ELECTROMAGNETIC THEORY

Transcription

1 ELECTROMAGNETIC THEORY A. B. Lahanas University of Athens, Physics Department, Nuclear and Particle Physics Section, Athens , 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: ffl Maxwell Equations ffl Conservation of Energy - The Poynting vector ffl Propagation of Electromagnetic (EM) Waves ffl Power Absorption by Conducting Surfaces ffl Propagation of EM Waves in Wave Guides ffl Energy Flow and Power Losses in Wave Guides ffl 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,, 3] ). In writing this lecture I have benefited from lectures given in previous CERN schools ( see [4] ).. MAXWELL EQUATIONS We start by reviewing Maxwell s equations : Electric charges whose density is ρ are the sources of the electric field E. ~ In the MKSA system this is expressed by Gauss s law ~r ~E = 1 ρ: (1) ffl 0 Electric currents with density J ~ = ρ ~u are the sources of the magnetic induction field B ~. This is expressed by Ampere s law ~r ~B = μ 0 ~J : () Field lines of ~ B 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 ~ B through the circuit (Faraday s law ). In differential form this law is expressed by the following formula ~r ~ E ~ B : (4) Within material media having polarization ~ P and magnetization ~ M the above laws still hold with the following replacements ρ ) ρ ~ r ~P ; ~ J ) ~ J + ~ r ~ M ~ P + ffl ~ E : 1

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 ~ ffl0 E ~ + P ~ and H ~ 1 ~ μ 0 B M ~ respectively, Maxwell equations can be brought into the following form ~r ~D = ρ ~r E ~ B ~ ~r H ~ = J ~ D ~ ~r ~B = 0 In some materials ( Linear media ) it happens that ~ D = ffl ~ E; ~ B = μ ~ H; where the quantities ffl; μ are called the dielectric constant and magnetic permeability of the medium respectively..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 = 0 (5) where ρ is the charge density and J ~ = ρ~u is the current density. This equation follows from Maxwell equations and it is not an independent hypothesis. The quantity R J ~ ~ S 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 that r ~ ~J = 0. In this case we say that we have steady currents. For the steady current case the integral H J ~ ds ~ over any closed surface S vanishes. In order to see its S consequence, consider the case of a surface crosing N wires carrying currents I 1 ;I ; :::; 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 + ::: + I N = 0, which is the well known Kirchoff s current conservation law.. The Lorentz force The force acting on a charge e, which is at rest within an electric field, is F ~ = e E ~. Also the force acting on a small wire element dl, ~ carrying electric current I, which is placed in a magnetic field, is ~F = I dl ~ 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 wellknown Lorentz force. For a continuous charge and current distribution, ρ and ~ J = ρ~u, it is convenient to define the force density ~ f, that is force per unit volume. On account of eq. (6) this is given by ~f = ρ ~ 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 distribution is given by f ~ ~u = ρ ( 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 r ~ H ~ ) = E ~D H ~ ~r E ~ + r ( ~ E ~ H) ~ = E ~D + ~B + r ( ~ E ~ H) ~ In Linear Media this takes the + ~ r ~N = ~ J ~E : (9) In (9) the quantities W ; ~ N are the Energy density and the Poynting vector respectively defined by W 1 ( ~ E ~D + ~ B ~H ) (10) ~N ~ E ~ H : (11) Integrating eq. (9) over an arbitrary volume V, whose boundary is S(V ), we get Z Z de V + ~N ds dt ~ = ~J ~E dv : (1) S(V ) In this equation ffl The first term R is the rate of change of the Electromagnetic Energy (EM) energy E V = 1 V ( E ~ ~D + B ~ ~H ) dv within the volume V. ffl The quantity R ~N ds ~ gives the flux of EM energy across the boundary S(V). S(V ) ffl R J ~ ~E dv is the power dissipated, or generated, within the volume V. V As an example, within a conductor of given conductivity ff the current density is J ~ = ff E ~ and therefore R J ~ ~E dv becomes R J dv. Furthermore if the conductor is a wire of constant cross V V ff section S and length L, the electric current is I = J S and the previous integral receives the wellknown form I R, where R is the resistance of the wire element given by R = L ffs. One can easily verify that except its sign, this is indeed the l.h.s. of eq. (1). V 4. PROPAGATION OF ELECTROMAGNETIC WAVES 4.1 Propagation in nonconducting media (ff =0) In a medium with values ffl; μ, for the dielectric constant and the magnetic permeability respectively, we can derive from Maxwell laws the following equations fr g E ~ = J ~ + 1 ffl ~ rρ fr g B ~ = μ 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 velocity u given by r 1 u = fflμ : (14) 3

