Single-Particle Electrodynamics - Assassination Science

any solution to our problem: if A and B are rigidly connected, then surely
their respective motions, “some time ago”, were identical?
The crucial realisation is that the retarded time of constituent A, as
seen by constituent B, is not, in general, the same as the retarded time of
constituent B, as seen by constituent A, if the body is being accelerated. This
fact can be recognised from first principles: Imagine that the spherical body,
of radius ε, is being accelerated in the positive-x direction, which we shall
describe as being “to the right” of the origin; let us take the body to be
stationary, centred on the origin, at t = 0, which we shall describe as “now”;
and let us take, for simplicity, the points A and B at diametrically opposite
ends of the sphere, along the x-axis, with A at the position x = +ε, and
B at the position x = −ε. Clearly, in the past, the centre of the sphere
was always to the right of the origin (since, for nonrelativistic motion, we
have x centre = 1 2 at2 ; the sphere has reached the minimum of its trajectory,
along the x-axis); thus, the points A and B were, in the past, to the right of
where they are now. But this means that, for constituent B to send a light
beam to constituent A, the beam only has to travel a shorter distance than
their separation (now) of 2ε; and for constituent A to send a light beam to
constituent B, the beam has to travel a longer distance than 2ε. But the
retarded time, in naturalised units, is simply equal to the distance the light
has to travel; hence, A “sees” B as he was a short time ago, but B “sees”
A as he was a longer time ago. This seemingly counterintuitive result is, of
course, simply another consequence of Einstein’s second postulate: that the
speed of light is an invariant, independent of the velocity of the sender.
Now, since the magnitude of the electric field generated by a charge is
determined by the velocity and acceleration of the charge as “seen” at the retarded
time, as well as directly depending (as 1/R or 1/R 2 ) on this retarded
distance itself,—and since the velocity and acceleration are themselves changing
with time,—then it is clear that the retarded Coulomb force of A on B
will, in the general case, be different to that of B on A.
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