Single-Particle Electrodynamics - Assassination Science

5.3.4 The fields for a point particle
The result (5.13) holds for arbitrary electric charge distributions J µ (x ′ ). We
now determine the retarded fields from the electric charge of a point particle
by substituting the current density expression for a point charge, equation
(5.4):
A µ LG(x) = q ∫ ∫
dτ U µ (τ) d 4 x ′ ϑ(x 0 −x ′ 0 ) δ [ (x − x ′ ) 2] δ (4) [x ′ − z(τ)] . (5.14)
2π
Integrating this expression over d 4 x ′ , we find
A µ LG(x) = q ∫
dτ U µ (τ) ϑ [ x 0 − z 0 (τ) ] δ { [x − z(τ)] 2} . (5.15)
2π
Clearly, this integral will only contain a contribution from z(τ) on the backwards
light cone of x, i.e., the particular τ = τ ret for which [x − z(τ ret )] 2 = 0
and z 0 (τ ret ) < x 0 .
We now discard the Lorentz-gauge four-potential A µ LG in favour of the
physically meaningful field strengths F αβ , using the definition (5.2). Let us
first concentrate on the first term on the right-hand side of (5.2), namely,
∂ α A β . The partial derivative ∂ α , acting on the Heaviside step function, will
give us a Dirac delta function at the four-position of the charge; since we
shall be treating the fields on the worldline of the particle separately, in
Section 5.5, we can ignore this contribution for the current analysis. We are
thus left with
∂ α A β = q ∫
dτ U β (τ) ϑ [ x 0 − z 0 (τ) ] ∂ α δ { [x − z(τ)] 2} . (5.16)
2π
To proceed from here, one needs to recall the chain rule of differentiation,
namely
∂ u g[f(u, τ)] ≡ ∂ u f · d f g[f(u, τ)] . (5.17)
Applying this to the case of the partial derivative ∂ α (for which the variable
u in (5.17) is the component x α ), and noting that we ultimately wish to
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