RADIATION OF ELECTROMAGNETIC WAVES

Transcription

1 Chapter RADIATION OF ELECTROMAGNETIC WAVES. Introduction We know that a charge q creates the Coulomb field given by E c 4πɛ 0 q r 2 e r, but a stationary charge cannot radiate electromagnetic waves which are necessarily accompanied by energy flow in the form of the Poynting vector S E H whose magnitude is E 2 /Z or ZH 2. For a stationary charge, the magnetic field is absent. Even a charge drifting with a constant velocity (not speed cannot radiate, since the electric field due to a drifting charge is still of Coulombic nature being proportional to /r 2. Because of energy conservation, the radiation electric field due to a localized source (such as point charge must be proportional to /r, so that the radiation power through a spherical surface with radius r is independent of the radius r, P Z E 2 r 2 dω constant where dω is the differential solid angle. Therefore, the radiation electric field should be inversely proportional to the distance r and it is fundamentally different from the familiar Coulomb field ( /r 2.

2 Electromagnetic radiation in free space occurs when charges are under acceleration or deceleration. In antennas, electrons are forced to oscillate back and forth by a generator, and they are under periodic acceleration and deceleration. In this Chapter, radiation of electromagnetic waves from an accelerated charge will be discussed first. This is followed by analysis on radiation from a macroscopic object (such as antennas in which many charges are collectively involved. In material media, a charged particle can have a velocity larger than the velocity of electromagnetic waves. In this case, Cherenkov radiation, which does not require acceleration on charges, occurs..2 Qualitative Picture of Radiation from an Accelerated Charge Let a charge q be accelerated from rest with an acceleration a (m/sec 2 for a short duration τ. The charge acquires a velocity v a τ after the acceleration, and starts drifting with the velocity. After t seconds, Before and after the acceleration, the electric field due to the charge is of Coulombic nature and radially outward from the charge. However, the electric field lines before acceleration are radially outward from the original stationary position of the charge, while those after acceleration originate from the position at vt (a tt from the origin. From the continuity of the electric flux (Gauss law, the electric field lines before and after acceleration must be somehow connected. The only way to make such a connection is to bend the field line at the radial position r ct, the distance travelled by the disturbance in the electric field lines at the speed of light, c. At the kink, there is indeed an electric field component perpendicular to the distance r, as well as the radial Coulomb field, E c. The tangential component, E t, is the desired radiation electric field E R. The ratio between the two fields is E R E C Since the Coulomb field at the kink is given by E C we find the following for the radiation electric field vt at sin θ sin θ (. c τ c q 4πε 0 (ct 2 q 4πε 0 r 2 (.2 E R qa 4πε 0 c 2 sin θ (.3 r The radiation field is maximum in the direction perpendicular to the acceleration a. Indeed, at θ 0 and π, there are no kinks in the electric field lines. Vectorially, the radiation electric field E R 2

3 due to an acceleration a can be written as E R q 4πε 0 c 2 n (n a (.4 r where n r r is the unit vector in the radial direction. Note that n (n a (n a n (n n a a a a where a is the component of acceleration perpendicular to the radius r. The magnitude of a is a sin θ where θ is the angle between the acceleration and radial vector r. It should be cautioned that the radiation field given in Eq. (.3 is valid only if the charge is nonrelativistic, v c. Also, the acceleration a appearing in Eqs. (.3 and (.4 is the acceleration r/c seconds earlier than the observing time t, because it takes the electromagnetic disturbance r/c seconds to travel over the distance r. The acceleration at t (r/c is called the acceleration at the retarded time and denoted by a ret Similarly, other variables, r and n, are, to be precise, those at the retarded time. If a charge undergoes harmonic oscillation (continuous acceleration and deceleration, the radiation field is also harmonic with the same frequency. (Again, this is valid only in non-relativistic cases. The radiation magnetic field associated with the radiation electric field is perpendicular to both E R and r which is the direction of the Poynting vector or energy flow, B R c n E R (.5 The magnitude of the Poynting vector is S r Z E2 R q 2 a 2 sin 2 θ ε 0 /µ 0 (4πε 0 c 2 r 2 q 2 a 2 sin 2 θ 4πε 0 4πc r 2 (W/m 2 (.6 3

4 Then, the total radiation power can be readily found, P r 2 S r dω (.7 q 2 a 2 4πε 0 4πc 3 sin 2 θdω (.8 where dω sin θdθdφ is the differential solid angle. Performing integration, we find P 2π q 2 a 2 π 4πε 0 4πc 3 sin 3 θdθ dφ 0 0 2q 2 a 2 4πε 0 3c 3 (W (.9 This is known as the Larmor s formula for radiation power emitted by nonrelativistic charge v c under acceleration. Example Short Dipole Antenna In antennas used for broadcasting and communication, a large number of conduction electrons are collectively accelerated by a harmonic generator. In fact, any unshielded transmission lines can effectively become an antenna and they radiate electromagnetic waves. For signal and power transmission purposes, this is an undesirable feature since energy loss inevitably occurs. If, however, a transmission line is carefully shielded except at its end, the open end becomes an effective antenna. At an open end, current standing waves are formed. For an antenna much shorter than the wavelength, the radiation electric field can be found from that due to an accelerated charge E The magnitude of the radiation electric field is q 4πε 0 c 2 n (n a (.0 r E qa sin θ 4πε 0 c 2 r (. where θ is the angle between the radial vector n r/r. Consider a cylindrical conductor of length l ( λ and cross-section A. The total charge in the conductor is denoted by q. If the charge move collectively at a velocity v along l, the current is I qv l 4

