In my previous post, I partially presented the first steps towards explaining the force, resulting from acceleration, exerted by a charge q1 on a charge q2 where q2 moves at the constant velocity v and q1 moved at the velocity [u] and with the acceleration [a] at the retarded time. Here, I will conclude this explanation. Let me first repeat the basic image that describes the situation.
The goal is to derive the ‘frequency’ with which [a]
is applied to q2, to be able to compare a situation in which
both q1 and q2 are static and one in which q1
moved at u at the retarded time, when it had the acceleration [a],
and q2 moves at v. I derived a vector 
and a number d1,2
that will help me to derive this frequency (see my previous post for more
details). First, I will consider a plane vertical to and at a distance d1
from (0, 0, 0), which is the current location of q2:
I will now work out in what time t1 that
plane and q2 meet if the plane moves at c towards (0, 0,
0) and q2 moves at v along the x-axis. The
equation for the plane S1 is
or
We can now insert
as well as what we previously calculated
for and d1:
We obtain
Hence
Now consider that the equivalent time for
a=v=u=0 is
and we obtain the first frequency factor
We now have to perform the same
calculation for a time t2, where we use d2
and 
. This is a bit different
from the case of uniform velocities because n> does not
feature here. We obtain:
This results in
We finally obtain:
