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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 particularly important 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 Guides

 Potentials - 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 Maxwell’s equations :

Electric charges whose density isare the sources of the electric fieldE~. In theMKSAsystem this is expressed by Gauss’s law

~

r

~

E = 1



0

: (1)

Electric currents with densityJ~ =  ~u are the sources of the magnetic induction fieldB~ . This is expressed by Ampere’s law

~

r

~

B = 

0

~

J : (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 fieldB~ through the circuit (Faraday’s law ). In differential form this law is expressed by the following formula

~

r

~

E =

@

~

B

@t

: (4)

Within material media having polarizationP~ and magnetizationM~ the above laws still hold with the following replacements

 ) 

~

r

~

P ;

~

J )

~

J+

~

r

~

M +

@

~

P

+

0

@

~

E

:

(2)

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

0

~

B

~

M respectively, Maxwell equations can be brought into the following form

~

r

~

D = 

~

r

~

E =

@

~

B

@t

~

r

~

H =

~

J +

@

~

D

@t

~

r

~

B = 0

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 charge density andJ~ = ~u is the current density. This equation follows from Maxwell equations and it is not an independent hypothesis.

The quantity

R

S

~

J 

~

dS represents the charge flowing out of surface S per unit time ( this is measured in Amperes in the system MKSA ). If the charge density is time independent then from the continuity equation it follows thatr~ J~ = 0 . In this case we say that we have steady currents. For the steady current case the integral

H

S

~

J 

~

dS over any closed surfaceS vanishes. In order to see its consequence, consider the case of a surface crosing N wires carrying currentsI1

;I

2

;:::;I

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

I

1 +I

2

+:::+I

N

= 0 , which is the well known Kirchoff’s current 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 elementdl~, carrying electric current I, which is placed in a magnetic field, is

~

F = I

~

dl

~

B . These two suggest that for a chargee moving with velocityu, the total force acting on it is

~

F = e(

~

E + ~u

~

B ): (6)

This is the wellknown Lorentz 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)

Using Maxwell equations, and after some trivial mathematical manipulations, the right hand side of this equation can be written as

~

J 

~

E =

~

E(

@

~

D

@t

~

r

~

H )

=

~

E

@

~

D

@t

~

H

~

r

~

E +

~

r(

~

E

~

H)

=

~

E

@

~

D

@t +

~

H

@

~

B

@t +

~

r(

~

E

~

H)

In Linear Media this takes the form

@W

@t +

~

r

~

N =

~

J

~

E : (9)

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

W  1

2 (

~

E

~

D +

~

B

~

H) (10)

~

N 

~

E

~

H : (11)

Integrating eq. (9) over an arbitrary volumeV , whose boundary isS(V), we get

dE

V

dt +

Z

S(V)

~

N

~

dS = Z

V

~

J

~

E dV : (12)

In this equation

 The first term is the rate of change of the Electromagnetic Energy (EM) energy

E

V

= 1

2 R

V (

~

E

~

D +

~

B

~

H)dV within the volumeV.

 The quantity

R

S(V)

~

N

~

dS gives the flux of EM energy across the boundaryS(V).

 R

V

~

J 

~

EdV is the power dissipated, or generated, within the volume V.

As an example, within a conductor of given conductivity  the current density isJ~ =  E~ and therefore

R

V

~

J 

~

E dV becomes

R

V J

2



dV . Furthermore if the conductor is a wire of constant cross section S and length L , the electric current is I = J S and the previous integral receives the wellknown form I2 R , where R is the resistance of the wire element given by R = L

S

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

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

fr 2



@ 2

@t 2

g

~

E = 

@

~

J

@t +

1



~

r

fr 2



@ 2

@t 2

g

~

B = 

~

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~ ; B~ , satisfy the free wave equations. The waves travel with velocityu given by

u = r

1

: (14)

(4)

In vacuo this is usually denoted by the symbol c and has the valuec = 1

00

'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 expi(

~

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!its frequency. This monochromatic pulse is a solution when the frequency is linearly related to the magnitude k j~kj of the wave vector ~k,

! = uk :

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

k = 2

 :

Using Gauss’s, r~ E~ = 0, and Faraday’s law, r~ E~ + @B~

@t

= 0, one can immediately arrive at the following relations for the wave number and the amplitudes of the electric and magnetic components:

~

k

~

E

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

k

in the sense shown in figure 1 .

