By R. Magyar

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**Additional resources for Companion to J.D. Jackson's Classical Electrodynamics**

**Example text**

The Kramer-Kr¨onig relation states: ℜ ǫ(ω) ǫ0 2 P π =1+ ǫ(ω ′ ) ω′ dω ′ ℑ ω ′2 − ω 2 ǫ0 ∞ 0 a. ℑ ǫ(ω) = λ [Θ(ω − ω1 ) − Θ(ω − ω2 )]. ǫ0 Plug this into the Kramer-Kronig relationship. ℜ ǫ(ω) ǫ0 =1+ 2λ π ω2 ω1 ω′ dω ′ + 0 ω ′2 − ω 2 Notice that the real part of ǫ(ω) depends on an integral over the entire frequency range for the imaginary part! Here, we will use a clever trick. ǫ(ω) ǫ0 ℜ = 1+ λ π ω22 ω12 d(ω ′2 ) ω ′2 − ω 2 And this integral is easy to do! ℜ ǫ(ω) ǫ0 λ λ ω22 − ω 2 ω2 ln ω ′2 − ω 2 |ω22 = 1 + ln 1 π π ω12 − ω 2 =1+ = (ω2 −ωλγω b.

A. Find the effective transitional “magnetization”, calculate ∇ · M , and evaluate all the non-vanishing radiation multi-poles in the longwavelength limit. The magnetization is 1 M = (r × J) 2 J can be broken up into Jr and Jz components. We take the cross product of the two components with r. r × Jr = 0 And −x y yˆ + xˆ a0 ρ z z To make things easier, we’ll use angles. tan θ = zr , sin φ = yr , cos φ = r × Jz = −iv0 r × J = −ia0 ρv0 (tan θ sin φˆ x − tan θ cos φˆ y) 60 x r Then, Don’t forget v0 = αc, so M = −i Let χ = −iαca0 2 αca0 tan θ(sin φˆ x − cos φˆ y )ρ 2 tan θ(sin φˆ x − cos φˆ y) then M = ρχ Now, we take the divergence.

Calculate the force in Newton’s acting a Dirac mono-pole of the minimum magnetic charge located a distance 21 an Angstrom from and in the median plane of a magnetic dipole with dipole moment equal to one nuclear magneton. q¯ h A magnetic dipole, m = 2m , creates a magnetic field, B. p B= µ0 3n(n · m) − m 4π |x|3 Along the meridian plane, n · m = 0 so B=− µ0 m 4π |x|3 h n. Suppose this field is acting on a magnetic mono-pole with charge g = 2π¯ q Where n is some quantum number which we’ll suppose to be 1.