4 q 1 In vacuo this is usually denoted by the symbol c and has the value c = ffl 0 μ 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. following we shall use complex notation and write the electric component of a plane wave as ~E = ~ E 0 exp i( ~ k ~x!t) : In the 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 by ~ B; ~ B 0 respectively. In this expression ~ E 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 : k is called the wave number and is related to the wave length by the relation k = ß : Using Gauss s, r ~ E ~ = 0, and Faraday s law, r ~ E ~ B ~ = 0, one can immediately arrive at the following relations for the wave number and the amplitudes of the electric and magnetic components: ~ k ~ E0 = 0 ; ~ B0 = 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. k E 0 k B = _ 1 0 k X E0 ω Fig. 1: A plane wave propagating along ~ k. The wave front is the plane formed by the amplitudes ~ E 0 ; ~ B0 of the electric and magnetic fields respectively. 4. Propagation within a conductor (ff 6= 0) Within a conductor the electric current density and the electric field are related by J ~ = ff E ~, from which it follows that r ~ ~J = ff r ~ ~E = ff ffl ρ:then from the continuity equation one + ff ffl ρ = 0 4

5 which is immediately solved to yield ff ρ(~x; t) = ρ(~x; 0) exp ( ff ffl t ) : (16) For good conductors ß sec 1 so that from the eq. (16) we conclude that charges move ffl almost instantly to the surface of the conductor. The ratio fi = ffl is called the relaxation time of the ff conducting medium. For perfect conductors, ff = 1, so that the relaxation time is vanishing. For good, but not perfect, conductors fi is small of the order of sec 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 field E, ~ see eq. (13), becomes fr g E ~ = 0 : Notice the appearance of a friction term which was absent in the free wave equation. If we seek for monochromatic solutions of the form E ~ = E ~ (~x) exp ( i!t), then the equation above takes on the form fr + K g E ~ (~x) = 0 where K = μ!(!ffl + iff). This can be immediately solved to yield, for a plane wave solution travelling along an arbitrary direction ~n, ~E = E ~ 0 e i(ffο!t) e fiο ; (17) where ο ~n ~x. The constants ff ; fi, appearing in (17), have dimensions of length 1 and are functions of ff. Their analytic expresions are not presented here. These can be traced in any standard book of Electromagnetic Theory (see for instance [1,, 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 ff = 0 and ff = k; fi = 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 conductivity is large so that the range of frequencies with ff fl ffl!is quite broad. In this case the constants ff; fi are given by ff ' fi ' ffi 1, where ffi is a constant called the Skin Depth, given by the following expression r ffi = μff! : (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 ο exp( ο ffi ) as it penetrates into the conductor. The depth of the penetration is set by ffi and is smaller the higher the conductivity, the higher the permeability and the frequency. As an example for copper ff = 5: mho m 1 and the skin depth is ffi ' 0: cm for 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 ~H = 1+i p r ff μ! ~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 45 0 from its corresponding electric component E ~, due to the appearance of the prefactor 1+i in eq. (19). 5

6 . A A A T n^ "1" "".. σ =oo H H σ < oo E T E T HT E HT E. Fig. : 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 ".If~n is a unit vector normal to the surface ( with direction from! 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 jj respectively, D 1 T D T = ± ; ~ E 1 jj = ~ E jj ~H 1 jj ~ H jj = ~ K ; B 1 T = B T : (0) In (0) ± is the surface charge density and K ~ is the surface current density. For the derivation of these boundary conditions see [1,, 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 ", is a conductor while the other, 1", is a nonconducting material. In this case direct application of these conditions yields : ffl If " is a perfect conductor ( ff = 1 ) then within it H ~ c ; Ec ~ = 0 in which case on its surface E ~ k ; H T = 0. Thus only E T ; Hjj ~ 6= 0. ffl If " is a good, but not perfect, conductor ( ff = large 6= 1 ) then within it H ~ c ; Ec ~ 6= 0 and attenuated. On its surface E ~ k ; H T 6= 0 but they are much smaller in comparison with their corresponding E T ; Hjj ~ components. In this case only E T, 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.. 6