E

B

k

0

0 = _ ω

1 k X E0

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 ( 6=0)

Within a conductor the electric current density and the electric field are related by J~ = E~ , from which it follows thatr~ J~ = r~ E~ = 



:Then from the continuity equation one has

@

+



 = 0

(5)

which is immediately solved to yield

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





t): (16)

For good conductors 



 10 14

sec

1 so that from the eq. (16) we conclude that charges move 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 the relaxation time is vanishing. For good, 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

fr 2



@ 2

@t 2



@

@t g

~

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

fr 2

+ K 2

g

~

E (~x) = 0

where K2 = ! (! + i ). This can be immediately solved to yield, for a plane wave solution travelling along an arbitrary direction ~n,

~

E =

~

E

0 e

i(  !t)

e 

; (17)

where   ~n~x . The constants ; , appearing in (17), have dimensions oflength 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 =~nk.

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   !is quite broad. In this case the constants

; are given by ' 'Æ 1 , whereÆis a constant called the Skin Depth , given by the following expression

Æ = r

2

!

: (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(



Æ

) as it penetrates into the conductor. The depth of the penetration is set by Æ and is smaller the higher the conductivity, the higher the permeability and the frequency.

As an example for copper  = 5:8107 mho m 1 and the skin depth is Æ ' 0:710 3 cm for a frequency ! = 100MHz .

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

~

H = 1+i

p

2 r



!

~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)

A

^ A

A||

T

H

E E

H

||

T

||

n

. .

.

.

T H

ET

H

ET

|| ||

oo

<

σ σ = oo

"2"

"1"

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~nis a unit vector normal to the surface ( with direction from2!1 ) then from Maxwell equations one can derive the following boundary conditions for the normal ( = vertical to the surface) and the parallel components of the fields involved, denoted by

T and jjrespectively,

D 1

T D

2

T

=  ;

~

E 1

jj

=

~

E 2

jj

~

H 1

jj

~

H 2

jj

=

~

K ; B 1

T

= B 2

T

: (20)

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 where one of the media, say the medium\2", 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

;

~

E

c

= 0 in which case on its surfaceE~k

; H

T

= 0. Thus onlyET ; H~jj

6= 0 .

 If\2" is a good, but not perfect, conductor(  = large6= 1 )then within itH~c

;

~

E

c 6= 0

and attenuated. On its surfaceE~k

; H

T

6= 0 but they are much smaller in comparison with their 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)

SinceH~ is continuous andjH~jj jH

T on the surface, we conclude that within a good conductor (see previous section)

~

H

c '

~

H

jj

exp i(=Æ !t) exp(  =Æ); (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 '

1 i

p

2 r

!



~n

~

H

jj

: (22)

In eq. (22) bothH~jj

;

~

E

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

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 >= 1

2 R e(

~

E

~

H



)

whereR e(:::)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

dP

l oss

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 by1

2 R e(

~

E

jj



~

H



jj

). Then using (22) we get from (23)

dP

l oss

dS

= 1

2 r

!

2 j

~

H

jj j

2

: (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 thezdirection. Then for monochromatic waves of given frequency! travelling alongzwe can write

~

E(~x;t) =

~

E(x;y)e i(k

g

z !t)

(25)

~

B(~x;t) =

~

B(x;y)e i(k

g

z !t)

(26) In eqs. (25 , 26) the quantitykg is called the wave propagation constant. When these are plugged into the free wave equations they yield



@ 2

@x 2

+

@ 2

@y 2

+ k 2

k 2

g



~

E(x;y)

~

B(x;y)

!

= 0 (27)

where k2 = !2   !2 =u2 .

We shall distinguish the following special modes of propagation:

(8)

Fig. 3: A rectangular wave guide.

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

Besides havingEz

= 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

E

z

= 0 everywhere;

@H

z

@n

S

= 0

 Transverse Magnetic (TM), in which case there is no longitudinal component of the magnetic field. In this case we have

H

z

= 0 everywhere; E

z j

S

= 0

 Transverse ElectroMagnetic (TEM) in which both electric and magnetic components are trans- verse to the wave guide axis. Thus

E

z

; H

z

= 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^A

z

;

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 Maxwell’s equations, 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

TEmodes

~

E

t

=

i!

(k 2

k 2

g )

^ z

~

r

t H

z

;

~

H

t

=

ik

g

(k 2

k 2

g )

~

r

t H

z

TMmodes

~

H

t

=

i!

(k 2

k 2

g )

^ z

~

r

t E

z

;

~

E

t

=

ik

(k 2

k 2

g )

~

r

t E

z

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)

modes. In this caseEz

= 0and we only need calculate Hz. From (27 ) we see that this satisfies the wave equation



@ 2

@x 2

+

@ 2

@y 2

+ k 2

t



H

z

= 0 ; (k 2

t

 ! 2

=u 2

k 2

g )

subject to the appropriate boundary conditions for the TE modes

@H

z

@x

x=0;a

=

@H

z

@y

y=0;b

= 0:

The solutions are:

H

z

= H

0 cos(

x

a

m) cos(

y

b n)

k 2

t

=  2

( m

2

a 2

+ n

2

b 2

) m; n = Integers :

The last relation yields

! 2

u 2

= k 2

g + 

2

( m

2

a 2

+ n

2

b 2

)

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

!

c

 u r

m 2

a 2

+ n

2

b 2

:

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

uk

g

!