5 or qv Il If the velocity and current are oscillating at frequency ω, the time derivative is q dv dt l di dt qa jωil Magnitude-wise, we have qa Iωl that is, the quantity qa can be replaced by Iωl for a short antenna. The radiation power can then be found readily from the Larmor s formula, P The radiation resistance may be defined by 2 q 2 a 2 4πε 0 3 c 3 2 (Iωl 2 4πε 0 3 c 3 6π Z 0 (kl 2 I 2 (W (.2 P R rad I 2, (.3 and in the case of short antenna kl under consideration, it is given by R rad 6π Z 0 (kl 2, (Ω (.4.3 Wave Equations for the Scalar and Vector Potentials Φ and A The four Maxwell s equations E ρ ε 0 or D ρ free (.5 E B t (.6 B 0 (.7 5

6 ( E B µ 0 J + ε 0 t (.8 are suffi cient to describe electromagnetic fields under any circumstances. In electrostatics, a scalar potential Φ was introduced E Φ (static (.9 and in magnetostatics, the vector potential was introduced B A (.20 The static potentials satisfy Poisson s equations 2 Φ ρ ε 0 (.2 whose solutions were Φ (r 4πε 0 A (r µ 0 4π 2 A µ 0 J (.22 ρ (r r r dv (static J (r r r dv (static For time varying sources, the electric field has additional term due to Faraday s law E Φ A t (.23 The curl of the above equation yields E B t since Φ 0 identically. Substituting E Φ A t into E ρ/ε 0, Substituting both B A and E Φ A t into 2 Φ + t A ρ ε 0 (.24 ( E B µ 0 J + ε 0 t we obtain ( A 2 A µ 0 J c 2 Φ t 2 A c 2 t 2 6

7 or 2 A c 2 2 A t 2 µ 0J + ( A + c 2 Φ t (.25 At this stage, let us recall the Helmholtz s theorem: For a vector to be uniquely defined, both its divergence and curl must be specified. The curl of A is B, B A. which satisfies B 0 since A 0 identically. What about the divergence of A? There are no known physical laws to define A and we have a freedom to assign any scalar function for A without affecting the electromagnetic fields E and B. The choice A+ c 2 Φ t 0, (Lorenz gauge (.26 is called Lorenz gauge and the choice A 0, (Coulomb gauge (.27 is called Coulomb gauge. The merit of the Lorenz gauge is that both potentials Φ and A satisfy similar wave equations 2 Φ c 2 2 Φ t 2 ρ ε 0 (.28 2 A c 2 2 A t 2 µ 0J (.29 In Coulomb gauge, such separation cannot be achieved. The scalar potential continues to satisfy the Poisson s equation even in time varying fields, while the vector potential is coupled to the scalar potential 2 ρ (r, t Φ (r, t (.30 ε 0 2 A c 2 2 A t 2 µ 0J + c 2 Φ t (.3 The scalar potential in Coulomb gauge is not subject to retardation and propagation is instantaneous. Of course, all electromagnetic fields must be retarded and this puzzling aspect of Coulomb gauge has been a subject of controversy. The solutions for the wave equations can be constructed as follows. We know the solution for static vector Poisson s equation for the vector potential 2 A µ 0 J (.32 7

8 is A (r µ 0 J (r 4π r r dv (.33 For time varying current J (r, t, the solution for the wave equation 2 A c 2 2 A t 2 µ 0J (.34 introduces retardation, that is, the field to be observed at distance r r from the current is due to the current at the instant t r r c namely, r r /c seconds earlier than the observing time t. Then A (r,t µ J (r, t r r 0 c 4π r r dv (.35 This intuitive solution is in fact correct and agrees with that based on the Green s function for wave equation. Likewise, the solution for the scalar potential is Φ (r, t ρ (r, t r r c 4πε 0 r r dv (.36 Example 2 Find the vector potential and radiation magnetic field due to a short antenna carrying a current I 0 e jωt in z direction. Since the current is in z direction, so is the vector potential. For a filamentary short current, JdV can be replaced by Idz. Then ( A z µ 0 I 0 l r t 4π r ejω c µ 0 I 0 l 4π r ej(ωt kr (.37 provided the antenna length l is much shorter than the wavelength λ, kl. Note that for r l, t r r c t r c + n z c (.38 where l/2 < z < l/2. The factor ω n z c kz cos θ (.39 8