7 Since H ~ is continuous and j H ~ jj jflh T on the surface, we conclude that within a good conductor (see previous section) ~H c ' H ~ jj exp i ( =ffi!t) exp( =ffi) ; (1) where H ~ 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 r μ! ff ~n ~ H jj : () In eq. () both H ~ jj ; Ejj ~ refer to values on the surface. Ejj ~ is small due to the largeness of the conductivity ff. 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 Re ( ~ E ~ H Λ ) where Re(:::) denotes the real part of the expression (:::) while ~ H Λ stands for the complex congugate of ~H. Therefore the time averaged power absorbed per unit area is dp loss ds = ~n < ~ N > : (3) From this we see that only the component of ~ N normal to the surface is responsible for power losses to the walls. This is given by 1 Re ( ~ E jj ~ H Λ jj ). Then using () we get from (3) dp loss ds = 1 r μ! ff j ~ H jj j : (4) One immediately observes from eq. (4) that ffl For perfect conductors, ff = 1, and no power is absorbed. ffl ~ 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 ffl and magnetic permeability μ. Suppose that the axis of the guide lies along the z direction. Then for monochromatic waves of given frequency! travelling along z we can write ~E(~x; t) = E(x; ~ y) e i ( kg z!t) (5) ~B(~x; t) = B(x; ~ y) e i ( kg z!t) (6) In eqs. (5, 6) the quantity k g is called the wave propagation constant. When these are plugged into the free wave equations they yield ρ E(x; ~ + k kg = 0 (7) ~B(x; y) where k =! μffl! =u. We shall distinguish the following special modes of propagation: 7

8 Fig. 3: A rectangular wave guide. ffl Transverse Electric (TE), in which there is no longitudinal component, E z, of the electric field. Besides having E z = 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 z E z = 0 everywhere ; = S ffl 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 ffl Transverse ElectroMagnetic (TEM) in which both electric and magnetic components are transverse 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 propagation of TEM waves. Any vector field A ~ 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 axis z. Splitting 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 (7), 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!μ ~E t = ( k kg ) ^z r ~ t H z ; Ht ~ = TM modes i!ffl ~H t = ( k kg ) ^z r ~ t E z ; Et ~ = ik g ( k k g ) ~ rt H z ik ( k k g ) ~ rt 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 8

9 modes. In this case E z = 0 and we only need calculate H z. From (7 ) we see that this satisfies the wave equation @y + k t H z = 0 ; ( kt! =u kg ) subject to the appropriate boundary conditions for the TE modes The solutions are: The last relation z H z = H 0 cos ( ßx a k t = ß ( m a! y=0;b ßy m ) cos ( b = 0 : n ) + n ) m; n = Integers : b u = k g + ß ( m + n a b ) 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 r m uß + n a b : Thus in each TE mode, labelled by ( m; n ), we have that! >! c. We further observe that ffl The relation between the wave propagation constant k g uk g! = (1! 1= c! ) and the frequency is < 1 : Thus for a given frequency the wavelength is larger than its free space value. ffl The phase velocity u p =! k g is larger than u, that is larger than its free space value. ffl The group velocity is u group d! dk g = u (1! 1= c! ) This is frequency dependent, therefore the guide behaves like a dispersive medium. The situation is best displayed in figure 4 where we plot the ukg 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 = ß!,is easily found to be given by U = 1 Z ( ffl E ~ ~E Λ + μ H ~ ~H Λ ) dx dy : 4 S 0 In this equation the integration is over the cross-sectional area of the guide, S 0. The x; y axes are vertical to the axis of the guide, and hence parallel to the surface S 0. On the other hand the rate of flow of energy transmitted through this area is Z P trans = < N z > ds S 0 : 9

10 1 u k g ω ω c Fig. 4: uk g =! versus frequency for the various TE modes allowed in the rectangular wave guide. ω where < N z > = 1 Re ( E x Hy Λ E y Hx Λ vector. From these two one can find that ) is the time-averaged z-component of the Poynting that is power is transmitted with the group velocity! P trans = U u group (8) 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 walls ) =P trans : (9) dz In eq. (9) the numerator expresses the power absorbed per unit lenght of the wall which using eq. (4) is found to be du walls dz = 1 r μ! ff Z S c j ~ H jj j ds : In this equation the integration is over a wall stripe S c of unit width. Therefore, using energy conservation, one finds that the power transmitted down the guide at the point having coordinate z + dz is related to the corresponding quantity at z through the relation P trans ( z + dz ) P trans ( z ) = du walls dz dz : (30) The situation is graphically represented in figure 5 where for a slice of width dz, of arbitrary crosssectional 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 10

