direction ~n,

E~ =E~0 ei( !t) e ; (17)

where ~n ~x . The constants ; , appearing in (17), have dimensions of length 1 and are

functions of . Their analytic expresions are not presented here. These can be traced in any standard

book of Electromagnetic Theory (see for instance [1, 2, 3]). However we can distinguish two particular

cases in which their forms are simplified a great deal. These regard the case of an isolator and the case

of a very good conductor respectively.

For an isolator = 0 and = k ; = 0 . In this case (17) reduces to an ordinary plane

wave which is propagating with wave vector~k = ~n k .

For a very good conductor, and certainly this includes the case of a perfect conductor, the conduc-tivity is large so that the range of frequencies with ! quite broad. In this case the constants

; are given by ' ' 1 , where is a constant called the Skin Depth , given by the followingexpression2

=!

: (18)

Therefore we see from eq. (17) that inside a good conductor :

The field is attenuated in the direction of the propagation and its magnitude decreases exponentially

exp( ) as it penetrates into the conductor. The depth of the penetration is set by and is smaller

the higher

the conductivity, the higher the permeability and the frequency.

As an example for copper = 5:8 107 mho m 1 and the skin depth is ' 0:7 10 3 cmfor a frequency ! = 100 MHz .

In order to close this section, we point out that the magnetic field within the conductor is related

to the electric field by the following relation r1 + iH =

2~n E~

!: (19)

As in the case of the nonconducting materials both E~ ; H~ are perpendicular to each other and to the

direction of propagation ~n . However now the magnetic field has a phase difference of 450 from its

corresponding electric component E~ , due to the appearance of the prefactor 1 + i in eq. (19).


6/16

.

AA ||

AT

n

"1" "2"

.

.

H ||

= oo

H||

< oo

ETE

T

E ||

HT

E ||

HT

.

Fig. 2: Fields near the surface of a perfect (left) and a good (right) conductor. The components are as

shown on the top figure.

5. POWER ABSORPTION BY CONDUCTING SURFACES

Consider a surface separating two media \1" and \2" . If ~n is a unit vector normal to the surface ( withdirection from 2!1 ) then from Maxwell equations one can derive the following boundary conditionsfor the normal ( = vertical to the surface) and the parallel components of the fields involved, denoted by

T and jj respectively,T DT = ; E~jj =E~jj

D1 2 1 2

jj

H~jj = K~ ; BT = BT : (20)

H~1 2 1 2

In (20) is the surface charge density and K~ is the surface current density. For the derivation of these

boundary conditions see [1, 2, 3]. These conditions are extremely useful in order to know how the fields

behave near the surface separating the two media. In particular we will be interested in the case whereone of the media, say the medium , is a conductor while the other,

\1", is a nonconducting material.In this case direct application of these conditions yields :

If \2" is a perfect conductor ( = 1 ) then within itH~ c ;surfaceE~k ; HT = 0 . Thus onlyET ; H~jj 6= 0 .

E~c = 0 in which case on its

If \2" is a good, but not perfect, conductor ( = large 6= 1) then within itH~c ;

E~c 6= 0

and attenuated. On its surfaceE~k ; HT 6= 0 but they are much smaller in comparison withtheir

correspondingET ;H~jj components. In this case only ET , the normal electric component, is

discontinuous across the surface.

The conditions on the surface of a perfect and a good conductor are shown in fig. 2.


7/16

Since H~ is continuous andjH~jjj HT on the surface, we conclude that within a good conductor(see previous section)

H~c 'H~jj exp i ( = !t) exp( = ) ; (21)whereH~jj is the value on the surface of the conductor and is the distance from the surface. Then from

eq. (19), relating the electric and magnetic fields within a conductor, we get

E~jj '1p i ~n H~jj : (22)

2

In eq. (22) bothH~jj ;

conductivity .

E~jj refer to values on the surface. E~jj is small due to the largeness of the

This small tangential component of the electric field on the surface of a good conductor is responsible

for power flow into the conductor !

In order to calculate the power absorbed by the walls of a conductor we first need calculate the

value of the Poynting vector. Its time average over a cycle, is found to be given by

< N~ > = Re ( E~ H~ )2

whereRe(:::) denotes the real part of the expression (:::) whileH~ stands for the complex congugate of

H~ . Therefore the time averaged power absorbed per unit area is

dPloss

dS= ~n < N~ > : (23)

From this we see that only the component ofN~ normal to the surface is responsible for power losses to

the walls. This is given by 1 Re ( E~jj H~ ) . Trhen using (22) we get from (23)

2 jj

dPloss 1=

dS 2

!jH~jjj2

: (24)

One immediately observes from eq. (24) that

For perfect conductors, = 1 , and no power is absorbed .H~jj on the surface is only needed to calculate the power absorbed by the walls of the conductor.

