r,t r R Z j ³ 0 1 4π² 0 r,t) = 4π

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1 5.4 Lienard-Wiechert Potential and Consequent Fields Potential and Fields (chapter 10) Lienard-Wiechert potential In the previous section, we studied the radiation from an electric dipole, a λ/2 antenna and a magnetic dipole. The source of radiation is due to the collectively accelerated charges (contained in the factor cos(ωt)). In this section, we shall study the radiation from an accelerated charge. In order to study the radiation from an accelerated charge, we have to find out how the scalar and vector potentials of a point charge are affected by its motion (particularly at relativistic velocities). Let s examine the apparent length ofamovingrod(onlyforintegration purposes. This is different from Einstein s relativistic theory which was developed later than the introduction of Lienard-Wiechert potential) for the following retarded potentials Φ ( r,t) 1 4π² 0 A ( r,t) µ 0 4π Z ρ ³ 0 r,t r R Z j ³ 0 r,t r Rodmovestowardstheobserverwith a speed v. The retarded time for integral in the retarded potentials are different from the front end and rear end. This makes the apparent length for integration dependent of the velocity. We have to make sure that the signals from both the front and rear ends arrive at the observation point at the same time. R dτ 0 dτ 0 Actual length l: measuredwhenv 0 (in a rest frame). Apparent length l 0 for integral 1/note19

2 Assumethelightsleftthefrontendatthetimet 1. If the light left the rear end at t 1 t and arrives the front end also at the time t 1, both signals (light) will then arrive at the observer at o atthesametimet. The extra time t is t l0 c During this time, the rod moved a distance l 0 l. So t l0 l v Equating the RHS of above two equations yields l 0 l 0 c l0 l v l 1 (v/c) The rod appears longer for the integration purposes. If the rod is moving away from the observer, t l0 c, l 0 l l0 t v l 1+(v/c) The rod appears shorter. Generally, if the rod moves in a direction with an angle θ with respect to the observation direction, 2/note19

3 l 0 l 1 (v/c)cosθ Or l 0 l 1 ~n ~β where β ~ ~v/c and ~n is an unit vector pointing from the source (it will be a point charge) to the observer. Because the change in the appeared length, the charge density and current density are modified. dv 0 dv 1 ~n ~β apparent volume element dq 0 ρdv 1 ~n ~β dq 1 ~n ~β charge element Current density ~j ρ~v ~jdv 0 ~ jdv 1 ~n ~β current element This consideration should be also applied to the (idealized) point charge. Considering also theretardedtimeeffect,wecanwritenowthesocalledlienard-wiechert potential for a point charge φ 1 q 4π² 0 (1 ~n ~β) ~r ~r p (t r ) ret. time ~A µ 0 q~v 4π (1 ~n ~β) ~r ~r p (t r ) ret. time 3/note19

4 where ~r p (t r ) is the location of the charge at retarded time t r t ~r ~r p(t r ) c Electric and Magnetic Fields In order to calculate electric and magnetic fields, we need to calculate the operators,, which are not trivial. Let s consider φ. Thepotentialφ is an explicit function of ~r and t r : φ φ(~r, t r ), But t r t ~r ~r p(t r ) c In principle, we can solve this equation for t r,sothatt r t r (~r, t). φ ~r φ(~r, t r )+ φ r t r ~r φ(~r, t r )+ φ t r r 4/note19

5 where and where As an example substitute R r R ~r ~r p (t r ) R into r r r Ã t ~r ~r! p(t r ) c 1 1 R c q [x x p (t r )] 2 +[y y p (t r )] 2 +[y y p (t r )] 2 R r ( ( 1) [x x p (t r )] x p(t r ) R r +[y y p (t r )] y p(t r ) r +[z z p (t r )] z p(t r ) r r ~n ~v expression ~n ~ R R, ~v ~r p(t r ) r r 1+1 r c ~n ~v 1+ r ~n ~β ( ~ β ~v/c) The equation yields r 1 1 ~n ~β 1 κ (κ 1 ~n ~β) (A) After lengthy derivations, we can get the expression for electric and magnetic fields: ~E ½ q 1 4π² 0 κ 3 R (~n β)(1 ~ β 2 )+ 1 2 cκ 3 R ~n [(~n β) ~ ~ ¾ β] (10.65) E ~ coulomb + E ~ rad ~B ~ A 1 c ~n ~ E (10.66) The electric field consists of two terms: (1) Generalized Coulomb field term The first term R 2,contains β. ~ It dominants for the near-field. It is essentially the Coulomb field corrected for relativistic effect (β) ) 5/note19