9 is ignorable for short antenna. The magnetic field is B A z µ 0I 0 le jωt 4π µ 0I 0 le jωt 4π ( e jkr r ( jk r + r 2 e z e jkr sin θe φ (.40 In source free region (far away from the antenna, the electric field can be found from E c 2 t B as E r c2 jω µ 0 I 0 le j(ωt kr 4π Z 0 I 0 e j(ωt kr 2π ( jk 2 r 2 + r 3 cos θ ( r 2 j kr 3 cos θ (.4 Derivation is left for exercise. I 0 le j(ωt kr ( k E θ Z 0 j 4π Z 0 I 0 le j(ωt kr 4π r j r 2 kr 3 ( jk r + r 2 j kr 3 sin θ sin θ (.42 In both the magnetic field and electric field, the dominant radiation terms are those proportional to /r. Higher order terms proportional to /r 2 and /r 3. do not contribute to radiation of energy. However, in radiation of angular momentum, the radial electric and magnetic field proportional to /r 2 play a crucial role as we will see later..4 Lienard-Wiechert Potentials for Single Charge and Consequent Fields If the particle velocity is large and approaches the speed of light, the radiation electric field in Eq. (.4 will be modified significantly. To see how relativistic effects modifies the radiation field, we need to find how the scalar and vector potentials are affected by relativistic velocity. As a preparation, let us first convince ourselves that a rod moving toward (away from us appears longer 9

10 (shorter than its actual length l. (This has nothing to do with the celebrated relativistic length contraction, which was formulated by Lorentz well after the work by Lienard-Wiechert. This is because light emitted from the rear end of the rod takes a longer time than that emitted from the front end to reach an observer, and for an observer to be able to measure the length of the rod, he/she needs the two signals arriving at the same instant. Let the rod move toward an observer with a velocity v. Light emitted by the front end at the instant when the front end is at a distance x from the observer reaches the observer after x/c sec. Light emitted by the rear end at the same instant reaches the observer at (x + l/c sec, that is l/c sec later. If we denote the apparent length seen by the observer by l, the extra time needed for the light leaving the rear end is During this interval, the rod has moved a distance l l with a velocity v. Therefore, l c l c l l v. Solving for l, we find l l (.43 v c In general, if an object is moving with a velocity v, its dimension toward an observer appears to change by a factor n β (.44 where β v/c, and n is the unit vector toward the observer. If the object has a volume dv, the apparent volume is dv dv n β (.45 For a charge density ρ, the apparent differential charge is therefore dq ρdv n β dq n β (.46 The current density J ρv is also modified as JdV JdV n β (.47 0

11 c τ E R E C q vt a τt r ct θ z Figure.: The Coulomb field E C which is in radial direction and the radiation field E R due to an accelerated charge which is transverse to the radius r. The asymmetric shell radially expands at speed c. A v B pulse A pulse B l l' Figure.2: A rod of length l moving toward an observer appears to be longer l should not be confused with relativity effect. l β This

12 Therefore, the scalar and vector potentials due to a charge moving at velocity v (t are to be evaluated according to Φ q 4πε 0 κ r r p (t A µ 0 qv (t 4π κ r r p (t (.48 (.49 where r p (t is the instantaneous location of the charge at the retarded time and v (t dr p (t /dt is the particle velocity at the retarded time t. The dimensionless quantity κ κ ( t n β (.50 also depends on time t. The potentials given in Eqs. (.48,.49 are called the Lienard-Wiechert potentials which were formulated in The electromagnetic fields to be derived from these potentials properly contain all relativistic effects. It should be noted that all variables in those potentials are to be evaluated at the retarded time which is related to the observing time t through t t r r p(t c (.5 where r (t is the instantaneous position of the charge. For example, the gradient operation with respect to the observing coordinates r is to be performed as follows: R + t t R + t t t t (.52 where R r r p (t, R R, n R/R. Similarly, the time derivative follows the chain rule, t t t t (.53 where In this derivation, note that t t t + n v c ( t r r p(t c t t v(t r t 2

13 Then, we find an important relationship between t and t, t t n β (.54 Figure.3: The change in the unit vector n is caused by the perpendicular velocity v. The gradient of the retarded time t can be calculated similarly, ( t t r r p (t c c r r p ( t n c + (n β t from which it follows t n n β c n cκ (.55 The transformations in Eqs. (.54 and (.55 allow us to evaluate the electric and magnetic fields due to a moving charge. For example, the electric field is to be found from E Φ A t (.56 3

14 The gradient of the scalar potential becomes ( q Φ R + t 4πε 0 t n β R ( ( q [ R + t 4πε 0 n β R t q β n(n β 4πε 0 ( n β 2 R n 2 n β q 4πε 0 R 2 n β ] R n c ( n β [ { β n n(n β n n ( n β 2 R 2 ( n β 3 cr t β + n β }] t n t β + n β t ( n β 2 R + n β However, the time variation of the unit vector n R/R can be caused only by the velocity component perpendicular to n, as seen in Fig..3. From the two similar triangles in the figure, we find or dn v dt R n t v dt R Therefore, Φ q [ (n β ( n β + n ( n β (n β nβ 2 4πε 0 where The time derivative of the vector potential is κ 3 R 2 + n cκ 3 R n v R 2 (.57 (.58 ( n β ] (.59 β β t (.60 A t t A t t µ 0q ( v 4π κ t ( κr q β 4πε 0 cκ κr β (κr 2 t (κr [ ] q β 4πε 0 cκ κr + κ 2 R β(n β cβ (κr 2 (β2 n β (.6 where µ 0 has been eliminated in favour of ɛ 0 through c 2 /ɛ 0 µ 0. Substituting Eqs. (.59 and (.6 into (.56, we thus find E (r, t q [ 4πε 0 κ 3 R 2 (n β ( β 2 ] ret + q [ [ ] ] 4πε 0 cκ 3 R n (n β β ret (.62 4