6. PROPAGATION OF EM WAVES IN WAVE GUIDES

A wave guide is a metalic open ended tube of arbitrary cross sectional shape. Under certain conditions

EM waves can propagate along its axis. A rectangular wave guide is shown in fig. 3. The tube can be

filled with a nondissipative medium characterized by dielectric constant and magnetic permeability

. Suppose that the axis of the guide lies along thez direction. Then for monochromatic waves of given

frequency ! travelling alongz we can write

E~ (~x; t) =

B~ (~x; t) =

E~ (x; y) e i ( kgz ! t) (25)

B~ (x; y) e i ( kgz ! t) (26)

In eqs. (25 , 26) the quantity kg is called the wave propagation constant. When these are plugged into

the free wave equations th ey yield

@2

@x2+

where k2 = !2 !2 = u2 .

@ 2 2

@y2+ k

k2

!E~ (x; y)

B~ (x; y)= 0

(27)

We shall distinguish the following special modes of propagation:


8/16

( k2 k2 )~ r~

~( k k )

r~

Fig. 3: A rectangular wave guide.

Transverse Electric (TE), in which there is no longitudinal component,Ez , of the electric field.

Besides having Ez = 0 , the appropriate boundary conditions on the walls of the guide dictate

that the directional derivative of the z-components of the magnetic field on the conducting wall

vanishes. Thus for the TE modes we have

Ez = 0 everywhere ;@Hz

= 0

@n S

Transverse Magnetic (TM), in which case there is no longitudinal component of the magnetic

field. In this case we have

Hz = 0 everywhere ; EzjS = 0

Transverse ElectroMagnetic (TEM) in which both electric and magnetic components are trans-

verse to the wave guide axis. Thus

Ez ; Hz = 0 everywhere

It can be proven that a hollow wave guide, whose walls are perfect conductors, cannot support propaga-

tion of TEM waves.

Any vector fieldA~ can be written as

A~ =A~ t +z Az ;

that is it can be decomposed into its parallel and its transverse component with respect the axisz. Split-

ting the electric and magnetic field in this way, and using Maxwellsequations, it can be shown that

in the TE and TM modes the transverse components of the EM fields are expressed in terms of their

longitudinal compoments alone . The latter are determined from the wave equations (27), subject to the

appropriate boundary conditions as given before. The explicit formulae relating the transverse to the

longitudinal components, in the TE and TM modes, are as given below

TE modes

i!

Et = z tHz ;g

H~ t = ikg

( k2 k2 ) r~ tHz

TM modes

i !Ht = 2 2 z tEz ;

g

E~ t =i k

( k2 k2 )r~ tEz

In the following we shall work out a particular example, that of the rectangular wave guide with

transverse dimensions a ; b, as shown in figure 3. Suppose that we want to find the TE propagation


9/16

g

g

Z

wave equation

modes. In this caseEz = 0 and we only need calculate Hz. From (27 ) we see that this satisfies the@2 @2 2

2 2 2 2

@x2+

@y2+ kt Hz = 0 ; ( kt ! =u kg )

subject to the appropriate boundary conditions for the TE modes

The solutions are:

@Hz=

@x x=0;a

@Hz

@y y=0;b= 0 :

Hz = H0 cos (x y

m ) cos ( n )a b

k2 2m n

The last relation yields

t = (a2

+b2

) m ; n = Integers :

!2 2

2 m2 n2

u2= kg + ( a2

+b2

)from which it is seen that there is cut-off frequency !c for each mode characterized by the integers

( m ; n ). The cut-off frequency is given by

r!c u m +n

:a2 b2

Thus in each TE mode, labelled by ( m ; n ), we have that ! > !c .

We further observe that

The relation between the wave propagation constant kg and the frequency is

u kg

!= ( 1

2 1=2

c )!2

< 1 :

Thus for a given frequency the wavelength is larger than its free space value.

The phase velocity up =!

is larger than u , that is larger than its free space value.

