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Electromagnetic Theory: PHAS3201, Winter 2008 9. Emission of Radiation 1 Retarded Potentials We will consider the emission of electromagnetic waves from sources. To start with, we will solve for the scalar and vector potentials in terms of charge and current densities. Fields • How are ﬁelds determined by potentials ? • B is easy (from · B = 0): B= ×A (1) • We get E from Faraday’s law: ∂A E=− φ− (2) ∂t • Use these to relate A and φ to J and ρ TAKE NOTES Summary • We wish to solve for ﬁelds E(r, t) and B(r, t) • It is easier to work in terms of potentials: B = ×A (3) ∂A E = − φ− (4) ∂t • From Maxwell’s equations, we ﬁnd: 2 ∂ ρ φ + ( · A) = − (5) ∂t 0 2 ∂2A ∂φ − A + µ 2 + ( · A) + µ = µJ (6) ∂t ∂t • We can use gauges to simplify Lorenz Condition • We will impose a condition on our potentials: ∂φ ·A+ µ =0 (7) ∂t • This is known as the Lorenz condition. • Notice that we can always write A → A + Λ and φ → φ − ∂Λ/∂t • So: 2 ∂2Λ Λ− µ =0 (8) ∂t2 • All potentials belonging to the Lorenz gauge satisfy this condition PHAS3201 Winter 2008 Section IX. Emission of Radiation 1 PHAS3201: Electromagnetic Theory Wave Equations • The vector potential now satisﬁes: 2 ∂2A − A+ µ = µJ (9) ∂t2 • The scalar potential can be shown to satisfy: 2 ∂2φ ρ − φ+ µ = (10) ∂t2 • How do we solve these equations ? TAKE NOTES Retarded Time • We write t = t − r/c • This is called retarded time • We choose: q(t − r/c) f (r − ct) = (11) 4π 0 • This means that we can write: q(t − r/c) φ(r, t) = (12) 4π 0 r • This solves for a point charge at origin. If we now apply this solution to Eq. (10) we can write: Form • Retarded scalar potential: 1 ρ (r , t ) φ(r, t) = dv (13) 4π 0 V |r − r | • We have t = t − |r − r |/c • By considering the components of A we can write: µ0 J (r , t ) A(r, t) = dv (14) 4π V |r − r | TAKE NOTES 2 Hertzian Dipole Geometry • Two small spheres connected by a short wire • Each sphere has charge q(t) • Wire length l, no capacitance • Spherical polars shown TAKE NOTES PHAS3201 Winter 2008 Section IX. Emission of Radiation 2 PHAS3201: Electromagnetic Theory Figure 1: Potentials • The vector potential: µ0 l I(t − r/c) Az (r, t) = (15) 4π r • The scalar potential: l z q(t − r/c) I(t − r/c) φ(r, t) = + (16) 4π 0 r2 r c • Choose: q(t − r/c) = q0 cos ω(t − r/c) (17) TAKE NOTES Important Points • These do not depend on φ • E and B are perpendicular • The power is radially outwards • The average power is: ¯ l2 ω 2 I02 P = 3 2 (18) 6π 0 c • The radial components of E and B are zero TAKE NOTES PHAS3201 Winter 2008 Section IX. Emission of Radiation 3

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PHAS3201 EM Theory Radiation Emission (UCL), astronomy, astrophysics, cosmology, general relativity, quantum mechanics, physics, university degree, lecture notes, physical sciences

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posted: | 11/8/2012 |

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PHAS3201 EM Theory Radiation Emission (UCL), astronomy, astrophysics, cosmology, general relativity, quantum mechanics, physics, university degree, lecture notes, physical sciences

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