15 and the vector identity A (B C B (A C C (A B (.63 has been used. Eq. (.28 is the desired general expression for the electric field due to a moving charge, and valid for arbitrarily large particle velocity. The electric field above was first formulated by Heaviside. [ ] ret means that all quantities in the brackets are retarded, that is, for evaluation of the electric field at time t, [ ] evaluated at t t c r rp ( t should be used. The reason the formula, Eq. (.28, remains valid at relativistic velocities even though it has been derived from the Lienard-Wiechert potentials formulated before the discovery of the relativity theory (Einstein 905 is due to the obvious fact that electromagnetic waves propagate at the speed c irrespective of the source speed once they are emitted. Sound waves in air also propagate with the sound velocity after being emitted irrespective of source speed. A major difference between electromagnetic waves in vacuum and sound waves in air is that the speed of electromagnetic waves remains c even when the observer is moving, while the sound speed appears to change if the observer is moving relative to the wave medium, that is, the air. The magnetic field is to be calculated from B A c n E (.64 where E is the electric field given in Eq. (.28. Derivation of Eq. (.32 is left for an exercise. The electric field in Eq. (.28 has two terms. The first term is inversely proportional to R 2, and does not contain the acceleration β. This is essentially the Coulomb field corrected for relativistic effects (β. The second term is inversely proportional to R and proportional to the acceleration β. This term is the desired radiation electric field. Note that at large R, the radiation field ( /R becomes predominant over the Coulomb field ( /R 2. In nonrelativistic limit, β, we recover the radiation electric field worked out in Eq. (.4 from qualitative arguments where all quantities, n, R, v are to be evaluated at the retarded time. We will return to radiation problems associated with relativistic particles in Sec..5. 5

16 .5 Radiation from a Charge under Linear Acceleration If the acceleration is parallel (or anti-parallel to the velocity, β β, the particle trajectory remains linear, as in linear accelerators. (The reason high energy electron accelerators are linear rather than circular as for proton accelerators is because radiation loss in circular electron accelerators becomes intolerably large. The radiation electric field in this case is given by ( E (r, t q n n β 4πε 0 cr κ 3 t (.65 where all quantities at the retarded time should be used. If the angle between β and n is θ, the Poynting flux is S (r, t Z 0 E (r, t 2 4πε 0 r 2 q 2 2 β sin 2 θ 4πc ( β cos θ 6 t (.66 If the Poynting flux is integrated over the spherical surface of radius r, one gets a radiation power at the observing time t. However, what is more meaningful is the radiation power at the retarded time. Since the Poynting flux at the retarded time t is dt dt β cos θ Integration over the solid angle yields S ( r, t ( β cos θ S (r, t q 2 2 β sin 2 θ 4πε 0 r 2 4πc ( β cos θ 5. (.67 P ( t 2 q 2 β 4πε 0 4πc 2π 2 4πε 0 3 π 0 sin 2 θ 5 sin θdθ ( β cos θ q 2 β2 γ 6 (.68 c where γ β 2 (.69 6

17 E r q θ v, dv/dt Figure.4: Linear acceleration β β with nonrelativistic velcoity β. In highly relativistic case, γ, the radiation is emitted predominantly along the beam velocity. (Fig. 7 is the relativity factor. Relevant integral is x 2 ( βx 5 dx 4 ( 3 β 2 3 (.70 The radiation power is independent of the particle energy γmc 2 since the parallel acceleration is inversely proportional to γ 3 as can be seen from the equation of motion mc d dt β mc β 2 β β 2 F β 2 β + ( β 2 mcγ 3 β F 3/2 F Therefore, the radiation power in Eq. (.68 is independent of the particle energy. This is the main advantage of linear electron accelerators. The angular distribution of the radiation power is proportional to the function f (θ sin 2 θ ( β cos θ 5 (.7 In nonrelativistic limit β, the radiation intensity peaks in the direction θ π/2 (perpendicular to the velocity β and acceleration β. In relativistic case β, γ, the radiation intensity profile becomes narrow with an angular spread about the velocity of order θ /γ. The angle at which the radiation intensity peaks can be found from df (θ dθ 0 7

18 which yields In the limit β (γ, 3β cos 2 θ + 2 cos θ 5β 0 cos θ ( 5β 2 + 3β cos θ 3β ( 3 ( 5β ( γ 2 + Since θ, cos θ θ 2 /2, we find 5 8γ 2 5 θ 4 γ The radiation is essentially confined in the angle θ /γ about the velocity vector. This alignment with the velocity is in fact independent of the acceleration direction and θ /γ about the velocity holds even for acceleration perpendicular to the velocity. sin 2 θ ( 0. cos θ 5 y x.0 sin 2 θ ( 0.9 cos θ 5 8

19 y x β 0.9. sin 2 θ ( 0.99 cos θ 5 y e+6 e+6 5e+6 e+7.5e+7 x Figure.5: Radiation pattern when, from top, β 0., β 0.90 and β Note the large radiation intensity as β appoaches unity...6 Radiation due to Acceleration Perpendicular to the Velocity β β In this case, the radiation electric field is E (r, t e [ ] 4πε 0 cκ 3 r n (n β β (.72 9