= (1

! 2

c

! 2

) 1=2

< 1:

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

 The phase velocityup = !

k

g

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

 The group velocity is

u

group

 d!

d k

g

= u(1

! 2

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 theukg

!

as function of the frequencies for the various TE modes allowed. For each mode there is a cut-off frequency!c. The value of ukg

!

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 by

U = 1

4 Z

S

0 (

~

E

~

E



+ 

~

H

~

H



)dxdy:

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

P

trans

= Z

S

< N

z

> dS

(10)

1

ω u kg

ω

ωc

Fig. 4:ukg

=! versus frequency for the various TE modes allowed in the rectangular wave guide.

where< Nz

> = 1

2

R e ( E

x H



y E

y H



x

) is the time-averaged z-component of the Poynting vector. From these two one can find that

P

trans

= U u

group (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  ( dU

wal l s

dz

)=P

trans

: (29)

In eq. (29) the numerator expresses the power absorbed per unit lenght of the wall which using eq. (24) is found to be

dU

wal l s

dz

= 1

2 r

!

2 Z

Sc j

~

H

jj j

2

dS:

In this equation the integration is over a wall stripeScof unit width . Therefore, using energy conserva- tion, one finds that the power transmitted down the guide at the point having coordinatez+dzis related to the corresponding quantity atzthrough the relation

P

trans

(z+dz) P

trans

(z) = dU

wal l s

dz

dz: (30)

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

P

trans

(z) = P

trans (0)e

Kz

(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)

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 beK  10 4 !c=c. Thus the power transmitted is decreased to its

30%after 200 400m.

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;ycomponents of the electric and magnetic fields. B) The power transmitted down the guide. C) The attenuation constantK.

A) From the formulae relating Hz(x;y) to the remaining components ofE(x;~ y) ; H(x;~ y) and using the solutionsHzfound in the previous section we find in a straightforward manner that

E

x

= 0; E

y

= iH

0

! u 2

a! 2

c

sin(

x

a )

H

y

= 0; H

x

= iH

0

 k

g u

2

a! 2

c

sin(

x

a )

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

< N

z

>= jH

0 j

2

( u

2

k

g

!

2! 2

c

) sin 2

(

x

a )

which when integrated over the cross sectional area yields

P

trans

= Z

b

0 Z

a

0

< N

z

> dxdy = jH

0 j

2

(ab)( u

2

k

g

!

4! 2

c )

The averaged power per unit slice isU = jH0j2 (ab) ( !2

4! 2

c

)and thus the ratio ofPtrans toU is indeedugroupas expected.

C) At the walls located atx = 0; x = a, we have that jH~jj j

2

= jH

z j

2 and the power dissipated at these walls is

dU

wal l s

dz

(x=0; a) = jH

0 j

2

( b

 Æ )

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

2

= jH

z j

2

+ jH

x j

2

and thus

dU

wal l s

dz

(y=0; b) = jH

0 j

2

( b

 Æ

)(1 + a

2b

! 2

! 2

)

(12)

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

K = r



 1

Æ

c (

C

S

0

)[ + (

!

c

! )

2

] (

!

!

c )

1

2

(1

! 2

c

! 2

) 1

2

In this expression :

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

 S

0 is its cross sectional area, ab for the rectangular wave guide under consideration.

 ;  are dimensionless numbers, equal to a

a+b

and 2b

a+b

respectively for the particular guide.

 Æ

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

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 like K !1=2, hence larger frequencies result to greater power losses. For a given geometry the value of the frequency !min minimizing K yields the frequency for which power losses are the least possible. For the TM modes the value of!min is independent of the shape and equal to!min

= p

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 equationr~ B~ = 0which implies thatB~ = r~ A~ . Using this one obtains from the Faraday’s lawE~ = r~ @A~

@t

. ; A~ are called the “scalar ” and “vector” potentials respectively. These are not uniquely defined. Equivalents descriptions can be also obtained if one uses a new set of potentials 0;A~0 that are related to ;A~ by the following transformations known as Gauge transformations

~

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

c 2

@

@t

= 0 ;

where c2  1



. This is called Lorentz Gauge. In this gauge the potentials satisfy the following “wave equations”



r 2

1

c 2

@ 2

@t 2

 



~

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

c

 ;

~

A )is a “four vector”

that is transforms like the space - time coordinates(ct; ~x)under Lorentz transformations.

For a chargeqmoving 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

[ q

] ;

~

A(~x;t) =



0

[ q~u

]

References

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