Single-Particle Electrodynamics - Assassination Science

6.8.8 Magnetic contact force contribution
We now consider the contribution to the radiation reaction force equation
of motion of the contact force on the magnetic dipole moments of the constituents.
Since this force depends on the current J, it may be wondered why we
are considering it at all: where would this current come from? The answer,
of course, is found in the considerations of Chapter 5: the magnetic dipole
moment itself has a current sheet surrounding it, which is actually responsible
for generating the moment. The current density J(r) is given by
J(r) = 3µ σ×n δ(r − ε). (6.126)
4πε3 Again, we can verify (6.126) in four steps. Firstly, we note that the direction
of the cross-product is such that the sheet of current is circulating in the
correct sense to produce a magnetic field in the direction of σ. Secondly,
we note that the magnitude of the cross-product (namely, of the form sin θ,
where θ is the “latitude” on the sphere, if σ points to the “North Pole”)
is maximum perpendicular to σ, and vanishes in the directions parallel and
antiparallel to σ, in agreement with the analysis of Chapter 5. Thirdly, the
coëfficient of (6.126) is the correct stepping of the magnetic field from its
internal value to its external value; this may be verified quickly by noting
that, before we added the Maxwell field, the field matched smoothly around
the “Equator”; the step in the field around this circular boundary is therefore
just that of the Maxwell field. Fourthly, the delta function reflects the fact
that it is a sheet current.
Now, the contact force contribution of each constituent will be
λ(r) σ×J(r) = 3µ λ(r) σ×(σ×n) δ(r − ε). (6.127)
4πε3 Because (6.126) is odd in r, we need to use the odd part r(n· ˙v) of λ(r) to
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