A charge q executes simple harmonic motion with amplitude A and angular
frequency ω.
Calculate

(a) the far field components of the electric and magnetic fields, and

(b) the rate of energy loss.

Solution:

- Concepts:

The radiation field of a point charge moving non-relativistically, the Larmor formula - Reasoning:

Accelerating charges radiate. At large distances the radiation fields dominate, since they decrease inversely with distance. -
Details of the calculation:

(a) The radiation field**E**(**r**,t) of a point charge moving non-__relativistical__ly is (in SI units)

**E**(**r**,t) = -(q/(4πε_{0}c^{2}r''))**a**_{⊥}(t - r''/c),**r**'' =**r**-**r**'(t - |**r**-**r**'|/c).

r' ≈ 0, r'' ≈ r for a charge oscillating about the origin.

**a**_{⊥}(t - r/c) = -a(t - r/c)sinθ (**θ**/θ), z = A cos(ωt), a = -ω^{2}A cos(ωt).

**E**(**r**,t) = -[q/(4πε_{0}c^{2}r)]ω^{2}A cos(ω(t - r/c))sinθ(**θ**/θ).

**B**(**r**,t) = (**r**/(rc))×**E**(**r**,t) = -[q/(4πε_{0}c^{3}r)]ω^{2}A cos(ω(t - r/c))sinθ (**φ**/φ).

(b)**S**(**r**,t) = (1/μ_{0})**E**(**r**,t)×**B**(**r**,t) = [q^{2}/(4πε_{0})^{2}][1/(c^{5}r^{2}μ_{0})]ω^{4}A^{2}cos^{2}(ω(t - r/c))sin^{2}θ**(r**/r).

Now let us integrate over a spherical surface of radius r.

P = ∫**S∙**d**A**= (2πr^{2})[q^{2}/(4π)^{2}][1/(c^{3}r^{2}ε_{0})]ω^{4}A^{2}cos^{2}(ω(t - r/c))∫sin^{3}θdθ

= [q^{2}/(6πc^{3}ε_{0})]ω^{4}A^{2}cos^{2}(ω(t - r/c)).

<P> = [q^{2}ω^{4}A^{2}/(12πc^{3}ε_{0})] = rate of energy loss.

This is also what we get from the Larmor formula,

P =∮_{A}**S∙**d**A**= ⅔e^{2}a^{2}/c^{2}, e^{2}= q^{2}/(4πε_{0}).