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 region K turns out to be K ο 10 4! c =c. Thus the power transmitted is decreased to its 30% after ο 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 relating H z (x; y) to the remaining components of E(x; ~ y) ; H(x; ~ y) and using the solutions H z found in the previous section we find in a straightforward manner that ßμ!u E x = 0 ; E y = ih 0 a! c ßk g u H y = 0 ; H x = ih 0 a! c sin ( ßx a ) sin ( ßx a ) B) The time-averagez z-component of the Poynting vector is easily found to be < N z > = μ jh 0 j ( u k g!! c which when integrated over the cross sectional area yields P trans = Z b 0 Z a 0 ) sin ( ßx a ) < N z > dx dy = μ jh 0 j ( ab)( u k g! 4! c The averaged power per unit slice is U = μ jh 0 j ( ab)( 4! ) and thus the ratio of P trans to U is c indeed u group as expected. C) At the walls located at x = 0; x = a, we have that j H ~ jj j = jh z j and the power dissipated at these walls is du walls b ( x =0; a ) = jh 0 j ( dz ffffi ) On the other hand at the walls which are located at y =0; y = b,wehavej H ~ jj j = jh z j + jh x j and thus du walls dz ( y =0; b ) = jh 0 j (! b ffffi )(1 + a b!! c ) ) 11

12 From these and P trans found previously, one arrives at the following conclusion for the attenuation constant r ffl 1 K = ( C )[ο + (! c (! μ ffffi c S 0! )! ] c ) 1 (1! c 1! ) In this expression : ffl C is the circumference of the guide, ( a + b) for the guide at hand. ffl S 0 is its cross sectional area, ab for the rectangular wave guide under consideration. ffl ο; are dimensionless numbers, equal to a a + b and b a + b respectively for the particular guide. ffl ffi 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; S 0 ; ο; depend on the particular characteristics of the wave guide under consideration. Especially for the TM modes we have = 0. For large frequencies K behaves like K ο! 1=, 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! p 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 that B ~ = r ~ A ~. Using this one obtains from the Faraday s law E ~ = rφ A ~. Φ; ~ 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 Φ 0 ; A ~ 0 that are related to Φ; A ~ by the following transformations known as Gauge transformations ~A 0 = A ~ + rλ ~ ; Φ 0 = : In these the function Λ is an arbitrary function of space and time. Exploiting this gauge freedom, one can choose the potentials in such a way that they satisfy where c 1 equations μffl ~r ~A + = 0 ; c. This is called Lorentz Gauge. In this gauge the potentials satisfy the following wave ρ r 1 Φ c A ~ = ρ=ffl μ ~ J Given the charge and current densities these can be solved to yield Φ ; A ~ and from these one can derive the Electromagnetic fields. We should perhaps point out that the quartet ( 1 c Φ ; A ~ ) is a four vector that is transforms like the space - time coordinates (c t; ~x ) under Lorentz 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 : Φ(~x; t ) = ffl ß [ q s ] ; A ~ μ 0 (~x; t ) = 4 ß [ q~u s ] : 1

13 x, t Observer s position / time x R u R(t ) R ψ x (t R) u(t ) R Virtual position of particle at the time t Particle position at retarded time t R Particle trajectory Fig. 6: Position parameters for the field of a charge in arbitrary motion. ~u ~R In these equations the quantity s is given by s = R. These solutions are called Lienard - c Wiechert potentials. Before proceeding we should draw the reader s attention to the following points ( See figure 6 ): by ffl t; ~x refer to the observation time and point. ffl The symbol [:::] means that the quantities within the brackets are evaluated at the retarded time t R = t R(t R ) =c, that is the time the signal was emitted from the moving charge. ffl In terms of R ~ q u, [s] = R u 1 ( u(t R ) =c ) sin ψ. From these potentials one can derive the expressions of the Electromagnetic fields which are given ~E = q 1 4 ßffl 0 s f R ~ 3 u (1 u c ) + 1 R ~ ( Ru ~ c ~u _ )g (3) ~B = 1 R cr ~ E: ~ (33) In these equations the quantities s; ~u ; _ ~u; ~ R are calculated at the retarded time tr. The electric component consists of two terms: The first term varies like ο 1 for large distances giving rise to a Poynting vector that behaves r like N ~ ο 1=r4. Therefore the energy flux due to this first term, over a spherical surface of large radius falls like R ~ S r N ds ~ ο 1=r and hence vanishes as the radius r tends to 1. In this case we have No radiation. The second term behaves like ο 1 at large distances. Unlike the previous case its contribution r to the Poynting vector is now N ~ ο 1 and the energy flow over an infinitely distant surface is r nonvanishing. Therefore the second term results to Radiation. Notice that this term is absent when _~u = 0. Thus only accelerated charges can radiate electromagnetic energy. 13

