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b sin k0 x) flows in the plane z = 0. Find the magnetic 2. The surface current J = J0 (b x sin k0 y + y field in the whole space. Making use of the superposition principle, one can conveniently find separately the fields b sin k0 y is created by the currents directed in the x and y direction. The field of the current J0 x described by the vector potential A = (Ax , 0, 0). The Coulomb gauge condition, ∇ · A = 0, yields ∂Ax = 0. Then the potential obeys the equation ∂x ∂ 2 Ax ∂ 2 Ax + = 0. ∂y 2 ∂z 2 (1) Separation of variables at z > 0: Ax = ψ(y)f (z); ψ 00 f 00 + = 0; ψ f (2) f = C1 e−kz . ψ = sin(ky + φ); (3) B = ∇ × A = −C1 k [ŷ sin(ky + φ) + ẑ cos(ky + φ)] e−kz ; z>0 (4) Analogously B = C2 k [ŷ sin(ky + φ) − ẑ cos(ky + φ)] ekz ; z<0 (5) Continuity of Bz yields C1 = C2 . The boundary condition, By (z = +0) − By (z = −0) = −(4π/c)J0 sin k0 y yields k0 = k, φ = 0 and C1 = 2πj0 . ck0 (6) Now one gets 2π B= J0 c − (ŷ sin k0 y + ẑ cos k0 y) e−kz ; z > 0 (ŷ sin k0 y − ẑ cos k0 y) ekz ; z < 0. (7) The field of the current flowing in the y direction is found analogously. Finally one gets the total field 2π [x̂ sin k0 x − ŷ sin k0 y + ẑ(cos k0 x − cos k0 y)] e−kz ; z > 0 J0 Btot = (8) [−x̂ sin k0 x + ŷ sin k0 y + ẑ(cos k0 x − cos k0 y)] ekz ; z < 0. c 1 3. A photon with the frequency ω propagates in the positive x direction and is scattered off an electron moving in the negative x direction with the energy E. Find the frequency of the photon, the energy and the direction of the electron after the scattering as a function of the scattering angle (the angle between the photon directions after and before the scattering) The 4-momentums of the electron and the photon before the scattering are P = √ 1 E, − E 2 − m2 c4 , 0, 0 ; c Pγ = h̄ω (1, 1, 0, 0) ; c (9) where the identity P 2 = (E/c)2 − p2 = m2 c2 was used in order to express p via E. After the scattering, P0 = √ 1 0 √ 02 E , E − m2 c4 cos ψ, E 02 − m2 c4 sin ψ, 0 ; c Pγ0 = h̄ω 0 (1, cos θ, sin θ, 0) ; c (10) where θ and ψ are angles between the x axis and the directions of motion of the scattered photon and electron, correspondingly. We have to express ω 0 , E 0 and ψ via ω, E and θ. In order to eliminate the parameters of the scattered electron, let us write the conservation of 4-momentum, 0 2 0 0 P + Pγ = P + Pγ , as P + Pγ − Pγ = P 02 and take into account that P 2 = m2 c2 , Pγ2 = 0. This yields P · Pγ − P · Pγ0 − Pγ · Pγ0 = 0, (11) √ where P · Pγ = (h̄/c2 ) Eω + ω E 2 − m2 c4 etc. Now one gets an equation for ω 0 , √ √ Eω + ω E 2 − m2 c4 = ω 0 E + ω 0 cos θ E 2 − m2 c4 + h̄ωω 0 (1 − cos θ), (12) which is immediately resolved √ E + E 2 − m2 c4 √ ω. ω = E + cos θ E 2 − m2 c4 + h̄ω(1 − cos θ) 0 (13) The energy of the scattered electron is found the from the energy conservation: E 0 = E + h̄ω − h̄ω 0 . (14) Conservation of the y component of the momentum vector implies p0y = −h̄ω 0 /c. (15) √ Taking into account cp0 = E 02 − m2 c4 , one finds the angle between the electron direction after the scattering and the x axis p0y h̄ω 0 sin ψ = 0 = − √ . (16) p E 02 − m2 c4 2 4. A sphere of the radius r is filled homogeneously with the electric charge with the density ρ. The sphere executes torsional oscillations around the diameter, ϕ = ϕ0 sin ωt. Find the timeaveraged angular distribution and the power of the radiation. The charges move in the ϕ direction with the velocity v = r0 sin θϕ̇ = ϕ0 ωr0 sin θ cos ωt, where 0 r Rand θ are the spherical coordinates. The current j = ρv creates the magnetic moment m = 1 r0 × jdV . It follows from the symmetry, that m has only z component 2c Z Z 1 1 0 m=b z (r × j)z dV = b z r0 sin θjdV 2c 2c Z πρϕ0 ω 4π =b z cos ωt r04 sin2 θd(cos θ)dr0 = b z ρϕ0 ω cos ωtr5 . c 15c The electric field of the wave emitted in the direction n is E= 1 c2 r 0 [nm̈] = sin θ m̈ϕ. b c2 r 0 (17) Now the radiation power in the direction n is 2πρ2 ϕ20 ω 6 r10 dP E2 = r02 c = sin2 θ. dΩ 4π 225c5 (18) The total radiation power is 2πρ2 ϕ20 ω 6 r10 P = 225c5 Z sin2 θdΩ = 3 16π 2 ρ2 ϕ20 ω 6 r10 . 675c5 (19)