20 We consider a charge undergoing circular motion with radius ρ and normalized velocity β in the x z plane. At t 0, the charge passes the origin. At this instant, β βe z and β β e x Substituting e x x (r sin θ cos φ sin θ cos φn + cos θ cos φe θ sin φe φ (.73 and e z (r cos θ cos θn sin θe θ (.74 into n [(n βe z β ] e x we obtain and n [(n βe z β ] e x β [(β cos θ cos φeθ + ( β cos θ sin φe φ ] (.75 The retarded differential power is E (r, t e β 4πε 0 cκ 3 r [(β cos θ cos φe θ + ( β cos θ sin φe φ ] (.76 dp (t dω r 2 Z 0 E 2 β 2 e2 4πε 0 4πc [ ( β cos θ 5 ( β cos θ 2 ] γ 2 sin2 θ cos 2 φ (.77 and the radiation power is P ( t dp (t dω dω e2 β 2 γ 4 (.78 4πε 0 4πc 20

21 Relevant integrals are dx ( βx 3 2 ( β 2 2 2γ4 (.79 x 2 ( βx 5 dx 4 3 ( β γ6 (.80 Example: Radiation power emitted by 3 GeV electron undergoing circular motion with radius R 0 m. The relativity factor is The acceleration is Then the radiation power is P γ E mc 2 3 GeV 0.52 MeV 5900 a v2 R c2 R 9 05 m/s 2 e2 β 2 γ 4 4πε 0 4πc e 2 a 2 4πε 0 4πc 3 γ 4 ( ( π ( ( W ev/s/electron Electron rapidly loses its energy to radiation (synchrotron radiation. To reduce radiation power, the orbit radius R must be increased and γ must be decreased. Circular particle accelerators are therefore practical only for protons. (For proton energy of 00 GeV, the relativity factor is rather mild, γ Synchrotron Radiation Charged particles emit radiation whenever they are subjected to acceleration. Synchrotron radiation is emitted by relativistic electrons. The classical radiation mechanism is simply bending electron trajectory by a magnetic field. The acceleration due to trajectory bending is a c2 R 2

22 where R is the curvature radius of the trajectory. The radiation power due to perpendicular acceleration is with peak frequency components around where P 2 4πε 0 3 (ea 2 c 3 γ 4 ω γ 3 eb γm e γ 2 eb m e ω ce eb γm e is the relativistic cyclotron frequency. If B 0.5 T and γ 5000, the dominant radiation frequency is Hz. Modern synchrotron light sources are equipped with wigglers and undulators to cover wider radiation spectrum and provide higher radiation intensities. In wigglers, pairs of NS magnets are placed periodically and electron beam going through such structure experiences periodic kicks in the direction perpendicular to the beam. The wavelength of emitted radiation is approximately given by λ λ w 2γ 2 where λ w is the spatial period of the wiggler. In contrast to the radiation by bending magnet, wiggler radiation is a result of maser or laser action, namely, amplification of electromagnetic waves in an electron beam. Wiggler radiation is thus more coherent than bending magnet radiation..8 Radiation by Macroscopic Sources (Antennas, Apertures Radiation by antennas can be analyzed by solving the wave equation for the vector potential A ( 2 c 2 2 t 2 A µ 0 J (.8 provided the Lorenz gauge is adopted, A + c 2 Φ t 0 (.82 The scalar potential Φ obeys a similar wave equation ( 2 c 2 2 t 2 Φ ρ (.83 ε 0 22

23 Figure.6: In a wiggler, an electron beam is modulated by a periodic magnetic field. Electrons acquire spatially oscillating perpendicular displacement x(z and velocity v x (z which together with the radiation magnetic field B Ry produces a ponderomotive force v x (z B Ry (z directed in the z direction. The force acts to cause electron bunching required for coherent radiation (as in lasers. Since the charge density ρ and the current density J are related through the charge conservation law, ρ t + J 0 (.84 for a given current J, the charge density can be found in principle and the scalar potential found from the wave equation should be consistent with that calculated through the Lorenz condition in terms of the vector potential A. In practice, finding the vector potential alone should be suffi cient, since all electric field and magnetic fields can be derived from the vector potential as follows. The magnetic field is B A (.85 In source free region (ρ 0, J 0, yields the electric field B c 2 E t (.86 E c2 jω B c2 jω A j c2 ω [ ( A 2 A] (.87 Since in the Lorenz gauge, A + c 2 Φ t 0 23

24 and in source free region, the vector potential satisfies the wave equation ( 2 c 2 2 t 2 A 0 (.88 the electric field above reduces to E Φ A t.8. Short Dipole Antenna (revisit In Section.2, we applied the Larmor s formula to calculate the radiation power emitted by a short dipole antenna kl 2πl. In this case, electrons oscillate up and down collectively without λ any phase difference along the antenna. Here we directly solve the wave equation for the vector potential to see if the same result can be recovered. We assume an antenna of length l carries a current I 0 e jωt in z direction. The vector potential has only z component since the current density is unidirectional in z direction. The wave equation for A z ( 2 c 2 2 can be integrated as t 2 A z (r, t µ 0 4πr A z µ 0 J z e jωt (.89 ( J z r e jω(t r r c dv (.90 where retardation due to finite propagation velocity c is taken into account. position r is much larger than the antenna length and we have The observation A z (r, t µ 0 4πr ej(wt kr I 0 l (.9 where for a filamentary current, JdV has been replaced with Idl. The radiation magnetic field is B A jk A j µ 0I 0 l 4πr ej(wt kr ke r ( sin θe θ + cos θe r j µ 0I 0 l 4πr ej(wt kr k sin θe φ (.92 The Poynting vector is S r Z 0 H 2 Z 0 (I 0 l 2 6π 2 r 2 k2 sin 2 θ (.93 24