6 (2) Radiation term The second term R 1, contains the acceleration β ~ ~a/c (and also velocity β). ~ It dominants for the far-field, or in the radiation zone. If β 1, then κ ' 1and~n ~ β ' ~n ~E rad q 4π² 0 1 cκ 3 R ~n [(~n ~ β) ~ β] ~E rad q 1 ~n (~n ~a) 4π² 0 c 2 R This is the formula to calculate the radiation of an accelerated non-relativistic charge. Note, ~n, ~v and R are to be evaluated at the retarded time, i.e., R/c seconds earlier than the observation time Linear Acceleration For linear acceleration (or charge is at rest at retarded time) ~β k ~ β, ~ β ~β 0 So ~E rad q 1 1 ~n (~n ~ β) 4π² 0 c ~r ~r p (1 ~n ~β) 3 ~n (~n ~ β) ~β sin θ ~n ~β β cos θ 6/note19

7 So E ~ rad q 1 ~ β sin θ 4π² 0 c ~r ~r p (1 β cos θ) 3 The radial component of the Poynting vector is S r 1 s Z E ~ 2 ²0 q 2 1 β2 sin 2 θ µ 0 (4π² 0 c) 2 R 2 (1 β cos θ) 6 1 q 2 β2 sin 2 θ 4π² 0 4πcR 2 (1 β cos θ) 6 On a patch of a spherical surface of radius R dp S r da S r R 2 sin θdθdφ S r R 2 dω where dω isthesolidangleintowhichthepowerisradiated. The power distribution (Watt per unit solid angle) at time t for the observer is dp dω d Ã! dw S r R 2 1 q 2 β 2 sin 2 θ dω dt 4π² 0 4πc (1 β cos θ) 6 independent of the radius (energy conservation) The directionality function is If β 0then f(θ) sin 2 θ (1 β cos θ) 6 f(θ) sin 2 θ 7/note19

8 as we encountered before for non-relativistic case (electric dipole, antennas). If β 6 0,wecanfind the angle θ m at which f(θ m ) is the maximum. Use substitution x cosθ f(θ) F (x) 1 x2 (1 βx) 6 At the maximum, or So which have solutions or If - sign is chosen Not a physical solution Chose the + sign df dθ df dx dx dθ 0 df dx 0 df dx d " # 1 x 2 dx (1 βx) 6 1 (1 βx) [( 2x)(1 12 βx)6 (1 x 2 )( 6β)(1 βx) 5 ] 2 (1 βx) [x(1 βx) 3β(1 7 x2 )] x(1 βx) 3β(1 x 2 )2βx 2 + x 3β 0 x m 1 ± 1+24β 4β x m β 4β 1+24β +1 16β x m > 4β 4β 1 > 1 β cos θ m x m > 1 x m cosθ m 1+24β 1 4β β cos θ m θ m f(θ m ) o o /note19

9 When β 1, θ m approaches zero in a manner The total Radiated Power P dw dt θ m 1 γ, γ 1 1 β 2 Z R 2 S r dω (dω sinθdθdφ) 1 q 2 4π² 0 4πc β Z π 2 sin 3 θ 2π 0 (1 β cos θ) dθ 6 1 q 2 β2 4π² 0 2c 1 q 2 β2 4π² 0 2c Z t 2 dt (t cosθ) (1 βt) 6 µ β2 1 (1 β 2 ) 4 {z } γ 8 1 2q 2 v 2 Ã! 1+ β2 4π² 0 3c 3 5 If β 1, then γ ' 1and P 1 2q 2 v 2 4π² 0 3c 3 This is the Larmor formula (cf. eq. (11.70)) for nonrelativistic case. Energy Radiation Rate at the Retarded Time The power we examined is the power the observer measured at the observation time t. The power emitted by the particle at the retarded time t r is different from the above power. Considering a gun is moving towards a target at a velocity v shooting a stream of bullets (at velocity c ) at a rate N g. The rate N t at which the target receives the bullets is larger (read p. 462). γ 8 9/note19

10 The power radiated at time t r is We have derived earlier in eq (A) P r dw dt r dw dt dt dt r r 1 1 ~n ~β 1 κ (κ 1 ~n ~β) (A) So dp r dω d Ã! dw 1 dω dt r Z E2 κr 2 1 q 2 4π² 0 4πc β 2 sin 2 θ (1 β cos θ) 5 The Power emitted by the charged particle at t r and that received by the observer at time t may be different for a highly relativistic particle. It turns out that P (t r ) 1 2q 2 v 2 4π² 0 3c 3 γ6 β 1 ~ ~ 2 β (11.73) β For linear acceleration ~ β ~ β 0 P (t r ) 1 2q 2 v 2 4π² 0 3c 3 Note that the difference between γ 6 in P (t r )andγ 8 in P (t) can be significant when β 1 (γ ). Example: Acceleration in a constant electric field Suppose a high voltage V is applied between two parallel plate with a separation d. The electric field is, of course, E V/d. γ6 Phys 463, E & M III, C. Xiao 10 / note19