The group velocity isugroup

d !

d kg

= u ( 1!c

!2

1=2

) :

This is frequency dependent, therefore the guide behaves like a dispersive medium.The situation is best displayed in figure 4 where we plot the

u kg

!as function of the frequencies for the

various TE modes allowed. For each mode there is a cut-off frequency !c. The value ofu k

is always

less than unity and frequency dependent.!

7. ENERGY FLOW AND POWER LOSSES IN WAVE GUIDES

The time averaged Electromagnetic Energy per unit length of the guide, over a period T = 2 , is

easily found to be given by1

U =4

!

( E~ E~ + H~ H~ ) dx dy :S0

In this equation the integration is over the cross-sectional area of the guide, S0 . The x; y axes are

vertical to the axis of the guide, and hence parallel to the surface S0 . On the other hand the rate of flow

of energy transmitted through this area is

Ptrans = < Nz > dSS0


10/16

that

1

u kg

c

Fig. 4: u kg = ! versus frequency for the various TE modes allowed in the rectangular wave guide.

where < Nz > = 1Re ( Ex H EyH )isthetime-averagedz-componentofthePoynting

vector. From these two

2

one can find

y x

Ptrans = U ugroup (28)

that is power is transmitted with the group velocity !

For perfectly conducting walls all energy is transmitted down the guide. However for good - but

not perfect - conducting walls energy is dissipated in Ohmic losses and flow is attenuated ! A useful

quantity which can be used to describe Ohmic losses is the attenuation constant defined as the ratio of

the power loss per unit length of the guide to power transmitted through the guide. In formula this is

given by

K (dUwalls

) =Ptrans : (29)dz

is found to be

r ZIn eq. (29) the numerator expresses the power absorbed per unit lenght of the wall which using eq. (24)dUwalls 1

=dz 2

!

2 ScjH~jjj dS :

In this equation the integration is over a wall stripe Sc of unit width . Therefore, using energy conserva-

tion, one finds that the power transmitted down the guide at the point having coordinatez + dz is related

to the corresponding quantity atz through the relation

Ptrans ( z + dz ) Ptrans ( z ) =dUwalls

dz : (30)dz

The situation is graphically represented in figure 5 where for a slice of width dz, of arbitrary cross-

sectional shape, we show the energy flow and the power absorbed by the conducting walls.

Eq. (30) is easily solved to yield

Ptrans ( z ) = Ptrans ( 0 ) eK z (31)

from which the physical meaning of the attenuation constant becomes manifest. From (31) we see that

the energy flow through the guide is attenuated exponentially and the attenuation is governed by the

parameter K . To have an estimate of the magnitude of K, for copper guides for instance in the


11/16

Z

Fig. 5: Slice of width dz of a wave guide of arbitrary cross-sectional shape. P(z), P(z+dz) represent the

energy flow entering and leaving the surfaces located at the points z and z+dz respectively. dU is the

energy absorbed by the conducting walls of the slice.

microwave regionK turns out to be K 10 4 !c=c . Thus the power transmitted is decreased to its

30% after 200 400 m .

The following example may be instructive in order to understand the basic notions given in this

section :Example

For the rectangular wave guide of figure 3 and for the (m,n)=(1,0) TE mode calculate : A) The

x; y components of the electric and magnetic fields. B) The power transmitted down the guide. C) The

attenuation constant K.

A) From the formulae relatingHz(x; y) to the remaining components ofE~(x; y) ; H~ (x; y) and

using the solutionsHz found in the previous section we find in a straightforward manner that

Ex = 0 ; Ey = i H0!u2

a !22

xsin ( )

a

Hy = 0 ; Hx = i H0kg u

a !2

xsin ( )

a

B) The time-averagez z-component of the Poynting vector is easily found to be

< Nz > = jH0j2 (u

kg ! x

) sin ( )2 !2 a

which when integrated over the cross sectional area yields

c

Ptrans = < Nz > dx dy = jH0j2 ( a b ) (u

kg !

)0 0 4 !2

The averaged power per unit slice is U = jH0j2 ( a b ) ( ! ) and thus the ratio ofPtrans to U isindeed ugroup as expected.

4 !2

jHzj2the power dissipated atC) At the walls located atx = 0; x = a , we have thatjH~jjj =

these walls is anddUwalls

dz( x = 0; a ) =jH0j2 (

b)

On the other hand at the walls which are located aty = 0; y = b , we havejH~jjjand thus

= jHzj2 +jHxj2

dUwalls ( y = 0; b ) =jH0j2 (b a !

2

) ( 1 + )dz 2 b c


12/16

( !