25 daz r z'cosθ feeder I(z'dz' z' θ I cos(kz' 0 r Figure.7: Center-fed half wavelength dipole antenna. and the radiation power is P r 2 S r dω (I 0 l 2 π 2π Z 0 6π 2 k2 sin 3 θdθ dφ 0 0 µ0 /ε 0 (kl 2 I0 2 (W (.94 6π in agreement with the earlier result based on equivalent acceleration. Remember this is subject to the condition of short antenna, kl. In practice, a stand alone current segment cannot exist physically. What is happening is that the charge oscillates along the antenna and at the top and bottom, opposite charges (dipole appear to satisfy charge conservation law. Charges create scalar potential Φ. However, as long as the radiation power is concerned, the scalar potential does not contribute and can be ignored. In analyzing fields near the antenna, scalar potential does play roles as we will see in the case of half wavelength dipole antenna..8.2 Half Wavelength (λ/2 Dipole Antenna Figure.7 shows the case of center-fed half wavelength long (l λ/2 antenna. Since the antenna length is comparable with the wavelength, the radiation field (vector potential should be calculated taking into account the phase difference of the antenna current. To find the radiation field, we 25

26 assume the following standing wave, I (z, t I 0 cos (kz e jωt, λ 4 < z < λ 4. (.95 The retarded radiation vector potential at kr is A z (r, θ µ 0 I 0 e j(ωt kr λ/4 4π r µ 0 4π and the radiation magnetic field is µ cos 0I 0 2πr λ/4 λ/4 I 0 e j(ωt kr 2 r 0 ( π 2 cos θ k sin 2 θ H µ 0 jk A z e jkz cos θ cos kz dz cos ( kz cos θ cos ( kz dz e j(ωt kr (.96 which yields The radiation Poynting flux is ( π H φ j I cos 0 2 cos θ e j(ωt kr (.97 2πkr sin θ and the radiation power is S r Z 0 H 2 ( π I0 2 cos2 Z 2 cos θ 0 (2πr 2 sin 2, W/m 2 (.98 θ I 2 π 0 P Z 0 (2π 2 0 I 2 π 0 Z 0 2π 0 cos 2 ( π 2 cos θ sin θ ( π cos 2 2 cos θ sin θ 2π dθ dφ 0 dθ (.99 The integral is approximately.22 and we find the radiation resistance of the half wavelength dipole antenna, R rad Z Ω (.00 2π If the Poynting flux on the antenna surface is directly integrated, the reactive power can be estimated as well. When the antenna length is λ/2, the reactance is about j40 Ω (inductive. However, it sensitively depends on the length. For l 0.49 λ, the reactance vanishes. 26

27 Figure.8: The Poynting flux on the surface of λ/2 antenna of finite radius a is approximated by that on the cylindrical surface of radius a surrounding a thin antenna. The calculation presented above entirely ignores the reactive power which may exist due to storage of electric and/or magnetic energy in the vicinity of the antenna. To account for such reactive power, we must deviate from the far-field analysis and integrate the Poynting flux directly on a surface close to the antenna surface. Instead of calculating the Poynting flux on the antenna surface of finite radius a, we calculate the Poynting flux on a cylindrical surface of radius a surrounding an ideally thin antenna of length λ/2 as shown in Fig..8. In this approximation, the magnetic field on the antenna surface may be replaced by the static form without retardation, H φ (z I(z 2πa I 0 2πa cos(kz, λ 4 < z < λ 4. (.0 The electric field on the antenna surface is zero within our assumption of ideally conducting antenna except at the gap at the midpoint. However, the electric field due to the current filament assumed at the axis is finite at the surface a distance a away. It can be calculated from c 2 t E B ( A 2 A, (.02 where A is the vector potential on the surface. It can be found from the integral, A z (z µ λ/4 0 I 0 cos kz 4π λ/4 R(z, z e jkr(z,z dz, (.03 27

28 where R(z, z (z z 2 + a 2, (.04 is the distance between a point on the surface (ρ a, z and a current segment I(z dz at z. With this approximation, Eq. (.02 reduces to j ω c 2 E z(z 2 A z z 2 + k2 A z, (.05 since in current-free region the vector potential satisfies the Helmholtz equation ( 2 + k 2 A z 0. (.06 The integral in can be performed by noting E z (z j c2 µ 0 I 0 4πω λ/4 λ/4 e jkr(z,z z R(z, z and by integrating by parts twice with the result E z (z j Z ( 0I 0 e jkr 4π where R ( cos(kz 2 e jkr(z,z z 2 + k2 R(z, z dz, (.07 z e jkr(z,z R(z, z, (.08 R ( z λ 2 + a 4 2, R 2 + e jkr2, (.09 R 2 ( z + λ 2 + a 4 2. (.0 The radial outward Poynting flux on the antenna surface is therefore given by S ρ E z Hφ j Z 0I0 2 ( e jkr 8π 2 a R + e jkr2 cos(kz, (. R 2 and the power leaving through the antenna surface is λ/4 P 2πa j Z 0I 2 0 4π λ/4 λ/4 λ/4 S ρ dz ( e jkr R + e jkr2 cos(kzdz. (.2 R 2 The integral has to be performed numerically. Introducing x 4z λ, A 4a λ, 28