11 If the electric field is high, an electron can be accelerated to a high speed. At high speeds, the mass of the electron is larger than its rest mass (we shall learn that later): The momentum is also larger Newton s law (equation of motion) becomes p mv m m 0 1 β 2 m 0v 1 β 2 F dp dt m 0cβ 1 β 2 Ã! d m0 cβ ee dt r 1 β 2 The equation of motion becomes LHS Ã m0 cβ β 1 β 2! dβ dt r {z} β v/c " # m 0 c + m β 1 β 2 0cβ (1 β 2 ) 3/2 m 0 c β (1 β 2 ) (1 3/2 β2 + β 2 ) m 0 c β 1 ( 1 β 2 ) m 0c βγ 3 3 β m 0 c βγ 3 ee Solve for β The radiated power is β ee m 0 cγ 3 P (t r ) 1 2 e 2 β2 4π² 0 3 c γ6 1 2 e 2 Ã! 2 ee γ 6 4π² 0 3 c m 0 cγ e 2 µ ee 2 4π² 0 3 c 3 m 0 Phys 463, E & M III, C. Xiao 11 / note19

12 This shows that the energy loss due to radiation during the acceleration is independent of γ. The electrons can be accelerated to a speed close to the light speed without significant radiation loss. Example: A linear accelerator with following parameters: 300 MeV, d 30m. Calculate the radiation power and the acceleration power Electric field: E dc V MV/m d 30 Energy gradient along the field direction (z-direction): dε dz Radiated power (power loss): 300 MeV 30m 10(MeV/m) 1 2 e 2 µ ee 2 P rad. (t r ) 1 µ 2 ee 2 4π² 0 3 c 3 m 0 4π² 0 3 e2 c m 0 c π ( ) 2 µ 10 2 ( ) (W) 110 (ev/sec) where m 0 c MeV. Acceleration time: Total radiated Energy: t ' d c (sec) Acceleration power: ε rad. P (t r ) t (ev) P accel. (t r ) dε dz v ' dε dz c (ev/m) (m/sec) (ev/sec) As we can see P rad. (t r ) P accel. (t r ) Circular Acceleration (Synchrotron Radiation) The general expression of radiated electric field is ~E q 4π² 0 ½ 1 κ 3 R 2 (~n ~ β)(1 β 2 )+ 1 cκ 3 R ~n [(~n ~ β) ~ β] ¾ In case of circular acceleration (ignore the first coulomb field term), Phys 463, E & M III, C. Xiao 12 / note19

13 ~v ~v ~ β β~ez ~β β~ex ( β a/c) n ~r r (sinθ cos φ~e x +sinθ sin φ~e y +cosθ~e z ) It can be shown that the radiation per unit solid angle is (Problem 11.16, page 465) dp (t r ) dω 1 4π² 0 q 2 a 2 4πc 3 (1 β cos θ) 2 (1 β 2 )sin 2 θ cos 2 φ (1 β cos θ) 5 The total radiated power is (Problem 11.16, page 465) P (t r ) 1 2q 2 a 2 4π² 0 3c 3 For nonrelativistic case ( β 1, γ 1), the formula reduces to the Larmor formula again γ4 P (t r ) 1 2q 2 a 2 4π² 0 3c 3 Again, the angular distribution of the radiated power is forward peaked with respect to ~v. Shown in the diagram is the power distribution on the plane defined by ~ β and β ~ (φ 0) Phys 463, E & M III, C. Xiao 13 / note19

14 The angular spread is q θ 0 1/γ γ 1/ 1 β 2 An accelerated electron can be bent by a magnetic field through ~F e~v ~ B (ma ebv) The radiation at any instant from a single electron is peaked forward along the velocity direction. If we observe the radiation at a fixed location, we would see a series of radiation bursts. This type of radiation is called synchrotron radiation. Thedurationoftheburstatthetimeofradiation 2π dt r θ 0 ' 1 2π ω 0 γ ω 0 where ω 0 is the angular frequency of a single rotating electron. Duration of the burst at the time of observation As derived earlier dt r dt 1 κ dt r 1 κ dt dt 1 ~n ~β dt 1 β γ2 dt (~n k ~ β) γ 2 dt 1 γ 2π ω 0 Phys 463, E & M III, C. Xiao 14 / note19

15 Figure 1: dt 2π γ 3 ω 0 β 1 dt 0 (Sharp pulse) The spectrum of these periodic pulses consists of harmonics of the fundamental frequency ω 0 and the spectrum peaks at the frequency ω peak 2π dt γ3 ω 0 Phys 463, E & M III, C. Xiao 15 / note19

16 Of course, in the synchrotron ring, there are many electron forming a constant current. The radiation power is also constant for the observer. Phys 463, E & M III, C. Xiao 16 / note19

RADIATION OF ELECTROMAGNETIC WAVES

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