( 1 c

r2

From these and Ptrans found previously, one arrives at the following conclusion for the attenuation

constant

1K =

C( ) [ + (

!c 2) ]

1

2

!c1

In this expression :

c S0 ! 2 2

!2 )

C is the circumference of the guide, 2 ( a + b) for the guide at hand.

S0 is its cross sectional area, a b for the rectangular wave guide under consideration.

; are dimensionless numbers, equal to aa

b and2 b

respectively for the particular guide.

c is the skin depth at the cut-off frequency !c .a + b

We should point out that the above expression for the attenuation constant is a general result valid for any

wave guide of arbitrary cross sectional shape. Only the values of the parameters C; S0; ; depend

on the particular characteristics of the wave guide under consideration. Especially for the TM modes we

have = 0 .

For large frequenciesKbehaves likeK !1=2, hence larger frequencies result to greater power

losses. For a given geometry the value of the frequency !min minimizingK yields the frequency for

which power losses are thepleast possible. For the TM modes the value of !min is independent of theshape and equal to !min = 3 !c, due to the vanishing of the parameter in these modes.

8. POTENTIALS - RADIATION BY MOVING CHARGES

Knowledge of the radiation emitted by moving electric charges is of utmost importance for particle

accelerator physicists ( and not only!). A relatively easy way to obtain the fields of the moving charges

is through the definition of the potentials that they produce.

We start from the equation r~ B~ = 0 which implies thatB~ = r~ A~ .Usingthisoneobtainsfrom the Faradays lawE~ = r~ @A~ . ; A~ are called the scalarand vectorpotentialsrespectively. These are not uniquely defined

@

.t

Equivalents descriptions can be also obtained if one uses a

new set of potentials 0; A~0 that are related to ; A~

transformations

by the following transformations known as Gauge

A~0 = A~ + r~ ;0

=

@

:@t

In these the function is an arbitrary function of space and time. Exploiting this gauge freedom, one

can choose the potentials in such a way that they satisfy

r~ A~ 1 @c2 @t

= 0 ;

where c2 1 . This is calledLorentz Gauge. In this gauge the potentials satisfy the following wave

equations

1

c2@2

@t2

=

A~ J~:

Given the charge and current densities these can be solved to yield ; A~ and from these one can derive

the Electromagnetic fields. We should perhaps point out that the quartet (1 ; A~ ) is a fourvector

that is transforms like the space - time coordinates (c t ; ~x )underLorentzc

transformations.

For a charge q moving in free space on a given trajectory ~x (t) with velocity ~u (t) the solutions

for the scalar and vector potentials are :

(~x; t ) =0

1

4

q[ ] ;s

A~ (~x; t ) =0

4

q~u[ ]

s


13/16

x , t Observersposition / time

x

Ru

R(tR)

x (tR)

Particle position at

u(tR) Virtual positionof particle at thetime t

retarded time tR

Particle trajectory

Fig. 6: Position parameters for the field of a charge in arbitrary motion.

In these equations the quantity s is given by s = R ~u R~

. These solutions are called Lienard -

Wiechertpotentials.c

Before proceeding we should draw the readersattention to the following points ( See figure 6 ):

t ; ~x refer to the observation time and point.

The symbol [:::] means that the quantities within the brackets are evaluated at the retarded time

tR = t R(tR) = c , that is the time the signal was emitted from the moving charge.

In terms of R~ u , [s] = Ru 1 ( u(tR)2=c2 ) sin2 .

From these potentials one can derive the expressions of the Electromagnetic fields which are givenby

E~ =1fR~u ( 1

u) +

1 R~ (R~u ~u_ )g (32)

4 0 s3 c2 c2

B~ =c R

R~ E~ : (33)

In these equations the quantities s ; ~u ; ~u_ ; R~

The electric component consists of two terms:

are calculated at the retarded time tR.

The first term varies like 1r2

for large distances giving rise to aPoyntingvectorthat behaves

like N~

1=r4

. Therefore the energy flux due to this first term, over a spherical surface of large radiusfalls like N dS 1=r2 and hence vanishes as the radius r tends to 1. In this case we have NoSr

~ ~radiation.

The second term behaves like 1r

at large distances. Unlike the previous case its contribution

to the Poyntingvectoris now N~ 1 and the energy flow over an infinitely distant surface is

nonvanishing. Therefore the second termr

results to Radiation. Notice that this term is absent when

~u_ = 0. Thus only accelerated charges can radiate electromagnetic energy.