29 we rewrite the integral in the form ( j f(a Re exp π ( x A 2 + ( x 2 + A 2 ( exp j π ( + x A 2 cos ( + x 2 + A 2 ( π 2 x dx (.3 which is shown in Fig..9 as a function of the normalized antenna radius A 4a/λ. For A < 0.0, y x 2 3 Figure.9: Real part (red and imaginary part (black of the integral in Eq..3 as functions of A 4a/λ. the radiation impedance is constant and approximately equal to Z rad j42.0 (Ω which is inductive. The real part agrees with the radiation resistance calculated earlier. The reactive part of the impedance is inductive due to dominant magnetic energy compared with the electric capacitive energy. However, the reactance is a very sensitive function of the antenna length. It vanishes if the antenna length is chosen at l 0.49λ and further decrease in the length makes the reactance capacitive. Radiation from a center-fed antenna can be analyzed by assuming a standing wave form, I(z I 0 sin[k(l z ], and is left for exercise. 29

30 The axial electric field E z (z in Eq. (.09 can be alternatively (perhaps more conveniently found from E z z Φ t A z, (.4 where Φ is the retarded scalar potential given by Φ (z e j r r 4πε 0 r r ρ ( r dv e jr 4πε 0 R ρ ( l z dz, (.5 with R (z z 2 + a 2, and ρ l the linear charge density that can be found as ρ l t + I (z z 0, ρ l (z j c I 0 sin kz, (C m. (.6 Then, E z z Φ t A z ji 0 4πε 0 c ji 0 4πε 0 c j Z 0I 0 4π λ/4 λ/4 z ( e jkr R ( e jkr R ( e jkr R + e jkr2 R 2 + e jkr2 R 2 sin ( kz dz jωµ 0I 0 4π λ/4 λ/4 cos kz R e jkr dz, (.7 which agrees with Eq. (.09. In radiation zone, the scalar potential is irrelevant but in near field region, it should be considered together with the vector potential in a self consistent manner. The radiation impedance Z rad can be defined by Z rad j Z λ/4 ( 0 e jkr 4π λ/4 R + e jkr2 cos(kzdz (.8 R 2 Remember that this is for a center-fed λ/2 antenna. For an antenna of arbitrary length 2l with a current standing wave I (z I m sin [k (l z ] (.9 30

31 the impedance is modified as Z rad j Z ( 2 l 0 Im 4π I (0 j Z 0 4π sin 2 (kl l l l ( e jkr R ( e jkr R + e jkr2 R 2 + e jkr2 R 2 2 cos (kl e jkr R 2 cos (kl e jkr R sin [k (l z ] dz sin [k (l z ] dz (.20 where R,2 (z l 2 + a 2, R z 2 + a 2 (.2 Note that the current seen by the generator is I (0 I m sin (kl..9 Radiation by Small Sources (Multipole Radiation The retarded vector potential due to a nonrelativistic (β, κ charged particle A (r, t µ 0 4π ev(t r r p (t (.22 ret can be generalized for a collection of moving charges as ( A (r, t µ J r, t r r 0 c 4π r r dv (.23 where J (r, t is the current density. If the current is oscillating at ω, we have A (r, t µ 0 4πr ej(ωt kr where the following approximation is used, ( e jω t r r c e j(ωt kr+jk r J ( r e jk r dv (.24 Note that k r r kr k r, r r If the radiation source is small compared with the wavelength λ, that is, if ωr /c, or kr Eq. (.24 may be approximated by A (r, t µ 0 4πr ej(ωt kr J ( r ( + jk r 2 ( k r 2 dv (.25 3

32 Note that in the limit of dc (or very low frequency current (ω 0, k 0, Eq. (.25 does reduce to the vector potential in magnetostatics. By assumption, k r. As we will see, each term in this series expansion can be identified as electric and magnetic multipole radiation fields. The lowest order radiation vector potential is A (r, t µ 0 4πr ej(ωt kr J ( r dv (.26 The integral of the current density can be calculated as follows. The x component of the integral is J x dv x JdV (xj dv x JdV (.27 The first integral vanishes, (xj dv xj ds 0 because the closed surface S is at infinity where all sources vanish. Then J x dv S x JdV x ρ t dv d dt p x where p x xρdv is the x component of the electric dipole moment. In general, JdV d dt p (.28 The term of next order J (r (k rdv (.29 can be calculated in a similar manner. J (r (k rdv r (k J dv k r JdV and r (k J dv k k d dt rr ρ t dv (k r JdV Q (k r JdV 32

