### Point charges, periodic motion

#### Problem:

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-relativistically is (in SI units)
E(r,t) = -(q/(4πε0c2r''))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 = -ω2A cos(ωt).
E(r,t) = -[q/(4πε0c2r)]ω2A cos(ω(t - r/c))sinθ(θ/θ).
B(r,t) = (r/(rc))×E(r,t) = -[q/(4πε0c3r)]ω2A cos(ω(t - r/c))sinθ (φ/φ).
(b)  S(r,t) = (1/μ0)E(r,t)×B(r,t) = [q2/(4πε0)2][1/(c5r2μ0)]ω4A2 cos2(ω(t - r/c))sin2θ (r/r).
Now let us integrate over a spherical surface of radius r.
P = ∫S∙dA = (2πr2)[q2/(4π)2][1/(c3r2ε0)]ω4A2 cos2(ω(t - r/c))∫sin3θdθ
= [q2/(6πc3ε0)]ω4A2 cos2(ω(t - r/c)).
<P> = [q2ω4A2/(12πc3ε0)] = rate of energy loss.
This is also what we get from the Larmor formula,
P =∮A S∙dA = ⅔e2a2/c2, e2 = q2/(4πε0).