14 E ψ 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. (3, 33). The electric component is given by ~E(~x; t) = q 4 ßffl 0 ~ R(t) R(t) 3 (1 u =c ) (1 u =c sin ψ) 3= : In this R(t) ~ is the the vector pointing from the particle s true position, at the observation time t, to the observation point and ψ is the angle between R(t) ~ and particle s velocity ~u. In figure 7 we show the electric field lines of a uniformly moving charge whose velocity is ~u. We observe that ffl The transverse component ET ~ is larger than its longitudinal component ( the one parallel to ~u ) ~E jj. E ffl For ultrarelativistic particles ~ jj E T fi 1, that is the field is almost vertical to its direction of motion. q ffl For the nonrelativistic particle, u<<c, the electric field is R ~ 4 ßffl 0. That is its form is exactly R 3 like that of the Coulomb field produced by a point charge. Another way of obtaining the Electromagnetic fields in this case, is by applying a Lorentz transformation L( ~u), specified by velocity ~u, to the fields of a static charge. In that case we need know how the Electromagnetic fields are transformed when passing from one coordinate system to the other. 8. The energy radiated by an accelerated charge For an accelerated charged particle the total amount of energy lost per unit particle s time is found to be P = q 6 ßcffl 0 fl 6 [ _ ~ fi ( ~ fi _ ~ fi ) ] (34) ~fi = ~u=c ; fl = (1 fi ) 1 14

15 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 fi ~ jj fi ~ _. b) Motion in a circular orbit as it occurs for instance in Circular Accelerators. In that case fi ~? fi ~ _. The power radiated in the two cases a and b can be written, on account of eq. (34), as shown below P a = q 6 ßffl 0 q m 0 c3 ( dp dt ) (35) P b = q 6 ßffl 0 q m 0 c3 fl ( d~p dt ) : (36) From eqs. (35) and (36) we see that : ffl For given magnitude of applied force the radiated power in circular orbit is a factor of fl larger! ffl 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 = m 0 fl c is the energy of the charge and p = m 0 ufl 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 G = P a (de=dt) ß 1 q =m 0 c 6 ßffl 0 m 0 c ( de dx ) : (37) If the energy is supplied by an external electric field E, then de=dx is q E and G is written as G = 3 r 0 m 0 c ( q E ) : (38) In (38) r 0 is the charge s classical radius given by r 0 q = (4ßffl 0 m 0 c ). From this equation we see that the radiation loss will be unimportant unless the gain in energy is of the order of m 0 c in a distance of r 0. For an electron for instance r 0 = : cm ; m 0 c = :511 MeV while typical energy gains are less than 10 8 ev=meter. Therefore the value of G for an electron is G electron < 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 d~p dt = m 0 fla = m 0 flu R where a = u =R is the acceleration and R the orbit radius. Then the energy lost per revolution is U = T P b = f 4 ß 3 r 0 R fi3 fl 4 g m 0 c (39) where T is the period of the revolution T = ßR=u. The constant fl is fl = E=( m 0 c ) and thus for relativistic motion we get from eq. (39) U ο E 4 =R : 15

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 U e (GeV ) = 8: [ E (GeV )] 4 =R(meters) which for a circular accelerator of radius R ß 4; 500 m, that is close to LEP size at CERN, and for beam energies E = 1 TeV, yields U e ß 10 4 GeV. For protons the corresponding energy loss is much smaller since U protons = ( m e ) 4 U e ß 1: GeV : m p 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). [] 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