33 we find where J (r (k rdv k m + 2 k d Q (.30 dt Q rrρdv (.3 is the quadrupole moment and m 2 r JdV (.32 is the magnetic dipole moment. Therefore, to order kr, the vector potential is given by and the radiation magnetic field H is A (r,t µ ( 0 4πr ej(ωt kr ṗ jk m + j 2 k d Q dt (.33 H µ 0 A j µ 0 k A 4πcr ej(ωt kr ( p n+ c n (n m 2c n ( n... Q (.34 where n r r is the radial unit vector. Example 3 Radiation by a Nonrelativistic Charge undergoing Circular Motion: Electric dipole radiation We assume a charge q is undergoing circular motion with a radius a and angular frequency ω, as shown in Fig.. The dipole moment is p (t qa (cos (ωt e x + sin (ωt e y Then the radiation magnetic field is H (r, t 4πcr ej(ωt kr p n ω2 4πcr ej(ωt kr n p 33

34 z x ωt a q y Figure.0: Charge in circular motion. and the radiation power is given by P Z 0 r 2 H 2 dω 2 4πε 0 3c 3 ω4 p 2 2 4πε 0 3c 3 p 2 (.35 The radiation magnetic field may be written as H (r, t ω2 4πcr ej(ωt kr n p H (r, t j aqω2 4πcr ej(ωt kr 2 (e θ j cos θe φ (.36 The field is plane polarized at θ π/2 and circularly polarized at θ 0. The radiation magnetic filed due to an electric dipole is proportional to p, that is, acceleration of the charge. Example 4 Radiation by a Small Current Loop: Magnetic dipole radiatio If the loop radius is a (ka, the magnetic dipole moment is m z πa 2 I 0 e jωt The radiation magnetic field is H 4πc 2 r ej(ωt kr n (n m ω2 4πc 2 r ej(ωt kr n (n m 34

35 The radiation power is P r 2 Z 0 H 2 dω Z 0k 4 ( πa 2 2 (4π 2 I 0 Z 0 I0 2 (ka 4 6π sin 2 θdω The radiation resistance is (ka 4 R rad Z 0 6π This is smaller than the radiation resistance of electric dipole antenna of length l ( λ (kl 2 R rad Z 0 6π if l and a are comparable..0 Radiation of Angular Momentum The Poynting vector S E H (.37 is the energy flux density. Since the electromagnetic energy is carried at the velocity c, the momentum flux density may be defined by and the momentum density by Similarly, the angular momentum flux density is given by and the angular momentum density by c E H (.38 c 2 E H (.39 r (E H (.40 c c 2 r (E H ε 0r (E B ε 0 r (E ( A (.4 35

36 The vector E ( A can be expanded as E A E i A i (E A (.42 and we have r (E B r (E i A i r [(E A] (.43 However, r [(E A] i (E i r A E A ( E (r A i (E i r A E A since E 0 in source free region. Then the total angular momentum is ε 0 r (E BdV ε 0 r (E i A i dv + ε 0 (E AdV (.44 where use is made of i (E i r A dv r A(E ds 0 It is evident that in Eq. (.44 the first term containing the factor r can be identified as the orbital angular momentum. Then the last term can be interpreted as the spin angular momentum. Consider a circularly polarized plane wave propagating in the z direction. If the field has positive helicity, the electric fields components are Corresponding vector potentials are The spin momentum density is For negative helicity wave, E x (z, t E 0 cos (ωt kz E y (z, t E 0 sin (ωt kz A x (z, t E 0 ω A y (z, t + E 0 ω sin (ωt kz cos (ωt kz ε 0 E A ω ε 0E 2 0e z (.45 E x (z, t E 0 cos (ωt kz E y (z, t E 0 sin (ωt kz 36

37 A x (z, t E 0 sin (ωt kz ω A y (z, t E 0 ω cos (ωt kz the spin direction is reversed, ε 0 E A ω ε 0E 2 0e z (.46 as expected. Example 5 Radiation of angular momentum by an electric dipole Electric multipoles radiate Transverse Magnetic (TM modes having no radial component of magnetic field, H r 0. Then the angular momentum flux density becomes c r (E H (r E H c The radiation vector potential of an electric dipole is Corresponding magnetic field is where A µ 0 4πr ej(ωt kr ṗ (.47 H A j ej(ωt kr k ṗ µ 0 4πr ej(ωt kr n p (.48 4πcr jk n jω c n c t The electric field can be found from the Maxwell s equation ε 0 µ 0 E t B ( A ( A 2 A ( A + k 2 A A is A µ 0 4π µ 0 4π ( d e j(ωt kr n ṗ dr r ( j kr r 2 e j(ωt kr n ṗ 37

38 Then the radial component of the electric field is The rate of angular momentum radiation is When applied to a charge undergoing circular motion, E r ej(ωt kr 2πε 0 cr 2 n ṗ (.49 r2 (r E H dω c 8π 2 ε 0 c 3 (n ṗ (n p dω (.50 p x eρ cos (ωt, p y eρ sin (ωt we find n ṗ eρω sin θ sin (φ ωt and e z (n p eρω 2 sin θ sin (φ ωt Then dl dt 8π 2 ε 0 c 3 e2 ρ 2 ω 3 8π 2 ε 0 c 3 e2 ρ 2 ω 3 8π 2 ε 0 c π (n ṗ (n p dω sin 2 θ sin 2 (φ ωt dω e2 ρ 2 ω 3 6πε 0 c 3 e z P ω e z (.5 where P e2 ρ 2 ω 4 6πε 0 c 3 e2 (ρω 2 2 6πε 0 c 3 is the radiation power with a ρω 2 the acceleration. 38