14/16

,

E

u

Fig. 7: The Electric field lines of a uniformly moving charge.

8.1 The fields of a uniformly moving charge

For a charge moving with constant velocity u, the electromagnetic fields produced can be calculated

using eqs. (32 , 33). The electric component is given by

E~ (~x; t) =

R~(t) (1 u =c )

2 24 0R(t)3 ( 1 u2=c2 sin2 )

3=2:

In thisR~ (t) is the the vector pointing from the particles true position, at the observation time t, to the

observation point and is the angle betweenR~ (t) and particles velocity ~ . In figure 7 we show the

electric field lines of a uniformly moving charge whose velocity is ~u. We observe that

The transverse componentE~T is larger than its longitudinal component ( the one parallel to ~u )

E~jj .

For ultrarelativistic particlesE~jj

ET1,thatisthefieldisalmostverticaltoitsdirectionofmotion.

For the nonrelativistic particle, u


15/16

In the following we shall discuss two cases which are of of relevance to particle accelerators.

a)Motion of the particle in a linear orbit as it occurs in Linear Accelerators. In this case ~jj ~_ .b)Motion in a circular orbit as it occurs for instance in Circular Accelerators. In that case ~

_.

The power radiated in the two cases a and b can be written, on account of eq. (34), as shown below

Pa =q 2

6 0

q 2

q 2

m2 c3

q 2

dp 2( )dt

2

(35)

Pb =

From eqs. (35) and (36) we see that :

6 0 m2 c32

(dp~

)

dt : (36)

For given magnitude of applied force the radiated power in circular orbit is a factor of 2 larger !

The power emitted is inverse proportional to the mass squared of the radiating particle. Therefore

heavier particles radiate less !

For the linear motion dp=dt = dE=dx where E = m0 c2 is the energy of the charge and

p = m0 u its linear momentum. Since the particle looses energy, in order to maintain the particle

at constant velocity external forces should supply these power losses. The ratio of the power radiated to

the power supplied, for relativistic motion, is found to be

PaG =

1 q2 = m0 c2 dE( ) : (37)

(dE=dt) 6 0 m0 c2 dx

If the energy is supplied by an external electric fieldE, then dE=dx is qE and G is written as2 r0

G =3 m0 c2

( qE ) : (38)

In (38) r0 is the chargesclassicalradiusgiven by r0 q2= ( 4 0 m0 c2 ) . From this equation

we see that the radiation loss will be unimportant unless the gain in energy is of the order of m0 c2 in a

distance of r0. For an electron for instance

r0 = 2:82 10

13

cm ; m0 c

2

= :511 M eV

while typical energy gains are less than 108 eV =meter . Therefore the value of G for an electron is

Gelectron < 1013

that is quite small.

The situation is quite different in circular accelerators. When the charge moves in a circular orbit

the magnitude of the force acting on it is given by

dp~

dt= m0 a =

m0 u2

R

where a = u2=R is the acceleration and R the orbit radius. Then the energy lost per revolution is

U = T Pb =f4

3

3 4gm0 c2 (39)R

where T is the period of the revolution T = 2 R=u . The constant is = E=( m0c2 ) and thus

for relativistic motion we get from eq. (39)

U E4=R :


16/16

,

Thus energy losses in circular orbits are proportional to the fourth power of the energy and inverse

proportional to the radius. For electrons the energy lost per revolution is found to be

Ue (GeV ) = 8:85 105 [E (GeV 4 = R (meters)

which for a circular accelerator of radiusR 4; 500 m that is close to LEP size at CERN, and for beam

energiesE = 1 T eV , yields Ue 2 104 GeV . For protons the corresponding energy loss is

much smaller sincemeUprotons = (

mp

) Ue 1:76 109 GeV :

These energy losses should be compared to the power gained per turn, and the available radiofrequency

to overcome such losses is a dominant factor which should be taken into consideration.

ACKNOWLEDGEMENTS

I would like to thank the organizers for giving me the opportunity to participate and enjoy the pleasant

atmosphere of the school.

References[1] J. D. Jackson, Classical Electrodynamics John Wiley, New York (Second Edition 1975).

[2] W. Panofsky and M. Phillips, Classical Electricity and Magnetism, Addison - Wesley (1964).

[3] J. A. Stratton,Electromagnetic Theory, McGraw - Hill (1941).

[4] C. Prior, Lectures given at the CERN Introductory Accelerator School, OXFORD, September 1998.

Electromiagnetic theory

Documents

Transcript of Electromiagnetic theory