Single-Particle Electrodynamics - Assassination Science

This asserted result of the author’s may be verified in four steps. Firstly, we
note that the magnitude of (6.125), 3µ/4πε 3 , is simply given by the magnitude
of the Maxwell field, since the step is from this value to zero. Secondly,
the factor (σ·n) in (6.125) is unity when n is in the direction of σ, negative
unity when n is antiparallel to σ, and zero when n is perpendicular to σ;
this factor embodies the fact that the step occurs as we move radially outwards.
Thirdly, the minus sign of (6.125) reflects the fact that the step is
from B M down to the value zero. Fourthly, the vector σ is the direction of
the Maxwell field itself.
To compute the self-force on the body as a whole, we multiply (6.125) by
the redshift factor λ(r), the step function ϑ(ε − r) that cuts off the receiving
moment density at the surface, and integrate over all space; since (6.125) is
odd, we must select the odd part r(n· ˙v) of λ(r):
( ) µ 2 ∫
F (M)
µµ = −
4πε 3
d 3 r r(n· ˙v)(σ·n)σ δ(r − ε) ϑ(ε − r).
In performing this integral, we note that the angular integration will yield a
factor of 1/3, and the presence of the delta and step functions a factor of 1/2;
the remaining factors in the integrand contribute 4πε 3 ( ˙v·σ)σ, and hence
F (M)
µµ = − 1 3 · 1
2 · 4πε3 ( µ
4πε 3 ) 2(
˙v·σ)σ
= − 3 2 µ2 η ′ 3( ˙v·σ)σ.
For the “moment-arm” torque following from this same force, we must of
course now choose the even part (unity) of λ(r), and then cross the vector r
into the result. However, we shall then be integrating an expression involving
the factor (n·σ)n×σ, which, after angular integration, yields σ ×σ, and
hence vanishes.
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