PHYS 4011 – HEA Lec. 5 Lecture 5: Radiation from Moving Charges In the previous lectures, we found that particles interacting with scattering centres (i.e. other particles or fields) can be accelerated to very high energies. When these accelerated particles are charges, they produce electromagnetic waves. Radiation is an irreversible flow of electromagnetic energy from the source (charges) to infinity. This is possible only because the 2 electromagnetic fields associated with accelerating charges fall off as 1/r instead of 1/r as is the case for charges at rest or moving uniformly. So the total energy flux obtained from the Poynting flux is finite at infinity. Note: The material in this lecture is largely a review of some of the content presented in the Advanced Electromagentic Theory courses in 3rd and 4th years. As such, it is non-examinable for this course. 40 PHYS 4011 – HEA Lec. 5 5.1 Overview of the Radiation Field of Single Moving Charges Consider a radiating charge moving along a trajectory r0(t). Suppose we wish to measure the radiation field at a point P at a time t. Let the location of this field point be r(t). At time t, the charge is at point S, located at r0(t). But the radiation measured at P was actually emitted by ! ! the particle when it was at point S at an earlier time t . This is because an EM wave has a ! finite travel time |r(t) − r0(t )|/c before arriving at point P . Thus, the radiation field at P ! needs to be specified in terms of the time of emission t , referred to as the retarded time: ! ! |r(t) − r0(t )| t = t − (1) c Information from the charged particle’s trajectory arriving at field point P has propagated a finite dis- tance and taken a finite time to reach there at time t. At most, only one source point on the trajectory is in communication with P at time t. 41 PHYS 4011 – HEA Lec. 5 The radiation field at P at time t is calculated from the retarded scalar and vector potentials V and A using E = −∇V − ∂A/∂t and B = ∇ × A. But Maxwell’s equations define these in terms of the continuous charge and current densities ρ and J. So to evaluate V and A for a point charge, it is necessary to integrate over the volume distribution at one instant in time taking the limit as the size of the volume goes to zero. Formally, this can be done by taking ρ(r, t) = qδ(r − r0(t)) and J(r, t) = qv(t)δ(r − r0(t)), where v(t) = r˙ 0(t) is the velocity of the charge. Similarly, another delta function is introduced to single out only the source point at the retarded time that we are interested in. After integrating over the volume, we have ! r r ! r q δ(t − t + | (t) − 0(t )|/c) ! V ( , t) = ! dt (2) 4π%0 |r(t) − r0(t )| " ! r r ! A r µ0q v ! δ(t − t + | (t) − 0(t )|/c) ! ( , t) = (t ) ! dt 4π |r(t) − r0(t )| " After introducing the simplifying notation ! ! ! ! ! R(t ) R(t ) = r(t) − r0(t ) ,R(t ) = |r(t) − r0(t )| , Rˆ = R(t!) 42 PHYS 4011 – HEA Lec. 5 the integral can be solved with a change of variables giving ! µ0 qv(t ) v A(r, t) = = V (r, t) Lienard–Wiechart´ potentials (3) 4π (1 − Rˆ · v(t!)/c)R c2 These are the famous retarded potentials for a moving point charge. Points to note: ! 1. The factor (1 − Rˆ · v(t )/c) implies geometrical beaming. It means that the potentials ! are strongest at field points lying ahead of the source point S and closely aligned with the particle’s trajectory. The effect is enhanced when the particle speed becomes relativistic. 2. Retardation is what makes it possible for a charged particle to radiate. To see why, note that the potentials fall off as 1/r. Differentiation to retrieve the fields would yield a 1/r2 fall-off if there were no other r-dependence in the potentials. This does not give rise to a net electromagnetic energy flux as r → ∞ and hence, no radiation field. (Recall that the rate of change of EM energy goes as S · dA). However, there is an implicit r-dependence in the retarded time that leads! to a 1/r-dependence in the fields, upon differentiation of the potentials. This does result in a net flow of EM energy towards infinity. 43 PHYS 4011 – HEA Lec. 5 The radiation field The differentiation of the Lienard–Weichart´ potentials to obtain the radiation field of a single moving charge is lengthly, but straightforward (see Jackson for details). Writing the charged ! particle’s velocity at the retarded time as βc = r˙ 0(t ) and its corresponding acceleration as ! β˙ c = ¨r0(t ), the fields are 2 q (Rˆ − β)(1 − β ) Rˆ 1 ˙ E(r, t) = + × (Rˆ − β) × β (4) 4π%0 # (1 − Rˆ · β)3R2 (1 − Rˆ · β)3R c & $ % 1 B(r, t) = Rˆ × E(r, t) (5) c The first term on the RHS of the E-field is the velocity field. It falls off as 1/R2 and is just the generalisation of Coulomb’s Law to uniformly moving charges. The second term is the acceleration field contribution. It falls off as 1/R, is proportional to the particle’s acceleration and is perpendicular to Rˆ . 44 PHYS 4011 – HEA Lec. 5 This electric field, along with the corresponding magnetic field, constitute the radiation field of a moving charge: q Rˆ 1 ˙ Erad(r, t) = × (Rˆ − β) × β 4π%0 #(1 − Rˆ · β)3R c & $ % 1 Brad(r, t) = Rˆ × Erad(r, t) (6) c Note that Erad, Brad and Rˆ are mutually perpendicular. 45 PHYS 4011 – HEA Lec. 5 5.2 Radiation from Nonrelativistic Charged Particles The Larmor formula When β & 1, the radiation fields simplify to Rˆ E r q 1 Rˆ ˙ rad( , t) = × 2 × v (7) 4π%0 ' R c * ( ) and Brad(r, t) follows from (6). Note that Erad lies in the plane containing Rˆ and v˙ (i.e. the plane of polarisation) and Brad is perpendicular to this plane. If we let θ be the angle between Rˆ and v˙ , then E B q v˙ | rad| = c| rad| = 2 sin θ 4π%0 Rc The Poynting vector is in the direction of Rˆ and has the magnitude 2 2 1 2 µ0 q v˙ 2 S = Erad = 2 2 sin θ (8) µ0c 16π c R We now want to express this as an emission coefficient. 46 PHYS 4011 – HEA Lec. 5 Since S is the EM energy dW emitted per unit time dt per unit area dA (i.e. S = dW/(dtdA)), we can write dA = R2dΩ , where dΩ is the solid angle about the direction Rˆ of S. So the power emitted per solid angle is dW µ0 = SR2 = q2v˙2 sin2 θ dtdΩ 16π2c 2 Note the characteristic dipole pattern ∝ sin θ: there is no emission in the direction of acceleration and the maximum radiation is emitted perpendicular to the acceleration. The total electromagnetic power emitted into all angles is obtained by integrating this: 2π π +1 dW µ0 2 2 2 µ0 2 2 2 P = = 2 q v˙ sin θ sin θ dθ dφ = q r¨0 (1 − µ ) dµ dt 16π c 8πc − "0 "0 " 1 The integral gives a factor of 4/3, whence we arrive at the Larmor formula for the electromagnetic power emitted by an accelerating charge: 2 2 µ0q r¨0 P = Larmor formula (9) 6πc 47 PHYS 4011 – HEA Lec. 5 The dipole approximation To calculate the radiation field from a system of many moving charges, we must keep track of the phase relations between the radiating sources, because the retarded times will differ for each charge. In some situations, however, it is possible to neglect this complication and use the principle of superposition to determine the properties of the radiation field at large r. Suppose we have a collection of particles with positions ri, velocities vi and charges qi. Suppose further that these particles are confined to a region of size L and that the typical timescale over which the system changes is τ . If τ is much longer than the light travel time across the system (i.e. if τ ( L/c) then the differences in retarded time across the source are negligible. The timescale τ is also the characteristic timescale over which Erad varies, the above condition is equivalent to λ ( L, where λ ∼ cτ is the characteristic wavelength of the emitted radiation. The timescale τ also represents the characteristic time a particle takes to change its velocity substantially. Then τ ( L/c implies v & c, so we can use the nonrelativistic limits of the radiation fields derived above. 48 PHYS 4011 – HEA Lec. 5 Applying the superposition rule, we have Rˆ E qi i 1 Rˆ r rad = × 2 i × ¨i 4π%0 R c i ' i * + ( ) where Ri is the distance between each source point ri and the field point r. But the differences between the Ri are negligible, particularly as r → ∞. So we can just keep ! R = |r − r0(t )| as a characteristic distance and use the definition for the dipole moment, viz. d = qiri (10) i + to get Rˆ E 1 1 Rˆ d¨ rad = × 2 × (11) 4π%0 ' R c * ( ) Then, as before, we find dW µ0 = d¨2 sin2 θ dtdΩ 16π2c 49 PHYS 4011 – HEA Lec.